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(2014-2016)

JEE Main Question Papers and Answer Keys

IntroductionCareers360 brings you a compilation of JEE Main Question Papers and Answer Keys from the year 2014 to 2016. This e-book helps you master questions which appeared in the national level engi-neering entrance examination in the last 3 years.

This booklet contains 40 printed pages.

ß‚ ¬ÈÁSÃ∑§Ê ◊¥ ◊ÈÁŒ˝Ã ¬Îc∆ 40 „Ò¥–

Do not open this Test Booklet until you are asked to do so.

ß‚ ¬⁄ËˇÊÊ ¬ÈÁSÃ∑§Ê ∑§Ê Ã’ Ã∑§ Ÿ πÊ‹¥ ¡’ Ã∑§ ∑§„Ê Ÿ ¡Ê∞–Read carefully the Instructions on the Back Cover of this Test Booklet.

ß‚ ¬⁄ËˇÊÊ ¬ÈÁSÃ∑§Ê ∑§ Á¬¿‹ •Êfl⁄áÊ ¬⁄ ÁŒ∞ ª∞ ÁŸŒ¸‡ÊÊ¥ ∑§Ê äÿÊŸ ‚ ¬…∏¥–

Name of the Candidate (in Capital letters ) :

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•ŸÈ∑§◊Ê¥∑§ — •¥∑§Ê¥ ◊¥: in words — ‡ÊéŒÊ¥ ◊¥

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2. Invigilator’s Signature : ÁŸ⁄ËˇÊ∑§ ∑§ „SÃÊˇÊ⁄ —

No. :

Important Instructions :

1. Immediately fill in the particulars on this page of the TestBooklet with only Blue / Black Ball Point Pen provided bythe Board.

2. The Answer Sheet is kept inside this Test Booklet. When youare directed to open the Test Booklet, take out the AnswerSheet and fill in the particulars carefully.

3. The test is of 3 hours duration.4. The Test Booklet consists of 90 questions. The maximum

marks are 360.5. There are three parts in the question paper A, B, C

consisting of Physics, Chemistry and Mathematics having30 questions in each part of equal weightage. Each questionis allotted 4 (four) marks for correct response.

6. Candidates will be awarded marks as stated above in instructionNo. 5 for correct response of each question. ¼ (one fourth) markswill be deducted for indicating incorrect response of each question.No deduction from the total score will be made if no response isindicated for an item in the answer sheet.

7. There is only one correct response for each question. Fillingup more than one response in any question will be treated aswrong response and marks for wrong response will bededucted accordingly as per instruction 6 above.

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13. Do not fold or make any stray mark on the Answer Sheet.

◊„ûfl¬ÍáÊ¸ ÁŸŒ¸‡Ê —

1. ¬⁄UËˇÊÊ ¬ÈÁSÃ∑§Ê ∑§ ß‚ ¬Îc∆U ¬⁄U •Êfl‡ÿ∑§ Áflfl⁄UáÊ ∑§fl‹ ’Ê«¸U mÊ⁄UÊ©¬‹éœ ∑§⁄UÊÿ ªÿ ŸË‹ / ∑§Ê‹ ’ÊÚ‹ åflÊß¥≈U ¬Ÿ ‚ Ãà∑§Ê‹ ÷⁄¥–

2. ©ûÊ⁄U ¬òÊ ß‚ ¬⁄UËˇÊÊ ¬ÈÁSÃ∑§Ê ∑§ •ãŒ⁄U ⁄UπÊ „Ò– ¡’ •Ê¬∑§Ê ¬⁄UËˇÊÊ ¬ÈÁSÃ∑§ÊπÊ‹Ÿ ∑§Ê ∑§„Ê ¡Ê∞, ÃÊ ©ûÊ⁄U ¬òÊ ÁŸ∑§Ê‹ ∑§⁄U ‚ÊflœÊŸË¬Ífl∑§ Áflfl⁄UáÊ ÷⁄¥U–

3. ¬⁄UËˇÊÊ ∑§Ë •flÁœ 3 ÉÊ¥≈U „Ò–4. ß‚ ¬⁄UËˇÊÊ ¬ÈÁSÃ∑§Ê ◊¥ 90 ¬˝‡Ÿ „Ò¥– •Áœ∑§Ã◊ •¥∑§ 360 „Ò¥–5. ß‚ ¬⁄UËˇÊÊ ¬ÈÁSÃ∑§Ê ◊¥ ÃËŸ ÷Êª A, B, C „Ò¥, Á¡‚∑§ ¬˝àÿ∑§ ÷Êª ◊¥

÷ÊÒÁÃ∑§ ÁflôÊÊŸ, ⁄U‚ÊÿŸ ÁflôÊÊŸ ∞fl¥ ªÁáÊÃ ∑§ 30 ¬˝‡Ÿ „Ò¥ •ÊÒ⁄U ‚÷Ë¬˝‡ŸÊ¥ ∑§ •¥∑§ ‚◊ÊŸ „Ò¥– ¬˝àÿ∑§ ¬˝‡Ÿ ∑§ ‚„Ë ©ûÊ⁄U ∑§ Á‹∞ 4 (øÊ⁄U)•¥∑§ ÁŸœÊ¸Á⁄UÃ Á∑§ÿ ªÿ „Ò¥–

6. •èÿÁÕ¸ÿÊ¥ ∑§Ê ¬˝àÿ∑§ ‚„Ë ©ûÊ⁄U ∑§ Á‹∞ ©¬⁄UÊÄÃ ÁŸŒ¸‡ÊŸ ‚¥ÅÿÊ 5 ∑§ÁŸŒ¸‡ÊÊŸÈ‚Ê⁄U •¥∑§ ÁŒÿ ¡Êÿ¥ª– ¬˝àÿ∑§ ¬˝‡Ÿ ∑§ ª‹Ã ©ûÊ⁄U ∑§ Á‹ÿ¼ flÊ¥ ÷Êª ∑§Ê≈U Á‹ÿÊ ¡ÊÿªÊ– ÿÁŒ ©ûÊ⁄U ¬òÊ ◊¥ Á∑§‚Ë ¬˝‡Ÿ ∑§Ê ©ûÊ⁄UŸ„Ë¥ ÁŒÿÊ ªÿÊ „Ê ÃÊ ∑È§‹ ¬˝Ê#Ê¥∑§ ‚ ∑§Êß¸ ∑§≈UÊÒÃË Ÿ„Ë¥ ∑§Ë ¡ÊÿªË–

7. ¬˝àÿ∑§ ¬˝‡Ÿ ∑§Ê ∑§fl‹ ∞∑§ „Ë ‚„Ë ©ûÊ⁄U „Ò– ∞∑§ ‚ •Áœ∑§ ©ûÊ⁄U ŒŸ ¬⁄U©‚ ª‹Ã ©ûÊ⁄U ◊ÊŸÊ ¡ÊÿªÊ •ÊÒ⁄U ©¬⁄UÊÄÃ ÁŸŒ¸‡Ê 6 ∑§ •ŸÈ‚Ê⁄U •¥∑§ ∑§Ê≈UÁ‹ÿ ¡Êÿ¥ª–

8. ©ûÊ⁄U ¬òÊ ∑§ ¬Îc∆U-1 ∞fl¥ ¬Îc∆U-2 ¬⁄U flÊ¥Á¿UÃ Áflfl⁄UáÊ ∞fl¥ ©ûÊ⁄U •¥Á∑§Ã∑§⁄UŸ „ÃÈ ’Ê«¸U mÊ⁄UÊ ©¬‹éœ ∑§⁄UÊÿ ªÿ ∑§fl‹ ŸË‹/∑§Ê‹ ’ÊÚ‹ åflÊß¥≈UU¬Ÿ ∑§Ê „Ë ¬˝ÿÊª ∑§⁄¥U–

9. ¬⁄UËˇÊÊÕË mÊ⁄UÊ ¬⁄UËˇÊÊ ∑§ˇÊ/„ÊÚ‹ ◊¥ ¬fl‡Ê ∑§Ê«U ∑§ •‹ÊflÊ Á∑§‚Ë ÷Ë ¬∑§Ê⁄U∑§Ë ¬Ê∆Uÿ ‚Ê◊ªË, ◊ÈÁŒÃ ÿÊ „SÃÁ‹ÁπÃ, ∑§Êª¡ ∑§Ë ¬ÁøÿÊ°, ¬¡⁄U, ◊Ê’Êß‹»§ÊŸ ÿÊ Á∑§‚Ë ÷Ë ¬˝∑§Ê⁄U ∑§ ß‹Ä≈UÊÚÁŸ∑§ ©¬∑§⁄UáÊÊ¥ ÿÊ Á∑§‚Ë •ãÿ ¬˝∑§Ê⁄U ∑§Ë‚Ê◊ªË ∑§Ê ‹ ¡ÊŸ ÿÊ ©¬ÿÊª ∑§⁄UŸ ∑§Ë •ŸÈ◊ÁÃ Ÿ„Ë¥ „Ò–

10. ⁄U»§ ∑§Êÿ¸ ¬⁄UËˇÊÊ ¬ÈÁSÃ∑§Ê ◊¥ ∑§fl‹ ÁŸœÊ¸Á⁄UÃ ¡ª„ ¬⁄U „Ë ∑§ËÁ¡∞– ÿ„¡ª„ ¬˝àÿ∑§ ¬Îc∆U ¬⁄U ŸËø ∑§Ë •Ê⁄U •ÊÒ⁄U ¬ÈÁSÃ∑§Ê ∑§ •¥Ã ◊¥ ∞∑§ ¬Îc∆U ¬⁄U(¬Îc∆U 39) ŒË ªß¸ „Ò–

11. ¬⁄UËˇÊÊ ‚◊ÊåÃ „ÊŸ ¬⁄U, ¬⁄UËˇÊÊÕË¸ ∑§ˇÊ/„ÊÚ‹ ¿UÊ«∏Ÿ ‚ ¬Ífl¸ ©ûÊ⁄U ¬òÊ ∑§ˇÊÁŸ⁄UËˇÊ∑§ ∑§Ê •fl‡ÿ ‚ÊÒ¥¬ Œ¥– ¬⁄UËˇÊÊÕË¸ •¬Ÿ ‚ÊÕ ß‚ ¬⁄UËˇÊÊ ¬ÈÁSÃ∑§Ê∑§Ê ‹ ¡Ê ‚∑§Ã „Ò¥–

12. ß‚ ¬ÈÁSÃ∑§Ê ∑§Ê ‚¥∑§Ã E „Ò– ÿ„ ‚ÈÁŸÁ‡øÃ ∑§⁄U ‹¥ Á∑§ ß‚ ¬ÈÁSÃ∑§Ê ∑§Ê‚¥∑§Ã, ©ûÊ⁄U ¬òÊ ∑§ ¬Îc∆U-2 ¬⁄U ¿U¬ ‚¥∑§Ã ‚ Á◊‹ÃÊ „Ò •ÊÒ⁄U ÿ„ ÷Ë‚ÈÁŸÁ‡øÃ ∑§⁄U ‹¥ Á∑§ ¬⁄UËˇÊÊ ¬ÈÁSÃ∑§Ê •ÊÒ⁄U ©ûÊ⁄U ¬òÊ ∑§Ë ∑˝§◊ ‚¥ÅÿÊÁ◊‹ÃË „Ò– •ª⁄U ÿ„ Á÷ÛÊ „Ê ÃÊ ¬⁄UËˇÊÊÕË¸ ŒÍ‚⁄UË ¬⁄UËˇÊÊ ¬ÈÁSÃ∑§Ê •ÊÒ⁄U©ûÊ⁄U ¬òÊ ‹Ÿ ∑§ Á‹∞ ÁŸ⁄UËˇÊ∑§ ∑§Ê ÃÈ⁄UãÃ •flªÃ ∑§⁄UÊ∞°–

13. ©ûÊ⁄U ¬òÊ ∑§Ê Ÿ ◊Ê«∏¥ ∞fl¥ Ÿ „Ë ©‚ ¬⁄U •ãÿ ÁŸ‡ÊÊŸ ‹ªÊ∞°–

Test Booklet Code

¬⁄ËˇÊÊ ¬ÈÁSÃ∑§Ê ‚¥∑§Ã

E

PAPER - 1 : PHYSICS, CHEMISTRY & MATHEMATICS

¬˝‡Ÿ¬ÈÁSÃ∑§Ê - 1 : ÷ÊÒÁÃ∑§ ÁflôÊÊŸ, ⁄U‚ÊÿŸ ÁflôÊÊŸ ÃÕÊ ªÁáÊÃ

S S O

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Rohit

Text Box

JEE Main 2016 Question Paper 1 Offline (April 3, 2016) - SET E

E/Page 2 SPACE FOR ROUGH WORK / ⁄U»§ ∑§Êÿ¸ ∑§ Á‹∞ ¡ª„

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PART A — PHYSICS

ALL THE GRAPHS GIVEN ARE SCHEMATIC

AND NOT DRAWN TO SCALE.

1. A student measures the time period of 100

oscillations of a simple pendulum four

times. The data set is 90 s, 91 s, 95 s and

92 s. If the minimum division in the

measuring clock is 1 s, then the reported

mean time should be :

(1) 92±2 s

(2) 92±5.0 s

(3) 92±1.8 s

(4) 92±3 s

2. A particle of mass m is moving along the

side of a square of side ‘a’, with a uniform

speed v in the x-y plane as shown in the

figure :

Which of the following statements is false

for the angular momentum →

L about the

origin ?

(1)→ ∧

2

mvL R k=− when the particle is

moving from A to B.

(2)→ ∧

2

RL mv a k= − when the

particle is moving from C to D.

(3)→ ∧

2

RL mv a k= + when the

particle is moving from B to C.

(4)→ ∧

2

mvL R k= when the particle is

moving from D to A.

÷Êª A — ÷ÊÒÁÃ∑§ ÁflôÊÊŸÁŒ∞ ªÿ ‚÷Ë ª˝Ê»§ •Ê⁄UπËÿ „Ò¥ •ÊÒ⁄US∑§‹ ∑§ •ŸÈ‚Ê⁄U ⁄UπÊ¥Á∑§Ã Ÿ„Ë¥ „Ò–

1. ∞∑§ ¿UÊòÊ ∞∑§ ‚⁄U‹-•ÊflÃ¸-ŒÊ‹∑§ ∑§ 100 •ÊflÎÁûÊÿÊ¥∑§Ê ‚◊ÿ 4 ’Ê⁄U ◊Ê¬ÃÊ „Ò •ÊÒ⁄U ©Ÿ∑§Ê 90 s, 91 s,

95 s •ÊÒ⁄U 92 s ¬ÊÃÊ „Ò– ßSÃ◊Ê‹ ∑§Ë ªß¸ ÉÊ«∏Ë ∑§ÊãÿÍŸÃ◊ •À¬Ê¥‡Ê 1 s „Ò– Ã’ ◊Ê¬ ªÿ ◊Êäÿ ‚◊ÿ ∑§Ê©‚ Á‹πŸÊ øÊÁ„ÿ —

(1) 92±2 s

(2) 92±5.0 s

(3) 92±1.8 s

(4) 92±3 s

2. ÁøòÊ ◊¥ ÷È¡Ê ‘a’ ∑§Ê flª¸ x-y Ã‹ ◊¥ „Ò– m Œ˝√ÿ◊ÊŸ∑§Ê ∞∑§ ∑§áÊ ∞∑§‚◊ÊŸ ªÁÃ, v ‚ ß‚ flª¸ ∑§Ë ÷È¡Ê¬⁄U ø‹ ⁄U„Ê „Ò ¡Ò‚Ê Á∑§ ÁøòÊ ◊¥ Œ‡ÊÊ¸ÿÊ ªÿÊ „Ò–

Ã’ ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ‚Ê ∑§ÕŸ, ß‚ ∑§áÊ ∑§ ◊Í‹Á’¥ŒÈ

∑§ ÁªŒ¸ ∑§ÊáÊËÿ •ÊÉÊÍáÊ¸ →

L ∑§ Á‹ÿ, ª‹Ã „Ò?

(1)→ ∧

2

mvL R k=− , ¡’ ∑§áÊ A ‚ B ∑§Ë

•Ê⁄U ø‹ ⁄U„Ê „Ò–

(2)→ ∧

2

RL mv a k= − , ¡’ ∑§áÊ C ‚

D ∑§Ë •Ê⁄U ø‹ ⁄U„Ê „Ò–

(3)→ ∧

2

RL mv a k= + , ¡’ ∑§áÊ B ‚

C ∑§Ë •Ê⁄U ø‹ ⁄U„Ê „Ò–

(4)→ ∧

2

mvL R k= , ¡’ ∑§áÊ D ‚ A ∑§Ë •Ê⁄U

ø‹ ⁄U„Ê „Ò–

SPACE FOR ROUGH WORK / ⁄U»§ ∑§Êÿ¸ ∑§ Á‹∞ ¡ª„E/Page 3

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3. ‘m’ Œ˝√ÿ◊ÊŸ ∑§Ê ∞∑§ Á’¥ŒÈ ∑§áÊ ∞∑§ πÈ⁄UŒ⁄U ¬Õ PQR

(ÁøòÊ ŒÁπÿ) ¬⁄U ø‹ ⁄U„Ê „Ò– ∑§áÊ •ÊÒ⁄U ¬Õ ∑§ ’ËøÉÊ·¸áÊ ªÈáÊÊ¥∑§ µ „Ò– ∑§áÊ P ‚ ¿UÊ«∏ ¡ÊŸ ∑§ ’ÊŒ R ¬⁄U¬„È°ø ∑§⁄U L§∑§ ¡ÊÃÊ „Ò– ¬Õ ∑§ ÷Êª PQ •ÊÒ⁄U QR ¬⁄Uø‹Ÿ ◊¥ ∑§áÊ mÊ⁄UÊ πø¸ ∑§Ë ªß¸ ™§¡Ê¸∞° ’⁄UÊ’⁄U „Ò¥–PQ ‚ QR ¬⁄U „ÊŸ flÊ‹ ÁŒ‡ÊÊ ’Œ‹Êfl ◊¥ ∑§Êß¸ ™§¡Ê¸πø¸ Ÿ„Ë¥ „ÊÃË–Ã’ µ •ÊÒ⁄U ŒÍ⁄UË x(=QR) ∑§ ◊ÊŸ ‹ª÷ª „Ò¥ ∑˝§◊‡Ê— —

(1) 0.2 •ÊÒ⁄U 6.5 m

(2) 0.2 •ÊÒ⁄U 3.5 m

(3) 0.29 •ÊÒ⁄U 3.5 m

(4) 0.29 •ÊÒ⁄U 6.5 m

3. A point particle of mass m, moves along

the uniformly rough track PQR as shown

in the figure. The coefficient of friction,

between the particle and the rough track

equals µ. The particle is released, from rest,

from the point P and it comes to rest at a

point R. The energies, lost by the ball, over

the parts, PQ and QR, of the track, are

equal to each other, and no energy is lost

when particle changes direction from PQ

to QR.

The values of the coefficient of friction µ

and the distance x(=QR), are, respectively

close to :

(1) 0.2 and 6.5 m

(2) 0.2 and 3.5 m

(3) 0.29 and 3.5 m

(4) 0.29 and 6.5 m


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4. ∞∑§ ÷Ê⁄UÊûÊÊ‹∑§ ÷Ê⁄U ∑§Ê ¬„‹ ™§¬⁄U •ÊÒ⁄U Á»§⁄U ŸËøÃ∑§ ‹ÊÃÊ „Ò– ÿ„ ◊ÊŸÊ ¡ÊÃÊ „Ò Á∑§ Á‚»¸§ ÷Ê⁄U ∑§Ê™§¬⁄U ‹ ¡ÊŸ ◊¥ ∑§Êÿ¸ „ÊÃÊ „Ò •ÊÒ⁄U ŸËø ‹ÊŸ ◊¥ ÁSÕÁÃ¡™§¡Ê¸ ∑§Ê OÊ‚ „ÊÃÊ „Ò– ‡Ê⁄UË⁄U ∑§Ë fl‚Ê ™§¡Ê¸ ŒÃË „Ò ¡ÊÿÊ¥ÁòÊ∑§Ëÿ ™§¡Ê¸ ◊¥ ’Œ‹ÃË „Ò– ◊ÊŸ ‹¥ Á∑§ fl‚Ê mÊ⁄UÊŒË ªß¸ ™§¡Ê¸ 3.8×107 J ¬˝ÁÃ kg ÷Ê⁄U „Ò, ÃÕÊ ß‚∑§Ê◊ÊòÊ 20% ÿÊ¥ÁòÊ∑§Ëÿ ™§¡Ê¸ ◊¥ ’Œ‹ÃÊ „Ò– •’ ÿÁŒ∞∑§ ÷Ê⁄UÊûÊÊ‹∑§ 10 kg ∑§ ÷Ê⁄U ∑§Ê 1000 ’Ê⁄U 1 m

∑§Ë ™°§øÊß¸ Ã∑§ ™§¬⁄U •ÊÒ⁄U ŸËø ∑§⁄UÃÊ „Ò Ã’ ©‚∑§‡Ê⁄UË⁄U ‚ fl‚Ê ∑§Ê ˇÊÿ „Ò — (g=9.8 ms−2 ‹¥)

(1) 2.45×10−3 kg

(2) 6.45×10−3 kg

(3) 9.89×10−3 kg

(4) 12.89×10−3 kg

5. ŒÊ ‡Ê¥∑È§ ∑§Ê ©Ÿ∑§ ‡ÊË·¸ O ¬⁄U ¡Ê«∏∑§⁄U ∞∑§ ⁄UÊ‹⁄U’ŸÊÿÊ ªÿÊ „Ò •ÊÒ⁄U ©‚ AB fl CD ⁄U‹ ¬⁄U •‚◊Á◊Ã⁄UπÊ ªÿÊ „Ò (ÁøòÊ ŒÁπÿ)– ⁄UÊ‹⁄U ∑§Ê •ˇÊ CD ‚‹ê’flÃ „Ò •ÊÒ⁄U O ŒÊŸÊ¥ ⁄U‹ ∑§ ’ËøÊ’Ëø „Ò– „À∑§ ‚œ∑§‹Ÿ ¬⁄U ⁄UÊ‹⁄U ⁄U‹ ¬⁄U ß‚ ¬˝∑§Ê⁄U ‹È…∏∑§ŸÊ •Ê⁄Uê÷∑§⁄UÃÊ „Ò Á∑§ O ∑§Ê øÊ‹Ÿ CD ∑§ ‚◊Ê¥Ã⁄U „Ò (ÁøòÊŒÁπÿ)– øÊÁ‹Ã „Ê ¡ÊŸ ∑§ ’ÊŒ ÿ„ ⁄UÊ‹⁄U —

(1) ’Ê°ÿË¥ •Ê⁄U ◊È«∏ªÊ–

(2) ŒÊÿË¥ •Ê⁄U ◊È«∏ªÊ–

(3) ‚ËœÊ ø‹ÃÊ ⁄U„ªÊ–

(4) ’Êÿ¥ ÃÕÊ ŒÊÿ¥ ∑˝§◊‡Ê— ◊È«∏ÃÊ ⁄U„ªÊ–

4. A person trying to lose weight by burning

fat lifts a mass of 10 kg upto a height of

1 m 1000 times. Assume that the potential

energy lost each time he lowers the mass

is dissipated. How much fat will he use

up considering the work done only when

the weight is lifted up ? Fat supplies

3.8×107 J of energy per kg which is

converted to mechanical energy with a

20% efficiency rate. Take g=9.8 ms−2 :

(1) 2.45×10−3 kg

(2) 6.45×10−3 kg

(3) 9.89×10−3 kg

(4) 12.89×10−3 kg

5. A roller is made by joining together two

cones at their vertices O. It is kept on two

rails AB and CD which are placed

asymmetrically (see figure), with its axis

perpendicular to CD and its centre O at

the centre of line joining AB and CD (see

figure). It is given a light push so that it

starts rolling with its centre O moving

parallel to CD in the direction shown. As

it moves, the roller will tend to :

(1) turn left.

(2) turn right.

(3) go straight.

(4) turn left and right alternately.


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6. ¬ÎâflË ∑§Ë ‚Ã„ ‚ ‘h’ ™°§øÊß¸ ¬⁄U ∞∑§ ©¬ª˝„ flÎûÊÊ∑§Ê⁄U¬Õ ¬⁄U øÄ∑§⁄U ∑§Ê≈U ⁄U„Ê „Ò (¬ÎâflË ∑§Ë ÁòÊíÿÊ R ÃÕÊh<<R)– ¬ÎâflË ∑§ ªÈL§àfl ˇÊòÊ ‚ ¬‹ÊÿŸ ∑§⁄UŸ ∑§Á‹ÿ ß‚∑§Ë ∑§ˇÊËÿ ªÁÃ ◊¥ •Êfl‡ÿ∑§ ãÿÍŸÃ◊ ’Œ‹Êfl„Ò — (flÊÿÈ◊¥«U‹Ëÿ ¬˝÷Êfl ∑§Ê Ÿªáÿ ‹ËÁ¡∞–)

(1) 2 gR

(2) gR

(3) 2gR /

(4) ( ) 2 1gR −

7. ∞∑§ ¬ãU«ÈU‹◊ ÉÊ«∏Ë 408C ÃÊ¬◊ÊŸ ¬⁄U 12 s ¬˝ÁÃÁŒŸœË◊Ë „Ê ¡ÊÃË „Ò ÃÕÊ 208C ÃÊ¬◊ÊŸ ¬⁄U 4 s ¬˝ÁÃÁŒŸÃ$¡ „Ê ¡ÊÃË „Ò– ÃÊ¬◊ÊŸ Á¡‚ ¬⁄U ÿ„ ‚„Ë ‚◊ÿŒ‡ÊÊ¸ÿªË ÃÕÊ ¬ãU«ÈU‹◊ ∑§Ë œÊÃÈ ∑§Ê ⁄UπËÿ-¬˝‚Ê⁄U ªÈáÊÊ¥∑§(α) ∑˝§◊‡Ê— „Ò¥ —

(1) 258C; α=1.85×10−5/8C

(2) 608C; α=1.85×10−4/8C

(3) 308C; α=1.85×10−3/8C

(4) 558C; α=1.85×10−2/8C

6. A satellite is revolving in a circular orbit at

a height ‘h’ from the earth’s surface (radius

of earth R ; h<<R). The minimum increase

in its orbital velocity required, so that the

satellite could escape from the earth’s

gravitational field, is close to : (Neglect

the effect of atmosphere.)

(1) 2 gR

(2) gR

(3) 2gR /

(4) ( ) 2 1gR −

7. A pendulum clock loses 12 s a day if the

temperature is 408C and gains 4 s a day if

the temperature is 208C. The temperature

at which the clock will show correct time,

and the co-efficient of linear expansion

(α) of the metal of the pendulum shaft are

respectively :

(1) 258C; α=1.85×10−5/8C

(2) 608C; α=1.85×10−4/8C

(3) 308C; α=1.85×10−3/8C

(4) 558C; α=1.85×10−2/8C


E E

E E

8. ∞∑§ •ÊŒ‡Ê¸ ªÒ‚ ©à∑˝§◊áÊËÿ SÕÒÁÃ∑§-∑§À¬ ¬˝∑˝§◊ ‚ªÈ$¡⁄UÃË „Ò ÃÕÊ ©‚∑§Ë ◊Ê‹⁄U-™§c◊Ê-œÊÁ⁄UÃÊ C ÁSÕ⁄U⁄U„ÃË „Ò– ÿÁŒ ß‚ ¬˝∑˝§◊ ◊¥ ©‚∑§ ŒÊ’ P fl •ÊÿÃŸV ∑§ ’Ëø ‚¥’¥œ PVn=constant „Ò– (C

P ÃÕÊ

CV ∑˝§◊‡Ê— ÁSÕ⁄U ŒÊ’ fl ÁSÕ⁄U •ÊÿÃŸ ¬⁄U ™§c◊Ê-

œÊÁ⁄UÃÊ „Ò) Ã’ ‘n’ ∑§ Á‹ÿ ‚◊Ë∑§⁄UáÊ „Ò —

(1) P

V

Cn

C=

(2)

P

V

C Cn

C C

−

=

−

(3)

P

V

C Cn

C C

−

=

−

(4)

V

P

C Cn

C C

−

=

−

9. ‘n’ ◊Ê‹ •ÊŒ‡Ê¸ ªÒ‚ ∞∑§ ¬˝∑˝§◊ A→B ‚ ªÈ$¡⁄UÃË „Ò(ÁøòÊ ŒÁπÿ)– ß‚ ¬˝∑˝§◊ ∑§ ŒÊÒ⁄UÊŸ ©‚∑§Ê •Áœ∑§Ã◊ÃÊ¬◊ÊŸ „ÊªÊ —

(1) 0 09

4

P V

nR

(2) 0 03

2

P V

nR

(3) 0 09

2

P V

nR

(4) 0 09 P V

nR

8. An ideal gas undergoes a quasi static,

reversible process in which its molar heat

capacity C remains constant. If during this

process the relation of pressure P and

volume V is given by PVn=constant, then

n is given by (Here CP and C

V are molar

specific heat at constant pressure and

constant volume, respectively) :

(1) P

V

Cn

C=

(2)

P

V

C Cn

C C

−

=

−

(3)

P

V

C Cn

C C

−

=

−

(4)

V

P

C Cn

C C

−

=

−

9. ‘n’ moles of an ideal gas undergoes a

process A→B as shown in the figure. The

maximum temperature of the gas during

the process will be :

(1) 0 09

4

P V

nR

(2) 0 03

2

P V

nR

(3) 0 09

2

P V

nR

(4) 0 09 P V

nR


E E

E E

10. ∞∑§ ∑§áÊ ‘A’ •ÊÿÊ◊ ‚ ‚⁄U‹-•ÊflÃ¸ ŒÊ‹Ÿ ∑§⁄U ⁄U„Ê

„Ò– ¡’ ÿ„ •¬Ÿ ◊Í‹-SÕÊŸ ‚ 23

A ¬⁄U ¬„È°øÃÊ „Ò

Ã’ •øÊŸ∑§ ß‚∑§Ë ªÁÃ ÁÃªÈŸË ∑§⁄U ŒË ¡ÊÃË „Ò– Ã’

ß‚∑§Ê ŸÿÊ •ÊÿÊ◊ „Ò —

(1) 413

A

(2) 3A

(3) 3A

(4)7

3

A

11. 20 m ‹ê’Êß¸ ∑§Ë ∞∑§‚◊ÊŸ «UÊ⁄UË ∑§Ê ∞∑§ ŒÎ…∏ •ÊœÊ⁄U‚ ‹≈U∑§ÊÿÊ ªÿÊ „Ò– ß‚∑§ ÁŸø‹ Á‚⁄U ‚ ∞∑§ ‚Íˇ◊Ã⁄¥Uª-S¬¥Œ øÊÁ‹Ã „ÊÃÊ „Ò– ™§¬⁄U •ÊœÊ⁄U Ã∑§ ¬„È°øŸ◊¥ ‹ªŸ flÊ‹Ê ‚◊ÿ „Ò —(g = 10 ms−2 ‹¥)

(1) 2 2 sπ

(2) 2 s

(3) 2 2 s

(4) 2 s

10. A particle performs simple harmonic

motion with amplitude A. Its speed is

trebled at the instant that it is at a distance

2

3

A from equilibrium position. The new

amplitude of the motion is :

(1) 413

A

(2) 3A

(3) 3A

(4)7

3

A

11. A uniform string of length 20 m is

suspended from a rigid support. A short

wave pulse is introduced at its lowest end.

It starts moving up the string. The time

taken to reach the support is :

(take g = 10 ms−2)

(1) 2 2 sπ

(2) 2 s

(3) 2 2 s

(4) 2 s


E E

E E

12. ÁòÊíÿÊ ‘a’ ÃÕÊ ‘b’ ∑§ ŒÊ ∞∑§-∑§ãŒ˝Ë ªÊ‹Ê¥ ∑§ (ÁøòÊŒÁπÿ) ’Ëø ∑§ SÕÊŸ ◊¥ •ÊÿÃŸ •Êfl‡Ê-ÉÊŸàfl

A

r =ρ „Ò, ¡„Ê° A ÁSÕ⁄UÊ¥∑§ „Ò ÃÕÊ r ∑§ãŒ˝ ‚ ŒÍ⁄UË „Ò–

ªÊ‹Ê¥ ∑§ ∑§ãŒ˝ ¬⁄U ∞∑§ Á’ãŒÈ-•Êfl‡Ê Q „Ò– ‘A’ ∑§Êfl„ ◊ÊŸ ’ÃÊÿ¥ Á¡‚‚ ªÊ‹Ê¥ ∑§ ’Ëø ∑§ SÕÊŸ ◊¥∞∑§‚◊ÊŸ flÒlÈÃ-ˇÊòÊ „Ê —

(1)2

2

Q

aπ

(2)( )2 2

2

Q

b a−π

(3)( )2 2

2

Q

a b−π

(4)2

2Q

aπ

12. The region between two concentric spheres

of radii ‘a’ and ‘b’, respectively (see figure),

has volume charge density A

r =ρ , where

A is a constant and r is the distance from

the centre. At the centre of the spheres is

a point charge Q. The value of A such

that the electric field in the region between

the spheres will be constant, is :

(1)2

2

Q

aπ

(2)( )2 2

2

Q

b a−π

(3)( )2 2

2

Q

a b−π

(4)2

2Q

aπ


E E

E E

13. ‚¥œÊÁ⁄UòÊÊ¥ ‚ ’Ÿ ∞∑§ ¬Á⁄U¬Õ ∑§Ê ÁøòÊ ◊¥ ÁŒπÊÿÊ ªÿÊ„Ò– ∞∑§ Á’ãŒÈ-•Êfl‡Ê Q (Á¡‚∑§Ê ◊ÊŸ 4 µF ÃÕÊ9 µF flÊ‹ ‚¥œÊÁ⁄UòÊÊ¥ ∑§ ∑È§‹ •Êfl‡ÊÊ¥ ∑§ ’⁄UÊ’⁄U „Ò)∑§ mÊ⁄UÊ 30 m ŒÍ⁄UË ¬⁄U flÒlÈÃ-ˇÊòÊ ∑§Ê ¬Á⁄U◊ÊáÊ „ÊªÊ —

(1) 240 N/C

(2) 360 N/C

(3) 420 N/C

(4) 480 N/C

14. ÃÊ°’Ê ÃÕÊ •◊ÊÁŒÃ (undoped) Á‚Á‹∑§ÊŸ ∑§¬˝ÁÃ⁄UÊœÊ¥ ∑§Ë ©Ÿ∑§ ÃÊ¬◊ÊŸ ¬⁄U ÁŸ÷¸⁄UÃÊ, 300-400 K

ÃÊ¬◊ÊŸ •¥Ã⁄UÊ‹ ◊¥, ∑§ Á‹ÿ ‚„Ë ∑§ÕŸ „Ò —

(1) ÃÊ°’Ê ∑§ Á‹ÿ ⁄UπËÿ ’…∏Êfl ÃÕÊ Á‚Á‹∑§ÊŸ ∑§Á‹ÿ ⁄UπËÿ ’…∏Êfl–

(2) ÃÊ°’Ê ∑§ Á‹ÿ ⁄UπËÿ ’…∏Êfl ÃÕÊ Á‚Á‹∑§ÊŸ ∑§Á‹ÿ ø⁄UUÉÊÊÃÊ¥∑§Ë ’…∏Êfl–

(3) ÃÊ°’Ê ∑§ Á‹ÿ ⁄UπËÿ ’…∏Êfl ÃÕÊ Á‚Á‹∑§ÊŸ ∑§Á‹ÿ ø⁄UUÉÊÊÃÊ¥∑§Ë ÉÊ≈UÊfl–

(4) ÃÊ°’Ê ∑§ Á‹ÿ ⁄UπËÿ ÉÊ≈UÊfl ÃÕÊ Á‚Á‹∑§ÊŸ ∑§Á‹ÿ ⁄UπËÿ ÉÊ≈UÊfl–

13. A combination of capacitors is set up as

shown in the figure. The magnitude of

the electric field, due to a point charge Q

(having a charge equal to the sum of the

charges on the 4 µF and 9 µF capacitors),

at a point distant 30 m from it, would

equal :

(1) 240 N/C

(2) 360 N/C

(3) 420 N/C

(4) 480 N/C

14. The temperature dependence of resistances

of Cu and undoped Si in the temperature

range 300-400 K, is best described by :

(1) Linear increase for Cu, linear

increase for Si.

(2) Linear increase for Cu, exponential

increase for Si.

(3) Linear increase for Cu, exponential

decrease for Si.

(4) Linear decrease for Cu, linear

decrease for Si.


E E

E E

15. ŒÊ ∞∑§‚◊ÊŸ ÃÊ⁄U A fl B ¬˝àÿ∑§ ∑§Ë ‹ê’Êß¸ ‘l’, ◊¥‚◊ÊŸ œÊ⁄UÊ I ¬˝flÊÁ„Ã „Ò– A ∑§Ê ◊Ê«∏∑§⁄U R ÁòÊíÿÊ ∑§Ê∞∑§ flÎûÊ •ÊÒ⁄U B ∑§Ê ◊Ê«∏∑§⁄U ÷È¡Ê ‘a’ ∑§Ê ∞∑§ flª¸’ŸÊÿÊ ¡ÊÃÊ „Ò– ÿÁŒ B

A ÃÕÊ B

B ∑˝§◊‡Ê— flÎûÊ ∑§

∑§ãŒ˝ ÃÕÊ flª¸ ∑§ ∑§ãŒ˝ ¬⁄U øÈê’∑§Ëÿ ˇÊòÊ „Ò ¥,

Ã’ •ŸÈ¬ÊÃ A

B

B

B

„ÊªÊ —

(1)2

8

π

(2)2

16 2

π

(3)2

16

π

(4)2

8 2

π

16. ŒÊ øÈê’∑§Ëÿ ¬ŒÊÕ¸ A ÃÕÊ B ∑§ Á‹ÿ Á„S≈⁄UÁ‚‚-‹Í¬ ŸËø ÁŒπÊÿ ªÿ „Ò¥ —

ßŸ ¬ŒÊÕÊZ ∑§Ê øÈê’∑§Ëÿ ©¬ÿÊª ÁfllÈÃ-¡Ÿ⁄U≈U⁄U ∑§øÈê’∑§, ≈˛UÊã‚»§ÊÚ◊¸⁄U ∑§Ë ∑˝§Ê«U ∞fl¥ ÁfllÈÃ-øÈê’∑§ ∑§Ë∑˝§Ê«U •ÊÁŒ ∑§ ’ŸÊŸ ◊¥ Á∑§ÿÊ ¡ÊÃÊ „Ò– Ã’ ÿ„ ©ÁøÃ„Ò Á∑§ —

(1) A ∑§Ê ßSÃ◊Ê‹ ÁfllÈÃ-¡Ÿ⁄U≈U⁄U ÃÕÊ ≈˛UÊã‚»§ÊÚ◊¸⁄UŒÊŸÊ¥ ◊¥ Á∑§ÿÊ ¡Ê∞–

(2) A ∑§Ê ßSÃ◊Ê‹ ÁfllÈÃ-øÈê’∑§ ◊¥ ÃÕÊ B ∑§ÊÁfllÈÃ-¡Ÿ⁄U≈U⁄U ◊¥ Á∑§ÿÊ ¡Ê∞–

(3) A ∑§Ê ßSÃ◊Ê‹ ≈˛UÊã‚»§ÊÚ◊¸⁄U ◊¥ ÃÕÊ B ∑§ÊÁfllÈÃ-¡Ÿ⁄U≈U⁄U ◊¥ Á∑§ÿÊ ¡Ê∞–

(4) B ∑§Ê ßSÃ◊Ê‹ ÁfllÈÃ-øÈê’∑§ ÃÕÊ ≈˛UÊã‚»§ÊÚ◊¸⁄UŒÊŸÊ¥ ◊¥ Á∑§ÿÊ ¡Ê∞–

15. Two identical wires A and B, each of length

‘l’, carry the same current I. Wire A is bent

into a circle of radius R and wire B is bent

to form a square of side ‘a’. If BA

and BB

are the values of magnetic field at the

centres of the circle and square

respectively, then the ratio A

B

B

B

is :

(1)2

8

π

(2)2

16 2

π

(3)2

16

π

(4)2

8 2

π

16. Hysteresis loops for two magnetic materials

A and B are given below :

These materials are used to make magnets

for electric generators, transformer core

and electromagnet core. Then it is proper

to use :

(1) A for electric generators and

transformers.

(2) A for electromagnets and B for

electric generators.

(3) A for transformers and B for electric

generators.

(4) B for electromagnets and

transformers.


E E

E E

17. ∞∑§ •Ê∑¸§ ‹Òê¬ ∑§Ê ¬˝∑§ÊÁ‡ÊÃ ∑§⁄UŸ ∑§ Á‹ÿ 80 V ¬⁄U10 A ∑§Ë ÁŒc≈U œÊ⁄UÊ (DC) ∑§Ë •Êfl‡ÿ∑§ÃÊ „ÊÃË „Ò–©‚Ë •Ê∑¸§ ∑§Ê 220 V (rms) 50 Hz ¬˝àÿÊflÃË¸ œÊ⁄UÊ(AC) ‚ ø‹ÊŸ ∑§ Á‹ÿ üÊáÊË ◊¥ ‹ªŸ flÊ‹ ¬˝⁄U∑§àfl∑§Ê ◊ÊŸ „Ò —

(1) 80 H

(2) 0.08 H

(3) 0.044 H

(4) 0.065 H

18. ÁŸêŸ ¬˝ÁÃ ÄflÊ¥≈U◊ flÒlÈÃ-øÈê’∑§Ëÿ ÁflÁ∑§⁄UáÊÊ¥ ∑§Ê ©Ÿ∑§Ë™§¡Ê¸ ∑§ ’…∏Ã „È∞ ∑˝§◊ ◊¥ ‹ªÊÿ¥ —

A : ŸË‹Ê ¬˝∑§Ê‡Ê B : ¬Ë‹Ê ¬˝∑§Ê‡Ê

C : X - Á∑§⁄UáÊ¥ D : ⁄UÁ«UÿÊ Ã⁄¥Uª

(1) D, B, A, C

(2) A, B, D, C

(3) C, A, B, D

(4) B, A, D, C

19. ŒÍ⁄U ÁSÕÃ 10 m ™°§ø ¬«∏ ∑§Ê ∞∑§ 20 •Êflœ¸Ÿ ˇÊ◊ÃÊflÊ‹ ≈UÁ‹S∑§Ê¬ ‚ ŒπŸ ¬⁄U ÄÿÊ ◊„‚Í‚ „ÊªÊ?

(1) ¬«∏ 10 ªÈŸÊ ™°§øÊ „Ò–

(2) ¬«∏ 10 ªÈŸÊ ¬Ê‚ „Ò–

(3) ¬«∏ 20 ªÈŸÊ ™°§øÊ „Ò–

(4) ¬«∏ 20 ªÈŸÊ ¬Ê‚ „Ò–

17. An arc lamp requires a direct current of

10 A at 80 V to function. If it is connected

to a 220 V (rms), 50 Hz AC supply, the

series inductor needed for it to work is

close to :

(1) 80 H

(2) 0.08 H

(3) 0.044 H

(4) 0.065 H

18. Arrange the following electromagnetic

radiations per quantum in the order of

increasing energy :

A : Blue light B : Yellow light

C : X-ray D : Radiowave.

(1) D, B, A, C

(2) A, B, D, C

(3) C, A, B, D

(4) B, A, D, C

19. An observer looks at a distant tree of

height 10 m with a telescope of magnifying

power of 20. To the observer the tree

appears :

(1) 10 times taller.

(2) 10 times nearer.

(3) 20 times taller.

(4) 20 times nearer.


E E

E E

20. ∞∑§ Á¬Ÿ-„Ê‹ ∑Ò§◊⁄UÊ ∑§Ë ÀÊê’Êß¸ ‘L’ „Ò ÃÕÊ Á¿UŒ˝ ∑§ËÁòÊíÿÊ a „Ò– ©‚ ¬⁄U λ Ã⁄¥UªŒÒÉÿ¸ ∑§Ê ‚◊Ê¥Ã⁄U ¬˝∑§Ê‡Ê•Ê¬ÁÃÃ „Ò– Á¿UŒ˝ ∑§ ‚Ê◊Ÿ flÊ‹Ë ‚Ã„ ¬⁄U ’Ÿ S¬ÊÚ≈U∑§Ê ÁflSÃÊ⁄U Á¿UŒ˝ ∑§ íÿÊÁ◊ÃËÿ •Ê∑§Ê⁄U ÃÕÊ ÁflflÃ¸Ÿ∑§ ∑§Ê⁄UáÊ „È∞ ÁflSÃÊ⁄U ∑§Ê ∑È§‹ ÿÊª „Ò– ß‚ S¬ÊÚ≈U ∑§ÊãÿÍŸÃ◊ •Ê∑§Ê⁄U b

min Ã’ „ÊªÊ ¡’ —

(1)2

aL=

λÃÕÊ b

min=

22

L

λ

(2) a L= λ ÃÕÊ bmin

=

22

L

λ

(3) a L= λ ÃÕÊ bmin

= 4 Lλ

(4)2

aL=

λÃÕÊ b

min= 4 Lλ

21. ∞∑§ »§Ê≈UÊ-‚‹ ¬⁄U λ Ã⁄¥UªŒÒÉÿ¸ ∑§Ê ¬˝∑§Ê‡Ê •Ê¬ÁÃÃ„Ò– ©à‚Á¡¸Ã ß‹Ä≈˛UÊÚŸ ∑§Ë •Áœ∑§Ã◊ ªÁÃ ‘v’ „Ò–

ÿÁŒ Ã⁄¥UªŒÒÉÿ¸ 3

4

λ „Ê Ã’ ©à‚Á¡¸Ã ß‹Ä≈˛UÊÚŸ ∑§Ë

•Áœ∑§Ã◊ ªÁÃ „ÊªË —

(1)

1

24

3v

>

(2)

1

24

3v

<

(3)

1

24

3v =

(4)

1

23

4v =

20. The box of a pin hole camera, of length L,

has a hole of radius a. It is assumed that

when the hole is illuminated by a parallel

beam of light of wavelength λ the spread

of the spot (obtained on the opposite wall

of the camera) is the sum of its geometrical

spread and the spread due to diffraction.

The spot would then have its minimum

size (say bmin

) when :

(1)2

aL=

λand b

min=

22

L

λ

(2) a L= λ and bmin

=

22

L

λ

(3) a L= λ and bmin

= 4 Lλ

(4)2

aL=

λand b

min= 4 Lλ

21. Radiation of wavelength λ, is incident on

a photocell. The fastest emitted electron

has speed v. If the wavelength is changed

to 3

4

λ, the speed of the fastest emitted

electron will be :

(1)

1

24

3v

>

(2)

1

24

3v

<

(3)

1

24

3v =

(4)

1

23

4v =


E E

E E

22. ŒÊ ⁄UÁ«UÿÊœ◊Ë¸ Ãàfl A ÃÕÊ B ∑§Ë •h¸•ÊÿÈ ∑˝§◊‡Ê—20 min ÃÕÊ 40 min „Ò¥– ¬˝Ê⁄¥U÷ ◊¥ ŒÊŸÊ¥ ∑§ Ÿ◊ÍŸÊ¥ ◊¥ŸÊÁ÷∑§Ê¥ ∑§Ë ‚¥ÅÿÊ ’⁄UÊ’⁄U „Ò– 80 min ∑§ ©¬⁄UÊ¥ÃA ÃÕÊ B ∑§ ˇÊÿ „È∞ ŸÊÁ÷∑§Ê¥ ∑§Ë ‚¥ÅÿÊ ∑§Ê •ŸÈ¬ÊÃ„ÊªÊ —

(1) 1 : 16

(2) 4 : 1

(3) 1 : 4

(4) 5 : 4

23. ∞∑§ ª≈U ◊¥ a, b, c, d ßŸ¬È≈U „Ò¥ •ÊÒ⁄U x •Ê™§≈U¬È≈U „Ò–Ã’ ÁŒÿ ªÿ ≈UÊß◊-ª˝Ê»§ ∑§ •ŸÈ‚Ê⁄U ª≈U „Ò —

(1) NOT

(2) AND

(3) OR

(4) NAND

22. Half-lives of two radioactive elements

A and B are 20 minutes and 40 minutes,

respectively. Initially, the samples have

equal number of nuclei. After 80 minutes,

the ratio of decayed numbers of A and B

nuclei will be :

(1) 1 : 16

(2) 4 : 1

(3) 1 : 4

(4) 5 : 4

23. If a, b, c, d are inputs to a gate and x is its

output, then, as per the following time

graph, the gate is :

(1) NOT

(2) AND

(3) OR

(4) NAND


E E

E E

24. ‚„Ë ∑§ÕŸ øÈÁŸÿ —(1) •ÊÿÊ◊ ◊Ê«È‹Ÿ ◊¥ ©ìÊ •ÊflÎÁûÊ ∑§Ë flÊ„∑§U

Ã⁄¥Uª ∑§ •ÊÿÊ◊ ◊¥ ’Œ‹Êfl äflÁŸ Á‚ÇŸ‹ ∑§•ÊÿÊ◊ ∑§ •ŸÈ¬ÊÃË „Ò–

(2) •ÊÿÊ◊ ◊Ê«ÈU‹Ÿ ◊¥ ©ìÊ •ÊflÎÁûÊ ∑§Ë flÊ„∑§Ã⁄¥Uª ∑§Ë •ÊflÎÁûÊ ◊¥ ’Œ‹Êfl äflÁŸ Á‚ÇãÊ‹ ∑§•ÊÿÊ◊ ∑§ •ŸÈ¬ÊÃË „Ò–

(3) •ÊflÎÁûÊ ◊Ê«ÈU‹Ÿ ◊¥ ©ìÊ •ÊflÎÁûÊ ∑§Ë flÊ„∑§Ã⁄¥Uª ∑§ •ÊÿÊ◊ ◊¥ ’Œ‹Êfl äflÁŸ Á‚ÇãÊ‹ ∑§•ÊÿÊ◊ ∑§ •ŸÈ¬ÊÃË „Ò–

(4) •ÊflÎÁûÊ ◊Ê«ÈU‹Ÿ ◊¥ ©ìÊ-•ÊflÎÁûÊ ∑§Ë flÊ„∑§Ã⁄¥Uª ∑§Ë •ÊÿÊ◊ ◊¥ ’Œ‹Êfl äflÁŸ Á‚ÇŸ‹ ∑§Ë•ÊflÎÁûÊ ∑§ •ŸÈ¬ÊÃË „Ò–

25. ∞∑§ S∑Í§-ª¡ ∑§Ê Á¬ø 0.5 mm „Ò •ÊÒ⁄U ©‚∑§ flÎûÊËÿ-S∑§‹ ¬⁄U 50 ÷Êª „Ò¥– ß‚∑§ mÊ⁄UÊ ∞∑§ ¬Ã‹Ë•ÀÿÈ◊ËÁŸÿ◊ ‡ÊË≈U ∑§Ë ◊Ê≈UÊß¸ ◊Ê¬Ë ªß¸– ◊Ê¬ ‹Ÿ ∑§¬Ífl¸ ÿ„ ¬ÊÿÊ ªÿÊ Á∑§ ¡’ S∑˝Í§-ª¡ ∑§ ŒÊ ¡ÊÚflÊ¥ ∑§ÊSÊê¬∑¸§U ◊¥ ‹ÊÿÊ ¡ÊÃÊ „Ò Ã’ 45 flÊ¥ ÷Êª ◊ÈÅÿ S∑§‹‹Êß¸Ÿ ∑§ ‚¥¬ÊÃË „ÊÃÊ „Ò •ÊÒ⁄U ◊ÈÅÿ S∑§‹ ∑§Ê ‡ÊÍãÿ (0)

◊ÈÁ‡∑§‹ ‚ ÁŒπÃÊ „Ò– ◊ÈÅÿ S∑§‹ ∑§Ê ¬Ê∆KÊ¥∑§ ÿÁŒ0.5 mm ÃÕÊ 25 flÊ¥ ÷Êª ◊ÈÅÿ S∑§‹ ‹Êß¸Ÿ ∑§‚¥¬ÊÃË „Ê, ÃÊ ‡ÊË≈U ∑§Ë ◊Ê≈UÊß¸ ÄÿÊ „ÊªË?

(1) 0.75 mm

(2) 0.80 mm

(3) 0.70 mm

(4) 0.50 mm

24. Choose the correct statement :

(1) In amplitude modulation the

amplitude of the high frequency

carrier wave is made to vary in

proportion to the amplitude of the

audio signal.

(2) In amplitude modulation the

frequency of the high frequency



audio signal.

(3) In frequency modulation the




audio signal.

(4) In frequency modulation the



proportion to the frequency of the

audio signal.

25. A screw gauge with a pitch of 0.5 mm and

a circular scale with 50 divisions is used to

measure the thickness of a thin sheet of

Aluminium. Before starting the

measurement, it is found that when the

two jaws of the screw gauge are brought

in contact, the 45th division coincides with

the main scale line and that the zero of

the main scale is barely visible. What is

the thickness of the sheet if the main scale

reading is 0.5 mm and the 25th division

coincides with the main scale line ?

(1) 0.75 mm

(2) 0.80 mm

(3) 0.70 mm

(4) 0.50 mm


E E

E E

26. ŒÊŸÊ¥ Á‚⁄UÊ¥ ¬⁄U πÈ‹ ∞∑§ ¬Êß¬ ∑§Ë flÊÿÈ ◊¥◊Í‹-•ÊflÎÁûÊ ‘f ’ „Ò– ¬Êß¬ ∑§Ê ™§äflÊ¸œ⁄U ©‚∑§Ë•ÊœË-‹ê’Êß¸ Ã∑§ ¬ÊŸË ◊¥ «ÈU’ÊÿÊ ¡ÊÃÊ „Ò– Ã’ ß‚◊¥’ø flÊÿÈ-∑§Ê‹◊ ∑§Ë ◊Í‹ •ÊflÎÁûÊ „ÊªË —

(1)2

f

(2)3

4

f

(3) 2 f

(4) f

27. ∞∑§ ªÒÀflŸÊ◊Ë≈U⁄U ∑§ ∑§Êß‹ ∑§Ê ¬˝ÁÃ⁄UÊœ 100 Ω „Ò–1 mA œÊ⁄UÊ ¬˝flÊÁ„Ã ∑§⁄UŸ ¬⁄U ß‚◊¥ »È§‹-S∑§‹ ÁflˇÊ¬Á◊‹ÃÊ „Ò– ß‚ ªÒÀflŸÊ◊Ë≈U⁄U ∑§Ê 10 A ∑§ ∞◊Ë≈U⁄U ◊¥’Œ‹Ÿ ∑§ Á‹ÿ ¡Ê ¬˝ÁÃ⁄UÊœ ‹ªÊŸÊ „ÊªÊ fl„ „Ò —

(1) 0.01 Ω

(2) 2 Ω

(3) 0.1 Ω

(4) 3 Ω

28. ∞∑§ ¬˝ÿÊª ∑§⁄U∑§ ÃÕÊ i− δ ª˝Ê»§ ’ŸÊ∑§⁄U ∞∑§ ∑§Ê°ø‚ ’Ÿ Á¬˝ï◊ ∑§Ê •¬flÃ¸ŸÊ¥∑§ ÁŸ∑§Ê‹Ê ¡ÊÃÊ „Ò– ¡’∞∑§ Á∑§⁄UáÊ ∑§Ê 358 ¬⁄U •Ê¬ÁÃÃ ∑§⁄UŸ ¬⁄U fl„ 408 ‚ÁfløÁ‹Ã „ÊÃË „Ò ÃÕÊ ÿ„ 798 ¬⁄U ÁŸª¸◊ „ÊÃË „Ò– ß‚ÁSÕÁÃ ◊¥ ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ‚Ê ◊ÊŸ •¬flÃ¸ŸÊ¥∑§ ∑§•Áœ∑§Ã◊ ◊ÊŸ ∑§ ‚’‚ ¬Ê‚ „Ò?

(1) 1.5

(2) 1.6

(3) 1.7

(4) 1.8

26. A pipe open at both ends has a

fundamental frequency f in air. The pipe

is dipped vertically in water so that half of

it is in water. The fundamental frequency

of the air column is now :

(1)2

f

(2)3

4

f

(3) 2 f

(4) f

27. A galvanometer having a coil resistance

of 100 Ω gives a full scale deflection, when

a current of 1 mA is passed through it.

The value of the resistance, which can

convert this galvanometer into ammeter

giving a full scale deflection for a current

of 10 A, is :

(1) 0.01 Ω

(2) 2 Ω

(3) 0.1 Ω

(4) 3 Ω

28. In an experiment for determination of

refractive index of glass of a prism by

i− δ, plot, it was found that a ray incident

at angle 358, suffers a deviation of 408 and

that it emerges at angle 798⋅ Ιn that case

which of the following is closest to the

maximum possible value of the refractive

index ?

(1) 1.5

(2) 1.6

(3) 1.7

(4) 1.8


E E

E E

29. ÁøòÊ (a), (b), (c), (d) Œπ∑§⁄U ÁŸœÊ¸Á⁄UÃ ∑§⁄¥U Á∑§ ÿÁøòÊ ∑˝§◊‡Ê— Á∑§Ÿ ‚◊Ë∑§ã«UÄ≈U⁄U Á«UflÊß¸‚ ∑§•Á÷‹ˇÊÁáÊ∑§ ª˝Ê»§ „Ò¥?

(1) ‚ÊœÊ⁄UáÊ «UÊÿÊ«U, ¡ËŸ⁄U «UÊÿÊ«, ‚Ê‹⁄U ‚‹,LDR (‹Êß¸≈U Á«U¬ã«Uã≈U ⁄UÁ¡S≈Uã‚)

(2) ¡ËŸ⁄U «UÊÿÊ«U, ‚ÊœÊ⁄UáÊ «UÊÿÊ«U, LDR (‹Êß¸≈UÁ«U¬ã«Uã≈U ⁄UÁ¡S≈Uã‚), ‚Ê‹⁄U ‚‹

(3) ‚Ê‹⁄U ‚‹, LDR (‹Êß≈U Á«U¬ã«Uã≈U ⁄UÁ¡S≈Uã‚),¡ËŸ⁄U «UÊÿÊ«U, ‚ÊœÊ⁄UáÊ «UÊÿÊ«U

(4) ¡ËŸ⁄U «UÊÿÊ«U, ‚Ê‹⁄U ‚‹, ‚ÊœÊ⁄UáÊ «UÊÿÊ«U,LDR (‹Êß¸≈U Á«U¬ã«Uã≈U ⁄UÁ¡S≈Uã‚)

30. ©÷ÿÁŸc∆U-©à‚¡¸∑§ ÁflãÿÊ‚ ∑§ Á‹ÿ α ÃÕÊ β ∑§’Ëø ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ‚Ê ‚¥’¥œ ª‹Ã „Ò? α ÃÕÊ βÁøq ‚Ê◊Êãÿ ◊Ã‹’ flÊ‹ „Ò¥ —

(1)1 1

1= +

α β

(2) 1

=

−

βα

β

(3) 1

=

+

βα

β

(4)2

2

1

=

+

βα

β

29. Identify the semiconductor devices whose

characteristics are given below, in the

order (a), (b), (c), (d) :

(1) Simple diode, Zener diode, Solar cell,

Light dependent resistance

(2) Zener diode, Simple diode, Light

dependent resistance, Solar cell

(3) Solar cell, Light dependent

resistance, Zener diode, Simple

diode

(4) Zener diode, Solar cell, Simple diode,

Light dependent resistance

30. For a common emitter configuration, if

α and β have their usual meanings, the

incorrect relationship between α and β

is :

(1)1 1

1= +

α β

(2) 1

=

−

βα

β

(3) 1

=

+

βα

β

(4)2

2

1

=

+

βα

β


E E

E E

÷Êª B — ⁄U‚ÊÿŸ ÁflôÊÊŸ

31. 300 K ÃÕÊ 1 atm ŒÊ’ ¬⁄ U, 15 mL ªÒ‚Ëÿ„Êß«˛UÊ∑§Ê’¸Ÿ ∑§ ¬ÍáÊ¸ Œ„Ÿ ∑§ Á‹ÿ 375 mL flÊÿÈÁ¡‚◊¥ •ÊÿÃŸ ∑§ •ÊœÊ⁄U ¬⁄U 20% •ÊÚÄ‚Ë¡Ÿ „Ò, ∑§Ë•Êfl‡ÿ∑§ÃÊ „ÊÃË „Ò– Œ„Ÿ ∑§ ’ÊŒ ªÒ‚¥ 330 mL

ÉÊ⁄UÃË „Ò– ÿ„ ◊ÊŸÃ „È∞ Á∑§ ’ŸÊ „È•Ê ¡‹ Œ˝fl M§¬ ◊¥„Ò ÃÕÊ ©‚Ë ÃÊ¬◊ÊŸ ∞fl¥ ŒÊ’ ¬⁄U •ÊÿÃŸÊ¥ ∑§Ë ◊Ê¬ ∑§Ëªß¸ „Ò ÃÊ „Êß«˛UÊ∑§Ê’¸Ÿ ∑§Ê »§Ê◊¸Í‹Ê „Ò —

(1) C3H

6

(2) C3H

8

(3) C4H

8

(4) C4H

10

32. ‚◊ÊŸ •ÊÿÃŸ (V) ∑§ ŒÊ ’¥Œ ’À’, Á¡Ÿ◊¥ ∞∑§ •ÊŒ‡Ê¸ªÒ‚ ¬˝Ê⁄UÁê÷∑§ ŒÊ’ pi ÃÕÊ ÃÊ¬ T

1 ¬⁄U ÷⁄UË ªß¸ „Ò, ∞∑§Ÿªáÿ •ÊÿÃŸ ∑§Ë ¬Ã‹Ë ≈˜UÿÍ’ ‚ ¡È«∏ „Ò¥ ¡Ò‚Ê Á∑§ŸËø ∑§ ÁøòÊ ◊¥ ÁŒπÊÿÊ ªÿÊ „Ò– Á»§⁄U ßŸ◊¥ ‚ ∞∑§’À’ ∑§Ê ÃÊ¬ ’…∏Ê∑§⁄U T

2 ∑§⁄U ÁŒÿÊ ¡ÊÃÊ „Ò– •¥ÁÃ◊ŒÊ’ pf „Ò —

(1)1 2

1 2

i

T Tp

T T

+

(2)1

1 2

2 i

Tp

T T

+

(3)2

1 2

2 i

Tp

T T

+

(4)1 2

1 2

2 i

T Tp

T T

+

PART B — CHEMISTRY

31. At 300 K and 1 atm, 15 mL of a gaseous

hydrocarbon requires 375 mL air

containing 20% O2 by volume for complete

combustion. After combustion the gases

occupy 330 mL. Assuming that the water

formed is in liquid form and the volumes

were measured at the same temperature

and pressure, the formula of the

hydrocarbon is :

(1) C3H

6

(2) C3H

8

(3) C4H

8

(4) C4H

10

32. Two closed bulbs of equal volume (V)

containing an ideal gas initially at pressure

pi and temperature T1 are connected

through a narrow tube of negligible

volume as shown in the figure below. The

temperature of one of the bulbs is then

raised to T2. The final pressure pf is :

(1) 1 2

1 2

i

T Tp

T T

+

(2)1

1 2

2 i

Tp

T T

+

(3)2

1 2

2 i

Tp

T T

+

(4)1 2

1 2

2 i

T Tp

T T

+


E E

E E

33. ∞∑§ ª◊¸ Á»§‹Ê◊¥≈U ‚ ÁŸ∑§‹Ë ß‹Ä≈˛UÊÚŸ œÊ⁄UÊ ∑§ÊV esu ∑§ Áfl÷flÊãÃ⁄U ¬⁄ ⁄Uπ ŒÊ •ÊflÁ‡ÊÃ åÀÊ≈UÊ¥ ∑§’Ëø ‚ ÷¡Ê ¡ÊÃÊ „Ò– ÿÁŒ ß‹Ä≈˛UÊÚŸ ∑§ •Êfl‡Ê ÃÕÊ‚¥„ÁÃ ∑˝§◊‡Ê— e ÃÕÊ m „Ê¥ ÃÊ h/λ ∑§Ê ◊ÊŸ ÁŸêŸ ◊¥ ‚Á∑§‚∑§ mÊ⁄UÊ ÁŒÿÊ ¡ÊÿªÊ? (¡’ ß‹Ä≈˛UÊÚŸ Ã⁄¥Uª ‚‚ê’ÁãœÃ Ã⁄¥UªŒÒäÿ¸ λ „Ò)

(1) meV

(2) 2meV

(3) meV

(4) 2 meV

34. fl„ S¬Ë‡ÊË$¡, Á¡‚◊¥ N ¬⁄U◊ÊáÊÈ sp ‚¥∑§⁄UáÊ ∑§Ë •flSÕÊ◊¥ „Ò, „ÊªË —

(1)2

NO+

(2)2

NO−

(3)3NO−

(4) NO2

35. ∑§Ê’Ÿ ÃÕÊ ∑§Ê’Ÿ ◊ÊŸÊÄ‚ÊÚß«U ∑§Ë Œ„Ÿ ™§c◊Êÿ¥ ∑§◊‡Ê—−393.5 ÃÕÊ −283.5 kJ mol−1 „Ò¥– ∑§Ê’¸Ÿ◊ÊŸÊÄ‚Êß«U ∑§Ë ‚¥÷flŸ ™§c◊Ê (kJ ◊¥) ¬ÁÃ ◊Ê‹ „ÊªË —

(1) 110.5

(2) 676.5

(3) −676.5

(4) −110.5

33. A stream of electrons from a heated

filament was passed between two charged

plates kept at a potential difference V esu.

If e and m are charge and mass of an

electron, respectively, then the value of

h/λ (where λ is wavelength associated

with electron wave) is given by :

(1) meV

(2) 2meV

(3) meV

(4) 2 meV

34. The species in which the N atom is in a

state of sp hybridization is :

(1)2

NO+

(2)2

NO−

(3)3NO−

(4) NO2

35. The heats of combustion of carbon and

carbon monoxide are −393.5 and

−283.5 kJ mol−1, respectively. The heat

of formation (in kJ) of carbon monoxide

per mole is :

(1) 110.5

(2) 676.5

(3) −676.5

(4) −110.5


E E

E E

36. 18 g Ç‹È∑§Ê‚ (C6H

12O

6) ∑§Ê 178.2 g ¬ÊŸË ◊¥

Á◊‹ÊÿÊ ¡ÊÃÊ „Ò– ß‚ ¡‹Ëÿ Áfl‹ÿŸ ∑§ Á‹∞ ¡‹∑§Ê flÊc¬ ŒÊ’ (torr ◊¥) „ÊªÊ —

(1) 7.6

(2) 76.0

(3) 752.4

(4) 759.0

37. ÃÊ¬◊ÊŸ 298 K ¬⁄U, ∞∑§ •Á÷Á∑˝§ÿÊ A+BC+D

∑§ Á‹∞ ‚Êêÿ ÁSÕ⁄UÊ¥∑§ 100 „Ò– ÿÁŒ ¬˝Ê⁄UÁê÷∑§ ‚ÊãŒ˝ÃÊ‚÷Ë øÊ⁄UÊ¥ S¬Ë‡ÊË¡ ◊¥ ‚ ¬˝àÿ∑§ ∑§Ë 1 M „ÊÃË, ÃÊ D∑§Ë ‚Êêÿ ‚ÊãŒ˝ÃÊ (mol L−1 ◊¥) „ÊªË —

(1) 0.182

(2) 0.818

(3) 1.818

(4) 1.182

38. ªÒÀflŸÊß¡‡ÊŸ ÁŸêŸ ◊¥ ‚ Á∑§‚∑§ ∑§Ê≈U ‚ „ÊÃÊ „Ò?

(1) Pb

(2) Cr

(3) Cu

(4) Zn

36. 18 g glucose (C6H

12O

6) is added to

178.2 g water. The vapor pressure of

water (in torr) for this aqueous solution

is :

(1) 7.6

(2) 76.0

(3) 752.4

(4) 759.0

37. The equilibrium constant at 298 K for a

reaction A+BC+D is 100. If the initial

concentration of all the four species were

1 M each, then equilibrium concentration

of D (in mol L−1) will be :

(1) 0.182

(2) 0.818

(3) 1.818

(4) 1.182

38. Galvanization is applying a coating of :

(1) Pb

(2) Cr

(3) Cu

(4) Zn


E E

E E

39. H2O

2 ∑§Ê ÁflÉÊ≈UŸ ∞∑§ ¬˝Õ◊ ∑§ÊÁ≈U ∑§Ë •Á÷Á∑˝§ÿÊ

„Ò– ¬øÊ‚ Á◊Ÿ≈U ◊¥ ß‚ ¬˝∑§Ê⁄U ∑§ ÁflÉÊ≈UŸ ◊¥ H2O

2

∑§Ë ‚ÊãŒ˝ÃÊ ÉÊ≈U∑§⁄U 0.5 ‚ 0.125 M „Ê ¡ÊÃË „Ò– ¡’H

2O

2 ∑§Ë ‚ÊãŒ˝ÃÊ 0.05 M ¬„È°øÃË „Ò, ÃÊ O

2 ∑§

’ŸŸ ∑§Ë Œ⁄U „ÊªË —

(1) 6.93×10−2 mol min−1

(2) 6.93×10−4 mol min−1

(3) 2.66 L min−1 (STP ¬⁄U)

(4) 1.34×10−2 mol min−1

40. » ˝ § Ê Úÿã« U Á‹∑§ •Áœ‡ÊÊ ·áÊ ‚◊ÃÊ¬Ë fl∑ ˝ § ◊ ¥log (x/m) ÃÕÊ log p ∑§ ’Ëø πË¥ø ªÿ ⁄UπËÿå‹Ê≈U ∑§ Á‹∞ ÁãÊêŸ ◊¥ ‚ ∑§ÊÒŸ ‚Ê ∑§ÕŸ ‚„Ë „Ò?(k ÃÕÊ n ÁSÕ⁄UÊ¥∑§ „Ò¥)

(1) k ÃÕÊ 1/n ŒÊŸÊ¥ „Ë S‹Ê¬ ¬Œ ◊¥ •ÊÃ „Ò¥–

(2) 1/n ßã≈U⁄U‚å≈U ∑§ M§¬ •ÊÃÊ „Ò–

(3) ◊ÊòÊ 1/n S‹Ê¬ ∑§ M§¬ ◊¥ •ÊÃÊ „Ò–

(4) log (1/n) ßã≈U⁄U‚å≈U ∑§ M§¬ ◊¥ •ÊÃÊ „Ò–

41. ÁŸêŸ ¬⁄U◊ÊáÊÈ•Ê¥ ◊¥ Á∑§‚∑§Ë ¬˝Õ◊ •ÊÿŸŸ ™§¡Ê¸ ©ëøÃ◊„Ò?

(1) Rb

(2) Na

(3) K

(4) Sc

39. Decomposition of H2O

2 follows a first

order reaction. In fifty minutes the

concentration of H2O

2 decreases from

0.5 to 0.125 M in one such decomposition.

When the concentration of H2O

2 reaches

0.05 M, the rate of formation of O2 will

be :

(1) 6.93×10−2 mol min−1

(2) 6.93×10−4 mol min−1

(3) 2.66 L min−1 at STP

(4) 1.34×10−2 mol min−1

40. For a linear plot of log (x/m) versus log p

in a Freundlich adsorption isotherm,

which of the following statements is

correct ? (k and n are constants)

(1) Both k and 1/n appear in the slope

term.

(2) 1/n appears as the intercept.

(3) Only 1/n appears as the slope.

(4) log (1/n) appears as the intercept.

41. Which of the following atoms has the

highest first ionization energy ?

(1) Rb

(2) Na

(3) K

(4) Sc


E E

E E

42. »˝§ÊÚÕ ç‹Ê≈U‡ÊŸ ÁflÁœ mÊ⁄UÊ ÁŸêŸ ◊¥ ‚ fl„ ∑§ÊÒŸ ‚Ê•ÿS∑§ ‚flÊ¸Áœ∑§ M§¬ ‚ ‚ÊÁãŒ˝Ã Á∑§ÿÊ ¡Ê ‚∑§ÃÊ„Ò?

(1) ◊ÒÇŸ≈UÊß≈U

(2) Á‚«U⁄UÊß≈U

(3) ªÒ‹ŸÊ

(4) ◊Ò‹Ê∑§Êß≈U

43. ¡‹ ∑§ ‚ê’ãœ ◊¥ ÁŸêŸ ∑§ÕŸÊ¥ ◊¥ ‚ ∑§ÊÒŸ ‚Ê ∞∑§ª‹Ã „Ò?

(1) ¬˝∑§Ê‡Ê‚¥‡‹·áÊ ◊¥ ¡‹ •ÊÄ‚Ë∑Î§Ã „Ê∑§⁄U•ÊÄ‚Ë$¡Ÿ ŒÃÊ „Ò–

(2) ¡‹, •ê‹ ÃÕÊ ˇÊÊ⁄U∑§ ŒÊŸÊ¥ „Ë M§¬ ◊¥ ∑§Êÿ¸∑§⁄U ‚∑§ÃÊ „Ò–

(3) ß‚∑§ ‚¥ÉÊÁŸÃ ¬˝ÊflSÕÊ ◊¥ ÁflSÃËáÊ¸ •¥Ã—•áÊÈ∑§„Êß«˛UÊ¡Ÿ •Ê’ãœ „ÊÃ „Ò¥–

(4) ÷Ê⁄UË ¡‹ mÊ⁄UÊ ’ŸÊ ’»¸§ ‚Ê◊Êãÿ ¡‹ ◊¥ «ÍU’ÃÊ„Ò–

44. „flÊ ∑§ •ÊÁœÄÿ ◊¥ Li, Na •ÊÒ⁄U K ∑§ Œ„Ÿ ¬⁄U’ŸŸflÊ‹Ë ◊ÈÅÿ •ÊÄ‚Êß«¥U ∑˝§◊‡Ê— „Ò¥ —

(1) Li2O, Na

2O ÃÕÊ KO

2

(2) LiO2, Na

2O

2 ÃÕÊ K

2O

(3) Li2O

2, Na

2O

2 ÃÕÊ KO

2

(4) Li2O, Na

2O

2 ÃÕÊ KO

2

42. Which one of the following ores is best

concentrated by froth floatation method ?

(1) Magnetite

(2) Siderite

(3) Galena

(4) Malachite

43. Which one of the following statements

about water is FALSE ?

(1) Water is oxidized to oxygen during

photosynthesis.

(2) Water can act both as an acid and

as a base.

(3) There is extensive intramolecular

hydrogen bonding in the condensed

phase.

(4) Ice formed by heavy water sinks in

normal water.

44. The main oxides formed on combustion of

Li, Na and K in excess of air are,

respectively :

(1) Li2O, Na

2O and KO

2

(2) LiO2, Na

2O

2 and K

2O

(3) Li2O

2, Na

2O

2 and KO

2

(4) Li2O, Na

2O

2 and KO

2


E E

E E

45. ÃŸÈ ÃÕÊ ‚ÊãŒ˝ ŸÊßÁ≈˛U∑§ ∞Á‚«U ∑§ ‚ÊÕ Á¡¥∑§ ∑§Ë•Á÷Á∑˝§ÿÊ mÊ⁄UÊ ∑˝§◊‡Ê— ©à¬ãŸ „ÊÃ „Ò¥ —

(1) N2O ÃÕÊ NO

2

(2) NO2 ÃÕÊ NO

(3) NO ÃÕÊ N2O

(4) NO2 ÃÕÊ N

2O

46. fl„ ÿÈÇ◊ Á¡Ÿ◊¥ »§ÊS»§Ê⁄U‚ ¬⁄U◊ÊáÊÈ•Ê¥ ∑§Ë »§Ê◊¸‹•ÊÚÄ‚Ë∑§⁄UáÊ •flSÕÊ +3 „Ò, „Ò —

(1) •ÊÕÊ¸»§ÊS»§Ê⁄U‚ ÃÕÊ ¬Êÿ⁄UÊ»§ÊS»§Ê⁄U‚ ∞Á‚«U

(2) ¬Êÿ⁄UÊ»§ÊS»§Ê⁄U‚ ÃÕÊ „Êß¬Ê»§ÊS»§ÊÁ⁄U∑§ ∞Á‚«U

(3) •ÊÕÊ¸»§ÊS»§Ê⁄U‚ ÃÕÊ „Êß¬Ê»§ÊS»§ÊÁ⁄U∑§ ∞Á‚«U

(4) ¬Êÿ⁄UÊ»§ÊS»§Ê⁄U‚ ÃÕÊ ¬Êÿ⁄UÊ»§ÊS»§ÊÁ⁄U∑§ ∞Á‚«U

47. ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ ‚Ê ÿÊÒÁª∑§ œÊÁàfl∑§ ÃÕÊ »§⁄UÊ◊ÒªŸÁ≈U∑§(‹ÊÒ„ øÈê’∑§Ëÿ) „Ò?

(1) TiO2

(2) CrO2

(3) VO2

(4) MnO2

45. The reaction of zinc with dilute and

concentrated nitric acid, respectively,

produces :

(1) N2O and NO

2

(2) NO2 and NO

(3) NO and N2O

(4) NO2 and N

2O

46. The pair in which phosphorous atoms

have a formal oxidation state of +3 is :

(1) Orthophosphorous and

pyrophosphorous acids

(2) Pyrophosphorous and

hypophosphoric acids

(3) Orthophosphorous and

hypophosphoric acids

(4) Pyrophosphorous and

pyrophosphoric acids

47. Which of the following compounds is

metallic and ferromagnetic ?

(1) TiO2

(2) CrO2

(3) VO2

(4) MnO2


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48. ∞∑§„Ë øÈê’∑§Ëÿ •ÊÉÊÍáÊ¸ ∑§Ê ÿÈÇ◊ „Ò —

[At. No. : Cr=24, Mn=25, Fe=26, Co=27]

(1) [Cr(H2O)

6]2+ ÃÕÊ [CoCl

4]2−

(2) [Cr(H2O)

6]2+ ÃÕÊ [Fe(H

2O)

6]2+

(3) [Mn(H2O)

6]2+ ÃÕÊ [Cr(H

2O)

6]2+

(4) [CoCl4]2− ÃÕÊ [Fe(H

2O)

6]2+

49. ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ ‚Ê ∑§ÊÚêå‹Ä‚ ¬˝∑§ÊÁ‡Ê∑§ ‚◊ÊflÿflÃÊ¬˝ŒÁ‡Ê¸Ã ∑§⁄UªÊ?

(1) [Co(NH3)3Cl

3]

(2) cis[Co(en)2Cl

2]Cl

(3) trans[Co(en)2Cl

2]Cl

(4) [Co(NH3)4Cl

2]Cl

(en=ethylenediamine)

50. ÷ÍÁ◊ªÃ ¤ÊË‹ ‚ ¬˝ÊåÃ ¡‹ ¬˝ÁÃŒ‡Ê¸ ◊¥ ç‹Ê⁄UÊß«U, ‹«U,ŸÊß≈˛U≈U ÃÕÊ •Êÿ⁄UŸ ∑§Ë ‚ÊãŒ˝ÃÊ ∑˝§◊‡Ê— 1000 ppb,

40 ppb, 100 ppm ÃÕÊ 0.2 ppm ¬Êß¸ ªß¸– ÿ„¡‹ ÁŸêŸ ◊¥ ‚ Á∑§‚∑§Ë ©ìÊ ‚ÊãŒ˝ÃÊ ‚ ¬ËŸ ÿÊÇÿŸ„Ë¥ „Ò?

(1) ç‹Ê⁄UÊß«U

(2) ‹«U

(3) ŸÊß≈˛U≈U

(4) •Êÿ⁄UŸ

48. The pair having the same magnetic

moment is :

[At. No. : Cr=24, Mn=25, Fe=26, Co=27]

(1) [Cr(H2O)

6]2+ and [CoCl

4]2−

(2) [Cr(H2O)

6]2+ and [Fe(H

2O)

6]2+

(3) [Mn(H2O)

6]2+ and [Cr(H

2O)

6]2+

(4) [CoCl4]2− and [Fe(H

2O)

6]2+

49. Which one of the following complexes

shows optical isomerism ?

(1) [Co(NH3)3Cl

3]

(2) cis[Co(en)2Cl

2]Cl

(3) trans[Co(en)2Cl

2]Cl

(4) [Co(NH3)4Cl

2]Cl

(en=ethylenediamine)

50. The concentration of fluoride, lead, nitrate

and iron in a water sample from an

underground lake was found to be

1000 ppb, 40 ppb, 100 ppm and 0.2 ppm,

respectively. This water is unsuitable for

drinking due to high concentration of :

(1) Fluoride

(2) Lead

(3) Nitrate

(4) Iron


E E

E E

51. ‚Ê’ÈŸ ©lÊª ◊¥ ÷ÈÄÃ‡Ê· ‹Êß (S¬ã≈U ‹Êß) ‚ ÁÇ‹‚⁄UÊÚ‹¬ÎÕ∑§ ∑§⁄UŸ ∑§ Á‹∞ ‚’‚ ©¬ÿÈÄÃ •Ê‚flŸ ÁflÁœ„Ò —

(1) ‚Ê◊Êãÿ •Ê‚flŸ

(2) ¬˝÷Ê¡Ë •Ê‚flŸ

(3) ’Êc¬ •Ê‚flŸ

(4) ‚◊ÊŸËÃ ŒÊ’ ¬⁄U •Ê‚flŸ

52. ŸËø ŒË ªß¸ •Á÷Á∑˝§ÿÊ ∑§ Á‹∞ ©à¬ÊŒ „ÊªÊ —

(1)

(2)

(3)

(4)

51. The distillation technique most suited for

separating glycerol from spent-lye in the

soap industry is :

(1) Simple distillation

(2) Fractional distillation

(3) Steam distillation

(4) Distillation under reduced pressure

52. The product of the reaction given below

is :

(1)

(2)

(3)

(4)


E E

E E

53. ÁŒ∞ ªÿ ÿÊÒÁª∑§ ∑§Ê ÁŸ⁄U¬ˇÊ ÁflãÿÊ‚ „Ò —

CO2H

OHH

ClH

CH3

(1) (2R, 3S)

(2) (2S, 3R)

(3) (2S, 3S)

(4) (2R, 3R)

54. ◊ÕŸÊÚ‹ ◊ ¥ 2-Ä‹Ê⁄UÊ -2-◊ÁÕ‹¬ã≈ UŸ, ‚ÊÁ«Uÿ◊◊ÕÊÄ‚Êß«U ∑§ ‚ÊÕ •Á÷Á∑˝§ÿÊ ∑§⁄U∑§ ŒÃË „Ò —

(a)

(b)

(c)

(1) ßŸ◊¥ ‚ ‚÷Ë

(2) (a) ÃÕÊ (c)

(3) ◊ÊòÊ (c)

(4) (a) ÃÕÊ (b)

53. The absolute configuration of

CO2H

OHH

ClH

CH3

is :

(1) (2R, 3S)

(2) (2S, 3R)

(3) (2S, 3S)

(4) (2R, 3R)

54. 2-chloro-2-methylpentane on reaction

with sodium methoxide in methanol

yields :

(a)

(b)

(c)

(1) All of these

(2) (a) and (c)

(3) (c) only

(4) (a) and (b)


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E E

55. ¬˝Ê¬ËŸ ∑§Ë HOCl (Cl2+H

2O) ∑§ ‚ÊÕ •Á÷Á∑˝§ÿÊ

Á¡‚ ◊äÿflÃË¸ ‚ „Ê∑§⁄U ‚ê¬ãŸ „ÊÃË „Ò, fl„ „Ò —

(1) CH3−CH+−CH

2−OH

(2) CH3−CH+−CH

2−Cl

(3)3 2CH CH(OH) CH+− −

(4)3 2

CH CHCl CH+

− −

56. „Ê»§◊ÊŸ ’˝Ê◊Ê◊Êß«U ÁŸêŸË∑§⁄UáÊ •Á÷Á∑˝§ÿÊ ◊¥, NaOH

ÃÕÊ Br2 ∑§ ¬˝ÿÈÄÃ ◊Ê‹Ê¥ ∑§Ë ‚¥ÅÿÊ ¬˝ÁÃ◊Ê‹ •◊ËŸ

∑§ ’ŸŸ ◊¥ „ÊªË —

(1) ∞∑§ ◊Ê‹ NaOH ÃÕÊ ∞∑§ ◊Ê‹ Br2–

(2) øÊ⁄U ◊Ê‹ NaOH ÃÕÊ ŒÊ ◊Ê‹ Br2–

(3) ŒÊ ◊Ê‹ NaOH ÃÕÊ ŒÊ ◊Ê‹ Br2–

(4) øÊ⁄U ◊Ê‹ NaOH ÃÕÊ ∞∑§ ◊Ê‹ Br2–

57. ÁŸêŸ ÉÊãÊàfl ∑§ ¬Ê‹ËÕËŸ ∑§ ‚ê’ãœ ◊¥ ÁŸêŸ ◊¥ ‚∑§ÊÒŸ ‚Ê ∑§ÕŸ ª‹Ã „Ò?

(1) ß‚∑§ ‚¥‡‹·áÊ ◊¥ ©ìÊ ŒÊ’ ∑§Ë •Êfl‡ÿ∑§ÃÊ„ÊÃË „Ò–

(2) ÿ„ ÁfllÈÃ ∑§Ê „ËŸ øÊ‹∑§ „Ò–

(3) ß‚◊¥ «UÊß¸•ÊÄ‚Ë¡Ÿ •ÕflÊ ¬⁄U•ÊÄ‚Êß«UßŸËÁ‚ÿ≈U⁄ (¬˝Ê⁄Uê÷∑§) ©à¬˝⁄U∑§ ∑§ M§¬ ◊¥øÊÁ„∞–

(4) ÿ„ ’∑§≈U (’ÊÀ≈UË), «US≈U-Á’Ÿ, •ÊÁŒ ∑§©à¬ÊŒŸ ◊¥ ¬˝ÿÈÄÃ „ÊÃË „Ò–

55. The reaction of propene with HOCl

(Cl2+H

2O) proceeds through the

intermediate :

(1) CH3−CH+−CH

2−OH

(2) CH3−CH+−CH

2−Cl

(3)3 2CH CH(OH) CH+− −

(4)3 2

CH CHCl CH+

− −

56. In the Hofmann bromamide degradation

reaction, the number of moles of NaOH

and Br2

used per mole of amine produced

are :

(1) One mole of NaOH and one mole of

Br2

.

(2) Four moles of NaOH and two moles

of Br2

.

(3) Two moles of NaOH and two moles

of Br2

.

(4) Four moles of NaOH and one mole

of Br2

.

57. Which of the following statements about

low density polythene is FALSE ?

(1) Its synthesis requires high pressure.

(2) It is a poor conductor of electricity.

(3) Its synthesis requires dioxygen or a

peroxide initiator as a catalyst.

(4) It is used in the manufacture of

buckets, dust-bins etc.


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58. ÕÊÿÊ‹ ª˝È¬ Á¡‚◊¥ ©¬ÁSÕÃ „Ò, fl„ „Ò —

(1) ‚Êß≈UÊ‚ËŸ

(2) Á‚ÁS≈UŸ (Cystine)

(3) Á‚S≈UËŸ (Cysteine)

(4) ◊ÕÊß•ÊŸËŸ

59. ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ ‚Ê ∞ŸÊßÁŸ∑§ Á«U≈U⁄U¡¥≈U „Ò?

(1) ‚ÊÁ«Uÿ◊ S≈UË•⁄U≈U

(2) ‚ÊÁ«Uÿ◊ ‹ÊÁ⁄U‹ ‚À»§≈U

(3) ‚Á≈U‹≈˛UÊß◊ÁÕ‹ •◊ÊÁŸÿ◊ ’˝Ê◊Êß«U

(4) ÁÇ‹‚Á⁄U‹ •ÊÁ‹∞≈U

60. ŸËø ŒË ªß¸ Á»§ª⁄U ◊¥ ’Èã‚Ÿ ç‹◊ ∑§Ê ‚flÊ¸Áœ∑§ ª◊¸÷Êª „Ò —

(1) ⁄UË¡Ÿ 1

(2) ⁄UË¡Ÿ 2

(3) ⁄UË¡Ÿ 3

(4) ⁄UË¡Ÿ 4

58. Thiol group is present in :

(1) Cytosine

(2) Cystine

(3) Cysteine

(4) Methionine

59. Which of the following is an anionic

detergent ?

(1) Sodium stearate

(2) Sodium lauryl sulphate

(3) Cetyltrimethyl ammonium bromide

(4) Glyceryl oleate

60. The hottest region of Bunsen flame shown

in the figure below is :

(1) region 1

(2) region 2

(3) region 3

(4) region 4


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÷Êª C — ªÁáÊÃ

61. ÿÁŒ f (x)+2f

1

x

=3x, 0x ≠ „Ò, ÃÕÊ

S : ( ) ( )x f x f xR= = − „Ò ; ÃÊ S :

(1) ∞∑§ Á⁄UÄÃ ‚◊ÈìÊÿ „Ò–

(2) ◊¥ ∑§fl‹ ∞∑§ •flÿfl „Ò–

(3) ◊¥ ÃâÿÃ— ŒÊ •flÿfl „Ò¥–

(4) ◊¥ ŒÊ ‚ •Áœ∑§ •flÿfl „Ò¥–

62. θ ∑§Ê fl„ ∞∑§ ◊ÊŸ Á¡‚∑§ Á‹∞ 2 3 sin

1 2 sin

i

i

+

−

θ

θ

¬ÍáÊ¸Ã—

∑§ÊÀ¬ÁŸ∑§ „Ò, „Ò —

(1)3

π

(2)6

π

(3)1 3

sin4

−

(4)1 1

sin3

−

63. x ∑§ ©Ÿ ‚÷Ë flÊSÃÁfl∑§ ◊ÊŸÊ¥ ∑§Ê ÿÊª ¡Ê ‚◊Ë∑§⁄UáÊ

( )2

4 602

5 5 1x x

x x

+ −

− + = ∑§Ê ‚¥ÃÈc≈U ∑§⁄UÃ

„Ò¥, „Ò —

(1) 3

(2) −4

(3) 6

(4) 5

PART C — MATHEMATICS

61. If f (x)+2f

1

x

=3x, 0x ≠ , and

S : ( ) ( )x f x f xR= = − ; then S :

(1) is an empty set.

(2) contains exactly one element.

(3) contains exactly two elements.

(4) contains more than two elements.

62. A value of θ for which 2 3 sin

1 2 sin

i

i

+

−

θ

θ

is

purely imaginary, is :

(1)3

π

(2)6

π

(3)1 3

sin4

−

(4)1 1

sin3

−

63. The sum of all real values of x satisfying

the equation

( )2

4 602

5 5 1x x

x x

+ −

− + = is :

(1) 3

(2) −4

(3) 6

(4) 5


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64. ÿÁŒ 5

A 3 2

a b

−

= ÃÕÊ A adj A=A AT „Ò¥,

ÃÊ 5a+b ’⁄UÊ’⁄U „Ò —

(1) −1

(2) 5

(3) 4

(4) 13

65. ⁄ÒUÁπ∑§ ‚◊Ë∑§⁄UáÊ ÁŸ∑§Êÿ

x+λy−z=0

λx−y−z=0

x+y−λz=0

∑§Ê ∞∑§ •ÃÈë¿U „‹ „ÊŸ ∑§ Á‹∞ —

(1) λ ∑§ •Ÿ¥Ã ◊ÊŸ „Ò¥–

(2) λ ∑§Ê ÃâÿÃ— ∞∑§ ◊ÊŸ „Ò–

(3) λ ∑§ ÃâÿÃ— ŒÊ ◊ÊŸ „Ò¥–

(4) λ ∑§ ÃâÿÃ— ÃËŸ ◊ÊŸ „Ò¥–

66. ‡ÊéŒ SMALL ∑§ •ˇÊ⁄UÊ¥ ∑§Ê ¬˝ÿÊª ∑§⁄U∑§, ¬Ê°ø •ˇÊ⁄UÊ¥flÊ‹ ‚÷Ë ‡ÊéŒÊ¥ (•Õ¸¬ÍáÊ¸ •ÕflÊ •Õ¸„ËŸ) ∑§Ê‡ÊéŒ∑§Ê‡Ê ∑§ ∑˝§◊ÊŸÈ‚Ê⁄U ⁄UπŸ ¬⁄U, ‡ÊéŒ SMALL ∑§ÊSÕÊŸ „Ò —

(1) 46 flÊ¥

(2) 59 flÊ¥

(3) 52 flÊ¥

(4) 58 flÊ¥

64. If 5

A 3 2

a b

−

= and A adj A=A AT, then

5a+b is equal to :

(1) −1

(2) 5

(3) 4

(4) 13

65. The system of linear equations

x+λy−z=0

λx−y−z=0

x+y−λz=0

has a non-trivial solution for :

(1) infinitely many values of λ.

(2) exactly one value of λ.

(3) exactly two values of λ.

(4) exactly three values of λ.

66. If all the words (with or without meaning)

having five letters, formed using the letters

of the word SMALL and arranged as in a

dictionary; then the position of the word

SMALL is :

(1) 46th

(2) 59th

(3) 52nd

(4) 58th


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67. ÿÁŒ 2

2 41 , 0

n

x

x x

≠

− + ∑§ ¬˝‚Ê⁄U ◊¥ ¬ŒÊ¥

∑§Ë ‚¥ÅÿÊ 28 „Ò, ÃÊ ß‚ ¬˝‚Ê⁄U ◊¥ •ÊŸ flÊ‹ ‚÷Ë ¬ŒÊ¥

∑§ ªÈáÊÊ¥∑§Ê¥ ∑§Ê ÿÊª „Ò —

(1) 64

(2) 2187

(3) 243

(4) 729

68. ÿÁŒ ∞∑§ •ø⁄UÃ⁄U ‚◊Ê¥Ã⁄U üÊ…∏Ë ∑§Ê ŒÍ‚⁄UÊ, 5 flÊ¥ ÃÕÊ9 flÊ¥ ¬Œ ∞∑§ ªÈáÊÊûÊ⁄U üÊ…∏Ë ◊¥ „Ò¥, ÃÊ ©‚ ªÈáÊÊûÊ⁄U üÊ…∏Ë∑§Ê ‚Êfl¸ •ŸÈ¬ÊÃ „Ò —

(1)8

5

(2)4

3

(3) 1

(4)7

4

69. ÿÁŒ üÊáÊË

2 2 2 223 2 1 4

1 2 3 4 4 ...... ,5 5 5 5

+ + + + +

∑§ ¬˝Õ◊ Œ‚ ¬ŒÊ¥ ∑§Ê ÿÊª 16

5m „Ò, ÃÊ m ’⁄UÊ’⁄U

„Ò —

(1) 102

(2) 101

(3) 100

(4) 99

67. If the number of terms in the expansion of

2

2 41 , 0,

n

x

x x

≠

− + is 28, then the sum

of the coefficients of all the terms in this

expansion, is :

(1) 64

(2) 2187

(3) 243

(4) 729

68. If the 2nd, 5th and 9th terms of a

non-constant A.P. are in G.P., then the

common ratio of this G.P. is :

(1)8

5

(2)4

3

(3) 1

(4)7

4

69. If the sum of the first ten terms of the series

2 2 2 223 2 1 4

1 2 3 4 4 ...... ,5 5 5 5

+ + + + +

is 16

5m , then m is equal to :

(1) 102

(2) 101

(3) 100

(4) 99


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70. ◊ÊŸÊ ( )1

22

0

lim 1 tanx

x

p x→ += + „Ò, ÃÊ log p

’⁄UÊ’⁄U „Ò —

(1) 2

(2) 1

(3)1

2

(4)1

4

71. x e R ∑ § Á‹∞ f (x)=?log2−sinx? ÃÕÊg(x)=f (f (x)) „Ò¥, ÃÊ —

(1) x=0 ¬⁄U g •fl∑§‹ŸËÿ Ÿ„Ë¥ „Ò–

(2) g9(0)=cos(log2) „Ò–

(3) g9(0)=−cos(log2) „Ò–

(4) x=0 ¬⁄ U g •fl∑§‹ŸËÿ „ Ò ÃÕÊg9(0)=−sin(log2) „Ò–

72.1 1 sin

( ) tan , 0, 1 sin 2

xf x x

x

−+

=

−

π

¬⁄U ÁfløÊ⁄U ∑§ËÁ¡∞– y=f (x) ∑§ Á’¥ŒÈ 6

x=

π ¬⁄U

πË¥øÊ ªÿÊ •Á÷‹¥’ ÁŸêŸ Á’¥ŒÈ ‚ ÷Ë „Ê∑§⁄U ¡ÊÃÊ „Ò —

(1) (0, 0)

(2)2

0, 3

π

(3) , 06

π

(4) , 04

π

70. Let ( )1

22

0

lim 1 tanx

x

p x→ += + then log p

is equal to :

(1) 2

(2) 1

(3)1

2

(4)1

4

71. For x e R, f (x)=?log2−sinx? and

g(x)=f (f (x)), then :

(1) g is not differentiable at x=0

(2) g9(0)=cos(log2)

(3) g9(0)=−cos(log2)

(4) g is differentiable at x=0 and

g9(0)=−sin(log2)

72. Consider

1 1 sin( ) tan , 0,

1 sin 2

xf x x .

x

−+

=

−

π

A normal to y=f (x) at 6

x=

π

also passes

through the point :

(1) (0, 0)

(2)2

0, 3

π

(3) , 06

π

(4) , 04

π


E E

E E

73. 2 ß∑§Êß¸ ‹¥’Ë ∞∑§ ÃÊ⁄U ∑§Ê ŒÊ ÷ÊªÊ¥ ◊¥ ∑§Ê≈U ∑§⁄U ©ã„¥∑˝§◊‡Ê— x ß∑§Êß¸ ÷È¡Ê flÊ‹ flª¸ ÃÕÊ r ß∑§Êß¸ ÁòÊíÿÊflÊ‹ flÎûÊ ∑§ M§¬ ◊¥ ◊Ê«∏Ê ¡ÊÃÊ „Ò– ÿÁŒ ’ŸÊÿ ªÿ flª¸ÃÕÊ flÎûÊ ∑§ ˇÊòÊ»§‹Ê¥ ∑§Ê ÿÊª ãÿÍŸÃ◊ „Ò, ÃÊ —

(1) 2x=(π+4)r

(2) (4−π)x=πr

(3) x=2r

(4) 2x=r

74. ‚◊Ê∑§‹ ( )

12 9

35 3

2 5

1

x xdx

x x

+

+ +

’⁄UÊ’⁄U „Ò —

(1)

( )

5

25 3

1

xC

x x

−+

+ +

(2)

( )

10

25 3

2 1

xC

x x

+

+ +

(3)

( )

5

25 3

2 1

xC

x x

+

+ +

(4)

( )

10

25 3

2 1

xC

x x

−+

+ +

¡„Ê° C ∞∑§ Sflë¿U •ø⁄U „Ò–

73. A wire of length 2 units is cut into two

parts which are bent respectively to form

a square of side=x units and a circle of

radius=r units. If the sum of the areas of

the square and the circle so formed is

minimum, then :

(1) 2x=(π+4)r

(2) (4−π)x=πr

(3) x=2r

(4) 2x=r

74. The integral

( )

12 9

35 3

2 5

1

x xdx

x x

+

+ +

is equal

to :

(1)

( )

5

25 3

1

xC

x x

−+

+ +

(2)

( )

10

25 3

2 1

xC

x x

+

+ +

(3)

( )

5

25 3

2 1

xC

x x

+

+ +

(4)

( )

10

25 3

2 1

xC

x x

−+

+ +

where C is an arbitrary constant.


E E

E E

75. ( ) ( )1

2

1 2 . . . 3lim

n

nn

n n n

n→

∞

+ + ’⁄UÊ’⁄U „Ò —

(1) 4

18

e

(2) 2

27

e

(3) 2

9

e

(4) 3 log3−2

76. ˇÊòÊ

2 2 2( , ): 2 4 0, 0x y y x x y x, x y + ≤ ÃÕÊ

∑§Ê ˇÊòÊ»§‹ (flª¸ ß∑§ÊßÿÊ¥ ◊¥) „Ò —

(1)4

3−π

(2)8

3−π

(3)4 2

3

−π

(4)2 2

2 3−

π

75. ( ) ( )1

2

1 2 . . . 3lim

n

nn

n n n

n→

∞

+ + is equal

to :

(1) 4

18

e

(2) 2

27

e

(3) 2

9

e

(4) 3 log3−2

76. The area (in sq. units) of the region

2 2 2( , ): 2 and 4 0, 0x y y x x y x, x y + ≤

is :

(1)4

3−π

(2)8

3−π

(3)4 2

3

−π

(4)2 2

2 3−

π


E E

E E

77. ÿÁŒ ∞∑§ fl∑˝§ y=f (x) Á’¥ŒÈ (1, −1) ‚ „Ê∑§⁄U ¡ÊÃÊ„Ò ÃÕÊ •fl∑§‹ ‚◊Ë∑§⁄UáÊ y(1+xy) dx=x dy

∑§Ê ‚¥ÃÈc≈U ∑§⁄UÃÊ „Ò, ÃÊ 1 2

f − ’⁄UÊ’⁄U „Ò —

(1)2

5−

(2)4

5−

(3)2

5

(4)4

5

78. ÿÁŒ ∞∑§ ‚◊øÃÈ÷¸È¡ ∑§Ë ŒÊ ÷È¡Ê∞°, ⁄UπÊ•Ê¥x−y+1=0 ÃÕÊ 7x−y−5=0 ∑§Ë ÁŒ‡ÊÊ ◊¥ „Ò¥ÃÕÊ ß‚∑§ Áfl∑§áÊ¸ Á’¥ŒÈ (−1, −2) ¬⁄U ¬˝ÁÃë¿UŒ∑§⁄UÃ „Ò¥, ÃÊ ß‚ ‚◊øÃÈ÷¸È¡ ∑§Ê ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ-‚Ê‡ÊË·¸ „Ò?

(1) (−3, −9)

(2) (−3, −8)

(3)

1 8,

3 3

−

(4)

10 7,

3 3

− −

79. ©Ÿ flÎûÊÊ¥ ∑§ ∑§ãŒ˝, ¡Ê flÎûÊ x2+y2−8x−8y−4=0

∑§Ê ’Ês M§¬ ‚ S¬‡Ê¸ ∑§⁄UÃ „Ò¥ ÃÕÊ x-•ˇÊ ∑§Ê ÷ËS¬‡Ê¸ ∑§⁄UÃ „Ò¥, ÁSÕÃ „Ò¥ —

(1) ∞∑§ flÎûÊ ¬⁄U–

(2) ∞∑§ ŒËÉÊ¸flÎûÊ ¬⁄U ¡Ê flÎûÊ Ÿ„Ë¥ „Ò–

(3) ∞∑§ •ÁÃ¬⁄Ufl‹ÿ ¬⁄U–

(4) ∞∑§ ¬⁄Ufl‹ÿ ¬⁄U–

77. If a curve y=f (x) passes through the point

(1, −1) and satisfies the differential

equation, y(1+xy) dx=x dy, then 1

2

f −

is equal to :

(1)2

5−

(2)4

5−

(3)2

5

(4)4

5

78. Two sides of a rhombus are along the lines,

x−y+1=0 and 7x−y−5=0. If its

diagonals intersect at (−1, −2), then

which one of the following is a vertex of

this rhombus ?

(1) (−3, −9)

(2) (−3, −8)

(3)

1 8,

3 3

−

(4)

10 7,

3 3

− −

79. The centres of those circles which touch

the circle, x2+y2−8x−8y−4=0,

externally and also touch the x-axis, lie

on :

(1) a circle.

(2) an ellipse which is not a circle.

(3) a hyperbola.

(4) a parabola.


E E

E E

80. ÿÁŒ ‚◊Ë∑§⁄UáÊ x2+y2−4x+6y−12=0 mÊ⁄UÊ¬˝ŒûÊ ∞∑§ flÎûÊ ∑§Ê ∞∑§ √ÿÊ‚ ∞∑§ •ãÿ flÎûÊ S, Á¡‚∑§Ê∑§ãŒ˝ (−3, 2) „Ò, ∑§Ë ¡ËflÊ „Ò, ÃÊ flÎûÊ S ∑§Ë ÁòÊíÿÊ„Ò —

(1) 5 2

(2) 5 3

(3) 5

(4) 10

81. ◊ÊŸÊ ¬⁄Ufl‹ÿ y2=8x ∑§Ê P ∞∑§ ∞‚Ê Á’¥ŒÈ „Ò ¡ÊflÎûÊ x2+(y+6)2=1, ∑§ ∑§ãŒ˝ C ‚ ãÿÍŸÃ◊ ŒÍ⁄UË¬⁄U „Ò, ÃÊ ©‚ flÎûÊ ∑§Ê ‚◊Ë∑§⁄UáÊ ¡Ê C ‚ „Ê∑§⁄U ¡ÊÃÊ„Ò ÃÕÊ Á¡‚∑§Ê ∑§ãŒ˝ P ¬⁄U „Ò, „Ò —

(1) x2+y2−4x+8y+12=0

(2) x2+y2−x+4y−12=0

(3) x2+y2−4

x

+2y−24=0

(4) x2+y2−4x+9y+18=0

82. ©‚ •ÁÃ¬⁄Ufl‹ÿ, Á¡‚∑§ ŸÊÁ÷‹¥’ ∑§Ë ‹¥’Êß¸ 8 „ÒÃÕÊ Á¡‚∑§ ‚¥ÿÈÇ◊Ë •ˇÊ ∑§Ë ‹¥’Êß¸ ©‚∑§Ë ŸÊÁ÷ÿÊ¥∑§ ’Ëø ∑§Ë ŒÍ⁄UË ∑§Ë •ÊœË „Ò, ∑§Ë ©à∑§ãŒ˝ÃÊ „Ò —

(1)4

3

(2)4

3

(3)2

3

(4) 3

80. If one of the diameters of the circle, given

by the equation, x2+y2−4x+6y−12=0,

is a chord of a circle S, whose centre is at

(−3, 2), then the radius of S is :

(1) 5 2

(2) 5 3

(3) 5

(4) 10

81. Let P be the point on the parabola, y2=8x

which is at a minimum distance from the

centre C of the circle, x2+(y+6)2=1.

Then the equation of the circle, passing

through C and having its centre at P is :

(1) x2+y2−4x+8y+12=0

(2) x2+y2−x+4y−12=0

(3) x2+y2−4

x

+2y−24=0

(4) x2+y2−4x+9y+18=0

82. The eccentricity of the hyperbola whose

length of the latus rectum is equal to 8 and

the length of its conjugate axis is equal to

half of the distance between its foci, is :

(1)4

3

(2)4

3

(3)2

3

(4) 3


E E

E E

83. Á’¥ŒÈ (1, −5, 9) ∑§Ë ‚◊Ã‹ x−y+z=5 ‚ fl„ŒÍ⁄UË ¡Ê ⁄UπÊ x=y=z ∑§Ë ÁŒ‡ÊÊ ◊¥ ◊Ê¬Ë ªß¸ „Ò, „Ò —

(1) 3 10

(2) 10 3

(3)10

3

(4)20

3

84. ÿÁŒ ⁄UπÊ 2 3 4

2 1 3

yx z+− += =

−

, ‚◊Ã‹

lx+my−z=9 ◊¥ ÁSÕÃ „Ò, ÃÊ l2+m2 ’⁄UÊ’⁄U „Ò —

(1) 26

(2) 18

(3) 5

(4) 2

85. ◊ÊŸÊ , a b→ →

ÃÕÊ c→

ÃËŸ ∞‚ ◊ÊòÊ∑§ ‚ÁŒ‡Ê „Ò¥ Á∑§

3 .

2a b c b c

→ → → → → × × = + „ Ò– ÿÁŒ

b→

, c→

∑§ ‚◊Ê¥Ã⁄U Ÿ„Ë¥ „Ò, ÃÊ a→

ÃÕÊ b→

∑§ ’Ëø

∑§Ê ∑§ÊáÊ „Ò —

(1)3

4

π

(2)2

π

(3)2

3

π

(4)5

6

π

83. The distance of the point (1, −5, 9) from

the plane x−y+z=5 measured along the

line x=y=z is :

(1) 3 10

(2) 10 3

(3)10

3

(4)20

3

84. If the line, 2 3 4

2 1 3

yx z+− += =

−

lies in

the plane, lx+my−z=9, then l2+m2 is

equal to :

(1) 26

(2) 18

(3) 5

(4) 2

85. Let , and a b c→ → →

be three unit vectors such

that 3

.2

a b c b c

→ → → → → × × = + If

b→

is not parallel to c→

, then the angle

between a→

and b→

is :

(1)3

4

π

(2)2

π

(3)2

3

π

(4)5

6

π


E E

E E

86. ÿÁŒ ‚¥ÅÿÊ•Ê¥ 2, 3, a ÃÕÊ 11 ∑§Ê ◊ÊŸ∑§ Áflø‹Ÿ3.5 „Ò, ÃÊ ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ-‚Ê ‚àÿ „Ò?

(1) 3a2−26a+55=0

(2) 3a2−32a+84=0

(3) 3a2−34a+91=0

(4) 3a2−23a+44=0

87. ◊ÊŸÊ ŒÊ •ŸÁ÷ŸÃ ¿U— »§‹∑§Ëÿ ¬Ê‚ A ÃÕÊ B ∞∑§‚ÊÕ ©¿UÊ‹ ªÿ– ◊ÊŸÊ ÉÊ≈UŸÊ E

1 ¬Ê‚ A ¬⁄U øÊ⁄U

•ÊŸÊ Œ‡ÊÊ¸ÃË „Ò, ÉÊ≈UŸÊ E2 ¬Ê‚ B ¬⁄U 2 •ÊŸÊ Œ‡ÊÊ¸ÃË

„Ò ÃÕÊ ÉÊ≈UŸÊ E3 ŒÊŸÊ¥ ¬Ê‚Ê¥ ¬⁄U •ÊŸ flÊ‹Ë ‚¥ÅÿÊ•Ê¥

∑§Ê ÿÊª Áfl·◊ Œ‡ÊÊ¸ÃË „Ò, ÃÊ ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ-‚Ê∑§ÕŸ ‚àÿ Ÿ„Ë¥ „Ò?

(1) E1 ÃÕÊ E

2 SflÃ¥òÊ „Ò¥–

(2) E2 ÃÕÊ E


(3) E1 ÃÕÊ E


(4) E1, E2 ÃÕÊ E


88. ÿÁŒ 0≤x<2π „Ò, ÃÊ x ∑§ ©Ÿ flÊSÃÁfl∑§ ◊ÊŸÊ¥ ∑§Ë‚¥ÅÿÊ ¡Ê ‚◊Ë∑§⁄UáÊcosx+cos2x+cos3x+cos4x=0 ∑§Ê ‚¥ÃÈc≈U∑§⁄UÃ „Ò¥, „Ò —

(1) 3

(2) 5

(3) 7

(4) 9

86. If the standard deviation of the numbers

2, 3, a and 11 is 3.5, then which of the

following is true ?

(1) 3a2−26a+55=0

(2) 3a2−32a+84=0

(3) 3a2−34a+91=0

(4) 3a2−23a+44=0

87. Let two fair six-faced dice A and B be

thrown simultaneously. If E1 is the event

that die A shows up four, E2 is the event

that die B shows up two and E3 is the event

that the sum of numbers on both dice is

odd, then which of the following

statements is NOT true ?

(1) E1 and E

2 are independent.

(2) E2 and E

3 are independent.

(3) E1 and E

3 are independent.

(4) E1, E2 and E

3 are independent.

88. If 0≤x<2π, then the number of real values

of x, which satisfy the equation

cosx+cos2x+cos3x+cos4x=0, is :

(1) 3

(2) 5

(3) 7

(4) 9


E E

E E

SPACE FOR ROUGH WORK / ⁄U»§ ∑§Êÿ¸ ∑§ Á‹∞ ¡ª„

89. ∞∑§ √ÿÁÄÃ ∞∑§ ™§äflÊ¸œ⁄U π¥÷ ∑§Ë •Ê⁄U ∞∑§ ‚Ëœ ¬Õ¬⁄U ∞∑§ ‚◊ÊŸ øÊ‹ ‚ ¡Ê ⁄U„Ê „Ò– ⁄UÊSÃ ¬⁄U ∞∑§ Á’¥ŒÈA ‚ fl„ π¥÷ ∑§ Á‡Êπ⁄U ∑§Ê ©ÛÊÿŸ ∑§ÊáÊ 308 ◊Ê¬ÃÊ„Ò– A ‚ ©‚Ë ÁŒ‡ÊÊ ◊¥ 10 Á◊Ÿ≈U •ÊÒ⁄U ø‹Ÿ ∑§ ’ÊŒÁ’¥ŒÈ B ‚ fl„ π¥÷ ∑§ Á‡Êπ⁄U ∑§Ê ©ÛÊÿŸ ∑§ÊáÊ 608

¬ÊÃÊ „Ò, ÃÊ B ‚ π¥÷ Ã∑§ ¬„È°øŸ ◊¥ ©‚ ‹ªŸ flÊ‹Ê‚◊ÿ (Á◊Ÿ≈UÊ¥ ◊¥) „Ò —

(1) 6

(2) 10

(3) 20

(4) 5

90. ’Í‹ ∑§ √ÿ¥¡∑§ (Boolean Expression)

(p∧~q)∨q∨(~p∧q) ∑§Ê ‚◊ÃÈÀÿ „Ò —

(1) ~ p ∧ q

(2) p ∧ q

(3) p ∨ q

(4) p ∨ ~ q

- o O o -

89. A man is walking towards a vertical pillar

in a straight path, at a uniform speed. At

a certain point A on the path, he observes

that the angle of elevation of the top of

the pillar is 308. After walking for 10

minutes from A in the same direction, at a

point B, he observes that the angle of

elevation of the top of the pillar is 608.

Then the time taken (in minutes) by him,

from B to reach the pillar, is :

(1) 6

(2) 10

(3) 20

(4) 5

90. The Boolean Expression (p∧~q)∨q∨(~p∧q)

is equivalent to :

(1) ~ p ∧ q

(2) p ∧ q

(3) p ∨ q

(4) p ∨ ~ q

- o O o -

Read the following instructions carefully :

1. The candidates should fill in the required particularson the Test Booklet and Answer Sheet (Side–1) withBlue/Black Ball Point Pen.

2. For writing/marking particulars on Side–2 of theAnswer Sheet, use Blue/Black Ball Point Pen only.

3. The candidates should not write their Roll Numbersanywhere else (except in the specified space) on theTest Booklet/Answer Sheet.

4. Out of the four options given for each question, onlyone option is the correct answer.

5. For each incorrect response, one–fourth (¼) of the totalmarks allotted to the question would be deducted fromthe total score. No deduction from the total score,however, will be made if no response is indicated foran item in the Answer Sheet.

6. Handle the Test Booklet and Answer Sheet with care,as under no circumstances (except for discrepancy inTest Booklet Code and Answer Sheet Code), another setwill be provided.

7. The candidates are not allowed to do any rough workor writing work on the Answer Sheet. All calculations/writing work are to be done in the space provided forthis purpose in the Test Booklet itself, marked ‘Spacefor Rough Work’. This space is given at the bottom ofeach page and in one page (i.e. Page 39) at the end ofthe booklet.

8. On completion of the test, the candidates must handover the Answer Sheet to the Invigilator on duty in theRoom/Hall. However, the candidates are allowed totake away this Test Booklet with them.

9. Each candidate must show on demand his/her AdmitCard to the Invigilator.

10. No candidate, without special permission of theSuperintendent or Invigilator, should leave his/herseat.

11. The candidates should not leave the Examination Hallwithout handing over their Answer Sheet to theInvigilator on duty and sign the Attendance Sheetagain. Cases where a candidate has not signed theAttendance Sheet second time will be deemed not tohave handed over the Answer Sheet and dealt with asan unfair means case. The candidates are also requiredto put their left hand THUMB impression in the spaceprovided in the Attendance Sheet.

12. Use of Electronic/Manual Calculator and anyElectronic device like mobile phone, pager etc. isprohibited.

13. The candidates are governed by all Rules andRegulations of the JAB/Board with regard to theirconduct in the Examination Hall. All cases of unfairmeans will be dealt with as per Rules and Regulationsof the JAB/Board.

14. No part of the Test Booklet and Answer Sheet shall bedetached under any circumstances.

15. Candidates are not allowed to carry any textualmaterial, printed or written, bits of papers, pager,mobile phone, electronic device or any other materialexcept the Admit Card inside the examinationroom/hall.

ÁŸêŸÁ‹ÁπÃ ÁŸŒ¸‡Ê äÿÊŸ ‚ ¬…∏¥ —

1. ¬⁄UËˇÊÊÁÕ¸ÿÊ¥ ∑§Ê ¬⁄UËˇÊÊ ¬ÈÁSÃ∑§Ê •ÊÒ⁄U ©ûÊ⁄U ¬òÊ (¬ÎD -1) ¬⁄U flÊ¥Á¿UÃÁflfl⁄UáÊ ŸË‹/∑§Ê‹ ’ÊÚ‹ åflÊß¥≈U ¬Ÿ ‚ „Ë ÷⁄UŸÊ „Ò–

2. ©ûÊ⁄U ¬òÊ ∑§ ¬ÎD-2 ¬⁄U Áflfl⁄UáÊ Á‹πŸ/•¥Á∑§Ã ∑§⁄UŸ ∑§ Á‹∞ ∑§fl‹ŸË‹/∑§Ê‹ ’ÊÚ‹ åflÊß¥≈U ¬Ÿ ∑§Ê ¬˝ÿÊª ∑§⁄¥U–

3. ¬⁄UËˇÊÊ ¬ÈÁSÃ∑§Ê/©ûÊ⁄U ¬òÊ ¬⁄U ÁŸœÊ¸Á⁄UÃ SÕÊŸ ∑§ •‹ÊflÊ ¬⁄UËˇÊÊÕË¸•¬ŸÊ •ŸÈ∑˝§◊Ê¥∑§ •ãÿ ∑§„Ë¥ Ÿ„Ë¥ Á‹π¥–

4. ¬˝àÿ∑§ ¬˝‡Ÿ ∑§ Á‹ÿ ÁŒÿ ªÿ øÊ⁄U Áfl∑§À¬Ê¥ ◊¥ ‚ ∑§fl‹ ∞∑§ Áfl∑§À¬‚„Ë „Ò–

5. ¬˝àÿ∑§ ª‹Ã ©ûÊ⁄U ∑§ Á‹∞ ©‚ ¬˝‡Ÿ ∑§ Á‹∞ ÁŸœÊ¸Á⁄UÃ ∑È§‹ •¥∑§Ê¥◊¥ ‚ ∞∑§-øÊÒÕÊß¸ (¼) •¥∑§ ∑È§‹ ÿÊª ◊¥ ‚ ∑§Ê≈U Á‹∞ ¡Ê∞°ª–ÿÁŒ ©ûÊ⁄U ¬òÊ ◊¥ Á∑§‚Ë ¬˝‡Ÿ ∑§Ê ∑§Êß¸ ©ûÊ⁄U Ÿ„Ë¥ ÁŒÿÊ ªÿÊ „Ò, ÃÊ∑È§‹ ÿÊª ◊¥ ‚ ∑§Êß¸ •¥∑§ Ÿ„Ë¥ ∑§Ê≈U ¡Ê∞°ª–

6. ¬⁄UËˇÊÊ ¬ÈÁSÃ∑§Ê ∞fl¥ ©ûÊ⁄U ¬òÊ ∑§Ê äÿÊŸ¬Ífl¸∑§ ¬˝ÿÊª ∑§⁄¥U ÄÿÊ¥Á∑§Á∑§‚Ë ÷Ë ¬Á⁄UÁSÕÁÃ ◊¥ (∑§fl‹ ¬⁄UËˇÊÊ ¬ÈÁSÃ∑§Ê ∞fl¥ ©ûÊ⁄U ¬òÊ ∑§‚¥∑§Ã ◊¥ Á÷ÛÊÃÊ ∑§Ë ÁSÕÁÃ ∑§Ê ¿UÊ«∏∑§⁄U), ŒÍ‚⁄UË ¬⁄UËˇÊÊ ¬ÈÁSÃ∑§Ê©¬‹éœ Ÿ„Ë¥ ∑§⁄UÊÿË ¡Ê∞ªË–

7. ©ûÊ⁄U ¬òÊ ¬⁄U ∑§Êß¸ ÷Ë ⁄U»§ ∑§Êÿ¸ ÿÊ Á‹πÊß¸ ∑§Ê ∑§Ê◊ ∑§⁄UŸ ∑§Ë•ŸÈ◊ÁÃ Ÿ„Ë¥ „Ò– ‚÷Ë ªáÊŸÊ ∞fl¥ Á‹πÊß¸ ∑§Ê ∑§Ê◊, ¬⁄UËˇÊÊ ¬ÈÁSÃ∑§Ê◊¥ ÁŸœÊ¸Á⁄UÃ ¡ª„ ¡Ê Á∑§ “⁄U»§ ∑§Êÿ¸ ∑§ Á‹∞ ¡ª„” mÊ⁄UÊ ŸÊ◊Ê¥Á∑§Ã„Ò, ¬⁄U „Ë Á∑§ÿÊ ¡Ê∞ªÊ– ÿ„ ¡ª„ ¬˝àÿ∑§ ¬ÎD ¬⁄U ŸËø ∑§Ë •Ê⁄U •ÊÒ⁄U¬ÈÁSÃ∑§Ê ∑§ •¥Ã ◊¥ ∞∑§ ¬ÎD ¬⁄U (¬ÎD 39) ŒË ªß¸ „Ò–

8. ¬⁄ËˇÊÊ ‚ê¬ÛÊ „ÊŸ ¬⁄U, ¬⁄UËˇÊÊÕË¸ ∑§ˇÊ/„ÊÚ‹ ¿UÊ«∏Ÿ ‚ ¬Ífl¸ ©ûÊ⁄U ¬òÊ∑§ˇÊ ÁŸ⁄UËˇÊ∑§ ∑§Ê •fl‡ÿ ‚ÊÒ¥¬ Œ¥– ¬⁄UËˇÊÊÕË¸ •¬Ÿ ‚ÊÕ ß‚¬⁄UËˇÊÊ ¬ÈÁSÃ∑§Ê ∑§Ê ‹ ¡Ê ‚∑§Ã „Ò¥–

9. ◊Ê¥ª ¡ÊŸ ¬⁄U ¬˝àÿ∑§ ¬⁄UËˇÊÊÕË¸ ÁŸ⁄UËˇÊ∑§ ∑§Ê •¬ŸÊ ¬˝fl‡Ê ∑§Ê«¸ •fl‡ÿÁŒπÊ∞°–

10. •œËˇÊ∑§ ÿÊ ÁŸ⁄UËˇÊ∑§ ∑§Ë Áfl‡Ê· •ŸÈ◊ÁÃ ∑§ Á’ŸÊ ∑§Êß¸ ¬⁄UËˇÊÊÕË¸•¬ŸÊ SÕÊŸ Ÿ ¿UÊ«∏¥–

11. ∑§Êÿ¸⁄UÃ ÁŸ⁄UËˇÊ∑§ ∑§Ê •¬ŸÊ ©ûÊ⁄U ¬òÊ ÁŒ∞ Á’ŸÊ ∞fl¥ ©¬ÁSÕÁÃ ¬òÊ¬⁄U ŒÈ’Ê⁄UÊ „SÃÊˇÊ⁄U Á∑§∞ Á’ŸÊ ∑§Êß¸ ¬⁄UËˇÊÊÕË¸ ¬⁄UËˇÊÊ „ÊÚ‹ Ÿ„Ë¥ ¿UÊ«∏¥ª–ÿÁŒ Á∑§‚Ë ¬⁄UËˇÊÊÕË¸ Ÿ ŒÍ‚⁄UË ’Ê⁄U ©¬ÁSÕÁÃ ¬òÊ ¬⁄U „SÃÊˇÊ⁄U Ÿ„Ë¥Á∑§∞ ÃÊ ÿ„ ◊ÊŸÊ ¡Ê∞ªÊ Á∑§ ©‚Ÿ ©ûÊ⁄U ¬òÊ Ÿ„Ë¥ ‹ÊÒ≈UÊÿÊ „Ò Á¡‚•ŸÈÁøÃ ‚ÊœŸ ¬˝ÿÊª üÊáÊË ◊¥ ◊ÊŸÊ ¡Ê∞ªÊ– ¬⁄UËˇÊÊÕË¸ •¬Ÿ ’Êÿ¥„ÊÕ ∑§ •¥ªÍ∆U ∑§Ê ÁŸ‡ÊÊŸ ©¬ÁSÕÁÃ ¬òÊ ◊¥ ÁŒ∞ ª∞ SÕÊŸ ¬⁄U•fl‡ÿ ‹ªÊ∞°–

12. ß‹Ä≈˛UÊÚÁŸ∑§/„SÃøÊÁ‹Ã ¬Á⁄U∑§‹∑§ ∞fl¥ ◊Ê’Êß‹ »§ÊŸ, ¬¡⁄U ßàÿÊÁŒ¡Ò‚ Á∑§‚Ë ß‹Ä≈˛UÊÚÁŸ∑§ ©¬∑§⁄UáÊ ∑§Ê ¬˝ÿÊª flÁ¡¸Ã „Ò–

13. ¬⁄UËˇÊÊ „ÊÚ‹ ◊¥ •Êø⁄UáÊ ∑§ Á‹∞ ¬⁄UËˇÊÊÕË¸ ¡.∞.’./’Ê«¸U ∑§ ‚÷ËÁŸÿ◊Ê¥ ∞fl¥U ÁflÁŸÿ◊Ê¥ mÊ⁄UÊ ÁŸÿÁ◊Ã „Ê¥ª– •ŸÈÁøÃ ‚ÊœŸ ¬˝ÿÊª ∑§‚÷Ë ◊Ê◊‹Ê¥ ∑§Ê »Ò§‚‹Ê ¡.∞.’./’Ê«¸U ∑§ ÁŸÿ◊Ê¥ ∞fl¥ ÁflÁŸÿ◊Ê¥ ∑§•ŸÈ‚Ê⁄U „ÊªÊ–

14. Á∑§‚Ë ÷Ë ÁSÕÁÃ ◊¥ ¬⁄UËˇÊÊ ¬ÈÁSÃ∑§Ê ÃÕÊ ©ûÊ⁄U ¬òÊ ∑§Ê ∑§Êß¸ ÷Ë ÷Êª•‹ª Ÿ„Ë¥ Á∑§ÿÊ ¡Ê∞ªÊ–

15. ¬⁄UËˇÊÊÕË¸ mÊ⁄UÊ ¬⁄UËˇÊÊ ∑§ˇÊ/„ÊÚ‹ ◊¥ ¬˝fl‡Ê ∑§Ê«¸U ∑§ •‹ÊflÊÁ∑§‚Ë ÷Ë ¬˝∑§Ê⁄U ∑§Ë ¬Ê∆˜Uÿ ‚Ê◊ª˝Ë, ◊ÈÁŒ˝Ã ÿÊ „SÃÁ‹ÁπÃ,∑§Êª¡ ∑§Ë ¬Áø¸ÿÊ°, ¬¡⁄U, ◊Ê’Êß‹ »§ÊŸ ÿÊ Á∑§‚Ë ÷Ë ¬˝∑§Ê⁄U∑§ ß‹Ä≈˛UÊÚÁŸ∑§ ©¬∑§⁄UáÊÊ¥ ÿÊ Á∑§‚Ë •ãÿ ¬˝∑§Ê⁄U ∑§Ë ‚Ê◊ª˝Ë∑§Ê ‹ ¡ÊŸ ÿÊ ©¬ÿÊª ∑§⁄UŸ ∑§Ë •ŸÈ◊ÁÃ Ÿ„Ë¥ „Ò–

E/Page 40

E E

E E

Question and Answer KeyQuestion No. Answer Key Question No. Answer Key Question No. Answer Key

Q1 1 Q31 2 Q61 3Q2 8 Q32 3 Q62 4Q3 3 Q33 4 Q63 1Q4 4 Q34 1 Q64 2Q5 1 Q35 4 Q65 4Q6 4 Q36 3 Q66 4Q7 1 Q37 3 Q67 4Q8 2 Q38 4 Q68 2Q9 1 Q39 2 Q69 2

Q10 4 Q40 3 Q70 3Q11 3 Q41 4 Q71 2Q12 1 Q42 3 Q72 2Q13 3 Q43 3 Q73 3Q14 3 Q44 4 Q74 2Q15 4 Q45 1 Q75 2Q16 4 Q46 1 Q76 2Q17 4 Q47 2 Q77 4Q18 1 Q48 2 Q78 3Q19 4 Q49 2 Q79 4Q20 3 Q50 3 Q80 2Q21 1 Q51 4 Q81 1Q22 4 Q52 2 Q82 3Q23 3 Q53 2 Q83 2Q24 1 Q54 1 Q84 4Q25 2 Q55 2 Q85 4Q26 4 Q56 4 Q86 2Q27 1 Q57 4 Q87 4Q28 1 Q58 3 Q88 3Q29 1 Q59 2 Q89 4Q30 8 Q60 2 Q90 3

Note:- 8 indicates that answer option 2 and 4 both are correct.

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JEE Main 2016 Answer Key Paper 1 Offline (April 3) - SET E

SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI

Set - 04 1

1. In the following ‘I’ refers to current and

other symbols have their usual meaning.

Choose the option that corresponds to the

dimensions of electrical conductivity :

(1) ML−3

T−3

I2

(2) M−1

L3

T3

I

(3) M−1

L−3

T3

I2

(4) M−1

L−3

T3

I

1. ÁŸêŸÁ‹ÁπÃ ◊¥ ‘I’ ÁfllÈÃ œÊ⁄UÊ ∑§Ê ∞fl¥ •ãÿ Áøq•¬Ÿ ‚Ê◊Êãÿ •Õ¸ ∑§Ê ß¥ÁªÃ ∑§⁄UÃ „Ò¥– ÁŸêŸÁ‹ÁπÃ ◊¥‚ ∑§ÊÒŸ-‚Ê Áfl∑§À¬ flÒlÈÃ øÊ‹∑§ÃÊ ∑§Ë ‚„Ë Áfl◊Ê ∑§Ê’ÃÊÃÊ „Ò?(1) ML

−3 T

−3 I2

(2) M−1

L3

T3

I

(3) M−1

L−3

T3

I2

(4) M−1

L−3

T3

I

1. r_ç_rgrMsdp„ ‘I’ A¡ rhÛysâhpl v$ip®h¡ R>¡ s\p AÞek„¿epAp¡ s¡_p¡ âQrgs A\®. _uQ¡ Ap ¡g rhL$ë`p¡dp„\urhÛyshplL$sp_y„ kpQy„ `qfdpZ v$ip®hsp¡ rhL$ë` `k„v$L$fp¡.(1) ML

−3 T

−3 I2

(2) M−1

L3

T3

I

(3) M−1

L−3

T3

I2

(4) M−1

L−3

T3

I

JEE Main 2016 Question Paper 1 Online (April 9, 2016)JEE Main 2016 Question Paper 1 Online (April 9, 2016)JEE Main 2016 Question Paper 1 Online (April 9, 2016)

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Typewritten Text

JEE Main 2016 Question Paper 1 Online (April 9, 2016)

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Typewritten Text


Set - 04 2

2. Which of the following option correctly

describes the variation of the speed v and

acceleration ‘a’ of a point mass falling

vertically in a viscous medium that applies

a force F =−kv, where ‘k’ is a constant, on

the body ? (Graphs are schematic and not

drawn to scale)

(1)

(2)

(3)

(4)

2. ÁŸêŸÁ‹ÁπÃ ◊¥ ‚ ∑§ÊÒŸ-‚Ê Áfl∑§À¬ ©‚ Á’ãŒÈ-Œ˝√ÿ◊ÊŸ∑§Ë ªÁÃ ‘v’ •ÊÒ⁄U àfl⁄UáÊ ‘a’ ∑§ ’Œ‹Êfl ∑§Ê ‚„Ë Ã⁄U„ ‚Œ‡ÊÊ¸ÃÊ „Ò ¡Ê Á∑§ Á∑§‚Ë ‡ÿÊŸ ◊Êäÿ◊ ◊¥ ™§äflÊ¸œ⁄U ÁŒ‡ÊÊ◊¥ ŸËø ∑§Ë •Ê⁄U Áª⁄UÃ „È∞ ◊Êäÿ◊ ∑§ ∑§Ê⁄UáÊ ∞∑§ ’‹F =−kv, ¡„Ê° ¬⁄U ‘k’ ∞∑§ ÁŸÿÃÊ¥∑§ „Ò, ∑§Ê •ŸÈ÷fl∑§⁄UÃÊ „Ò– (ª˝Ê»§Ê¥ ∑§Ê √ÿflSÕÊà◊∑§ ÁŸM§¬áÊ ◊Ê¬ ∑§•ŸÈ‚Ê⁄U Ÿ„Ë¥ „Ò–)

(1)

(2)

(3)

(4)

2. r_ç_rgrMs rhL$ë`p¡dp„\u L$ep¡ rhL$ë` tbvy$ Öìedp__uNrs v A_¡ âh¡N a _p¡ kpQp¡ k„b„ hZ®h¡ R>¡. Alutbvy$ Öìedp_ A¡ õ\p_ dpÝed L¡$ S>¡ F =−kv S>¡V$gp¡`v$p\® f bm gNpX¡$ R>¡ s¡hp dpÝeddp„ DÝh® _uQ¡ sfa`X¡$ R>¡. ‘k’ A¡ AQmp„L$ R>¡. (N°pa_y„ ìehõ\pÐdL$ r_ê$ ZR>¡ A_¡ dp` A_ykpf _\u.)

(1)

(2)

(3)

(4)


Set - 04 3

3. A rocket is fired vertically from the earth

with an acceleration of 2g, where g is the

gravitational acceleration. On an inclined

plane inside the rocket, making an angle θ

with the horizontal, a point object of mass

m is kept. The minimum coefficient of

friction µmin

between the mass and the

inclined surface such that the mass does not

move is :

(1) tanθ

(2) 2 tanθ

(3) 3 tanθ

(4) tan2θ

4. A car of weight W is on an inclined road

that rises by 100 m over a distance of 1 km

and applies a constant frictional force • ••

on the car. While moving uphill on the road

at a speed of 10 ms−1

, the car needs power

P. If it needs power ••

while moving

downhill at speed v then value of v is :

(1) 20 ms−1

(2) 15 ms−1

(3) 10 ms−1

(4) 5 ms−1

3. ∞∑§ ⁄UÊÚ∑§≈U ∑§Ê ¬ÎâflË ‚ ™§äflÊ¸œ⁄U ÁŒ‡ÊÊ ◊¥ 2g ∑§ àfl⁄UáÊ‚ ¬˝ˇÊÁ¬Ã Á∑§ÿÊ ªÿÊ „Ò– ß‚ ⁄UÊÚ∑§≈U ∑§ •¥Œ⁄ U ˇÊÒÁÃ¡‚ θ ∑§ÊáÊ ’ŸÊÃ „È∞ ∞∑§ •ÊŸÃ-Ã‹ ¬⁄U ∞∑§m Œ˝√ÿ◊ÊŸ ∑§Ê Á’¥ŒÈ ∑§áÊ ÁSÕÃ „Ò– ÿÁŒ ⁄UÊÚ∑§≈U ∑§¬˝ˇÊÁ¬Ã „ÊŸ ¬⁄U Á’¥ŒÈ-∑§áÊ ÁSÕ⁄U •flSÕÊ ◊¥ „Ë ⁄U„ÃÊ „ÒÃ’ Œ˝√ÿ◊ÊŸ ∞fl¥ •ÊŸÃ Ã‹ ∑§ ’Ëø ÉÊ·¸áÊ-ªÈáÊÊ¥∑§µ

min ∑§Ê ◊ÊŸ ÄÿÊ „ÊªÊ? (‘g’ ªÈL§àflËÿ àfl⁄UáÊ „Ò) —

(1) tanθ

(2) 2 tanθ

(3) 3 tanθ

(4) tan2θ

4. ∞∑§ W ÷Ê⁄U ∑§Ë ∑§Ê⁄U ∞∑§ ∞‚Ë •ÊŸÃ-‚«∏∑§ ¬⁄U ø‹⁄U„Ë „Ò ¡Ê Á∑§ 1 km ŒÍ⁄UË ¬⁄U 100 m ™°§øË „Ê ¡ÊÃË „Ò,

•ÊÒ⁄U ∑§Ê⁄U ¬⁄U • ••

◊ÊŸ ∑§Ê ÁŸÿÃ ÉÊ·¸áÊ ’‹ ‹ªÊÃË „Ò–

ÿÁŒ ∑§Ê⁄U ∑§Ê ‚«∏∑§ ¬⁄U ™§¬⁄U ∑§Ë •Ê⁄U 10 ms−1 ∑§Ë

ªÁÃ „ÃÈ P ‡ÊÁÄàÊ ∑§Ë •Êfl‡ÿ∑§ÃÊ „Ò ∞fl¥ ŸËø ∑§Ë •Ê⁄U

v ªÁÃ ‚ ø‹ÊŸ „ÃÈ ••

‡ÊÁÄàÊ ∑§Ë •Êfl‡ÿ∑§ÃÊ ¬«∏ÃË

„Ò, ÃÊ v ∑§Ê ◊ÊŸ „ÊªÊ —(1) 20 ms

−1

(2) 15 ms−1

(3) 10 ms−1

(4) 5 ms−1

3. A¡L$ fp¡L¡$V$_¡ ©Õhu\u DÝh® qv$ipdp„ 2g âh¡N\u ân¡r`sL$fhpdp„ Aph¡ R>¡. Al] g NyfyÐhâh¡N R>¡. Ap fp¡L¡$V$dp„kdrnrsS> kp\¡ θ S>¡V$gp¡ MyZp¡ b_phsp Y$msp pqV$ep`f m - v$m_p¡ A¡L$ tbvy$hs `v$p\® dyL¡$g R>¡. Ap v$m[õ\f fl¡ (Mk¡ _lu) s¡ dpV¡$_p¡ v$m A_¡ Y$msp pqV$ep_uk`pV$u hÃQ¡_p¡ gOyÑd ^j®Zp¯L$ µ

min R>¡.

(1) tanθ

(2) 2 tanθ

(3) 3 tanθ

(4) tan2θ

4. 1 km _p A„sf ky udp„ 100 m S>¡V$gp¡ KQp¡ \C S>spY$p¡mphhpmp fp¡X$ D`f W hS>__u A¡L$ L$pf R>¡.

Y$p¡mphhpmp¡ Ap fp¡X$ L$pf `f • ••

S>¡V$gy„ AQm Oj®Z

bm gNpX¡$ R>¡. 10 ms−1 _u Nrs\u fp¡X$ `f D`f

sfa S>sp„ L$pf_¡ phf P _u S>ê$f X¡$ R>¡. Å¡ L$pf Y$p¡mph

`f\u v S>¡V$gu TX$`\u _uQ¡ Dsfsp phf ••

_u S>ê$f

`X¡$ sp¡ v R>¡ :(1) 20 ms

−1

(2) 15 ms−1

(3) 10 ms−1

(4) 5 ms−1


Set - 04 4

5. A cubical block of side 30 cm is moving

with velocity 2 ms−1

on a smooth

horizontal surface. The surface has a bump

at a point O as shown in figure. The angular

velocity (in rad/s) of the block immediately

after it hits the bump, is :

(1) 5.0

(2) 6.7

(3) 9.4

(4) 13.3

6. Figure shows elliptical path abcd of a planet

around the sun S such that the area of

triangle csa is

•• the area of the ellipse. (See

figure) With db as the semimajor axis, and

ca as the semiminor axis. If t1

is the time

taken for planet to go over path abc and t2

for path taken over cda then :

(1) t1=t

2

(2) t1=2t

2

(3) t1=3t

2

(4) t1=4t

2

5. ∞∑§ 30 cm ÷È¡Ê flÊ‹Ê ÉÊŸËÿ é‹ÊÚ∑§ ∞∑§ Áø∑§Ÿ ÊÒÁÃ¡Ã‹ ¬⁄U 2 ms

−1 ∑§ flª ‚ ªÁÃ◊ÊŸ „Ò– ¡Ò‚Ê Á∑§ ÁøòÊ◊¥ ÁŒπÊÿÊ ªÿÊ „Ò, O ¬⁄U ∞∑§ •fl⁄UÊœ ÁSÕÃ „Ò– •fl⁄UÊœ‚ ≈U∑§⁄UÊŸ ∑§ ÃÈ⁄¥UÃ ’ÊŒ é‹ÊÚ∑§ ∑§Ê ∑§ÊáÊËÿ flª (⁄UÁ«UÿŸ/

‚∑¥§«U ◊¥) „ÊªÊ —

(1) 5.0

(2) 6.7

(3) 9.4

(4) 13.3

6. ∞∑§ ª˝„ ‚Íÿ¸ S ∑§ øÊ⁄UÊ¥ •Ê⁄U ∞∑§ ŒËÉÊ¸flÎûÊËÿ ∑§ˇÊabcd ◊¥ ß‚ Ã⁄U„ ‚ øÄ∑§⁄U ‹ªÊÃÊ „Ò Á∑§ csa ÁòÊ÷È¡∑§Ê ˇÊòÊ»§‹ ŒËÉÊ¸flÎûÊ ∑§ ˇÊòÊ»§‹ ∑§Ê ∞∑§-øÊÒÕÊß¸ „Ò(ÿ„Ê° ¬⁄U ac ‹ÉÊÈ-•ˇÊ ∞fl¥ bd ŒËÉÊ¸-•ˇÊ „Ò)– ÿÁŒª˝„ abc ÃÕÊ cda ∑§ˇÊËÿ ¬ÕÊ¥ ∑§ Á‹∞ ∑˝§◊‡Ê—t1 ÃÕÊ t

2 ∑§Ê ‚◊ÿ ‹ÃÊ „Ò, Ã’ —

(1) t1=t

2

(2) t1=2t

2

(3) t1=3t

2

(4) t1=4t

2

5. 30 cm bpSy>hpmp¡ A¡L$ kdO_ 2 ms−1 _p h¡N\u

kdrnrsS> k`pV$u `f Nrs L$f¡ R>¡. ApL©$rÑdp„ v$ip®ìepâdpZ¡ O tbvy$ pk¡ A¡L$ bç` R>¡. Ap bç`_¡ A\X$peL¡$ sfs S> kdO__p¡ L$p¡Zueh¡N (rad/sdp„) R>¡ :

(1) 5.0

(2) 6.7

(3) 9.4

(4) 13.3

6. k|e® (S) t1 _u afs¡ Nrs L$fsp A¡L$ D`N°l_p¡ D`hgeu

dpN® abcd A¡ ApL©$rsdp„ v$ip®h¡g R>¡. Äep„ rÓcyS> csa

_y„ n¡Óam A¡ D`hge_p n¡Óam_y„ •• R>¡, db A¡ s¡_u

v$uO®-An s\p ca s¡_u gOy-An R>¡. Å¡ D`N°l s¡_pL$nue \ abc A_¡ cda dpV¡$ g¡hpsp¡ kde A_y¾$d¡ t

1

A_¡ t2 lp¡e sp¡ :

(1) t1=t

2

(2) t1=2t

2

(3) t1=3t

2

(4) t1=4t

2


Set - 04 5

7.

Consider a water jar of radius R that has

water filled up to height H and is kept on a

stand of height h (see figure). Through a

hole of radius r (r << R) at its bottom, the

water leaks out and the stream of water

coming down towards the ground has a

shape like a funnel as shown in the figure.

If the radius of the cross-section of water

stream when it hits the ground is x. Then :

(1)

•• •••• • •

•

•• •

(2)

•••• •••

• • ••

•• •

(3)

•••• •••

• • ••

•• •

(4)

••• •••• • •

•

•• •

7.

∞∑§ R ÁòÊíÿÊ ∑§ ¬ÊŸË ∑§ ¡Ê⁄, Á¡‚ ¬ÊŸË ‚ H ™°§øÊß¸Ã∑§ ÷⁄UÊ ªÿÊ „Ò, ∑§Ê h ™°§øÊß¸ ∑§ S≈Ò¥U«U ¬⁄U ⁄UπÊ ªÿÊ „Ò(ÁøòÊ Œπ¥)– Ã‹ ◊¥ ∞∑§ ¿UÊ≈U Á¿UŒ˝, Á¡‚∑§Ë ÁòÊíÿÊ r„Ò (r << R), ‚ ŸËø Áª⁄UÃ „È∞ ¬ÊŸË ∑§Ë œÊ⁄U ∞∑§“∑§Ë¬” ∑§Ê •Ê∑§Ê⁄U œÊ⁄UáÊ ∑§⁄UÃË „Ò– ÿÁŒ ÷ÍÁ◊ ∑§ Ã‹¬⁄U ¬ÊŸË ∑§Ë œÊ⁄U ∑§ •ŸÈ¬˝SÕ ∑§Ê≈U ∑§Ë ÁòÊíÿÊ x „Ò,Ã’ —

(1)

•• •••• • •

•

•• •

(2)

•••• •••

• • ••

•• •

(3)

•••• •••

• • ••

•• •

(4)

••• •••• • •

•

•• •

7.

R rÓÄep fphsp A¡L$ pZu_p Åfdp„ H KQpC ky u_y„`pZu cf¡gy A_¡ s¡_¡ h KQpC_p õV¡$ÞX$ `f fpM¡g R>¡s¡d ^pfp¡. (ApL©$rs Sy>Ap¡) Åf_p smue¡ r -rÓÄep_pA¡L$ L$pZpdp„\u (r << R) `pZu r_L$m¡ R>¡ S>¡ N°pDÞX$sfa _pmQp_p ApL$pfdp„ Å¡hp dm¡ R>¡. Ap `pZu_u^pfp Äepf¡ S>du__¡ AX¡$ R>¡ Ðepf¡ s¡_p ApX$R>¡v$_u rÓÄepx R>¡ sp¡ :

(1)

•• •••• • •

•

•• •

(2)

•••• •••

• • ••

•• •

(3)

•••• •••

• • ••

•• •

(4)

••• •••• • •

•

•• •


Set - 04 6

8. 200 g water is heated from 40C to 60C.

Ignoring the slight expansion of water, the

change in its internal energy is close to

(Given specific heat of

water=4184 J/kg/K) :

(1) 8.4 kJ

(2) 4.2 kJ

(3) 16.7 kJ

(4) 167.4 kJ

9. The ratio of work done by an ideal

monoatomic gas to the heat supplied to it

in an isobaric process is :

(1)

••

(2)

••

(3)

••

(4)

••

8. ¬ÊŸË ∑§ ÁflSÃÊ⁄U ∑§Ê Ÿªáÿ ◊ÊŸÃ „È∞, 200 g ¬ÊŸË ∑§Ê40C ‚ 60C Ã∑§ ª⁄U◊ ∑§⁄UŸ ¬⁄U ©‚∑§Ë •Ê¥ÃÁ⁄U∑§™§¡Ê¸ ◊¥ •ŸÈ◊ÊÁŸÃ ¬Á⁄UflÃ¸Ÿ „ÊªÊ (¬ÊŸË ∑§Ê ÁflÁ‡Êc≈UÃÊ¬ = 4184 J/kg/K ‹¥) —(1) 8.4 kJ

(2) 4.2 kJ

(3) 16.7 kJ

(4) 167.4 kJ

9. Á∑§‚Ë ‚◊÷ÊÁ⁄U∑§ ¬˝Á∑˝§ÿÊ ◊¥ ∞∑§ •ÊŒ‡Ê¸ ∞∑§¬⁄U◊ÊáÊÈ∑§ªÒ‚ ∑§ mÊ⁄UÊ Á∑§ÿ ª∞ ∑§Êÿ¸ ÃÕÊ ©‚ ŒË ªß¸ ™§c◊Ê ∑§Ê•ŸÈ¬ÊÃ „ÊªÊ —

(1)

••

(2)

••

(3)

••

(4)

••

8. 200 g `pZu_¡ 40C \u 60C ky u Nfd L$fhpdp„Aph¡ R>¡. `pZu_y„ rhõsfZ AhNZsp, s¡_u Ap„sqfL$EÅ®dp„ \sp¡ a¡fapf A¡ gNcN li¡ :

(`pZu_u rhriô$ Dódp=4184 J/kg/K)(1) 8.4 kJ

(2) 4.2 kJ

(3) 16.7 kJ

(4) 167.4 kJ

9. L$p¡C kdcpqfL$ âq¾$epdp„ A¡L$ Apv$i® A¡L$ `fdpÎhuehpey Üpfp \sp L$pe® s\p s¡_¡ Ap`hpdp„ Aphsu Dódp_p¡NyZp¡Ñf R>¡ :

(1)

••

(2)

••

(3)

••

(4)

••


Set - 04 7

10. Two particles are performing simple

harmonic motion in a straight line about

the same equilibrium point. The amplitude

and time period for both particles are same

and equal to A and T, respectively. At time

t=0 one particle has displacement A while

the other one has displacement •••

and

they are moving towards each other. If they

cross each other at time t, then t is :

(1) ••

(2)

•••

(3) ••

(4) ••

11. Two engines pass each other moving in

opposite directions with uniform speed of

30 m/s. One of them is blowing a whistle

of frequency 540 Hz. Calculate the

frequency heard by driver of second engine

before they pass each other. Speed of sound

is 330 m/sec :

(1) 450 Hz

(2) 540 Hz

(3) 648 Hz

(4) 270 Hz

10. ŒÊ ∑§áÊ ∞∑§ ‚⁄U‹ ⁄UπËÿ ¬Õ ¬⁄U ÁSÕÃ ∞∑§ „Ë ◊ÊäÿÁ’¥ŒÈ ∑§ ‚Ê¬ Ê ß‚ Ã⁄U„ ‚ ‚⁄U‹ •ÊflÃ¸ ªÁÃ◊ÊŸ •flSÕÊ◊¥ „Ò Á∑§ ©Ÿ∑§ •ÊÿÊ◊ (A) ÃÕÊ •ÊflÃ¸-∑§Ê‹ (T) ∞∑§‚◊ÊŸ „Ò¥– ÿÁŒ t=0 ‚◊ÿ ¬⁄U ∞∑§-ŒÍ‚⁄U ∑§Ë Ã⁄U»§•ÊÃ „È∞, ∞∑§ ∑§áÊ ∑§Ê ÁflSÕÊ¬Ÿ A „Ò ÃÕÊ ŒÍ‚⁄U ∑§áÊ

∑§Ê ÁflSÕÊ¬Ÿ •••

„Ê, ÃÊ t ‚◊ÿ ¬⁄U fl ∞∑§ ŒÍ‚⁄U ∑§Ê

¬Ê⁄U ∑§⁄UÃ „Ò¥– t ∑§Ê ◊ÊŸ „ÊªÊ —

(1) ••

(2)

•••

(3) ••

(4) ••

11. ŒÊ ⁄U‹-ß¥¡Ÿ ∞∑§-ŒÍ‚⁄U ∑§Ê ¬Ê⁄U ∑§⁄UÃ „È∞ Áfl¬⁄UËÃÁŒ‡ÊÊ ◊¥ 30 m/s ∑§Ë ∞∑§ ‚◊ÊŸ ªÁÃ ‚ ø‹ ⁄U„ „Ò¥–©Ÿ◊¥ ‚ ∞∑§ ß¥¡Ÿ ÿÁŒ 540 Hz •ÊflÎÁûÊ ‚ ‚Ë≈UË ’¡Ê⁄U„Ê „Ò, ÃÊ ŒÍ‚⁄U ß¥¡Ÿ ∑§ «˛UÊß¸fl⁄U mÊ⁄UÊ ‚ÈŸË ªß¸ äflÁŸ ∑§Ë•Êfl Î Áû Ê „Ê ªË (äflÁŸ ∑§Ë ªÁÃ ∑§Ê ◊ÊŸ330 m/sec ‹¥) —(1) 450 Hz

(2) 540 Hz

(3) 648 Hz

(4) 270 Hz

10. A¡L$S> kdsygus tbvy$_u kp ¡n¡ b¡ L$Zp¡ ku^u f¡Mpdp„kfm Aphs®Nrs L$f¡ R>¡. Ap b„_¡ L$Zp¡ dpV¡$ L„$`rhõspfs\p Aphs® L$pm kdp_ R>¡ A_¡ s¡ ¾$di: A s\p T R>¡.t=0 kde¡, A¡L$buÅ sfa Aphsp A¡L$ L$Z_y„ õ\p_p„sf

A s\p buÅ_y„ õ\p_p„sf •••

R>¡. Å¡ t kde¡ s¡

A¡L$buÅ_¡ `pf L$f¡ sp¡ t R>¡ :

(1) ••

(2)

•••

(3) ••

(4) ••

11. b¡ A¡ [ÞS>_ A¡L $buÅ_¡ `pf L $fsp rh`fusqv$ipdp„ 30 m/s _u kdp_ Nrs\u Qpg¡ R> ¡.Apdp_y „ A¡L$ A¡[ÞS>_ Å¡ 540 Hz Aph©rs\ukuV$u hNpX$u füy „ lp ¡e, sp ¡ buÆ A¡[ÞS>__pX² $pehf hX¡$ k„cmpsp AhpS>_u Aph©rs li¡ :(AhpS>_u TX$` 330 m/sec gp¡)(1) 450 Hz

(2) 540 Hz

(3) 648 Hz

(4) 270 Hz


Set - 04 8

12. The potential (in volts) of a charge

distribution is given by

V(z)=30−5z2 for z 1 m

V(z)=35−10 z for z 1 m.

V(z) does not depend on x and y. If this

potential is generated by a constant charge

per unit volume ρ0 (in units of

0 ) which is

spread over a certain region, then choose

the correct statement.

(1) ρ0=10

0 for z 1 m and ρ

0=0

elsewhere

(2) ρ0=20

0 in the entire region

(3) ρ0=40

0 in the entire region

(4) ρ0=20

0 for z 1 m and ρ

0=0

elsewhere

13. Three capacitors each of 4 µF are to be

connected in such a way that the effective

capacitance is 6 µF. This can be done by

connecting them :

(1) all in series

(2) two in series and one in parallel

(3) all in parallel

(4) two in parallel and one in series

12. ∞∑§ •Êfl‡Ê-ÁflÃ⁄UáÊ ∑§ mÊ⁄UÊ ÁŸêŸÁ‹ÁπÃ Áfl÷fl (flÊÀ≈U◊¥) ©à¬ÛÊ „ÊÃÊ „Ò —

V(z)=30−5z2, z 1 m ◊¥

V(z)=35−10 z , z 1 m ◊¥

V(z), x ∞fl¥ y ¬⁄U ÁŸ÷¸⁄U Ÿ„Ë¥ ∑§⁄UÃÊ– ÿÁŒ ÿ„ Áfl÷fl∞∑§ ÁŸÿÃ •Êfl ‡Ê ¡Ê ¬ ˝ ÁÃ ß∑§Êß ¸ •ÊÿÃŸρ

0 (

0 ß∑§ÊßÿÊ¥ ◊¥) „Ò ÃÕÊ ∞∑§ ÁŒÿ „È∞ ˇÊòÊ ◊¥ »Ò§‹Ê

„È•Ê „Ò, ‚ ©à¬ÊÁŒÃ „Ò, Ã’ ÁŸêŸÁ‹ÁπÃ ◊¥ ‚ ‚„ËÁfl∑§À¬ ∑§Ê øÿŸ ∑§⁄¥U —

(1) ρ0=10

0, z1 m ◊¥ ÃÕÊ ρ

0=0 •ãÿòÊ

(2) ρ0=20

0, ‚fl¸òÊ

(3) ρ0=40

0, ‚fl¸òÊ

(4) ρ0=20

0, z1 m ◊¥ ÃÕÊ ρ

0=0 •ãÿòÊ

13. 4 µF œÊÁ⁄UÃÊ ∑§ ÃËŸ ‚¥œÊÁ⁄UòÊÊ¥ ‚ ß‚ Ã⁄U„ ‚ ‚¥ÿÊ¡Ÿ’ŸÊŸÊ „Ò Á∑§ ¬˝÷ÊflË œÊÁ⁄UÃÊ 6 µF „Ê ¡Ê∞– ÿ„ ÁŸêŸ‚¥ÿÊ¡Ÿ ‚ ¬˝ÊåàÊ Á∑§ÿÊ ¡Ê ‚∑§ÃÊ „Ò —

(1) ÃËŸÊ¥ üÊáÊË ∑˝§◊ ◊¥

(2) ŒÊ üÊáÊË ∑˝§◊ ◊¥ ÃÕÊ ÃË‚⁄UÊ ¬Ê‡fl¸∑˝§◊ ◊¥

(3) ÃËŸÊ¥ ¬Ê‡fl¸∑˝§◊ ◊¥

(4) ŒÊ ¬Ê‡fl¸∑˝§◊ ◊¥ ÃÕÊ ÃË‚⁄UÊ üÊáÊË ∑˝§◊ ◊¥

12. A¡L$ rhÛyscpf rhsfZdp„ rhch (volts dp„)V(z)=30−5z

2, z 1 m dpV¡$

V(z)=35−10 z, z 1 m dpV¡$ Ap`hpdp„ Aph¡R>¡.

V(z) A¡ x A_¡ y `f Ap^pf fpMsp¡ _\u. Å¡ Ap[õ\rsdp_ A¡ r_es rhS>cpf ârs L$v$ ρ

0 (

0 _p

A¡L$ddp„) hX¡$ DÐ`Þ_ L$fhpdp„ Aph¡ L¡$ S>¡ L$p¡C A¡L$Qp¡½$k n¡Ódp„ rhõsf¡g lp¡e, sp¡ kpQy„ rh^p_ `k„v$L$fp¡.

(1) ρ0=10

0, z1 m dpV¡$ s\p ρ

0=0 AÞeÓ

(2) ρ0=20

0 kh®n¡Ódp„

(3) ρ0=40

0 kh®n¡Ódp„

(4) ρ0=20

0, z1 m dpV¡$ A_¡ ρ

0=0 AÞeÓ

13. 4 µF _p¡ A¡L$ A¡hp ÓZ L¸$`p rkV$fp¡ A¡ fus¡ Å¡X$hpdp„Aph¡g R>¡ L¡$ s¡d_p¡ kdsyëe L¸$`prkV$Þk 6 µF R>¡.Aphy„ s¡d_¡ :

(1) î¡Zudp„ Å¡X$hp\u \C iL¡$

(2) b¡_¡ î¡Zudp„ A_¡ A¡L$_¡ kdp„sf Å¡X$hp\u \CiL¡$

(3) kdp„sf Å¡X$hp\u \C iL¡$

(4) b¡_¡ kdp„sf A_¡ A¡L$_¡ î¡Zudp„ Å¡X$hp\u \CiL¡$


Set - 04 9

14.

In the circuit shown, the resistance r is a

variable resistance. If for r=f R, the heat

generation in r is maximum then the value

of f is :

(1)

••

(2)

••

(3)

••

(4) 1

15. A magnetic dipole is acted upon by two

magnetic fields which are inclined to each

other at an angle of 75. One of the fields

has a magnitude of 15 mT. The dipole

attains stable equilibrium at an angle of 30

with this field. The magnitude of the other

field (in mT ) is close to :

(1) 11

(2) 36

(3) 1

(4) 1060

14.

ÁŒÿ „È∞ ¬Á⁄U¬Õ ◊¥ r ∞∑§ ø⁄U-¬˝ÁÃ⁄UÊœ „Ò– ÿÁŒr=f R, Ã’ r ◊¥ ™§c◊Ê ©à¬ÊŒŸ •Áœ∑§Ã◊ „ÊŸ ∑§Á‹ÿ f ∑§Ê ◊ÊŸ „ÊªÊ —

(1)

••

(2)

••

(3)

••

(4) 1

15. ∞∑§ øÈê’∑§Ëÿ Ámœ˝Èfl ¬⁄U ŒÊ øÈê’∑§Ëÿ ˇÊòÊ, ¡Ê •Ê¬‚◊¥ 75 ∑§ÊáÊ ’ŸÊÃ „Ò¥, ∞∑§ ‚ÊÕ Á∑˝§ÿÊ ∑§⁄UÃ „Ò¥– ÿÁŒÿ„ Ámœ˝Èfl ‚¥ÃÈ‹Ÿ ∑§Ë •flSÕÊ ◊¥ øÈê’∑§Ëÿ ¬˝⁄UáÊ15 mT ∑§ ∞∑§ øÈê’∑§Ëÿ ˇÊòÊ ‚ 30 ∑§Ê ∑§ÊáÊ ’ŸÊÃÊ„Ò, ÃÊ ŒÍ‚⁄U øÈê’∑§Ëÿ ÊòÊ ∑§ øÈê’∑§Ëÿ ¬⁄UáÊ ∑§Ê ‹ª÷ª◊ÊŸ ( mT ◊¥ ) „ÊªÊ —(1) 11

(2) 36

(3) 1

(4) 1060

14.

Ap ¡g `qf`\dp„ r A¡L$ Qg Ahfp¡ R>¡. Å¡ r=f R

dpV¡$ r dp„ dlÑd Dódp DÐ`Þ_ \pe sp¡ f _u qL„$ds\i¡ :

(1)

••

(2)

••

(3)

••

(4) 1

15. A¡L$buÅ_u kpd¡ 75 _p MyZp¡ Y$msp b¡ Qy„bL$ue n¡Óp¡`f A¡L$ Qy„bL$ue qÜ- y°h (Dipole) L$pe®fs R>¡. L$p¡CA¡L$ n¡Ó_y„ dyëe 15 mT R>¡ A_¡ Ap n¡Ó kp\¡ 30 _pMyZp\u X$pe`p¡g õ\peu k„sygus Ahõ\p âpá L$f¡ R>¡.buÅ n¡Ó_y„ dyëe (mT dp„) Apif¡ li¡ :(1) 11

(2) 36

(3) 1

(4) 1060


Set - 04 10

16. A 50 Ω resistance is connected to a battery

of 5 V. A galvanometer of resistance

100 Ω is to be used as an ammeter to

measure current through the resistance, for

this a resistance rs

is connected to the

galvanometer. Which of the following

connections should be employed if the

measured current is within 1% of the

current without the ammeter in the

circuit ?

(1) rs=0.5 Ω in parallel with the

galvanometer

(2) rs=0.5 Ω in series with the

galvanometer

(3) rs

= 1 Ω in series with galvanometer

(4) rs=1 Ω in parallel with galvanometer

16. ∞∑§ 50 Ω ∑§Ê ¬˝ÁÃ⁄UÊœ ∞∑§ 5 V ∑§Ë ’Ò≈U⁄UË ‚ ¡È«∏Ê „È•Ê„Ò– ∞∑§ ªÒÀflŸÊ◊Ë≈U⁄U Á¡‚∑§Ê ¬˝ÁÃ⁄UÊœ 100 Ω „Ò, ∑§Ê∞ê¬Ëÿ⁄U◊Ë≈U⁄U ∑§ M§¬ ◊¥ ¬˝ÿÊª Á∑§ÿÊ ¡ÊŸÊ „Ò–ªÒÀflŸÊ◊Ë≈U⁄U ∑§ ‚ÊÕ ∞∑§ ¬˝ÁÃ⁄UÊœ r

s ‚¥ÿÊÁ¡Ã „Ò– ÿÁŒ

ß‚ ‚¥ÿÊ¡Ÿ ◊¥ ◊ÊÁ¬Ã œÊ⁄UÊ ∞ê¬Ëÿ⁄U◊Ë≈U⁄U ∑§Ê „≈UÊŸ ¬⁄U◊ÊÁ¬Ã œÊ⁄UÊ ∑§ ◊ÊŸ ‚ 1% ∑§ ÷ËÃ⁄U „Ê ÃÊ ÁŸêŸÁ‹ÁπÃ◊¥ ‚ ∑§ÊÒŸ-‚Ê ‚¥ÿÊ¡Ÿ ©ÁøÃ „ÊªÊ?

(1) rs=0.5 Ω ªÒÀflŸÊ◊Ë≈U⁄U ∑§ ‚ÊÕ ¬Ê‡fl¸∑˝§◊ ◊¥

(2) rs=0.5 Ω ªÒÀflŸÊ◊Ë≈U⁄U ∑§ ‚ÊÕ üÊáÊË ∑˝§◊ ◊¥

(3) rs

= 1 Ω ªÒÀflŸÊ◊Ë≈U⁄U ∑§ ‚ÊÕ üÊáÊË ∑˝§◊ ◊¥

(4) rs=1 Ω ªÒÀflŸÊ◊Ë≈U⁄U ∑§ ‚ÊÕ ¬Ê‡fl¸∑˝§◊ ◊¥

16. 5 V _u b¡V$fu kp\¡ 50 Ω _p¡ A¡L$ Ahfp¡ gNpX¡$g R>¡.Ap Ahfp ¡^dp „\u `kpf \sp âhpl dp`hp100 Ω _p Ahfp¡ fphsy„ N¡ëh¡_p¡duV$f_¡ A¡rdV$f sfuL¡$hp`fhpdp„ Aph¡ R>¡. Ap dpV¡$ A¡L$ Ahfp¡ r

s N¡ëh¡_p¡duV$f

kp\¡ Å¡X$hpdp„ Aph¡ R>¡.

_uQ¡ Ap ¡g L$ey Å¡X$pZ hp`fhy„ Å¡CA¡ L¡$ S>¡\u dp ¡gâhpl A¡ A¡rdV$f hNf `qf`\_p dp„_p âhpl_u1% _u Ahr^dp„ lp¡e.

(1) rs=0.5 Ω N¡ëh¡_p¡duV$f_¡ kdp„sf

(2) rs=0.5 Ω N¡ëh¡_p¡duV$f_u î¡Zudp„

(3) rs

= 1 Ω N¡ëh¡_p¡duV$f_u î¡Zudp„

(4) rs=1 Ω N¡ëh¡_p¡duV$f_¡ kdp„sf


Set - 04 11

17. A series LR circuit is connected to a voltage

source with V(t)=V0

sinΩt. After very

large time, current I(t) behaves as

••••

>>

:

(1)

(2)

(3)

(4)

17. ∞∑§ üÊ áÊË LR ¬Á⁄U¬Õ ∑§Ê ∞∑§ flÊÀ≈UËÿ dÊÃV(t)=V

0 sinΩt ‚ ¡Ê«∏Ê ¡ÊÃÊ „Ò ∑§Ê»§Ë ‹¥’ ‚◊ÿ

’ÊŒ ÁfllÈÃ œÊ⁄UÊ I(t) ∑§Ê ‚„Ë ÁøòÊáÊ Á∑§‚ Ã⁄U„ ∑§Ê

„ÊªÊ? ••••

>>

¡„Ê°

(1)

(2)

(3)

(4)

17. A¡L$ LR î¡Zu qf`\ V(t)=V0

sinΩt hp¡ëV¡$S> öp¡skp\¡ Å¡X¡$g R>¡. Ap gp„bp kde A„sfpg bpv$, rhÛys

âhpl I(t) hs®i¡ ••••

>>

:

(1)

(2)

(3)

(4)


Set - 04 12

18. Microwave oven acts on the principle of :

(1) transferring electrons from lower to

higher energy levels in water

molecule

(2) giving rotational energy to water

molecules

(3) giving vibrational energy to water

molecules

(4) giving translational energy to water

molecules

18. ◊Êß∑˝§Êflfl •ÊflŸ Á∑§‚ ¬˝Á∑˝§ÿÊ ¬⁄U •ÊœÊÁ⁄UÃ „Ò?

(1) ¡‹ •áÊÈ•Ê¥ ◊¥ ÁSÕÃ ß‹Ä≈˛UÊÚŸÊ¥ ∑§ ∑§◊ ™§¡Ê¸‚ •Áœ∑§ ™§¡Ê¸ flÊ‹ ‹fl‹ ◊¥ SÕÊŸÊ¥ÃÁ⁄UÃ ∑§⁄UŸ∑§Ë ¬˝Á∑˝§ÿÊ ¬⁄U–

(2) ¡‹ •áÊÈ•Ê¥ ∑§Ê ÉÊÍáÊ¸Ÿ ™§¡Ê¸ ¬˝ŒÊŸ ∑§⁄UŸ ∑§Ë¬˝Á∑˝§ÿÊ ¬⁄U–

(3) ¡‹ •áÊÈ•Ê¥ ∑§Ê ∑¥§¬Ÿ ™§¡Ê¸ ¬˝ŒÊŸ ∑§⁄UŸ ∑§Ë¬˝Á∑˝§ÿÊ ¬⁄U–

(4) ¡‹ •áÊÈ•Ê¥ ∑§Ê SÕÊŸÊ¥Ã⁄UË ™§¡Ê¸ ¬˝ŒÊŸ ∑§⁄UŸ∑§Ë ¬˝Á∑˝§ÿÊ ¬⁄U–

18. dpC¾$p¡h¡h Ap¡h_ L$ep rkÝ^p„s `f L$pe® L$f¡ R>¡ ?

(1) `pZu_p AÏ„Ap¡dp„ Ap¡R>u \u h y EÅ® õsfsfa Cg¡¼V²$p¡__y„ õ\p_p„sfZ

(2) `pZu_p AÏ„Ap¡_¡ O|Z®_ EÅ® Ap`hp_y„

(3) `pZu_p AÏ„Ap¡_¡ L„$`_ EÅ® Ap`hp_y„

(4) `pZu_p AÏ„Ap¡_¡ õ\p_p„sqfs EÅ® Ap`hp_y„


Set - 04 13

19. A convex lens, of focal length 30 cm, a

concave lens of focal length 120 cm, and a

plane mirror are arranged as shown. For

an object kept at a distance of 60 cm from

the convex lens, the final image, formed by

the combination, is a real image, at a

distance of :

(1) 60 cm from the convex lens

(2) 60 cm from the concave lens

(3) 70 cm from the convex lens

(4) 70 cm from the concave lens

19. ∞∑§ ©ûÊ‹ ‹Ò¥‚ fl •flÃ‹ ‹Ò¥‚, Á¡Ÿ∑§Ë $»§Ê∑§‚ ŒÍ⁄UË∑˝§◊‡Ê— 30 cm ∞fl¥ 120 cm „Ò, ÃÕÊ ‚◊Ã‹ Œ¬¸áÊÁŸêŸ ÁøòÊ ∑§ •ŸÈ‚Ê⁄U ⁄Uπ ªÿ „Ò– ∞∑§ Á’ê’ ©ûÊ‹‹Ò¥‚ ‚ 60 cm ∑§Ë ŒÍ⁄UË ¬⁄U ÁSÕÃ „Ò– ß‚ ‚¥ÿÊ¡Ÿ mÊ⁄UÊÁŸÁ◊¸Ã •¥ÁÃ◊ ¬˝ÁÃÁ’ê’ ∞∑§ flÊSÃÁfl∑§ ¬˝ÁÃÁ’ê’ „ÒÁ¡‚∑§Ë ÁSÕÁÃ ÁŸêŸÁ‹ÁπÃ „ÊªË —

(1) ©ûÊ‹ ‹Ò¥‚ ‚ 60 cm ∑§Ë ŒÍ⁄UË ¬⁄U–

(2) •flÃ‹ ‹Ò¥‚ ‚ 60 cm ∑§Ë ŒÍ⁄UË ¬⁄U–

(3) ©ûÊ‹ ‹Ò¥‚ ‚ 70 cm ∑§Ë ŒÍ⁄UË ¬⁄U–

(4) •flÃ‹ ‹Ò¥‚ ‚ 70 cm ∑§Ë ŒÍ⁄UË ¬⁄U–

19. 30 cm L¡$ÞÖg„bpC ^fphsp¡ A¡L$ brlNp£m L$pQ,120 cm L¡$ÞÖg„bpC ^fphsp¡ A¡L$ A„sNp£m L$pQ A_¡A¡L$ kpv$p¡ Afukp¡ ApL©$rÑdp„ bspìep âdpZ¡ Np¡W$$h¡gR>¡.

brlNp£m L$pQ\u 60 cm vy$f fpM¡g hõsy_y„ _uQ¡ v$ip®h¡gL$C [õ\rsdp„ Ap Np¡W$hhp\u L¡$V$gp A„sf¡ hõsy_y„ hpõsrhL$ârstbb dmi¡.

(1) brlNp£m L$pQ\u 60 cm

(2) A„sNp£m L$pQ\u 60 cm

(3) brlNp£m L$pQ\u 70 cm

(4) A„sNp£m L$pQ\u 70 cm


Set - 04 14

20. In Young’s double slit experiment, the

distance between slits and the screen is

1.0 m and monochromatic light of 600 nm

is being used. A person standing near the

slits is looking at the fringe pattern. When

the separation between the slits is varied,

the interference pattern disappears for a

particular distance d0 between the slits. If

the angular resolution of the eye is

•• •

•,

the value of d0 is close to :

(1) 1 mm

(2) 2 mm

(3) 4 mm

(4) 3 mm

20. ÿ¥ª ∑§ Ám-Á! Ê⁄UË ¬˝ÿÊª ◊¥, Á¡‚◊¥ ¬Œ¸ ∞fl¥ Á! Ê⁄UË ∑§’Ëø ∑§Ë ŒÍ⁄UË 1.0 m ÃÕÊ 600 nm Ã⁄¥UªŒÒäÿ¸ ∑§∞∑§fláÊË¸ÿ ¬˝∑§Ê‡Ê ∑§Ê ©¬ÿÊª Á∑§ÿÊ ªÿÊ „Ò– Á! ÊÁ⁄UÿÊ¥∑§ ‚◊Ë¬ π«∏Ê „È•Ê ∞∑§ √ƒÊÁÄàÊ √ÿÁÃ∑§⁄UáÊ ¬Ò≈UŸ¸ ∑§ÊŒπ ⁄U„Ê „Ò– ŒÊŸÊ¥ Á! ÊÁ⁄UÿÊ¥ ∑§ ’Ëø ∑§Ë ŒÍ⁄UË ∑§Ê ¬Á⁄UflÁÃ¸Ã∑§⁄Ÿ ¬⁄U ∞∑§ Áfl‡Ê· ŒÍ⁄UË d

0 ¬⁄U √ÿÁàÊ∑§⁄UáÊ ¬Ò≈UŸ¸ ‹ÈåàÊ

„Ê ¡ÊÃÊ „Ò– ÿÁŒ √ÿÁÄàÊ ∑§Ë •Ê°π ∑§Ê ∑§ÊáÊËÿ ÁflÿÊ¡Ÿ

•• •

•„Ê, ÃÊ d

0 ∑§Ê ◊ÊŸ ‹ª÷ª „ÊªÊ —

(1) 1 mm

(2) 2 mm

(3) 4 mm

(4) 3 mm

20. e„Ó_p X$bg [õgV$_p âep¡N, [õgV$$ A_¡ `X$v$p hÃQ¡_y„A„sf 1.0 m R>¡ s\p 600 nm A¡L$ f„Nue âL$piD`ep¡Ndp„ g¡hpdp„ Aph¡g R>¡. [õgV$_u _ÆL$ Dc¡g A¡L$ìe[¼s igpL$pcps sfa Å¡h¡ R>¡. Äepf¡ [õgV$ hÃQ¡_y„A„sf bv$ghpdp„ Aph¡ R>¡ Ðepf¡ b¡ [õgV$ hÃQ¡_p rhi¡jA„sf d

0 dpV¡$ ìersL$fZ cps gyá \pe R>¡. Å¡ Ap„M_y

L$p¡Zue rhc¡v$_ •• •

• R>¡, sp¡ d

0 _y„ dyëe _________

_u _ÆL$_y„ \i¡.(1) 1 mm

(2) 2 mm

(3) 4 mm

(4) 3 mm


Set - 04 15

21. When photons of wavelength λ1

are

incident on an isolated sphere, the

corresponding stopping potential is found

to be V. When photons of wavelength λ2

are used, the corresponding stopping

potential was thrice that of the above value.

If light of wavelength λ3 is used then find

the stopping potential for this case :

(1)

• • •

• • •• ••• •• •• •

•• • •• •• • •

(2)

• • •

• • •• ••• •• •• •

•• • •• •• • •

(3)

• • •

• • •• •

• ••• •• •• •

•• • •• •• • •

(4)

• • •

• • ••

• ••• •• •• •

•• • •• •• • •

21. ¡’ λ1 Ã⁄¥UªŒÒäÿ¸ ∑§ »§Ê≈ÊŸ ∞∑§ Áfl‹ÁªÃ ªÊ‹ ∑§Ê

¬˝ŒËåàÊ ∑§⁄UÃ „Ò¥, ÃÊ ‚¥ªÃ “ÁŸ⁄UÊœË-Áfl÷fl” ∑§Ê ◊ÊŸ V¬ÊÿÊ ¡ÊÃÊ „Ò– ¡’ λ

2 Ã⁄¥UªŒÒäÿ¸ ∑§ »§Ê≈ÊŸ ©¬ÿÊª ◊¥

‹Êÿ ¡ÊÃ „Ò¥ ÃÊ ÁŸ⁄UÊœË-Áfl÷fl ∑§Ê ◊ÊŸ ÁÃªÈŸÊ (3V)

„Ê ¡ÊÃÊ „Ò– •ª⁄U λ3 Ã⁄¥UªŒÒäÿ¸ ∑§ »§Ê≈UÊŸ ‚ ªÊ‹ ∑§Ê

¬˝ŒËåàÊ Á∑§ÿÊ ¡Ê∞ ÃÊ ÁŸ⁄ÊœË-Áfl÷fl ∑§Ê ◊ÊŸ „ÊªÊ —

(1)

• • •

• • •• ••• •• •• •

•• • •• •• • •

(2)

• • •

• • •• ••• •• •• •

•• • •• •• • •

(3)

• • •

• • •• •

• ••• •• •• •

•• • •• •• • •

(4)

• • •

• • ••

• ••• •• •• •

•• • •• •• • •

21. Äepf¡ λ1 sf„Ng„bpC_p ap¡V$p¡Þk_¡ A¡L$ AgN L$f¡g Np¡mp

`f Ap`ps L$fhpdp„ Aph¡ R>¡ Ðepf¡ A_yê$` r_fp¡ u-rhch (õV$p¡t`N `p¡V¡$[Þieg) V S>¡V$gp¡ \pe R>¡. Äepf¡λ

2 sf„Ng„bpC_p ap¡V$p¡Þk_p¡ D`ep¡N L$fhpdp„ Aph¡ R>¡

Ðepf¡ s¡_¡ A_yê$` r_fp¡ u-rhch l¡gp L$fsp„ ÓZ NZp¡\pe R>¡. Äepf¡ λ

3 sf„Ng„bpC_p¡ âL$pi hp`fhpdp„ Aph¡

R>¡ Ðepf¡ r_fp¡ u-rhch_y„ dyëe \i¡ :

(1)

• • •

• • •• ••• •• •• •

•• • •• •• • •

(2)

• • •

• • •• ••• •• •• •

•• • •• •• • •

(3)

• • •

• • •• •

• ••• •• •• •

•• • •• •• • •

(4)

• • •

• • ••

• ••• •• •• •

•• • •• •• • •


Set - 04 16

22. A hydrogen atom makes a transition from

n=2 to n=1 and emits a photon. This

photon strikes a doubly ionized lithium

atom (z=3) in excited state and completely

removes the orbiting electron. The least

quantum number for the excited state of the

ion for the process is :

(1) 2

(2) 3

(3) 4

(4) 5

23. The truth table given in fig. represents :

• • •• • •• • •• • •• • •

(1) AND - Gate

(2) OR - Gate

(3) NAND - Gate

(4) NOR - Gate

22. ∞∑§ „Êß«˛UÊ¡Ÿ ¬⁄U◊ÊáÊÈ n=2 Ä√ÊÊ¥≈U◊ ‚¥ÅÿÊ flÊ‹ ™§¡Ê¸‹fl‹ ‚ n=1 Ä√ÊÊ¥≈U◊ ‚¥ÅÿÊ flÊ‹ ™§¡Ê¸ ‹fl‹ ◊¥‚¥∑˝§◊áÊ ∑§⁄UŸ ¬⁄U ∞∑§ »§Ê≈UÊŸ ©à‚Á¡¸Ã ∑§⁄UÃÊ „Ò– ÿ„»§Ê≈UÊŸ ∞∑§ Ám-•ÊÿÁŸÃ Á‹ÁÕÿ◊ ¬⁄U◊ÊáÊÈ (z=3) (¡ÊÁ∑§ ©ûÊÁ¡Ã •flSÕÊ ◊¥ „Ò) ‚ ≈U∑§⁄UÊÃÊ „Ò •ÊÒ⁄U ∑§ˇÊËÿß‹Ä≈˛UÊÚŸ (orbiting electron) ∑§Ê ¬Í⁄UË Ã⁄U„ ‚ ’Ê„⁄UÁŸ∑§Ê‹ ŒÃÊ „Ò– ß‚ ¬˝Á∑˝§ÿÊ ∑§ Á‹∞ •ÊÿŸ ∑§Ë ©ûÊÁ¡Ã•flSÕÊ ∑§Ë ãÿÍŸÃ◊ Ä√ÊÊ¥≈U◊ ‚¥ÅÿÊ „ÊªË —(1) 2

(2) 3

(3) 4

(4) 5

23. ÁøòÊ ◊¥ ÁŒπÊß¸ ªß¸ ‚àÿ◊ÊŸ-‚Ê⁄UáÊË ÁŸêŸÁ‹ÁπÃ ◊¥ ‚Á∑§‚ ª≈U ∑§Ê Œ‡ÊÊ¸ÃË „Ò?

• • •• • •• • •• • •• • •

(1) AND ª≈U

(2) OR ª≈U

(3) NAND ª≈U

(4) NOR ª≈U

22. A¡L$ lpCX²$p¡S>_ `fdpÏ„ n=2 \u n=1 EÅ® õsfdp„k„¾$dZ L$f¡ R>¡ A_¡ A¡L$ ap¡V$p¡_ DÐkÆ®s L$f¡ R>¡. Apap¡V$p¡_ qÜ-Apep¡_pCÈX$ rg\ued `fdpÏ„ (z=3) _¡s¡_u DÑ¡Æs Ahõ\pdp„ A\X$pe R>¡ A_¡ L$nueCg¡¼V²$p¡__¡ k„ |Z® blpf r_L$pm¡ R>¡. Ap âq¾$ep dpV¡$Ape__u DÑ¡Æs Ahõ\p_u Þe|_sd ¼hpÞV$d _„bf\i¡ :(1) 2

(2) 3

(3) 4

(4) 5

23. ApL©$rsdp„ Ap`¡g V³$\ V¡$bg _uQ¡ Ap`¡g L$ey N¡V$ bsph¡R>¡ ?

• • •• • •• • •• • •• • •

(1) AND - Gate

(2) OR - Gate

(3) NAND - Gate

(4) NOR - Gate


Set - 04 17

24. A¡L$ Ýhr_ rkÁ_g b¡ Sy>v$p Sy>v$p AhÅ¡ ^fph¡ R>¡. A¡L$dp_h-cprjs rkÁ_g R>¡ S>¡ 200 Hz \u 2700 Hz

Aph©rÑ A„sfpgdp„ R>¡. Äepf¡ buSy> A¡ DÃQ Aph©rÑk„Nus_y„ rkÁ_g R>¡ S>¡ 10200 Hz \u 15200 Hz

Aph©rÑ A„sfpgdp„ R>¡. bÞ_¡ rkÁ_gp¡_¡ kp\¡ k„QfhpdpV¡$ S>ê$fu AM rkÁ_g_u b¢X$ rhX¹$\ s\p a¼s dp_h-cprjs rkÁ_g k„QfZ dpV¡$ S>ê$fu AM rkÁ_g b¢X$rhX¹$\_p¡ NyZp¡Ñf R>¡.(1) 3

(2) 5

(3) 6

(4) 2

25. m Öìedp_ hpmp Np¡mp ^fphsp s\p Öìedp_ frls^psy_p spf\u b_¡g A¡L$ kpv$p gp¡gL$_p¡ T=0C A¡Aphs®L$pm 2 s R>¡. Å¡ spf_p sp`dp_dp„ h^pfp¡ L$fsp,Aphs®L$pmdp„ \sp a¡fapf_¡ Apg¡M Üpfp v$ip®hpe, sp¡`qfZpdu Apg¡M A¡ A¡L$ f¡Mp R>¡ S>¡_p¡ Y$pm S R>¡. Å¡^psy_p¡ f¡Mue-âkfZp„L$ α lp¡e, sp¡ S _y„ dyëe R>¡ :(1) α

(2) ••

(3) 2α

(4)

••

24. ∞∑§ äflÁŸ-Á‚ÇŸ‹ ŒÊ S¬c≈U äflÁŸÿÊ¥ ‚ ÁŸÁ◊¸Ã „Ò–ßŸ◊¥ ‚ ∞∑§ ◊ŸÈcÿ mÊ⁄UÊ ÷ÊÁ·Ã Á‚ÇãÊ‹ „Ò ¡Ê200 Hz ‚ 2700 Hz ∑§Ë •ÊflÎÁûÊ •¥Ã⁄UÊ‹ ∑§Ê „Ò, ÃÕÊŒÍ‚⁄UÊ Á‚ÇŸ‹ 10200 Hz ‚ 15200 Hz ©ìÊ •ÊflÎÁûÊflÊ‹ ‚¥ªËÃ ∑§Ê „Ò– ŒÊŸÊ¥ Á‚ÇŸ‹Ê¥ ∑§ ‚¥ø⁄UáÊ ∑§ Á‹∞•Êfl‡ÿ∑§ AM Á‚ÇãÊ‹ ∑§Ë ’Ò¥«U-øÊÒ«∏Êß¸ •ÊÒ⁄U ∑§fl‹◊ŸÈcÿ mÊ⁄UÊ ÷ÊÁ·Ã Á‚ÇŸ‹ ∑§ ‚¥ø⁄UáÊ ∑§ Á‹∞ •Êfl‡ÿ∑§AM Á‚ÇãÊ‹ ∑§Ë ’Ò¥«U-øÊÒ«∏Êß¸ ∑§Ê •ŸÈ¬ÊÃ ÄÿÊ „ÊªÊ?(1) 3

(2) 5

(3) 6

(4) 2

25. T=0C ¬⁄U ∞∑§ ‚⁄U‹-‹Ê‹∑§, ¡Ê Á∑§ m Œ˝√ÿ◊ÊŸ∑§ ªÊ‹∑§ •ÊÒ⁄U Œ˝√ÿ◊ÊŸ ⁄UÁ„Ã œÊÃÈ ∑§ ÃÊ⁄U ‚ ÁŸÁ◊¸Ã „Ò,∑§Ê •ÊflûÊ¸-∑§Ê‹ 2 s „Ò– •ª⁄U ÃÊ⁄U ∑§ ÃÊ¬◊ÊŸ ∑§Ê’…∏ÊŸ ‚, •ÊflûÊ¸-∑§Ê‹ ◊¥ „Èß¸ flÎÁh ∑§Ê ª˝Ê»§ mÊ⁄UÊŒ‡ÊÊ¸ÿÊ ¡Êÿ, ÃÊ ¬Á⁄UáÊÊ◊Ë ª˝Ê»§ ∑§Ë …Ê‹-◊Ê¬ ( slope)

S „Ò– ÿÁŒ ÃÊ⁄U ∑§Ê ⁄ÒUÁπ∑§-¬˝‚Ê⁄U ªÈáÊÊ¥∑§ α „Ò ÃÊ S ∑§Ê◊ÊŸ „ÊªÊ —(1) α

(2) ••

(3) 2α

(4)

••

24. An audio signal consists of two distinct

sounds : one a human speech signal in the

frequency band of 200 Hz to 2700 Hz, while

the other is a high frequency music signal

in the frequency band of 10200 Hz to

15200 Hz. The ratio of the AM signal

bandwidth required to send both the

signals together to the AM signal

bandwidth required to send just the human

speech is :

(1) 3

(2) 5

(3) 6

(4) 2

25. A simple pendulum made of a bob of

mass m and a metallic wire of negligible

mass has time period 2 s at T=0C. If the

temperature of the wire is increased and the

corresponding change in its time period is

plotted against its temperature, the

resulting graph is a line of slope S. If the

coefficient of linear expansion of metal is α

then the value of S is :

(1) α

(2) ••

(3) 2α

(4)

••


Set - 04 18

26. A¡L$ eyr_ap¡d® i„Ly$ ApL$pf_p¡ spf Y e„N dp¡X$éygk hpmpÖìedp„\u b_ph¡g R>¡ S>¡_u sZphdy¼s g„bpC L R>¡.Ap i„Ly$ ApL$pf_p spf_p D`f_p R>¡X$p_u rÓÄep R A_¡_uQ¡_p R>¡X$p_u rÓÄep 3R R>¡ D`f_p¡ R>¡X$p¡ ×Y$ Ap^pfkp\¡ Å¡X¡$g R>¡ A_¡ _uQ¡_p R>¡X¡$ M v$m Å¡X¡$g R>¡.k„s ygus Ahõ\pdp „ Ap spf_u sZph g„bpC\i¡ :

(1) ••• • • ••

• ••

• •

•π

(2) ••• • • • ••

• ••

• •

•π

(3) ••• • • • ••

• ••

• •

•π

(4) ••• • • • ••

• ••

• •

•π

27. A ®Aphs®__u fus¡ N¡ëh¡_p¡duV$f_p¡ Ahfp¡ G ÅZhp,V

E fphsu b¡V$fu A_¡ Ahfp¡ R. A¡ N¡ëh¡_p¡duV$fdp„

θ M|Zp_y„ Aphs®_ d¡mhhp h`fpe R>¡. Å¡ S Ahfp¡ _p¡i„V$ A¡ A ® Aphs®_ dpV¡$ S>ê$fu lp¡e, sp¡ G, R A_¡ SA¡ hÃQ¡_p¡ k„b„ R>¡ :(1) 2S (R+G)=RG

(2) S (R+G)=RG

(3) 2S=G

(4) 2G=S

26. •ÁflSÃÊÁ⁄UÃ L ‹ê’Êß¸ ∑§Ë ∞∑§‚◊ÊŸ ‡Ê¥∑È§ŸÈ◊Ê ÃÊ⁄U ∑§Á‚⁄UÊ¥ ∑§Ë ÁòÊíÿÊ ∑˝§◊‡Ê— R ÃÕÊ 3R „Ò¥– ©‚∑§Ë œÊÃÈ ∑§Êÿ¥ª-◊Ê«ÈUU‹‚ Y „Ò– R ÁòÊíÿÊ flÊ‹ Á‚⁄U ∑§Ê ∞∑§ ŒÎ…∏•ÊœÊ⁄U ¬⁄U ¡Á«∏Ã Á∑§ÿÊ ªÿÊ „Ò ÃÕÊ ŒÍ‚⁄U Á‚⁄U ¬⁄U MŒ˝√ÿ◊ÊŸ ‹≈U∑§ÊÿÊ ªÿÊ „Ò– ‚¥ÃÈ‹Ÿ-•flSÕÊ ◊¥ ÃÊ⁄U∑§Ë ÀÊê’Êß¸ „ÊªË —

(1) ••• • • ••

• ••

• •

•π

(2) ••• • • • ••

• ••

• •

•π

(3) ••• • • • ••

• ••

• •

•π

(4) ••• • • • ••

• ••

• •

•π

27. ∞∑§ ªÒÀflŸÊ◊Ë≈U⁄U ∑§Ê ¬˝ÁÃ⁄UÊœ G ◊Ê¬Ÿ ∑§ Á‹ÿ •h¸-ÁflˇÊ¬ Ã⁄UË∑§ ∑§Ê ßSÃ◊Ê‹ Á∑§ÿÊ ªÿÊ Á¡‚◊¥ ’Ò≈U⁄UË ∑§Ëemf V

E „Ò– ¬˝ÁÃ⁄UÊœ R ∑§ Á‹ÿ θ ÁflˇÊ¬ Á◊‹Ê–

‡Ê¥≈U-¬˝ÁÃ⁄UÊœ S ∑§ Á‹ÿ •ÊœÊ ÁflˇÊ¬ Á◊‹Ê– Ã’G, R ÃÕÊ S Á∑§‚ ‚◊Ë∑§⁄UáÊ ‚ ‚¥’¥ÁœÃ „Ò¥?(1) 2S (R+G)=RG

(2) S (R+G)=RG

(3) 2S=G

(4) 2G=S

26. A uniformly tapering conical wire is made

from a material of Young’s modulus Y and

has a normal, unextended length L. The

radii, at the upper and lower ends of this

conical wire, have values R and 3R,

respectively. The upper end of the wire is

fixed to a rigid support and a mass M is

suspended from its lower end. The

equilibrium extended length, of this wire,

would equal :

(1) ••• • • ••

• ••

• •

•π

(2) ••• • • • ••

• ••

• •

•π

(3) ••• • • • ••

• ••

• •

•π

(4) ••• • • • ••

• ••

• •

•π

27. To know the resistance G of a galvanometer

by half deflection method, a battery of emf

VE

and resistance R is used to deflect the

galvanometer by angle θ. If a shunt of

resistance S is needed to get half deflection

then G, R and S are related by the

equation :

(1) 2S (R+G)=RG

(2) S (R+G)=RG

(3) 2S=G

(4) 2G=S


Set - 04 19

28. brlNp£m Afukp_u L¡$ÞÖg„bpC dp`hp, A¡L$ rhÛp\_uQ¡_p Ahgp¡L$_p¡ _p¢ ¡ R>¡.

hõsy u_ brlNp£m L$pQ brlNp£m Afukp¡ ârstbb u_• • •• •• • • • •• •• • • • •• •• • • • •• •• •

brlNp£m L$pQ_u L¡$ÞÖg„bpC f1 A_¡ Afukp_u L¡$ÞÖg„bpC

f2 R>¡. index correction _NÎe g¡sp f

1 A_¡ f

2 _y„

dyëe_u _ÆL$ li¡ :(1) f

1=12.7 cm f

2=7.8 cm

(2) f1=7.8 cm f

2=12.7 cm

(3) f1=7.8

cm f

2=25.4 cm

(4) f1=15.6 cm f

2=25.4 cm

29. T¡_fX$pep¡X$_u I - V gpnrZL$sp_p Aæepk dpV¡$ A¡L$âep¡N L$fhpdp„ Aph¡ R>¡. Äepf: `p¡V¡$[Þiep¡duV$f L¡$ S>¡_p¡Äep„ Ahfp¡ R=100 Ω R>¡ A_¡ 1 W dlÑd `phfX¡$ku`¡i_ R>¡. Ap `qf`\dp„ gNphhpdp„ Aphsp DC

õÓp¡s_u Þe|_sd hp¡ëV¡$S>_u Ahr^ R>¡ :(1) 0 – 5 V

(2) 0 – 8 V

(3) 0 – 12 V

(4) 0 – 24 V

28. ©ûÊ‹-Œ¬¸áÊ ∑§Ë »§Ê∑§‚ ŒÍ⁄UË ÁŸ∑§Ê‹Ÿ ∑§ ∞∑§ ¬˝ÿÊª◊¥ ÁŸêŸ «UÊ≈UÊ ¬˝ÊåàÊ „È•Ê

Á’¥’ ©ûÊ‹ ‹Ò¥‚ ©ûÊ‹ Œ¬áÊ ¬ÁÃÁ’¥’

• • •• •• • • • •• •• • • • •• •• • • • •• •• •

©ûÊ‹ ‹Ò¥‚ ∑§Ë »§Ê∑§‚ ŒÍ⁄UË f1 ÃÕÊ ©ûÊ‹-Œ¬¸áÊ ∑§Ë

»§Ê∑§‚ ŒÍ⁄UË f2 „Ò– index correction Ÿªáÿ „Ò–

Ã’ —(1) f

1=12.7 cm f

2=7.8 cm

(2) f1=7.8 cm f

2=12.7 cm

(3) f1=7.8

cm f

2=25.4 cm

(4) f1=15.6 cm f

2=25.4 cm

29. ∞∑§ ¡ËŸ⁄U «UÊÿÊ«U ∑§Ê •Á÷‹ˇÊÁáÊ∑§ I - V ª˝Ê»§ ’ŸÊŸ∑§ Á‹ÿ ∞∑§ ¬˝ÿÊª Á∑§ÿÊ ªÿÊ Á¡‚◊¥ R=100 Ω ∑§Ê¬˝Ê≈UÁÄ≈Ufl ¬˝ÁÃ⁄UÊœ •ÊÒ⁄U •Áœ∑§Ã◊ ¬Êfl⁄U 1 W ŒË ªß¸–Ã’ ¬Á⁄U¬Õ ◊¥ ‹ªÊÿ ªÿ DC dÊÃ ∑§Ë ãÿÍŸÃ◊ flÊÀ≈UÃÊ„Ò —(1) 0 – 5 V

(2) 0 – 8 V

(3) 0 – 12 V

(4) 0 – 24 V

28. To find the focal length of a convex mirror,

a student records the following data :

Object Pin Convex Lens Convex Mirror Image Pin

22.2 cm 32.2 cm 45.8 cm 71.2 cm

The focal length of the convex lens is f1 and

that of mirror is f2. Then taking index

correction to be negligibly small, f1 and f

2

are close to :

(1) f1=12.7 cm f

2=7.8 cm

(2) f1=7.8 cm f

2=12.7 cm

(3) f1=7.8

cm f

2=25.4 cm

(4) f1=15.6 cm f

2=25.4 cm

29. An experiment is performed to determine

the I - V characteristics of a Zener diode,

which has a protective resistance of

R=100 Ω, and a maximum power of

dissipation rating of 1 W. The minimum

voltage range of the DC source in the circuit

is :

(1) 0 – 5 V

(2) 0 – 8 V

(3) 0 – 12 V

(4) 0 – 24 V


Set - 04 20

30. ∞∑§ •ôÊÊÃ ≈˛UÊ¥Á¡S≈U⁄U ∑§Ê npn •ÕflÊ pnp ∑§ ¬˝∑§Ê⁄U◊¥ ¬„øÊŸ ∑§⁄UŸÊ „Ò– ∞∑§ pnp ≈˛UÊ¥Á¡S≈U⁄U ∑§Ê ≈UÁ◊¸Ÿ‹ 2©‚∑§Ê ’‚ „Ò– ∞∑§ êÊÀ≈UË◊Ë≈U⁄U ∑§ +ve fl −ve

≈UÁ◊¸Ÿ‹ ß‚ ≈˛UÊ¥Á¡S≈U⁄U ∑§ ÁflÁ÷ÛÊ ≈UÁ◊¸Ÿ‹Ê¥ 1, 2 ÿÊ 3 ∑§’Ëø ‹ªÊ∑§⁄U ¬˝ÁÃ⁄UÊœ ◊Ê¬ ªÿ Ã’ ß‚ ≈˛UÊ¥Á¡S≈U⁄U ∑§Á‹∞ ∑§ÊÒŸ-‚Ê ÁŸêŸ ∑§ÕŸ ‚àÿ „Ò?

(1) +ve ‚ ≈UÁ◊Ÿ‹ 1, −ve ‚ ≈UÁ◊Ÿ‹ 2, ¬ÁÃ⁄UÊœíÿÊŒÊ–



(4) +ve ‚ ≈UÁ◊Ÿ‹ 2, −ve ‚ ≈UÁ◊Ÿ‹ 3, ¬ÁÃ⁄UÊœ∑§◊

31. 35.5 g •Ê⁄U‚ÁŸ∑§ •ê‹ ∑§Ê, ‚Ê¥Œ˝ HCl ∑§Ë ©¬ÁSÕÁÃ◊¥ H

2S ∑§Ë •Áœ∑§ ◊ÊòÊÊ ‚ ÁflfløŸ ∑§⁄UŸ ¬⁄U •Ê⁄U‚ÁŸ∑§

¬ã≈UÊ‚À$»§Êß«U ∑§Ë ¬˝Ê# „ÊŸ flÊ‹Ë ◊ÊòÊÊ „Ò (ÿÁŒ 100%

¬Á⁄UflÃ¸Ÿ ◊ÊŸ¥ ÃÊ)—

(1) 0.50 ◊Ê‹

(2) 0.25 ◊Ê‹

(3) 0.125 ◊Ê‹

(4) 0.333 ◊Ê‹

30. An unknown transistor needs to be

identified as a npn or pnp type. A

multimeter, with +ve and −ve terminals,

is used to measure resistance between

different terminals of transistor. If terminal

2 is the base of the transistor then which of

the following is correct for a pnp

transistor ?

(1) +ve terminal 1, −ve terminal

2, resistance high


1, resistance high


2, resistance high


3, resistance low

31. The amount of arsenic pentasulphide that

can be obtained when 35.5 g arsenic acid is

treated with excess H2S in the presence of

conc. HCl ( assuming 100% conversion)

is :

(1) 0.50 mol

(2) 0.25 mol

(3) 0.125 mol

(4) 0.333 mol

30. A¡L$ npn A\hp pnp V²$p[ÞTõV$f_¡ Ap¡mMhp_p¡ R>¡.Ap dpV¡$ V²$p[ÞTõV$f_p AgN AgN V$rd®_ëk hÃQ¡_p¡Ahfp¡ +ve A_¡ −ve dëV$uduV$f\u dp`hpdp„ Aph¡R>¡. Å¡ V$rd®_g 2 A¡ V²$p[ÞTõV$f_p¡ b¡T lp¡e sp¡ pnp

V²$p[ÞTõV$f dpV¡ L$ey„ rh^p_ kpQy„ R>¡ ?(1) +ve terminal 1, −ve terminal

2, Ahfp¡ h^pf¡(2) +ve terminal 2, −ve terminal



3, Ahfp¡ Ap¡R>p¡

31. 35.5 g Apk£r_L$ A¡rkX$_¡, kp„Ö HCl _u lpS>fudp„h^pf¡ `X$sp H

2S kp\¡ âq¾$ep L$fsp„ Apk£r_L$

`¡ÞV$pkëapCX$ L¡ $V$gu dpÓpdp„ âpá \C iL$i¡ ?(100% `qfhs®_ dp_u gp¡)

(1) 0.50 dp¡g

(2) 0.25 dp¡g

(3) 0.125 dp¡g

(4) 0.333 dp¡g


Set - 04 21

32. M|b S> KQp v$bpZ¡, A¡L$ dp¡g hpey_p¡ v$b_ue Aheh(compressibility factor) _uQ¡ Ap ¡gdp„\u S>Zphp¡.

(1)

• •• •

(2) • •• •• •

• •

(3) • •• •• •

• •

(4) • • • ••• • • •

−

33. dy¿e ¼hp¡ÞV$d Ap„L$ 5 kp\¡ k„L$mpe¡g L$nL$p¡_u Ly$gk„¿ep ip¡ p¡.(1) 5

(2) 10

(3) 20

(4) 25

34. _uQ¡_pdp„\u L$ep Ap„sfApÎhue bmp¡ L¡$ S>¡ T¡_p¡_ hpey_pâhpluL$fZ dpV¡$ kp¥\u h^pf¡ S>hpbv$pf R>¡ ?

(1) qÜ y°h - qÜ °yh

(2) Ape_ - qÜ °yh

(3) Ðhqfs qÜ y°h - â¡qfs qÜ y°h (Instantaneous

dipole - induced dipole)

(4) Aper_L$

32. •àÿÁœ∑§ ŒÊ’ ¬⁄U ∞∑§ ◊Ê‹ ªÒ‚ ∑§Ê ‚¥¬Ë«˜ÿÃÊ ªÈáÊÊ¥∑§„ÊªÊ —

(1)

• •• •

(2) • •• •• •

• •

(3) • •• •• •

• •

(4) • • • ••• • • •

−

33. ◊ÈÅÿ ÄflÊ¥≈U◊ •¥∑§ 5 ‚ ¡È«∏ „È∞ ∑§ˇÊ∑§Ê¥ (•ÊÚÁ’¸≈U‹Ê¥)∑§Ë ∑È§‹ ‚¥ÅÿÊ „Ò —(1) 5

(2) 10

(3) 20

(4) 25

34. ßŸ◊¥ ‚ ∑§ÊÒŸ‚Ë •¥Ã⁄UÊ-•ÊÁáfl∑§ ’‹ ¡ËŸÊÚŸ ∑§Œ˝flË∑§⁄UáÊ ∑§ Á‹∞ ‚’‚ •Áœ∑§ ©ûÊ⁄UŒÊÿË „Ò ?

(1) ÁmœÈ˝fl - Ámœ˝Èfl

(2) •ÊÿŸ - Ámœ˝Èfl

(3) ÃÊà∑§ÊÁ‹∑§ Ámœ˝Èfl - ¬˝Á⁄UÃ Ámœ˝Èfl

(4) •ÊÿÁŸ∑§

32. At very high pressures, the compressibility

factor of one mole of a gas is given by :

(1)

• •• •

(2) • •• •• •

• •

(3) • •• •• •

• •

(4) • • • ••• • • •

−

33. The total number of orbitals associated with

the principal quantum number 5 is :

(1) 5

(2) 10

(3) 20

(4) 25

34. Which intermolecular force is most

responsible in allowing xenon gas to

liquefy ?

(1) Dipole - dipole

(2) Ion - dipole

(3) Instantaneous dipole - induced dipole

(4) Ionic


Set - 04 22

35. 1 bar f A¡L$ fpkperZL$ âq¾$ep _uQp sp`dp_¡ Ap`d¡m¡(õhe„c|) \su _\u `f„sy KQp sp`dp_¡ Ap`d¡m¡(õhe„c|) \pe R>¡. âq¾$ep A„N¡ _uQ¡ Ap ¡gp rh^p_p¡dp„\ukpQy„ rh^p_ ip¡ p¡.

(1) ∆H s\p ∆S b„_¡ F>Z R>¡.

(2) ∆H s\p ∆S b„_¡ ^_ R>¡.

(3) ∆H ^_ R>¡ Äepf¡ ∆S F>Z R>¡.

(4) ∆H F>Z R>¡ Äepf¡ ∆S ^_ R>¡.

36. 300 K A_¡ 500 torr (V$p¡f) Ap„riL$ v$bpZ¡ N2 _u

Öpìesp 0.01 g L−1 R>¡. 750 V$p¡f (torr) Ap„riL$

v$bpZ `f Öpìesp (g L−1 dp„) iy„ li¡ ?

(1) 0.0075

(2) 0.015

(3) 0.02

(4) 0.005

37. âq¾$ep dpV¡$, A(g)+B(g) → C(g)+D(g) 298 K

`f, ∆H A_¡ ∆S A_y¾$d¡,−29.8 kJ mol−1 A_¡

−0.100 kJ K−1

mol−1 R>¡. 298 K `f âq¾$ep

dpV¡$_p¡ k„syg_ AQmp„L$ _uQ¡_pdp„\u ip¡ p¡.(1) 1.0×10

−10

(2) 1.0×1010

(3) 10

(4) 1

35. ∞∑§ ⁄UÊ‚ÊÿÁŸ∑§ •Á÷Á∑§ÿÊ ÁŸêŸ ÃÊ¬ ¬⁄U •SflÃ— ¬flÁÃÃ„Ò Á∑§ãÃÈ ©ìÊ ÃÊ¬ ¬⁄U SflÃ— ¬˝flÁÃ¸Ã „Ê ¡ÊÃË „Ò– ß‚•Á÷Á∑˝§ÿÊ ∑§ ’Ê⁄U ◊¥ ÁŸêŸÁ‹ÁπÃ ∑§ÕŸÊ¥ ◊¥ ‚ ‚„Ë∑§ÕŸ ∑§Ê ¬„øÊÁŸÿ —

(1) ∆H ÃÕÊ ∆S, ŒÊŸÊ¥ ´§áÊÊà◊∑§ „Ò¥–

(2) ∆H ÃÕÊ ∆S, ŒÊŸÊ¥ œŸÊà◊∑§ „Ò¥–

(3) ∆H œŸÊà◊∑§ ÃÕÊ ∆S ´§áÊÊà◊∑§ „Ò–

(4) ∆H ´§áÊÊà◊∑§ ÃÕÊ ∆S œŸÊà◊∑§ „Ò–

36. N2 ∑§Ë ¡‹ ◊¥ Áfl‹ÿÃÊ 300 K ÃÕÊ 500 torr •Ê¥Á‡Ê∑§

ŒÊ’ ¬⁄U 0.01 g L−1 „Ò– ß‚∑§Ë Áfl‹ÿÃÊ (g L

−1

◊¥) 750 torr •Ê¥Á‡Ê∑§ ŒÊ’ ¬⁄U „ÊªË —(1) 0.0075

(2) 0.015

(3) 0.02

(4) 0.005

37. ⁄UÊ‚ÊÿÁŸ∑§ •Á÷Á∑˝§ÿÊA(g)+B(g) → C(g)+D(g), ∑ Á‹∞ 298 K ¬⁄U∆H ÃÕÊ ∆S ∑§ ◊ÊŸ ∑˝§◊‡Ê— −29.8 kJ mol

−1

ÃÕÊ−0.100 kJ K−1

mol−1 „Ò¥– ß‚ •Á÷Á∑˝§ÿÊ ∑§Ê

298 K ¬⁄U ‚Êêÿ ÁSÕ⁄UÊ¥∑§ „Ò —(1) 1.0×10

−10

(2) 1.0×1010

(3) 10

(4) 1

35. A reaction at 1 bar is non-spontaneous at

low temperature but becomes spontaneous

at high temperature. Identify the correct

statement about the reaction among the

following :

(1) Both ∆H and ∆S are negative.

(2) Both ∆H and ∆S are positive.

(3) ∆H is positive while ∆S is negative.

(4) ∆H is negative while ∆S is positive.

36. The solubility of N2 in water at 300 K and

500 torr partial pressure is 0.01 g L−1

. The

solubility (in g L−1

) at 750 torr partial

pressure is :

(1) 0.0075

(2) 0.015

(3) 0.02

(4) 0.005

37. For the reaction,

A(g)+B(g) → C(g)+D(g), ∆H and ∆S are,

respectively, −29.8 kJ mol−1

and

−0.100 kJ K−1

mol−1

at 298 K. The

equilibrium constant for the reaction at

298 K is :

(1) 1.0×10−10

(2) 1.0×1010

(3) 10

(4) 1


Set - 04 23

38. Å¡ L$p¡ f ^psy_p¡ A¡L$ ågp¡L$ (block) A¡L$ buL$fdp„ `X$uÅe L¡$ S>¡ 1M ZnSO

4 _y„ ÖphZ ^fph¡ R>¡ sp¡ iy„

b_i¡ ?

(1) L$p¡ f ^psy Ap¡Nmu S>i¡ A_¡ T]L$ ^psy S>dp\i¡.

(2) lpCX²$p¡S>_ hpey _uL$mhp_u kp\¡ L$p¡`f ^psyAp¡Nmu S>i¡.

(3) Ap¡[¼kS>_ hpey _uL$mhp_u kp\¡ L$p¡`f ^psyAp¡Nmu S>i¡.

(4) L$p¡C âq¾$ep \i¡ _l].

39. ¼gp¡qf_ `fdpÏAp¡_u lpS>fudp„, Ap¡Tp¡_ kp\¡_uAp¡[¼kS>_ `fdpÏAp¡_u âq¾$ep, _uQ¡ âdpZ¡ b¡sb½$pdp„ â¾$d v$ip®h¡g R>¡ :

O3(g)+Cl

•(g) → O

2(g)+ClO

•(g) ____ (i)

ki= 5.2×10

9 L mol

−1 s

−1

ClO•(g)+O

•(g) → O

2(g)+Cl

•(g) ____ (ii)

kii= 2.6×10

10 L mol

−1 s

−1

Ly$g âq¾$ep O3(g)+O

•(g) → 2 O

2(g) _p¡ kp¥\u

_ÆL$_p¡ h¡N AQmp„L$ ip¡ p¡.(1) 5.2×10

9 L mol

−1 s

−1

(2) 2.6×1010

L mol−1

s−1

(3) 3.1×1010

L mol−1

s−1

(4) 1.4×1020

L mol−1

s−1

38. ÿÁŒ ∑§ÊÚ¬⁄U ∑§ ∞∑§ é‹ÊÚ∑§ (block) ∑§Ê ∞∑§ ’Ë∑§⁄U ◊¥«UÊ‹Ê ¡Êÿ Á¡‚◊¥ 1M ZnSO

4 ∑§Ê Áfl‹ÿŸ „Ê ÃÊ

ÄÿÊ „ÊªÊ ?

(1) ∑§ÊÚ¬⁄U œÊÃÈ ÉÊÈ‹ ¡ÊÿªË ÃÕÊ Á$¡¥∑§ œÊÃÈ ÁŸˇÊÁ¬Ã„Ê ¡ÊÿªË–

(2) „Êß«˛UÊ¡Ÿ ªÒ‚ ∑§ ÁŸ∑§‹Ÿ ∑§ ‚ÊÕ-‚ÊÕ ∑§ÊÚ¬⁄UœÊÃÈ ÉÊÈ‹ ¡ÊÿªË–

(3) •ÊÚÄ‚Ë¡Ÿ ªÒ‚ ∑§ ÁŸ∑§‹Ÿ ∑§ ‚ÊÕ-‚ÊÕ ∑§ÊÚ¬⁄UœÊÃÈ ÉÊÈ‹ ¡ÊÿªË–

(4) ∑§Êß¸ •Á÷Á∑˝§ÿÊ Ÿ„Ë¥ „ÊªË–

39. Ä‹Ê⁄ËUŸ ¬⁄U◊ÊáÊÈ•Ê¥ ∑§Ë ©¬ÁSÕÁÃ ◊¥, •Ê$¡ÊŸ ∑§Ë•ÊÚÄ‚Ë¡Ÿ ¬⁄U◊ÊáÊÈ•Ê¥ ‚ •Á÷Á∑˝§ÿÊ ÁŸêŸÁ‹ÁπÃÁm¬ŒËÿ ¬˝∑˝§◊ mÊ⁄UÊ „ÊÃË „Ò —

O3(g)+Cl

•(g) → O

2(g)+ClO

•(g) ____ (i)

ki= 5.2×10

9 L mol

−1 s

−1

ClO•(g)+O

•(g) → O

2(g)+Cl

•(g) ____ (ii)

kii= 2.6×10

10 L mol

−1 s

−1

∑È§‹ •Á÷Á∑˝§ÿÊ O3(g)+O

•(g) → 2 O

2(g) ∑§Ê

ÁŸ∑§≈UÃ◊ flª ÁŸÿÃÊ¥∑§ „Ò —(1) 5.2×10

9 L mol

−1 s

−1

(2) 2.6×1010

L mol−1

s−1

(3) 3.1×1010

L mol−1

s−1

(4) 1.4×1020

L mol−1

s−1

38. What will occur if a block of copper metal

is dropped into a beaker containing a

solution of 1M ZnSO4 ?

(1) The copper metal will dissolve and

zinc metal will be deposited.

(2) The copper metal will dissolve with

evolution of hydrogen gas.

(3) The copper metal will dissolve with

evolution of oxygen gas.

(4) No reaction will occur.

39. The reaction of ozone with oxygen atoms

in the presence of chlorine atoms can occur

by a two step process shown below :

O3(g)+Cl

•(g) → O

2(g)+ClO

•(g) ____ (i)

ki= 5.2×10

9 L mol

−1 s

−1

ClO•(g)+O

•(g) → O

2(g)+Cl

•(g) ____ (ii)

kii= 2.6×10

10 L mol

−1 s

−1

The closest rate constant for the overall

reaction O3(g)+O

•(g) → 2 O

2(g) is :

(1) 5.2×109 L mol

−1 s

−1

(2) 2.6×1010

L mol−1

s−1

(3) 3.1×1010

L mol−1

s−1

(4) 1.4×1020

L mol−1

s−1


Set - 04 24

40. A¡L$ rhi¡j Arîp¡jZ â¾$d_¡ _uQ¡_u gpnrZL$spAp¡R>¡ : (i) s¡ hpÞX$fhpg bmp¡_¡ L$pfZ¡ Dv¹$ch¡ R>¡. A_¡(ii) s¡ ârshse R>¡.D`f hZ®h¡g Arîp¡jZ â¾$d dpV¡$ kpQy„ rh^p_ip¡^p¡.

(1) Arîp¡jZ_u A¡Þ\pëù 100 kJ mol−1 \u

h^pf¡ lp¡e R>¡.

(2) kq¾$eL$fZ EÅ® (i[¼s) _uQu R>¡.

(3) Arîp¡jZ A¡ A¡L$ õsfue R>¡.

(4) sp`dp_ h^hp_u kp\¡ Arîp¡jZ h ¡ R>¡.

41. A¡L$ A^psy S>¡ ^_ (positive) Ap¡[¼kX¡$i_ Ahõ\pv$ip®hsu _\u s¡ ip¡ p¡.

(1) Ap¡[¼kS>_

(2) Apep¡qX$_

(3) ¼gp¡qf_

(4) ãgp¡qf_

40. ∞∑§ Áfl‡Ê· •Áœ‡ÊÊ·áÊ ¬Á∑§ÿÊ ∑§ Áfl‡Ê· ªÈáÊœ◊ „Ò¥ —(i) ÿ„ flÊ¥«U⁄U flÊÀ‚ ’‹ ∑§ ∑§Ê⁄áÊ „ÊÃË „Ò ÃÕÊ (ii) ÿ„©à∑˝§◊áÊËÿ „Ò– ÁŸêŸÁ‹ÁπÃ ◊¥ ‚ fl„ ‚„Ë ∑§ÕŸ¬„øÊÁŸÿ ¡Ê ß‚ •Áœ‡ÊÊ·áÊ ¬˝Á∑˝§ÿÊ ∑§Ê ‚„Ë fláÊ¸Ÿ∑§⁄UÃÊ „Ò —

(1) •Áœ‡ÊÊ·áÊ ∑§Ë ∞ãÕÒÀ¬Ë 100 kJ mol−1 ‚

•Áœ∑§ „Ò–

(2) ‚Á∑˝§ÿáÊ ™§¡Ê¸ ÁŸêŸ „Ò–

(3) •Áœ‡ÊÊ·áÊ ∞∑§‹ •áÊÈ∑§ ¬⁄UÃËÿ „Ò–

(4) ÃÊ¬ ’…∏Ÿ ¬⁄U •Áœ‡ÊÊ·áÊ ’…∏ÃÊ „Ò–

41. fl„ •œÊÃÈ ¡Ê œŸÊà◊∑§ •ÊÚÄ‚Ë∑§⁄UáÊ •flSÕÊ Ÿ„Ë¥Œ‡ÊÊ¸ÃË, „ÊªË —

(1) •ÊÚÄ‚Ë¡Ÿ

(2) •ÊÿÊ«UËŸ

(3) Ä‹Ê⁄UËŸ

(4) ç‹È•Ê⁄UËŸ

40. A particular adsorption process has the

following characteristics : (i) It arises due

to van der Waals forces and (ii) it is

reversible. Identify the correct statement

that describes the above adsorption

process :

(1) Enthalpy of adsorption is greater than

100 kJ mol−1

.

(2) Energy of activation is low.

(3) Adsorption is monolayer.

(4) Adsorption increases with increase in

temperature.

41. The non-metal that does not exhibit

positive oxidation state is :

(1) Oxygen

(2) Iodine

(3) Chlorine

(4) Fluorine


Set - 04 25

42. Apg¡Mdp„ _uQ¡ Ap ¡g b¡ âq¾$epAp¡ dpV¡$ −ln Kp

rhf yÝ^ sp`dp__u rcÞ_sp v $ip ®h ¡g R> ¡ .

••• •••• •• •• •• •• • ••••

→• • A_¡

••• •••• •• •• •• •• • ••••

→• •

kpQy„ rh^p_ ip¡ p¡.

(1) T>1200 K `f L$pb®_, MO(s) dp„\u M(s)

dp„ qfX$¼n_ L$fi¡.

(2) T<1200 K `f, âq¾$ep MO(s)+C(s) →

M(s)+CO(g) õhe„c| R>¡.

(3) T<1200 K `f, L$pb®__y„ Ap¡[¼kX¡$i_ ârsL|$m(unfavourable) R>¡.

(4) b^p S> sp`dp_ `f L$pb®__y„ Ap¡[¼kX¡$i_A_yLy$m (favourable) R>¡.

42. ‚ÊÕ ÁŒÿ „Èÿ •Ê‹π ◊¥ ÁŸêŸÁ‹ÁπÃ ŒÊ •Á÷Á∑˝§ÿÊ•Ê¥∑§ Á‹ÿ −ln K

p ∑§Ê ÃÊ¬ ∑§ ‚ÊÕ ¬Á⁄UflÃ¸Ÿ Œ‡ÊÊ¸ÿÊ

ªÿÊ „Ò–

••• •••• •• •• •• •• • ••••

→• ÃÕÊ

•• •••••

• O2(g) → CO(s)

ÁŸêŸÁ‹ÁπÃ ∑§ÕŸÊ¥ ◊¥ ‚ ‚„Ë ∑§ÕŸ ¬„øÊÁŸÿ —

(1) T>1200 K, ¬⁄U ∑§Ê’¸Ÿ MO(s) ∑§Ê •¬øÁÿÃ∑§⁄U∑§ M(s) ŒªÊ–

(2) T<1200 K, ¬⁄U •Á÷Á∑˝§ÿÊ, MO(s)+C(s)

→ M(s)+CO(g) SflÃ— ¬˝flÁÃ¸Ã „Ò –

(3) T<1200 K ¬⁄U ∑§Ê’¸Ÿ ∑§Ê ©¬øÿŸ ¬˝ÁÃ∑Í§‹„Ò–

(4) ∑§Ê’¸Ÿ ∑§Ê ©¬øÿŸ ‚÷Ë ÃÊ¬ ¬⁄U •ŸÈ∑Í§‹ „Ò–

42. The plot shows the variation of −ln Kp

versus temperature for the two reactions.

→

→

•

•

•• •••• •• •• •• •• • ••••• • • •••• •••• •• •• •• •• • ••••

• •

• •

Identify the correct statement :

(1) At T>1200 K, carbon will reduce

MO(s) to M(s).

(2) At T<1200 K, the reaction

MO(s)+C(s) → M(s)+CO(g) is

spontaneous.

(3) At T<1200 K, oxidation of carbon is

unfavourable.

(4) Oxidation of carbon is favourable at

all temperatures.


Set - 04 26

43. cpf¡ `pZu k„b„r^s AkÐe rh^p_ ip¡ p¡.(1) s¡ Al

4C

3 kp\¡ âq¾$ep L$fu, CD

4 A_¡

Al(OD)3 b_ph¡ R>¡.

(2) L¡$ÞÖue fuA¡¼V$fp¡dp„ s¡_p¡ iusgL$ (coolant)

sfuL¡$ D`ep¡N \pe R>¡.(3) s¡ CaC

2 kp\¡ âq¾$ep L$fu, C

2D

2 A_¡

Ca(OD)2 b_ph¡ R>¡.

(4) s¡ SO3 kp\ ¡ âq¾ $ep L $fu_ ¡, X $é yV ¡ $ qfs ¹

(deuterated) këãeyqfL$ A¡rkX$ (D2SO

4)

b_ph¡ R>¡.$

44. ApëL$pgpC_-A\® psy këa¡V$p¡_u pZudp„_u Öpìesp_p¡kpQp¡ ¾$d ip¡ p¡.(1) Mg < Ca < Sr < Ba

(2) Mg < Sr < Ca < Ba

(3) Mg > Sr > Ca > Ba

(4) Mg > Ca > Sr > Ba

45. L$p¡gd - I dp„ Ap ¡gu ApCV$dp¡ (items) _¡ s¡_pL$p¡gd - II dp„ _p¢ ¡g dy¿e D`ep¡Np¡ kp\¡ Å¡X$p¡.

(A) rkrgL$p S>¡g (i) V²$p[ÞTõV$f(B) rkrgL$_•(Silicon) (ii) Ape_-rhr_deL$(C) rkrgL$p¡_•(Silicone) (iii) iyóL$ L$sp®(D) rkrgL¡$V$ (iv) kugÞV

L$ p ¡gd•••• L$ p ¡gd•••••

(1) (A)-(iii), (B)-(i), (C)-(iv), (D)-(ii)

(2) (A)-(iv), (B)-(i), (C)-(ii), (D)-(iii)

(3) (A)-(ii), (B)-(iv), (C)-(i), (D)-(iii)

(4) (A)-(ii), (B)-(i), (C)-(iv), (D)-(iii)

43. ÷Ê⁄UË ¬ÊŸË ∑§ ’Ê⁄U ◊¥ ÁŒÿ ªÿ ∑§ÕŸÊ¥ ◊¥ ‚ •‚àÿ∑§ÕŸ ¬„øÊÁŸÿ —(1) ÿ„ Al

4C

3 ‚ •Á÷Á∑˝§ÿÊ ∑§⁄U∑§ CD

4 ÃÕÊ

Al(OD)3 ’ŸÊÃÊ „Ò –

(2) ß‚∑§Ê ©¬ÿÊª ŸÊÁ÷∑§Ëÿ Á⁄U∞Ä≈U⁄U ◊¥ ‡ÊËÃ‹∑§∑§ M§¬ ◊¥ Á∑§ÿÊ ¡ÊÃÊ „Ò–

(3) ÿ„ CaC2 ‚ •Á÷Á∑˝§ÿÊ ∑§⁄U∑§ C

2D

2 ÃÕÊ

Ca(OD)2’ŸÊÃÊ „Ò–

(4) ÿ„ SO3 ‚ •Á÷Á∑˝§ÿÊ ∑§⁄U∑§ «KÍ≈UÁ⁄UÃ

‚ÀçÿÍÁ⁄U∑§ •ê‹ (D2SO

4) ’ŸÊÃÊ „Ò–

44. ˇÊÊ⁄UËÿ ◊ÎŒÊ œÊÃÈ ‚À$»§≈UÊ¥ ∑§Ë ¡‹ ◊¥ Áfl‹ÿÃÊ ∑§Ê ‚„Ë∑˝§◊ „Ò —(1) Mg < Ca < Sr < Ba

(2) Mg < Sr < Ca < Ba

(3) Mg > Sr > Ca > Ba

(4) Mg > Ca > Sr > Ba

45. ∑§ÊÚ‹◊ I ◊¥ ÁŒÿ ªÿ ¬ŒÊÕÊZ (items) ∑§Ê ∑§ÊÚ‹◊ II

◊¥ ÁŒÿ ªÿ ©¬ÿÊªÊ¥ ‚ ‚È◊Á‹Ã ∑§ËÁ¡ÿ —

•• • Á‚Á‹∑§Ê ¡ÒÒ‹ ••• ≈UUÊ¥Á‚S≈U⁄U•• • Á‚Á‹∑§Ÿ •••• •ÊÿŸ ÁflÁŸ◊ÿ∑§•• • Á‚Á‹∑§ÊŸ ••••• ‡ÊÈc∑§Ÿ ∑§◊∑§•• • Á‚Á‹∑§≈U ••• • ‚Ë‹∑§ ••• • •• • ••

∑§ÊÚ‹◊ • ∑§ÊÚ‹◊ •••

(1) (A)-(iii), (B)-(i), (C)-(iv), (D)-(ii)

(2) (A)-(iv), (B)-(i), (C)-(ii), (D)-(iii)

(3) (A)-(ii), (B)-(iv), (C)-(i), (D)-(iii)

(4) (A)-(ii), (B)-(i), (C)-(iv), (D)-(iii)

43. Identify the incorrect statement regarding

heavy water :

(1) It reacts with Al4C

3 to produce CD

4

and Al(OD)3.

(2) It is used as a coolant in nuclear

reactors.

(3) It reacts with CaC2 to produce C

2D

2

and Ca(OD)2.

(4) It reacts with SO3 to form deuterated

sulphuric acid (D2SO

4).

44. The correct order of the solubility of

alkaline-earth metal sulphates in water is :

(1) Mg < Ca < Sr < Ba

(2) Mg < Sr < Ca < Ba

(3) Mg > Sr > Ca > Ba

(4) Mg > Ca > Sr > Ba

45. Match the items in Column I with its main

use listed in Column II :

(A) Silica gel (i) Transistor

(B) Silicon (ii) Ion-exchanger

(C) Silicone (iii) Drying agent

(D) Silicate (iv) Sealant

• • •• • • •• • • •• • • •••

(1) (A)-(iii), (B)-(i), (C)-(iv), (D)-(ii)

(2) (A)-(iv), (B)-(i), (C)-(ii), (D)-(iii)

(3) (A)-(ii), (B)-(iv), (C)-(i), (D)-(iii)

(4) (A)-(ii), (B)-(i), (C)-(iv), (D)-(iii)


Set - 04 27

46. L$ep AÏAp¡_p¡ kd|l kh®kd A\hp ArcÞ_ ApL$pf(Identical shape) ^fph¡ R>¡ ?(1) SF

4

, XeF4

, CCl4

(2) ClF3

, XeOF2

, •• • ••

(3) BF3

, PCl3

, XeO3

(4) PCl5

, IF5

, XeO2F

2

47. _uQ¡ Ap`¡gp [õ`rkTp¡ (ÅrsAp¡) dp„\u L$C A¡L$ S>gueÖphZdp„ õ\peu R>¡ ?(1) Cr

2+

(2) Cu+

(3)••• • • •

(4)••• • • •

48. _uQ¡ Ap ¡gp k„L$uZp£dp„\u L$ey„ A¡L$ Ag(NO3) _p S>gue

ÖphZ_u Ar^L$ syëesp_p¡ D`ep¡N L$fi¡ ?(1) Na

3[CrCl

6]

(2) [Cr(H2O)

5Cl]Cl

2

(3) [Cr(H2O)

6]Cl

3

(4) Na2[CrCl

5(H

2O)]

46. ∑§ÊÒŸ ‚Ê ‚◊Í„ ‚◊M§¬ •áÊÈ•Ê¥ ∑§Ê ‚◊Í„ „Ò —(1) SF

4

, XeF4

, CCl4

(2) ClF3

, XeOF2

, •• • ••

(3) BF3

, PCl3

, XeO3

(4) PCl5

, IF5

, XeO2F

2

47. ÁŸêŸÁ‹ÁπÃ ◊¥ ‚ ¡‹Ëÿ Áfl‹ÿŸ ◊¥ SÕÊÿË S¬Ë‡ÊË$¡∑§ÊÒŸ ‚Ë „Ò?(1) Cr

2+

(2) Cu+

(3)••• • • •

(4)••• • • •

48. ÁŸêŸÁ‹ÁπÃ ‚¥∑È§‹Ê¥ ◊¥ ‚ ∑§ÊÒŸ ‚Ê ‚¥∑È§‹ Ag(NO3)

∑§ ¡‹Ëÿ Áfl‹ÿŸ ∑§ •Áœ∑§ ‚◊ÃÈÀÿ π¬ÊÿªÊ ?(1) Na

3[CrCl

6]

(2) [Cr(H2O)

5Cl]Cl

2

(3) [Cr(H2O)

6]Cl

3

(4) Na2[CrCl

5(H

2O)]

46. The group of molecules having identical

shape is :

(1) SF4

, XeF4

, CCl4

(2) ClF3

, XeOF2

, •• • ••

(3) BF3

, PCl3

, XeO3

(4) PCl5

, IF5

, XeO2F

2

47. Which one of the following species is stable

in aqueous solution ?

(1) Cr2+

(2) Cu+

(3)••• • • •

(4)••• • • •

48. Which one of the following complexes will

consume more equivalents of aqueous

solution of Ag(NO3) ?

(1) Na3[CrCl

6]

(2) [Cr(H2O)

5Cl]Cl

2

(3) [Cr(H2O)

6]Cl

3

(4) Na2[CrCl

5(H

2O)]


Set - 04 28

49. _uQ¡ Ap`¡gpdp„\u kpQu qv$ip (trend) ip¡^p¡.(`fdpÎhue ¾$dp„L$=Ti : 22, Cr=24 A_¡ Mo=42)

(1) ∆o _p¡ [Cr(H2O)

6]2+

>

[Mo(H2O)

6]2+

A_¡∆o _p¡ [Ti(H

2O)

6]3+

> [Ti(H2O)

6]2+

(2) ∆o _p¡ [Cr(H2O)

6]2+

> [Mo(H2O)

6]2+ A_¡

∆o _p¡ [Ti(H2O)

6]3+

< [Ti(H2O)

6]2+

(3) ∆o _p¡ [Cr(H2O)

6]2+

< [Mo(H2O)

6]2+

A_¡∆o _p¡ [Ti(H

2O)

6]3+

> [Ti(H2O)

6]2+

(4) ∆o _p¡ [Cr(H2O)

6]2+

< [Mo(H2O)

6]2+

A_¡∆o _p¡ [Ti(H

2O)

6]3+

< [Ti(H2O)

6]2+

50. BOD v$ip®h¡ R>¡.

(1) bpep¡gp¡ÆL$g Ap¡[¼kS>_ qX$dpÞX$

(2) b¡¼V$¡qfAg Ap¡[¼kX¡$i_ qX$dpÞX$

(3) bpep¡L¡$rdL$g Ap¡[¼kS>_ qX$dpÞX$

(4) bpep¡L¡$rdL$g Ap¡[¼kX¡$i_ qX$dpÞX$

49. ÁŸêŸÁ‹ÁπÃ ◊¥ ‚ ‚„Ë ¬˝flÎÁûÊ ¬„øÊÁŸÿ —(¬⁄U◊ÊáÊÈ ∑˝§◊Ê¥∑§=Ti : 22, Cr : 24 ÃÕÊ Mo : 42)

(1) [Cr(H2O)

6]2+

∑§Ê

∆o > [Mo(H2O)

6]2+

ÃÕÊ

[Ti(H2O)

6]3+

∑§Ê ∆o > [Ti(H2O)

6]2+

(2) [Cr(H2O)

6]2+

∑§Ê ∆o >

[Mo(H2O)

6]2+

ÃÕÊ [Ti(H2O)

6]3+∑§Ê

∆o < [Ti(H2O)

6]2+

(3) [Cr(H2O)

6]2+

∑§Ê

∆o < [Mo(H2O)

6]2+

ÃÕÊ

[Ti(H2O)

6]3+

∑§Ê ∆o > [Ti(H2O)

6]2+

(4) [Cr(H2O)

6]2+

∑§Ê

∆o < [Mo(H2O)

6]2+

ÃÕÊ

[Ti(H2O)

6]3+ ∑§Ê ∆o < [Ti(H

2O)

6]2+

50. ’Ë.•Ê.«UË. (BOD) Œ‡ÊÊ¸ÃÊ „Ò —

(1) ’ÊÿÊ‹Ê¡Ë∑§‹ •ÊÚÄ‚Ë¡Ÿ Á«U◊Ê¥«U

(2) ’ÒÄ≈UËÁ⁄Uÿ‹ •ÊÚÄ‚Ë«U‡ÊŸ Á«U◊Ê¥«U

(3) ’ÊÿÊ∑Ò§Á◊∑§‹ •ÊÚÄ‚Ë¡Ÿ Á«U◊Ê¥«U

(4) ’ÊÿÊ∑Ò§Á◊∑§‹ •ÊÚÄ‚Ë«U‡ÊŸ Á«U◊Ê¥«U

49. Identify the correct trend given below :

(Atomic No.=Ti : 22, Cr : 24 and Mo : 42)

(1) ∆o of [Cr(H2O)

6]2+

>

[Mo(H2O)

6]2+

and

∆o of [Ti(H2O)

6]3+

> [Ti(H2O)

6]2+

(2) ∆o of [Cr(H2O)

6]2+

> [Mo(H2O)

6]2+

and

∆o of [Ti(H2O)

6]3+

< [Ti(H2O)

6]2+

(3) ∆o of [Cr(H2O)

6]2+

< [Mo(H2O)

6]2+

and

∆o of [Ti(H2O)

6]3+

> [Ti(H2O)

6]2+

(4) ∆o of [Cr(H2O)

6]2+

< [Mo(H2O)

6]2+

and

∆o of [Ti(H2O)

6]3+

< [Ti(H2O)

6]2+

50. BOD stands for :

(1) Biological Oxygen Demand

(2) Bacterial Oxidation Demand

(3) Biochemical Oxygen Demand

(4) Biochemical Oxidation Demand


Set - 04 29

51. A¡L$ L$pb®r_L$ k„ep¡S>_ C, H A_¡ S ^fph¡ R>¡. Å¡ s¡dp„8% këaf lp¡e sp¡ k„ep¡S>__p¡ Þe|_sd (minimum)

AÏcpfip¡ p¡. (`fdpÎhue hS>_ S _y„ =32 amu)(1) 200 g mol

−1

(2) 400 g mol−1

(3) 600 g mol−1

(4) 300 g mol−1

52. kps L$pb®_ `fdpÏAp¡hpmp A¡L$ lpCX²$p¡L$pb®_ L$ep¡ li¡L¡$ S>¡dp„ A¡L$ r_ep¡`¡ÞV$pCg A_¡ A¡L$ rh_pCg kd|l^fph¡ R>¡ ?

(1) 2, 2-X$perd\pCg-4-`¡ÞV$u_

(2) ApCkp¡âp¡`pCg-2-åe|V$u_

(3) 4, 4-X$perd\pCg`¡ÞV$u_

(4) 2, 2-X$perd\pCg-3-`¡ÞV$u_

53. 5 L A¡L$ ApëL¡$__p k„ |Z® v$l_ dpV¡$ 25 L Ap¡[¼kS>__uAphíeL$sp R>¡. b^p S> L$v$p¡ (volumes) AQmsp`dp_ A_¡ v$bpZ dp`hpdp„ Aph¡g lp¡e sp¡ ApëL¡$_L$ep¡ R>¡ ?

(1) C\¡_

(2) âp¡`¡_

(3) åe|V¡$_

(4) ApCkp¡åe|V¡_

51. ∞∑§ ∑§Ê’¸ÁŸ∑§ ÿÊÒÁª∑§ ◊¥ C, H ÃÕÊ S Áfll◊ÊŸ „Ò¥–ÿÁŒ ß‚ ÿÊÒÁª∑§ ◊¥ 8% ‚À$»§⁄U „Ê ÃÊ ß‚∑§Ê ãÿÍŸÃ◊•áÊÈ ÷Ê⁄U „ÊªÊ —

( S ∑§Ê ¬⁄U◊ÊáÊÈ ÷Ê⁄U =32 amu)

(1) 200 g mol−1

(2) 400 g mol−1

(3) 600 g mol−1

(4) 300 g mol−1

52. ‚ÊÃ ∑§Ê’¸Ÿ ¬⁄U◊ÊáÊÈ•Ê¥ flÊ‹Ê ∞∑§ „Êß«˛UÊ∑§Ê’¸Ÿ ∑§ÊÒŸ„ÊªÊ Á¡‚◊¥ ∞∑§ ÁŸ•Ê¬Áã≈U‹ ‚◊Í„ ÃÕÊ ∞∑§ flÊßÁŸ‹‚◊Í„ „Ê —

(1) 2, 2-«UÊß¸◊ÁÕ‹-4-¬ã≈UËŸ

(2) •Êß‚Ê¬˝ÊÁ¬‹-2-éÿÍÁ≈UŸ

(3) 4, 4-«UÊß¸◊ÁÕ‹¬ã≈UËŸ

(4) 2, 2-«UÊß¸◊ÁÕ‹-3-¬ã≈UËŸ

53. ∞∑§ ∞À∑§Ÿ ∑§Ë 5 L ◊ÊòÊÊ ∑§ ¬ÍáÊ¸ Œ„Ÿ ∑§ Á‹ÿ 25 L

•ÊÚÄ‚Ë¡Ÿ ∑§Ë •Êfl‡ÿ∑§ÃÊ „ÊÃË „Ò– ÿÁŒ ‚÷Ë •ÊÿÃŸ◊ÊŸ∑§ ÃÊ¬ ÃÕÊ ŒÊ’ ¬⁄U ◊Ê¬ ªÿ „Ê¥, ÃÊ ∞À∑§Ÿ„ÊªË —

(1) ∞ÕŸ

(2) ¬˝Ê¬Ÿ

(3) éÿÍ≈UŸ

(4) •Êß‚ÊéÿÍ≈UŸ

51. An organic compound contains C, H and

S. The minimum molecular weight of the

compound containing 8% sulphur is :

(atomic weight of S=32 amu)

(1) 200 g mol−1

(2) 400 g mol−1

(3) 600 g mol−1

(4) 300 g mol−1

52. The hydrocarbon with seven carbon atoms

containing a neopentyl and a vinyl group

is :

(1) 2, 2-dimethyl-4-pentene

(2) Isopropyl-2-butene

(3) 4, 4-dimethylpentene

(4) 2, 2-dimethyl-3-pentene

53. 5 L of an alkane requires 25 L of oxygen for

its complete combustion. If all volumes are

measured at constant temperature and

pressure, the alkane is :

(1) Ethane

(2) Propane

(3) Butane

(4) Isobutane


Set - 04 30

54. CH3MgBr _¡ rd\¡_p¡gdp„ Nfd L$fsp„ DÐ`Þ_ \sp¡

hpey ip¡ p¡.(1) HBr

(2) rd\¡_

(3) C\¡_

(4) âp¡`¡_

55. b|hp¡-åg¡ÞL$ (Bouveault-Blanc) fuX$n_ âq¾$epdp„\pe R>¡ :

(1) A¡kpCg l¡gpCX$_y„ H2/Pd kp\¡ qfX$n_.

(2) A¡õV$f_y„ Na/C2H

5OH kp\¡ qfX$n_.

(3) L$pbp¡®r_g k„ep¡S>__y„ Na/Hg A_¡ HCl kp\¡qfX$n_.

(4) A¡_lpCX²$pCX$_y„ LiAlH4 kp\¡ qfX$n_.

56. âp\rdL$, qÜrseL$ A_¡ s©rseL$ A¡dpC__¡ âc¡qv$s L$fsuL$kp¡V$u _uQ¡_pdp„\u ip¡^p¡.

(1) L$pbp®gA¡dpC_ âq¾$ep(2) C

6H

5SO

2Cl

(3) k¡ÞXd¡ef âq¾$ep

(4) dõV$X®$ Ap¡Cg L$kp¡V$u

54. CH3MgBr ∑§Ê ◊ÕÒŸÊÚ‹ ◊¥ ª◊¸ ∑§⁄UŸ ¬⁄U ©à¬ÛÊ „ÊŸ

flÊ‹Ë ªÒ‚ „Ò —(1) HBr

(2) ◊ÕÒŸ

(3) ∞ÕŸ

(4) ¬˝Ê¬Ÿ

55. ’ÍflÊ-é‹Ê¥∑§ Á⁄U«UÄ‡ÊŸ ¬˝Á∑˝§ÿÊ ◊¥ „ÊÃÊ „Ò —

(1) ∞Á‚‹ „Ò‹Êß«U ∑§Ê H2/Pd ‚ •¬øÿŸ–

(2) ∞S≈U⁄U ∑§Ê Na/C2H

5OH ‚ •¬øÿŸ–

(3) ∑§Ê’Ê¸ÁŸ‹ ÿÊÒÁª∑§ ∑§Ê Na/Hg ÃÕÊ HCl ‚•¬øÿŸ–

(4) ∞∑§ ∞Ÿ„Êß«˛UÊß«U ∑§Ê LiAlH4 ‚ •¬øÿŸ–

56. ¬˝ÊÕÁ◊∑§, ÁmÃËÿ∑§ ÃÕÊ ÃÎÃËÿ∑§ ∞◊ËŸÊ¥ ◊¥ •ãÃ⁄U ∑§⁄UŸ∑§ Á‹ÿ ¬˝ÿÈÄÃ „ÊŸ flÊ‹Ê ¬⁄UËˇÊáÊ „Ò —

(1) ∑§ÊÁ’¸‹∞◊ËŸ •Á÷Á∑˝§ÿÊ(2) C

6H

5SO

2Cl

(3) ‚Òã«U◊Êÿ⁄U •Á÷Á∑˝§ÿÊ

(4) ◊S≈U«¸U •ÊÚÿ‹ ¬⁄UËˇÊáÊ

54. The gas evolved on heating CH3MgBr in

methanol is :

(1) HBr

(2) Methane

(3) Ethane

(4) Propane

55. Bouveault-Blanc reduction reaction

involves :

(1) Reduction of an acyl halide with

H2/Pd.

(2) Reduction of an ester with

Na/C2H

5OH.

(3) Reduction of a carbonyl compound

with Na/Hg and HCl.

(4) Reduction of an anhydride with

LiAlH4.

56. The test to distinguish primary, secondary

and tertiary amines is :

(1) Carbylamine reaction

(2) C6H

5SO

2Cl

(3) Sandmeyer’s reaction

(4) Mustard oil test


Set - 04 31

57. L$\_ : f¡ep¡_ A¡L$ A^®kp„ïg¡rjs blºgL$ R>¡. S>¡_pNyZ^dp£ Ly$v$fsu L$`pk (cotton) L$fsp h y kpfpR>¡.

L$pfZ : A¡rkqV$g¡i_ hX¡$ k¡ëeygp¡T_p ep„rÓL$ A_¡ kp¦v$e®^fphsp NyZ^dp£_¡ h y kpfp iL$pe R>¡.

(1) L$\_ A_¡ L$pfZ b„_¡ kpQp R>¡. L$pfZ A¡ L$\_dpV¡$_u kpQu kdS|>su R>¡.

(2) L$\_ A_¡ L$pfZ b„_¡ kpQp R>¡, `Z L$pfZ A¡L$\_ dpV¡$_u kpQu kdS|>su _\u.

(3) L$\_ A¡ AkÐe rh^p_ R>¡ `Z L$pfZ A¡ kÐeR>¡.

(4) L$\_ A_¡ L$pfZ b„_¡ AkÐe R>¡.

58. A¡õ`pqV®$L$ A¡rkX$ dpV¡$ _uQ¡_p¡ ¾$d Ýep_dp„ gp¡.

A¡õ`pqV®$L$ A¡rkX$_y„ pI (kdrhch tbvy$) ip¡ p¡.(1) 1.88

(2) 3.65

(3) 5.74

(4) 2.77

57. ∑§ÕŸ : ⁄UÿÊÚŸ ∞∑§ •h‚¥Á‡‹C ’„È‹∑§ „Ò Á¡‚∑§ªÈáÊœ◊ ¬Ê∑Î§ÁÃ∑§ ∑§¬Ê‚ ‚ •Áœ∑§ •ë¿U„Ò¥–

∑§Ê⁄UáÊ : ∞‚ËÁ≈U‹Ë∑§⁄UáÊ ‚ ‚‹È‹Ê‚ ∑§ ÿÊ¥ÁòÊ∑§fl ‚ÊÒ¥Œÿ¸¬⁄U∑§ ªÈáÊœ◊ÊZ ∑§Ê ‚ÈœÊ⁄UÊ ¡Ê‚∑§ÃÊ „Ò–

(1) “∑§ÕŸ” fl “∑§Ê⁄UáÊ” ŒÊŸÊ¥ ‚„Ë „Ò¥ ÃÕÊ “∑§Ê⁄áÊ”,“∑§ÕŸ” ∑§Ë ‚„Ë √ÿÊÅÿÊ „Ò–

(2) “∑§ÕŸ” fl “∑§Ê⁄UáÊ” ŒÊŸÊ¥ ‚„Ë „Ò¥ Á∑§ãÃÈ “∑§Ê⁄UáÊ”,“∑§ÕŸ” ∑§Ë ‚„Ë √ÿÊÅÿÊ Ÿ„Ë¥ „Ò–

(3) “∑§ÕŸ” ª‹Ã „Ò Á∑§ãÃÈ “∑§Ê⁄UáÊ” ‚„Ë „Ò–

(4) “∑§ÕŸ” fl “∑§Ê⁄UáÊ” ŒÊŸÊ¥ ª‹Ã „Ò¥–

58. ∞S¬ÊÁ≈¸U∑§ •ê‹ ∑§ ÁŸêŸÁ‹ÁπÃ •ŸÈ∑˝§◊ ¬⁄U ÁfløÊ⁄U∑§ËÁ¡∞ —

∞S¬ÊÁ≈U¸∑§ •ê‹ ∑§Ê pI (‚◊Áfl÷fl Á’¥ŒÈ ) „Ò —(1) 1.88

(2) 3.65

(3) 5.74

(4) 2.77

57. Assertion : Rayon is a semisynthetic

polymer whose properties

are better than natural cotton.

Reason : Mechanical and aesthetic

properties of cellulose can be

improved by acetylation.

(1) Both assertion and reason are correct,

and the reason is the correct

explanation for the assertion.


but the reason is not the correct


(3) Assertion is incorrect statement, but

the reason is correct.

(4) Both assertion and reason are

incorrect.

58. Consider the following sequence for

aspartic acid :

The pI (isoelectric point) of aspartic acid

is :

(1) 1.88

(2) 3.65

(3) 5.74

(4) 2.77


Set - 04 32

59. _uQ¡_pdp„\u L$ep L©$rÓd Nþep `v$p\p£_p¡ Nm`Z Ap„L$(sweetness value) L¡$_ kyNf (cane sugar) _usyg_pdp„ kp¥\u h^pf¡ R>¡.

(1) A¡õ`pV£_

(2) k¡L¡$fu_

(3) ky¾$pgp¡T

(4) A¡rgV¡$d

60. A¡N-Apëbydu_ kp¡g (egg-albumin sol) b_phhpdpV¡$_u kp¥\u kpfu b„ b¡ksu Ý^rs _uQ¡_pdp„\u ip¡ p¡.

(1) IX$p_¡ kpQhu_¡ sp¡X$p¡ A_¡ s¡_p `pfv$i cpN_¡100 mL 5% w/V _p ghZue (saline)

ÖphZdp„ d¡mhu_¡ kpfu fus¡ lgphp¡.

(2) IX$p_¡ kpQhu_¡ sp¡X$p¡ A_¡ s¡_p a¼s ump cpN_¡100 mL 5% w/V _p ghZue (saline)

ÖphZdp„ d¡mhu_¡ kpfu fus¡ lgphp¡.

(3) IX$p_¡ DL$msp `pZudp„ 10 rdr_V$ dpV¡$ fpMp¡s¡_u R>pg v|$f L$ep® bpv$ s¡_p¡ ka¡v$ cpN100 mL 5% w/V _p ghZue (saline)

ÖphZdp„ ep„rÓL$ i¡L$f (shaker) hX¡$ kdp„NuL©$s(homogenize) L$fp¡.

(4) IX$p_¡ DL$msp `pZudp„ 10 rdr_V$ dpV¡$ fpMp¡s¡_u R>pg v|$f L$ep® bpv$ s¡_p `ump¡ cpN100 mL 5% w/V _p ghZue (saline)

ÖphZdp„ ep„rÓL$ i¡L$f (shaker) hX¡$ kdp„NuL©$s(homogenize) L$fp¡.

59. fl„ ∑Î§ÁòÊ◊ ◊œÈ⁄U∑§ Á¡‚∑§Ê ßˇÊÈ ‡Ê∑¸§⁄UÊ ∑§Ë ÃÈ‹ŸÊ ◊¥◊ÊœÈÿ¸◊ÊŸ ‚’‚ •Áœ∑§ „Ò —

(1) ∞‚¬Ê≈¸UŸ

(2) ‚Ò∑§⁄UËŸ

(3) ‚È∑˝§Ê‹Ê‚

(4) ∞‹Ë≈U◊

60. •¥«U ∞ÀéÿÍÁ◊Ÿ ∑§Ê ‚ÊÚ‹ ’ŸÊŸ ∑§Ë ‚’‚ ©ÁøÃ ÁflÁœ„Ò —

(1) •¥«U ∑§Ê äÿÊŸ¬Ífl¸∑§ ÃÊ«∏¥ •ÊÒ⁄U ©‚∑§ ¬Ê⁄UŒ‡ÊË¸÷Êª ∑§Ê 100 mL 5% w/V ‹fláÊ ¡‹ ◊¥Á◊‹Ê ∑§⁄U •ë¿UË Ã⁄U„ Á„‹Êÿ¥–

(2) •¥«U ∑§Ê äÿÊŸ¬Ífl¸∑§ ÃÊ«∏¥ •ÊÒ⁄U ©‚∑§ ¬Ë‹ ÷Êª∑§Ê 100 mL 5% w/V ‹fláÊ ¡‹ ◊¥ Á◊‹Ê∑§⁄U •ë¿UË Ã⁄U„ Á„‹Êÿ¥–

(3) •¥«U ∑§Ê ©’‹Ã ¡‹ ◊¥ 10 Á◊Ÿ≈U Ã∑§ ⁄Uπ¥ –©‚∑§Ê Á¿U‹∑§Ê ©ÃÊ⁄UŸ ∑§ ¬‡øÊÃ˜ ©‚∑§ ‚»§Œ÷Êª ∑§Ê 100 mL 5% w/V ‹fláÊ ¡‹ ◊¥Á◊‹Êÿ¥ •ÊÒ⁄U ÿÊ¥ÁòÊ∑§ „ÁÀ‹òÊ ◊¥ ‚◊Ê¥ªË∑Î§Ã ∑§⁄¥U–

(4) •¥«U ∑§Ê ©’‹Ã ¡‹ ◊¥ 10 Á◊Ÿ≈U Ã∑§ ⁄Uπ¥–©‚∑§Ê Á¿U‹∑§Ê ©ÃÊ⁄UŸ ∑§ ¬‡øÊÃ˜ ©‚∑§ ¬Ë‹÷Êª ∑§Ê 100 mL 5% w/V ‹fláÊ ¡‹ ◊¥Á◊‹Êÿ¥ •ÊÒ⁄U ÿÊ¥ÁòÊ∑§ „ÁÀ‹òÊ ◊¥ ‚◊Ê¥ªË∑Î§Ã ∑§⁄¥U–

59. The artificial sweetener that has the highest

sweetness value in comparison to cane

sugar is :

(1) Aspartane

(2) Saccharin

(3) Sucralose

(4) Alitame

60. The most appropriate method of making

egg-albumin sol is :

(1) Break an egg carefully and transfer

the transparent part of the content to

100 mL of 5% w/V saline solution

and stir well.

(2) Break an egg carefully and transfer

only the yellow part of the content to

100 mL of 5% w/V saline solution

and stir well.

(3) Keep the egg in boiling water for 10

minutes. After removing the shell,

transfer the white part of the content

to 100 mL of 5% w/V saline solution

and homogenize with a mechanical

shaker.

(4) Keep the egg in boiling water for 10

minutes. After removing the shell,

transfer the yellow part of the content

to 100 mL of 5% w/V saline solution

and homogenize with a mechanical

shaker.


Set - 04 33

61. ^pfp¡ L¡$ x R, x ≠ 0, x ≠ 1 dpV¡$ ••• • ••

• •• •

•• •

• •A_¡ f

n+1 (x)=f

0 (f

n(x)), n=0, 1, 2, . . . sp¡ lh¡

• • • • •• ••• ••• •

• • •

• • •• • =

(1)

••

(2)

••

(3)

••

(4)

••

62. ApN®ÞX$ kdsgdp„ 2+i \u v$ip®h¡g tbvy$ A¡L$ A¡L$d|h® sfa Mk¡ R>¡, Ðepfbpv$ 2 A¡L$d DÑf sfa Mk¡ R>¡

A_¡ Ðep„\u R>¡hV¡$ • • A¡L$d _¥F>Ðe sfa Mk¡ R>¡.ApN®ÞX$ kdsgdp„ Ap tbvy$_y„ _hy„ õ\p_ L$C k„¿ep\uv$ip®hhpdp„ Aph¡ ?(1) 2+2i

(2) 1+i

(3) −1−i

(4) −2−2i

61. x R, x ≠ 0, x ≠ 1 ∑§ Á‹∞ ◊ÊŸÊ ••• • ••

• •• •

•• •

• •ÃÕÊ f

n+1 (x)=f

0 (f

n(x)), n=0, 1, 2, . . . „Ò, ÃÊ

• • • • •• ••• ••• •

• • •

• • •• • ’⁄UÊ’⁄U „Ò —

(1)

••

(2)

••

(3)

••

(4)

••

62. •Ê⁄Uªá«U ‚◊Ã‹ ◊¥ 2+i mÊ⁄UÊ ÁŸÁŒ¸c≈U Á’¥ŒÈ, 1 ß∑§Êß¸¬Ífl¸ ÁŒ‡ÊÊ ◊¥ ø‹ÃÊ „Ò •ÊÒ⁄U Á»§⁄U 2 ß∑§Êß¸ ©ûÊ⁄U ÁŒ‡ÊÊ ◊¥

ø‹ÃÊ „Ò ÃÕÊ •ãÃ ◊¥ • • ß∑§Êß¸ ŒÁˇÊáÊ-¬Á‡ø◊ÁŒ‡ÊÊ ◊¥ ¡ÊÃÊ „Ò– ÃÊ •Ê⁄Uªá«U ‚◊Ã‹ ◊¥ ß‚∑§Ê ŸÿÊSÕÊŸ Á¡‚ Á’¥ŒÈ mÊ⁄UÊ ÁŸÁŒ¸c≈U „ÊÃÊ „Ò, fl„ „Ò —(1) 2+2i

(2) 1+i

(3) −1−i

(4) −2−2i

61. For x R, x ≠ 0, x ≠ 1, let ••• • ••

• •• •

•• •

• • and

fn+1

(x)=f0

(fn(x)), n=0, 1, 2, . . . Then the

value of • • • • •• ••• ••• •

• • •

• • •• • is equal

to :

(1)

••

(2)

••

(3)

••

(4)

••

62. The point represented by 2+i in the

Argand plane moves 1 unit eastwards, then

2 units northwards and finally from there

• • units in the south-westwards

direction. Then its new position in the

Argand plane is at the point represented

by :

(1) 2+2i

(2) 1+i

(3) −1−i

(4) −2−2i


Set - 04 34

63. Å¡ kduL$fZp¡ x2+bx−1=0 A_¡ x

2+x+b=0

_¡ −1 rkhpe_y„ A¡L$ kpdpÞe buS> lp¡e sp¡ |b|

bfpbf :

(1) •(2) 2

(3) 3

(4) •

64. Å¡

• •• •• •• ••• •

• •• •• •• •• •• •

• ••

,

• •• •

• •• •• •• •

• • A_¡

Q=PAPT lp¡e sp¡ PT

Q2015

P bfpbf :

(1)

• • • • •• •• •• •• •

(2)

• • • • •• • • • •

• •• •• •

(3)

• • • • •• • • • •

• •• •• •

(4)

• • • • •• •• •• •• •

63. ÿÁŒ ‚◊Ë∑§⁄UáÊÊ¥ x2+bx−1=0 ÃÕÊ x2

+x+b=0

∑§Ê −1 ‚ Á÷ãŸ ∞∑§ ‚Ê¥! ÊÊ ◊Í‹ „Ò, ÃÊ |b| ’⁄UÊ’⁄U„Ò —

(1) •(2) 2

(3) 3

(4) •

64. ÿÁŒ

• •• •• •• ••• •

• •• •• •• •• •• •

• ••

,

• •• •

• •• •• •• •

• • ÃÕÊ

Q=PAPT „Ò, ÃÊ PT

Q2015

P „Ò —

(1)

• • • • •• •• •• •• •

(2)

• • • • •• • • • •

• •• •• •

(3)

• • • • •• • • • •

• •• •• •

(4)

• • • • •• •• •• •• •

63. If the equations x2+bx−1=0 and

x2+x+b=0 have a common root different

from −1, then |b| is equal to :

(1) •(2) 2

(3) 3

(4) •

64. If

• •• •• •• ••• •

• •• •• •• •• •• •

• ••

,

• •• •

• •• •• •• •

• • and

Q=PAPT, then P

T Q

2015 P is :

(1)

• • • • •• •• •• •• •

(2)

• • • • •• • • • •

• •• •• •

(3)

• • • • •• • • • •

• •• •• •

(4)

• • • • •• •• •• •• •


Set - 04 35

65. kduL$fZ •• •• ••• • ••• •••• • •• •• ••• • •••• • ••• • •• ••

• • •• • •• • •

•• • _p A„sfpg

•• ••• •

• •• •π π

•• dp„ Aph¡g rcÞ_ buÅ¡_u k„¿ep L¡$V$gu

li¡ ?(1) 4

(2) 3

(3) 2

(4) 1

66. iåv$ “MEDITERRANEAN” _p„ Anfp¡_p¡ D`ep¡NL$fu Qpf Anfp¡ hpmp (A\® lp¡hp¡ S>ê$fu _\u) L¡$ S>¡_p¡â\d Anf R A_¡ Qp¡\p¡ Anf E lp¡e s¡hp sdpdiåv$p¡_u k„¿ep L¡$V$gu \pe ?

(1)( )

•• ••• •

(2) 110

(3) 56

(4) 59

65. ‚◊Ë∑§⁄UáÊ •• •• ••• • ••• •••• • •• •• ••• • •••• • ••• • •• ••

• • •• • •• • •

•• • , ∑§ •¥Ã⁄UÊ‹

•• ••• •

• •• •π π

•• ◊¥ Á÷ãŸ flÊSÃÁfl∑§ ◊Í‹Ê¥ ∑§Ë ‚¥ÅÿÊ „Ò —

(1) 4

(2) 3

(3) 2

(4) 1

66. ‡ÊéŒ “MEDITERRANEAN” ∑§ •ˇÊ⁄UÊ¥ ‚ øÊ⁄U•ˇÊ⁄UÊ¥ ∑§ ∞‚ ‡ÊéŒ (øÊ„ •Õ¸„ËŸ „Ê¥) ’ŸÊŸ „Ò¥ Á¡Ÿ∑§Ê¬„‹Ê •ˇÊ⁄U R ÃÕÊ øÊÒÕÊ •ˇÊ⁄U E „Ê, ÃÊ ∞‚ ‚÷Ë‡ÊéŒÊ¥ ∑§Ë ∑È§‹ ‚¥ÅÿÊ „Ò —

(1)( )

•• ••• •

(2) 110

(3) 56

(4) 59

65. The number of distinct real roots of the

equation,

•• •• ••• • ••• •••• • •• •• ••• • •••• • ••• • •• ••

• • •• • •• • •

•• • in the

interval •• ••• •

• •• •π π

•• is :

(1) 4

(2) 3

(3) 2

(4) 1

66. If the four letter words (need not be

meaningful ) are to be formed using the

letters from the word

“MEDITERRANEAN” such that the first

letter is R and the fourth letter is E, then

the total number of all such words is :

(1)( )

•• ••• •

(2) 110

(3) 56

(4) 59


Set - 04 36

67. Å¡ (1+x)2016

+x(1+x)2015

+x2(1+x)

2014

+......+x2016

=

• • • •

• •• •

••

• ••• •

, x R, x ≠−1 ; sp¡

a17

=

(1)

• • • • •• • •••• • • • •

(2)

• • • • •• • •••• • • • •

(3)

• • • • •• • • • •

(4)

• • • • •• • •

68. ^pfp¡ L¡$ _ hpõsrhL$ k„¿epAp¡ x, y, z A¡hu R>¡ L¡$ S>¡\ux+y+z=12 A_¡ x

3y

4z

5=(0.1) (600)

3. sp¡

x3+y

3+z

3=

(1) 270

(2) 258

(3) 342

(4) 216

67. x R, x ≠−1 ∑§ Á‹∞, ÿÁŒ

(1+x)2016

+x(1+x)2015

+x2(1+x)

2014

+......+x2016

=

• • • •

• •• •

••

• ••• •

„Ò, ÃÊ a17

’⁄UÊ’⁄U „Ò —

(1)

• • • • •• • •••• • • • •

(2)

• • • • •• • •••• • • • •

(3)

• • • • •• • • • •

(4)

• • • • •• • •

68. ◊ÊŸÊ x, y, z ∞‚Ë œŸÊà◊∑§ flÊSÃÁfl∑§ ‚¥ÅÿÊ∞° „Ò¥ Á∑§,x+y+z=12 ÃÕÊ x

3y

4z

5=(0.1) (600)

3 „Ò, ÃÊ

x3+y

3+z

3 ’⁄UÊ’⁄U „Ò —

(1) 270

(2) 258

(3) 342

(4) 216

67. For x R, x ≠−1, if

(1+x)2016

+x(1+x)2015

+x2(1+x)

2014

+......+x2016

=

• • • •

• •• •

••

• ••• •

, then a17

is equal

to :

(1)

• • • • •• • •••• • • • •

(2)

• • • • •• • •••• • • • •

(3)

• • • • •• • • • •

(4)

• • • • •• • •

68. Let x, y, z be positive real numbers

such that x+y+z=12 and

x3y

4z

5=(0.1) (600)

3. Then x

3+y

3+z

3 is

equal to :

(1) 270

(2) 258

(3) 342

(4) 216


Set - 04 37

69.

• •• ••

• ••• •

••

•

•••

•−=

=

(1) 560

(2) 680

(3) 1240

(4) 1085

70. Å¡ •

•••••• • •

•

•

• • •• •→∞

+ − = sp¡ ‘a’ =

(1) 2

(2)

••

(3)

••

(4)

••

69.

• •• ••

• ••• •

••

•

•••

•−=

∑§Ê ◊ÊŸ „Ò —

(1) 560

(2) 680

(3) 1240

(4) 1085

70. ÿÁŒ

••

••••• • •

•

•

• •• •→∞

+ − = „Ò, ÃÊ ‘a’ ’⁄UÊ’⁄U

„Ò —(1) 2

(2)

••

(3)

••

(4)

••

69. The value of

• •• ••

• ••• •

••

•

•••

•−=

is equal

to :

(1) 560

(2) 680

(3) 1240

(4) 1085

70. If

••

••••• • •

•

•

• • •• •→∞

+ − = then ‘a’ is

equal to :

(1) 2

(2)

••

(3)

••

(4)

••


Set - 04 38

71. Å¡ rh ¡e

f(x)=

( )•

••••••••••••••••••••••••••••• •

• •• • • •• •

• • •

• • • • •

<•••

≤ ≤•••

•

• • • •

x=1 ApNm rhL$g_ue lp¡e sp¡ •• =

(1)

••

π •• •

(2)

••

π • • • •

(3)

••

π • •

(4) −1−cos−1

(2)

72. Å¡ h¾$ x=4t2+3, y=8t

3−1, t R, _p¡ P tbvy$

ApNm_p¡ õ`i®L$, âpQg t kp\¡, h¾$_¡ afuhpf Q tbvy$A¡dm¡ R>¡ sp¡ Q tbvy$_p„ epd R>¡:(1) (t

2+3, −t

3−1)

(2) (4t2+3, −8t

3−1)

(3) (t2+3, t

3−1)

(4) (16t2+3, −64t

3−1)

71. ÿÁŒ »§‹Ÿ

f(x)=

( )•

••••••••••••••••••••••••••••• •

• •• • • •• •

• • •

• • • • •

<•••

≤ ≤•••

•

• • • •

x=1 ¬⁄U •fl∑§‹ŸËÿ „Ò, ÃÊ •• ∑§Ê ◊ÊŸ „Ò —

(1)

••

π •• •

(2)

••

π • • • •

(3)

••

π • •

(4) −1−cos−1

(2)

72. ÿÁŒ fl∑˝§ x=4t2+3, y=8t

3−1, t R ∑§ Á’¥ŒÈ P,

t ¬˝Êø‹ ∑§ ‚ÊÕ, ¬⁄U S¬‡Ê¸ ⁄UπÊ, fl∑˝§ ∑§Ê ŒÈ’Ê⁄UÊ Á’¥ŒÈQ ¬⁄U Á◊‹ÃË „Ò, ÃÊ Q ∑§ ÁŸŒ¸‡ÊÊ¥∑§ „Ò¥ —(1) (t

2+3, −t

3−1)

(2) (4t2+3, −8t

3−1)

(3) (t2+3, t

3−1)

(4) (16t2+3, −64t

3−1)

71. If the function

f(x)=

( )•

••••••••••••••••••••••••••••• •

• •• • • •• •

• • •

• • • • •

<•••

≤ ≤•••

•

• • • •

is differentiable at x=1, then

•• is equal

to :

(1)

••

π •• •

(2)

••

π • • • •

(3)

••

π • •

(4) −1−cos−1

(2)

72. If the tangent at a point P, with parameter

t, on the curve x=4t2+3, y=8t

3−1, t R,

meets the curve again at a point Q, then the

coordinates of Q are :

(1) (t2+3, −t

3−1)

(2) (4t2+3, −8t

3−1)

(3) (t2+3, t

3−1)

(4) (16t2+3, −64t

3−1)


Set - 04 39

73. h¾$ y=x2−4 D`f_p„ tbvy$Ap¡\u ENdtbvy$_y„ Þe|_Ñd

A„sf L¡$V$gy„ \pe ?

(1)

• ••

(2)

• ••

(3)

• ••

(4)

• ••

74. Å¡

( ) ( )• •

• •• • • •• • ••• • • •••• •

• • • • •• •

• • •

Äep„ k A¡ k„L$g__p¡ AQmp„L$ R>¡, sp¡ A+B+C=

(1)

• ••

(2)

• ••

(3)

•• •

(4)

• •• •

73. fl∑˝§ y=x2−4 ∑§ ∞∑§ Á’¥ŒÈ ‚ ◊Í‹ Á’¥ŒÈ ∑§Ë ãÿÍŸÃ◊

ŒÍ⁄UË „Ò —

(1)

• ••

(2)

• ••

(3)

• ••

(4)

• ••

74. ÿÁŒ

( ) ( )• •

• •• • • •• ••• • • •••• •

• • • • •• •

• • •

„Ò, ¡„Ê° k ‚◊Ê∑§‹Ÿ •ø⁄U „Ò, ÃÊ A+B+C ’⁄UÊ’⁄U„Ò —

(1)

• ••

(2)

• ••

(3)

•• •

(4)

• •• •

73. The minimum distance of a point on the

curve y=x2−4 from the origin is :

(1)

• ••

(2)

• ••

(3)

• ••

(4)

• ••

74. If

( ) ( )• •

• •• • • •• • ••• • • •••• •

• • • • •• •

• • •

where k is a constant of integration, then

A+B+C equals :

(1)

• ••

(2)

• ••

(3)

•• •

(4)

• •• •


Set - 04 40

75. Å¡

• • • • • •• •• • •• •

• • •• • •• • •• •• • • • • • • •• •• •• • •

sp¡ • • •••• • •• ••• • • •• • • • =

(1) log4

(2) •• • ••

+π

(3) log2

(4) •• • ••π •• •

76. âv$¡iA=(x, y)|y ≥ x

2−5x+4, x+y ≥ 1, y ≤ 0 _y„

n¡Óam (Qp¡.A¡L$ddp„) L¡$V$gy„ \i¡ ?

(1)

••

(2)

• ••

(3)

• ••

(4)

• ••

75. ÿÁŒ

• • • • • •• •• • •• •

• • •• • •• • •• •• • • • • • •• •• •= − + „Ò,

ÃÊ

• • •••• • •• ••• • • •• • • • ’⁄UÊ’⁄U „Ò —

(1) log4

(2) •• • ••

+π

(3) log2

(4) •• • ••π •• •

76. A = (x, y)|y ≥ x2−5x+4, x+y ≥ 1, y ≤ 0

mÊ⁄UÊ ÁŸœÊ¸Á⁄UÃ ˇÊòÊ ∑§Ê ˇÊòÊ»§‹ (flª¸ ß∑§ÊßÿÊ¥ ◊¥) „Ò —

(1)

••

(2)

• ••

(3)

• ••

(4)

• ••

75. If • • • • • •• •• • •• •

• • •• • •• • •• •• • • • • • • •• •• •• • •

then

• • •••• • •• ••• • • •• • • • is equal to :

(1) log4

(2) •• • ••

+π

(3) log2

(4) •• • ••π •• •

76. The area (in sq. units) of the region

described by

A=(x, y)|y ≥ x2−5x+4, x+y ≥ 1, y ≤ 0

is :

(1)

••

(2)

• ••

(3)

• ••

(4)

• ••


Set - 04 41

77. Å¡ rh ¡e f (x) A„sfpg (0, ∞) D`f rhL$g_ue lp¡e

s\p f (1) = 1 A_¡ •••• •→

• •• •• • • • • ••• •

• • • • • •• •• ••

âÐe¡L$ x>0 dpV¡$ lp¡e sp¡ ( )••• =

(1)

• ••

(2)

• •• •

(3)

• ••

(4)

• •• •

78. f ¡MpAp ¡ • • • ••• •

•• • • A_¡ • • • ••• •

•• • • _p „

R>¡v$dp„\u kpf \su A¡L$ Qrgs f¡Mp epdpnp¡_¡ tbvy$Ap¡A A_¡ B, (A ≠ B) dp„ dm¡ R>¡. f¡MpM„X$ AB _pdÝetbvy$_p¡ tbvy$`\ L$ep kduL$fZ_y„ kdp^p_ L$fi¡ ?(1) 6xy=7(x+y)

(2) 4(x+y)2−28(x+y)+49=0

(3) 7xy=6(x+y)

(4) 14(x+y)2−97(x+y)+168=0

77. ÿÁŒ f (x), •¥Ã⁄UÊ‹ (0, ∞) ◊¥ ∞∑§ ∞‚Ê •fl∑§‹ŸËÿ»§‹Ÿ „Ò Á∑§ f (1) = 1 ÃÕÊ ¬˝àÿ∑§ x>0 ∑§ Á‹∞,

•••• •→

• •• •• • • • • ••• •

• • • • • •• •• ••

„Ò, ÃÊ ( )••• ’⁄UÊ’⁄U

„Ò —

(1)

• ••

(2)

• •• •

(3)

• ••

(4)

• •• •

78. ÿÁŒ ⁄ UπÊ•Ê ¥ • • • ••• •

•• • • ÃÕÊ • • • ••• •

•• • •

∑§ ¬˝ÁÃë¿UŒŸ ‚ „Ê∑§⁄U ¡ÊŸ flÊ‹Ë ∞∑§ ø⁄U ⁄UπÊ ß‚¬˝∑§Ê⁄U πË¥øË ªß¸ „Ò Á∑§ ÿ„ ÁŸŒ¸‡ÊÊ¥∑§ •ˇÊÊ¥ ∑§ÊA ÃÕÊ B, (A ≠ B) ¬⁄U Á◊‹ÃË „Ò, ÃÊ AB ∑§ ◊äÿÁ’¥ŒÈ∑§Ê Á’¥ŒÈ¬Õ „Ò —(1) 6xy=7(x+y)

(2) 4(x+y)2−28(x+y)+49=0

(3) 7xy=6(x+y)

(4) 14(x+y)2−97(x+y)+168=0

77. If f (x) is a differentiable function in the

interval (0, ∞) such that f (1) = 1 and

•••• •→

• •• •• • • • • ••• •

• • • • • •• •• ••

, for each x>0,

then ( )••• is equal to :

(1)

• ••

(2)

• •• •

(3)

• ••

(4)

• •• •

78. If a variable line drawn through the

intersection of the lines • • • ••• •

•• • • and

• • • ••• •

•• • • , meets the coordinate axes at

A and B, (A ≠ B), then the locus of the

midpoint of AB is :

(1) 6xy=7(x+y)

(2) 4(x+y)2−28(x+y)+49=0

(3) 7xy=6(x+y)

(4) 14(x+y)2−97(x+y)+168=0


Set - 04 42

79. tbvy$ (2, 1) _y„ • • A¡L$d\u f¡Mp L : x−y=4 _¡

kdp„sf õ\p_p„sfZ L$fhpdp„ Aph¡ R>¡. Å¡ Ap _hy„ tbvy$Q ÓuÅ QfZdp„ Aph¡g lp¡e sp¡ Q dp„\u `kpf \suA_¡ L _¡ g„b lp¡e s¡hu f¡Mp_y„ kduL$fZ R>¡:

(1) x+y= • ••

(2) x+y= • • ••

(3) x+y= • • ••

(4) 2x+2y= • ••

80. A¡L$ hsy®m (−2, 4) dp„\u `kpf \pe R>¡ A_¡y-An_¡ tbvy$ (0, 2) ApNm õ`i£ R>¡. _uQ¡_pdp„\uL$ey„ kduL$fZ Ap hsy®m_p„ ìepk_¡ kdphsu f¡Mpv$ip®hi¡ ?(1) 4x+5y−6=0

(2) 2x−3y+10=0

(3) 3x+4y−3=0

(4) 5x+2y+4=0

81. ^pfp¡ L¡$ a A_¡ b A_y¾$d¡ A¡ Arshge_p„ A ® dy¿eAnA_¡ A ® A_ybÝ^ An_p„ dp` R>¡ S>¡_u DÐL¡$ÞÖspkduL$fZ 9e

2−18e+5=0 _y„ kdp^p_ L$f¡ R>¡. Å¡

S(5, 0) A¡ _prc A_¡ A_yê$` r_eprdL$p 5x=9 lp¡esp¡ Ap Arshge dpV¡$ a2

−b2 bfpbf :

(1) 7

(2) −7

(3) 5

(4) −5

79. Á’¥ŒÈ (2, 1) ∑§Ê ⁄UπÊ L : x−y=4 ∑§ ‚◊Ê¥Ã⁄U,

• • ß∑§Êß¸ SÕÊŸÊãÃÁ⁄UÃ Á∑§ÿÊ ªÿÊ– ÿÁŒ ŸÿÊ Á’¥ŒÈ

Q ÃË‚⁄U øÃÈÕÊZ‡Ê ◊¥ ÁSÕÃ „Ò, ÃÊ Á’¥ŒÈ Q ‚ „Ê∑§⁄U ¡ÊŸflÊ‹Ë ÃÕÊ L ∑§ ‹¥’flÃ ⁄UπÊ ∑§Ê ‚◊Ë∑§⁄UáÊ „Ò —

(1) x+y= • ••

(2) x+y= • • ••

(3) x+y= • • ••

(4) 2x+2y= • ••

80. ∞∑§ flÎûÊ Á’¥ŒÈ (−2, 4) ‚ „Ê ∑§⁄U ¡ÊÃÊ „Ò ÃÕÊ y-•ˇÊ∑§Ê (0, 2) ¬⁄U S¬‡Ê¸ ∑§⁄UÃÊ „Ò– ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ ‚Ê ∞∑§‚◊Ë∑§⁄UáÊ ß‚ flÎûÊ ∑§ √ÿÊ‚ ∑§Ê ÁŸM§Á¬Ã ∑§⁄UÃÊ „Ò?(1) 4x+5y−6=0

(2) 2x−3y+10=0

(3) 3x+4y−3=0

(4) 5x+2y+4=0

81. ◊ÊŸÊ a ÃÕÊ b ∑˝§◊‡Ê—, ∞∑§ •ÁÃ¬⁄Ufl‹ÿ Á¡‚∑§Ë©à∑¥§Œ˝ÃÊ ‚◊Ë∑§⁄UáÊ 9e

2−18e+5=0 ∑§Ê ‚¥ÃÈc≈U ∑§⁄UÃË

„Ò, ∑§ •œ¸•ŸÈ¬˝SÕ •ˇÊ ÃÕÊ •œ¸‚¥ÿÈÇ◊Ë •ˇÊ „Ò¥–ÿÁŒ S(5, 0) ß‚ •ÁÃ¬⁄Ufl‹ÿ ∑§Ë ∞∑§ ŸÊÁ÷ ÃÕÊ5x=9 ‚¥ªÃ ÁŸÿãÃÊ (directrix) „Ò, ÃÊ a2

−b2 ’⁄UÊ’⁄U

„Ò —(1) 7

(2) −7

(3) 5

(4) −5

79. The point (2, 1) is translated parallel to the

line L : x−y=4 by • • units. If the new

point Q lies in the third quadrant, then the

equation of the line passing through Q and

perpendicular to L is :

(1) x+y= • ••

(2) x+y= • • ••

(3) x+y= • • ••

(4) 2x+2y= • ••

80. A circle passes through (−2, 4) and touches

the y-axis at (0, 2). Which one of the

following equations can represent a

diameter of this circle ?

(1) 4x+5y−6=0

(2) 2x−3y+10=0

(3) 3x+4y−3=0

(4) 5x+2y+4=0

81. Let a and b respectively be the semi-

transverse and semi-conjugate axes of a

hyperbola whose eccentricity satisfies the

equation 9e2−18e+5=0. If S(5, 0) is a focus

and 5x=9 is the corresponding directrix of

this hyperbola, then a2−b

2 is equal to :

(1) 7

(2) −7

(3) 5

(4) −5


Set - 04 43

82. Å¡ D`hge ••

• • • ••• • •

•• • • _p¡ õ`i®L$, epdpnp¡_¡

A A_¡ B tbvy$dp„ dmsp¡ lp¡s s\p Å¡ O ENdtbvy$lp¡e sp¡ rÓL$p¡Z OAB _y„ gOyÑd n¡Óam (Qp¡.A¡L$ddp„)L¡$V$gy„ \pe ?

(1)

••

(2) • •

(3) • •(4) 9

83. f¡MpAp¡ • • • •• • •

•• •• • A_¡

• ••• •• • ••• • • •• •• • •

•• •−+ −

= =

−

hÃQ¡_y„ gOysd A„sf

L$ep A„sfpgdp„ Aph¡g R>¡ ?(1) [0, 1)

(2) [1, 2)

(3) (2, 3]

(4) (3, 4]

82. ÿÁŒ ŒËÉÊ¸flÎûÊ ••

• • • ••• • •

•• • • ∑§ ∞∑§ Á’¥ŒÈ ¬⁄U πË¥øË

ªß¸ S¬‡Ê¸ ⁄UπÊ, ÁŸŒ¸‡ÊÊ¥∑§ •ˇÊÊ¥ ∑§Ê A ÃÕÊ B ¬⁄U Á◊‹ÃË„Ò ÃÕÊ O ◊Í‹ Á’¥ŒÈ „Ò, ÃÊ ÁòÊ÷È¡ OAB ∑§Ê ãÿÍŸÃ◊ˇÊòÊ»§‹ (flª¸ ß∑§ÊßÿÊ¥ ◊¥) „Ò —

(1)

••

(2) • •

(3) • •(4) 9

83. ⁄UπÊ•Ê¥ • • • •• • •

•• •• • ÃÕÊ

• ••• •• • ••• • • •• •• • •

•• •−+ −

= =

−

∑§ ’Ëø ∑§Ë ãÿÍŸÃ◊

ŒÍ⁄UË, Á¡‚ •¥Ã⁄UÊ‹ ◊¥ „Ò, fl„ „Ò —(1) [0, 1)

(2) [1, 2)

(3) (2, 3]

(4) (3, 4]

82. If the tangent at a point on the ellipse

••• • • ••

• • ••• • • meets the coordinate axes at

A and B, and O is the origin, then the

minimum area (in sq. units) of the triangle

OAB is :

(1)

••

(2) • •

(3) • •(4) 9

83. The shortest distance between the lines

• • • •• • •

•• •• • and

• ••• •• • ••• • • •• •• • •

•• •−+ −

= =

−

lies in the interval :

(1) [0, 1)

(2) [1, 2)

(3) (2, 3]

(4) (3, 4]


Set - 04 44

84. tbvy$ (1, 2, 2) dp„\u `kpf \sp„ A_¡ kdsgp¡x−y+2z=3 A_¡ 2x−2y+z+12=0 _¡ g„blp¡e s¡hp kdsg_y„ tbv$y$ (1, −2, 4) \u A„sf L¡$V$gy„\pe ?

(1) • •(2) 2

(3) •

(4)

••

85. ^pfp¡ L¡$ rÓL$p¡Z ABC dp„ rifp¡tbvy$ A ApNm L$pV$M|Zp¡fQpe R>¡. Å¡ tbvy$ A, B A_¡ C _p„ õ\p_ kqv$ip¡

A_ y¾ $d ¡ • • •i j k i j p k

∧ ∧ ∧ ∧ ∧ ∧

• • • • • • • •• • •• • • A_¡

• •i q j k

∧ ∧ ∧

•• • •• • lp¡e sp¡, tbvy$ (p, q) S>¡ f¡Mp D`f

Aph¡g R>¡ s¡

(1) x-An_¡ kdp„sf R>¡.

(2) y-An_¡ kdp„sf R>¡.

(3) ^_ x-An kp\¡ gOyL$p¡Z b_ph¡ R>¡.

(4) ^_ x-An kp\¡ NyfyL$p¡Z b_ph¡ R>¡.

84. Á’¥ŒÈ (1, −2, 4) ∑§Ë ©‚ ‚◊Ã‹ ‚ ŒÍ⁄UË, ¡Ê Á’¥ŒÈ(1, 2, 2) ‚ „Ê ∑§⁄ U ¡ÊÃÊ „ Ò ÃÕÊ ‚◊Ã‹Ê ¥x−y+2z=3 ÃÕÊ 2x−2y+z+12=0 ∑§ ‹¥’flÃ„Ò, „Ò —

(1) • •(2) 2

(3) •

(4)

••

85. ∞∑§ ÁòÊ÷È¡ ABC, ¡Ê Á∑§ ‡ÊË·¸ A ¬⁄U ‚◊∑§ÊáÊ „Ò, ◊¥A, B ÃÕÊ C ∑ § ÁSÕÁÃ ‚ÁŒ‡Ê ∑ ˝ §◊‡Ê—

• • •i j k i j p k

∧ ∧ ∧ ∧ ∧ ∧

• • • • • • • •• • •• • • ÃÕÊ

• •i q j k

∧ ∧ ∧

•• • •• • „Ò¥, ÃÊ Á’¥ŒÈ (p, q) Á¡‚ ⁄UπÊ ¬⁄U

ÁSÕÃ „Ò, fl„ —

(1) x-•ˇÊ ∑§ ‚◊Ê¥Ã⁄U „Ò–

(2) y-•ˇÊ ∑§ ‚◊Ê¥Ã⁄U „Ò–

(3) x-•ˇÊ ∑§Ë œŸÊà◊∑§ ÁŒ‡ÊÊ ‚ ãÿÍŸ ∑§ÊáÊ ’ŸÊÃË„Ò–

(4) x-•ˇÊ ∑§Ë œŸÊà◊∑§ ÁŒ‡ÊÊ ‚ •Áœ∑§ ∑§ÊáÊ’ŸÊÃË „Ò–

84. The distance of the point (1, −2, 4) from

the plane passing through the point

(1, 2, 2) and perpendicular to the planes

x−y+2z=3 and 2x−2y+z+12=0, is :

(1) • •(2) 2

(3) •

(4)

••

85. In a triangle ABC, right angled at the vertex

A, if the position vectors of A, B and C are

respectively • • •i j k i j p k

∧ ∧ ∧ ∧ ∧ ∧

• • • • • • • •• • •• • •

and • • •i q j k

∧ ∧ ∧

•• • •• • then the point (p, q) lies

on a line :

(1) parallel to x-axis.

(2) parallel to y-axis.

(3) making an acute angle with the

positive direction of x-axis.

(4) making an obtuse angle with the

positive direction of x-axis.


Set - 04 45

86. If the mean deviation of the numbers

1, 1+d, ..., 1+100d from their mean is 255,

then a value of d is :

(1) 10.1

(2) 20.2

(3) 10

(4) 5.05

87. If A and B are any two events such that

P(A)=•• and

•• •• • •• • • •∩ • , then the

conditional probability, P(A(A∪B)),

where A denotes the complement of A, is

equal to :

(1)••

(2)•• •

(3)•• •

(4)• •• •

88. The number of x [0, 2π] for which

• • • •

• • • •• ••• • • •• • • •• • • • •••• • • •• •• • •

=1 is :

(1) 2

(2) 4

(3) 6

(4) 8

86. Å¡ k„¿epAp¡ 1, 1+d, ..., 1+100d _y„ dÝeL$ \udÝeL$-rhQg_ (kf¡fpi rhQg_) 255 lp¡e sp¡ d _uqL„$ds L¡$V$gu \pe ?(1) 10.1

(2) 20.2

(3) 10

(4) 5.05

87. Å¡ OV$_pAp¡ A A_¡ B dpV¡$ P(A)=•• A_¡

•• •• • •• • • •∩ • lp ¡e sp ¡ ifsu k „cph_p

P(A(A∪B)) L¡$V$gu \pe ? Al] A A¡ A _p¡ |fL$v$ip®h¡ R>¡ ?

(1)••

(2)•• •

(3)•• •

(4)• •• •

88. x [0, 2π] A_¡

• • • •

• • • •• ••• • • •• • • •• • • • •••• • • •• •• • •

=1 lp¡e s¡hp L¡$V$gp x dmi¡ ?(1) 2

(2) 4

(3) 6

(4) 8

86. ÿÁŒ ‚¥ÅÿÊ•Ê¥ 1, 1+d, ..., 1+100d ∑§ ◊Êäÿ ‚◊Êäÿ-Áflø‹Ÿ 255 „Ò, ÃÊ d ∑§Ê ∞∑§ ◊ÊŸ „Ò —(1) 10.1

(2) 20.2

(3) 10

(4) 5.05

87. ÿÁŒ A ÃÕÊ B ŒÊ ∞‚Ë ÉÊ≈UŸÊ∞° „Ò¥ Á∑§ P(A)=••

ÃÕÊ •• •• • •• • • •∩ • „Ò, ÃÊ ¬˝ÁÃ’¥ÁœÃ ¬˝ÊÁÿ∑§ÃÊ

P(A(A∪B)), ¡„Ê° A, A ∑§ ¬Í⁄U∑§ ‚◊ÈìÊÿ ∑§ÊÁŸÁŒ¸c≈U ∑§⁄UÃÊ „Ò, ’⁄UÊ’⁄U „Ò —

(1)••

(2)•• •

(3)•• •

(4)• •• •

88. x [0, 2π] ∑§Ë ‚¥ÅÿÊ, Á¡Ÿ∑§ Á‹∞

• • • •

• • • •• ••• • • •• • • •• • • • •••• • • •• •• • •

=1, „Ò —(1) 2

(2) 4

(3) 6

(4) 8


Set - 04 46

89. If m and M are the minimum and the

maximum values of

• ••• • • ••• • • •• •• • ••

• •• • x R, then

M−m is equal to :

(1)

• ••

(2)

••

(3)

••

(4)

••

90. Consider the following two statements :

P : If 7 is an odd number, then 7 is

divisible by 2.

Q : If 7 is a prime number, then 7 is an

odd number.

If V1 is the truth value of the contrapositive

of P and V2

is the truth value of

contrapositive of Q, then the ordered pair

(V1, V

2) equals :

(1) (T, T)

(2) (T, F)

(3) (F, T)

(4) (F, F)

- o O o -

89. Å¡ m A_¡ M A¡ • ••• • • ••• • • •• •• • ••

• •• •

x R _p„ Þe|_sd A_¡ dlÑd d|ëep¡ lp¡e sp¡ M−m

_u qL„$ds L¡$V$gu \i¡ ?

(1)

• ••

(2)

••

(3)

••

(4)

••

90. _uQ¡_p b¡ rh^p_p¡ Ap ¡g R>¡ :

P : Å¡ 7 A¡ AeyÁd k„¿ep lp¡e sp¡ 7 _¡ 2 hX¡$r_:i¡j cpNu iL$pe R>¡.

Q : Å¡ 7 A¡ ArhcpÄe k„¿ep lp¡e sp¡ 7 AeyÁdR>¡.

Å¡ V1 A¡ P dp„ kdp_p\ â¡fZ (contrapositive)

_y„ kÐep\®sp d|ëe A_¡ V2 A¡ Q _p kdp_p\ â¡fZ_p„

kÐep\®sp d|ëe lp¡e sp¡ ¾$dey¼s Å¡X$ (V1, V

2) bfpbf:

(1) (T, T)

(2) (T, F)

(3) (F, T)

(4) (F, F)

- o O o -

89. ÿÁŒ m ÃÕÊ M, √ÿ¥¡∑§

• ••• • • ••• • • •• •• • ••

• •• • x R ∑§ ∑ ˝§◊‡Ê—

ãÿÍŸÃ◊ ÃÕÊ •Áœ∑§Ã◊ ◊ÊŸ „Ò¥, ÃÊ M−m ’⁄UÊ’⁄U„Ò —

(1)

• ••

(2)

••

(3)

••

(4)

••

90. ÁŸêŸ ŒÊ ∑§ÕŸÊ¥ ¬⁄U ÁfløÊ⁄U ∑§ËÁ¡∞ —

P : ÿÁŒ 7 ∞∑§ Áfl·◊ ‚¥ÅÿÊ „Ò, ÃÊ 7, 2 ‚ ÷Êíÿ„Ò–

Q : ÿÁŒ 7 ∞∑§ •÷Êíÿ ‚¥ÅÿÊ „Ò, ÃÊ 7 ∞∑§ Áfl·◊‚¥ÅÿÊ „Ò–

ÿÁŒ V1, P ∑§ ¬˝ÁÃœŸÊà◊∑§ ∑§Ê ‚àÿ◊ÊŸ „Ò ÃÕÊ

V2, Q ∑§ ¬˝ÁÃœŸÊà◊∑§ ∑§Ê ‚àÿ◊ÊŸ „Ò, ÃÊ ∑˝§Á◊Ã ÿÈÇ◊

(V1, V

2) ’⁄UÊ’⁄U „Ò —

(1) (T, T)

(2) (T, F)

(3) (F, T)

(4) (F, F)

- o O o -

Question and Answer Key - April 9 OnlineQuestion No. Answer Key Question No. Answer Key Question No. Answer Key



Note:-

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JEE Main 2016 Answer Key Paper 1 Online (April 9,2016)


Set - 03 1

1. A, B, C and D are four different physical

quantities having different dimensions.

None of them is dimensionless. But we

know that the equation AD=C ln(BD)

holds true. Then which of the combination

is not a meaningful quantity ?

(1) A2

− B2C

2

(2)

• • • •• •••

(3) • •• •••

(4)

•• •• • •

• • ••

2. A particle of mass M is moving in a circle

of fixed radius R in such a way that its

centripetal acceleration at time t is given

by n2

R t2 where n is a constant. The power

delivered to the particle by the force acting

on it, is :

(1) M n2

R2

t

(2) M n R

2 t

(3) M n R2

t2

(4)

•• M n

2 R

2 t2

1. A, B, C ÃÕÊ D øÊ⁄U Á÷ÛÊ ◊ÊòÊÊ∞° „Ò¥ Á¡Ÿ∑§Ë Áfl◊Ê∞¥Á÷ãŸ „Ò¥– ∑§Êß¸ ÷Ë ◊ÊòÊÊ Áfl◊Ê-⁄UÁ„Ã ◊ÊòÊÊ Ÿ„Ë¥ „Ò¥,‹Á∑§Ÿ AD=C ln(BD) ‚àÿ „Ò– Ã’ ÁŸêŸ ◊¥ ‚∑§ÊÒŸ •Ê‡Êÿ-⁄UÁ„Ã ◊ÊòÊÊ „Ò?(1) A

2 − B

2C

2

(2)

• • • •• •••

(3) • •• •••

(4)

•• •• • •

• • ••

2. Œ˝√ÿ◊ÊŸ M ∑§Ê ∞∑§ ∑§áÊ ÁŸÁ‡øÃ ÁòÊíÿÊ R ∑§ flÎûÊËÿ¬Õ ¬⁄U-ß‚ ¬∑§Ê⁄U ø‹ ⁄U„Ê „Ò Á∑§ ‚◊ÿ ‘t’ ¬⁄U •Á÷∑§ãŒËàfl⁄UáÊ n2

R t2 mÊ⁄UÊ ÁŒÿÊ ¡Ê ‚∑§ÃÊ „Ò, ÿ„Ê° ‘n’ •ø⁄U

„Ò– Ã’ ∑§áÊ ¬⁄U ‹ª ⁄U„ ’‹ mÊ⁄UÊ ©‚∑§Ê ŒË ªß¸ ‡ÊÁÄÃ„Ò —(1) M n

2 R

2 t

(2) M n R

2 t

(3) M n R2

t2

(4)

•• M n

2 R

2 t2

1. A, B, C A_¡ D A¡ Qpf Sy>v$p-Sy>v$p `qfdpZ ^fphsuSy>v$u-Sy>v$u cp¥rsL$fpriAp¡ R>¡. s¡dp„_u L$p¡C`Z `qfdpZfrls _\u. Å¡ AD=C ln(BD) kduL$fZ kpQy„ lp¡esp¡ _uQ¡ Ap ¡g ¥L$u L$ey„ k„ep¡S>_ A¡ A\®kcf (kpQu)fpri v$ip®hsy„ _\u ?(1) A

2 − B

2C

2

(2)

• • • •• •••

(3) • •• •••

(4)

•• •• • •

• • ••

2. M v$m ^fphsp¡ L$Z R S>¡V$gu AQm rÓÄep ^fphsphsy®mpL$pf dpN® f A¡hu fus¡ Nrs L$f¡ R>¡ L¡$ t kde¡ s¡_p¡L¡$ÞÖNpdu âh¡N n2

R t2 hX¡$ Ap`u iL$pe; Äep„ n A¡

AQmp„L$ R>¡. sp¡ L$Z `f gpNsp bm hX¡$ L$Z_¡ dmsp¡`phf (L$pe®v$nsp) __________ \i¡.(1) M n

2 R

2 t

(2) M n R

2 t

(3) M n R2

t2

(4)

•• M n

2 R

2 t2

JEE Main 2016 Question Paper 1 Online (April 10, 2016)JEE Main 2016 Question Paper 1 Online (April 10, 2016)JEE Main 2016 Question Paper 1 Online (April 10, 2016)

Rohit

Typewritten Text



Set - 03 2

3. Concrete mixture is made by mixing

cement, stone and sand in a rotating

cylindrical drum. If the drum rotates too

fast, the ingredients remain stuck to the

wall of the drum and proper mixing of

ingredients does not take place. The

maximum rotational speed of the drum in

revolutions per minute(rpm) to ensure

proper mixing is close to :

(Take the radius of the drum to be 1.25 m

and its axle to be horizontal) :

(1) 0.4

(2) 1.3

(3) 8.0

(4) 27.0

3. ∑¥§∑˝§Ë≈U Á◊Ä‚ø⁄U ’ŸÊŸ ∑§ Á‹ÿ ‚Ë◊¥≈U, ⁄UÃ ÃÕÊ ⁄UÊ«∏Ë∑§Ê ∞∑§ ÉÊÍáÊË¸ÿ ’‹ŸÊ∑§Ê⁄U «˛U◊ ◊¥ «UÊ‹Ê ¡ÊÃÊ „Ò– ÿÁŒ«˛U◊ ∑§Ë ÉÊÍáÊ¸Ÿ-ªÁÃ ’„ÈÃ Ã¡ „Ê ÃÊ ‚¥ÉÊ≈U∑§ «˛U◊ ∑§ËŒËflÊ⁄U ‚ Áø¬∑§ ⁄U„Ã „Ò¥ •ÊÒ⁄U Á◊Ä‚ø⁄U ∆UË∑§ ‚ Ÿ„Ë¥’ŸÃÊ– ÿÁŒ «˛U◊ ∑§Ë ÁòÊíÿÊ 1.25 m „Ò •ÊÒ⁄U ß‚∑§ËœÈ⁄UË ˇÊÒÁÃ¡ „Ò, Ã’ •ë¿UË Ã⁄U„ Á◊Ä‚ „ÊŸ ∑§ Á‹ÿ¡M§⁄UË •Áœ∑§Ã◊ ÉÊÍáÊË¸ÿ-ªÁÃ rpm ◊¥ „Ò —(1) 0.4

(2) 1.3

(3) 8.0

(4) 27.0

3. A¡L$ QpL$Nrs L$fsp _mpL$pfue X²$ddp„ rkd¡ÞV$, `Õ\fA_¡ f¡su_¡ c¡Np L$fu_¡ L$p¢q¾$V$ rdîZ s¥epf L$fhpdp„Aph¡ R>¡. lh¡ Å¡ X²$d M|b S> TX$`\u QpL$Nrs L$f¡ sp¡rdîZ L$f¡gp sÒhp¡ X²$d_u qv$hpg_¡ Qp¢V$u Åe R>¡ A_¡s¡d_y„ ep¡Áe rdîZ b_phu iL$psy„ _\u. sp¡ `qfc°dZârs rd_uV$ (rpm) _p `v$dp„, ep¡Áe rdîZ b_hpdpV¡$_u X²$d_u dlÑd L$p¡Zue TX$` __________ _u_ÆL$_u li¡. (X²$d_u rÓÄep 1.25 m A_¡ s¡_u AnkdrnrsS> R>¡ s¡d ^pfp¡).(1) 0.4

(2) 1.3

(3) 8.0

(4) 27.0


Set - 03 3

4. Velocity-time graph for a body of mass

10 kg is shown in figure. Work-done on

the body in first two seconds of the motion

is :

(1) 12000 J

(2) −12000 J

(3) −4500 J

(4) −9300 J

5. In the figure shown ABC is a uniform wire.

If centre of mass of wire lies vertically below

point A, then

• •• •

is close to :

(1) 1.85

(2) 1.37

(3) 1.5

(4) 3

4. 10 kg Œ˝√ÿ◊ÊŸ ∑§ Á¬¥«U ∑§ Á‹ÿ flª-‚◊ÿ ª˝Ê»§ ÁøòÊ◊¥ ÁŒÿÊ „Ò– Á¬¥«U ¬⁄U ¬„‹ 2 ‚. ◊¥ Á∑§ÿÊ ªÿÊ ∑§Êÿ¸„Ò —

(1) 12000 J

(2) −12000 J

(3) −4500 J

(4) −9300 J

5. ÁŒÿ ªÿ ÁøòÊ ◊¥ ÃÊ⁄U ABC ∞∑§ ‚◊ÊŸ „Ò– ÿÁŒ‚¥„ÁÃ-∑§ãŒ˝ Á’¥ŒÈ A ∑ ™§äflÊ¸œ⁄U ŸËø ÁSÕÃ „Ò, Ã’

• •• •

‹ª÷ª „Ò —

(1) 1.85

(2) 1.37

(3) 1.5

(4) 3

4. 10 kg v$m fphsp v$p\® dpV¡$_p¡ h¡N-kde_p¡ Apg¡MApL©$rÑdp„ v$ip®h¡g R>¡. v$p\® f â\d 2 k¡L$ÞX$dp„ \suNrs v$fçep_ \sy„ L$pe® __________\i¡.

(1) 12000 J

(2) −12000 J

(3) −4500 J

(4) −9300 J

5. ApL©$rÑdp„ v$ip®ìep dyS>b ABC A¡ kdp_ spf R>¡. Å¡spf_y„ Öìedp_ L¡$ÞÖ tbvy$ A _u bfp¡bf _uQ¡ Aphsy„

lp¡e sp¡ • •• •

_y„ d|ëe _________ _u _ÆL$_y„ \i¡.

(1) 1.85

(2) 1.37

(3) 1.5

(4) 3


Set - 03 4

6. An astronaut of mass m is working on a

satellite orbiting the earth at a distance h

from the earth’s surface. The radius of the

earth is R, while its mass is M. The

gravitational pull FG

on the astronaut is :

(1) Zero since astronaut feels weightless

(2)• •• • •• • ••

• • •••

(3) •• • •• •• •• • • ••

• • • • • ••• • ••

(4)• • •• • • •

•• • ••• •

••

6. ¬ÎâflË ∑§Ë ‚Ã„ ‚ ‘h’ ŒÍ⁄UË ¬⁄U ÁSÕÃ ∞∑§ ©¬ª˝„ ¬⁄U∞∑§ ‘m’ Œ˝√ÿ◊ÊŸ ∑§Ê •¥ÃÁ⁄UˇÊ-ÿÊòÊË ∑§Ê◊ ∑§⁄U ⁄U„Ê „Ò–¬ÎâflË ∑§Ê Œ˝√ÿ◊ÊŸ ‘M’ ÃÕÊ ÁòÊíÿÊ ‘R’ „Ò– Ã’ ©‚ÿÊòÊË ¬⁄U ‹ª ⁄U„Ê ªÈL§àflËÿ ’‹ F

G „Ò —

(1) ‡ÊÍãÿ, ÄÿÊ¥Á∑§ fl„ ÿÊòÊË ÷Ê⁄U„ËŸÃÊ ◊„‚Í‚ ∑§⁄UÃÊ„Ò–

(2)• •• • •• • ••

• • •••

(3) •• • •• •• •• • • ••

• • • • • ••• • ••

(4)• • •• • • •

•• • ••• •

••

6. `©Õhu_u k`pV$u\u h KQpC A¡ `qfc°dZ L$fspk¡V¡gpCV$dp„ m v$m ^fphsp¡ MNp¡mipõÓu L$pe® L$f¡ R>¡.©Õhu_u rÓÄep R R>¡ Äepf¡ s¡_y„ v$m M R>¡. MNp¡mipõÓu

`f gpNsy„ NyfyÐhpL$j M¢QpZ FG

__________ li¡.

(1) i|Þe, L$pfZ L¡$ MNp¡mipõÓu hS>_frls[õ\rsdp„ R>¡.

(2)• •• • •• • ••

• • •••

(3) •• • •• •• •• • • ••

• • • • • ••• • ••

(4)• • •• • • •

•• • ••• •

••


Set - 03 5

7. A bottle has an opening of radius a and

length b. A cork of length b and radius

(a+∆a) where (∆a<<a) is compressed to fit

into the opening completely (See figure).

If the bulk modulus of cork is B and

frictional coefficient between the bottle and

cork is µ then the force needed to push the

cork into the bottle is :

(1) (πµB b) ∆a

(2) (2πµB b) ∆a

(3) (πµB b) a

(4) (4πµB b) ∆a

7. ∞∑§ ’ÊÃ‹ ∑§ ◊È°„ ∑§Ë ÁòÊíÿÊ ‘a’ „ÒÒ ÃÕÊ ‹ê’Êß¸ ‘b’ „Ò–∞∑§ ‘b’ ‹ê’Êß¸ •ÊÒ⁄U (a+∆a) ÁòÊíÿÊ (∆a<<a) flÊ‹∑§Ê∑¸§ ∑§Ê ©‚∑§ ◊È°„ ◊¥ ¬Í⁄UË Ã⁄U„ ∆Í°U‚ ÁŒÿÊ ªÿÊ „Ò (ÁøòÊŒÁπÿ)– ÿÁŒ ∑§Ê∑¸§ ∑§Ê •ÊÿÃŸ ¬˝àÿÊSÕÃÊ ªÈáÊÊ¥∑§ B„Ò ÃÕÊ ’ÊÃ‹ •ÊÒ⁄U ∑§Ê∑¸§ ∑§ ’Ëø ÉÊ·¸áÊ-ªÈáÊÊ¥∑§ µ „Ò,Ã’ ∑§Ê∑§ ∑§Ê ◊È°„ ◊¥ ÉÊÈ‚ÊŸ ∑§ Á‹ÿ •Êfl‡ÿ∑§ ’‹ „Ò —

(1) (πµB b) ∆a

(2) (2πµB b) ∆a

(3) (πµB b) a

(4) (4πµB b) ∆a

7. A¡L$ bp¡V$g_p Nmp_u rÓÄep a A_¡ g„bpC b R>¡.ApL©$rÑdp„ v$ip®ìep dyS>b, b g„bpC_p A_¡ (a+∆a)

Äep„ (∆a<<a) _u rÓÄep ^fphsp b|Q_¡ v$bpZ |h®L$hX¡$ bp¡V$g_p Nmphpmp cpN_¡ k„ |Z® b„ L$fhpdp„ Aph¡R>¡. Å¡ b|Q_p¡ L$v$ [õ\rsõ\p`L$spA„L$ B lp¡e A_¡bp¡V$g A_¡ b|Q hÃQ¡_p¡ OjZp¯L$ µ lp¡e sp¡ b|Q_¡bp¡V$gdp„ O|kpX$hp S>ê$fu bm ________ \i¡.

(1) (πµB b) ∆a

(2) (2πµB b) ∆a

(3) (πµB b) a

(4) (4πµB b) ∆a


Set - 03 6

8. A Carnot freezer takes heat from water at

0C inside it and rejects it to the room at a

temperature of 27C. The latent heat of ice

is 336×103

J kg−1

. If 5 kg of water at 0C is

converted into ice at 0C by the freezer, then

the energy consumed by the freezer is close

to :

(1) 1.67×105

J

(2) 1.68×106

J

(3) 1.51×105

J

(4) 1.71×107

J

8. ∞∑§ ∑§ÊŸÊ≈U »˝§Ë¡⁄U •¬Ÿ •¥Œ⁄U 0C ¬⁄U ⁄Uπ „È∞ ¡‹ ‚™§c◊Ê ‹∑§⁄U ©‚ ∑§◊⁄U ∑§ ÃÊ¬◊ÊŸ 27C ¬⁄U ÁŸc∑§ÊÁ‚Ã∑§⁄UÃÊ „Ò– ’»¸§ ∑§Ë ªÈåÃ ™§c◊Ê 336×10

3 J kg

−1 „Ò–ÿÁŒ »˝§Ë¡⁄U ◊¥ ⁄UπÊ 0C ¬⁄U 5 kg ¡‹, 0C ¬⁄U ’»¸§ ◊¥’Œ‹ÃÊ „Ò Ã’ »˝§Ë¡⁄U mÊ⁄UÊ π¬ÊßZ ªß¸ ™§¡Ê¸ ‹ª÷ª „Ò —(1) 1.67×10

5 J

(2) 1.68×106

J

(3) 1.51×105

J

(4) 1.71×107

J

8. A¡L$ L$p_p£V$ f¡qäS>f¡V$f `pZudp„\u 0C sp`dp_¡ Dódpip¡ju 27C sp`dp_¡ fl¡gp hpsphfZdp„ a¢L¡$ R>¡. bfa_ug¡V¡$ÞV$ (Nyá) Dódp 336×10

3 J kg

−1 R>¡. Å¡f¡qäS>f¡V$fdp„ 0C _y„ 5 kg pZu 0C bfadp„ ê$`p„sfusL$fhy„ lp¡e sp¡ f¡qäS>f¡V$f¡ hp`f¡g EÅ® __________ _u_ÆL$_y„ d|ëe ^fphi¡.(1) 1.67×10

5 J

(2) 1.68×106

J

(3) 1.51×105

J

(4) 1.71×107

J


Set - 03 7

9. Which of the following shows the correct

relationship between the pressure ‘P’ and

density ρ of an ideal gas at constant

temperature ?

(1)

(2)

(3)

(4)

9. Á∑§‚Ë •ÊŒ‡Ê¸ ªÒ‚ ∑§ Á‹ÿ ÁSÕ⁄U ÃÊ¬◊ÊŸ ¬⁄U ©‚∑§ ŒÊ’‘P’ ÃÕÊ ÉÊŸàfl ‘ρ’ ∑§ ’Ëø ‚¥’¥œ ∑§ Á‹ÿ ÁŸêŸ ◊¥ ‚∑§ÊÒŸ-‚Ê ÁøòÊ ‚„Ë „Ò?

(1)

(2)

(3)

(4)

9. _uQ¡ Ap ¡g ApL©$rsAp¡ ¥L$u L$C ApL©$rÑ AQm sp`dp_¡fl¡g Apv$i® hpey dpV¡$ v$bpZ ‘P’ A_¡ O_sp ρ hÃQ¡_p¡kpQp¡ k„b„ v$ip®h¡ R>¡.

(1)

(2)

(3)

(4)


Set - 03 8

10. In an engine the piston undergoes vertical

simple harmonic motion with amplitude

7 cm. A washer rests on top of the piston

and moves with it. The motor speed is

slowly increased. The frequency of the

piston at which the washer no longer stays

in contact with the piston, is close to :

(1) 0.1 Hz

(2) 1.2 Hz

(3) 0.7 Hz

(4) 1.9 Hz

11. A toy-car, blowing its horn, is moving with

a steady speed of 5 m/s, away from a wall.

An observer, towards whom the toy car is

moving, is able to hear 5 beats per second.

If the velocity of sound in air is 340 m/s,

the frequency of the horn of the toy car is

close to :

(1) 680 Hz

(2) 510 Hz

(3) 340 Hz

(4) 170 Hz

10. ∞∑§ ß¥¡Ÿ ∑§Ê Á¬S≈UŸ 7 cm •ÊÿÊ◊ ∑§Ë ‚⁄U‹-•ÊflÃ¸-ªÁÃ ™§äflÊ¸œ⁄U ◊¥ ∑§⁄U ⁄U„Ê „Ò– Á¬S≈UŸ ∑§ ™§¬⁄U ∞∑§flÊ‡Ê⁄U ⁄UπÊ „Ò ¡Ê ©‚∑§ ‚ÊÕ ø‹ÃÊ „Ò– ◊Ê≈U⁄U ∑§Ë ªÁÃœË⁄U-œË⁄U ’…∏Êß¸ ¡ÊÃË „Ò ÃÊ Á¬S≈UŸ ∑§Ë •ÊflÎÁûÊ Á¡‚ ¬⁄UflÊ‡Ê⁄U Á¬S≈UŸ ∑§Ê ‚ÊÕ ¿UÊ«∏ ŒÃÊ „Ò, fl„ ‹ª÷ª „Ò —(1) 0.1 Hz

(2) 1.2 Hz

(3) 0.7 Hz

(4) 1.9 Hz

11. ∞∑§ Áπ‹ÊÒŸÊ ∑§Ê⁄U „ÊŸ¸ ’¡ÊÃË „Èß¸ 5 m/s ∑§Ë ÁSÕ⁄UªÁÃ ‚ ∞∑§ ŒËflÊ⁄U ‚ ŒÍ⁄U ÃÕÊ ∞∑§ √ÿÁÄÃ ∑§Ë •Ê⁄U ¡Ê⁄U„Ë „Ò– ©‚ √ÿÁÄÃ ∑§Ê 5 ’Ë≈U/‚¥. ‚ÈŸÊß¸ ŒÃË „Ò¥– ÿÁŒ„flÊ ◊¥ äflÁŸ ∑§Ë ªÁÃ 340 m/s „Ò, Ã’ „ÊŸ¸ ∑§Ë•ÊflÎÁûÊ ‹ª÷ª „Ò —

(1) 680 Hz

(2) 510 Hz

(3) 340 Hz

(4) 170 Hz

10. A¡L$ A¡[ÞS>_dp„ r`õV$_ DÝh® qv$ipdp„ 7 cm _pL„$`rhõspf kp\¡ SHM (k.Ap.N) L$f¡ R>¡. r`õV$__uD`f fl¡g hp¸if s¡_u kp\¡ S> Nrs L$f¡ R>¡. r`õV$__¡Nrs L$fphsu dp¡V$f_u TX$` ud¡\u h^pfhpdp„ Aph¡ R>¡.hp¸if r`õV$__u k`pV$u_¡ R>p¡X$u v$¡ s¡ r`õV$__u S>ê$fuAph©rÑ __________ _u _ÆL$_y„ d|ëe li¡.(1) 0.1 Hz

(2) 1.2 Hz

(3) 0.7 Hz

(4) 1.9 Hz

11. A¡L$ lp¡_® hNpX$su fdL$X$p„_u L$pf qv$hpg \u v|$f sfa5 m/s _u AQm TX$`\u Nrs L$f ¡ R> ¡. A¡L$Ahgp¡L$_L$pf, L¡$ S>¡_u sfa fdL$X$p„_u L$pf Nrs L$f¡ R>¡,A¡L$ k¡L$ÞX$dp„ 5 õ „v$ (beats) kp„cm¡ R>¡. Å¡ lhpdp„Ýhr__p¡ h¡N 340 m/s lp¡e sp¡ L$pf_p lp¡_®_u Aph©qÑ_______ _u _ÆL$_u li¡.(1) 680 Hz

(2) 510 Hz

(3) 340 Hz

(4) 170 Hz


Set - 03 9

12. Within a spherical charge distribution of

charge density ρ(r), N equipotential

surfaces of potential V0, V

0+∆V, V

0+2∆V,

.......... V0+N∆V (∆V>0), are drawn and

have increasing radii

r0, r

1, r

2,..........r

N, respectively. If the

difference in the radii of the surfaces is

constant for all values of V0 and ∆V then :

(1) ρ (r) α r

(2) ρ (r) = constant

(3) ( )•• • •••

• •

(4) ( ) ••• • •••

• •

12. •Êfl‡Ê-ÉÊŸàfl ρ (r) ∑§ Á∑§‚Ë ªÊ‹Ëÿ-•Êfl‡Ê-ÁflÃ⁄UáÊ,∑§ •ãŒ⁄U N ‚◊Áfl÷fl-¬Îc∆U, Á¡Ÿ∑§Ë Áfl÷fl „ÒV

0, V

0+∆V, V

0+2∆V,

........ V

0+N∆V

(∆V>0), •Ê⁄UÁπÃ Á∑§ÿ ªÿ „Ò¥ •ÊÒ⁄U ©Ÿ∑§Ë ÁòÊíÿÊ∞¥∑˝§◊‡Ê— r

0, r

1, r

2,..........r

N „Ò¥– ÿÁŒ ÁòÊíÿÊ•Ê¥ ∑§Ê

•ãÃ⁄UÊ‹, ‚÷Ë V0 ÃÕÊ ∆V ∑§ ◊ÊŸÊ¥ ∑§ Á‹ÿ, ÁSÕ⁄U „Ò

Ã’ —(1) ρ (r) α r

(2) ρ (r)=•ø⁄U

(3) ( )•• • •••

• •

(4) ( ) ••• • •••

• •

12. ρ(r) S>¡V$gu Np¡gue rhÛyscpf rhsfZ dpV¡$, A_y¾$d¡r

0, r

1, r

2,..........r

N rÓÄep ^fphsu A_¡

V0, V

0+∆V, V

0+2∆V, .........., V

0+N∆V

(∆V>0) S>¡V$gy„ [õ\rsdp_ ^fphsu N kd[õ\rsdp_k`pV$u ( ©›$) v$p¡fhpdp„ Aph¡ R>¡. V

0 A_¡ ∆V _p

b^pS> d|ëe dpV¡$ Å¡ k`pV$uAp¡_u rÓÄep hÃQ¡_p¡ saphsAQm fl¡sp¡ lp¡e sp¡ ________.

(1) ρ (r) α r

(2) ρ (r) = AQm

(3) ( )•• • •••

• •

(4) ( ) ••• • •••

• •


Set - 03 10

13. Figure shows a network of capacitors where

the numbers indicates capacitances in

micro Farad. The value of capacitance C if

the equivalent capacitance between point

A and B is to be 1 µF is :

(1)

• • • •• •

•

(2)• • • •• •

•

(3)• • • •• •

•

(4)

• • • •• •

•

13. ÁøòÊ ‚¥œÊÁ⁄UòÊÊ¥ ∑§Ê ÁŸ∑§Êÿ Œ‡ÊÊ¸ÃÊ „Ò, ¡„Ê° •¥∑§ µF ◊¥œÊÁ⁄UÃÊ Œ‡ÊÊ¸Ã „Ò¥– A fl B ∑§ ’Ëø ¬˝÷ÊflË œÊÁ⁄UÃÊ1 µF „ÊŸ ∑§ Á‹ÿ C ∑§Ë œÊÁ⁄UÃÊ „ÊŸË øÊÁ„ÿ —

(1)

• • • •• •

•

(2)• • • •• •

•

(3)• • • •• •

•

(4)

• • • •• •

•

13. ApL©$rÑdp„ k„OpfL$p¡_p¡ b_¡gy„ _¡V$hL®$ v$ip®h¡ R>¡ Äep„L¡ $`¡kuV$f_u bpSy>dp„ gM¡g _„bf s¡ L¡ $`¡kuV$f_y „dpC¾$p¡a¡fpX$dp„ d|ëe v$ip®h¡ R>¡. Å¡ A A_¡ B hÃQ¡_y„`qfZpdu L¡$`¡kuV$Þk (k„OpfL$sp) 1 µF Å¡Csy„ lp¡e sp¡k„OpfL$ C _y„ d|ëe ________ Å¡Ci¡.

(1)

• • • •• •

•

(2)• • • •• •

•

(3)• • • •• •

•

(4)

• • • •• •

•


Set - 03 11

14. The resistance of an electrical toaster has a

temperature dependence given by

R(T)=R0

[1+α(T−T0)] in its range of

operation. At T0=300 K, R=100 Ω and at

T=500 K, R=120 Ω. The toaster is

connected to a voltage source at 200 V and

its temperature is raised at a constant rate

from 300 to 500 K in 30 s. The total work

done in raising the temperature is :

(1)

• •• • • • • ••• •••••

(2)

•• • • • • •••

••

(3)

•• • • • • •••

••

(4) 300 J

14. Á’¡‹Ë ‚ ø‹Ÿ flÊ‹ ≈UÊS≈U⁄U ∑§ ¬˝ÁÃ⁄UÊœ ∑§Ê ÃÊ¬◊ÊŸ ‚’Œ‹Êfl R(T)=R

0 [1+α(T−T

0)] mÊ⁄UÊ ÁŒÿÊ ªÿÊ

„ Ò– T0=300 K ¬⁄ U R=100 Ω „ Ò ÃÕÊ

T=500 K ¬⁄U R=120 Ω „Ò– ≈UÊS≈U⁄U 200 VU ∑§dÊÃ ‚ ¡È«∏Ê „Ò, ÃÕÊ ©‚∑§Ê ÃÊ¬◊ÊŸ 300 K ‚ ∞∑§‚◊ÊŸ Œ⁄U ¬⁄U ’…∏∑§⁄U 30 s ◊¥ 500 K „Ê ¡ÊÃÊ „Ò– Ã’ß‚ ¬˝∑˝§◊ ◊¥ Á∑§ÿÊ ªÿÊ ∑È§‹ ∑§Êÿ¸ „Ò —

(1)

• •• • • • • ••• •••••

(2)

•• • • • • •••

••

(3)

•• • • • • •••

••

(4) 300 J

14. A¡L$ Cg¡¼V²$uL$ V$p¡õV$f_p¡ s¡_p D`ep¡N v$fçep_, sp`dp_kp\¡_p¡ k„b„^ R(T)=R

0 [1+α(T−T

0)] hX¡$

Ap`hpdp „ Aph ¡ R> ¡ . T0=300 K dpV ¡ $

R=100 Ω A_¡ T=500 K dpV¡$ R=120 Ω R>¡.V$p¡õV$f_¡ A¡L$ 200 V _p hp¡ëV¡$S> Dv¹$Nd kp\¡ Å¡X$u s¡_y„sp`dp_ AQm v$f¡ 300 \u 500 K ky^u gC S>hpdp„Aph¡ R>¡. s¡ dpV¡$ gpNsp¡ kdeNpmp¡ 30 s R>¡. sp¡sp`dp__p¡ h^pfp ¡ L$fhp dpV¡ $ S>ê$fu L y $g L$pe®__________.

(1)

• •• • • • • ••• •••••

(2)

•• • • • • •••

••

(3)

•• • • • • •••

••

(4) 300 J


Set - 03 12

15. Consider a thin metallic sheet

perpendicular to the plane of the paper

moving with speed ‘v’ in a uniform

magnetic field B going into the plane of the

paper (See figure). If charge densities σ1 and

σ2 are induced on the left and right surfaces,

respectively, of the sheet then (ignore

fringe effects) :

(1) σ1=

0 v B, σ

2=−

0 v B

(2)• •

• ••• • •• • • •

• •• ••• •• • −

= =

(3) σ1=σ

2=

0 vB

(4)• •

• •• •• • • • •

• •• ••• •• •−

= =

15. ∞∑§ ¬Ã‹Ë œÊÃÈ ‡ÊË≈U ¬Îc∆U ∑§ ‹ê’flÃ ⁄UπË „Ò •ÊÒ⁄UÁøòÊ ◊¥ ÁŒπÊß ÁŒ‡ÊÊ ◊¥ flª ‘v’ ‚ ∞∑§ ‚◊ÊŸ øÈê’∑§Ëÿ-ˇÊòÊ B ◊¥ ø‹ ⁄U„Ë „Ò– øÈê’∑§Ëÿ-ˇÊòÊ ß‚ ‚◊Ã‹ ¬Îc∆U◊¥ ¬˝fl‡Ê ∑§⁄U ⁄U„Ê „Ò– ÿÁŒ ß‚ ‡ÊË≈U ∑§Ë ’ÊßZ •ÊÒ⁄U ŒÊßZ‚Ã„Ê¥ ¬⁄U ∑˝§◊‡Ê— ¬Îc∆U-•Êfl‡Ê-ÉÊŸàfl σ

1 ÃÕÊ σ

2 ¬Á⁄UÃ

„ÊÃ „Ò¥, Ã’ ©¬Ê¥Ã-¬˝÷Êfl ∑§Ê Ÿªáÿ ◊ÊŸÃ „È∞σ

1 ÃÕÊ σ

2 ∑§ ◊ÊŸ „Ê¥ª —

(1) σ1=

0 v B, σ

2=−

0 v B

(2)• •

• ••• • •• • • •

• •• ••• •• • −

= =

(3) σ1=σ

2=

0 vB

(4)• •

• •• •• • • • •

• •• ••• •• •−

= =

15. kdp„Nu Qy„bL$ue n¡Ó B dp„, yõsL$_p ¡S>_¡ g„b A_¡‘v’ S>¡V$gu AQm TX$`\u `¡S>_u A„v$f v$pMg \su^psy_u `psmu s[¼s (`©›$) _¡ Ýep_dp „ gp ¡.(ApL©$rÑ Sy>Ap¡) Å¡ s¡_u X$pbu A_¡ S>dZu k`pV$u `fA_y¾$d¡ σ

1 A_¡ σ

2 S>¡V$gu ©›$rhÛyscpf O_sp Dv¹$ch¡

sp¡ __________ \i¡. (qäÞS> Akf AhNZp¡.)

(1) σ1=

0 v B, σ

2=−

0 v B

(2)• •

• ••• • •• • • •

• •• ••• •• • −

= =

(3) σ1=σ

2=

0 vB

(4)• •

• •• •• • • • •

• •• ••• •• •−

= =


Set - 03 13

16. A fighter plane of length 20 m, wing span

(distance from tip of one wing to the tip of

the other wing) of 15 m and height

5 m is flying towards east over Delhi. Its

speed is 240 ms−1

. The earth’s magnetic

field over Delhi is 5×10−5

T with the

declination angle ~0 and dip of θ such that

sin θ =

•• . If the voltage developed is V

B

between the lower and upper side of the

plane and VW

between the tips of the wings

then VB

and V

W are close to :

(1) VB=45 mV; V

W=120 mV with right

side of pilot at higher voltage

(2) VB=45 mV; V

W=120 mV with left


(3) VB=40 mV; V

W=135 mV with right

side of pilot at high voltage

(4) VB=40 mV; V

W=135 mV with left


16. ∞∑§ ‹«∏Ê∑Í§ ¡„Ê¡ ∑§Ë ‹ê’Êß¸ 20 m, ¬¥πÊ¥ ∑§ Á‚⁄UÊ¥ ∑§’Ëø ŒÍ⁄UË 15 m ÃÕÊ ™°§øÊß¸ 5 m „Ò, •ÊÒ⁄U ÿ„ ÁŒÀ‹Ë∑§ ™§¬⁄U ¬Ífl¸-ÁŒ‡ÊÊ ◊¥ 240 ms

−1 ªÁÃ ‚ ©«∏ ⁄U„Ê „Ò–ÁŒÀ‹Ë ∑§ ™§¬⁄U ¬ÎâflË ∑§Ê øÈê’∑§Ëÿ-ˇÊòÊ 5×10

−5 T

„Ò, Á«UÁÄ‹Ÿ‡ÊŸ ∑§ÊáÊ ~0 „Ò, ÃÕÊ Á«U¬ ∑§ÊáÊ θ ∑§ Á‹ÿ

sin θ =

•• „Ò– ÿÁŒ ¬Á⁄UÃ-Áfl÷fl „Ò¥ — V

B ¡„Ê¡ ∑§

™§¬⁄U fl ŸËø ∑§ ’Ëø ; VW

¬¥πÊ¥ ∑§ Á‚⁄UÊ¥ ∑§ ’Ëø–Ã’ —

(1) VB=45 mV ; V

W=120 mV ŒÊÿÊ¥ ¬¥π-

Á‚⁄UÊ+ve

(2) VB=45 mV ; V

W=120 mV ’ÊÿÊ¥ ¬¥π-

Á‚⁄UÊ−ve

(3) VB=40 mV ; V

W=135 mV ŒÊÿÊ¥ ¬¥π-

Á‚⁄UÊ+ve

(4) VB=40 mV ; V

W=135 mV ’ÊÿÊ¥ ¬¥π-

Á‚⁄UÊ−ve

16. 20 m g„bpC fphsy„, 15 m p„Muep_u lp¡mpC (A¡L$bpSy>_u `p„Muep_p R>¡X$p\u buÆ bpSy>_p `p„rMep_pR>¡X$p ky u) A_¡ 5 m S>¡V$gu KQpC fphsy„ A¡L$ gX$peL$rhdp_ qv$ëlu_p |h® sfa EX$u füy„ R>¡. s¡_u TX$`240 ms

−1 R>¡. qv$ëlu D`f ©Õhu_y„ Qy„bL$ue n¡Ó_y„d|ëe 5×10

−5 T d¡Á_¡V$uL$ X¡$[¼g_¡i_ ~0 A_¡ X$u`

A¡ÞNg θ A¡hp¡ R>¡ L¡$ S>¡\u sin θ =

••

\pe. Å¡

àg¡__p _uQ¡_p A_¡ D`f_p R>¡X$p hÃQ¡ VB S>¡V$gp¡

hp¡ëV¡$S> A_¡ VW

S>¡V$gp¡ hp¡ëV¡$S> b¡ p„Mp¡_p R>¡X$p hÃQ¡DÑ`Þ_ \sp¡ lp¡e sp¡ V

B A_¡ V

W _y „ d|ëe

__________ _ÆL$_y„ li¡..

(1) VB=45 mV A_¡ V

W= `pegp¡V$_u S>dZu

bpSy> KQp hp¡ëV¡$S>¡ lp¡e s¡ fus¡ 120 mV

(2) VB=45 mV; V

W= `pegp¡V$_u X$pbu bpSy>

KQp hp¡ëV¡$S>¡ lp¡e s¡ fus¡ 120 mV

(3) VB=40 mV; V

W= pegp¡V$_u S>dZu bpSy>


(4) VB=40 mV; V

W= `pegp¡V$_u X$pbu bpSy>



Set - 03 14

17. A conducting metal circular-wire-loop of

radius r is placed perpendicular to a

magnetic field which varies with time as

•• ••

• • ••• • , where B

0 and τ are constants,

at time t=0. If the resistance of the loop is

R then the heat generated in the loop after

a long time • •• →• is :

(1)

• • ••

•• ••

π

τ

(2)

• • ••

•• ••

π

τ

(3)

• • •• •• • •π

τ

(4)

• • ••• •

•π

τ

17. ‘r’ ÁòÊíÿÊ ∑§ œÊÃÈ flÎûÊËÿ-ÃÊ⁄U-‹Í¬ ∑§Ê ¬Îc∆U,

•• ••

• • ••• • mÊ⁄UÊ ’Œ‹Ã „È∞ øÈê’∑§Ëÿ-ˇÊòÊ ∑§

‹ê’flÃ ⁄UπÊ „Ò– ¡„Ê° ‚◊ÿ t=0 ¬⁄U B0 ÃÕÊ τ •ø⁄U

„Ò¥– ÿÁŒ ‹Í¬ ∑§Ê ¬˝ÁÃ⁄UÊœ R „Ò, Ã’ ∑§Ê»§Ë íÿÊŒÊ ‚◊ÿ

• •• →• ªÈ¡⁄UŸ ∑§ ’ÊŒ ©‚ ‹Í¬ ◊¥ ¬ÒŒÊ „Èß¸ ™§¡Ê¸ „Ò —

(1)

• • ••

•• ••

π

τ

(2)

• • ••

•• ••

π

τ

(3)

• • •• •• • •π

τ

(4)

• • ••• •

•π

τ

17. A¡L$ kyhplL$ ^psy_y„ hsy®mpL$pf N|„Qmy„ L¡$ S>¡_u rÓÄep rlp¡e s¡_¡ Qy„bL$ue n¡Ó_¡ g„bê$ ¡ d|L$hpdp„ Aph¡ R>¡. Ap

Qy„bL$ue n¡Ó kde kp\¡ •• ••

• • ••• • âdpZ¡ bv$gpe

R>¡, Äep„ B0 A_¡ τ t=0 kde¡ AQmp„L$ R>¡. Å¡ N|„Qmp_p¡

Ahfp¡ R lp¡e sp¡ M|b S> gp„bp kde • •• →• _¡

A„s¡ N|„Qmpdp„ DÐ`Þ_ Dódp _______ \i¡.

(1)

• • ••

•• ••

π

τ

(2)

• • ••

•• ••

π

τ

(3)

• • •• •• • •π

τ

(4)

• • ••• •

•π

τ


Set - 03 15

18. Consider an electromagnetic wave

propagating in vacuum. Choose the correct

statement :

(1) For an electromagnetic wave

propagating in +x direction the

electric field is

( )•

•• • •• • •• • ••

• • • • • • •• •→ ∧ ∧

= − and the

magnetic field is

( )• • •• • •• • ••

B B x, t y zyz

→ ∧ ∧

•• • •


propagating in +x direction the

electric field is

( )•• • • •• • • •• • ••

• • • • •• • • •• •→ ∧ ∧

• • and

the magnetic field is

( )•• • • •• • • •• • ••

• • • • •• • • •• •→ ∧ ∧

• •


propagating in +y direction the

electric field is •• • • •• • •••

• • • • • •• •→ ∧

•

and the magnetic field is

•• • • •• • •••

B B x, t zyz

→ ∧

•

18. √ÿÊ◊ ◊¥ ø‹ ⁄U„Ë flÒlÈÃ-ëÊÈê’∑§Ëÿ Ã⁄¥Uª ∑§ Á‹∞ ‚„ËÁfl∑§À¬ øÈÁŸ∞–

(1) +x ÁŒ‡ÊÊ ◊¥ øÊÁ‹Ã flÒlÈÃ-øÈê’∑§Ëÿ Ã⁄¥Uª ∑§

Á‹ÿ ( )•

•• • •• • •• • ••

• • • • • • •• •→ ∧ ∧

= −,

( )• • •• • •• • ••

B B x, t y zyz

→ ∧ ∧

•• • •

(2) +x ÁŒ‡ÊÊ ◊¥ øÊÁ‹Ã flÒlÈÃ-øÈê’∑§Ëÿ Ã⁄¥Uª ∑§

Á‹ÿ ( )•• • • •• • • •• • ••

• • • • •• • • •• •→ ∧ ∧

• • ,

( )•• • • •• • • •• • ••

• • • • •• • • •• •→ ∧ ∧

• •

(3) +y ÁŒ‡ÊÊ ◊¥ ø‹ ⁄U„Ë flÒlÈÃ-øÈê’∑§Ëÿ Ã⁄¥Uª ∑§

Á‹ÿ •• • • •• • •••

• • • • • •• •→ ∧

• ,

•• • • •• • •••

B B x, t zyz

→ ∧

•

18. i|ÞephL$pidp„ âkfsp rhÛysQy„bL$ue sf„N_¡ Ýep_dp„gp¡. _uQ¡dp„\u kpQy„ rh^p_ `k„v$ L$fp¡.

(1) +x -qv$ipdp„ Nrs L$fsp rhÛysQy„bL$ue sf„NdpV¡$ rhÛys n¡Ó

( )•

•• • •• • •• • ••

• • • • • • •• •→ ∧ ∧

= − A_¡

Qy„bL$ue n¡Ó

( )• • •• • •• • ••

B B x, t y zyz

→ ∧ ∧

•• • • \i¡.

(2) +x -qv$ipdp„ Nrs L$fsp rhÛysQy„bL$ue sf„NdpV¡$ rhÛys n¡Ó

( )•• • • •• • • •• • ••

• • • • •• • • •• •→ ∧ ∧

• • A_¡

Qy„bL$ue n¡Ó

( )•• • • •• • • •• • ••

• • • • •• • • •• •→ ∧ ∧

• • \i¡.

(3) +y - qv$ipdp„ Nrs L$fsp rhÛysQy„bL$ue sf„N

dpV¡$ rhÛys n¡Ó •• • • •• • •••

• • • • • •• •→ ∧

•

A_¡ Qy„bL$ue n¡Ó •• • • •• • •••

B B x, t zyz

→ ∧

•

\i¡.


Set - 03 16


propagating in +y direction the

electric field is •• • • •• • •••

• • • • • •• •→ ∧

•

and the magnetic field is

•• • • •• • •••

B B x, t yz

→ ∧

•

19. A hemispherical glass body of radius

10 cm and refractive index 1.5 is silvered

on its curved surface. A small air bubble is

6 cm below the flat surface inside it along

the axis. The position of the image of the

air bubble made by the mirror is seen :

(1) 14 cm below flat surface




(4) +y ÁŒ‡ÊÊ ◊¥ ø‹ ⁄U„Ë flÒlÈÃ-øÈê’∑§Ëÿ Ã⁄¥Uª ∑§

Á‹ÿ •• • • •• • •••

• • • • • •• •→ ∧

• ,

•• • • •• • •••

B B x, t yz

→ ∧

•

19. ∞∑§ ∑§Ê°ø ∑§ •h¸ªÊ‹Ëÿ ∆UÊ‚ ∑§Ë ÁòÊíÿÊ 10 cm ÃÕÊ•¬flÃ¸ŸÊ¥∑§ 1.5 „Ò– ©‚∑§Ë fl∑˝§Ëÿ ‚Ã„ ¬⁄U øÊ°ŒË ∑§Ë¬⁄UÃ ø…∏Êß¸ ªß¸ „Ò– ‚◊Ã‹ ¬Îc∆U ∑§ 6 cm ŸËø ÃÕÊ•ˇÊ ¬⁄U, ∞∑§ ‚Í ◊ „flÊ ∑§Ê ’È‹’È‹Ê ÁSÕÃ „Ò– Ã’fl∑˝§Ëÿ-Œ¬¸áÊ ‚ ’Ÿ ⁄U„ ’È‹’È‹ ∑§Ë ¬˝ÁÃÁ’ê’ ŒÍ⁄UË „Ò —

(1) ‚◊Ã‹ ‚Ã„ ‚ 14 cm ŸËø

(2) ‚◊Ã‹ ‚Ã„ ‚ 30 cm ŸËø

(3) ‚◊Ã‹ ‚Ã„ ‚ 20 cm ŸËø

(4) ‚◊Ã‹ ‚Ã„ ‚ 16 cm ŸËø

(4) +y -qv$ipdp„ Nrs L$fsp rhÛysQy„bL$ue sf„N

dpV¡ $ rhÛysn¡Ó •• • • •• • •••

• • • • • •• •→ ∧

•

A_¡ Qy„bL$ue n¡Ó •• • • •• • •••

B B x, t yz

→ ∧

•

\i¡.

19. 10 cm _u rÓÄep fphsu A_¡ 1.5 h¾$uch_p„L$ fphspÁgpk_p A¡L$ A ®Np¡mpL$pf cpN_u h¾$ k`pV$u f Qp„v$u_p¡Y$p¡m QY$phhpdp„ Aph¡g R>¡. s¡_u ku^u k`pV$u\u 6 cm

_uQ¡, A ®Np¡mpL$pf_u A„v$f_p cpNdp„, A¡L$ lhp_p¡ _p_p¡`f`p¡V$p¡ s¡_u An `f fl¡g R>¡. Afukp Üpfp lhp_p`f`p¡V$p_p„ ârstbb_y„ õ\p_ _________ v$¡Mpi¡.

(1) ku^u k`pV$u_u _uQ¡ 14 cm A„sf¡


(3) ku^¡ k`pV$u_u _uQ¡ 20 cm A„sf¡



Set - 03 17

20. Two stars are 10 light years away from the

earth. They are seen through a telescope

of objective diameter 30 cm. The

wavelength of light is 600 nm. To see the

stars just resolved by the telescope, the

minimum distance between them should

be (1 light year=9.46× 1015

m) of the order

of :

(1) 106 km

(2) 108 km

(3) 1011

km

(4) 1010

km

21. A photoelectric surface is illuminated

successively by monochromatic light of

wavelengths λ and ••

. If the maximum

kinetic energy of the emitted

photoelectrons in the second case is

3 times that in the first case, the work

function of the surface is :

(1) •• ••

(2) •• ••

(3)

• ••

(4)

• •• ••

20. ŒÊ ÃÊ⁄U ¬ÎâflË ‚ 10 ¬˝∑§Ê‡Ê-fl·¸ ∑§Ë ŒÍ⁄UË ¬⁄U „Ò¥– ©Ÿ∑§Ê∞∑§ ≈UÁ‹S∑§Ê¬ mÊ⁄UÊ ŒπÊ ¡ÊÃÊ „Ò, Á¡‚∑§Ê •Á÷ŒÎ‡ÿ∑§30 cm √ÿÊ‚ ∑§Ê „Ò– ¬˝∑§Ê‡Ê ∑§Ë Ã⁄¥UªŒÒÉÿ¸ 600 nm

„Ò– (1 ¬˝∑§Ê‡Ê-fl·¸ =9.46× 1015

m) „Ò– ≈UÁ‹S∑§Ê¬•ª⁄U ©Ÿ ÃÊ⁄UÊ¥ ∑§Ê ‹ª÷ª Áfl÷ÁŒÃ Œπ ¬Ê ⁄U„Ê „Ò, Ã’©Ÿ∑§ ’Ëø ∑§Ë ŒÍ⁄UË ∑§Ê order „Ò —(1) 10

6 km

(2) 108 km

(3) 1011

km

(4) 1010

km

21. ∞∑§ ¬˝∑§Ê‡Ê-flÒlÈÃ ‚Ã„ ¬⁄U ¬„‹Ë ’Ê⁄U λ ÃÕÊ ŒÍ‚⁄UË

’Ê⁄U ••

Ã⁄¥UªŒÒÉÿ¸ ∑§Ê ¬˝∑§Ê‡Ê «UÊ‹Ê ¡ÊÃÊ „Ò– ÿÁŒ

©à‚Á¡¸Ã ¬˝∑§Ê‡Ê-ß‹Ä≈˛UÊÚŸ ∑§Ë •Áœ∑§Ã◊ ªÁÃ¡-™§¡Ê¸ŒÍ‚⁄UË ’Ê⁄U ◊¥ ¬„‹Ë ’Ê⁄U ∑§Ë ÁÃªÈŸË „Ê, Ã’ ©‚ ‚Ã„∑§Ê ∑§Êÿ¸-»§‹Ÿ „Ò —

(1) •• ••

(2) •• ••

(3)

• ••

(4)

• •• ••

20. b¡ spfpAp¡ `©Õhu\u 10 âL$pihj® v|$f R>¡. s¡Ap¡_¡30 cm ìepk fphsp Ap¡bS>¡¼¹$V$uh (hõsy-g¡Þk) hpmpV¡$guõL$p¡`\u Å¡hpdp„ Aph¡ R>¡. âL$pi_u sf„Ng„bpC600 nm R>¡. Ap b¡ spfp_¡ just R|>V$p `X¡ $gp(rhc¡qv$s \e¡gp) Å¡hp dpV¡$ s¡d_u hÃQ¡_y„ A„sf_________ _p ¾$d_y„ lp¡hy„ Å¡CA¡.

(1 âL$pi hj®=9.46× 1015

m)

(1) 106 km

(2) 108 km

(3) 1011

km

(4) 1010

km

21. A¡L$ ap¡V$p¡Cg¡¼V²$uL$ k`pV$u_¡ hpfpasu A_y¾$d¡ λ A_¡ ••

sf„Ng„bpC ^fphsp A¡L$f„Nu âL$pi\u âL$pris L$fhpdp„Aph¡ R>¡. buÅ qL$õkpdp„ Å¡ DÐkÅ®sp ap¡V$p¡Cg¡¼V²$p¡__udlÑd Nrs EÅ® â\d qL$õkp L$fsp„ ÓZ NZu dmsulp¡e sp¡ k`pV$u_y„ hL®$a„¼i_ _________ \i¡.

(1) •• ••

(2) •• ••

(3)

• ••

(4)

• •• ••


Set - 03 18

22. ‘v’ S>¡V$gu TX$`\u Nrs L$fsp¡ ÞeyV²$p¡_ A¡L$ [õ\f A_¡^fp[õ\rsdp„ fl¡gp lpCX²$p¡S>_ `fdpÏ kp\¡ ku^uA\X$pdZ A_ych¡ R>¡. ÞeyV²$p¡__u A[õ\rsõ\p`L$A\X $ pdZ \pe s ¡ dpV ¡ $_u dlÑd NrsEÅ®__________ \i¡.(1) 10.2 eV

(2) 16.8 eV

(3) 12.1 eV

(4) 20.4 eV

23. ApL©$rÑdp„ v$ip®h¡g qf`\ dpV¡$ ApDV$ yV$ 1 dm¡ s¡ dpV¡$S>ê$fu C_`yV$ __________ \i¡.

(1) a=0, b=1, c=0

(2) a=1, b=0, c=0

(3) a=1, b=0, c=1

(4) a=0, b=0, c=1

22. A neutron moving with a speed ‘v’ makes

a head on collision with a stationary

hydrogen atom in ground state. The

minimum kinetic energy of the neutron for

which inelastic collision will take place is :

(1) 10.2 eV

(2) 16.8 eV

(3) 12.1 eV

(4) 20.4 eV

23. To get an output of 1 from the circuit shown

in figure the input must be :

(1) a=0, b=1, c=0

(2) a=1, b=0, c=0

(3) a=1, b=0, c=1

(4) a=0, b=0, c=1

22. ªÁÃ ‘v’ ‚ ø‹ÃÊ „È•Ê ∞∑§ ãÿÍ≈˛UÊÚŸ ∞∑§ ÁSÕ⁄U „Êß¸«˛UÊ¡Ÿ¬⁄U◊ÊáÊÈ, ¡Ê •¬ŸË •Êl-•flSÕÊ ◊¥ „Ò, ‚ ‚ê◊Èπ≈UÄ∑§⁄U ∑§⁄UÃÊ „Ò– ãÿÈ≈˛UÊÚŸ ∑§Ë fl„ ãÿÍŸÃ◊ ªÁÃ¡ ™§¡Ê¸’ÃÊÿ¥ Á¡‚ ∑§ „ÊŸ ¬⁄U ÿ„ ≈UÄ∑§⁄U •¬˝àÿÊSÕ „ÊªË —(1) 10.2 eV

(2) 16.8 eV

(3) 12.1 eV

(4) 20.4 eV

23. ÁŒÿ ªÿ ¬Á⁄U¬Õ ‚ 1 ÁŸª◊ ¬ÊåÃ ∑§⁄UŸ ∑§ Á‹ÿ •Êfl‡ÿ∑§ÁŸfl‡Ê „ÊŸÊ øÊÁ„ÿ —

(1) a=0, b=1, c=0

(2) a=1, b=0, c=0

(3) a=1, b=0, c=1

(4) a=0, b=0, c=1


Set - 03 19

24. A modulated signal Cm

(t) has the form

Cm

(t)=30 sin 300πt+10 (cos 200πt

−cos 400πt). The carrier frequency fc, the

modulating frequency (message frequency)

fω

, and the modulation index µ are

respectively given by :

(1) fc=200 Hz ; f

ω=50 Hz ;

•• ••

• •

(2) fc=150 Hz ; f

ω=50 Hz ;

•• ••

• •

(3) fc=150 Hz ; f

ω=30 Hz ;

•• ••

• •

(4) fc=200 Hz ; f

ω=30 Hz ;

•• ••

• •

25. A particle of mass m is acted upon by a force

F given by the empirical law •• • ••• • ••••

• .

If this law is to be tested experimentally by

observing the motion starting from rest, the

best way is to plot :

(1) v(t) against t2

(2) log v(t) against •••

(3) log v(t) against t

(4) log v(t) against

••

24. Cm

(t)=30 sin 300πt+10 (cos 200πt

−cos 400πt) ∞∑§ ◊Ê«ÈUÁ‹Ã Á‚ÇŸ‹ ∑§Ê Œ‡ÊÊ¸ÃÊ „Ò–Ã’ flÊ„∑§ •ÊflÎÁûÊ f

c, ◊Ê«ÈU‹∑§ •ÊflÎÁûÊ f

ω ÃÕÊ ◊Ê«ÈU‹∑§

ßŸ«UÄ‚ µ ∑˝§◊‡Ê— „Ò¥ —

(1) fc=200 Hz ; f

ω=50 Hz ;

•• ••

• •

(2) fc=150 Hz ; f

ω=50 Hz ;

•• ••

• •

(3) fc=150 Hz ; f

ω=30 Hz ;

•• ••

• •

(4) fc=200 Hz ; f

ω=30 Hz ;

•• ••

• •

25. m Œ˝√ÿ◊ÊŸ ∑§ ∑§áÊ ¬⁄U F ’‹ ‹ª ⁄U„Ê „Ò, •ÊÒ⁄U ©‚∑§

Á‹ÿ •ÊŸÈ÷Áfl∑§ ‚ê’¥œ „Ò •• • ••• • ••••

• ß‚ ‚ê’¥œ

∑§ ‚àÿÊ¬Ÿ ∑§ Á‹∞ ÁSÕ⁄U •flSÕÊ ‚ ∑§áÊ ∑§Ë ªÁÃ ∑§Ê¬ ÊáÊ (Observation) ∑§⁄U ÁŸêŸÁ‹ÁπÃ ◊¥ ‚ ∑§ÊÒŸ‚Ê ª˝Ê»§ ‚flÊûÊ◊ „ÊªÊ?

(1) t2 ∑§ ÁflL§h v(t)

(2) ••• ∑§ ÁflL§h log v(t)

(3) t ∑§ ÁflL§h log v(t)

(4)

•• ∑§ ÁflL§h log v(t)

24. dp¡X$éygf (Ar^rdrîs) \e¡gp rkÁ_g Cm

(t) _uQ¡dyS>b Ap`u iL$pe R>¡. C

m(t)=30 sin 300πt+10

(cos 200πt−cos 400πt) L¡$fuef Aph©rÑ fc

, k„v$¡ip(dp¡X$éyg¡V$]N) Aph©rÑ f

ω, A_¡ dp¡X$éyg¡i_ A„L$

(index) µ, A_y¾$d¡ ______ hX¡$ Ap`u iL$pe.

(1) fc=200 Hz ; f

ω=50 Hz ;

•• ••

• •

(2) fc=150 Hz ; f

ω=50 Hz ;

•• ••

• •

(3) fc=150 Hz ; f

ω=30 Hz ;

•• ••

• •

(4) fc=200 Hz ; f

ω=30 Hz ;

•• ••

• •

25. m v$m ^fphsp¡ L$Z •• • ••• • ••••

• _¡ A_ykfsp

F bm_u Akf l¡W$m R>¡. Å¡ Ap bm-r_ed_¡ âpep¡rNL$fus¡, L$Z_u Nrs [õ\f [õ\rsdp„\u iê$ \pe s¡ fus¡QL$pkhp¡ lp¡e sp¡ __________ _p¡ N°pa A¡ kp¥\u kpfp¡rhL$ë` \i¡.

(1) v(t) rhfyÝ^ t2

(2) log v(t) rhfyÝ^ •••

(3) log v(t) rhfyÝ^ t

(4) log v(t) rhfyÝ^ ••


Set - 03 20

26. A thin 1 m long rod has a radius of 5 mm.

A force of 50 πkN is applied at one end to

determine its Young’s modulus. Assume

that the force is exactly known. If the least

count in the measurement of all lengths is

0.01 mm, which of the following statements

is false ?

(1)

•••

gets minimum contribution

from the uncertainty in the length.

(2) The figure of merit is the largest for

the length of the rod.

(3) The maximum value of Y that can be

determined is 1014

N/m2.

(4)

•••

gets its maximum contribution

from the uncertainty in strain.

26. 1 m ‹ê’Ë ¬Ã‹Ë ¿U«∏ ∑§Ë ÁòÊíÿÊ 5 mm „Ò– ÿ¥ª◊Ê«U‹‚ ÁŸ∑§Ê‹Ÿ ∑§ Á‹ÿ ß‚ ∑§ Á‚⁄U ¬⁄U 50 πkN

∑§Ê ’‹ ‹ªÊÿÊ ªÿÊ– ◊ÊŸ¥ Á∑§ ’‹ Á’‹∑È§‹ ∆UË∑§ ‚ôÊÊÃ „Ò– ÿÁŒ ‹ê’ÊßÿÊ ¥ ∑§ ◊Ê¬Ÿ ∑§ •À¬Ê¥‡Ê0.01 mm „Ò¥– Ã’ ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ ‚Ê ∑§ÕŸ ª‹Ã

„Ò?

(1)

•••

◊¥ ‹ê’Êß¸ ∑§Ë •ÁŸÁ‡øÃÃÊ ∑§Ê ÿÊªŒÊŸ

ãÿÍŸÃ◊ „Ò–

(2) ¿U«∏ ∑§Ë ‹ê’Êß¸ ∑§ Á‹ÿ ŒˇÊÃÊ¥∑§ ‚’‚ ’«∏Ê „Ò–

(3) Y ∑§Ê •Áœ∑§Ã◊ ¬˝Ê# „Ê ‚∑§Ÿ flÊ‹Ê ◊ÊŸ10

14 N/m

2.

(4)

•••

◊¥ Áfl∑Î§ÁÃ ∑§Ë •ÁŸÁ‡øÃÃÊ ∑§Ê ÿÊªŒÊŸ

•Áœ∑§Ã◊ „Ò–

26. 1 m gp„bp A_¡ 5 mm rÓÄep ^fphsp¡ A¡L$ `psmp¡krmep ¡ R> ¡ . s ¡_p ¡ ×Y $ [õ\rsõ\p`L $sp A„L $(e„N dp¡X$éygk) ip¡^hp dpV¡$ s¡_p A¡L$ R>¡X$p D`f50 πkN S>¡V$gy„ bm gNpX$hpdp„ Aph¡ R>¡. A¡hy„ ^pfp¡ L¡$Ap bm_y„ d|ëe kQp¡V$ fus¡ dpg|d R>¡. Å¡ g„bpC_pb^p S> dp`_dp„ gOyÑd dp` i[¼s 0.01 mm lp¡esp¡ _uQ¡ ¥L$u_y„ L$ey„ rh^p_ Mp¡Vy„$ li¡ ?

(1) g„bpC_u AQp¡½$kpC_¡ L$pfZ¡ •••

dp„ gOyÑd

apmp¡ Aphsp¡ li¡.

(2) krmep_u g„bpC dpV¡$ v$nsp„L$ (figure of

merit) kp¥\u h^pf¡ \i¡.

(3) d¡mhu iL$pe s¡hu Y _u dlÑd qL„ $ds10

14 N/m

2 \i¡.

(4)

•••

dp „ dlÑd apmp ¡ rhL © $ rsdp „ fl¡g

AQp¡½$kpC_¡ L$pfZ¡ li¡.


Set - 03 21

27. N¡ëh¡_p¡duV$fdp„ 50 L$p`p R>¡. b¡V$fu_p¡ Ap„sqfL$ Ahfp¡i|Þe R>¡. Äepf¡ =2400 Ω lp¡e R>¡ Ðepf¡ N¡ëh¡_p¡duV$fdp„ 40 L$p`p (divisions) ky u_y„ Aphs®_ dm¡ R>¡.Äepf¡ Ahfp¡ ¡V$udp„\u 4900 Ω S>¡V$gp¡ L$pY$hpdp„ Aph¡R>¡ Ðepf¡ 20 divisions S>¡V$gy„ Aphs®_ \pe R>¡. sp¡Ap`Z¡ spfZ Ap`u iL$uA¡ L¡$ __________.

(1) N¡ëh¡_p¡duV$f_p¡ Ahfp¡ 200 Ω li¡.

(2) |Z®-õL¡$g Aphs®_ dpV¡$ âhpl 2 mA li¡.

(3) N¡ëh ¡_p ¡duV $f_u ` °hpl k „h ¡ qv $sp20 µA/division \i¡.

(4) 10 divisions S>¡V$gy„ Aphs®_ d¡mhhp dpV¡$Ahfp¡ ¡V$udp„ Ahfp¡ _y„ d|ëe 9800 Ω Å¡Ci¡.

28. V²$ph¡g]N dpC¾$p¡õL$p¡`_u dv$v$\u Ágpk_p Qp¡kgp_p¡h¾$uch_p„L$ dp`hp dpV¡$ Ap¡R>pdp„ Ap¡R>p S>ê$fuAhgp¡L$_p¡_u k„¿ep __________ li¡.

(1) b¡

(2) ÓZ

(3) Qpf

(4) `p„Q

27. ∞∑§ ªÀflŸÊ◊Ë≈U⁄U ∑§Ë S∑§‹ 50 ÷ÊªÊ¥ ◊¥ ’¥≈UË „Ò– ’Ò≈U⁄UË∑§Ê •Ê ¥ÃÁ⁄ U∑§ ¬ ˝ ÁÃ⁄ U Ê œ ‡Ê Í ãÿ „ Ò– ÿÁŒR=2400 Ω „Ò ÃÊ ÁflˇÊ¬ =40 ÷Êª „Ò– ÿÁŒR=4900 Ω „Ò ÃÊ ÁflˇÊ¬ =20 ÷Êª „Ò– Ã’ „◊ÁŸœÊ¸Á⁄UÃ ∑§⁄U ‚∑§Ã „Ò¥ Á∑§ —

(1) ªÀflŸÊ◊Ë≈U⁄U ∑§Ê ¬˝ÁÃ⁄UÊœ 200 Ω „Ò–

(2) »È§‹-S∑§‹ ÁflˇÊ¬ ∑§ Á‹ÿ œÊ⁄UÊ 2 mA „Ò–

(3) ªÀflŸÊ◊Ë≈U⁄U ∑§Ë œÊ⁄UÊ-‚¥flŒŸ‡ÊË‹ÃÊ 20 µA

¬˝ÁÃ ÷Êª „Ò–

(4) ÁflˇÊ¬ =10 ÷Êª ∑§ Á‹ÿ R=9800 Ω.

28. ∑§Ê°ø ∑§Ë S‹Ò’ ∑§Ê ø‹-◊Êß¸∑˝§ÊS∑§Ê¬ mÊ⁄UÊ •¬flÃ¸ŸÊ¥∑§ÁŸ∑§Ê‹Ÿ ∑§ Á‹ÿ ¡M§⁄UË ¬Ê∆˜ÿÊ¥∑§Ê¥ ∑§Ë ãÿÍŸÃ◊ ‚¥ÅÿÊ„Ò —

(1) ŒÊ

(2) ÃËŸ

(3) øÊ⁄U

(4) ¬Ê°ø

27. A galvanometer has a 50 division scale.

Battery has no internal resistance. It is

found that there is deflection of 40 divisions

when R=2400 Ω. Deflection becomes

20 divisions when resistance taken from

resistance box is 4900 Ω. Then we can

conclude :

(1) Resistance of galvanometer is 200 Ω.

(2) Full scale deflection current is 2 mA.

(3) Current sensitivity of galvanometer

is 20 µA/division.

(4) Resistance required on R.B. for a

deflection of 10 divisions is 9800 Ω.

28. To determine refractive index of glass slab

using a travelling microscope, minimum

number of readings required are :

(1) Two

(2) Three

(3) Four

(4) Five


Set - 03 22

29. A¡L$ V²$p[ÞTõV$f_p common emitter (CE) k„fQ_pdp„C_ yV$ gpnrZL$sp dpV¡$ C_ yV$ Ahfp¡ _p ìeõs_p¡ kp¥\uh^y hpõsrhL$ N°pa __________ \i¡.

(1)

(2)

(3)

29. ∞∑§ ≈˛ U Ê ¥ Á¡S≈U⁄ U ∑§Ê common-emitter (CE)

•Á÷ÁflãÿÊ‚ ◊¥ ÁŸfl‡Ê-•Á÷‹ÊˇÊÁáÊ∑§ ◊Ê¬Ÿ Á∑§ÿÊ ªÿÊ–Ã’ ÁŸfl‡Ê-¬˝ÁÃ⁄UÊœ ∑§ √ÿÈà∑˝§◊ ∑§Ê ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ-‚Ê ª˝Ê»§ ©ÁøÃ „Ò?

(1)

(2)

(3)

29. A realistic graph depicting the variation of

the reciprocal of input resistance in an input

characteristics measurement in a common-

emitter transistor configuration is :

(1)

(2)

(3)


Set - 03 23

(4)

30. V²$p[ÞTõV$f_u C_`yV$ A_¡ ApDV$`yV$ gpnrZL$sp_pdp`_dp„ ApDV$ yV$ Ahfp¡ r

0 A_¡ C_ yV$ Ahfp¡ r

i

_p¡ NyZp¡Ñf (R) __________ S>¡V$gu f¡ÞS>dp„ lp¡e R>¡.(1) R~10

2−10

3

(2) R~1−10

(3) R~0.1−0.01

(4) R~0.1−1.0

31. 0.1N X$peb¡rTL$ A¡rkX$_y„ L$v$ iy„ li¡ L¡$ S>¡ 1 g b¡CT_psV$õ\uL$fZ L$fhp dpV¡$ ep®á lp¡e L¡$ S>¡_p S>gue ÖphZdp„0.04 dp¡g OH

− Aph¡gp R>¡ ?(1) 200 mL

(2) 400 mL

(3) 600 mL

(4) 800 mL

(4)

30. Á∑§‚Ë ≈˛UÊ¥Á¡S≈U⁄U ∑§Ë ÁŸfl‡Ê-ÁŸª¸◊ •Á÷‹ÊˇÊÁáÊ∑§◊Ê¬Ÿ ∑§ Á‹ÿ ¬˝ÿÈÄÃ ÁŸª¸◊-¬˝ÁÃ⁄UÊ œ (r

0) fl

ÁŸfl‡Ê-¬˝ÁÃ⁄UÊœ (ri) ∑§ •ŸÈ¬ÊÃ (R) ∑§Ê •ÊÿÊ◊

(range) „ÊªÊ?(1) R~10

2−10

3

(2) R~1−10

(3) R~0.1−0.01

(4) R~0.1−1.0

31. 0.1N ÁmˇÊÊ⁄UËÿ •ê‹ ∑§Ê •ÊÿÃŸ ÄÿÊ „ÊªÊ ¡Ê 1 ª˝Ê◊ˇÊÊ⁄U∑§ Á¡‚∑§ ¡‹Ëÿ Áfl‹ÿŸ ◊¥ 0.04 ◊Ê‹ OH

− „Ò∑§Ê ©ŒÊ‚ËŸ ∑§⁄UŸ ∑§ Á‹ÿ ¬ÿÊ¸# „Ò?(1) 200 mL

(2) 400 mL

(3) 600 mL

(4) 800 mL

(4)

30. The ratio (R) of output resistance r0, and

the input resistance ri in measurements of

input and output characteristics of a

transistor is typically in the range :

(1) R~102−10

3

(2) R~1−10

(3) R~0.1−0.01

(4) R~0.1−1.0

31. The volume of 0.1N dibasic acid sufficient

to neutralize 1 g of a base that furnishes 0.04

mole of OH−

in aqueous solution is :

(1) 200 mL

(2) 400 mL

(3) 600 mL

(4) 800 mL


Set - 03 24

32. r_ròs sp`dp_ `f, iê$Apsdp„, N2 AÏAp¡_p¡

kf¡fpi hN®d|m (root mean square) (rms) h¡N u

R>¡. Å¡ Ap sp`dp_ bdÏ„ (Doubled) L$fhpdp„ Aph¡sp¡ b^p S> _pCV²$p¡S>_ AÏAp¡_y„ rhep¡S>_ _pCV²$p¡S>_`fdpÏAp¡dp„ \pe sp¡, R>u _hy„ kf¡fpi hN®d|m (rms)

h¡N iy„ \i¡ ?(1) u/2

(2) 2u

(3) 4u

(4) 14u

33. Cg¡¼V²$p¡r_L$ rhÞepk 1s22s

22p

63s

23p

6 ^fphsp L$epnpf_y„ S>gue ÖphZ Ape_p¡ ^fphsp„ _\u ?(1) NaF

(2) NaCl

(3) KBr

(4) CaI2

34. L$ep k„ep¡S>_dp„ H-X-H b„ M|Zp¡ khp®r^L$ R>¡ ?(1) CH

4

(2) NH3

(3) H2O

(4) PH3

32. ∞∑§ Áfl‡Ê· ÃÊ¬ ¬⁄U ¬˝Ê⁄Uê÷ ◊¥ ŸÊß≈˛UÊ¡Ÿ •áÊÈ•Ê¥ (N2)

∑§Ê flª¸ ◊Êäÿ ◊Í‹ flª u „Ò– ÿÁŒ ß‚ ÃÊ¬ ∑§Ê ŒÈªÈŸÊ∑§⁄U ÁŒÿÊ ¡Êÿ •ÊÒ⁄U ‚÷Ë ŸÊß≈˛UÊ¡Ÿ •áÊÈ ÁflÿÊÁ¡Ã„Ê∑§⁄U ŸÊß≈˛UÊ¡Ÿ ¬⁄U◊ÊáÊÈ ’Ÿ ¡Ê∞ ÃÊ ŸÿÊ flª¸ ◊Êäÿ◊Í‹ flª „ÊªÊ —(1) u/2

(2) 2u

(3) 4u

(4) 14u

33. Á∑§‚ ‹fláÊ ∑§ ¡‹Ëÿ Áfl‹ÿŸ ◊¥ 1s22s

22p

63s

23p

6

ß‹Ä≈˛UÊÚÁŸ∑§ ÁflãÿÊ‚ ∑§ •ÊÿŸ Ÿ„Ë¥ „Ê¥ª?(1) NaF

(2) NaCl

(3) KBr

(4) CaI2

34. Á∑§‚ ÿÊÒÁª∑§ ◊¥ H-X-H •Ê’ãœ ∑§ÊáÊ ‚flÊ¸Áœ∑§ „Ò?(1) CH

4

(2) NH3

(3) H2O

(4) PH3

32. Initially, the root mean square (rms) velocity

of N2

molecules at certain temperature is u.

If this temperature is doubled and all the

nitrogen molecules dissociate into nitrogen

atoms, then the new rms velocity will be :

(1) u/2

(2) 2u

(3) 4u

(4) 14u

33. Aqueous solution of which salt will not

contain ions with the electronic

configuration 1s22s

22p

63s

23p

6 ?

(1) NaF

(2) NaCl

(3) KBr

(4) CaI2

34. The bond angle H-X-H is the greatest in the

compound :

(1) CH4

(2) NH3

(3) H2O

(4) PH3


Set - 03 25

35. Å¡ H2O

2 _p 100 dp¡g 1 bpf (bar) A_¡ 300 K f

rhOV$us \pe sp¡, 1 bpf (bar) v$bpZ_p rhfyÝ^ 1

dp¡g O2(g) _p rhõsfZ \hp_¡ gu ¡ \e¡g L$pe® (kJ)

ip¡^p¡.

• • • •• • • ••• • • • •••• •• •• •••

(R = 8.3 J K−1

mol−1

)

(1) 62.25

(2) 124.50

(3) 249.00

(4) 498.00

36. A¡L$ r_ròs sp`dp_ `f, npf MX2 _p S>gue

ÖphZ_p¡ hpÞV$-lp¡a Aheh 2 R>¡. sp¡ npf_p ApÖphZ_p¡ rhep¡S>_A„i ip¡ p¡.(1) 0.33

(2) 0.50

(3) 0.67

(4) 0.80

37. A¡L$ b„ (sealed) r_hp®rss (evacuated) `pÓdp„fpMhpdp„ Aph¡g O_ (solid) XY rhOV$us \C_¡ T

sp`dp_¡ hpeyAp¡_y„ rdîZ X A_¡ Y b_¡ R>¡. Ap pÓdp„k„syg_ v$bpZ 10 bar (bpf) R>¡. Ap âq¾$ep dpV¡$ K

p

iy„ R>¡ ?(1) 5

(2) 10

(3) 25

(4) 100

35. ÿÁŒ H2O

2 ∑§ 100 ◊Ê‹ 1 bar ÃÕÊ 300 K ¬⁄U

ÁflÿÊÁ¡Ã „Ê ÃÊ 1 bar ŒÊ’ ∑§ ÁflL§h 1 ◊Ê‹ •ÊÚÄ‚Ë¡Ÿ∑§ ÁflSÃÊÁ⁄UÃ „ÊŸ ¬⁄U Á∑§ÿÊ „È•Ê ∑§Êÿ¸ (kJ ◊¥) „ÊªÊ —

• • • •• • • ••• • • • •••• •• •• •••

(R = 8.3 J K−1

mol−1)

(1) 62.25

(2) 124.50

(3) 249.00

(4) 498.00

36. Á∑§‚Ë Áfl‡Ê· ÃÊ¬ ¬⁄U, ∞∑§ ‹fláÊ MX2

∑§ ¡‹ËÿÁfl‹ÿŸ ∑§Ê flÊã≈U •ÊÚ»§ »Ò§Ä≈U⁄U 2 „Ò– ‹fláÊ ∑§ ß‚Áfl‹ÿŸ ∑§ Á‹∞ ÁflÿÊ¡Ÿ ◊ÊòÊÊ „ÊªË —(1) 0.33

(2) 0.50

(3) 0.67

(4) 0.80

37. ∞∑§ ’¥Œ (‚ËÀ«U) ÁŸflÊ¸ÁÃÃ ¬ÊòÊ ◊¥ ⁄UπÊ ªÿÊ ∆UÊ‚ XY

ÁflÉÊÁ≈UÃ „Ê∑§⁄U ÃÊ¬ T ¬⁄U ŒÊ ªÒ‚¥ X ÃÕÊ Y ∑§Ê Á◊üÊáÊ’ŸÊÃÊ „Ò– ß‚ ¬ÊòÊ ◊¥ ‚Êêÿ ŒÊ’ 10 bar „Ò– ß‚•Á÷Á∑˝§ÿÊ ∑§ Á‹ÿ K

p „ÊªÊ —

(1) 5

(2) 10

(3) 25

(4) 100

35. If 100 mole of H2O

2 decompose at 1 bar and

300 K, the work done (kJ) by one mole of

O2(g) as it expands against 1 bar pressure

is :

• • • •• • • ••• • • • •••• •• •• •••(R = 8.3 J K

−1 mol

−1)

(1) 62.25

(2) 124.50

(3) 249.00

(4) 498.00

36. An aqueous solution of a salt MX2

at certain

temperature has a van’t Hoff factor of 2.

The degree of dissociation for this solution

of the salt is :

(1) 0.33

(2) 0.50

(3) 0.67

(4) 0.80

37. A solid XY kept in an evacuated sealed

container undergoes decomposition to

form a mixture of gases X and Y at

temperature T. The equilibrium pressure

is 10 bar in this vessel. Kp

for this reaction

is :

(1) 5

(2) 10

(3) 25

(4) 100


Set - 03 26

38. k[¼k_¡V$ Ape__p Ap¡[¼kX¡$i_\u Cr\gu_ A_¡ L$pb®_X$pep¡¼kpCX$ hpeyAp¡ DÐ`Þ_ \pe R>¡. `p¡V¡$riedk[¼k_¡V$_p S>gue ÖphZdp„ 0.2 a¡fpX¡$ rhÛysâhpl kpfL$fsp, STP `f (1 atm A_¡ 273 K) A¡ hpey_p¡ _y„Ly$g L$v$ (L¡$\p¡X$ A_¡ A¡_p¡X$ b„_¡ `f) ip¡ p¡.(1) 2.24 L

(2) 4.48 L

(3) 6.72 L

(4) 8.96 L

39. _uQ¡ Ap ¡g âq¾$ep dpV¡$_p¡ h¡N-r_ed k [A][B] ìe„S>L$\u ìe¼s (expression) L$f¡g R>¡.

A+B _u`S>

A _u kp„Ösp 0.1 mole (dp¡g) fpMuA¡ A_¡ Å¡ B _ukp„Ösp 0.1 \u h^pfu_¡ 0.3 dp¡g L$fhpdp„ Aph¡ sp¡h¡NAQmp„L$ iy„ li¡ ?(1) k

(2) k/3

(3) 3k

(4) 9k

40. L¡$V$gpL$ L$rggp¡_p õhZp¯L$ (Np¡ëX$ _„bf) Ap âdpZ¡ R>¡.Æg¡V$u_ (Gelatin) : 0.005 - 0.01, Nd Af¡rbL$(Gum Arabic) : 0.15 - 0.25; Ap ¡ rgA¡V $ $(Oleate) : 0.04 - 1.0; õV$pQ® (Starch) : 15 - 25.

Apdp„\u L$ep¡ kp¥\u h^pfp¡ kpfp¡ frns L$rgg R>¡ ?

(1) Æg¡V$u_ (Gelatin)

(2) Nd Af¡rbL$ (Gum Arabic)

(3) Ap¡rgA¡V$ (Oleate)

(4) õV$pQ® (Starch)

38. ‚ÁÄ‚Ÿ≈U •ÊÿŸ ∑§ •ÊÚÄ‚Ë∑§⁄UáÊ ‚ ∞ÁÕ‹ËŸ ÃÕÊ ∑§Ê’Ÿ«UÊß•ÊÚÄ‚Êß«U ªÒ‚¥ ’ŸÃË „Ò¥– ¬Ê≈Ò˜UÁ‡Êÿ◊ ‚ÁÄ‚Ÿ≈U ∑§¡‹Ëÿ Áfl‹ÿŸ ‚ 0.2 »Ò§⁄UÊ«U ÁfllÈÃ ¬˝flÊÁ„Ã ∑§⁄UŸ ¬⁄UªÒ‚Ê¥ ∑§Ê ∑È§‹ •ÊÿÃŸ (∑Ò§ÕÊ«U ÃÕÊ ∞ŸÊ«U ŒÊŸÊ¥ ¬⁄U)STP (1 atm ÃÕÊ 273 K) ¬⁄U „ÊªÊ —(1) 2.24 L

(2) 4.48 L

(3) 6.72 L

(4) 8.96 L

39. ŸËø ŒË ªß¸ •Á÷Á∑˝§ÿÊ ∑§ Á‹∞ Œ⁄U ÁŸÿ◊ k [A][B]

√ÿ¥¡∑§ ‚ √ÿÄÃ Á∑§ÿÊ ¡ÊÃÊ „Ò

A+B ©à¬ÊŒ

A ∑§Ë ‚ÊãŒ˝ÃÊ ∑§Ê ◊ÊŸ 0.1 ◊Ê‹ ¬⁄U ⁄UπÃ „È∞ ÿÁŒ B∑§Ë ‚ÊãŒ˝ÃÊ 0.1 ‚ ’…∏Ê∑§⁄U 0.3 ◊Ê‹ ∑§⁄U ŒË ¡ÊÃË „Ò ÃÊŒ⁄U ÁSÕ⁄UÊ¥∑§ „ÊªÊ —(1) k

(2) k/3

(3) 3k

(4) 9k

40. ∑È§¿U ∑§Ê ‹Êß«UÊ ¥ ∑ § SfláÊÊ Z∑§ „ Ò ¥ , Á¡‹Á≈ UŸ —0.005 - 0.01, ª◊ ∞⁄ U Á’∑§ — 0.15 - 0.25;

•ÊÁ‹∞≈U — 0.04 - 1.0; S≈UÊø¸ — 15 - 25, ßŸ◊¥ ∑§ÊÒŸ-‚Ê ’„Ã⁄U ⁄UˇÊË ∑§Ê‹Êÿ«U „ÊªÊ?

(1) Á¡‹Á≈UŸ

(2) ª◊ ∞⁄UÁ’∑§

(3) •ÊÁ‹∞≈U

(4) S≈UÊø¸

38. Oxidation of succinate ion produces

ethylene and carbon dioxide gases. On

passing 0.2 Faraday electricity through an

aqueous solution of potassium succinate,

the total volume of gases (at both cathode

and anode) at STP (1 atm and 273 K) is :

(1) 2.24 L

(2) 4.48 L

(3) 6.72 L

(4) 8.96 L

39. The rate law for the reaction below is given

by the expression k [A][B]

A+B Product

If the concentration of B is increased from

0.1 to 0.3 mole, keeping the value of A at

0.1 mole, the rate constant will be :

(1) k

(2) k/3

(3) 3k

(4) 9k

40. Gold numbers of some colloids

are : Gelatin : 0.005 - 0.01, Gum

Arabic : 0.15 - 0.25; Oleate : 0.04 - 1.0;

Starch : 15 - 25. Which among these is a

better protective colloid ?

(1) Gelatin

(2) Gum Arabic

(3) Oleate

(4) Starch


Set - 03 27

41. _uQ¡_p rh^p_p¡ Aphs®L$p¡ô$dp„_p sÒhp¡_¡ k„b„r^s R>¡._uQ¡_pdp„\u L$ey„ kpQy„ R>¡ ?

(1) kd|l 17 dp„ b^p S> sÒhp¡ hpeyAp¡ R>¡.

(2) kd|l 13 dp„ b^p S> sÒhp¡ ^psyAp¡ R>¡.

(3) Aphs®_¡ A_ygnu_¡ kd|l 15 _p sÒhp¡_usyg_pdp„ kd|l 16 _p sÒhp¡_u â\d Ape_uL$fZA¡Þ\pëù Ap¡R>u R>¡.

(4) kd|l 15 _p sÒhp¡ dpV¡$, kd|ldp„ _uQ¡ S>CA¡s¡d +5 Ap¡[¼kX¡$i_ Ahõ\p_u [õ\fsp h ¡R>¡.

42. âÖphZ (smelting) Üpfp L$p¡ f_y„ r_óL$j®Z L$fsu hMs¡Dd¡f¡g rkrgL$p_p¡ Dèp¡N _uQ¡_pdp„\u iy„ v|$f L$fhp\pe R>¡ ?(1) Cu

2S

(2) FeO

(3) FeS

(4) Cu2O

43. âq¾$ep Ap¡mMu bsphp¡ L¡$ S>¡dp„ lpCX²$p¡S>_ DÐkS>®_\sp¡ _\u.

(1) S>gue ApëL$gu_u T]L$ kp\¡_u âq¾$ep

(2) Pt Cg¡¼V²$p¡X$p¡_p¡ Dèp¡N L$fu_¡ A¡rkqX$L$ `pZu_y„rhÛysrhcpS>_

(3) âhplu A¡dp¡r_epdp„ kp¡qX$ed_p ÖphZ_¡ fpMu_¡R>p¡X$u v$¡hy„

(4) B2H

6 kp\¡ rgr\ed lpCX²$pCX$_u âq¾$ep

41. ÁŸêŸ ∑§ÕŸ •ÊflÃ ÃÊÁ‹∑§Ê ◊¥ ©¬ÁSÕÃ ÃàflÊ¥ ‚ ‚ê’¥ÁœÃ„Ò¥– ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ-‚Ê ‚àÿ „Ò?

(1) ªÈ¬ 17 ◊¥ ‚÷Ë Ãàfl ªÒ‚ „Ò¥–

(2) ªÈ¬ 13 ◊¥ ‚÷Ë Ãàfl œÊÃÈ „Ò¥–

(3) ªÈ¬ 15 ∑§ ÃàflÊ¥ ∑§Ë ÃÈ‹ŸÊ ◊¥ ‚¥ªÃ •ÊflÃ¸ ∑§ª˝È¬ 16 ∑§ ÃàflÊ¥ ◊¥ •ÊÿŸŸ ∞ãÕÒÀ¬Ë ∑§Ê ◊ÊŸ∑§◊ ⁄U„ÃÊ „Ò–

(4) ª˝È¬ 15 ∑§ ÃàflÊ¥ ∑§ Á‹∞, ª˝È¬ ◊¥ ŸËø ¡ÊŸ ¬⁄U+5 •ÊÚÄ‚Ë∑§⁄UáÊ •flSÕÊ ∑§Ê SÕÊÁÿàfl ’…∏ÃÊ„Ò–

42. S◊ÁÀ≈¥Uª mÊ⁄UÊ ∑§ÊÚ¬⁄U ∑§ ÁŸc∑§·¸áÊ ◊¥ Á‚Á‹∑§Ê ÿÊíÿ ∑§M§¬ ◊¥ ÁŸêŸ ◊¥ ‚ Á∑§‚∑§Ê „≈UÊŸ ∑§ Á‹∞ ∑§Ë ¡ÊÃË „Ò ?(1) Cu

2S

(2) FeO

(3) FeS

(4) Cu2O

43. ©‚ •Á÷Á∑˝§ÿÊ ∑§Ê ’ÃÊß∞ Á¡‚◊¥ „Êß«˛UÊ¡Ÿ ©à‚Á¡¸ÃŸ„Ë¥ „ÊÃË „Ò —

(1) ¡‹Ëÿ ˇÊÊ⁄U ∑§ ‚ÊÕ Á¡¥∑§ ∑§Ë •Á÷Á∑˝§ÿÊ

(2) å‹Á≈UŸ◊ ß‹Ä≈˛UÊ«UÊ¥ ∑§Ê ¬˝ÿÊª ∑§⁄U∑§ •ê‹Ë∑Î§Ã¡‹ ∑§Ê ÁfllÈÃ •¬ÉÊ≈UŸ

(3) Œ˝fl •◊ÊÁŸÿÊ ◊¥ ‚ÊÁ«Uÿ◊ ∑§ Áfl‹ÿŸ ∑§Ê ÁSÕ⁄U„ÊŸ ∑§ Á‹∞ ¿UÊ«∏ ŒŸÊ

(4) B2H

6 ∑§ ‚ÊÕ ‹ËÁÕÿ◊ „Êß«˛ UÊß«U ∑§Ë

•Á÷Á∑˝§ÿÊ

41. The following statements concern elements

in the periodic table. Which of the

following is true ?

(1) All the elements in Group 17 are

gases.

(2) The Group 13 elements are all metals.

(3) Elements of Group 16 have lower

ionization enthalpy values compared

to those of Group 15 in the

corresponding periods.

(4) For Group 15 elements, the stability

of +5 oxidation state increases down

the group.

42. Extraction of copper by smelting uses silica

as an additive to remove :

(1) Cu2S

(2) FeO

(3) FeS

(4) Cu2O

43. Identify the reaction which does not

liberate hydrogen :

(1) Reaction of zinc with aqueous alkali.

(2) Electrolysis of acidified water using

Pt electrodes.

(3) Allowing a solution of sodium in

liquid ammonia to stand.

(4) Reaction of lithium hydride with

B2H

6.


Set - 03 28

44. L¡$ëiued Ap¡¼kpCX$_y„ ìep`pfu ^p¡fZ¡ _pd Ap`p¡.

(1) v|$r^ep¡ Q|_p¡ (rdëL$ Ap¡a gpCd) (Milk of

lime)

(2) ap¡X¡$gp¡ Q|_p¡ (Slaked lime)

(3) Q|_p_p¡ `Õ\f (Limestone)

(4) L$mu Q|_p¡ (Quick lime)

45. L$\_ : L$pb®__p blºê$`p¡dp„, lufp¡ A¡ AhplL$ R>¡Äepf¡ N°¡apCV$ A¡ rhÛys_p kyhplL$ R>¡.

L$pfZ : lufpdp„ A_¡ N°¡apCV$dp„ L$pb®__y„ k„L$fZA_y¾d¡ sp

3 A_¡ sp2 R>¡.

(1) L$\_ A_¡ L$pfZ b„_¡ kpQp R>¡, s\p L$pfZ A¡L$\_ dpV¡$_u kpQu kdS|>su R>¡.

(2) L$\_ A_¡ L$pfZ b„_¡ kpQp R>¡, `Z L$pfZ A¡L$\_ dpV¡$_u kpQu kdS|>su _\u.

(3) L$\_ A¡ AkÐe rh^p_ R>¡. `Z L$pfZ kÐeR>¡.

(4) L$\_ A_¡ L$pfZ b„_¡ AkÐe R>¡.

46. Mp¡Vy„$ rh^p_ ip¡ u bsphp¡.

(1) S2 A¡ Ap¡[¼kS>__u dpaL$ A_yQy„bL$ue R>¡.

(2) f¹lp¡[çbL$ A_¡ dp¡_p¡[¼gr_L$ këafdp„ S8

AÏAp¡ R>¡.

(3) S8 hge_p¡ ApL$pf dyNV$ (¾$pD_) S>¡hp¡ R>¡.

(4) S8 A_¡ S

6 hgep¡dp„ S-S-S b„^ M|ZpAp¡

A¡L$kfMp R>¡.

44. ∑Ò§ÁÀ‡Êÿ◊ •ÊÚÄ‚Êß«U ∑§Ê √ÿÊfl‚ÊÁÿ∑§ ŸÊ◊ „Ò —

(1) Á◊À∑§ •ÊÚ»§ ‹Êß◊

(2) S‹ÒÄ«U ‹Êß◊

(3) ‹Êß◊S≈UÊŸ

(4) ÁÄfl∑§ ‹Êß◊

45. ∑§ÕŸ — ∑§Ê’Ÿ ∑§ •¬⁄UM§¬Ê¥ ◊¥, «UÊÿ◊¥«U ∑È§øÊ‹∑§„Ò ¡’ Á∑§ ª˝»§Êß≈U ∞∑§ ÁfllÈÃ ‚ÈøÊ‹∑§„Ò–

∑§Ê⁄UáÊ — «UÊÿ◊ã«U ÃÕÊ ª˝»§Êß≈U ◊¥ ∑§Ê’¸Ÿ ∑§Ê‚¥∑§⁄UáÊ ∑˝§◊‡Ê— sp

3 ÃÕÊ sp

2 „Ò–

(1) ∑§ÕŸ ÃÕÊ ∑§Ê⁄UáÊ ŒÊŸÊ¥ ‚„Ë „Ò¥ ÃÕÊ ∑§Ê⁄UáÊ ∑§ÕŸ∑§Ë ‚„Ë √ÿÊÅÿÊ „Ò–

(2) ∑§ÕŸ ÃÕÊ ∑§Ê⁄UáÊ ŒÊŸÊ¥ ‚„Ë „Ò¥ ¬⁄UãÃÈ ∑§Ê⁄UáÊ,∑§ÕŸ ∑§Ë ‚„Ë √ÿÊÅÿÊ Ÿ„Ë¥ „Ò–

(3) ∑§ÕŸ •‚àÿ „Ò ¬⁄UãÃÈ ∑§Ê⁄UáÊ ‚àÿ „Ò–

(4) ∑§ÕŸ ÃÕÊ ∑§Ê⁄UáÊ ŒÊŸÊ¥ „Ë •‚àÿ „Ò¥–

46. •‚àÿ ∑§ÕŸ ∑§Ê ¬„øÊÁŸ∞ —

(1) •ÊÚÄ‚Ë¡Ÿ ∑§Ë Ã⁄U„ S2 •ŸÈøÈê’∑§Ëÿ „Ò–

(2) ⁄UÊÁê’∑§ (Áfl·◊‹¥’ÊˇÊ) ÃÕÊ ◊ÊŸÊÄ‹ËÁŸ∑§‚À»§⁄U ◊¥ S

8 •áÊÈ „Ò¥–

(3) S8 fl‹ÿ ∑§Ê •Ê∑§Ê⁄U ◊È∑È§≈U ∑§Ë Ã⁄U„ „Ò–

(4) S8 ÃÕÊ S

6 fl‹ÿÊ¥ ◊¥ S-S-S •Ê’ãœ ∑§ÊáÊ ∞∑§

¡Ò‚ „Ò¥–

44. The commercial name for calcium oxide

is :

(1) Milk of lime

(2) Slaked lime

(3) Limestone

(4) Quick lime

45. Assertion : Among the carbon

allotropes, diamond is an

insulator, whereas,

graphite is a good

conductor of electricity.

Reason : Hybridization of carbon

in diamond and graphite

are sp3 and sp

2,

respectively.


and the reason is the correct



but the reason is not the correct


(3) Assertion is incorrect statement, but

the reason is correct.

(4) Both assertion and reason are

incorrect.

46. Identify the incorrect statement :

(1) S2 is paramagnetic like oxygen.

(2) Rhombic and monoclinic sulphur

have S8 molecules.

(3) S8 ring has a crown shape.

(4) The S-S-S bond angles in the S8 and

S6 rings are the same.


Set - 03 29

47. kpQy„ rh^p_ ip¡ p¡.

(1) Ape_® (gp¡M„X$) Ap¡[¼kS>_-dy¼s `pZudp„npqfs \pe R>¡.

(2) npfhpmp `pZudp„ Ape_® (gp¡M„X$) M|b S>TX$`\u npqfs \pe R> ¡ L $pfZ L ¡ $ s ¡_p ¡rhÛysfpkperZL$ `p¡V¡$[Þieg KQp¡ R>¡.

(3) Ape_® (gp¡M„X$) _y„ npfZ, Äepf¡ s¡_p¡ buÆKQp fuX$n_ `p¡V¡$[Þieg ^fphsu ^psy kp\¡k„`L®$ L$fsp Ap¡Ry>„ L$fu iL$pe R>¡.

(4) Ape_® (gp¡M„X$) _y„ npfZ, s¡_u k`pV$u `fA`pfNçe Ahfp¡ b_phu_¡ Ap¡Ry>„ L$fu iL$peR>¡.

48. _uQ¡ Ap` ¡gpAp ¡dp „\u L $e y „ A ¡L $ lp ¡dp ¡g ¡àV $ uL $(homoleptic) kdp_uL©$s k„L$uZ®_y„ Dv$plfZ R>¡ ?(1) [Co(NH

3)6]Cl

3

(2) [Pt(NH3)2Cl

2]

(3) [Co(NH3)4Cl

2]

(4) [Co(NH3)5Cl]Cl

2

49. dpZ¡L$ (Ruby) A_¡ `Þ_p (emerald) dp„ f„Np¡ dpV¡$S>hpbv$pf k„¾$p„rs ^psy Ape_p¡ A_y¾$d¡ _uQ¡_pdp„\uip¡^p¡.

(1) Cr3+

A_¡ Co3+

(2) Co3+

A_¡ Cr3+

(3) Co3+

A_¡ Co3+

(4) Cr3+

A_¡ Cr3+

47. ‚„Ë ∑§ÕŸ ∑§Ê ¬„øÊÁŸÿ —

(1) •Êÿ⁄UŸ •ÊÚÄ‚Ë¡Ÿ-◊ÈÄÃ ¡‹ ◊¥ ‚¥ˇÊÊÁ⁄UÃ „ÊÃÊ„Ò–

(2) ‹fláÊËÿ ¡‹ ◊¥ •Êÿ⁄UŸ ¡ÀŒË ‚ ‚¥ˇÊÊÁ⁄UÃ „ÊÃÊ„Ò ÄÿÊ¥Á∑§ ß‚∑§Ê ÁfllÈÃ ⁄UÊ‚ÊÿÁŸ∑§ Áfl÷fl ©ìÊ„Ò–

(3) •Êÿ⁄UŸ ∑§Ê ‚¥ ÊÊ⁄UáÊ ß‚∑§Ê ©ìÊ •¬øÿŸ Áfl÷flflÊ‹ œÊÃÈ ∑§ ‚ê¬∑¸§ ◊¥ ‹ÊŸ ¬⁄U ∑§◊ Á∑§ÿÊ ¡Ê‚∑§ÃÊ „Ò–

(4) •Êÿ⁄UŸ ∑§Ê ‚¥ˇÊÊ⁄UáÊ ß‚∑§ ‚Ã„ ¬⁄U •¬Ê⁄Uªêÿ•fl⁄UÊœ ’ŸÊ∑§⁄U ∑§◊ Á∑§ÿÊ ¡Ê ‚∑§ÃÊ „Ò–

48. ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ „Ê◊Ê‹Áå≈U∑§ (homoleptic) ‚¥∑È§‹∑§Ê ∞∑§ ©ŒÊ„⁄UáÊ „Ò?(1) [Co(NH

3)6]Cl

3

(2) [Pt(NH3)2Cl

2]

(3) [Co(NH3)4Cl

2]

(4) [Co(NH3)5Cl]Cl

2

49. M§’Ë ∞fl¥ ß◊⁄UÊÀ«U ◊¥ Á¡Ÿ ‚¥∑˝§◊áÊ œÊÃÈ•Ê¥ ∑§ •ÊÿŸÊ¥∑§Ë ©¬ÁSÕÁÃ ∑§ ∑§Ê⁄UáÊ ⁄¥Uª „ÊÃÊ „Ò, fl ∑˝§◊‡Ê— „Ò¥ —

(1) Cr3+

ÃÕÊ Co3+

(2) Co3+

ÃÕÊ Cr3+

(3) Co3+

ÃÕÊ Co3+

(4) Cr3+

ÃÕÊ Cr3+

47. Identify the correct statement :

(1) Iron corrodes in oxygen-free water.

(2) Iron corrodes more rapidly in salt

water because its electrochemical

potential is higher.

(3) Corrosion of iron can be minimized

by forming a contact with another

metal with a higher reduction

potential.

(4) Corrosion of iron can be minimized

by forming an impermeable barrier

at its surface.

48. Which of the following is an example of

homoleptic complex ?

(1) [Co(NH3)6]Cl

3

(2) [Pt(NH3)2Cl

2]

(3) [Co(NH3)4Cl

2]

(4) [Co(NH3)5Cl]Cl

2

49. The transition metal ions responsible

for color in ruby and emerald are,

respectively :

(1) Cr3+

and Co3+

(2) Co3+

and Cr3+

(3) Co3+

and Co3+

(4) Cr3+

and Cr3+


Set - 03 30

50. X²$pe¼gu_]Ndp„ _uQ¡ Ap ¡gp `v$p\p£dp„\u L$ep¡ A¡L$_p¡D`ep¡N hpsphfZue âv|$jZ_¡ r_e„ÓZ L$fhpdp„ h yL$pe®nd R>¡ ?

(1) V¡$V²$p¼gp¡fp¡Cr\rg_

(2) L$pb®_ X$pep¡¼kpCX

(3) këaf X$pep¡¼kpCX$

(4) _pCV²$p¡S>_ X$pep¡¼kpCX$

51. l¡gp¡S>_p¡_u L$kp¡V$u L$fsp l¡gp kp¡qX$ed AL®$ (extract)

_¡ kp„Ö HNO3 _u kp\¡ Nfd L$fhpdp„ Aph¡ R>¡ L$pfZ

L¡$...

(1) rkëhf l¡gpCX$p¡ _pCV²$uL$ A¡rkXdp„ k„`|Z® fus¡AÖpìe R>¡.

(2) A¡rkqX$L$ dpÝeddp„ Ag2S A_¡ AgCN Öpìe

\pe R>¡.

(3) Å¡ S2− A_¡ CN

− lpS>f lp¡e sp¡ kp„ÖHNO

3 \u rhOV$_ \pe R>¡. s¡\u `funZdp„

(L$kp¡V$udp„) v$MgNufu L$fsp„ _\u.

(4) A¡rkqX$L$ dpÝeddp„ Ag _u âq¾$ep l¡gpCX$p¡ kp\¡TX$`u \pe R>¡.

50. «˛UÊß¸Ä‹ËÁŸ¥ª ◊¥ ¬˝ÿÈÄÃ ÁŸêŸ ¬ŒÊÕÊZ ◊¥ ‚ Á∑§‚∑§Ê ¬˝ÿÊªflÊÃÊfl⁄UáÊ ¬˝ŒÍ·áÊ ∑§ ÁŸÿ¥òÊáÊ ∑§Ë ’„Ã⁄U ∑§Êÿ¸ ŸËÁÃ „Ò?

(1) ≈U≈˛UÊÄ‹Ê⁄UÊ∞ÁÕ‹ËŸ

(2) ∑§Ê’¸Ÿ «UÊß•ÊÚÄ‚Êß«U

(3) ‚À»§⁄U «UÊß•ÊÚÄ‚Êß«U

(4) ŸÊß≈˛UÊ¡Ÿ «UÊß•ÊÚÄ‚Êß«U

51. „Ò‹Ê¡Ÿ˜‚ ∑§Ë ¡Ê°ø ∑§ ¬„‹ ‚ÊÁ«Uÿ◊ ∞ÄS≈˛ÒUÄ≈U ∑§Ê‚ÊãŒ˝ HNO

3 ∑§ ‚ÊÕ ª◊¸ Á∑§ÿÊ ¡ÊÃÊ „Ò ÄÿÊ¥Á∑§ —

(1) Á‚Àfl⁄U „Ò‹Êß«U ŸÊßÁ≈˛U∑§ •ê‹ ◊¥ ¬ÍáÊ¸M§¬áÊ•ÉÊÈ‹Ÿ‡ÊË‹ „Ò¥–

(2) •ê‹Ëÿ ◊Êäÿ◊ ◊ ¥ Ag2S ÃÕÊ AgCN

ÉÊÈ‹Ÿ‡ÊË‹ „Ò¥–

(3) ÿÁŒ S2−

ÃÕÊ CN−

©¬ÁSÕÃ „Ò¥ ÃÊ ‚ÊãŒ˝HNO

3 ‚ ÁflÉÊÁ≈UÃ „Ê ¡ÊÃ „Ò¥ ß‚Á‹ÿ ¬⁄UËˇÊáÊ

◊¥ „SÃˇÊ¬ Ÿ„Ë¥ ∑§⁄UÃ–

(4) •ê‹Ëÿ ◊Êäÿ◊ ◊¥ Á‚Àfl⁄U, „Ò‹Êß«UÊ¥ ∑§ ‚ÊÕÃ¡ •Á÷Á∑˝§ÿÊ ∑§⁄UÃÊ „Ò–

50. Which one of the following substances used

in dry cleaning is a better strategy to control

environmental pollution ?

(1) Tetrachloroethylene

(2) Carbon dioxide

(3) Sulphur dioxide

(4) Nitrogen dioxide

51. Sodium extract is heated with concentrated

HNO3 before testing for halogens

because :

(1) Silver halides are totally insoluble in

nitric acid.

(2) Ag2S and AgCN are soluble in acidic

medium.

(3) S2−

and CN−

, if present, are

decomposed by conc. HNO3 and

hence do not interfere in the test.

(4) Ag reacts faster with halides in acidic

medium.


Set - 03 31

52. kpe¼gp ¡l ¡¼Tu__y „ b ° p ¡ rd_ ¡i_ Ap`¡g _uQ¡_u`qf[õ\rsAp¡dp„ L$C _u`S> Ap`i¡ ?

• •• • • •

(1)

• •

• •

(2)

• •

(3)

• •

• •

(4)

• •

• •

52. ŸËø ÁŒÿ ªÿ ¬ÁÃ’ãœÊ¥ ◊¥ ‚ÊßÄ‹Ê„Ä‚ËŸ ∑§Ê ’Ê◊ËŸ‡ÊŸŒÃÊ „Ò —

• •• • • •

(1)

• •

• •

(2)

• •

(3)

• •

• •

(4)

• •

• •

52. Bromination of cyclohexene under

conditions given below yields :

• •• • • •

(1)

• •

• •

(2)

• •

(3)

• •

• •

(4)

• •

• •


Set - 03 32

53. _uQ¡ Ap`¡g âq¾$ep_p¡ ¾$d Ýep_dp„ gp¡.

• • • •

X iy„ R>¡ ?

(1)

• •

• •

• •• •

(2)

• •

• •

• • • •

(3)

• • • •

53. ŸËø ŒË ªß¸ •Á÷Á∑˝§ÿÊ ∑˝§◊ ¬⁄U ÁfløÊ⁄U ∑§ËÁ¡∞ —

• • • •

X

„Ò —

(1)

• •

• •

• •• •

(2)

• •

• •

• • • •

(3)

• • • •

53. Consider the reaction sequence below :

• • • •

X

is :

(1)

• •

• •

• •• •

(2)

• •

• •

• • • •

(3)

• • • •


Set - 03 33

(4)

• •• •

54. _uQ¡ Ap`¡gpdp„\u L$ep¡ âq¾$eL$ A¡ rhgp¡`_ âq¾$ep dpV¡$kyk„Ns _\u ?

• •

(1) NaOH/H2O

(2) NaOEt/EtOH

(3) NaOH/H2O-EtOH

(4) NaI

55. PETN _u b_phV $dp „ h`fpsp ¡ Cfu\ ° uV $ p ¡g(C(CH

2OH)

4) _p k„ïg¡jZ dpV¡$ L$ey„ rh^p_ kpQy„

R>¡ ?

(1) k„ïg¡jZ dpV¡$ rd\¡_p¡g A_¡ C\¡_p¡g_u hÃQ¡Qpf ApëX$p¡g k„O___u AphíeL$sp lp¡e R>¡.

(2) k„ïg¡jZdp„ b¡ ApëX$p¡g k„O__ A_¡ b¡ L¡$r_Tpfp¡âq¾$ep_u AphíeL$sp R>¡.

(3) k„ïg¡jZdp„ ÓZ ApëX$p¡g k„O__ A_¡ A¡L$L¡$r_Tpfp¡ âq¾$ep_u AphíeL$sp R>¡.

(4) Ap âq¾$epdp„ C\¡_p¡g_p¡ Apëap lpCX²$p¡S>_ A_¡rd\¡_p¡g cpN g¡ R>¡.

(4)

• •• •

54. Áfl‹Ê¬Ÿ •Á÷Á∑˝§ÿÊ ∑§ Á‹∞ ßŸ •Á÷∑§Ê⁄U∑§Ê¥ ◊¥ ‚∑§ÊÒŸ-‚Ê ©¬ÿÈÄÃ Ÿ„Ë¥ „Ò?

• •

(1) NaOH/H2O

(2) NaOEt/EtOH

(3) NaOH/H2O-EtOH

(4) NaI

55. PETN ∑§ ’ŸÊŸ ◊¥ ¬˝ÿÈÄÃ ß⁄UËÁÕ˝≈UÊÚ‹ (C(CH2OH)

4)

∑§ ‚¥‡‹·áÊ ∑§ ‚ê’ãœ ◊¥ ‚„Ë ∑§ÕŸ „Ò —

(1) ‚¥‡‹·áÊ ◊¥ ◊ÕŸÊÚ‹ ÃÕÊ ∞ÕŸÊÚ‹ ∑§ ’Ëø øÊ⁄U∞À«UÊ‹ ‚¥ÉÊŸŸ ∑§Ë •Êfl‡ÿ∑§ÃÊ „ÊÃË „Ò–

(2) ‚¥‡‹·áÊ ◊¥ ŒÊ ∞À«UÊ‹ ‚¥ÉÊŸŸ ÃÕÊ ŒÊ ∑Ò§ÁŸ¡Ê⁄UÊ•Á÷Á∑˝§ÿÊ ∑§Ë ¡M§⁄UÃ „ÊÃË „Ò–

(3) ‚¥‡‹·áÊ ◊¥ ÃËŸ ∞À«UÊ‹ ‚¥ÉÊŸŸ ÃÕÊ ∞∑§∑Ò§ÁŸ¡Ê⁄UÊ •Á÷Á∑˝§ÿÊ ∑§Ë •Êfl‡ÿ∑§ÃÊ „ÊÃË „Ò–

(4) ß‚ •Á÷Á∑˝§ÿÊ ◊¥ ∞ÕŸÊÚ‹ ∑§ •À»§Ê „Êß«˛UÊ¡ŸÃÕÊ ◊ÕŸÊÚ‹ ÷Êª ‹Ã „Ò¥–

(4)

• •• •

54. Which one of the following reagents is not

suitable for the elimination reaction ?

• •

(1) NaOH/H2O

(2) NaOEt/EtOH

(3) NaOH/H2O-EtOH

(4) NaI

55. The correct statement about the synthesis

of erythritol (C(CH2OH)

4) used in the

preparation of PETN is :

(1) The synthesis requires four aldol

condensations between methanol

and ethanol.

(2) The synthesis requires two aldol

condensations and two Cannizzaro

reactions.

(3) The synthesis requires three aldol

condensations and one Cannizzaro

reaction.

(4) Alpha hydrogens of ethanol and

methanol are involved in this

reaction.


Set - 03 34

56. L$p¡C A¡fp¡d¡qV$L$ hge_y„ ãgp¡qf_¡i_ M|b S> kl¡gpC\uk„ch lp¡e R>¡, Äepf¡ s¡_p X$peTp¡_ued npf_¡ HBF

4

kp\¡ âq¾$ep L$fhpdp„ Aph¡ R>¡. D`f_u °q¾$ep dpV¡$ _uQ¡_pAp`¡gpdp„\u L$C `qf[õ\rs kpQu R>¡ ?

(1) a¼s Dódp(2) NaNO

2/Cu

(3) Cu2O/H

2O

(4) NaF/Cu

57. _uQ¡ Ap ¡gpdp„\u L$ep blºgL$_y„ k„ïg¡jZ A¡ dy¼s dygL$blºguL$fZ sL$_uL$_p¡ D`ep¡N L$fu_¡ L$fhpdp„ Aph¡ R>¡?

(1) V¡$agp¡_

(2) V¡$qfrg_

(3) d¡g¡dpC_ blºgL$

(4) _pegp¡_ 6,6

58. “N” L¡$ S>¡ Ap ¡g k„ep¡S>__u b¡rTL$spdp„ cpN g¡sp¡_\u s¡ bsphp¡.

•

• •

•

•

••

••

••

•• •

(1) N 7

(2) N 9

(3) N 1

(4) N 3

56. Á∑§‚Ë ∞⁄UÊÒ◊ÒÁ≈U∑§ fl‹ÿ ∑§Ê ç‹È•Ê⁄UË∑§⁄UáÊ •Ê‚ÊŸË ‚‚¥÷fl „ÊÃÊ „Ò ÿÁŒ ©‚∑§ «UÊß∞¡ÊÁŸÿ◊ ‹fláÊ ∑§ÊHBF

4 ∑§ ‚ÊÕ ©¬øÊÁ⁄UÃ Á∑§ÿÊ ¡Êÿ– ß‚ •Á÷Á∑˝§ÿÊ

∑§ ‚ê’ãœ ◊¥ ÁŸêŸÁ‹ÁπÃ ◊¥ ‚ ∑§ÊÒŸ‚Ë ¬Á⁄UÁSÕÁÃ©¬ÿÈÄÃ „Ò?

(1) ∑§fl‹ ™§c◊Ê(2) NaNO

2/Cu

(3) Cu2O/H

2O

(4) NaF/Cu

57. ÁŸêŸÁ‹ÁπÃ ◊¥ ‚ ∑§ÊÒŸ-‚Ê ’„È‹∑§ ◊ÈÄÃ ◊Í‹∑§’„È‹∑§Ë∑§⁄UáÊ ÁflÁœ mÊ⁄UÊ ‚¥‡‹Á·Ã Á∑§ÿÊ ¡ÊÃÊ „Ò?

(1) ≈Uç‹ÊÚŸ

(2) ≈ÒU⁄UË‹ËŸ

(3) ◊Ò‹Ò◊ÊßŸ ’„È‹∑§

(4) ŸÊÿ‹ÊÚŸ 6,6

58. fl„ “N” ¡Ê ÁŸêŸ ÿÊÒÁª∑§ ∑§Ë ÊÊ⁄UËÿ ¬˝flÎÁûÊ ◊¥ ÿÊªŒÊŸŸ„Ë¥ ŒÃÊ „Ò, fl„ „Ò —

•

• •

•

•

••

••

••

•• •

(1) N 7

(2) N 9

(3) N 1

(4) N 3

56. Fluorination of an aromatic ring is easily

accomplished by treating a diazonium salt

with HBF4. Which of the following

conditions is correct about this reaction ?

(1) Only heat

(2) NaNO2/Cu

(3) Cu2O/H

2O

(4) NaF/Cu

57. Which of the following polymers is

synthesized using a free radical

polymerization technique ?

(1) Teflon

(2) Terylene

(3) Melamine polymer

(4) Nylon 6,6

58. The “N” which does not contribute to the

basicity for the compound is :

•

• •

•

•

••

••

••

•• •

(1) N 7

(2) N 9

(3) N 1

(4) N 3


Set - 03 35

59. _uQ¡_pdp„\u L$ey„ A¡L$ b¡¼V¡$qfep_piL$ ârsÆhuAp¡(antibiotic) R>¡ ?

(1) Cqf\°p¡dpeku_

(2) V¡$V²$pkpe¼gu_

(3) ¼gp¡fA¡ça¡r_L$p¡g

(4) Ap¡ãgp¡¼k¡rk_

60. ""fyldp_ Å„byX$uep¡''(` ®g) (Rhumann’s purple)

Ahgp¡L$_ A¡ r_Zp®eL$ L$kp¡V$u _uQ¡_pdp„\u L$p¡_u lpS>fuk|Qh¡ R>¡ ?

(1) fuX$éyk]N iL®$fp

(2) ¼eyâuL$ Ape_

(3) âp¡V$u_

(4) õV$pQ®

61. ^pfp ¡ L ¡ $ P : sin cos 2 cos • • • •= − =

A_¡ Q : sin cos 2 sin • • • •••= + = b¡

NZp¡ R>¡. sp¡ :

(1) P ⊂ Q A_¡ Q−P ≠ φ

(2) Q ⊄ P

(3) P ⊄ Q

(4) P = Q

59. ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ-‚Ê ’ÒÄ≈UËÁ⁄UÿÊŸÊ‡ÊË ¬˝ÁÃ¡ÒÁfl∑§ „Ò?

(1) ∞Á⁄UÕ˝Ê◊Êß‚ËŸ

(2) ≈U≈˛UÊ‚ÊÿÄ‹ËŸ

(3) Ä‹Ê⁄U∞ê»Ò§ÁŸ∑§Ê‹

(4) •ÊÚ»˜§‹ÊÄ‚Ò‚ËŸ

60. L§„◊ÒŸ ŸË‹ ‹ÊÁ„Ã (¬¬‹) ∑§Ê ¬∑§≈U „ÊŸÊ ÁŸêŸÁ‹ÁπÃ◊¥ ‚ Á∑§‚∑§Ê ‚¥¬ÈÁc≈U ¬⁄UËˇÊáÊ „Ò?

(1) •¬øÊÿ∑§ ‡Ê∑¸§⁄UÊ

(2) ÄÿÍ¬Á⁄U∑§ •ÊÿŸ

(3) ¬˝Ê≈UËŸ

(4) S≈UÊø¸ (◊¥«U)

61. ◊ÊŸÊ P : sin cos 2 cos • • • •= − = ÃÕÊ

Q : sin cos 2 sin • • • •••= + = ŒÊ‚◊Èëøÿ „Ò¥, ÃÊ —

(1) P ⊂ Q ÃÕÊ Q−P ≠ φ

(2) Q ⊄ P

(3) P ⊄ Q

(4) P = Q

59. Which of the following is a bactericidal

antibiotic ?

(1) Erythromycin

(2) Tetracycline

(3) Chloramphenicol

(4) Ofloxacin

60. Observation of “Rhumann’s purple” is a

confirmatory test for the presence of :

(1) Reducing sugar

(2) Cupric ion

(3) Protein

(4) Starch

61. Let P : sin cos 2 cos • • • •= − = and

Q : sin cos 2 sin • • • •••= + = be two

sets. Then :

(1) P ⊂ Q and Q−P ≠ φ

(2) Q ⊄ P

(3) P ⊄ Q

(4) P = Q


Set - 03 36

62. Å¡ kduL$fZ

•• • • • • • • • • •• •• • ••

• • •

+ − − = _p¡ A¡L$

DL¡$g x R>¡, sp¡ •• • •• • • =__________ \pe.

(1)

••

(2)

••

(3) 2

(4) • •

63. ^pfp¡ L¡$ z=1+ai, a > 0, A¡L$ A¡hu k„L$f k„¿ep R>¡ L¡$S> ¡\u z

3 hpõsrhL$ k„¿ep \pe. sp ¡ kfhpmp ¡

1+z+z2+.....+z

11 =__________ \pe.

(1) • • • • • • ••

(2) • • • • • • •

(3) • • • • • • •

(4) • • • • • • ••

62. ÿÁŒ ‚◊Ë∑§⁄UáÊ

•• • • • • • • • • •• •• • • ••

• • •

+ − − = ∑§Ê x

∞∑§ „‹ „Ò, ÃÊ •• • •• • • ’⁄UÊ’⁄U „Ò —

(1)

••

(2)

••

(3) 2

(4) • •

63. ◊ÊŸÊ z=1+ai, a > 0 ∞∑§ ∞‚Ë ‚Áê◊üÊ ‚¥ÅÿÊ „Ò,Á∑§ z

3 ∞∑§ flÊSÃÁfl∑§ ‚ ¥ÅÿÊ „ Ò , ÃÊ ÿÊ ª

1+z+z2+.....+z

11 ’⁄UÊ’⁄U „Ò —

(1) • • • • • • ••

(2) • • • • • • •

(3) • • • • • • •

(4) • • • • • • ••

62. If x is a solution of the equation,

•

1

2 1 2 1 1, , then

2

4 1 is equal to :

• • •

•

+ − − =

−

(1)

••

(2)

••

(3) 2

(4) • •

63. Let z=1+ai be a complex number, a > 0,

such that z3 is a real number. Then the sum

1+z+z2+.....+z

11 is equal to :

(1) • • • • • • ••

(2) • • • • • • •

(3) • • • • • • •

(4) • • • • • • ••


Set - 03 37

64. ^pfp ¡ L ¡ $ A A¡L $ 3×3 î¡ rZL $ R> ¡ L ¡ $ S > ¡\uA

2−5A+7I=O.

rh^p_ - I :• •• ••• •• • ••

••

= −

rh^p_ - II : blº`v $ u A3−2A

2−3A+I _¡

5(A−4I) dp„ ê$`p„sqfs L$fu iL$pe R>¡.sp¡ __________.

(1) rh^p_-I kÐe R>¡, f„sy rh^p_-II AkÐe R>¡.

(2) rh^p_-I AkÐe R>¡, f„sy rh^p_-II kÐe R>¡.

(3) b„_¡ rh^p_p¡ kÐe R>¡.

(4) b„_¡ rh^p_p¡ AkÐe R>¡.

65. Å¡ • •

• •• •

• •• •• •

• •• , sp ¡ î ¡ rZL $

(A2016

−2A2015

−A2014

) _p ¡ r_òpeL $__________ R>¡.(1) 2014

(2) −175

(3) 2016

(4) −25

64. ◊ÊŸÊ A, 3×3 ∑§Ê ∞∑§ ∞‚Ê •Ê√ÿÍ„ „ Ò Á∑§A

2−5A+7I=O „Ò–

∑§ÕŸ - I : • •• ••• •• • ••

••

= −

∑§ÕŸ - II : ’„È¬Œ A3−2A

2−3A+I ∑§Ê

5(A−4I) ◊¥ ¬Á⁄UflÁÃÃ Á∑§ÿÊ ¡Ê ‚∑§ÃÊ„Ò–

ÃÊ,

(1) ∑§ÕŸ - I ‚àÿ „Ò ‹Á∑§Ÿ ∑§ÕŸ - II •‚àÿ „Ò–

(2) ∑§ÕŸ - I •‚àÿ „Ò ‹Á∑§Ÿ ∑§ÕŸ - II ‚àÿ „Ò–

(3) ŒÊŸÊ¥ ∑§ÕŸ ‚àÿ „Ò¥–

(4) ŒÊŸÊ¥ ∑§ÕŸ •‚àÿ „Ò¥–

65. ÿÁŒ • •

• •• •

• •• •• •

− −

= „ Ò , ÃÊ •Ê√ÿ Í„

(A2016

−2A2015

−A2014

) ∑§Ê ‚Ê⁄UÁáÊ∑§ „Ò —(1) 2014

(2) −175

(3) 2016

(4) −25

64. Let A be a 3×3 matrix such that

A2−5A+7I=O.

Statement - I :

• •• ••• •• • •••

• = −

Statement - II : The polynomial

A3−2A

2−3A+I can be

reduced to 5(A−4I).

Then :

(1) Statement-I is true, but Statement-II

is false.

(2) Statement-I is false, but Statement-II

is true.

(3) Both the statements are true.

(4) Both the statements are false.

65. If

• •• •

• •• •• •• •

− −

= , then the determinant of

the matrix (A2016

−2A2015

−A2014

) is :

(1) 2014

(2) −175

(3) 2016

(4) −25


Set - 03 38

66. Å¡ •

••

•

• • •• • ••

•

•

•

• • sp¡ n _uQ¡_pdp„\u L$ey„ kduL$fZ

k„`p¡j¡?(1) n

2+3n−108=0

(2) n2+5n−84=0

(3) n2+2n−80=0

(4) n2+n−110=0

67. Å¡ ••

••

• ••

••

•

•• • , (x > 0) _p rhõsfZdp„ x−2

A_¡ x−4 _p klNyZL$p¡ A_y¾$d¡ m A_¡ n lp¡e sp¡,

•• =__________ \pe.

(1) 182

(2)

••

(3)

••

(4) 27

68. ^pfp¡ L¡$ a1, a

2, a

3, ......, a

n, ..... kdp„sf î¡Zudp„ R>¡.

Å¡ a3+a

7+a

11+a

15=72, sp¡ s¡_p â\d 17 v$p¡_p¡

kfhpmp¡ __________ \pe.(1) 306

(2) 153

(3) 612

(4) 204

66. ÿÁŒ •

••

•

• • •• • ••

•

•

•

• • „Ò, ÃÊ n ÁŸêŸ ◊¥ ‚ Á∑§‚

‚◊Ë∑§⁄UáÊ ∑§Ê ‚¥ÃÈc≈U ∑§⁄UÃÊ „Ò?(1) n

2+3n−108=0

(2) n2+5n−84=0

(3) n2+2n−80=0

(4) n2+n−110=0

67. ÿÁŒ ••

••

• ••

••

•

•• • , (x > 0), ∑§ ¬˝‚Ê⁄U ◊¥ x−2

ÃÕÊ x−4 ∑§ ªÈáÊÊ¥∑§ ∑˝§◊‡Ê— m ÃÕÊ n „Ò¥, ÃÊ

••

’⁄UÊ’⁄U „Ò —(1) 182

(2)

••

(3)

••

(4) 27

68. ◊ÊŸÊ a1, a

2, a

3, ......, a

n, ..... ∞∑§ ‚◊Ê¥Ã⁄U üÊ…∏Ë ◊¥ „Ò¥–

ÿÁŒ a3+a

7+a

11+a

15=72 „Ò, ÃÊ ©‚∑§ ¬˝Õ◊ 17

¬ŒÊ¥ ∑§Ê ÿÊª ’⁄UÊ’⁄U „Ò —(1) 306

(2) 153

(3) 612

(4) 204

66. If

••

••

• • •• • ••

•

•

•

• • then n satisfies the

equation :

(1) n2+3n−108=0

(2) n2+5n−84=0

(3) n2+2n−80=0

(4) n2+n−110=0

67. If the coefficients of x−2

and x−4

in the

expansion of

••

••

• ••

••

•

•• • , (x > 0), are

m and n respectively, then

•• is equal to :

(1) 182

(2)

••

(3)

••

(4) 27

68. Let a1, a

2, a

3, ......, a

n, ..... be in A.P. If

a3+a

7+a

11+a

15=72, then the sum of its

first 17 terms is equal to :

(1) 306

(2) 153

(3) 612

(4) 204


Set - 03 39

69.( )

• ••

• •• • •• ••

•• ••

•• • • =__________.

(1) (11)!

(2) 10×(11!)

(3) 101×(10!)

(4) 11×(11!)

70.

( )•

•

• • •• • •••• •

• ••• • • ••• • ••

•• • • •→

• •• •

=__________.

(1) −2

(2)

••

• •

(3)

••

(4) 2

69. ÿÊª»§‹ ( )• •

•

• •• • •• ••

•• ••

•• • • ’⁄UÊ’⁄U „Ò —

(1) (11)!

(2) 10×(11!)

(3) 101×(10!)

(4) 11×(11!)

70.

( )•

•

• • •• • •••• •

• ••• • • ••• • ••

•• • • •→

• •• •

’⁄UÊ’⁄U „Ò —

(1) −2

(2)

••

• •

(3)

••

(4) 2

69. The sum ( )• •

•

• •• • •• ••

•• ••

•• • • is equal to :

(1) (11)!

(2) 10×(11!)

(3) 101×(10!)

(4) 11×(11!)

70.

( )•

•

• • •• • •••• •

• ••• • • ••• • ••

•• • • •→

• •• •

is :

(1) −2

(2)

••

• •

(3)

••

(4) 2


Set - 03 40

71. ^pfp¡ L¡$ a, b R, (a ≠ 0).

Å¡ rh^¡e

•

•

•

• • ••• •• • • • •• ••

• •• •••• •• • • • •• • • •

• • ••••• • • • •• •

• ••

• • • •

• • ••

••••••• ∞••

≤

= ≤

−

≤

A„sfpg [0, ∞) dp„ kss lp¡e, sp¡ ¾$dey¼s Å¡X$(a, b) = __________ R>¡.

(1) ( )• ••• • •• •

(2) ( )• ••• • •• • • •

(3) ( )• •• • • •• • •

(4) ( )• ••• • •• • • •

72. ^pfp¡ L¡$ f(x)=sin4x+cos

4x. sp¡ _uQ¡_pdp„\u L$ep

A„sfpgdp„ f h^sy„ rh ¡e R>¡ ?

(1) • •••

• •• •• •

π

(2) ••• •

• •• •• •π π

(3)

•••• •

• •• •• •π π

(4)

• •••• •

• •• •• •π π

71. ◊ÊŸÊ a, b R, (a ≠ 0)– ÿÁŒ »§‹Ÿ f ¡Ê, ÁŸêŸ mÊ⁄UÊ¬Á⁄U÷ÊÁ·Ã „Ò —

•

•

•

• • ••• •• • • • •• ••

• •• •••• •• • • • •• • • •

• • ••••• • • • •• •

• ••

• • • •

• • ••

••••••• ∞••

≤

= ≤

−

≤

•¥Ã⁄UÊ‹ [0, ∞) ◊¥ ‚ÃÃ „Ò, ÃÊ ∞∑§ ∑˝§Á◊Ã ÿÈÇ◊(a, b) „Ò —

(1) ( )• ••• • •• •

(2) ( )• ••• • •• • • •

(3) ( )• •• • • •• • •

(4) ( )• ••• • •• • • •

72. ◊ÊŸÊ f(x)=sin4x+cos

4x „Ò, ÃÊ ÁŸêŸ ◊¥ ‚ Á∑§‚

•¥Ã⁄UÊ‹ ◊¥ f ∞∑§ flœ¸◊ÊŸ »§‹Ÿ „Ò?

(1) • •••

• •• •• •

π

(2) ••• •

• •• •• •π π

(3)

•••• •

• •• •• •π π

(4)

• •••• •

• •• •• •π π

71. Let a, b R, (a ≠ 0). If the function f defined

as

•

•

•

• • ••• •• • • • •• ••

• •• •••• •• • • • •• • • •

• • ••••• • • • •• •

• ••

• • • •

• • ••

••••••• ∞••

≤

= ≤

−

≤

is continuous in the interval [0, ∞), then an

ordered pair (a, b) is :

(1) ( )• ••• • •• •

(2) ( )• ••• • •• • • •

(3) ( )• •• • • •• • •

(4) ( )• ••• • •• • • •

72. Let f(x)=sin4x+cos

4x. Then f is an

increasing function in the interval :

(1) • •••

• •• •• •

π

(2) ••• •

• •• •• •π π

(3)

•••• •

• •• •• •π π

(4)

• •••• •

• •• •• •π π


Set - 03 41

73. ^pfp¡ L¡$ A¡L$ h¾$ C A¡ •• •• • • • • • •• • •• • • • • •

••

• > Üpfp v$ip®h¡g R>¡. Å¡ C `f A¡L$ A¡hy„ tbvy$ P R>¡

L¡$ S>¡\u tbvy$ P ApNm_p õ`i®L$_p¡ Y$pm ••

\pe, sp¡ P

ApNm_p¡ Arcg„b __________ tbvy$dp„\u kpf \peR>¡.(1) (2, 3)

(2) (4, −3)

(3) (1, 7)

(4) (3, −4)

74.

( ) •• • • •

• •

• • •• • • •=__________.

(Äep„ C k„L$g__p¡ AQmp„L$ R>¡.)

(1)

• •• • •• •

••

• •• •• •• •

(2)

• •• • •• •

••

• •• •• •• •

(3)

• •• •• •

••

• •• •• •• •

(4)

• •• • •• •

••

• • •• •• •

73. ◊ÊŸÊ C ∞∑§ fl∑˝§ „Ò ¡Ê •• •• • • • • •• • •• • • • • • ,

••

• > mÊ⁄UÊ ¬˝ŒûÊ „Ò– ÿÁŒ C ¬⁄U P ∞∑§ ∞‚Ê Á’¥ŒÈ „Ò

Á∑§ P ¬⁄U πË¥øË ªß¸ S¬‡Ê¸ ⁄UπÊ ∑§Ë …Ê‹ ••

„Ò, ÃÊ fl„

Á’¥ŒÈ Á¡‚‚ P ¬⁄U πË¥øÊ ªÿÊ •Á÷‹¥’ ªÈ$¡⁄UÃÊ „Ò, „Ò —(1) (2, 3)

(2) (4, −3)

(3) (1, 7)

(4) (3, −4)

74. ‚◊Ê∑§‹ ( ) •• • • •

• •

• • •• • • • ’⁄UÊ’⁄U „Ò —

(¡„Ê° C ∞∑§ ‚◊Ê∑§‹Ÿ •ø⁄U „Ò–)

(1)

• •• • •• •

••

• •• •• •• •

(2)

• •• • •• •

••

• •• •• •• •

(3)

• •• •• •

••

• •• •• •• •

(4)

• •• • •• •

••

• • •• •• •

73. Let C be a curve given by

•• •• • • • • • •• • •= + −

••

• > . If P is a point

on C, such that the tangent at P has slope

••

, then a point through which the normal

at P passes, is :

(1) (2, 3)

(2) (4, −3)

(3) (1, 7)

(4) (3, −4)

74. The integral

( ) •• • • •

• •

• • •• • • • is equal

to :

(where C is a constant of integration.)

(1)

• •• • •• •

••

• •• •• •• •

(2)

• •• • •• •

••

• •• •• •• •

(3)

• • •• •

••

• •• •• •• •

(4)

• •• • •• •

••

• • •• •• •


Set - 03 42

75.

•

• •

•

• • • • • • •

• • •

• • •

• •• •

• • • •• • • •• • • • • •

=_________.

Äep„ [x] A¡ x \u _p_p A\hp x _¡ kdp_ sdpd|Zp¯L$p¡dp„ kp¥\u dp¡V$p¡ |Zp¯L$ v$ip®h¡ R>¡.

(1) 6

(2) 3

(3) 7

(4)

••

76. Å¡ x R, x ≠ 0 dpV¡$, y(x) A¡L$ rhL$g_ue rh^¡e R>¡ L¡$

S > ¡\u • ••• •• • • • • •• • •• ••

• •• • • • • • • • • • •• •• • • • , sp ¡

y(x)=__________.

(Äep„ C AQm R>¡.)

(1)

•• • •••

• •

(2)

•

•• •••

• •

(3)

•

•• • •••

• •

(4)

••• • •• •

75. ‚◊Ê∑§‹ •

• •

•

• • • • • • •

• • •

• • •

• •• •

• • • •• • • •• • • • • •

, ¡„Ê°

[x], x ‚ ∑§◊ ÿÊ x ∑§ ’⁄UÊ’⁄U ◊„ûÊ◊ ¬ÍáÊÊZ∑§ „Ò, ∑§Ê◊ÊŸ „Ò —(1) 6

(2) 3

(3) 7

(4)

••

76. x R, x ≠ 0, ∑§ Á‹∞, ÿÁŒ y(x) ∞∑§ ∞‚Ê •fl∑§‹ŸËÿ»§‹Ÿ „Ò Á∑§

• ••• •• • • • • •• • •• ••

• •• • • • • • • • • • •• •• • • • „Ò, ÃÊ y (x)

’⁄UÊ’⁄U „Ò —

(¡„Ê° C ∞∑§ •ø⁄U „Ò–)

(1)

•• • •••

• •

(2)

•

•• • •••

• •

(3)

•

•• • •••

• •

(4)

••• • •• •

75. The value of the integral

•

• •

•

• • • • • • •

• • •

• • •

• •• •

• • • •• • • •• • • • • •

, where [x]

denotes the greatest integer less than or

equal to x, is :

(1) 6

(2) 3

(3) 7

(4)

••

76. For x R, x ≠ 0, if y(x) is a differentiable

function such that

• ••• •• • • • • •• • •• ••

• •• • • • • • • • • • •• •• • • • , then y(x)

equals :

(where C is a constant.)

(1)

•• • •••

• •

(2)

•

•• • •••

• •

(3)

•

•• • •••

• •

(4)

••• • •• •


Set - 03 43

77. rhL$g kduL$fZ •• •••• • • • •

• • •• • • ••• • ••• • • Äep„

• • • •• ••

• π• , A_¡ y(0)=1, _p¡ DL¡$g _________

R>¡.

(1) • • ••• • • •• •

••• •

• • • •• •

(2)• • • •

•• • • •• •••

• •• • • •

• •

(3)• • • •

•• • • •• •••

• •• • • •

• •

(4) • • ••• • • •• •

••• •

• • • •• •

78. âL$pi_y„ A¡L$ qL$fZ f¡Mp 7x−y+1=0 `f Ap`ps\pe R>¡ S>¡ tbv$y$ (0, 1) ApNm dm¡ R>¡. Ðepfbpv$ ApqL$fZ Ap tbvy$dp„\u `fphrs®s \C f¡Mp y+2x=1

`f fl¡ R>¡. sp¡ Ap`ps qL$fZ_y„ kduL$fZ __________

R>¡.(1) 41x−38y+38=0

(2) 41x+25y−25=0

(3) 41x+38y−38=0

(4) 41x−25y+25=0

77. •fl∑§‹ ‚◊Ë∑§⁄UáÊ •• •••• • • • •

• • •• • • ••• • ••• • • ¡„Ê°

• • • •• ••

• π• „Ò ÃÕÊ y(0)=1 „Ò, ∑§Ê „‹ „Ò —

(1) • • ••• • • •• •

••• •

• • • •• •

(2)• • • •

•• • • •• •••

• •• • • •

• •

(3)• • • •

•• • • •• •••

• •• • • •

• •

(4) • • ••• • • •• •

••• •

• • • •• •

78. ¬˝∑§Ê‡Ê ∑§Ë ∞∑§ Á∑§⁄UáÊ ∞∑§ ⁄UπÊ ∑§Ë ÁŒ‡ÊÊ ◊¥ •Ê¬ÁÃÃ„Ò ¡Ê ∞∑§ •ãÿ ⁄UπÊ 7x−y+1=0 ∑§Ê Á’¥ŒÈ(0, 1) ¬⁄U Á◊‹ÃË „Ò– fl„ Á∑§⁄UáÊ Á»§⁄U ß‚ Á’¥ŒÈ ‚ ⁄UπÊy+2x=1 ∑§Ë ÁŒ‡ÊÊ ◊¥ ¬Á⁄UflÁÃ¸Ã „ÊÃË „Ò, ÃÊ •Ê¬ÁÃÃ¬˝∑§Ê‡Ê ∑§Ë Á∑§⁄UáÊ ∑§Ê ‚◊Ë∑§⁄UáÊ „Ò —(1) 41x−38y+38=0

(2) 41x+25y−25=0

(3) 41x+38y−38=0

(4) 41x−25y+25=0

77. The solution of the differential equation

•• •••• • • • •• • •

• • • ••• • ••• • • where • • • •• •

•• π• ,

and y(0)=1, is given by :

(1) • • ••• • • •• •

••• •

• • • •• •

(2)• • • •

•• • • •• •••

• •= +

+

(3)• • • •

•• • • •• •••

• •= −

+

(4) • • ••• • • •• •

••• •

• • • •• •

78. A ray of light is incident along a line which

meets another line, 7x−y+1=0, at the

point (0, 1). The ray is then reflected from

this point along the line, y+2x=1. Then

the equation of the line of incidence of the

ray of light is :

(1) 41x−38y+38=0

(2) 41x+25y−25=0

(3) 41x+38y−38=0

(4) 41x−25y+25=0


Set - 03 44

79. ENdtbvy$ O dp„\u `kpf \su A¡L$ f¡Mp, f¡MpAp¡3y=10−4x A_¡ 8x+6y+5=0 _¡ A_y¾$d¡tbvy$Ap¡ A A_¡ B dp„ dm¡ R>¡. sp¡ O A¡ f¡MpM„X$ AB _¡__________ NyZp¡Ñfdp„ rhcpS>_ L$f¡ R>¡.(1) 2 : 3

(2) 1 : 2

(3) 4 : 1

(4) 3 : 4

80. S>¡_y„ L¡$ÞÖ f¡MpAp¡ x−y=1 A_¡ 2x+y=3 _y„ R>¡v$tbvy$lp¡e s¡hp hsy®m_¡ tbvy$ (1, −1) ApNm_p õ`i®L$_y„kduL$fZ __________ R>¡.(1) 4x+y−3=0

(2) x+4y+3=0

(3) 3x−y−4=0

(4) x−3y−4=0

81. P A_¡ Q `fhge y2=4x `f Aph¡gp b¡ rcÞ_

tbv$y$Ap¡ R>¡ S>¡_p âpQgp¡ A_y¾$d¡ t A_¡ t1 R>¡. Å¡ P

ApNm_p¡ Arcg„b Q dp„\u `kpf \sp¡ lp¡e, sp¡ •••

_y„ Þe|_sd d|ëe __________ R>¡.(1) 2

(2) 4

(3) 6

(4) 8

79. ◊Í‹ Á’¥ŒÈ O ‚ „Ê∑§⁄U ¡ÊŸ flÊ‹Ë ∞∑§ ‚⁄U‹ ⁄UπÊ ⁄UπÊ•Ê¥3y=10−4x ÃÕÊ 8x+6y+5=0 ∑§Ê ∑˝§◊‡Ê—Á’¥ŒÈ•Ê¥ A ÃÕÊ B ¬⁄U Á◊‹ÃË „Ò¥, ÃÊ Á’¥ŒÈ O ⁄UπÊπ¥«UAB ∑§Ê Á¡‚ •ŸÈ¬ÊÃ ◊¥ Áfl÷ÊÁ¡Ã ∑§⁄UÃÊ „Ò, fl„ „Ò —(1) 2 : 3

(2) 1 : 2

(3) 4 : 1

(4) 3 : 4

80. ©‚ flÎûÊ Á¡‚∑§Ê ∑§ãŒ˝ ‚⁄U‹ ⁄UπÊ•Ê¥ x−y=1 ÃÕÊ2x+y=3 ∑§Ê ¬˝ÁÃë¿UŒ Á’¥ŒÈ „Ò, ∑§ Á’¥ŒÈ (1, −1) ¬⁄UπË¥øË ªß¸ S¬‡Ê¸ ⁄πÊ ∑§Ê ‚◊Ë∑§⁄UáÊ „Ò —(1) 4x+y−3=0

(2) x+4y+3=0

(3) 3x−y−4=0

(4) x−3y−4=0

81. P ÃÕÊ Q ¬⁄Ufl‹ÿ y2=4x ¬⁄U ÁSÕÃ ŒÊ Á÷ãŸ Á’¥ŒÈ „Ò

Á¡Ÿ∑§ ¬˝Êø‹ ∑˝§◊‡Ê— t ÃÕÊ t1 „Ò¥– ÿÁŒ P ¬⁄U πË¥øÊ

ªÿÊ •Á÷‹¥’ Q ‚ „Ê∑§⁄U ¡ÊÃÊ „Ò, ÃÊ ••• ∑§Ê ãÿÍŸÃ◊

◊ÊŸ „Ò —(1) 2

(2) 4

(3) 6

(4) 8

79. A straight line through origin O meets the

lines 3y=10−4x and 8x+6y+5=0 at

points A and B respectively. Then O

divides the segment AB in the ratio :

(1) 2 : 3

(2) 1 : 2

(3) 4 : 1

(4) 3 : 4

80. Equation of the tangent to the circle, at the

point (1, −1), whose centre is the point of

intersection of the straight lines x−y=1

and 2x+y=3 is :

(1) 4x+y−3=0

(2) x+4y+3=0

(3) 3x−y−4=0

(4) x−3y−4=0

81. P and Q are two distinct points on the

parabola, y2=4x, with parameters t and t

1

respectively. If the normal at P passes

through Q, then the minimum value of •••

is :

(1) 2

(2) 4

(3) 6

(4) 8


Set - 03 45

82. A¡L $ Arshge S> ¡_p ¡ d y¿e An ip „L $h••

•• •

•• •• • •• • _p â^p_ An `f R>¡ s\p s¡_p

rifp¡tbvy$Ap¡ Ap ip„L$h_u _prcAp¡ `f R>¡. Å¡

Arshge_u DÐL¡$ÞÖsp •• lp¡e, sp¡ _uQ¡_pdp„\u L$ey„

tbvy$ s¡_p `f Aph¡gy„ _l] lp¡e ?(1) (0, 2)

(2) ( )• ••• •

(3) ( )• • ••• •

(4) ( )• ••• •

83. A¡L$ kdsgdp„ A¡L$ rÓL$p¡Z ABC R>¡. S>¡_p rifp¡tbvy$Ap¡A(2, 3, 5), B(−1, 3, 2) A_¡ C(λ, 5, µ) R>¡. Å¡ Adp„\u _uL$msu dÝeNp epdpnp kp\¡ kdp_ fus¡ Y$m¡gR>¡, sp¡ (λ3

+µ3+5) _u qL„$ds __________ R>¡.

(1) 1130

(2) 1348

(3) 676

(4) 1077

82. ∞∑§ •ÁÃ¬⁄Ufl‹ÿ, Á¡‚∑§Ê •ŸÈ¬˝SÕ •ˇÊ ‡ÊÊ¥∑§fl••

•• •

•• •• • •• • ∑§ ŒËÉÊ¸ •ˇÊ ∑§Ë ÁŒ‡ÊÊ ◊¥ „Ò ÃÕÊ

Á¡‚∑§ ‡ÊË·¸ ß‚ ‡ÊÊ¥∑§fl ∑§Ë ŸÊÁ÷ÿÊ¥ ¬⁄U „Ò– ÿÁŒ

•ÁÃ¬⁄Ufl‹ÿ ∑§Ë ©à∑§ãŒ˝ÃÊ •• „Ò, ÃÊ ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ

‚Ê Á’¥ŒÈ ß‚ ¬⁄U ÁSÕÃ Ÿ„Ë¥ „Ò?(1) (0, 2)

(2) ( )• ••• •

(3) ( )• • ••• •

(4) ( )• ••• •

83. ∞∑§ ‚◊Ã‹ ◊¥ ∞∑§ ÁòÊ÷È¡ ABC „Ò Á¡‚∑§ ‡ÊË·¸A(2, 3, 5), B(−1, 3, 2) ÃÕÊ C(λ, 5, µ) „Ò¥– ÿÁŒA ‚ „Ê∑§⁄U ¡ÊÃË ◊ÊÁäÿ∑§Ê ÁŸŒ¸‡ÊÊ¥∑§ •ˇÊÊ¥ ¬⁄U ‚◊ÊŸM§¬ ‚ ! ÊÈ∑§Ë „Ò, ÃÊ (λ3

+µ3+5) ∑§Ê ◊ÊŸ „Ò —

(1) 1130

(2) 1348

(3) 676

(4) 1077

82. A hyperbola whose transverse axis is along

the major axis of the conic,

•••

• ••• •• • •• •

and has vertices at the foci of this conic. If

the eccentricity of the hyperbola is

•• , then

which of the following points does NOT

lie on it ?

(1) (0, 2)

(2) ( )• ••• •

(3) ( )• • ••• •

(4) ( )• ••• •

83. ABC is a triangle in a plane with vertices

A(2, 3, 5), B(−1, 3, 2) and C(λ, 5, µ). If the

median through A is equally inclined to the

coordinate axes, then the value of

(λ3+µ

3+5) is :

(1) 1130

(2) 1348

(3) 676

(4) 1077


Set - 03 46

84. S>¡_p dpV¡$ f¡MpAp¡ •• •• • • •

• ••• •

•− − +

= =

A_¡ •

• •• • • ••• •

•• ••− − −

= = kdsgue \pe

s¡hu λ _u rcÞ_ hpõsrhL$ qL „ $dsp ¡_u k„¿ep__________ R>¡.(1) 4

(2) 1

(3) 2

(4) 3

85. ^pfp¡ L¡$ ABC A¡L$ rÓL$p¡Z R>¡ S>¡_y„ `qfL¡$ÞÖ P ApNmR>¡. Å¡ A, B, C A_¡ P _p õ\p_kqv$ip¡ A_y¾$d¡

• •• • • • •→ → →

A_¡ • • • ••

• • •→ → →

• • lp¡e, sp¡ Ap

rÓL$p¡Z_p g„bL¡$ÞÖ_p¡ õ\p_kqv$i __________ R>¡.

(1) • • • •• • •→ → →

• •

(2)• • • ••

•• • •→ → →

• ••

(3) •→

(4) • • • •

•

• • •→ → →

• •

84. λ ∑§ fl„ Á÷ãŸ flÊSÃÁfl∑§ ◊ÊŸÊ¥ ∑§Ë ‚¥ÅÿÊ Á¡Ÿ∑§

Á‹∞ ⁄ πÊ∞° •• •• • • ••

• ••• •

•− − +

= = ÃÕÊ

•• •• • • ••

• ••• ••− − −

= = ‚◊Ã‹Ëÿ „Ò¥, „Ò —

(1) 4

(2) 1

(3) 2

(4) 3

85. ◊ÊŸÊ ABC ∞∑§ ÁòÊ÷È¡ „Ò Á¡‚∑§Ê ¬Á⁄U∑§ãŒ˝ P ¬⁄U „Ò–ÿÁŒ Á’¥ŒÈ•Ê¥ A, B, C ÃÕÊ P ∑§ ÁSÕÁÃ ‚ÁŒ‡Ê ∑˝§◊‡Ê—

• •• • • • •→ → →

ÃÕÊ • • • ••

• • •→ → →

• • „Ò¥, ÃÊ ß‚ ÁòÊ÷È¡

∑§ ‹¥’ - ∑§ãŒ˝ ∑§Ê ÁSÕÁÃ ‚ÁŒ‡Ê „Ò —

(1) • • • •• • •→ → →

• •

(2)• • • ••

•• • •→ → →

• ••

(3) •→

(4) • • • •

•

• • •→ → →

• •

84. The number of distinct real values of λ for

which the lines •• •• • • ••

• ••• •

•− − +

= =

and •

• •• • • ••• •

•• ••− − −

= = are

coplanar is :

(1) 4

(2) 1

(3) 2

(4) 3

85. Let ABC be a triangle whose circumcentre

is at P. If the position vectors of A, B, C

and P are • •• • • • •→ → →

and • • • •

•• • •→ → →

• •

respectively, then the position vector of the

orthocentre of this triangle, is :

(1) • • • •• • •→ → →

• •

(2)• • • ••

•• • •→ → →

• ••

(3) •→

(4) • • • •

•

• • •→ → →

• •


Set - 03 47

86. 5 Ahgp¡L$_p¡_p¡ dÝeL$ 5 A_¡ rhQfZ 124 R>¡. Å¡s¡dp„\u ÓZ Ahgp¡L$_p¡ 1, 2 A_¡ 6 lp¡e, sp¡ dprlsu_y„dÝeL$\u kf¡fpi rhQg_ __________ \pe.(1) 2.4

(2) 2.8

(3) 2.5

(4) 2.6

87. A¡L$ âep¡N S>¡V$gp hMs Akam \pe R>¡ s¡_p\u b¡NZp¡ kam \pe R>¡. Ap âep¡N_p R> âeÐ_p¡dp„\uAp¡R>pdp„ Ap¡R>u 5 kamsp dmhp_u k„cph_p__________ R>¡.

(1)

• • •• • •

(2)

• • •• • •

(3)

• • •• • •

(4)

• • •• • •

86. 5 ¬˝ˇÊáÊÊ¥ ∑§Ê ◊Êäÿ 5 „Ò ÃÕÊ ©Ÿ∑§Ê ¬˝‚⁄UáÊ 124 „Ò–ÿÁŒ ©Ÿ◊¥ ‚ ÃËŸ ¬˝ˇÊáÊ 1, 2 ÃÕÊ 6 „Ò¥, ÃÊ ßŸ •Ê°∑§«∏Ê¥∑§Ê ◊Êäÿ ‚ ◊Êäÿ Áflø‹Ÿ „Ò —(1) 2.4

(2) 2.8

(3) 2.5

(4) 2.6

87. ∞∑§ ¬˝ÿÊª ∑§ ‚»§‹ „ÊŸ ∑§Ê ‚¥ÿÊª ©‚∑§ Áfl»§‹„ÊŸ ∑§ ‚¥ÿÊª ∑§Ê ŒÈªÈŸÊ „Ò– ß‚ ¬˝ÿÊª ∑§ 6 ¬⁄UËˇÊáÊÊ¥◊¥ ‚ ∑§◊ ‚ ∑§◊ ¬Ê°ø ∑§ ‚»§‹ „ÊŸ ∑§Ë ¬˝ÊÁÿ∑§ÃÊ„Ò —

(1)

• • •• • •

(2)

• • •• • •

(3)

• • •• • •

(4)

• • •• • •

86. The mean of 5 observations is 5 and their

variance is 124. If three of the observations

are 1, 2 and 6 ; then the mean deviation

from the mean of the data is :

(1) 2.4

(2) 2.8

(3) 2.5

(4) 2.6

87. An experiment succeeds twice as often as

it fails. The probability of at least 5

successes in the six trials of this experiment

is :

(1)

• • •• • •

(2)

• • •• • •

(3)

• • •• • •

(4)

• • •• • •


Set - 03 48

88. If A>0, B>0 and • • •• • ••π• • , then the

minimum value of tanA+tanB is :

(1) • • • ••

(2) • • • ••

(3) • • •• ••

(4)••

89. The angle of elevation of the top of a

vertical tower from a point A, due east of it

is 45. The angle of elevation of the top of

the same tower from a point B, due south

of A is 30. If the distance between A and

B is • • • •• , then the height of the tower

(in metres), is :

(1) • • •(2) 54

(3) • • •(4) 108

88. ÿÁŒ A>0, B>0 ÃÕÊ • • •• • ••π• • „ Ò , ÃÊ

tanA+tanB ∑§Ê ãÿÍŸÃ◊ ◊ÊŸ „Ò —

(1) • • • ••

(2) • • • ••

(3) • • •• ••

(4)••

89. Á’¥ŒÈ A ‚, ¡Ê ∞∑§ ™§äflÊ¸œ⁄U ◊ËŸÊ⁄U ∑§ ¬Ífl¸ ∑§Ë •Ê⁄U„Ò, ◊ËŸÊ⁄U ∑§ ‡ÊË·¸ ∑§Ê ©ãŸÿŸ ∑§ÊáÊ 45 „Ò– Á’¥ŒÈ B,

¡Ê Á’¥ŒÈ A ∑§ ŒÁˇÊáÊ ◊¥ „Ò, ‚ ©‚Ë ◊ËŸÊ⁄U ∑§ ‡ÊË·¸ ∑§Ê©ãŸÿŸ ∑§ÊáÊ 30 „Ò– ÿÁŒ A ÃÕÊ B ∑§ ’Ëø ∑§Ë ŒÍ⁄UË

• • • ◊Ë. „Ò, ÃÊ ◊ËŸÊ⁄U ∑§Ë ™§°øÊß¸ (◊Ë. ◊¥) „Ò —

(1) • • •(2) 54

(3) • • •(4) 108

88. Å¡ A>0, B>0 A_¡ • • •• • ••π• • lp ¡e, sp ¡

tanA+tanB _y„ Þe|_sd d|ëe __________ R>¡.

(1) • • • ••

(2) • • • ••

(3) • • •• ••

(4)••

89. A¡L$ rifp¡g„b V$phf_u |h® sfa_p tbvy$ A \u V$phf_uV$p¡Q_p¡ DÐk¡ L$p¡Z 45 R>¡. tbvy$ A _u v$rnZ¡ Aph¡gtbvy$ B \u V$phf_u V$p¡Q_p¡ DÐk¡ L$p¡Z 30 R>¡. Å¡ A

A_¡ B hÃQ¡_y„ A„sf • • • •• lp¡e, sp¡ V$phf_uKQpC (duV$fdp„) __________ R>¡.

(1) • • •(2) 54

(3) • • •(4) 108


Set - 03 49

90. The contrapositive of the following

statement,

“If the side of a square doubles, then its area

increases four times”, is :

(1) If the side of a square is not doubled,

then its area does not increase four

times.

(2) If the area of a square increases four

times, then its side is doubled.

(3) If the area of a square increases four

times, then its side is not doubled.

(4) If the area of a square does not

increase four times, then its side is not

doubled.

- o O o -

90. ÁŸêŸ ∑§ÕŸ ∑§Ê ¬˝ÁÃœŸÊà◊∑§ (contrapositive)

„Ò —

““ÿÁŒ Á∑§‚Ë flª¸ ∑§Ë ÷È¡Ê ŒÈªÈŸË „Ê ¡Ê∞, ÃÊ ©‚∑§ÊˇÊòÊ»§‹ øÊ⁄U ªÈŸÊ ’…∏ ¡ÊÃÊ „Ò”” —

(1) ÿÁŒ ∞∑§ flª¸ ∑§Ë ÷È¡Ê ŒÈªÈŸË Ÿ ∑§Ë ¡Ê∞, ÃÊ©‚∑§Ê ˇÊòÊ»§‹ øÊ⁄U ªÈŸÊ Ÿ„Ë¥ ’…∏ÃÊ–

(2) ÿÁŒ Á∑§‚Ë flª¸ ∑§Ê ÊòÊ»§‹ øÊ⁄U ªÈŸÊ ’…∏ ¡Ê∞,ÃÊ ©‚∑§Ë ÷È¡Ê ŒÈªÈŸË „Ê ¡ÊÃË „Ò–

(3) ÿÁŒ Á∑§‚Ë flª¸ ∑§Ê ˇÊòÊ»§‹ øÊ⁄U ªÈŸÊ ’…∏ ¡ÊÃÊ„Ò, ÃÊ ©‚∑§Ë ÷È¡Ê ŒÈªÈŸË Ÿ„Ë¥ „ÊÃË–

(4) ÿÁŒ Á∑§‚Ë flª¸ ∑§Ê ÊòÊ»§‹ øÊ⁄U ªÈŸÊ Ÿ„Ë¥ ’…∏ÃÊ,ÃÊ ©‚∑§Ë ÷È¡Ê ŒÈªÈŸË Ÿ„Ë¥ „ÊÃË–

- o O o -

90. _uQ¡_p rh^p__y„ kdp_p\ â¡fZ L$ey„ R>¡ ?

""Å¡ L$p¡C Qp¡fk_u bpSy> bdZu L$fhpdp„ Aph¡, sp¡ s¡_y„n¡Óam QpfNÏ„ h ¡.''

(1) Å¡ L$p¡C Qp¡fk_u bpSy> bdZu L$fhpdp„ _ Aph¡,sp¡ s¡_y„ n¡Óam QpfNÏ„ h^i¡ _rl.

(2) Å¡ L$p¡C Qp¡fk_y„ n¡Óam QpfNÏ„ h^pfhpdp„ Aph¡,sp¡ s¡_u bpSy> bdZu \pe.

(3) Å¡ L$p¡C Qp¡fk_y„ n¡Óam QpfNÏ„ h^pfhpdp„Aph¡, sp¡ s¡_u bpSy> bdZu _ \pe.

(4) Å¡ L$p¡C Qp¡fk_y„ n¡Óam QpfNÏ„ h^pfhpdp„ _Aph¡, sp¡ s¡_u bpSy> bdZu _ \pe.

- o O o -

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PHYSICS : English & Hindi 091. x and y displacements of a particle aregiven as x(t)5a sinvt and y(t)5a sin2vt.Its trajectory will look like :

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1. ØçÎ ç·¤âè ·¤æ ·ð¤ x ÌÍæ y çßSÍæÂÙæð´ ·¤æð ·ý¤×àæÑ,x(t)5a sinvt ÌÍæ y(t)5a sin2vt, âð çÙM¤çÂÌç·¤Øæ ÁæÌæ ãñ Ìæð, §â·¤æ ÂýÿæðÂ - ÂÍ, çÙÙæ´ç·¤Ì ×ð´ âðç·¤â·ð¤ Áñâæ çÎ¹æ§ü Îð»æ?

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PHYSICS : English & Hindi 092. If a body moving in a circular pathmaintains constant speed of 10 ms21, thenwhich of the following correctly describesrelation between acceleration andradius ?

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2. ØçÎ ßëææ·¤æÚU ÂÍ ×ð´ »çÌ ·¤ÚUÌð ãé° ç·¤âè çÂ´ÇU (ßSÌé)·¤è ¿æÜ 10 ms21 ãñ ¥æñÚU Øã ¥¿ÚU ÕÙè ÚUãÌè ãñ Ìæð,çÙÙæ´ç·¤Ì ×ð´ âð ·¤æñÙâæ ¥æÜð¹, ßÚUæ ÌÍæ çæØæ ·ð¤Õè¿ âÕÏ ·¤æ ÆUè·¤ (âãè) ç¿ææ ·¤ÚUÌæ ãñ?

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PHYSICS : English & Hindi 093. A block of mass m510 kg rests on ahorizontal table. The coefficient of frictionbetween the block and the table is 0.05.When hit by a bullet of mass 50 g movingwith speed v, that gets embedded in it, theblock moves and comes to stop aftermoving a distance of 2 m on the table.

If a freely falling object were to acquirespeed

10

v after being dropped from heightH, then neglecting energy losses andtaking g510 ms22, the value of H is closeto :(1) 0.2 km(2) 0.3 km(3) 0.4 km(4) 0.5 km

4. A block of mass m50.1 kg is connected toa spring of unknown spring constant k. Itis compressed to a distance x from itsequilibrium position and released from rest.After approaching half the distance

2

x

from equilibrium position, it hits anotherblock and comes to rest momentarily, whilethe other block moves with a velocity3 ms21. The total initial energy of thespring is :(1) 1.5 J(2) 0.6 J(3) 0.3 J(4) 0.8 J

5. A uniform solid cylindrical roller of massm is being pulled on a horizontal surfacewith force F parallel to the surface andapplied at its centre. If the acceleration ofthe cylinder is a and it is rolling withoutslipping then the value of F is :(1) ma(2) 2 ma(3) 3

ma2

(4) 5 ma

3

3. ç·¤âè Üæò·¤ (»éÅU·ð¤) ·¤æ ÎýÃØ×æÙ m510 kg ãñÐØã °·¤ ÿæñçÌÁ ×ðÁ ÂÚU ÚU¹æ ãñÐ §Ù ÎæðÙæð´ ·ð¤ Õè¿æáüæ »éææ´·¤50.05 ãñÐ §â Üæò·¤ ÂÚU 50 g ÎýÃØ×æÙ·¤è °·¤ »æðÜè v ¿æÜ âð ÅU·¤ÚUæÌè ¥æñÚU §â×ð´ Ï´âÁæÌè ãñÐ §ââð Øã Üæò·¤, ×ðÁ ÂÚU 2 m çßSÍæçÂÌãæð·¤ÚU L¤·¤ ÁæÌæ ãñÐØçÎ, H ª¡¤¿æ§ü âð ×éÌ M¤Â âð ç»ÚUæÙð ·ð¤ Âà¿æÌ÷ ·¤æð§üßSÌé

10

v ¿æÜ ÂýæÌ ·¤ÚU ÜðÌè ãñ Ìæð, ª¤Áæü-ÿæØ ·¤æðÙ»Ø ×æÙÌð ãé°, H ·¤æ âçÙ·¤ÅU ×æÙ ãæð»æ Ñ(g510 ms22)(1) 0.2 km(2) 0.3 km(3) 0.4 km(4) 0.5 km

4. ç·¤âè Üæò·¤ ·¤æ ÎýÃØ×æÙ m50.1 kg ãñÐ Øã °·¤°ðâè ·¤×æÙè (çSÂý») âð ÁéÇæ ãñ çÁâ·¤æ ·¤×æÙè çSÍÚUæ·¤k ãñÐ §â·¤æð §â·¤è âæØæßSÍæ âð x ÎêÚUè Ì·¤ ÎÕæ·¤ÚUçßÚUæ×æßSÍæ âð ÀUæðÇ¸ çÎØæ ÁæÌæ ãñÐ âæØæßSÍæ âð

2

x

ÎêÚUè ÂÚU ¥æÙð ÂÚU Øã °·¤ ¥Ø Üæò·¤ âð ÅU·¤ÚUæÌæãñ ¥æñÚU ÿæçæ·¤ M¤Â âð L¤·¤ ÁæÌæ ãñ, ÁÕç·¤, ÎêâÚUæÜæò·¤ 3 ms21 ·ð¤ ßð» âð »çÌ ·¤ÚUÙð Ü»Ìæ ãñÐ Ìæð,·¤×æÙè ·¤è ·é¤Ü ÂýæÚ´UçÖ·¤ ª¤Áæü ãñ Ñ(1) 1.5 J(2) 0.6 J(3) 0.3 J(4) 0.8 J

5. m ÎýÃØ×æÙ ·ð¤ ç·¤âè °·¤â×æÙ ÆUæðâ çâçÜÇUÚU ·ð¤·ð¤Îý ÂÚU °·¤ ÕÜ F Ü»æ·¤ÚU, ©âð ç·¤âè â×ÌÜâÌã ÂÚU, ©â·ð¤ â×æÌÚU ¹è´¿æ Áæ ÚUãæ ãñÐ ØçÎ ØãçâçÜÇUÚU Õ»ñÚU (çÕÙæ) çÈ¤âÜð a ßÚUæ âð ÜéÉ¸·¤ÚUãæ ãñ Ìæð, F ·¤æ ×æÙ ãæð»æ Ñ(1) ma(2) 2 ma(3) 3

ma2

(4) 5 ma

3

PHYSICS : English & Hindi 096. Consider a thin uniform square sheet madeof a rigid material. If its side is a, mass mand moment of inertia I about one of itsdiagonals, then :

(1) 2maI >

12

(2) 2 2ma ma< I <

24 12

(3) 2maI

125

(4) 2maI

245

7. A very long (length L) cylindrical galaxyis made of uniformly distributed mass andhas radius R (R<<L). A star outside thegalaxy is orbiting the galaxy in a planeperpendicular to the galaxy and passingthrough its centre. If the time period of staris T and its distance from the galaxys axisis r, then :(1) T2 ; r3(2) T ; r2(3) T ; r(4) T r;

8. If it takes 5 minutes to fill a 15 litre bucketfrom a water tap of diameter 2

cmp

thenthe Reynolds number for the flow is(density of water5103 kg/m3 andviscosity of water51023 Pa.s) close to :(1) 5500(2) 11,000(3) 550(4) 1100

6. °·¤ ÂÌÜè, °·¤â×æÙ, ß»æü·¤æÚU, ¿æÎÚU (àæèÅU) ç·¤âèÎëÉ¸ ÂÎæÍü ·¤è ÕÙè ãñÐ ØçÎ §â·¤è °·¤ ÖéÁæ a,ÎýÃØ×æÙ m ÌÍæ ç·¤âè °·¤ çß·¤æü ·ð¤ ÂçÚUÌÑ, §â·¤æÁÇ¸ß ¥ææêæü I ãñ Ìæð Ñ(1) 2ma

I > 12

(2) 2 2ma ma< I <

24 12

(3) 2maI

125

(4) 2maI

245

7. °·¤ ÕãéÌ ÜÕè »ñÜðâè (×´Îæç·¤Ùè) (ÜÕæ§ü L)°·¤â×æÙ çßÌçÚUÌ ÎýÃØ ·¤è ÕÙè ãñ, §â·¤è çæØæR (R<<L) ãñÐ §â »ñÜðâè ·ð¤ ÕæãÚ °·¤ ÌæÚUæ,»ñÜðâè ·¤è ÂçÚU·ý¤×æ ·¤ÚU ÚUãæ ãñÐ §â·¤è ÂçÚU·ý¤×æ ·¤æâ×ÌÜ »ñÜðâè ·ð¤ â×ÌÜ ·ð¤ ÜÕßÌ÷ ãñ ÌÍæ §â·ð¤·ð¤Îý âð ãæð·¤ÚU »é$ÁÚUÌæ ãñÐ ØçÎ, ÌæÚðU ·¤è »ñÜðâè ·¤è¥ÿæ âð ÎêÚUè r ãñ ¥æñÚU ÌæÚðU ·¤æ ¥æßÌü ·¤æÜ T ãñ Ìæð ÑU(1) T2 ; r3(2) T ; r2(3) T ; r(4) T r;

8. ÂæÙè ·ð¤ ÙÜ ·¤è °·¤ ÅUæð´ÅUè ·¤æ ÃØæâ 2 cmp

ãñÐ§ââð 15 çÜÅUÚU ·¤è °·¤ ÕæËÅUè ·¤æð ÖÚUÙð ×ð´ 5 ç×ÙÅU·¤æ â×Ø Ü»Ìæ ã ñÐ (ØçÎ, ÁÜ ·¤ææÙß5103 kg/m3 ÌÍæ ÁÜ ·¤è àØæÙÌæ51023Pa.s ãñ) Ìæð, §â Âýßæã ·ð¤ çÜØð ÚðUÙËÇ÷Uâ â´Øæãæð»è Ñ(1) 5500(2) 11,000(3) 550(4) 1100

PHYSICS : English & Hindi 09

9. ØçÎ ·¤æ¡¿ ·¤è Îæð ÜðÅUæð´ ·ð¤ Õè¿ ×ð´ ÁÜ ·¤è °·¤ÂÌÜè ÂÚUÌ ãæð (¥æÚðU¹ Îðç¹Øð) Ìæð ©Ù ÜðÅUæð´ ·¤æð¹è´¿·¤ÚU ¥Ü» ·¤ÚUÙæ ÕãéÌ ·¤çÆUÙ ãæðÌæ ãñÐ

§â·¤æ ·¤æÚUæ Øã ãñ ç·¤, ÁÜ, ç·¤ÙæÚUæð´ ÂÚU çâçÜ´ÇUÚUè(ÕðÜÙæ·¤æÚU) âÌãð´ ÕÙæ ÎðÌæ ãñ çÁââð ßæØé×´ÇUÜ ·¤èÌéÜÙæ ×ð´ ßãæ¡ ÎæÕ ·¤× ãæð ÁæÌæ ãñÐ ØçÎ §â çâçÜ´ÇUÚUèâÌã (ÂëcÆU) ·¤è çæØæ R ãñ ÌÍæ ÁÜ ·¤æ ÂëcÆU ÌÙæßT ãñ Ìæð, Îæð ÜðÅUæð´ ·ð¤ Õè¿ ÁÜ ×ð´ ÎæÕ ç·¤ÌÙæ ·¤×ãæð»æ?(1) T

4R

(2) T

2R

(3) 4T

R

(4) 2T

R

9. If two glass plates have water betweenthem and are separated by very smalldistance (see figure), it is very difficult topull them apart. It is because the water inbetween forms cylindrical surface on theside that gives rise to lower pressure in thewater in comparison to atmosphere. If theradius of the cylindrical surface is R andsurface tension of water is T then thepressure in water between the plates islower by :

(1) T

4R

(2) T

2R

(3) 4T

R

(4) 2T

R

PHYSICS : English & Hindi 0910. An ideal gas goes through a reversible

cycle a®b®c®d has the V - T diagramshown below. Process d®a and b®c areadiabatic.

The corresponding P - V diagram for theprocess is (all figures are schematic andnot drawn to scale) :

(1)

(2)

(3)

(4)

10. °·¤ ¥æÎàæü »ñâ ·ð¤ ©·ý¤×æèØ ¿·ý¤ a®b®c®d ,·ð¤ çÜØð V - T ¥æÚðU¹ Øãæ¡ ÎàææüØæ »Øæ ãñÐ Âý·ý¤×d®a ÌÍæ b®c L¤Î÷Ïæðc× ãñ´Ð

Ìæð, §â Âý·ý¤× ·ð¤ çÜØð, â´»Ì P - V ¥æÚðU¹ ãæð»æ(âÖè ¥æÚðU¹ ÃØßSÍæ ¥æÚðU¹ ãñ ¥æñÚU S·ð¤Ü ·ð¤ ¥ÙéâæÚUÙãè´ ãñ´) Ñ

(1)

(2)

(3)

(4)

PHYSICS : English & Hindi 0911. In an ideal gas at temperature T, theaverage force that a molecule applies onthe walls of a closed container depends onT as Tq. A good estimate for q is :

(1) 2(2) 1(3) 1

2

(4) 1

4

12. A simple harmonic oscillator of angularfrequency 2 rad s21 is acted upon by anexternal force F5sint N. If the oscillatoris at rest in its equilibrium position at t50,its position at later times is proportionalto :(1) 1

sin sin 22

t t1

(2) 1sin cos 2

2t t1

(3) 1cos sin 2

2t t2

(4) 1sin sin 2

2t t2

13. A bat moving at 10 ms21 towards a wallsends a sound signal of 8000 Hz towardsit. On reflection it hears a sound offrequency f. The value of f in Hz is close to(speed of sound5320 ms21)(1) 8258(2) 8516(3) 8000(4) 8424

11. ç·¤âè ¥æÎàæü »ñâ ×ð´, ç·¤âè ¥æé mæÚUæ »ñâ ·ð¤ ÕÎÂææ ·¤è ÎèßæÚUæð´ ÂÚU Ü»æØæ »Øæ ¥æñâÌ ÕÜ, »ñâ ·ð¤ÌæÂ T ÂÚU, Tq ·ð¤ ¥ÙéâæÚU çÙÖüÚU ·¤ÚUÌæ ãñÐ Ìæð, q ·¤æâçÙ·¤ÅU ×æÙ ãñ Ñ(1) 2(2) 1(3) 1

2

(4) 1

4

12. ç·¤âè âÚUÜ ¥æßÌü ÎæðçÜæ ·¤è ·¤æðæèØ ¥æßëçæ2 rad s21 ãñÐ §â ÂÚU °·¤ Õæs ÕÜ, F5sintØêÅUÙ (N) Ü»Ìæ ãñÐ ØçÎ â×Ø t50 ÂÚU, Øã ÎæðçÜæ,¥ÂÙè âæØæßSÍæ ×ð´ çßÚUæ× çSÍçÌ ×ð´ ãñ Ìæð, §â·ð¤Âà¿æÌ÷ ·ð¤ ç·¤âè â×Ø ×ð´, §â·¤è çSÍçÌ çÙÙæ´ç·¤Ì×ð´ ç·¤â·ð¤ â×æÙéÂæÌè ãæð»è?(1) 1

sin sin 22

t t1

(2) 1sin cos 2

2t t1

(3) 1cos sin 2

2t t2

(4) 1sin sin 2

2t t2

13. 10 ms21 ·¤è ¿æÜ âð ç·¤âè ÎèßæÚU ·¤è ¥æðÚU ÁæÌæãé¥æ °·¤ ¿×»æÎÇ¸, ÎèßæÚU ·¤è ¥æðÚU 8000 Hz ·¤æßçÙ â´·ð¤Ì (çâÙÜ) ÂýðçáÌ ·¤ÚUÌæ (ÖðÁÌæ) ãñÐÎèßæÚU âð ÂÚUæßÌüÙ ·ð¤ Âà¿æÌ÷ ßã f ¥æßëçæ ·¤èßçÙ, âéÙÌæ ãñÐ ØçÎ, ßçÙ ·¤è ¿æÜ 320 ms21 ãñÌæð, Hz ×ð´ f ·¤æ âçÙ·¤ÅU ×æÙ ãæð»æ Ñ(1) 8258(2) 8516(3) 8000(4) 8424

PHYSICS : English & Hindi 0914. A thin disc of radius b52a has a concentrichole of radius a in it (see figure). It carriesuniform surface charge s on it. If theelectric field on its axis at height h (h<<a)from its centre is given as Ch then valueof C is :

(1)0a

s

e

(2)0

2a

se

(3)04a

s

e

(4)08a

s

e

14. °·¤ ÂÌÜè çÇUS·¤ (¿ç·ý¤·¤æ) ·¤è çæØæ b ãñÐ §â×ð´ÕÙð °·¤ â´·ð¤Îýè çÀUÎý (ÀðUÎ) ·¤è çæØæ a ãñÐ(b52a) Ð çÇS·¤ ÂÚU °·¤â×æÙ ÂëcÆU ¥æßðàæ s ãñÐØçÎ §â·¤è ¥ÿæ ÂÚU ÌÍæ §â·ð¤ ·ð¤Îý âð h ª¡¤¿æ§üÂÚU, (h<<a), çßléÌ ÿæðæ Ch ãæð Ìæð, C ·¤æ ×æÙãñ Ñ

(1)0a

s

e

(2)0

2a

se

(3)04a

s

e

(4)08a

s

e

PHYSICS : English & Hindi 0915. Shown in the figure are two point charges

1Q and 2Q inside the cavity of aspherical shell. The charges are kept nearthe surface of the cavity on opposite sidesof the centre of the shell. If s1 is the surfacecharge on the inner surface and Q1 netcharge on it and s2 the surface charge onthe outer surface and Q2 net charge on itthen :

(1) s1 ¹ 0, Q1 ¹ 0s2 ¹ 0, Q2 ¹ 0

(2) s1 ¹ 0, Q1 5 0s2 ¹ 0, Q2 5 0

(3) s1 ¹ 0, Q1 5 0s2 5 0, Q2 5 0

(4) s1 5 0, Q1 5 0s2 5 0, Q2 5 0

16. If the capacitance of a nanocapacitor ismeasured in terms of a unit u made bycombining the electronic charge e, Bohrradius a0, Plancks constant h and speedof light c then :(1) 2

0

e c

hau 5

(2) 2

0

e h

cau 5

(3) 20e a

hc

u 5

(4) 20

hc

e au 5

15. Øãæ¡ ¥æÚðU¹ ×ð´, ç·¤âè »æðÜæ·¤æÚU ·¤æðàæ (àæñÜ) ·ð¤ ·¤æðÅUÚU·ð¤ ÖèÌÚU Îæð çÕÎé-¥æßðàæ 1Q ÌÍæ 2Q ÎàææüØð »Øðãñ´Ð Øð ¥æßðàæ ·¤æðÅUÚU ·¤è âÌã ·ð¤ çÙ·¤Å U§â Âý·¤æÚUÚU¹ð »Øð ãñ´ ç·¤, °·¤ ¥æßðàæ ·¤æðàæ ·ð¤ ·ð¤Îý ·¤è °·¤¥æðÚU ãñ ¥æñÚU ÎêâÚUæ ·ð¤Îý ·ð¤ çßÂÚUèÌ ÎêâÚUè ¥æðÚUÐ ØçÎ,ÖèÌÚUè ÌÍæ ÕæãÚUè âÌãæð´ (ÂëcÆUæð´) ÂÚU, ÂëcÆU ¥æßðàæ·ý¤×àæÑ s1 ÌÍæ s2 ¥æñÚU ÙðÅU ¥æßðàæ ·ý¤×àæÑ Q1 ÌÍæQ2 ãæð Ìæð Ñ

(1) s1 ¹ 0, Q1 ¹ 0s2 ¹ 0, Q2 ¹ 0

(2) s1 ¹ 0, Q1 5 0s2 ¹ 0, Q2 5 0

(3) s1 ¹ 0, Q1 5 0s2 5 0, Q2 5 0

(4) s1 5 0, Q1 5 0s2 5 0, Q2 5 0

16. ØçÎ ç·¤âè ÙñÙæð â´ÏæçÚUæ ·¤è ÏæçÚUÌæ, °·¤ °ðâð ×ææ·¤u ×ð´ ×æÂè ÁæØ, Áæð §ÜðÅþUæòÙ ¥æßðàæ e , ÕæðÚU-çæØæa0, Üæ´·¤ çSÍÚUæ´·¤ h ÌÍæ Âý·¤æàæ ·¤è ¿æÜ c ·ð¤â´ØæðÁÙ âð ÕÙæ ãñ Ìæð Ñ(1) 2

0

e c

hau 5

(2) 2

0

e h

cau 5

(3) 20e a

hc

u 5

(4) 20

hc

e au 5

PHYSICS : English & Hindi 0917. Suppose the drift velocity vd in a materialvaried with the applied electric field E as

d Ev ; . Then V - I graph for a wiremade of such a material is best given by :

(1)

(2)

(3)

(4)

17. ØçÎ, ç·¤âè ÂÎæÍü ×ð ¥Âßæã ßð», vd ·¤æ ×æÙ, Ü»æØð»Øð çßléÌ ÿæðæ E ÂÚU §â Âý·¤æÚU çÙÖüÚU ·¤ÚUÌæ ãñ, ç·¤

d Ev ; Ð Ìæð çÙÙæ´ç·¤Ì ×ð´ âð ·¤æñÙ-âæ »ýæÈ¤(¥æÜð¹), §â ÂÎæÍü âð ÕÙð ÌæÚU ·ð¤ çÜØð, âçÙ·¤ÅUV - I »ýæÈ¤ ãæð»æ?

(1)

(2)

(3)

(4)


18. °·¤ ßæðËÅU×èÅUÚU âð â×æÌÚU ·ý¤× ×ð´, Îæð ÕñÅUçÚUØæ¡, ÁæðÇ¸è»§ü ãñÐ ÂãÜè, 10V ÌÍæ 1V ¥æÌçÚU·¤ ÂýçÌÚUæðÏ ·¤è¥æñÚU ÎêâÚUè, 15V ÌÍæ 0.6V ¥æÌçÚU·¤ ÂýçÌÚUæðÏ ·¤è(¥æÚðU¹ Îðç¹Øð) Ìæð, ßæðËÅU×èÅUÚU ·ð¤ ÂÆUÙ (ÚUèçÇ´U»)·¤æ âçÙ·¤ÅU ×æÙ ãæð»æ Ñ

(1) 11.9 V(2) 12.5 V(3) 13.1 V(4) 24.5 V

19. çßÖßæÌÚ V mæÚUæ ßçÚUÌ, °·¤ ÂýæðÅUæòÙ (ÎýÃØ×æÙ m),ç·¤âè ¥ÙéÂýSÍ °·¤â×æÙ ¿éÕ·¤èØ ÿæðæ B âð ãæð·¤ÚUÌèßý ¿æÜ âð »é$ÁÚUÌæ ãñÐ Øã ¿éÕ·¤èØ ÿæðæ d ¿æñÇ¸æ§üÌ·¤ çßSÌçÚUÌ ãñÐ ØçÎ, Øã ÂýæðÅUæòÙ, ¿éÕ·¤èØ ÿæððæ ·ð¤·¤æÚUæ ¥ÂÙè »çÌ ·¤è ÂýæÚ´UçÖ·¤ çÎàææ âð a ·¤æðæ âðçß¿çÜÌ ãæ ð ÁæÌæ ã ñ (¥æÚ ð U¹ Î ð ç¹Øð) Ìæ ð ,sin a ·¤æ ×æÙ ãæð»æ Ñ

(1) qdB

2 mV

(2) qB

d 2mV

(3) qBd

2mV

(4) BdqV

2m

18. A 10V battery with internal resistance1V and a 15V battery with internalresistance 0.6V are connected in parallelto a voltmeter (see figure). The reading inthe voltmeter will be close to :

(1) 11.9 V(2) 12.5 V(3) 13.1 V(4) 24.5 V

19. A proton (mass m) accelerated by apotential difference V flies through auniform transverse magnetic field B. Thefield occupies a region of space by widthd. If a be the angle of deviation of protonfrom initial direction of motion (seefigure), the value of sin a will be :

(1) qdB

2 mV

(2) qB

d 2mV

(3) qBd

2mV

(4) BdqV

2m


20. ç·¤âè ÂçÚUÙæçÜ·¤æ ·¤è ÜÕæ§ü 25 cm ÌÍæ çæØæ2 cm ãñ ¥æñÚU §â×ð´ ÌæÚU ·ð¤ ·é¤Ü 500 Èð¤ÚðU ÜÂðÅðU »Øðãñ´Ð §ââð 15 A ·¤è ÏæÚUæ ÂýßæçãÌ ãæð ÚUãè ãñÐ ØãÂçÚUÙæçÜ·¤æ, §âè âæ§Á ÌÍæ ¿éÕ·¤Ù M

→ (¿éÕ·¤èØ¥ææêæü/¥æØÌÙ), ·ð¤ ÌéËØ ãñ Ìæð , M

→? ? ãñ Ñ(1) 3p Am21(2) 30000 Am21(3) 300 Am21(4) 30000p Am21

21. ç·¤âè ·é´¤ÇUÜè âð ÂýßæçãÌ çßléÌ ÏæÚUæ ·¤æ ×æÙ, 0.1 s×ð´, 5 A âð 2 A ãæð ÁæÌæ ãñ çÁââð, 50 V ·¤è ¥æñâÌßæðËÅUÌæ ©ÂÙ ãæðÌè ãñÐ Ìæð, §â ·é¤ÇUÜè ·¤æ SßÂýðÚU·¤ßãñ Ñ(1) 0.67 H(2) 1.67 H(3) 3 H(4) 6 H

20. A 25 cm long solenoid has radius 2 cmand 500 total number of turns. It carries acurrent of 15 A. If it is equivalent to amagnet of the same size and magnetizationM→ (magnetic moment/volume), then M

→? ?is :(1) 3p Am21(2) 30000 Am21(3) 300 Am21(4) 30000p Am21

21. When current in a coil changes from 5 Ato 2 A in 0.1 s, an average voltage of 50 Vis produced. The self - inductance of thecoil is :(1) 0.67 H(2) 1.67 H(3) 3 H(4) 6 H


22. Îæð ÂçÚUÂÍæð´ (a) ÌÍæ (b) ×ð çSß¿ S1 ÌÍæ S2, t50ÂÚU ÕÎ ç·¤Øð ÁæÌð ãñ´ ¥æñÚU ÂØæüÌ ÜÕð â×Ø Ì·¤ÕÎ ÚU¹ð ÁæÌð ãñ´, Ìæð t ³ 0 ·ð¤ çÜØð, Îæð ÂçÚUÂÍæð´ ×ð´çßléÌ ÏæÚUæ¥æð´ ·ð¤ çß¿ÜÙ (ÂçÚUßÌüÙ) ·¤æð, ·¤æñÙâæ»ýæÈ¤ çÙ·¤ÅUÌ× Îàææ üÌæ ã ñ? (¥æÚ ð U¹ ·ð¤ßÜÃØßSÍæ×·¤ ãñ´ ¥æñÚU S·ð¤Ü ·ð¤ ¥ÙéâæÚU Ùãè´ ãñ´)

(1)

(2)

(3)

(4)

22. In the circuits (a) and (b) switches S1 andS2 are closed at t50 and are kept closedfor a long time. The variation of currentsin the two circuits for t ³ 0 are roughlyshown by (figures are schematic and notdrawn to scale) :

(1)

(2)

(3)

(4)


23. x-çÎàææ ×ð´ ¿ÜÌè ãé§ü ç·¤âè çßléÌ ¿éÕ·¤èØ ÌÚ´U» ·¤è¥æßëçæ 231014 Hz ãñ ÌÍæ §â·¤æ çßléÌ ÿæðæ27 Vm21 ãñÐ Ìæð, çÎØð »Øð çÙÙæ´ç·¤Ì çß·¤ËÂæð´ ×ð´âð ·¤æñÙ âæ çß·¤ËÂ, §â ÌÚ´U» ·ð¤ ¿éÕ·¤èØ ÿæðæ ·¤æðÂý·¤ÅU ·¤ÚUÌæ ãñ?(1) 8B ( , t) (3 10 T)x j

→ ∧25 3

8 14sin 2 (1.5 10 2 10 t)x

2p 3 23

(2) 8B ( , t) (9 10 T)x k→ ∧25 3

6 14sin 2 (1.5 10 2 10 t)x

2p 3 23

(3) 8B ( , t) (9 10 T)x i→ ∧25 3

8 14sin 2 (1.5 10 2 10 t)x

2p 3 23

(4) 8B ( , t) (9 10 T)x j→ ∧25 3

6 14sin 1.5 10 2 10 tx

23 23

24. ¥æÂ·¤æð °·¤ àæðçß´» ÎÂüæ ÕÙæÙð ·¤æð ·¤ãæ ÁæÌæ ãñÐØçÎ ·¤æð§ü ÃØçÌ §â ÎÂüæ ·¤æð ¥ÂÙð ¿ðãÚðU âð 10 cmÎêÚU ÚU¹Ìæ ãñ ¥æñÚU ¿ðãÚðU ·ð¤ ¥æßçÏüÌ ÂýçÌçÕÕ ·¤æð,âéçßÏæÁÙ·¤-ÎàæüÙ ·¤è çÙ·¤ÅUÌ× ÎêÚUè, 25 cm ÂÚUÎð¹Ìæ ãñ Ìæð, ÎÂüæ ·¤è ß·ý¤Ìæ çæØæ ãæð»è Ñ(1) 30 cm(2) 24 cm(3) 60 cm(4) 224 cm

23. An electromagnetic wave travelling in thex-direction has frequency of 231014 Hzand electric field amplitude of 27 Vm21.From the options given below, which onedescribes the magnetic field for thiswave ?(1) 8B ( , t) (3 10 T)x j

→ ∧25 3

8 14sin 2 (1.5 10 2 10 t)x

2p 3 23

(2) 8B ( , t) (9 10 T)x k→ ∧25 3

6 14sin 2 (1.5 10 2 10 t)x

2p 3 23

(3) 8B ( , t) (9 10 T)x i→ ∧25 3

8 14sin 2 (1.5 10 2 10 t)x

2p 3 23

(4) 8B ( , t) (9 10 T)x j→ ∧25 3

6 14sin 1.5 10 2 10 tx

23 23

24. You are asked to design a shaving mirrorassuming that a person keeps it 10 cm fromhis face and views the magnified image ofthe face at the closest comfortable distanceof 25 cm. The radius of curvature of themirror would then be :(1) 30 cm(2) 24 cm(3) 60 cm(4) 224 cm


25. x-çÎàææ ×ð´ »çÌ ·¤ÚUÌæ ãé¥æ °·¤ â×æÌÚU §ÜðÅþUæòÙ Âé´Ád ¿æñÇæ§ü ·¤è çÛæÚUè ÂÚU ¥æÂçÌÌ ãæðÌæ ãñ (¥æÚðU¹ Îðç¹Øð)ÐØçÎ §â çÛæÚUè âð çÙ·¤ÜÙð ·ð¤ Âà¿æÌ÷ §ÜðÅþUæòÙ,y-çÎàææ ×ð´, py â´ßð» ÂýæÌ ·¤ÚU ÜðÌð ãñ´ Ìæð, çSÜÅU âð»éÁÚUÙð ßæÜð ¥çÏ·¤æ´àæ §ÜðÅþUæòÙæð´ ·ð¤ çÜØð (ØçÎ hÜæ´·¤ çÙØÌæ´·¤ ãñ) Ñ

(1) ?py?d; h(2) ?py?d>h(3) ?py?d<h(4) ?py?d>>h

26. ç·¤âè ÎêÚUÎàæü·¤ ·ð¤ ¥çÖÎëàØ·¤ ÌÍæ Ùðçæ·¤æ ·¤è È¤æð·¤âÎêçÚUØæ¡ ·ý¤×àæÑ 150 cm ÌÍæ 5 cm ãñÐ ØçÎ 1 kmÎêÚU çSÍÌ ç·¤âè 50 m ª¡¤¿ð ÅUæßÚU (×èÙæÚU) ·¤æð, âæ×æØçßØæâ ×ð´, ÎêÚUÎàæü·¤ âð Îð¹Ùð ÂÚU, ÅUæßÚU ·ð¤ ÂýçÌçÕÕmæÚUæ ÕÙæØæ »Øæ ·¤æðæ, u ãæð Ìæð, u ·¤æ ×æÙ ãæð»æÜ»Ö» Ñ(1) 18(2) 158(3) 308(4) 608

27. 50 V ßæðËÅUÌæ mæÚUæ ßçÚUÌ §ÜðÅþUæòÙ ·¤è Îð-ÕýæòÜèÌÚ´U»ÎñØü ·¤æ çÙ·¤ÅUÌ× ×æÙ ãæð»æ Ñ(?e?51.6310219 C, me59.1310231 kg,h56.6310234 Js)(1) 0.5 Å(2) 1.2 Å(3) 1.7 Å(4) 2.4 Å

25. A parallel beam of electrons travelling inx-direction falls on a slit of width d (seefigure). If after passing the slit, an electronacquires momentum py in the y-directionthen for a majority of electrons passingthrough the slit (h is Plancks constant) :

(1) ?py?d; h(2) ?py?d>h(3) ?py?d<h(4) ?py?d>>h

26. A telescope has an objective lens of focallength 150 cm and an eyepiece of focallength 5 cm. If a 50 m tall tower at adistance of 1 km is observed through thistelescope in normal setting, the angleformed by the image of the tower is u, thenu is close to :(1) 18(2) 158(3) 308(4) 608

27. de-Broglie wavelength of an electronaccelerated by a voltage of 50 V is close to(?e?51.6310219 C, me59.1310231 kg,h56.6310234 Js) :(1) 0.5 Å(2) 1.2 Å(3) 1.7 Å(4) 2.4 Å


28. ØçÎ ¿éÕ·¤èØ ÿæðæ B ·ð¤ ÂýÖæß ×ð´, â×ÌÜ ×ð´ »çÌ·¤ÚUÌð ãé°, m ÎýÃØ×æÙ ÌÍæ q ¥æßðàæ ·ð¤ ·¤æ ÂÚU,ÕæðÚU ×æòÇUÜ Üæ»ê ç·¤Øæ (ÂýØæð» ×ð´ ÜæØæ) ÁæØ Ìæð,¥æßðçàæÌ ·¤æ ·¤è n ßè´ SÌÚU ×ð´ ª¤Áæü ãæð»è Ñ(1) hqB

n2 m

p

(2) hqBn

4 m

p

(3) hqBn

8 m

p

(4) hqBn

m

p

29. ç·¤âè ¥-ÕæØçâÌ n-p â´çÏ ×ð´ §ÜðÅþUæòÙ n - ÿæðæ âðp - ÿæðæ ·¤æð çßâçÚUÌ ãæðÌð ãñ´ Øæð´ç·¤ Ñ(1) p - ÿæðæ ×ð´ çÀUÎý (ãæðÜ) ©ãð´ ¥æ·¤çáüÌ ·¤ÚUÌð

ãñÐ(2) §ÜðÅþUæòÙ çßÖßæÌÚU ·ð¤ ·¤æÚUæ â´çÏ ·ð¤ ÂæÚU

¿Üð ÁæÌð ãñ´Ð(3) n - ÿæðæ ×ð´ §ÜðÅþUæòÙæð´ ·¤è âæ´ÎýÌæ p - ÿæðæ âð

¥çÏ·¤ ãæðÌè ãñÐ(4) §ÜðÅþUæòÙ ·ð¤ßÜ n âð p ÿæðæ ·¤æð ÁæÌð ã´ñ §â·ð¤

çßÂÚèUÌ (p âð n ·¤æð ) Ùãè´Ð

28. If one were to apply Bohr model to aparticle of mass m and charge q movingin a plane under the influence of amagnetic field B, the energy of thecharged particle in the nth level will be :(1) hqB

n2 m

p

(2) hqBn

4 m

p

(3) hqBn

8 m

p

(4) hqBn

m

p

29. In an unbiased n-p junction electronsdiffuse from n - region to p - regionbecause :(1) holes in p - region attract them(2) electrons travel across the junctiondue to potential difference(3) electron concentration in n - regionis more as compared to that inp - region(4) only electrons move from n to pregion and not the vice - versa

PHYSICS : English & Hindi 0930. Diameter of a steel ball is measured usinga Vernier callipers which has divisions of0.1 cm on its main scale (MS) and 10divisions of its vernier scale (VS) match 9divisions on the main scale. Three suchmeasurements for a ball are given as :

S.No. MS (cm) VS divisions

1. 0.5 8

2. 0.5 4

3. 0.5 6

If the zero error is 20.03 cm, then meancorrected diameter is :(1) 0.56 cm(2) 0.59 cm(3) 0.53 cm(4) 0.52 cm

- o 0 o -

30. SÅUèÜ ·¤è °·¤ »æðÜè ·¤æ ÃØæâ °·¤ °ðâð ßçÙüØÚU·ñ¤ÜèÂâü âð ÙæÂæ ÁæÌæ ãñ çÁâ·ð¤ ×éØ Âñ×æÙð ·¤æ °·¤Öæ» (MSD) 0.1 cm ãñ, ÌÍæ §â×ð´ ßçÙüØÚ Âñ×æÙðU(VS) ·ð¤ 10 Öæ», ×éØ Âñ×æÙð ·ð¤ 9 Öæ»æð´ ·ð¤ ÕÚUæÕÚUãñ´Ð »æðÜè ·ð¤ ÃØæâ ·ð¤ çÜØð ÌèÙ ÂæÆ÷UØæ´·¤ (ÚUèçÇU´»)Øãæ¡ çÎØð »Øð ãñ´ Ñ

â¼Ë ¼Î½ §Ö¼Ë¾Õ Í ¼Ë§ cm

ÄÌ¾á½¿U §Ö¼Ë¾Õ Õ »Ë

1. 0.5 8

2. 0.5 4

3. 0.5 6

ØçÎ ßçÙüØÚU ·ñ¤ÜèÂâü ·¤è àæêØ æéçÅU 20.03 cm, ãñÌæð, ÃØæâ ·¤æ ×æØ â´àææðçÏÌ ×æÙ ãæð»æ Ñ(1) 0.56 cm(2) 0.59 cm(3) 0.53 cm(4) 0.52 cm

- o 0 o -

Chemistry : English & Hindi 091. A sample of a hydrate of barium chlorideweighing 61 g was heated until all thewater of hydration is removed. The driedsample weighed 52 g. The formula of thehydrated salt is : (atomic mass,Ba5137 amu, Cl535.5 amu)

(1) BaCl2.H2O(2) BaCl2.2H2O(3) BaCl2.3H2O(4) BaCl2.4H2O

2. Which of the following is not anassumption of the kinetic theory of gases ?(1) A gas consists of many identicalparticles which are in continualmotion.(2) Gas particles have negligible volume.(3) At high pressure, gas particles aredifficult to compress.(4) Collisions of gas particles areperfectly elastic.

3. If the principal quantum number n56, thecorrect sequence of filling of electrons willbe :(1) ns®np®(n21)d®(n22)f(2) ns®(n22)f®(n21)d®np(3) ns®(n21)d®(n22)f®np(4) ns®(n22)f®np®(n21)d

1. 61 g ÕðçÚUØ× ÜæðÚUæ§ÇU ·ð¤ ãæ§ÇþðUÅU ·ð¤ °·¤ Ù×êÙð ·¤æð»ÚU× ·¤ÚU·ð¤ âé¹æØæ »ØæÐ âê¹ð Ù×êÙð ·¤æ ßÁÙ 52 gÍæÐ ãæ§ÇþðUÅU Üßæ ·¤æ âêæ ãñ Ñ (ÂÚU×ææé ÎýÃØ×æÙ,Ba5137 amu, Cl535.5 amu)(1) BaCl2.H2O(2) BaCl2.2H2O(3) BaCl2.3H2O(4) BaCl2.4H2O

2. çÙÙ ×ð´ âð ·¤æñÙ âè »ñâæð´ ·ð¤ ¥æé»çÌ·¤ çâhæ´Ì ·¤è¥ßÏæÚUææ Ùãè´ ãñ?(1) °·¤ »ñâ ÕãéÌ âæÚðU â×M¤Â ·¤ææð´ âð ÕÙÌè ãñ

Áæð Ü»æÌæÚU »çÌ·¤ ¥ßSÍæ ×ð´ ÚUãÌð ãñ´Ð(2) »ñâ ·ð¤ ·¤ææð´ ·¤æ ¥æØÌÙ Ù»Ø ãñÐ(3) ©æ ÎæÕ ÂÚU »ñâ ·¤ææð ·¤æ â´ÂèÇUÙ ·¤çÆUÙ ãñÐ(4) »ñâ ·ð¤ ·¤ææð´ ·ð¤ ×Ø â´æ^ ÂêæüÌÑ ÂýØæSÍ

ãæðÌð ãñÐ

3. ×éØ ßæ´ÅU× â´Øæ n56 ·ð¤ çÜ° §ÜðÅþUæòÙæð´ ·ð¤ ÖÚUÙð·¤æ âãè ·ý¤× ãñ Ñ(1) ns®np®(n21)d®(n22)f(2) ns®(n22)f®(n21)d®np(3) ns®(n21)d®(n22)f®np(4) ns®(n22)f®np®(n21)d

Chemistry : English & Hindi 094. After understanding the assertion andreason, choose the correct option.

Assertion : In the bonding molecularorbital (MO) of H2, electrondensity is increased betweenthe nuclei.Reason : The bonding MO is cA1cB,which shows destructiveinterference of the combiningelectron waves.(1) Assertion and reason are correct andreason is the correct explanation forthe assertion.(2) Assertion and reason are correct, butreason is not the correct explanationfor the assertion.(3) Assertion is correct, reason isincorrect.(4) Assertion is incorrect, reason iscorrect.

5. The heat of atomization of methane andethane are 360 kJ/mol and 620 kJ/mol,respectively. The longest wavelength oflight capable of breaking the C - C bond is(Avogadro number56.0231023,h56.62310234 J s) :(1) 1.493103 nm(2) 2.483103 nm(3) 2.483104 nm(4) 1.493104 nm

6. A solution at 208C is composed of 1.5 molof benzene and 3.5 mol of toluene. If thevapour pressure of pure benzene and puretoluene at this temperature are 74.7 torrand 22.3 torr, respectively, then the totalvapour pressure of the solution and thebenzene mole fraction in equilibrium withit will be, respectively :(1) 35.0 torr and 0.480(2) 38.0 torr and 0.589(3) 30.5 torr and 0.389(4) 35.8 torr and 0.280

4. ¥çÖ·¤ÍÙ ¥æñÚU Ì·ü¤ ·¤æð â×Ûæ·¤ÚU âãè çß·¤ËÂ ¿éçÙ°Ð¥çÖ·¤ÍÙ : ãæ§ÇþUæðÁÙ ·ð¤ ¥æÕ´Ïè ¥æçß·¤ ·¤ÿæ·¤

×ð´ §ÜðÅþUæòÙ æÙß ÙæçÖ·¤æð´ ·ð¤ Õè¿ÕÉ¸æ ãé¥æ ãæðÌæ ãñÐ

Ì·ü¤ : ¥æÕ´Ïè ¥æçß·¤ ·¤ÿæ·¤ cA1cB ãñÁæð â´ØæðÁè §ÜðÅþUæòÙ ÌÚ´U»æð´ ·¤æ çßÙæàæèÃØçÌ·¤ÚUæ ÎàææüÌæ ãñÐ

(1) ¥çÖ·¤ÍÙ ß Ì·ü¤ ÎæðÙæð´ âãè ãñ´ ¥æñÚU Ì·ü¤¥çÖ·¤ÍÙ ·¤æ âãè SÂcÅUè·¤ÚUæ ãñÐ

(2) ¥çÖ·¤ÍÙ ß Ì·ü¤ ÎæðÙæð´ âãè ãñ´ ×»ÚU Ì·ü¤¥çÖ·¤ÍÙ ·¤æ âãè SÂcÅUè·¤ÚUæ Ùãè´ ãñÐ

(3) ¥çÖ·¤ÍÙ âãè ãñ ×»ÚU Ì·ü¤ »ÜÌ ãñÐ(4) ¥çÖ·¤ÍÙ »ÜÌ ãñ ¥æñÚU Ì·ü¤ âãè ãñÐ

5. ç×ÍðÙ ÌÍæ §ÍðÙ ·ð¤ ·¤æè·¤ÚUæ ª¤c×æ ·ý¤×àæÑ360 kJ/mol ÌÍæ 620 kJ/mol ãñ´Ð C - C ¥æÕ´Ï·¤æð ÌæðÇ¸Ùð ·¤è ÿæ×Ìæ ÚU¹Ùð ßæÜè Âý·¤æàæ ·¤è ÎèæüÌ×ÌÚ´U»ÎñØü ãæð»è (¥æßæð»æÎýæð â´Øæ56.0231023,h56.62310234 J s) :(1) 1.493103 nm(2) 2.483103 nm(3) 2.483104 nm(4) 1.493104 nm

6. 208C ÂÚU °·¤ çßÜØÙ ×ð´ 1.5 ×æðÜ ÕðÁèÙ ¥æñÚU3.5 ×æðÜ ÅUæðÜé§Ù ãñ´Ð ¥»ÚU §â ÌæÂ ÂÚU àæéh ÕðÁèÙ¥æñÚU àæéh ÅUæðÜé§Ù ·ð¤ ßæcÂ ÎæÕ ·ý¤×àæÑ 74.7 torr¥æñÚU 22.3 torr ãñ, ÌÕ çßÜØÙ ·¤æ ·é¤Ü ßæcÂ ÎæÕ¥æñÚU ÕðÁèÙ ·¤æ ×æðÜ ¥´àæ §â·ð¤ âæØ ×ð´ ·ý¤×àæÑ ãñ´ Ñ(1) 35.0 torr ¥æñÚU 0.480(2) 38.0 torr ¥æñÚU 0.589(3) 30.5 torr ¥æñÚU 0.389(4) 35.8 torr ¥æñÚU 0.280

Chemistry : English & Hindi 097. Gaseous N2O4 dissociates into gaseousNO2 according to the reaction

N2O4(g)ì 2NO2(g)At 300 K and 1 atm pressure, the degreeof dissociation of N2O4 is 0.2. If one moleof N2O4 gas is contained in a vessel, thenthe density of the equilibrium mixture is :(1) 1.56 g/L(2) 3.11 g/L(3) 4.56 g/L(4) 6.22 g/L

8. A variable, opposite external potential(Eext) is applied to the cellZn|Zn21 (1 M) ?? Cu21 (1 M) | Cu, ofpotential 1.1 V. When Eext < 1.1 V andEext > 1.1 V, respectively electrons flowfrom :(1) anode to cathode and cathode toanode(2) cathode to anode and anode tocathode(3) cathode to anode in both cases(4) anode to cathode in both cases

9. The reaction2N2O5(g) ® 4NO2(g)1O2(g)follows first order kinetics. The pressureof a vessel containing only N2O5 wasfound to increase from 50 mm Hg to87.5 mm Hg in 30 min. The pressureexerted by the gases after 60 min. will be(Assume temperature remains constant) :(1) 106.25 mm Hg(2) 116.25 mm Hg(3) 125 mm Hg(4) 150 mm Hg

7. »ñâèØ N2O4 »ñâèØ NO2 ×ð´ çÙÙ ¥çÖç·ý¤Øæ ·ð¤¥ÙéâæÚU çßØæðçÁÌ ãæðÌæ ãñ ÑN2O4(g)ì 2NO2(g)300 K ¥æñÚU 1 atm ÎæÕ ÂÚU N2O4 ·¤è çßØæðÁÙ×æææ 0.2 ãñÐ ¥»ÚU °·¤ ×æðÜ N2O4 »ñâ ·¤æð °·¤ Âææ×ð´ çÜØæ Áæ° ÌÕ âæØ ç×ææ ·¤æ æÙß ãñ Ñ(1) 1.56 g/L(2) 3.11 g/L(3) 4.56 g/L(4) 6.22 g/L

8. 1.1 V çßÖß ·ð¤ âðÜZn|Zn21 (1 M) ?? Cu21 (1 M) | Cu, ×ð´ °·¤ÂçÚUßÌèü çßÂÚUèÌ Õæs çßÖß (Eext) Ü»æØæ »ØæÐÁÕ Eext < 1.1 V ÌÍæ Eext > 1.1 V, ãæð ÌÕ§ÜðÅþUæòÙæð´ ·¤æ Âýßæã ãæð»æ Ñ(1) °ÙæðÇU âð ·ñ¤ÍæðÇU ÌÍæ ·ñ¤ÍæðÇU âð °ÙæðÇU(2) ·ñ¤ÍæðÇU âð °ÙæðÇU ÌÍæ °ÙæðÇU âð ·ñ¤ÍæðÇU(3) ÎæðÙæð´ çSÍçÌ ×ð´ ·ñ¤ÍæðÇU âð °ÙæðÇU(4) ÎæðÙæð´ çSÍçÌ ×ð´ °ÙæðÇU âð ·ñ¤ÍæðÇU

9. çÙÙ ¥çÖç·ý¤Øæ °·¤ ÂýÍ× ·¤æðÅUè ÕÜ»çÌ·¤è ãñ Ñ2N2O5(g) ® 4NO2(g)1O2(g)30 min ×ð´, °·¤ Âææ çÁâ×ð´ ·ð¤ßÜ N2O5 Íæ, ©â·¤æÎæÕ 50 mm Hg âð 87.5 mm Hg ÕÉ¸ »ØæÐ60 min ·ð¤ ÕæÎ Øã ÎæÕ ãæð»æ (×æÙ ÜèçÁ° ·¤è ÌæÂçÙØÌ ãñ) :(1) 106.25 mm Hg(2) 116.25 mm Hg(3) 125 mm Hg(4) 150 mm Hg

Chemistry : English & Hindi 0910. The following statements relate to theadsorption of gases on a solid surface.Identify the incorrect statement amongthem :

(1) Enthalpy of adsorption is negative(2) Entropy of adsorption is negative(3) On adsorption, the residual forces onthe surface are increased(4) On adsorption decrease in surfaceenergy appears as heat

11. In the long form of the periodic table, thevalence shell electronic configuration of5s2 5p4 corresponds to the element presentin :(1) Group 16 and period 6(2) Group 17 and period 5(3) Group 16 and period 5(4) Group 17 and period 6

12. In the isolation of metals, calcinationprocess usually results in :(1) metal carbonate(2) metal oxide(3) metal sulphide(4) metal hydroxide

13. Permanent hardness in water cannot becured by :(1) Boiling(2) Ion exchange method(3) Calgons method(4) Treatment with washing soda

10. çÙÙ ·¤ÍÙ ÆUæðâ ÂëcÆU ÂÚU »ñâèØ ¥çÏàææðáæ ·ð¤ â´ÎÖü×ð´ ãñ´Ð §Ù×ð´ âð »ÜÌ ·¤ÍÙ ãñ Ñ(1) ¥çÏàææðáæ ·¤è °ÍñËÂè «¤ææ×·¤ ãñÐ(2) ¥çÏàææðáæ ·¤è °ÅþUæòÂè «¤ææ×·¤ ãñÐ(3) ¥çÏàææðáæ ÂÚU, ÂëcÆU ÂÚU ¥ßçàæcÅU ÕÜ ÕÉ¸Ìð

ãñÐ(4) ¥çÏàææðáæ ÂÚU, ÂëcÆU ª¤Áæü ·¤æ Oæâ ª¤c×æ ·ð¤

M¤Â ×ð´ Âý·¤ÅU ãæðÌæ ãñÐ

11. ¥æßÌü âæÚUæè ·ð¤ Îèæü SßM¤Â ×ð´, ¥»ÚU â´ØæðÁè ·¤æðàæ§ÜðÅþUæòÙ çßØæâ 5s2 5p4 ãñ ÌÕ ßã Ìß ©ÂçSÍÌãñ Ñ(1) ß»ü 16 ¥æñÚU ¥æßÌü 6 ×ð´(2) ß»ü 17 ¥æñÚU ¥æßÌü 5 ×ð´(3) ß»ü 16 ¥æñÚU ¥æßÌü 5 ×ð´(4) ß»ü 17 ¥æñÚU ¥æßÌü 6 ×ð´

12. ÏæÌé¥æð´ ·ð¤ çÙc·¤áüæ ×ð´, çÙSÌæÂÙ âð ¥·¤âÚU ÕÙÌðãñ´ Ñ(1) ÏæÌé ·¤æÕæðüÙðÅU(2) ÏæÌé ¥æòâæ§ÇU(3) ÏæÌé âËÈ¤æ§ÇU(4) ÏæÌé ãæ§ÇþUæòâæ§ÇU

13. ÁÜ ·¤è SÍæØè ·¤ÆUæðÚUÌæ ·¤æð §â Âýç·ý¤Øæ âð ÆUè·¤ Ùãè´ç·¤Øæ Áæ â·¤Ìæ ãñ Ñ(1) ©ÕæÜÙæ(2) ¥æØÙ çßçÙ×Ø çßçÏ(3) ·ð¤Ü»æòÙ çßçÏ(4) ÏæßÙ âæðÇUæ ·ð¤ ©Â¿æÚU âð

Chemistry : English & Hindi 09

14. ãæ§ÇþUæòâæ§ÇUæð´ ·¤æ ÌæÂèØ SÍæçØß ·¤æ âãè ·ý¤× ãñ Ñ(1) Ba(OH)2 < Sr(OH)2 < Ca(OH)2 <Mg(OH)2(2) Ba(OH)2 < Ca(OH)2 < Sr(OH)2 <Mg(OH)2(3) Mg(OH)2 < Ca(OH)2 < Sr(OH)2 <Ba(OH)2(4) Mg(OH)2 < Sr(OH)2 < Ca(OH)2 <Ba(OH)2

15. âÕâð ·¤× â´Øæ ·ð¤ ¥æâè¥Ü ÕÙæÌæ ãñ Ñ(1) Ùæ§ÅþUæðÁÙ(2) âËÈ¤ÚU(3) Üé¥æðÚUèÙ(4) ÜæðÚUèÙ

16. ßè.°â.§ü.Âè.¥æÚU. (VSEPR) çâhæ´Ì ·ð¤ ¥ÙéâæÚU,XeOF4 ·¤è Øæç×çÌ ãñ Ñ(1) çæ·¤æðæèØ çmçÂÚUæç×ÇUè(2) ß»ü çÂÚUæç×ÇUè(3) ¥cÅUÈ¤Ü·¤èØ(4) Â´¿·¤æðæèØ â×ÌÜèØ

17. Üßæ X ·¤æ ÁÜèØ çßÜØÙ SCN2 ·ð¤ âæÍ ¹êÙèÜæÜ Ú´U» ¥æñÚU K4[Fe(CN)6] ·ð¤ âæÍ ÙèÜæ Ú´U» ÎðÌæãñÐ X °·¤ â·¤æÚUæ×·¤ ·ý¤æðç×Ü ÜæðÚUæ§ÇU ÂÚUèÿææ ÖèÎðÌæ ãñÐ Üßæ X ãñ Ñ(1) CuCl2(2) FeCl3(3) Cu(NO3)2(4) Fe(NO3)3

18. çÙÙ ×ð âð ·¤æñÙ âæ ¥æé/¥æØÙ â´·é¤Ü Øæñç»·¤æð´ ×ð´´çÜ»ñÇU Ùãè´ ãæð â·¤Ìæ ãñ?(1) CO(2) CN2

(3) CH4(4) Br2

14. The correct order of thermal stability ofhydroxides is :(1) Ba(OH)2 < Sr(OH)2 < Ca(OH)2 <Mg(OH)2(2) Ba(OH)2 < Ca(OH)2 < Sr(OH)2 <Mg(OH)2(3) Mg(OH)2 < Ca(OH)2 < Sr(OH)2 <Ba(OH)2(4) Mg(OH)2 < Sr(OH)2 < Ca(OH)2 <Ba(OH)2

15. The least number of oxyacids are formedby :(1) Nitrogen(2) Sulphur(3) Fluorine(4) Chlorine

16. The geometry of XeOF4 by VSEPR theoryis :(1) trigonal bipyramidal(2) square pyramidal(3) octahedral(4) pentagonal planar

17. An aqueous solution of a salt X turns bloodred on treatment with SCN2 and blue ontreatment with K4[Fe(CN)6]. X also givesa positive chromyl chloride test. The saltX is :(1) CuCl2(2) FeCl3(3) Cu(NO3)2(4) Fe(NO3)3

18. Which molecule/ion among the followingcannot act as a ligand in complexcompounds ?(1) CO(2) CN2

(3) CH4(4) Br2

Chemistry : English & Hindi 0919. The correct statement on the isomerismassociated with the following complexions,

(a) [Ni(H2O)5NH3]21,(b) [Ni(H2O)4(NH3)2]21 and(c) [Ni(H2O)3(NH3)3]21 is :(1) (a) and (b) show only geometricalisomerism(2) (a) and (b) show geometrical andoptical isomerism(3) (b) and (c) show geometrical andoptical isomerism(4) (b) and (c) show only geometricalisomerism

20. Photochemical smog consists of excessiveamount of X, in addition to aldehydes,ketones, peroxy acetyl nitrile (PAN), andso forth. X is :(1) CH4(2) CO(3) CO2(4) O3

21. 1.4 g of an organic compound was digestedaccording to Kjeldahls method and theammonia evolved was absorbed in 60 mLof M/10 H2SO4 solution. The excesssulphuric acid required 20 mL of M/10NaOH solution for neutralization. Thepercentage of nitrogen in the compoundis :(1) 3(2) 5(3) 10(4) 24

22. The optically inactive compound from thefollowing is :(1) 2 - chloropropanal(2) 2 - chloropentane(3) 2 - chlorobutane(4) 2 - chloro - 2 - methylbutane

19. çÙÙ â´·é¤Ü ¥æØÙæð´ âð âÕçÏÌ â×æßØßÌæ ÂÚUâãè ·¤ÍÙ ãñ´(a) [Ni(H2O)5NH3]21,(b) [Ni(H2O)4(NH3)2]21 ¥æñÚU(c) [Ni(H2O)3(NH3)3]21 :(1) (a) ¥æñÚU (b) ·ð¤ßÜ Øæç×ÌèØ â×æßØßÌæ

ÎàææüÌð ãñ´Ð(2) (a) ¥æñÚU (b) Øæç×ÌèØ ¥æñÚU Ïýéßæ â×æßØßÌæ

ÎàææüÌð ãñ´Ð(3) (b) ¥æñÚU (c) Øæç×ÌèØ ¥æñÚU Ïýéßæ â×æßØßÌæ

ÎàææüÌð ãñ´Ð(4) (b) ¥æñÚU (c) ·ð¤ßÜ Øæç×ÌèØ â×æßØßÌæ

ÎàææüÌð ãñ´Ð

20. Âý·¤æàæ ÚUæâæØçÙ·¤ Ïê× ·¤æðãÚðU ×ð´ °ðçËÇUãæ§ÇU, ·¤èÅUæðÙ,ÂðÚUæðâè °çâÅUæ§Ü Ùæ§ÅþUæ§Ü (PAN) §ØæçÎ ·ð¤ ¥Üæßæ¥çÏ·¤ ×æææ ×ð´ X Öè ãæðÌæ ãñ, X ãñ Ñ(1) CH4(2) CO(3) CO2(4) O3

21. 1.4 g ·¤æÕüçÙ·¤ Øæñç»·¤ ·¤æð ·ñ¤ËÇUæÜ çßçÏ ·ð¤ ¥ÙéâæÚUÂæç¿Ì ç·¤Øæ »Øæ ÌÍæ çÙ·¤Üð ¥×æðçÙØæ ·¤æð 60 mL,M/10, H2SO4 çßÜØÙ ×ð ¥ßàææðçáÌ ç·¤Øæ »ØæÐ¥çÌçÚUÌ âËØêçÚU·¤ ¥Ü ·¤æð ©ÎæâèÙ ·¤ÚUÙð ·ð¤çÜ°, 20 mL, M/10, NaOH Ü»æÐ §â Øæñç»·¤×ð´ Ùæ§ÅþUæðÁÙ ·¤è ÂýçÌàæÌÌæ ãñ Ñ(1) 3(2) 5(3) 10(4) 24

22. çÙÙ ×ð´ âð Ïýéßæ ¥æêæü·¤ Øæñç»·¤ ãñ Ñ(1) 2 - ÜæðÚUæðÂýæðÂðÙÜ(2) 2 - ÜæðÚUæðÂðÅðUÙ(3) 2 - ÜæðÚUæðØêÅðUÙ(4) 2 - ÜæðÚUæð - 2 - ×ðçÍÜØêÅðUÙ

Chemistry : English & Hindi 09

23.

A is :(1)

(2)

(3)

(4)

24. A compound A with molecular formulaC10H13Cl gives a white precipitate onadding silver nitrate solution. A onreacting with alcoholic KOH givescompound B as the main product. B onozonolysis gives C and D. C givesCannizaro reaction but not aldolcondensation. D gives aldol condensationbut not Cannizaro reaction. A is :(1)

(2)

(3) C6H52CH22CH22CH22CH22Cl

(4)

23.

A ãñ Ñ(1)

(2)

(3)

(4)

24. Øæñç»·¤ A çÁâ·¤æ ¥æéâêæ C10H13Cl ãñ, çâËßÚUÙæ§ÅþðUÅU çßÜØÙ ç×ÜæÙð ÂÚU àßðÌ ¥ßÿæðÂ ÎðÌæ ãñÐA °ðË·¤æðãæòçÜ·¤ KOH ·ð¤ âæÍ ¥çÖç·ý¤Øæ ·¤ÚUÙðÂÚU ×éØ M¤Â âð Øæ ñç»·¤ B ÎðÌæ ãñÐ B ·¤æ¥æðÁæðÙ-¥ÂæÅUÙ ·¤ÚUÙð ÂÚU Øæñç»·¤ C ÌÍæ D ÂýæÌãæðÌð ãñ´Ð C ·ñ¤çÙÁæÚUæð ¥çÖç·ý¤Øæ ÎðÌæ ãñ, ÂÚUÌé °ðËÇUæÜâ´æÙÙ Ùãè´ ÎðÌæÐ D °ðËÇUæÜ â´æÙÙ ÎðÌæ ãñ, ÂÚUÌé·ñ¤çÙÁæÚUæð ¥çÖç·ý¤Øæ Ùãè´ ÎðÌæÐ A ãñ Ñ(1)

(2)

(3) C6H52CH22CH22CH22CH22Cl

(4)

Chemistry : English & Hindi 0925. In the presence of a small amount ofphosphorous, aliphatic carboxylic acidsreact with chlorine or bromine to yield acompound in which a - hydrogen has beenreplaced by halogen. This reaction isknown as :

(1) Wolff - Kischner reaction(2) Etard reaction(3) Hell - Volhard - Zelinsky reaction(4) Rosenmund reaction

26. Arrange the following amines in the orderof increasing basicity.

(1)

(2)

(3)

(4)

25. È¤æòSÈ¤æðÚUâ ·¤è ·¤× ×æææ ·¤è ©ÂçSÍçÌ ×ð´ °ÜèÈð¤çÅU·¤·¤æÕæðüçâçÜ·¤ ¥Ü ÜæðÚUèÙ ¥æñÚU Õýæð×èÙ ·ð¤ âæÍ¥çÖç·ý¤Øæ ·¤ÚUÌð ãé° ¥ÂÙð a - ãæ§ÇþUæðÁÙ ·¤æð ãñÜæðÁÙ×ð´ ÂçÚUßçÌüÌ ·¤ÚUÌð ãñ´Ð §â ¥çÖç·ý¤Øæ ·¤æ Ùæ× ãñ Ñ(1) ßæðËÈ¤-ç·¤àÙÚU ¥çÖç·ý¤Øæ(2) §üÅUæÇüU ¥çÖç·ý¤Øæ(3) ãðÜ-È¤æðÜæÇüU-ÁðçÜ´S·¤è ¥çÖç·ý¤Øæ(4) ÚUæðÁðÙ×é´ÇU ¥çÖç·ý¤Øæ

26. çÙÙ ¥×èÙæð´ ·¤æð ÿææÚU·¤Ìæ ·ð¤ ÕÉ¸Ìð ·ý¤× ×ð´ Ü»æ§°Ð

(1)

(2)

(3)

(4)

Chemistry : English & Hindi 0927. Match the polymers in column-A withtheir main uses in column-B and choosethe correct answer :

Column - A Column - B

(A) Polystyrene (i) Paints and

lacquers

(B) Glyptal (ii) Rain coats

(C) Polyvinyl

Chloride

(iii) Manufacture of

toys

(D) Bakelite (iv) Computer discs

(1) (A) - (ii), (B) - (i), (C) - (iii), (D) - (iv)(2) (A) - (iii), (B) - (i), (C) - (ii), (D) - (iv)(3) (A) - (ii), (B) - (iv), (C) - (iii), (D) - (i)(4) (A) - (iii), (B) - (iv), (C) - (ii), (D) - (i)

28. Complete hydrolysis of starch gives :(1) glucose and fructose in equimolaramounts(2) galactose and fructose in equimolaramounts(3) glucose only(4) glucose and galactose in equimolaramounts

29. is used as :

(1) Insecticide(2) Antihistamine(3) Analgesic(4) Antacid

30. The cation that will not be precipitated byH2S in the presence of dil HCl is :(1) Cu21

(2) Pb21

(3) As31

(4) Co21

- o O o -

27. ·¤æòÜ×-A ×ð´ çÎ° »° ÕãéÜ·¤æð´ ·¤æð ·¤æòÜ×-B ×ð´ ©Ù·ð¤Âý×é¹ ©ÂØæð» ·ð¤ âæÍ âé×ðçÜÌ ·¤Úð´U ÌÍæ âãè çß·¤ËÂ¿éÙð´ Ñ

Ë×Á¼ - A Ë×Á¼ - B

(A) §ËÁÍSªUË¿UÍ¾ (i) §âÁÕ§ ËÖ¿U §âÁËäË º¾Ë¾Õ ¼Õ

(B) ÌóÁåªUÁ (ii) º¿UÇËÌ±½Ë º¾Ë¾Õ ¼Õ(C) §Ë×ÁÍÄËÌ¾Á

þÁËÕ¿UË¬U(iii) ÌÁËÖ¾Õ º¾Ë¾Õ ¼Õ

(D) ºÖÕÁËªU (iv) å½ÏªU¿U Ì¬US º¾Ë¾Õ ¼Õ

(1) (A) - (ii), (B) - (i), (C) - (iii), (D) - (iv)(2) (A) - (iii), (B) - (i), (C) - (ii), (D) - (iv)(3) (A) - (ii), (B) - (iv), (C) - (iii), (D) - (i)(4) (A) - (iii), (B) - (iv), (C) - (ii), (D) - (i)

28. SÅUæ¿ü ·ð¤ Âêæü ÁÜ ¥ÂæÅUÙ âð ç×ÜÌæ ãñ Ñ(1) Üê·¤æðâ ¥æñÚU Èýé¤ÅUæðâ ·¤è â××æðÜ ×æææ(2) »ñÜðÅUæðâ ¥æñÚU Èýé¤ÅUæðâ ·¤è â××æðÜ ×æææ(3) ·ð¤ßÜ Üê·¤æðâ(4) Üê·¤æðâ ¥æñÚU »ñÜðÅUæðâ ·¤è â××æðÜ ×æææ

29. ·¤æ ÂýØæð» ç·¤â M¤Â ×ð´ ãæðÌæ ãñ?

(1) ·¤èÅUÙæàæ·¤(2) ÂýçÌçãSÅñUç×Ù(3) ÂèÇ¸æãæÚUè(4) ÂýçÌ¥Ü

30. ßã ÏÙæØÙ Áæð ÌÙé HCl ·ð¤ ©ÂçSÍçÌ ×ð´ H2S âð¥ßÿæðçÂÌ Ùãè´ ãæðÌæ ãñ, ßã ãñ Ñ(1) Cu21(2) Pb21(3) As31(4) Co21

- o O o -

MATHEMATICS : English & Hindi 091. In a certain town, 25% of the families owna phone and 15% own a car ; 65% familiesown neither a phone nor a car and 2,000families own both a car and a phone.Consider the following three statements :

(a) 5% families own both a car and aphone.(b) 35% families own either a car or aphone.(c) 40,000 families live in the town.Then,(1) Only (a) and (b) are correct.(2) Only (a) and (c) are correct.(3) Only (b) and (c) are correct.(4) All (a), (b) and (c) are correct.

2. The largest value of r for which the regionrepresented by the set v e C/?v242i?[ris contained in the region represented bythe set z e C/?z21?[?z1i?, is equal to :(1) 17

(2) 2 2

(3) 32

2

(4) 52

2

3. If 213i is one of the roots of the equation2x329x21kx21350, k e R, then the realroot of this equation :(1) does not exist.(2) exists and is equal to 1

2.

(3) exists and is equal to 1

22 .

(4) exists and is equal to 1.

1. ç·¤âè àæãÚU ×ð´, 25% ÂçÚUßæÚUæð´ ·ð¤ Âæâ È¤æðÙ ãñ ÌÍæ15% ·ð¤ Âæâ ·¤æÚU ãñ ; 65% ÂçÚUßæÚUæð´ ·ð¤ Âæâ Ù ÌæðÈ¤æðÙ ãñ ¥æñÚU Ù ãè ·¤æÚU ãñ, ÌÍæ 2,000 ÂçÚUßæÚUæð´ ·ð¤ ÂæâÈ¤æðÙ ÌÍæ ·¤æÚU ÎæðÙæð´ ãñ´Ð çÙÙ ÌèÙ ·¤ÍÙæð´ ÂÚU çß¿æÚU·¤èçÁ° Ñ(a) 5% ÂçÚUßæÚUæð´ ·ð¤ Âæâ ·¤æÚU ÌÍæ È¤æðÙ ÎæðÙæð´ ãñ´Ð(b) 35% ÂçÚUßæÚUæð´ ·ð¤ Âæâ Øæ Ìæð ·¤æÚU ãñ Øæ È¤æðÙ

ãñÐ(c) àæãÚU ×ð´ 40,000 ÂçÚUßæÚU ÚUãÌð ãñ´ÐÌæð,(1) ·ð¤ßÜ (a) ÌÍæ (b) âãè ãñ´Ð(2) ·ð¤ßÜ (a) ÌÍæ (c) âãè ãñ´Ð(3) ·ð¤ßÜ (b) ÌÍæ (c) âãè ãñ´Ð(4) (a), (b) ÌÍæ (c) âÖè âãè ãñÐ

2. r ·¤æ ßã ¥çÏ·¤Ì× ×æÙ çÁâ·ð¤ çÜ° â×éæØv e C/?v242i?[r mæÚUæ çÙÏæüçÚUÌ ÿæðæ, â×éæØz e C/?z21?[?z1i? mæÚUæ çÙÏæüçÚUÌ ÿæðæ ×ð ´âç×çÜÌ ãñ, ãñ Ñ(1) 17

(2) 2 2

(3) 32

2

(4) 52

2

3. ØçÎ 213i, â×è·¤ÚUæ 2x329x21kx21350,k e R ·¤æ °·¤ ×êÜ ãñ, Ìæð §â â×è·¤ÚUæ ·¤æ ßæSÌçß·¤×êÜ Ñ(1) çßl×æÙ Ùãè´ ãñÐ(2) çßl×æÙ ãñ ÌÍæ 1

2 ·ð¤ ÕÚUæÕÚU ãñÐ

(3) çßl×æÙ ãñ ÌÍæ 1

22 ·ð¤ ÕÚUæÕÚU ãñÐ

(4) çßl×æÙ ãñ ÌÍæ 1 ·ð¤ ÕÚUæÕÚU ãñÐ

MATHEMATICS : English & Hindi 094. The least value of the product xyz for

which the determinant 1 1

1 1

1 1

x

y

z

is

non-negative, is :(1) 2 22

(2) 16 22

(3) 28(4) 21

5. If 0 1A

1 0

25 , then which one of the

following statements is not correct ?(1) A42I5A21I(2) A32I5A(A2I)(3) A21I5A(A22I)(4) A31I5A(A32I)

6. The number of ways of selecting 15 teamsfrom 15 men and 15 women, such thateach team consists of a man and a woman,is :(1) 1120(2) 1240(3) 1880(4) 1960

7. Let the sum of the first three terms of anA.P. be 39 and the sum of its last four termsbe 178. If the first term of this A.P. is 10,then the median of the A.P. is :(1) 26.5(2) 28(3) 29.5(4) 31

4. »éæÙÈ¤Ü xyz ·¤æ ßã ØêÙÌ× ×êËØ çÁâ·ð¤ çÜ°

âæÚUçæ·¤1 1

1 1

1 1

x

y

z

«¤æðÌÚU ãñ, ãñ Ñ

(1) 2 22

(2) 16 22

(3) 28(4) 21

5. ØçÎ 0 1A

1 0

25 ãñ, Ìæð çÙÙ ×ð´ âð ·¤æñÙ-âæ °·¤

·¤ÍÙ âãè Ùãè´ ãñ?(1) A42I5A21I(2) A32I5A(A2I)(3) A21I5A(A22I)(4) A31I5A(A32I)

6. 15 ÂéL¤áæð´ ÌÍæ 15 ×çãÜæ¥æð´ ×ð´ âð °ðâè 15 ÅUè×ð´,çÁÙ×ð´ ÂýØð·¤ ×ð´ °·¤ ÂéL¤á ÌÍæ °·¤ ×çãÜæ ãæð, ¿éÙÙð·ð¤ ÌÚUè·¤æð´ ·¤è â´Øæ ãñ Ñ(1) 1120(2) 1240(3) 1880(4) 1960

7. ×æÙæ °·¤ â×æ´ÌÚU æðÉ¸è ·ð¤ ÂýÍ× ÌèÙ ÂÎæð´ ·¤æ Øæð» 39ãñ ÌÍæ §â·ð¤ ¥´çÌ× ¿æÚU ÂÎæð´ ·¤æ Øæð» 178 ãñÐ ØçÎ§â â×æ´ÌÚU æðÉ¸è ·¤æ ÂýÍ× ÂÎ 10 ãñ, Ìæð §â â×æ´ÌÚUæðÉ¸è ·¤æ ×æØ·¤ ãñ Ñ(1) 26.5(2) 28(3) 29.5(4) 31

MATHEMATICS : English & Hindi 098. If the coefficients of the three successiveterms in the binomial expansion of (11x)n

are in the ratio 1 : 7 : 42, then the first ofthese terms in the expansion is :(1) 6th(2) 7th(3) 8th(4) 9th

9. The value of 30

r 16

(r 2)(r 3)∑5

1 2 is equal to :(1) 7785(2) 7780(3) 7775(4) 7770

10.2

20

e cos

sin

x

x

xlim

x→

2 is equal to :(1) 3(2) 3

2

(3) 5

4

(4) 211. The distance, from the origin, of thenormal to the curve, x52 cost 1 2t sint,

y 5 2 sint22t cost at 4

tp

5 , is :(1) 4(2) 2 2

(3) 2(4) 2

8. ØçÎ (11x)n ·ð¤ çmÂÎ çßSÌæÚU ×ð´ ÌèÙ ·ý¤ç×·¤ ÂÎæð´·ð¤ »éææ´·¤æð´ ×ð´ 1 : 7 : 42 ·¤æ ¥ÙéÂæÌ ãñ, Ìæð §Ù ×ð´ âðçßSÌæÚU ×ð ÂãÜæ ÂÎ ãñ Ñ(1) ÀUÆUæ(2) âæÌßæ´(3) ¥æÆUßæ´(4) Ùæñßæ´

9.30

r 16

(r 2)(r 3)∑5

1 2 ·¤æ ×æÙ ÕÚUæÕÚU ãñ Ñ(1) 7785(2) 7780(3) 7775(4) 7770

10.2

20

e cos

sin

x

x

xlim

x→

2 ÕÚUæÕÚU ãñ Ñ(1) 3(2) 3

2

(3) 5

4

(4) 2

11. ß·ý¤ x52 cost12t sint, y52 sint22t cost ÂÚU

4t

p5 ÂÚU ¹è´¿ð »° ¥çÖÜ´Õ ·¤è ×êÜ çÕ´Îé âð

ÎêÚUè ãñ Ñ(1) 4(2) 2 2

(3) 2(4) 2

MATHEMATICS : English & Hindi 0912. If Rolles theorem holds for the function

f (x)52x31bx21cx, x e [21, 1], at thepoint 1

2

x 5 , then 2b1c equals :(1) 1(2) 21(3) 2(4) 23

13. Let the tangents drawn to the circle,x21y2516 from the point P(0, h) meet thex-axis at points A and B. If the area ofDAPB is minimum, then h is equal to :(1) 4 3

(2) 3 3

(3) 3 2

(4) 4 2

14. The integral 3 54 4

d

( 1) ( 2)

x

x x∫

1 2 is equal

to :

(1)14 1

4 C 2

x

x

1 12

(2)14 2

4 C 1

x

x

2 11

(3)144 1

C3 2

x

x

12 12

(4)144 2

C3 1

x

x

22 11

12. ØçÎ È¤ÜÙ f (x)52x31bx21cx, x e [21, 1]·ð¤ çÜ° çÕ´Îé 1

2

x 5 ÂÚU ÚUæðÜð ·¤æ Âý×ðØ Üæ»ê ãæðÌæ ãñ,Ìæð 2b1c ÕÚUæÕÚU ãñ Ñ(1) 1(2) 21(3) 2(4) 23

13. ×æÙæ çÕ´Îé P(0, h) âð ßëæ x21y2516 ÂÚU ¹è´¿è»§ü SÂàæü ÚðU¹æ°¡ x-¥ÿæ ·¤æð çÕ´Îé¥æð´ A ÌÍæ B ÂÚUç×ÜÌè ãñÐ ØçÎ DAPB ·¤æ ÿæðæÈ¤Ü ØêÙÌ× ãñ, Ìæðh ÕÚUæÕÚU ãñ Ñ(1) 4 3

(2) 3 3

(3) 3 2

(4) 4 2

14. â×æ·¤Ü 3 54 4

d

( 1) ( 2)

x

x x∫

1 2 ÕÚUæÕÚU ãñ Ñ

(1)14 1

4 C 2

x

x

1 12

(2)14 2

4 C 1

x

x

2 11

(3)144 1

C3 2

x

x

12 12

(4)144 2

C3 1

x

x

22 11

MATHEMATICS : English & Hindi 09

15. x > 0 ·ð¤ çÜ° ×æÙæ 1

log t( ) dt

1 t

x

f x ∫51

ãñ, Ìæð1

( ) f x fx

1 ÕÚUæÕÚU ãñ Ñ

(1) 21 (log )

4x

(2) 21 (log )

2x

(3) log x(4) 21

log 4

x

16. ß·ý¤æð´ y12x250 ÌÍæ y13x251 mæÚUæ ÂçÚUÕh ÿæðæ·¤æ ÿæðæÈ¤Ü (ß»ü §·¤æ§Øæð´ ×ð´) ÕÚUæÕÚU ãñ Ñ(1) 3

5

(2) 3

4

(3) 1

3

(4) 4

3

17. ØçÎ y(x), ¥ß·¤Ü â×è·¤ÚUæ

2d( 2) 4 9

d

yx x x

x1 5 1 2 , x ¹ 22 ¥æñÚU

y(0)50, ·¤æ ãÜ ãñ, Ìæð y(24) ÕÚUæÕÚU ãñ Ñ(1) 0(2) 1(3) 21(4) 2

15. For x > 0, let 1

log t( ) dt

1 t

x

f x ∫51

. Then1

( ) f x fx

1 is equal to :

(1) 21 (log )

4x

(2) 21 (log )

2x

(3) log x(4) 21

log 4

x

16. The area (in square units) of the regionbounded by the curves y12x250 andy13x251, is equal to :(1) 3

5

(2) 3

4

(3) 1

3

(4) 4

3

17. If y(x) is the solution of the differentialequation

2d( 2) 4 9

d

yx x x

x1 5 1 2 , x ¹ 22

and y(0)50, then y(24) is equal to :(1) 0(2) 1(3) 21(4) 2

MATHEMATICS : English & Hindi 09

18. The points 80,

3

, (1, 3) and (82, 30) :(1) form an obtuse angled triangle.(2) form an acute angled triangle.(3) form a right angled triangle.(4) lie on a straight line.

19. Let L be the line passing through the pointP(1, 2) such that its intercepted segmentbetween the co-ordinate axes is bisectedat P. If L1 is the line perpendicular to Land passing through the point (22, 1),then the point of intersection of L and L1is :(1) 4 12

, 5 5

(2) 11 29,

20 10

(3) 3 17,

10 5

(4) 3 23,

5 10

20. If y13x50 is the equation of a chord ofthe circle, x21y2230x50, then theequation of the circle with this chord asdiameter is :(1) x21y213x19y50(2) x21y223x19y50(3) x21y223x29y50(4) x21y213x29y50

18. çÕ´Îé 80,

3

, (1, 3) ÌÍæ (82, 30) Ñ(1) °·¤ ¥çÏ·¤·¤æðæ çæÖéÁ ÕÙæÌð ãñ´Ð(2) °·¤ ØêÙ·¤æðæ çæÖéÁ ÕÙæÌð ãñ´Ð(3) °·¤ â×·¤æðæ çæÖéÁ ÕÙæÌð ãñ´Ð(4) °·¤ âÚUÜ ÚðU¹æ ÂÚU çSÍÌ ãñ´Ð

19. ×æÙæ L, çÕ´Îé P(1, 2) âð ãæð·¤ÚU ÁæÙð ßæÜè ßã ÚðU¹æ ãñçÁâ·¤æ çÙÎðüàææ´·¤ ¥ÿææð´ ·ð¤ Õè¿ ·¤ÅUæ ÚðU¹æ¹ÇU P ÂÚUâ×çmÖæçÁÌ ãæðÌæ ãñÐ ×æÙæ L1 ßã ÚðU¹æ ãñ Áæð L ÂÚUÜ´ÕßÌ ãñ ÌÍæ çÕ´Îé (22, 1) âð ãæð·¤ÚU ÁæÌè ãñ, ÌæðL ÌÍæ L1 ·¤æ ÂýçÌÀðUÎÙ çÕ´Îé ãñ Ñ(1) 4 12

, 5 5

(2) 11 29,

20 10

(3) 3 17,

10 5

(4) 3 23,

5 10

20. ØçÎ y13x50, ßëæ x21y2230x50 ·¤è °·¤Áèßæ ·¤æ â×è·¤ÚUæ ãñ, Ìæð ©â ßëæ, çÁâ·¤æ ÃØæâ,Øã Áèßæ ãñ, ·¤æ â×è·¤ÚUæ ãñ Ñ(1) x21y213x19y50(2) x21y223x19y50(3) x21y223x29y50(4) x21y213x29y50

MATHEMATICS : English & Hindi 0921. If the tangent to the conic, y265x2 at(2, 10) touches the circle,

x21y218x22y5k (for some fixed k) at apoint (a, b) ; then (a, b) is :(1) 6 10

, 17 17

2

(2) 8 2,

17 17

2

(3) 4 1,

17 17

2

(4) 7 6,

17 17

2

22. An ellipse passes through the foci of thehyperbola, 9x224y2536 and its majorand minor axes lie along the transverse andconjugate axes of the hyperbolarespectively. If the product of eccentricitiesof the two conics is 1

2, then which of the

following points does not lie on theellipse ?(1) ( )13 , 0

(2) 39, 3

2

(3) 1 313 ,

2 2

(4) 13, 6

2

23. If the points (1, 1, l) and (23, 0, 1) areequidistant from the plane,3x14y212z11350, then l satisfies theequation :(1) 3x2210x1750(2) 3x2110x1750(3) 3x2110x21350(4) 3x2210x12150

21. ØçÎ àææ´·¤ß y265x2 ·ð¤ çÕ´Îé (2, 10) ÂÚU ¹è´¿è»§ü SÂàæü Úð U¹æ ßëæ x21y218x22y5k ·¤æ ð(ç·¤âè çÙçà¿Ì k ·ð¤ çÜ°) çÕ´Îé (a, b) ÂÚU SÂàæü·¤ÚUÌè ãñ, Ìæð (a, b) ãñ Ñ(1) 6 10

, 17 17

2

(2) 8 2,

17 17

2

(3) 4 1,

17 17

2

(4) 7 6,

17 17

2

22. °·¤ Îèæ üßëæ, ¥çÌÂÚUßÜØ 9x224y2536 ·ð¤ÙæçÖ·ð´¤Îýæð´ âð ãæð·¤ÚU ÁæÌæ ãñ ÌÍæ §â·ð¤ Îèæü ÌÍæ Üæé¥ÿæ ·ý¤×àæÑ ¥çÌÂÚUßÜØ ·ð¤ ¥ÙéÂýSÍ ÌÍæ â´Øé×è¥ÿææð ·ð¤ ¥ÙéçÎàæ ãñÐ ØçÎ §Ù Îæð àææ·¤ßæð ·¤è ©·ð¤ÎýÌæ¥æð·¤æ »éæÙÈ¤Ü 1

2 ãñ, Ìæð çÙÙ ×ð´ âð ·¤æñÙ-âæ çÕ´Îé

Îèæüßëæ ÂÚU çSÍÌ Ùãè´ ãñ?(1) ( )13 , 0

(2) 39, 3

2

(3) 1 313 ,

2 2

(4) 13, 6

2

23. ØçÎ çÕ´Î é (1, 1, l) ÌÍæ (23, 0, 1) â×ÌÜ3x14y212z11350 âð â×ÎêÚUSÍ ãñ´, Ìæð l, çÙÙâ×è·¤ÚUæ ·¤æð â´ÌécÅU ·¤ÚUÌæ ãñ Ñ(1) 3x2210x1750(2) 3x2110x1750(3) 3x2110x21350(4) 3x2210x12150

MATHEMATICS : English & Hindi 0924. If the shortest distance between the lines

1 1

1 1

yx z125 5

a 2, (a ¹21) and

x1y1z115052x2y1z13 is 1

3, then

a value of a is :(1) 16

192

(2) 19

162

(3) 32

19

(4) 19

32

25. Let a→ and b

→ be two unit vectors suchthat 5 3 . If

( )c a 2 b 3 a b→→ → →→

5 1 1 3 , then isequal to :(1) 55

(2) 51

(3) 43

(4) 37

26. Let X be a set containing 10 elements andP(X) be its power set. If A and B are pickedup at random from P(X), withreplacement, then the probability that Aand B have equal number of elements, is :(1) 20

10102

C

(2) ( )10

202 1

2

2

(3) ( )10

102 1

2

2

(4) 2010

202

C

24. ØçÎ ÚðU¹æ¥æð´ 1 1

1 1

yx z125 5

a 2, (a ¹21)

ÌÍæ x1y1z115052x2y1z13 ·ð¤ Õè¿·¤è ØêÙÌ× ÎêÚUè 1

3 ãñ, Ìæð a ·¤æ °·¤ ×æÙ ãñ Ñ

(1) 16

192

(2) 19

162

(3) 32

19

(4) 19

32

25. ×æÙæ a→ ÌÍæ b

→ °ðâð ×ææ·¤ âçÎàæ ã ñ ´ ç·¤5 3 ãñÐ ØçÎ

( )c a 2 b 3 a b→→ → →→

5 1 1 3 ãñ, Ìæð ÕÚUæÕÚUãñ Ñ(1) 55

(2) 51

(3) 43

(4) 37

26. ×æÙæ X °·¤ â×éæØ ãñ çÁâ×ð´ 10 ¥ßØß ãñ´ ÌÍæP(X) §â·¤æ ææÌ â×éæØ ãñÐ ØçÎ P(X) âð A ÌÍæB ØæÎëÀUØæ, ÂýçÌSÍæÂÙæ âçãÌ, çÜ° »° ãñ´, ÌæðA ÌÍæ B ×ð´ ÕÚUæÕÚU ¥ßØßæð´ ·ð¤ ãæðÙð ·¤è ÂýæçØ·¤Ìæ ãñ Ñ(1) 20

10102

C

(2) ( )10

202 1

2

2

(3) ( )10

102 1

2

2

(4) 2010

202

C

MATHEMATICS : English & Hindi 0927. A factory is operating in two shifts, dayand night, with 70 and 30 workersrespectively. If per day mean wage of theday shift workers is ` 54 and per day meanwage of all the workers is ` 60, then perday mean wage of the night shift workers(in `) is :

(1) 66(2) 69(3) 74(4) 75

28. In a DABC, a 2 3

b5 1 and ÐC5608.

Then the ordered pair (ÐA, ÐB) is equalto :(1) (158, 1058)(2) (1058, 158)(3) (458, 758)(4) (758, 458)

29. If 1 12

2( ) 2tan sin

1

xf x x

x

2 25 11

, x > 1,then f (5) is equal to :(1)

2

p

(2) p(3) 4 tan21(5)(4) 1 65

tan156

2

30. The contrapositive of the statementIf it is raining, then I will not come, is :(1) If I will come, then it is not raining.(2) If I will not come, then it is raining.(3) If I will not come, then it is notraining.(4) If I will come, then it is raining.

- o 0 o -

27. °·¤ Èñ¤ÅþUè Îæð ÂæçÚUØæð´, çÎÙ ÌÍæ ÚUæÌ, ×ð´ ¿ÜÌè ãñçÁÙ×ð´ ·ý¤×àæÑ 70 ÌÍæ 30 ·¤æ×»æÚU ·¤æØü ·¤ÚUÌð ãñ´ÐØçÎ çÎÙ ·¤è ÂæÚUè ·ð¤ ·¤æ×»æÚUæð´ ·¤æ ×æØ ÂýçÌçÎÙßðÌÙ ` 54 ãñ ÌÍæ âÖè ·¤æ×»æÚUæð´ ·¤æ ×æØ ÂýçÌçÎÙßðÌÙ ` 60 ãñ, Ìæð ÚUæÌ ×ð´ ·¤æØü ·¤ÚUÙð ßæÜð ·¤æ×»æÚUæð´ ·¤æ×æØ ÂýçÌçÎÙ ßðÌÙ (` ×ð´) ãñ Ñ(1) 66(2) 69(3) 74(4) 75

28. °·¤ çæÖéÁ ABC ×ð´, a 2 3

b5 1 ÌÍæ ÐC5608

ãñ, Ìæð ·ý¤ç×Ì Øé× (ÐA, ÐB) ÕÚUæÕÚU ãñ Ñ(1) (158, 1058)(2) (1058, 158)(3) (458, 758)(4) (758, 458)

29. ØçÎ 1 12

2( ) 2tan sin

1

xf x x

x

2 25 11

, x > 1ãñ, Ìæð f (5) ÕÚUæÕÚU ãñ Ñ(1)

2

p

(2) p(3) 4 tan21(5)(4) 1 65

tan156

2

30. ·¤ÍÙ ÑÒÒØçÎ ßáæü ãæð ÚUãè ãñ, Ìæð ×ñ´ Ùãè´ ¥æª´¤»æÓÓ·¤æ ÂýçÌÏÙæ×·¤ ·¤ÍÙ ãñ Ñ(1) ØçÎ ×ñ´ ¥æª´¤»æ, Ìæð ßáæü Ùãè´ ãæð ÚUãè ãñÐ(2) ØçÎ ×ñ´ Ùãè´ ¥æª´¤»æ, Ìæð ßáæü ãæð ÚUãè ãñÐ(3) ØçÎ ×ñ´ Ùãè´ ¥æª´¤»æ, Ìæð ßáæü Ùãè´ ãæð ÚUãè ãñÐ(4) ØçÎ ×ñ´ ¥æª´¤»æ, Ìæð ßáæü ãæð ÚUãè ãñÐ

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ANSWER KEY PAPER - 1 JEE (MAIN) - 2014

QUESTION

NO.

06/04/2014

SET E

06/04/2014

SET F

06/04/2014

SET G

06/04/2014

Book H

09/04/2014 11/04/2014 12/04/2014 19/04/2014

1. 1 4 2 3 2 3 2 1

2. 3 4 2 4 3 3 3 1

3. 2 4 3 1 2 1 4 4

4. 1 1 3 3 3 2 2 4

5. 3 3 1 3 3 2 4 3

6. 3 1 1 2 4 4 1 2

7. 4 4 4 3 3 1 2 *

8. 1 1 4 4 3 1 3 3

9. 3 2 1 2 3 2 2 1

10. 1 2 1 1 1 1 2 4

11. 3 1 3 2 1 2 2 3

12. 4 4 3 1 1 3 2 1

13. 1 3 2 3 2 2 1 4

14. 4 3 3 1 1 1 4 3

15. 3 2 2 1 1 4 2 2

16. 3 4 1 4 3 2 2 1

17. 1 1 3 2 4 3 3 4

18. 3 4 3 1 3 3 2 3

19. 2 1 3 4 4 1 3 2

20. 3 3 4 1 1 2 2 2

21. 3 3 3 1 3 1 2 1

22. 3 1 3 1 4 3 2 3

23. 2 2 3 4 4 4 1 3

24. 2 2 2 1 2 4 3 3

25. 4 4 2 4 4 1 2 3

26. 2 4 4 3 4 3 2 2

27. 3 1 4 1 2 2 2 3

28. 1 4 4 3 1 3 4 1

29. 4 3 3 2 2 2 2 2

30. 2 1 4 1 1 4 4 2

31. 1 1 4 1 4 3 4 4

32. 2 2 4 4 1 4 1 2

33. 3 2 4 1 3 4 4 2

34. 2 1 2 4 4 4 2 3

35. 1 3 4 2 2 2 2 2

36. 1 2 2 4 3 4 3 4

37. 3 3 4 4 3 2 2 2

38. 1 2 3 1 3 1 2 1

Rohit

Typewritten Text

Rohit

Text Box

JEE Main 2014 Answer key Paper 1 Offline and Online (All Answer keys)


QUESTION

NO.

06/04/2014

SET E

06/04/2014

SET F

06/04/2014

SET G

06/04/2014

Book H

09/04/2014 11/04/2014 12/04/2014 19/04/2014

39. 2 4 2 2 4 2 3 3

40. 4 2 4 2 3 1 3 4

41. 3 1 1 2 2 3 4 3

42. 2 1 4 4 3 3 3 2

43. 2 3 2 3 2 4 4 2

44. 1 3 2 4 3 2 2 3

45. 4 2 4 4 1 4 4 2

46. 2 1 3 3 2 2 4 2

47. 2 1 1 4 2 3 1 1

48. 1 1 3 1 2 2 2 4

49. 4 4 4 4 3 3 1 4

50. 2 4 3 1 1 2 3 4

51. 2 1 1 3 4 4 1 4

52. 4 1 1 2 3 3 2 4

53. 4 1 3 4 2 1 3 2

54. 3 4 2 4 1 3 1 3

55. 1 4 4 3 2 2 4 2

56. 1 1 2 4 2 3 4 3

57. 4 4 4 3 4 3 3 4

58. 1 2 4 1 2 3 1 4

59. 1 3 3 1 3 3 4 3

60. 3 1 1 3 2 3 4 2

61. 2 4 2 3 4 3 2 4

62. 4 2 2 4 4 1 3 2

63. 3 1 2 4 2 2 3 1

64. 2 2 2 3 1 1 3 2

65. 1 2 2 1 2 4 1 4

66. 4 1 3 3 1 4 2 2

67. 2 1 4 2 3 4 2 2

68. 1 4 3 4 2 3 1 2

69. 2 2 3 4 3 2 3 4

70. 2 1 4 4 1 4 2 4

71. 2 3 2 1 2 3 3 2

72. 2 4 2 4 4 3 2 4

73. 1 2 1 1 2 1 1 3

74. 4 3 4 3 2 3 2 2

75. 2 4 1 2 2 1 4 4

76. 3 3 1 4 3 3 1 1


QUESTION

NO.

06/04/2014

SET E

06/04/2014

SET F

06/04/2014

SET G

06/04/2014

Book H

09/04/2014 11/04/2014 12/04/2014 19/04/2014

77. 3 2 1 4 2 4 4 3

78. 4 2 3 2 3 2 2 4

79. 1 3 3 2 3 2 3 2

80. 1 2 3 2 4 2 4 1

81. 2 1 4 3 4 3 4 1

82. 3 4 3 3 1 4 4 1

83. 3 4 3 3 3 3 2 2

84. 3 3 1 1 3 2 3 3

85. 2 4 2 3 3 4 2 3

86. 1 1 4 4 1 2 2 1

87. 4 2 2 3 4 2 1 2

88. 2 2 2 1 3 2 4 3

89. 2 2 3 3 2 1 3 3

90. 3 2 1 2 2 3 2 3

Note :- * denotes that the question is dropped.

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