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Date: 03/April/2016 Time: 3 Hours. Max. Marks: 360 VERY IMPORTANT : A. The question paper consists of 3 parts (Mathematics, Physics & Chemistry). Please fill the OMR answer Sheet accordingly and carefully. B. Please ensure that the Question Paper you have received contains All the questions in each Section and Pages. If you found some mistake like missing questions or pages then contact immediately to the Invigilator. INSTRUCTIONS 1. All questions are single correct type questions. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. For each question, you will be awarded 4 marks if you have darkened only the bubble corresponding to the correct answer and zero mark if no bubble are darkened. In all other cases, minus one (–1) mark will be awarded. 2. Indicate the correct answer for each question by filling appropriate bubble in your answer sheet. 3. Use of Calculator, Log Table, Slide Rule and Mobile is not allowed. USEFUL DATA Atomic weights: Al = 27, Mg = 24, Cu = 63.5, Mn = 55, Cl = 35.5, O = 16, H = 1, P = 31, Ag = 108, N = 14, Li = 7, I = 127, Cr = 52, K=39, S = 32, Na = 23, C = 12, Br = 80, Fe = 56, Ca = 40, Zn = 65.5, Ti = 48, Ba = 137, U = 238, Co= 59, B =11, F = 19, He = 4, Ne = 20, Ar = 40 , Mo = 96, Ni = 58.5, Sr = 87.5, Hg = 200.5 , Tl = 204, Pb = 207 [Take : ln 2 = 0.69, ln 3 = 1.09, e = 1.6 × 10 –19 , m e = 9.1 × 10 –31 kg ] Take g = 10 m/s 2 unless otherwise stated H Code Paper JEE (Main) 2016

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Page 1: H JEE (Main) 2016 - Matrix JEE Academy · Matrix JEE Academy : Opposite Reliance Petrol Pump, Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999, 97836-31999 3 Matrix JEE (MAIN)

Date: 03/April/2016Time: 3 Hours. Max. Marks: 360

VERY IMPORTANT :A. The question paper consists of 3 parts (Mathematics, Physics & Chemistry). Please fill the OMR answer

Sheet accordingly and carefully.

B. Please ensure that the Question Paper you have received contains All the questions in each Section andPages. If you found some mistake like missing questions or pages then contact immediately to the Invigilator.

INSTRUCTIONS

1. All questions are single correct type questions. Each of these questions has four choices (A), (B), (C)and (D) out of which ONLY ONE is correct.For each question, you will be awarded 4 marks if you have darkened only the bubble corresponding tothe correct answer and zero mark if no bubble are darkened. In all other cases, minus one (–1) mark willbe awarded.

2. Indicate the correct answer for each question by filling appropriate bubble in your answer sheet.3. Use of Calculator, Log Table, Slide Rule and Mobile is not allowed.

USEFUL DATAAtomic weights: Al = 27, Mg = 24, Cu = 63.5, Mn = 55, Cl = 35.5, O = 16, H = 1, P = 31, Ag = 108, N = 14,Li = 7, I = 127, Cr = 52, K=39, S = 32, Na = 23, C = 12, Br = 80, Fe = 56, Ca = 40, Zn = 65.5, Ti = 48,Ba = 137, U = 238, Co= 59, B =11, F = 19, He = 4, Ne = 20, Ar = 40 , Mo = 96, Ni = 58.5, Sr = 87.5,Hg = 200.5 , Tl = 204, Pb = 207 [Take : ln 2 = 0.69, ln 3 = 1.09, e = 1.6 × 10–19, m

e= 9.1 × 10–31 kg ]

Take g = 10 m/s2 unless otherwise stated

HCode

Paper

JEE (Main) 2016

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PHYSICS

SECTION-ISINGLE CORRECT CHOICE TYPE

Q.1 to Q.30 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.

1. A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that it is a

distance 2A

3from equilibrium position. The new amplitude of the motion is :

,d d.k 'A' vk;ke ls ljy&vkorZ nksyu dj jgk gSA tc ;g vius ewy&LFkku ls 2A

3 ij igq¡prk gS rc vpkud

bldh xfr frxquh dj nh tkrh gSA rc bldk u;k vk;ke gS :

(1) A 3 (2) 7A

3(3)

A41

3(4) 3A

Sol.

2

2 2v A A

3

(1)

2

2 23v A ' A

3

(2)

Divide the two equations.

7A ' A

3

2. For a common emitter configuration, if and have their usual meanings, the uncorrect relationship

between and is :

mHk;fu"B&mRltZd foU;kl ds fy;s rFkk ds chp fuEu esa ls dkSulk laca/k xyr gS ? rFkk :

(1) 1

(2)

2

21

(3)

1 11

(4)

1

Sol.C

e

I

I

C

b

I

I

e b cI I I

e b

c c

I I1

I I

1 1

1

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3. 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 :

,d Nk=k ,d ljy&vkorZ&nksyd ds 100 vko`fr;ksa dk le; 4 ckj ekirk gS vkSj mudks 90 s, 91 s, 95 s rFkk 92

s ikrk gSA bLrseky dh xbZ ?kM+h dk U;wure vYika'k 1 s gSA rc ekis x;s ek/; le; dks mls fy[kuk pkfg;s :

(1) 92 ± 1.8 s (2) 92 ± 3 s (3) 92 ± 2 s (4) 92 ± 5.0 s

Sol.90 91 95 92

T4

= 92

1T | 90 – 92 | , 2T | 91– 92 | , 3T | 95 – 92 | , 4T | 92 – 92 |

T 1.5

Since least count of the measuring clock is 1 second therefore,

T = 92 ± 2

4. If a, b, c, d are inputs to a gate and x is the output, then, as per the following time graph, the gate is :

,d xsV esa a, b, c, d buiqV gSa vkSj x vkÅViqV gSA rc fn;s x;s Vkbe&xzkQ ds vuqlkj xsV gS :

(1) OR (2) NAND (3) NOT (4) AND

Sol. Output represents OR gate.

5. A particle of mass m is moving along the side of a square of side 'a', with a uniform speed in the x-y plane

as shown in the figure :

fp=k esa Hkqtk 'a' dk oxZ x-y ry esa gSA m nzO;eku dk ,d d.k ,dleku xfr] ls bl oxZ dh Hkqtk ij py jgk gS

tSlk fd fp=k esa n'kkZ;k x;k gS :

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Which of the following statements is false for the angular momentum L

about the origin ?

(1) R ˆL m a k2

when the particle is moving from B to C

(2) m ˆL Rk

2

when the particle is moving from D to AA

(3) m ˆL Rk

2

when the particle is moving from A to B

(4) R ˆL m a k2

when the particle is moving from C to D.

fp=k esa Hkqtk 'a' dk oxZ x-y ry esa gSA m nzO;eku dk ,d d.k ,dleku xfr] ls bl oxZ dh Hkqtk ij py jgk gS

tSlk fd fp=k esa n'kkZ;k x;k gS :

rc fuEu esa ls dkSulk dFku] bl d.k ds ewyfcUnq ds fxnZ dks.kh; vk?kw.kZ L

ds fy;s] xyr gS ?

(1) R ˆL m a k2

, tc d.k B ls C dh vksj py jgk gSA

(2) m ˆL Rk

2

, tc d.k D ls A dh vksj py jgk gSA

(3) m ˆL Rk

2

] tc d.k A ls B dh vksj py jgk gSA

(4) R ˆL m a k2

, tc d.k C ls D dh vksj py jgk gSA

Page 5: H JEE (Main) 2016 - Matrix JEE Academy · Matrix JEE Academy : Opposite Reliance Petrol Pump, Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999, 97836-31999 3 Matrix JEE (MAIN)

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Sol.m ˆL Rk

2

when particle is moving from D to A.

R ˆL m a k2

when the particle is moving from C to D.

6. Choose the correct statement :

(1) In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion

to the amplitude of the audio signal

(2) In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion

to the frequency of the audio singal

(3) In amplitude modulation the amplitude of the high frequency carrier wave is made to very in proportion

to the amplitude of the audio signal

(4) In amplitude modulation the frequency of the high frequency carrier wave is made to vary in proportion

to the amplitude of the audio signal

lgh dFku pqfu;s :

(1) vko`fr ekMqyu esa mPp vkof̀r dh okgd rjax ds vk;ke esa cnyko /ofu flXuy ds vk;ke ds vuqikrh gSA

(2) vko`fr ekMqyu esa mPp&vko`fr dh okgd rjax ds vk;ke esa cnyko /ofu flXuy dh vko`fr ds vuqikrh gSA

(3) vk;ke ekMqyu esa mPp vko`fr dh okgd rjax ds vk;ke esa cnyko /ofu flXuy ds vk;ke ds vuqikrh gSA

(4) vk;ke ekMqyu esa mPp vko`fr dh okgd rjax dh vko`fr esa cnyko /ofu flXuy ds vk;ke ds vuqikrh gSA

Sol. In amplitude modulation the amplitude of the high frequency carrier wave is made to very in proportion to

the amplitude of the audio signal

7. Radiation of wavelength , is incident on a photocell. The fastest emitted electron has speed . If the

wavelength is changed to 3

4

, the speed of the fastest emitted electron will be :

,d QksVks&lsy ij rjaxnS/;Z dk izdk'k vkifrr gSA mRlftZr bysDVªkWu dh vf/kdre xfr gSA ;fn rjaxnS/;Z

3

4

gks rc mRlftZr bysDVªkWu dh vf/kdre xfr gksxh &

(1)

1

24

3

(2)

1

23

4

(3)

1

24

3

(4)

1

24

3

Sol. 2hc 1mv

2

(1)

24hc 1

m v '3 2

(2)

Multiplying equation (1) by 4/3

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24 hc 1 4

m v3 3 2 3

..... (3)

From (2) and (3)

2 21 4 1m v m(v')

2 3 3 2

Hence v'

1

24v

3

8. 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 B

B are the values of magnetic field at the centres

of the circle and square respectively, then the ratio A

B

B

B is :

nks ,dleku rkj A o B izR;sd dh yEckbZ 'l', esa leku /kkjk I izokfgr gSA A dks eksM+dj R f=kT;k dk ,d o`r vkSj B

dks eksM+dj Hkqtk 'a' dk ,d oxZ cuk;k tkrk gSA ;fn BA rFkk B

B Øe'k% o`r ds dsUnz rFkk oxZ ds dsUnz ij pqEcdh;

{ks=k gSa] rc vuqikr A

B

B

B gksxk &

(1) 2

16

(2)

2

8 2

(3)

2

8

(4)

2

16 2

Sol.

0A

iB

2R

where R = L/2

0B

iB 4 sin 45 sin 45

a4

2

where a = L/4

2A

B

B

B 8 2

9. 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 :

nksuksa fljksa ij [kqys ,d ikbi dh ok;q esa ewy&vko`fr f gSA ikbi dks Å/okZ/kj mldh vk/kh&yEckbZ rd ikuh esa Mqck;k

tkrk gSA rc blesa cps ok;q&dkye dh ewy vko`fr gksxhA

(1) 2f (2) f (3) f

2(4)

3f

4

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Sol.

For the pipe open at both the ends, 0

vf

2L

For the pipe closed at one end and half length, 0

vf '

L4

2

0 0f ' f

10. The region between two concentric spheres of radii 'a' and 'b', respectively (see figure), has volume charge

density A

r , 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 :

f=kT;k 'a' rFkk 'b' ds nks ,d&dsUnzh xksyksa ds (fp=k nsf[k;s) chp ds LFkku esa vk;ru vkos'k&?kuRo A

r gS] tgk¡ A

fLFkjkad gS rFkk r dsUnz ls nwjh gSA xksyksa ds dsUnz ij ,d fcUnq&vkos'k Q gSA A dk og eku crk;sa ftlls xksyksa ds chp

ds LFkku esa ,dleku oS|qr&{ks=k gks &

(1) 2 2

2Q

a b (2) 2

2Q

a(3)

2

Q

2 a(4) 2 2

Q

2 b a

Sol. Electric field in the region between the 2 spheres

r

2

a

2

AK Q 4 r dr

rE

r

2 2

2

r aK Q 4 A

2

r

For E to be independent of r, 24 Aa

Q 02

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2

QA

2 a

11. 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 :

,d vkdZ ySEi dks izdkf'kr djus ds fy;s 80 V ij 10 A dh fn"V /kkjk (DC) dh vko';drk gksrh gSA mlh vkdZ

dh 220 V (rms), 50 Hz izR;korhZ /kkjk (AC) ls pykus ds fy, Js.kh esa yxus okys izsjdRo dk eku gS &

(1) 0.044 H (2) 0.065 H (3) 80 H (4) 0.08 H

Sol.80

R 810

22

220i 10

R L

where = 2 × 50

L = 0.065 H

12. '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

'n' eksy vkn'kZ xSl ,d izØe A B ls xqtjrh gS (fp=k nsf[k;s) A bl izØe ds nkSjku mldk vf/kdre rkieku

gksxk &

(1) 0 09P V

2nR(2)

0 09P V

nR(3)

0 09P V

4nR(4)

0 03P V

2nR

Sol. For the given P–V graph,

00

0

PP V 3P

V

0 20

0

PPV 1T 3P V V

nR V nR

For maximum temperature, dT

0dV

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Tmax

= 0 09P V

4nR

13. 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:

,d HkkjksÙkksyd Hkkj dks igys Åij vkSj fQj uhps rd ykrk gSA ;g ekuk tkrk gS fd flQZ Hkkj dks Åij ys tkus esa

dk;Z gksrk gS vkSj uhps ykus esa fLFkfrt ÅtkZ dk àkl gksrk gSA 'kjhj dh olk ÅtkZ nsrh gS tks ;kaf=kdh; ÅtkZ esa

cnyrh gSA eku ys fd olk }kjk nh xbZ ÅtkZ 3.8 × 107 J izfr kg Hkkj gS] rFkk bldk ek=k 20% ;kaf=kdh; ÅtkZ esa

cnyrk gSA vc ;fn ,d HkkjksÙkksyd 10 kg ds Hkkj dks 1000 ckj 1 m dh Å¡pkbZ rd Åij vkSj uhps djrk gS rc

mlds 'kjhj ls olk dk {k; gS % (g = 9.8 ms–2 ysa)

(1) 9.89 × 10–3 kg (2) 12.89 × 10–3 kg (3) 2.45 × 10–3 kg (4) 6.45 × 10–3 kg

Sol. Work done = 1000 × mgh = 9.8 × 104 J

Let x kg of fat is used.

x × 3.8 ×107 × 20

100= 9.8 × 104 J

x = 12.89 × 10–3 kg

14. A point particle of mass m, moves along the uniformly rough track PQR as shown in the figure. The coeffi-

cient of friction, between the particle and the rough track equals . 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 fricition and the distance x(= QR), are, respectively close to:

m nzO;eku dk ,d fcanq d.k ,d [kqjnjsa iFk PQR (fp=k nsf[k;s) ij py jgk gSA d.k vkSj iFk ds chp ?k"kZ.k xq.kkad

gSA d.k P ls NksM+s tkus ds ckn R ij igq¡p dj :d tkrk gSA iFk ds Hkkx PQ vkSj QR ij pyus esa d.k }kjk

[kpZ dh xbZ ÅtkZ,¡ cjkcj gSA PQ ls QR ij gksus okys fn'kk cnyko es dksbZ ÅtkZ [kpZ ugha gksrhA

rc vkSj nwjh x(= QR) ds eku yxHkx gSa Øe'k %

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(1) 0.29 and 3.5 m (2) 0.29 and 6.5 m (3) 0.2 and 6.5 m (4) 0.2 and 3.5 m

(1) 0.29 vkSj 3.5 m (2) 0.29 vkSj 6.5 m (3) 0.2 vkSj 6.5 m (4) 0.2 vkSj 3.5 m

Sol. Total energy lost from P to R = mg × 2

2 mg 2

mg cos30sin 30 2

mg 2mg x

2

(where x is distance travelled from Q to R)

0.29 & x 3.5m

15. 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, exponential decrease for Si

(2) Linear decrease for Cu, linear decrease for Si

(3) Linear increase for Cu, linear increase for Si

(4) Linear increase for Cu, exponential increase for Si

rk¡ck rFkk vekfnr (indoped) flfydku ds izfrjks/kksa dh muds rkieku ij fuHkZjrk 300-400 K rkieku varjky esa]

ds fy;s lgh dFku gS &

(1) rk¡ck ds fy;s js[kh; c<+ko rFkk flfydku ds fy;s pj?kkrkadh ?kVkoA

(2) rk¡ck ds fy;s js[kh; ?kVko rFkk flfydku ds fy;s js[kh; ?kVkoA

(3) rk¡ck ds fy;s js[kh; c<+ko rFkk flfydku ds fy;s js[kh; c<+koA

(4) rk¡ck ds fy;s js[kh; c<+ko rFkk flfydku ds fy;s pj?kkrkadh c<+koA

Sol. For metals,

0R R 1 T

Hence, for copper resistance increases linearly.

For silicon, since it is a semiconductor its resistance decreases exponentially

16. Arrange the following electromagnetic radiations per quantum in the order of increasing energy :

A : Blue light B : Yellow light C : X-ray D : Radiowave

fuEu izfr DokaVe oS|qr&pqEcdh; fofdj.kksa dks mudh ÅtkZ ds c<+rs gq, Øe esa yxk;sa &

A : uhyk izdk'k B : ihyk izdk'k C : X-fdj.ksa D :jsfM;ks rjax

(1) C, A, B, D (2) B, A, D, C (3) D, B, A, C (4) A, B, D, C

Sol. As decreases energy increases.

Increases order of energy will be radiowave < yellow light <blue light < X-ray

17. A galvanometer having a coil resistance of 100 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

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scale deflection for a current of 10 A, is :

,d xSYosuksehVj ds dkby dk izfrjks/k 100 gSA 1 mA /kkjk izokfgr djus ij blesa Qqy&Ldsy fo{ksi feyrk gSA

bl xSYosuksehVj dks 10 A ds ,ehVj esa cnyus ds fy;s tks izfrjks/k yxkuk gksxk og gS &

(1) 0.1 (2) 3 (3) 0.01 (4) 2

Sol.

Vg × R

g = I

s × R

s

3s1 10 A 100 9.999A R

Rs = 0.01

18. Half-lives of two radioactive elements A and B are 20 minutes and 40 mintutes, respectively. Initially, the

samples have equal number of nuclei. After 80 minutes the ratio of decayed numbers of A and B nuclei will

be

nks jsfM;ks/kehZ rRo A rFkk B dh v)Zvk;q Øe'k% 20 feuV rFkk 40 feuV gSA izkjaHk esa nksuksa ds uewuksa esa ukfHkdksa dh

la[;k cjkcj gSA 80 feuV ds mijkar A rFkk B ds {k; gq, ukfHkdksa dh la[;k dk vuqikr gksxk &

(1) 1 : 4 (2) 5 : 4 (3) 1 : 16 (4) 4 : 1

Sol. Number of half lives of A in 80 minutes = 4

Number of half lives of B in 80 minutes = 2

NN

decayed nuclei of A 516Ndecayed nucleiof B 4N4

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19. Identify the semiconductor devices whose characteristics are given below, in the order (a), (b), (c), (d) :

fp=k (a), (b), (c), (d) ns[kdj fu/kkZfjr djsa fd ;s fp=k Øe'k% fdu lsehdUMDVj fMokbZl ds vfHky{kf.kd xzkQ gSa \

(1) Solar cell, Light dependent resistance, Zener diole, Simple diode

(2) Zener diode, Solar cell, Simple diode, Light dependent resistance

(3) Simple diode, Zener diode, Solar cell, Light dependent resistance

(4) Zener diode, Simple diode, Light dependent resistance, Solar cell

(1) lksyj lsy] LDR (ykbZV fMisUMsUV jsftLVsUl), thuj Mk;ksM] lk/kkj.k Mk;ksM

(2) thuj Mk;ksM] lksyj lsy] lk/kkj.k Mk;ksM] LDR (ykbZV fMisUMsUV jsftLVsUl)

(3) lk/kkj.k Mk;ksM] thuj Mk;ksM] lksyj lsy] LDR (ykbZV fMisUMsUV jsftLVsUl)

(4) thuj Mk;ksM] lk/kkj.k Mk;ksM] LDR (ykbZV fMisUMsUV jsftLVsUl), lksyj lsy

Sol.

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(a) Simple diode(b) Zener diode(c) Solar cell

(d) Light dependent resistance

20. A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, dur to a

point charge Q (having a charge equal to the sum of the chargs on the 4 F and 9 F capacitors), at a point

distant 30 m from it, would equal :

la/kkfj=kksa ls cus ,d ifjiFk dks fp=k esa fn[kk;k x;k gSA ,d fcUnq&vkos'k Q (ftldk eku 4 F rFkk 9 F okys la/

kkfj=kksa ds dqy vkos'kksa ds cjkcj gSa) ds }kjk 30 m nwjh ij oS|qr&{ks=k dk ifjek.k gksxk &

(1) 420 N/C (2) 480 N/C (3) 240 N/C (4) 360 N/C

Sol. Charge on 4F capacitor = 24 C

Charge on 9F capacitor = 18 CSo total charge Q = 24 + 18 = 42 C

2

KQE

r , r = 30 m

= 420 N/C

21. 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)

i`Foh dh lrg ls 'h' Å¡pkbZ ij ,d mixzg or̀kdkj iFk ij pDdj dkV jgk gS (i`Foh dh f=kT;k R rFkk h < < R)A

i`Foh ds xq:Ro {ks=k ls iyk;u djus ds fy;s bldh d{kh; xfr esa vko';d U;wure cnyko gS : (ok;qeaMyh; izHkko dks

ux.; yhft,)

(1) gR / 2 (2) gR 2 1 (3) 2gR (4) gR

Sol.2

2

GMm mv

R R

GMv gR

R

escapev 2gR

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v gR 2 1

22. 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 ?

,d Ldw&xst dk fip 0.5 mm gS vkSj mlds o`rh; Ldsy ij 50 Hkkx gSA blds }kjk ,d iryh vY;qehfu;e 'khV dh

eksVkbZ ekih xbZA eki ysus ds iwoZ ;g ik;k x;k fd tc LØw&xst ds nks tkWoksa dks lEidZ esa yk;k tkrk gS rc 45ok

aHkkx eq[; Ldsy ykbZu ds laikrh gksrk gS vkSj eq[; Ldsy dk ikB;kad ;fn 0.5 mm rFkk 25 oka Hkkx eq[; Ldsy ds

laikrh gks] rks 'khV dh eksVkbZ D;k gksxh \

(1) 0.70 mm (2) 0.50 mm (3) 0.75 mm (4) 0.80 mm

Sol. Least count = 0.5

mm 0.01mm50

It has negative error = 5 × 0.01 = 0.05 mmReading = (0.5 + 25 × 0.01) + 0.05

0.80 mm

23. 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 paralel

to CD in the direction shwon. As it moves the roller will tend to :

nks 'kadq dks muds 'kh"kZ O ij tksM+dj ,d jksyj cuk;k x;k gS vkSj mls AB o CD jsy ij vlefer j[kk x;k gS

(fp=k nsf[k;s)A jksyj dk v{k CD ls yEcor gS vkSj O nksuksa jsy ds chpkschp gSA gYds ls /kdsyus ij jksyj jsy ij

bl izdkj yq<+duk vkjEHk djrk gS fd O dk pkyu CD ds lekarj gS (fp=k nsf[k;s)A pkfyr gks tkus ds ckn ;g

jksyj %

(1) go straight (2) turn left and right alternately

(3) turn left (4) turn right

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(1) lh/kk pyrk jgsxk (2) ck;sa rFkk nk;sa Øe'k% eqM+rk jgsxk

(3) ck¡;h vksj eqM+sxk (4) nk¡;h vksj eqM+sxk

Sol. The roller will trun left.

24. Hysteresis loops for two magnetic materials A and B are given below :

nks pqEcdh; inkFkZ A rFkk B ds fy;s fgLVsjsfll ywi uhps fn[kk;s x;s gS &

These materials are used to make magnets for electric generators, transformer core and electromagnet core.

Then it is proper to use :

bu inkFkksZa dk pqEcdh; mi;ksx fo|qr&tsusjsVj ds pqEcd] VªkUlQkWeZj dh ØksM ,oa fo|qr&pqEcd dh ØksM vkfn ds

cukus esa fd;k tkrk gSA rc ;g mfpr gS fd &

(1) A for transformers and B for electric generators

(2) B for electromagnets and transformers

(3) A for electric generators and transformers

(4) A for electromagnets and B for electric generators

(1) A dk bLrseky VªkUlQkWeZj esa rFkk B fo|qr&tsusjsVj esa fd;k tk,A

(2) B dk bLrseky fo|qr&pqEcd rFkk VªkUlQkWeZj nksuksa esa fd;k tk,A

(3) A dk bLrseky fo|qr&tsusjsVj rFkk VªkUlQkWeZj nksuksa esa fd;k tk,A

(4) A dl bLrseky fo|qr&pqEcd esa rFkk B dk fo|qr&tsusjsVj esa fd;k tk,A

Sol. For electromagnets and transformers, core should be quickly magnetized and demagnetized, therefore

B should be used.

25. The box of a pin hole camera, of length L, has a hole of radius a. It is assumed that when the hole is illumi-

nated by a parallel beam of light of wavelength 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 :

,d fiu gkWy dSejk dh yEckbZ L gS rFkk fNNz dh f=kT;k a gSA ml ij rjaxnS/;Z dk lekarj izdk'k vkifrr gSA

fNnz ds lkeus okyh lrg ij cus LikWV dk foLrkj fNnz ds T;kferh; vkdkj rFkk foorZu ds dkj.k gq, foLrkj dk

dqy ;ksx gSA bl LikWV dk U;wure vkdkj bmin

rc gksxk tc &

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(1) a L and minb 4 L (2)

2

aL

and

minb 4 L

(3) 2

aL

and

2

min

2b

L

(4) a L and 2

min

2b

L

(1) a L rFkk minb 4 L (2)

2

aL

rFkk

minb 4 L

(3) 2

aL

rFkk

2

min

2b

L

(4) a L rFkk 2

min

2b

L

Sol.

26. 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)

20 m yEckbZ dh ,dleku Mksjh dks ,d n`<+ vk/kkj ls yVdk;k x;k gSA blds fupys fljs esa ,d lw{e rjax&Lian

pkfyr gksrk gSA Åij vk/kkj rd igq¡pus esa yxus okyk le; gS &

(Take g = 10 ms–2)

(1) 2 2 s (2) 2 s (3) 2 2 s (4) 2 s

Sol. Tension at a distance x from lower end,

T= (x) g { is mass per unit length}

Tv xg

dxxg

dt

20 t

0 0

dxg dt

x

t 2 2

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27. An ideal gas undergoes a quasi static, revesible 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) :

,d vkn'kZ xSl mRØe.kh; LFkSfrd&dYi izØe ls xqtjrh gS rFkk mldh eksyj&Å"ek&/kkfjrk C fLFkj jgrh gSA ;fn

bl izØe esa mlds nkc P o vk;ru V ds chp laca/k PVn = constant gSA (CP rFkk C

V Øe'k% fLFkj nkc o fLFkj

vk;ru ij Å"ek /kkfjrk gS) rc n ds fy;s lehdj.k gS &

(1) P

V

C Cn

C C

(2) V

P

C Cn

C C

(3) P

V

Cn

C (4)

P

V

C Cn

C C

Sol. V

RC C

n 1

Find n & substitute R = CP – C

V

28. 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) 20 times taller (2) 20 times nearer

(3) 10 times taller (4) 10 times nearer

nwj fLFkr 10 m Å¡ps isM+ dks ,d 20 vko/kZu {kerk okys VsfyLdksi ls ns[kus ij D;k eglwl gksxk \

(1) isM+ 20 xquk Å¡pk gS (2) isM+ 20 xquk Å¡pk gS

(3) isM+ 10 xquk Å¡pk gS (4) isM+ 10 xquk ikl gS

Sol.I

o

h / x'm

h / x

Ih20

10

29. In an experiment for determination of refractive index of glass of a prism by i – , plot, it was found that a

ray incident at angle 35°, suffers a deviation of 40° and that it emerges at angle 79°. In that case which of

the following is closest to the maximum possible value of the refractive indes ?

,d iz;ksx djds rFkk i – xzkQ cukdj ,d dk¡p ls cus fizTe dk vuorZukad fudkyk tkrk gSA tc ,d fdj.k dks

35° ij vkifrr djus ij og 40° ls fopfyr gksrh gS rFkk ;g 79° ij fuxZe gksrh gSA bl fLFkfr esa fuEu esa ls

dkSulk eku vioZukad ds vf/kdre eku ds lcls ikl gS \

(1) 1.7 (2) 1.8 (3) 1.5 (4) 1.6

Sol.

minAsin

2

Asin

2

(1)

40 i e A A = 74°

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If we put min

= 40° (actually min

will definitely be less than 40°) & A = 74° then 1.4

min 40

1.4

30. A pendulum clock loses 12 s a day if the temperature is 40°C and gains 4 s a day if the temperature is

20°C. The temperature at which the clock will show correct time, and the co-efficient of linear expansion

() of the metal of the pendulum shaft are respectively :

,d isUMqye ?kM+h 40°C rkieku ij 12 s izfrfnu /kheh gks tkrh gS rFkk 20°C rkieku ij 4 s izfrfnu rst gks tkrh

gSA rkieku ftl ij ;g lgh le; n'kkZ;sxh rFkk isUMqye dh /kkrq dk js[kh; & izlkj xq.kkad () Øe'k% gS &

(1) 30°C; = 1.85 × 10–3/°C (2) 55°C; = 1.85 × 10–2/°C

(3) 25°C; = 1.85 × 10–5/°C (4) 60°C; = 1.85 × 10–4/°C

Sol.L

T 2g

T 1 L 1

T 2 L 2

1T T

2

1

12 T 40 T2

(1)

1

4 T T 202

(2)

Divide (1) by (2)T = 25°C = 1.85 × 10–5/°C

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1 2 3 4 5 6 7 8 9 10

2 2,4 3 1 2,4 3 3 2 2 3

11 12 13 14 15 16 17 18 19 20

2 3 2 1 1 3 3 2 3 1

21 22 23 24 25 26 27 28 29 30

2 4 3 2 1 1 4 1 3 3

31 32 33 34 35 36 37 38 39 40

4 1 4 4 2 3 4 1 2 1

41 42 43 44 45 46 47 48 49 50

4 4 4 2 2 4 2 2 2 4

51 52 53 54 55 56 57 58 59 60

1 1 3 2 2 4 2 4 1 1

61 62 63 64 65 66 67 68 69 70

1 1 2 2 1 2 1 4 1 2

71 72 73 74 75 76 77 78 79 80

4 3 1 1 2 4 4 4 Bonus 3

81 82 83 84 85 86 87 88 89 90

4 3 4 2 1 3 4 4 2 2

MATRIXANSWER-KEY JEE (Main) 2016 Code-H

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Date: 03/April/2016Time: 3 Hours. Max. Marks: 360

VERY IMPORTANT :A. The question paper consists of 3 parts (Mathematics, Physics & Chemistry). Please fill the OMR answer

Sheet accordingly and carefully.

B. Please ensure that the Question Paper you have received contains All the questions in each Section andPages. If you found some mistake like missing questions or pages then contact immediately to the Invigilator.

INSTRUCTIONS

1. All questions are single correct type questions. Each of these questions has four choices (A), (B), (C)and (D) out of which ONLY ONE is correct.For each question, you will be awarded 4 marks if you have darkened only the bubble corresponding tothe correct answer and zero mark if no bubble are darkened. In all other cases, minus one (–1) mark willbe awarded.

2. Indicate the correct answer for each question by filling appropriate bubble in your answer sheet.3. Use of Calculator, Log Table, Slide Rule and Mobile is not allowed.

USEFUL DATAAtomic weights: Al = 27, Mg = 24, Cu = 63.5, Mn = 55, Cl = 35.5, O = 16, H = 1, P = 31, Ag = 108, N = 14,Li = 7, I = 127, Cr = 52, K=39, S = 32, Na = 23, C = 12, Br = 80, Fe = 56, Ca = 40, Zn = 65.5, Ti = 48,Ba = 137, U = 238, Co= 59, B =11, F = 19, He = 4, Ne = 20, Ar = 40 , Mo = 96, Ni = 58.5, Sr = 87.5,Hg = 200.5 , Tl = 204, Pb = 207 [Take : ln 2 = 0.69, ln 3 = 1.09, e = 1.6 × 10–19, m

e= 9.1 × 10–31 kg ]

Take g = 10 m/s2 unless otherwise stated

ACode

Paper

JEE (Main) 2016

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SECTION-ISINGLE CORRECT CHOICE TYPE

Q.1 to Q.30 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.

31. The area (in sq. units) of the region {(x, y} : 2y 2x and x2 + y2 4x, x 0, y is :

{ks=k {(x, y} : 2y 2x rFkk x2 + y2 4x, x 0, y dk {ks=kQy (oxZ bdkbZ;ksa esa) gS :

(1) 4 2

3 (2)

2 2

2 3

(3)

4

3 (4*)

8

3

Sol.

Area = 2

0

1Area of circle 2xdx

4

8

3

32. If 1

f (x) 2f 3xx

, x 0 and S = {x R : f (x) = f(–x)}; then S :

(1*) contains exactly two elements (2) contains more than two elements

(3) is an empty set (4) contains exactly one element

;fn 1

f (x) 2f 3xx

, x 0 gS] rFkk S = {x R : f (x) = f(–x)} gS; rks S :

(1*) esa rF;r% nks vo;o gS (2) esa nks ls vf/kd vo;o gSaA

(3) ,d fjDr leqPp; gSA (4) esa dsoy ,d vo;o gSA

Sol. f(x) + 2f(1/x) = 3x (1)

f(1/x) + 2f(x) = 3

x(2)

From (1) & (2)

f(x) = 22 x

x

f(x) = f(–x)x2 = 2

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x= ± 2

S contains exactly two elements

33. The integral

12 9

35 3

2x 5xdx

x x 1

is equal to :

lekdy

12 9

35 3

2x 5xdx

x x 1

cjkcj gS :

(1)

5

25 3

xC

2 x x 1

(2)

10

25 3

xC

2 x x 1

(3)

5

25 3

xC

x x 1

(4*)

10

25 3

xC

2 x x 1

(where c is an arbitary constant)

(tgk¡ c ,d LosPN vpj gSA)

Sol.

12 9

315 2 5

2x 5x dx

x 1 x x

2 51 x x t

3

dtI

t

= 2

1C

2t

= 10

5 3 2

xC

2(x x 1)

34. For x R, f(x) = |log2–sinx| and g(x) = f(f(x)), then :

(1) g'(0) = – cos (log2) (2) g is differentiable at x = 0 and g'(0) = – sin(log 2)

(3) g is not differentiable at x = 0 (4*) g'(0) = cos(log 2)

x R ds fy, , f(x) = |log2–sinx| rFkk g(x) = f(f(x)) gSa, rks :

(1) g'(0) = – cos (log2) (2) x = 0 ij g vodyuh; gS rFkk g'(0) = – sin(log 2) gSA

(3) x = 0 ij g vodyuh; ugha gSA (4*) g'(0) = cos(log 2)

Sol. g(x) = |log2 – sin| log2 – sinx||

gx near by of x = 0g(x) = log2 – sin (log2 – sinx)g'(x) = – (cos(cos2–sinx)) (–cosx)

g'(0) = cos(log2)

35. The centres of those circles which touch the circle, x2 + y2 – 8x – 8y – 4 = 0, externally and also touch the

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x-axis, lie on :

(1) a hyperbola (2*) a parabola (3) a circle (4) an ellipse which is not a circle

mu o`Ùkksa ds dsUnz] tks o`Ùk x2 + y2 – 8x – 8y – 4 = 0 dks cká :i ls Li'kZ djrs gSa rFkk x-v{k dks Hkh Li'kZ djrs gSa]

fLFkr gSa :

(1) ,d vfrijoy; ij (2*) ,d ijoy; ij (3) ,d o`Ùk ij (4) ,d nh?kZo`Ùk ij tks o`Ùk ugha gSA

Sol.

C(4,4)

6r

Pr

6 y = -6

Distance of P from (4, 4) = Distance of P from y = –6

locus of P is parabola

36. The sum of all real values of x satisfying the equation :

2x 4x 602x 5x 5 1

is :

x ds mu lHkh okLrfod ekuksa dk ;ksx tks lehdj.k

2x 4x 602x 5x 5 1

dks lUrq"V djrs gSa] gS :

(1) 6 (2) 5 (3*) 3 (4) –4

Sol. x2 – 5x + 5 = 1

x = 1, 4x2 + 4x – 60 = 0x = –10, x = 6x2 – 5x + 5 = –1x = 2 x = 3 (Rejected)Sum = 1 + 4 + 6 – 10 +2

= 3

37. 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 :

;fn ,d vpjsrj lekUrj Js<+h dk 2nd , 5th rFkk 9th in ,d xq.kksÙkj Js<+h esa gSa] rks ml xq.kksÙkj Js<+h dk lkoZ vuqikr

gS :

(1) 1 (2) 7/4 (3) 8/5 (4*) 4/3

Sol. T2 = a

T5 = a + 3d

T9 = a + 7d

(a + 3d)2 = a(a + 7d)

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MATHS

ad

9

r = a 3d 3d

1a a

r = 3

19

=4

3

38. The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conju-

gate axis is equal to half of the distance between its foci, is :

ml vfrijoy;] ftlds ukfHkyEc dh yEckbZ 8 gS rFkk ftlds la;qXeh v{k dh yEckbZ mldh ukfHk;ksa ds chp dh nwjh

dh vk/kh gS] dh mRdsUnzrk gS :

(1*) 2

3(2) 3 (3)

4

3(4)

4

3

Sol. 2b = ae4b2 = a2e2

4a2(e2 – 1) = a2e2

e = 2

3

39. If the number of terms in the expansion of

n

2

2 41 , x 0

x x

, is 28, then the sum of the coefficients of

all the terms in this expansion, is :

;fn n

2

2 41 , x 0

x x

ds izlkj esa inksa dh la[;k 28 gS, rks bl izlkj esa vkus okys lHkh inksa ds xq.kkadksa dk

;ksx gS :

(1) 243 (2*) 729 (3) 64 (4) 2187

Sol. no of terms = n+2C2 = 28

n = 6Sum of coeff. = (1 – 2 +4)6

= 36 = 729

40. The Boolean Expression p ~ q q (~ p q) is equivalent to :

cwys ds O;atd (Boolean Expression) p ~ q q (~ p q) dk lerqY; gS :

(1*) p q (2) p ~ q (3) ~ p q (4) p q

Sol. (p ~q) (~p q) q

(~(p q)) (~(q p)) q~((p q) (q p)) q

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~ (p q) q((~p) q) q

Now make truth table

41. Consider 1 1 sin x

f (x) tan , x 0,1 sin x 2

. A normal to y = f(x) at x

6

also passes through the

point :

1 1 sin xf (x) tan , x 0,

1 sin x 2

ij fopkj dhft,A y = f(x) ds fcUnq x

6

ij [khapk x;k vfHkyEc

fuEu fcUnq ls Hkh gksdj tkrk gS :

(1) ,06

(2) ,0

4

(3) (0, 0) (4*)

20,

3

Sol. f(x) = 1 1 sin x

tancos x

1 xtan tan

4 2

xf (x)

4 2

1f '(x)

2

Eq. of normal y 2 x3 6

22x y

3

42.

1/n

2nn

(n 1)(n 2)....3nlim

n

is equal to :

1/n

2nn

(n 1)(n 2)....3nlim

n

cjkcj gS :

(1) 2

9

e(2) 3 log 3 – 2 (3) 4

18

e(4*) 2

27

e

Sol.

12n n

r 1

n rL

n

2n

r 1

1 rn L n 1

n n

l l

2

0

n 1 x dx l

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ln L = 3ln 3 – 2

2

27L

e

43. 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 :

;fn lehdj.k x2 + y2 – 4x + 6y – 12 = 0 }kjk iznÙk ,d o`Ùk dk ,d O;kl ,d vU; o`Ùk S, ftldk dsUnz

(–3, 2) gS, dh thok gS] rks o`Ùk S dh f=kT;k gS :

(1) 5 (2) 10 (3) 5 2 (4*) 5 3

Sol.

r1

(-2,3) d (-3,2)

r

r1 = 5

d = 50

2 21r r d

75 5 3

44. Let two fair six-faced dice A and B be thrown simultaneously. If E1 is the event that die A shows up four E

2

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

3 are independent (2*) E

1, E

2 and E

3 are independent

(3) E1 and E

2 are independent (4) E

2 and E

3 are independent

ekuk nks vufHkur N% Qydh; ikls A rFkk B ,d lkFk mNkys x;sA ekuk ?kVuk E1 ikls A ij pkj vkuk n'kkZrh gS]

?kVuk E2 ikls B ij 2 vkuk n'kkZrh gS rFkk ?kVuk E

3 nksuksa iklksa ij vkus okyh la[;kvksa dk ;ksx fo"ke n'kkZrh gS] rks

fuEu esa ls dkSu&lk dFku lR; ugha gS ?

(1) E1 rFkk E

3 Lora=k gSA (2*) E

1, E

2 rFkk E

3 Lora=k gSA

(3) E1 rFkk E

2 Lora=k gSA (4) E

2 rFkk E

3 Lora=k gSA

Sol. {(E1 E

2 E

3) = 0

and P(E1) = 1/6 P(E

2) = 1/6

P(E3) = 1/2

So E1, E

2 & E

3 are not independent

45. A value of for which 2 3isin

1 2isin

is purely imaginary, is :

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dk og ,d eku ftlds fy, 2 3isin

1 2isin

iw.kZr% dkYifud gS] gS :

(1) 1 3

sin4

(2*) 1 1

sin3

(3)

3

(4)

6

Sol.2 3isin

z1 2isin

z z 0 sin2 = 1/3

sin = 1

3

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

2 2 2 2

23 2 1 41 2 3 4 4 ......,

5 5 5 5

is 16

5m, then m is equal to :

;fn Js.kh 2 2 2 2

23 2 1 41 2 3 4 4 ......,

5 5 5 5

ds izFke nl inksa dk ;ksx 16

5m gS, rks m cjkcj gS :

(1) 100 (2) 99 (3) 102 (4*) 101

Sol.

211

r 2

16m 4r

5 5

112

r 2

16m 16 16r 505

5 25 25

m = 101

47. The system of linear equations

x + y – z = 0

y – y – z = 0

x + y – z = 0

has a non-trivial solution for :

(1) exactly two values of (2*) exactly three values of

(3) infinitely many values of (4) exactly one value of

jSf[kd lehdj.k fudk;

x + y – z = 0

y – y – z = 0

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MATHS

x + y – z = 0

dk ,d vrqPN gy gksus ds fy, :

(1) ds rF;r% nks eku gSA (2*) ds rF;r% rhu eku gSA

(3) ds vura eku gSA (4) dk rF;r% ,d eku gSA

Sol.

1 11 1 0

1 1

3 – = 0

= 0, = 1, = –1

48. If the line, x 3 y 2 z 4

2 1 3

lies in the plane, lx + my – z = 9, then l2 + m2 is equal to :

;fn js[kk x 3 y 2 z 4

2 1 3

, lery lx + my – z = 9 esa fLFkr gS, rks l2 + m2 cjkcj gS :

(1) 5 (2*) 2 (3) 26 (4) 18

Sol. (3, –2, –4) will lie on plane

3l – 2m + 4 = 6 (1)(2, –1, 3) is perpendicular to plane2l – m – 3 = 02l – m = 3 (2)

l = 1, m = –1

49. 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 :

'kCn SMALL ds v{kjksa dk iz;ksx djds] ik¡p v{kjksa okys lHkh 'kCnksa (vFkZiw.kZ vFkok vFkZghu) dks 'kCnksdks'k ds

Øekuqlkj j[kus ij] 'kCn SMALL dk LFkku gS :

(1) 52nd (2*) 58th (3) 46th (4) 59th

Sol. 4

A LLMS 122

L ALMS 4 24

4

M ALLS 122

3

SA LLM 32

SL ALM 3 6

SMALL 1

58

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MATHS

50. If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true ?

;fn la[;kvksa 2, 3, a rFkk 11 dk ekud fopyu 3.5 gS, rks fuEu esa ls dkSu&lk lR; gS ?

(1) 3a2 – 34a + 91 = 0 (2) 3a2 – 23a + 44 = 0 (3) 3a2 – 26a + 55 = 0 (4*) 3a2 – 32a + 84 = 0

Sol. 2 = (3.5)2 = 12.25

22 21

x xn

2

2 2 2 21 2 3 a 1112.25 2 3 a 11

4 4

3a2 – 32a + 84 = 0

51. 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 :

2 bdkbZ yEch ,d rkj dks nks Hkkxksa esa dkV dj mUgsa Øe'k% x bdkbZ Hkqtk okys oxZ rFkk r bdkbZ f=kT;k okys o`Ùk ds

:i esa eksM+k tkrk gSA ;fn cuk;s x;s oxZ rFkk o`Ùk ds {ks=kQyksa dk ;ksx U;wure gS] rks :

(1*) x = 2r (2) 2x = r (3) 2x = ( + 4)r (4) (4 –)x =r

Sol.4x 2-4x

2r = (2 – 4x)

2

2 2 4xA x

2

dA0

dx

2

2x 2 4x

2 2 r2x

x = 2r

52. Let 1

2 2x

x 0p lim 1 tan x

then log p is equal to :

ekuk 1

2 2x

x 0p lim 1 tan x

gS] rks log cjkcj gS :

(1*) 1

2(2)

1

4(3) 2 (4) 1

Sol. 1 form

211 tan x 1

2xP e

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1

2P e

log P = 1

2

53. 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 :

ekuk ijoy; y2 = 8x dk P dk ,slk fcUnq gS tks o`Ùk x2 + (y + 6)2 = 1, ds dsUnz C ls U;wure nwjh ij gS] rks ml o`Ùk

dk lehdj.k C ls gksdj tkrk gS rFkk ftldk dsUnz P ij gS] gS :

(1) 2 2 x

x y 2y 24 04

(2) 2 2x y 4x 9y 18 0

(3*) 2 2x y 4x 8y 12 0 (4) 2 2x y x 4y 12 0

Sol. P(2t2, 4t)

Normal at P passes through (0, –6)

tx + y = 4t + 2t3

–6 = 4t + 2t3

t3 + 2t + 3 = 0

t = –1

P(2, –4)

54. 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

f2

is equal to :

;fn ,d oØ y = f(x) fcUnq (1, –1) ls gksdj tkrk gS rFkk vody lehdj.k y(1 + xy)dx = x dy dks lUrq"V djrk

gS, rks 1

f2

cjkcj gS :

(1) 2/5 (2*) 4/5 (3) 2

5 (4)

4

5

Sol. y dx + xy2dx = xdy

ydx – xdy = – xy2dx

2

ydx xdyxdx

y

2x xC

y 2

2x 1(1, 1)

2

2

2xy

x 1

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55. Let a, b

and c

be three unit vectors such that 3a b c b c

2

. If b

is not parallel to c

, then the

angle between a

and b

is :

ekuk a, b

rFkk c

rhu ,sls ek=kd lfn'k gSa fd 3a b c b c

2

gSA ;fn b

, c

ds lekUrj ugha gS, rks

a

rFkk b

ds chp dk dks.k gS :

(1) 2

3

(2*)

5

6

(3)

3

4

(4)

2

Sol. 3 3a.c b a.b c b c

2 2

3a.b

2

3cos

2

5

6

56. If 5a b

A3 2

and A adj A = A AAT, then 5a + b is equal to :

;fn 5a b

A3 2

rFkk A adj A = A AAT gSa, rks 5a + b cjkcj gS :

(1) 4 (2) 13 (3) –1 (4*) 5

Sol. A adj A = AAT

adj A = AT

2 b 5a 33 5a b 2

2 = 5a, b = 3

57. 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 30º. 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 60º.

Then the time taken (in minutes) by him, from B to reach the pillar, is :

,d O;fDr ,d Å/okZ/kj [kaHks dh vksj ,d lh/ks iFk ij ,d leku pky ls tk jgk gSa jkLrs ij ,d fcUnq A ls og

[kaHks ds f'k[kj dk mUu;u dks.k 30º ekirk gSA A ls mlh fn'kk esa 10 feuV vkSj pyus ds ckn fcUnq B ls og [kaHks ds

f'k[kj dk mUu;u dks.k 60º ikrk gS] rks B ls [kaHks rd igqapus esa mls yxus okyk le; (feuVksa esa) gS :

(1) 20 (2*) 5 (3) 6 (4) 10

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Sol.

O B A

h60° 30°

OA = hcot30°

OB = hcot60°

AB h cot 30 h cot 60V

10 10

=h

5 3

OB h cot 60time 5

V h / 5 3

58. The distance of the point (1, –5, 9) from the plane x – y + z = 5 measured along the line x = y = z is :

fcUnq (1, –5, 9) dh lery x – y + z = 5 ls og nwjh tks js[kk x = y = z dh fn'kk esa ekih xbZ gS] gS :

(1) 10

3(2)

20

3(3) 3 10 (4*) 10 3

Sol.

P(1,-5,9)

xd

Q

sin (1, 1,1) (1,1,1)

(1 1 1) 1

3 3 3

1 5 9 5 10d

3 3

xsin = d

10x 3 10 3

3

59. 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 ?

;fn ,d leprqHkqZt dh nks Hkqtk,¡] js[kkvksa x – y + 1 = 0 rFkk 7x – y – 5 = 0 dh fn'kk esa gSa rFkk blds fod.kZ fcUnq

(–1, –2) ij izfrPNsn djrs gSa] rks bl leprqHkqZt dk fuEu esa ls dkSu&lk 'kh"kz gS ?

(1*) 1 8

,3 3

(2)

10 7,

3 3

(3) (–3, –9) (4) (–3, –8)

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Sol.

P

D C

BA

AB = 7x – y – 5 = 0AD = x – y + 1 = 0A(+1, 2)P(–1, – 2)

C(–3, –6)

BC = x – y – 3 = 0

CD = 7x – y + 15 = 0

1 8

B ,3 3

60. If x < 2, then the number of real values of x, which satisfy the equation

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

;fn x < 2gS, rks x ds mu okLrfod ekuksa dh la[;k tks lehdj.k

cosx + cos2x + cos3x + cos4x = 0 dks lUrq"V djrs gSa, gS :

(1*) 7 (2) 9 (3) 3 (4) 5

Sol. cosx + cos2x + cos3x + cos4x = 0

5x x2cos cos x cos 0

2 2

5xcos 0

2

3 7 9x , , , ,

5 5 5 5

cosx = 0

3x ,

2 2

xcos 0

2

x =

Total = 7 solution

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Date: 03/April/2016Time: 3 Hours. Max. Marks: 360

VERY IMPORTANT :A. The question paper consists of 3 parts (Mathematics, Physics & Chemistry). Please fill the OMR answer

Sheet accordingly and carefully.

B. Please ensure that the Question Paper you have received contains All the questions in each Section andPages. If you found some mistake like missing questions or pages then contact immediately to the Invigilator.

INSTRUCTIONS

1. All questions are single correct type questions. Each of these questions has four choices (A), (B), (C)and (D) out of which ONLY ONE is correct.For each question, you will be awarded 4 marks if you have darkened only the bubble corresponding tothe correct answer and zero mark if no bubble are darkened. In all other cases, minus one (–1) mark willbe awarded.

2. Indicate the correct answer for each question by filling appropriate bubble in your answer sheet.3. Use of Calculator, Log Table, Slide Rule and Mobile is not allowed.

USEFUL DATAAtomic weights: Al = 27, Mg = 24, Cu = 63.5, Mn = 55, Cl = 35.5, O = 16, H = 1, P = 31, Ag = 108, N = 14,Li = 7, I = 127, Cr = 52, K=39, S = 32, Na = 23, C = 12, Br = 80, Fe = 56, Ca = 40, Zn = 65.5, Ti = 48,Ba = 137, U = 238, Co= 59, B =11, F = 19, He = 4, Ne = 20, Ar = 40 , Mo = 96, Ni = 58.5, Sr = 87.5,Hg = 200.5 , Tl = 204, Pb = 207 [Take : ln 2 = 0.69, ln 3 = 1.09, e = 1.6 × 10–19, m

e= 9.1 × 10–31 kg ]

Take g = 10 m/s2 unless otherwise stated

HCode

Paper

JEE (Main) 2016

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CODE : H CHEMISTRY

SECTION-ISINGLE CORRECT CHOICE TYPE

Q.1 to Q.30 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.

61. Two closed bulbs of equal volume (V) containing an ideal gas initially at pressure pi and temperature T

1 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 p

f is :

leku vk;ru (V) ds nks can cYc] ftuesa ,d vkn'kZ xSl izkjfEHkd nkc pi rFkk rki T

1 ij Hkjh xbZ gS] ,d ux.;

vk;ru dh iryh V~;wc ls tqM+s gSa tSlk fd uhps ds fp=k esa fn[kk;k x;k gSA fQj buesa ls ,d cYc dk rki c<+kdj

T2 dj fn;k tkrk gSA vfUre nkc p

f gS :

(1) 2

i

1 2

T2p

T T

(2) 1 2i

1 2

T T2p

T T

(3) 1 2

i

1 2

T Tp

T T

(4) 1

i

1 2

T2p

T T

Sol. After raising temperature of one bulb

initial mole = final mole

i f f

1 1 2

p 2V p V p V

RT RT RT

if

1 1 2

2p 1 1p

T T T

1 2

f i

1 1 2

2T Tp p

T T T

2f i

1 2

Tp 2p

T T

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CODE : H PHYSICS

62. Which one of the following statements about water is FALSE ?

(1) There is extensive intramolecular hydrogen bonding in the condensed phase

(2) Ice formed by heavy water sinks in normal water

(3) Water is oxidized to oxygen during photosynthesis

(4) Water can act both as an acid and as a base

ty ds lEcU/k esa fuEu dFkuksa esa ls dkSu lk ,d xyr gS ?

(1) blds la?kfur izkoLFkk esa foLrh.kZ var% v.kqd gkbMªkstu vkcU/k gksrs gSaA

(2) Hkkjh ty }kjk cuk cQZ lkekU; ty esa Mwcrk gSA

(3) izdk'k la'ys"k.k esa ty vkWDlhÑr gksdj vkWDlhtu nsrk gSA

(4) ty] vEy rFkk {kkjd nksuksa gh :i esa dk;Z dj ldrk gS

Sol. There is extensive intermolecular not intramolecular hydrogen bonding in the condensed phase of water

63. In the Hofmann bromamide degradation reaction, the number of moles of NaOH and Br2 used per mole of

amine produced are :

(1) Two moles of NaOH and two moles of Br2

(2) Four moles of NaOH and one mole of Br2

(3) One mole of NaOH and one moles of Br2

(4) Four moles of NaOH and two moles of Br2

gkQeku czksekekbM fuEuhdj.k vfHkfØ;k esa] NaOH rFkk Br2 ds iz;qDr eksyksa dh la[;k izfreksy vehu ds cuus esa

gksxh :

(1) nks eksy NaOH rFkk nks eksy Br2

(2) pkj eksy NaOH rFkk ,d eksy Br2

(3) ,d eksy NaOH rFkk ,d eksy Br2

(4) pkj eksy NaOH rFkk nks eksy Br2

Sol.

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CODE : H PHYSICS

64.(PT)Which of the following atoms has the highest first ionization energy ?

fuEu ijek.kqvksa esa fdldh izFke vk;uu ÅtkZ mPpre gS ?

(1) K (2) Sc (3) Rb (4) Na

Sol. In general first ionization energy of d-block elements lies between s and p block elements

65. 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) Nitrate (2) Iron (3) Fluoride (4) Lead

Hkwfexr >hy ls izkIr ty izfrn'kZ esa ¶yksjkbM] ysM] ukbVªsV rFkk vk;ju dh lkUnzrk Øe'k% 1000 ppb, 40 ppb, 100

ppm rFkk 0.2 ppm ikbZ xbZA ;g ty fuEu esa ls fdldh mPp lkUnzrk ls ihus ;ksX; ugha gS :

(1) ukbVªsV (2) vk;ju (3) ¶yksjkbM (4) ysM

Sol. Ions Permissible limits

Iron 0.2 ppm

Fluoride 1.5 ppm

Lead 50 ppb

Nitrate 50 ppm

Hence the concentration of nitrate in a given water sample exceed the permissible limit.

66. 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 :

dkcZu rFkk dkcZu eksuksDlkbM dh ngu Å"ek;sa Øe'k% –393.5 rFkk –283.5 kJ mol–1 gSaA dkcZu eksuksDlkbM dh

laHkou Å"ek (kJ esa) izfr eksy gksxh :

(1) –676.5 (2) –110.5 (3) 110.5 (4) 676.5

Sol. C(s) + O2(g) CO

2(g) H

1 = –393.5 kJ mol

Co(g) + ½O2(g) CO

2(g) H

2 = –393.5 kJ mol

C(s) + ½O2(g) CO(g) H

f (Co, g)

Hf (Co, g) = H

1 – H

2

= –393.5 –(–283.5)

= 110 kJ

67. The equilibrium constant at 298 K for reaction A + B C + D is 100. If initial concentration of all the

four species were 1 M each, then equilibrium concentration of D (in mol L–1) will be :

rkieku 298 K ij] ,d vfHkfØ;k A + B C + D ds fy, lkE; fLFkjkad 100 gSA ;fn izkjfEHkd lkUnzrk lHkh

pkjksa Lih'kht esa ls izR;sd dh 1 M gksrh] rks D dh lkE; lkUnzrk (mol L–1 esa) gksxh :

(1) 1.818 (2) 1.182 (3) 0.182 (4) 0.818

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CODE : H PHYSICS

Sol. A + B C + D

t = 0 1 1 1 1 Q = 1 < Keq

t = teq

1–x 1 –x 1+x 1+x

(1 x)(1 x)100

(1 x)(1 x)

1 x10 10 10x 1 x

1 x

9x 0.818

11

68. The absolute configuration of fn;s x;s ;kSfxd dk fujis{k foU;kl gS %

(1) (2S, 3S) (2) (2R, 3R) (3) (2R, 3S) (4) (2S, 3R)

Sol.

69. For a linear plot of log (x/m) versus log p in a Freundlich adsorption isotherm, which of the following state-

ments is correct ? (k and n are constants)

(1) Only 1/n appears as the slope (2) log (1/n) appears as the intercept

(3) Both k and 1/n appear in the slope term (4) 1/n appears as the intercept

ÝkW;UMfyd vf/k'kks"k.k lerkih oØ esa log (x/m) rFkk log p ds chp [khaps x;s js[kh; IykV ds fy, fuEu esa ls dkSu

lk dFku lgh gS ? (k rFkk n fLFkjkad)

(1) ek=k 1/n Lyksi ds :i esa vkrk gSA (2) log (1/n) bUVjlsIV ds :i esa vkrk gSA

(3) k rFkk 1/n nksuksa gh Lyksi in esa vkrs gSaA (4) 1/n bUVjlsIV ds :i vkrk gSA

Sol.1/nx

kpm

x 1log log k log p

m n

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CODE : H PHYSICS

1slope

n

log k = intercept

70. The distillation technique most suited for separating glycerol from spent-lye in the soap industry is :

(1) Steam distillation (2) Distillation under reduced pressure

(3) Simple distillation (4) Fractional distillation

lkcqu m|ksx esa HkqDr'ks"k ykbZ (LisUV ykbZ) ls fXyljkWy i`Fkd djus ds fy, lcls mi;qDr vklou fof/k gS:

(1) ok"i vklou (2) lekuhr nkc ij vklou

(3) lkekU; vklou (4) izHkkth vklou

Sol. Distillation under reduced pressure.

Glycerol is separated from spent lye by distillation under reduced pressure, as for simple distillation very

high temperature is required which might decompose the component.

71. Which of the following is an anionic detergent ?

(1) Cetyltrimethyl ammonium bromide (2) Glyceryl oleate

(3) Sodium stearate (4) Sodium lauryl sulphate

fueu esa ls dkSu lk ,ukbfud fMVjtsaV gS ?

(1) lsfVyVªkbesfFky veksfu;e czksekbM (2) fXylfjy vksfy,V

(3) lksfM;e LVhvjsV (4) lksfM;e ykfjy lYQsV

Sol. Glyceryl oleate

Sodium sterate – C17

H35

COO–Na+

Sodium lauryl sulphate – CH3[CH

2]

11SO

4–Na+

Cetyltrimethyl ammonium bromide – CH3(CH

2)

15N+(CH

3)

3Br–

72. The species in which the N atom is in a state of sp hybridization is :

og Lih'kht] ftlesa N ijek.kq sp ladj.k dh voLFkk esa gS] gksxh :

(1) 3NO (2) NO2

(3) 2NO (4) 2NO

Sol.

73. Thiol group is present in :

Fkk;ksy xzqi ftlesa mifLFkr gS] og gS :

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(1) Cysteine (2) Methionine (3) Cytosine (4) Cystine

Sol.

74. Which one of the following ores is best concentrated by froth floatation method ?

ÝkWFk ¶yksVs'ku fof/k }kjk fuEu esa ls og dkSu lk v;Ld lokZf/kd :i ls lkfUnzr fd;k tk ldrk gS ?

(1) Galena (2) Malachite (3) Magnetite (4) Siderite

Sol. PbS = Galena is a sulphide ore hence froth floatation method is best suitable for its concentration.

75. Which of the following statements about low density polythene is FALSE ?

(1) Its synthesis requires dioxygen or a peroxide initiator as a catalyst

(2) It is used in the manufacture of buckets, dust-bins etc

(3) Its synthesis requires high pressure

(4) It is a poor conductor of electricity

fuEu ?kuRo ds ikyhFkhu ds lEcU/k esa fuEu esa ls dkSu lk dFku xyr gS ?

(1) buesa MkbZvkDlhtu vFkok ijvkWDlkbM buhfl;sVj (izkjEHkd) mRizsjd ds :i esa pkfg,A

(2) ;g cdsV (ckYVh), MLV&fcu] vkfn ds mRiknu esa iz;qDr gksrh gSA

(3) blds la'ys"k.k esa mPp nkc dh vko';drk gksrh gSA

(4) ;g fo|qr dk ghu pkyd gSA

Sol. High density polythene is used for buckets and dustbins

76. Which of the following compounds is metallic and ferromagnetic ?

fuEu esa ls dkSu lk ;kSfxd /kkfRod rFkk QsjkseSxusfVd (ykSg pqEcdh;) gS ?

(1) VO2

(2) MnO2

(3) TiO2

(4) CrO2

Sol. CrO2 fact

77. The product of the reaction given below is :

uhps nh xbZ vfHkfØ;k ds fy, mRikn gksxk &

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CODE : H

(1) (2) (3) (4)

Sol. 22 3 2 3H O CO OH H CO

78. The hottest region of Bunsen flame shown in the figure below is :

uhps nh xbZ fQxj esa cqUlu ¶yse dk lokZf/kd xeZ Hkkx gS %

(1) region 3 (2) region 4 (3) region 1 (4) region 2

Sol.

79. At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume

for complete conbustion. After combustion the gases occupy 330 mL. Assuming that the water formed is in

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CODE : H

liquid form and the volumes were measured at the same temperature and pressure, the formula of the

hydrocarbon is :

300 K rFkk 1 atm nkc ij, 15 mL xSlh; gkbMªksdkcZu ds iw.kZ ngu ds fy;s 375 mL ok;q ftlesa vk;ru ds vk/kkj

ij 20% O2 gS] dh vko';drk gksrh gSA ngu ds ckn xSlsa 330 mL ?ksjrh gSA ;g ekurs gq, fd cuk gqvk ty nzo

:i esa gS rFkk mlh rkieku ,oa nkc ij vk;ruksa dh eki dh xbZ gS rks gkbMªksdkcZu dk QkewZyk gS :

(1) C4H

8(2) C

4H

10(3) C

3H

6(4) C

3H

8

Sol.

80. The pair in which phosphorous atoms have a formal oxidation state of +3 is :

(1) Orthophosphorous and hypophosphoric acids

(2) Pyrophosphorous and pyrophosphoric acids

(3) Orthophosphorous and pyrophophorous acids

(4) Pyrophosphorous and hypophosphoric acids

og ;qXe ftuesa QkLQksjl ijek.kqvksa dh QkeZy vkWDlhdj.k voLFkk +3 gS] gS :

(1) vkFkksZQkLQksjl rFkk gkbiksQkLQksfjd ,flM

(2) ik;jksQkLQksjl rFkk ik;jksQkLQksfjd ,flM

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(3) vkFkksZQkLQksjl rFkk ik;jksQkLQksjl ,flM

(4) ik;jksQkLQksjl rFkk gkbiksQkLQksfjd ,flM

Sol. Ortho phosphorous acid = H3PO

3

Pyrophosphorous acid (H4P

2O

5)

81. The reaction of propene with HOCl (Cl2 + H

2O) proceeds through the intermediate :

izksihu dh HOCl (Cl2 + H

2O) ds lkFk vfHkfØ;k ftl e/;orhZ ls gksdj lEiUu gksrh gS] og gS :

(1) CH3 – CH(OH)–CH

2+ (2) CH

3 – CHCl–CH

2+

(3) CH3–CH+–CH

2–OH (4) CH

3 – CH+ – CH

2– Cl

Sol. 2 2Cl H O HOCl HO Cl

Cl

3 2 3 2CH CH CH CH CH CH Cl

82. 2-chloro-2-methylpentane on reaction with sodium methoxide in methanol yields :

esFksukWy esa 2-Dyksjks-2-esfFkyisUVsu] lksfM;e esFkkDlkbM ds lkFk vfHkfØ;k djds nsrh gS :

(a) (b)

(c)

(1) c only (2) a & b (3) All of these (4) a & c

(1) ek=k c (2) a rFkk b (3) buesa ls lHkh (4) a rFkk c

Sol.

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CODE : H

(3º)

83. Which one of the following complexes shows optiocal isomerism ?

fuEu esa ls dkSu lk dkWEIysDl izdkf'kd leko;ork iznf'kZr djsxk \

(1) trans[Co(en)2Cl

2]Cl (2) [Co(NH

3)

4Cl

2]Cl

(3) [Co(NH3)

3Cl

3] (4) cis[Co(en)

2Cl

2]Cl

(en = enthlenediamine)

Sol.

is optically active due to absence of plane of symmetry and centre of symmetry

84. The main oxides formed on combustion of Li, Na and K in excess of air are, respectively :

(1) Li2O

2, Na

2O

2 and KO

2(2) LI

2O, Na

2O

2 and KO

2

(3) Li2O, Na

2O and KO

2(4) LiO

2, Na

2O

2 and K

2O

gok ds vkf/kD; esa Li, Na rFkk K ds ngu ij cuus okyh eq[; vkWDlkbMsa Øe'k% gS :

(1) Li2O

2, Na

2O

2 rFkk KO

2(2) LI

2O, Na

2O

2 rFkk KO

2

(3) Li2O, Na

2O rFkk KO

2(4) LiO

2, Na

2O

2 rFkk K

2O

Sol. 2 24Li O 2Li O

2 2 22Na O Na O

2 2K O KO

85. 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 :

18 g Xyqdksl (C6H

12O

6) dks 178.2 g ikuh esa feyk;k tkrk gSA bl tyh; foy;u ds fy, ty dk ok"i nkc

(torr esa) gksxk :

(1*) 752.4 (2) 759.0 (3) 4.6 (4) 76.0

Sol.s

s

pº p n

p N

s

s

760 p 18 /180

178.2p18

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CODE : H

s

s

760 p 1 1

p 10 9.9

99 × 760 – 99 ps = p

s

s

99 760p

100

= 752.4

86. The reaction of zinc with dilute and concentrated nitric acid, respectively, produces :

(1) NO and N2O (2) NO

2 and N

2O (3) N

2O and NO

2(4) NO

2 and NO

ruq rFkk lkUnz ukbfVªd ,flM ds lkFk ftad dh vfHkfØ;k }kjk Øe'k% mRiUu gksrs gS :

(1) NO rFkk N2O (2) NO

2 rFkk N

2O (3) N

2O rFkk NO

2(4) NO

2 rFkk NO

Sol. Zn + 4HNO3(conc.) Zn(NO

3)

2 + 2NO

2 + 2H

2O

4Zn + 10HNO3(dilute) 4Zn(NO

3)

2 + 5H

2O + N

2O

87. Decomposition of H2O

2 follows a first order reaction. In fifty minutes the concentration of H

2O

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) 2.66 L min–1 at STP (2) 1.34 × 10–2 mol min–1

(3) 6.93 × 10–2 mol min–1 (4) 6.93 × 10–4 mol min–1

H2O

2 dk fo?kVu ,d izFke dksfV dh vfHkfØ;k gSA ipkl feuV esa bl izdkj ds fo?kVu esa H

2O

2 dh lkUnzrk ?kVdj

0.5 ls 0.125 M gks tkrh gSA tc H2O

2 dh lkUnzrk 0.05 M igqaprh gS, rks O

2 ds cuus dh nj gksxh :

(1) 2.66 L min–1 STP ij (2) 1.34 × 10–2 mol min–1

(3) 6.93 × 10–2 mol min–1 (4) 6.93 × 10–4 mol min–1

Sol.

1

2

0.5

t

0.25

t

0.125

2t1 = 50

t1 = 25 = t

1/2

25 = ln2/k

0.693k

25

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CODE : H

2 2 2 2

1H O H O O

2

2 2 2H O O

2 2

r rr k H O

1 1/ 2

2O 2 2

1r k H O

2

2O

1 0.693r 0.05

2 25

2

3Or 0.693 10

2

4Or 6.93 10

88. The pair having the same magnetic moment is :

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

(1) [Mn(H2O)

6]2+ and [Cr(H

2O)

6]2+ (2) [CoCl

4]2– and [Fe(H

2O)

6]2+

(3) [Cr(H2O)

6]2+ and [CoCl

4]2– (4) [Cr(H

2O)

6]2+ and [Fe(H

2O)

6]2+

,dgh pqEcdh; vk?kw.kZ dk ;qXe gS :

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

(1) [Mn(H2O)

6]2+ rFkk [Cr(H

2O)

6]2+ (2) [CoCl

4]2– rFkk [Fe(H

2O)

6]2+

(3) [Cr(H2O)

6]2+ rFkk [CoCl

4]2– (4) [Cr(H

2O)

6]2+ rFkk [Fe(H

2O)

6]2+

Sol. [CoCl4]2– has 3 unpaired electron

Co2+ 3d7 4sº

[Fe(H2O)

6]2+ has 4 unpaired electron

[Cr(H2O)

6]2+ has 4 unpaired electron

Cr2+ d4sº

Hence [Cr(H2O)

6]2+ and [Fe(H

2O)

6]2+

has same magnetic moment

89. Galvanization is applying a coating of :

xSYoukbts'ku fuEu esa ls fdlds dksV ls gksrk gS %

(1) Cu (2) Zn (3) Pb (4) Cr

Sol. Galvanization is the process of applying a protective zinc coating to prevent rusting.

90. A stream of electrons from a heated filament was passed between two charged plates kept at a potential

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difference V esu. If e and m are charge and mass of an electron, respectively, then the value of h/ (where

is wavelength associated with electron wave) is given by :

,d xeZ fQykesaV ls fudyh bysDVªkWu /kkjk dks V esu ds foHkokUrj ij j[ks nks vkosf'kr IysVksa ds chp ls Hkstk tkrk

gSA ;fn bysDVªkWu ds vkos'k rFkk lagfr Øe'k% e rFkk m gksa rks h/ dk eku fuEu esa ls fdlds }kjk fn;k tk;sxk ?

(tc bysDVªkWu rjax ls lEcfU/kr rjaxnS/;Z gS) :

(1) meV (2) 2meV (3) meV (4) 2meV

Sol.h

p

p 2m K.E.

p 2meV