ADVANCED PATTERN ONLINE TEST-1 (AONT-1) XI€¦ · code 0 ® advanced pattern online test-1...

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ADVANCED PATTERN ONLINE TEST-1 (AONT-1) XITARGET : JEE (MAIN+ADVANCED) 2017

SUBJECT : PHYSICSCOURSE : VIJETA (JP), VISHWAAS (JF), ANOOP (EP), AKHIL (EF)

Date (fnukad) : 05-01-2017 Time: 1 Hour (le; : 1 ?k.Vk) Max. Marks (egÙke vad) : 80

Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.

Ñi;k bu funsZ'kksa dks /;ku ls i<+saA vkidks 5 feuV fo'ks"k :i ls bl dke ds fy, fn;s x;s gSaA

INSTRUCTIONS / funsZ'k :A. General : A. lkekU; %1. This booklet is your Question Paper. Do not break

the seals of this booklet before being instructed to doso by the invigilators.

1. ;g iqfLrdk vkidk iz'u&i=k gSA bldh eqgjsa rc rd urksMsa tc rd fujh{kdksa ds }kjk bldk funsZ'k u fn;k tk;sA

2. Blank spaces and blank pages are provided in thequestion paper for your rough work. No additionalsheets will be provided for rough work.

2. dPps dke ds fy;s [kkyh i`"B vkSj [kkyh txg bliqfLrdk esa gh gSA dPps dke ds fy, dksbZ vfrfjDr dkxtugha fn;k tk;sxkA

3. Blank papers, clipboards, log tables, slide rules,calculators, cameras, cellular phones, pagers andelectronic gadgets are NOT allowed inside theexamination hall.

3. dksjs dkxt] fDyi cksMZ (CLIP BOARD)] ykWx rkfydk]LykbM:y] dSYdqysVj] dSejk] lsyQksu] istj vkSj fdlhizdkj ds bysDVªkfud midj.k ijh{kk d{k esa vuqefr ughagSA

4. Write you name and roll number in the spaceprovided on the back cover of this booklet.

4. bl iqfLrdk ds fiNys i`"B ij fn, x, LFkku esa viuk ukevkSj jksy uEcj fyf[k,A

5. Using a black ball point pen, darken the bubbleson the upper original sheet.

5. Åijh ewy i`"B ds cqycqyksa (BUBBLES) dks dkys ckWyIokbaV dye ls dkyk djsaA

6. DO NOT TAMPER WITH/MUTILATE THE ORS ORTHE BOOKLET.

6. vks-vkj-,l- (ORS) ;k bl iqfLrdk esa gsj&Qsj@foÑfr udjsaA

7. On breaking the seals of the booklet check that itcontains all the 20 questions and correspondinganswer choices are legible. Read carefully theInstructions printed at the beginning of each section.

7. bl iqfLrdk dh eqgjsa rksM+us ds i'pkr~ Ñi;k tk¡p ysa fd blesalHkh 20 iz'u vkSj muds mÙkj fodYi Bhd ls i<+s tk ldrs gSaAlHkh [kaMksa ds izkjaHk esa fn;s gq, funsZ'kksa dks /;ku ls i<+saA

B. Filling the ORSUse only Black ball point pen only for filling the ORS.

B. vks-vkj-,l (ORS) Hkjuk % ORS dks Hkjus ds fy, dsoy dkysck¡y iSu dk mi;ksx dhft,A

8. Write your Roll no. in the boxes given at the top leftcorner of your ORS with black ball point pen. Also,darken the corresponding bubbles with Black ballpoint pen only. Also fill your roll no on the back side ofyour ORS in the space provided (if the ORS is both sideprinted).

8. ORS ds lcls Åij cka;s dksus esa fn, x, ck¡Dl esa viukjksy uEcj dkys ck¡y ikbUV ls fyf[k, rFkk laxr xksys Hkhdsoy dkys isu ls Hkfj;sA ORS ds ihNs dh rjQ Hkh viukjksy uEcj fyf[k, (;fn ORS nksuksa rjQ Nih gqbZ gSA)

9. Fill your Paper Code as mentioned on the Test Paper anddarken the corresponding bubble with Black ball point pen.

9. ORS ij viuk isij dksM fyf[k, rFkk laxr xksyksa dks dkysck¡y isu ls dkys dhft,A

10. If student does not fill his/her roll no. and paper codecorrectly and properly, then his/her marks will not bedisplayed and 5 marks will be deducted (paper wise)from the total.

10. ;fn fo|kFkhZ viuk jksy uEcj rFkk isij dksM lgh vkSjmfpr rjhds ugha Hkjrk gS rc mldk ifj.kke jksd fy;ktkosxk rFkk iz'u&i=k esa izkIrkad ls 5 vad dkV fy, tkosaxsaA

(Please read the last page of this booklet for rest of the instructions)Ñi;k 'ks"k funsZ'kksa ds fy;s bl iqfLrdk ds vfUre i`"B dks i<+sA

Resonance Eduventures Ltd.CORPORATE OFFICE : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005

Ph.No. : +91-744-3012222, 6635555 | Toll Free : 258 5555Reg. Office : J-2, Jawahar Nagar, Main Road, Kota (Raj.)-324005 | Ph. No.: +91-744-3192222 | FAX No. : +91-022-39167222

Website : www.resonance.ac.in | E-mail : [email protected] | CIN: U80302RJ2007PLC024029

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Space for Rough Work (dPps dk;Z ds fy, LFkku)

Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005

Website : www.resonance.ac.in | E-mail : [email protected]®

Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029PJP/JFAONT1050117C0-1


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SECTION – 1 : (Maximum Marks : 40) This section contains TEN questions The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive For each question, darken the bubble corresponding to the correct integer in the ORS Marking scheme :

+4 If the bubble corresponding to the answer is darkened 0 If none of the bubbles is darkened–1 In all other cases

[kaM 1 : (vf/kdre vad : 40) bl [kaM esa nl iz'u gSaA izR;sd iz'u dk mÙkj 0 ls 9 rd, nksuksa 'kkfey] ds chp dk ,d ,dy vadh; iw.kk±d gSA izR;sd iz'u esa] vks- vkj- ,l- ij lgh iw.kk±d ds vuq:i cqycqys dks dkyk djsaA vadu ;kstuk :

+4 ;fn mÙkj ds vuq:i cqycqys dks dkyk fd;k tk,A 0 ;fn dksbZ Hkh cqycqyk dkyk u fd;k gksA–1 vU; lHkh voLFkkvksa esa

1. The drawing shows velocity (v) versus time (t) graphs for two cyclists moving along the same

straight segment of a highway from the same point. The first cyclist starts at t = 0 min and the

second cyclist starts moving at t = 3.0 min. Find the time at which the two cyclists will meet :

(Both velocity-time curves intersect at t = 4 min)

fp=kkuqlkj nks lkbZfdy pkydksa ds (v) osx dk le; (t) ds lkFk xzkQ fn;k x;k gS nksuksa lkbZfdy pkyd ,d

lh/kh js[kk esa leku fcUnq ls viuh xfr 'kq: djrs gSA igyk pkyd t = 0 feuV ij xfr izkjEHk esa djrk gS

rFkk nqljk pkyd t = 3.0 ij xfr izkjEHk esa djrk gSA og le; feuV esa Kkr djks tc nksuksa lkbZdhy pkyd

igyh ckj feyrs gSA(nksuksa xzkQ t = 4 feuV ij ,d nqljs dks dkVrs gSA )

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2. The graph shows the variation of1

v (where v is velocity of particle) with respect to time. Find the

instantaneous acceleration of the particle at time t = 3 sec.

çnf'kZr xzkQ esa 1

v (v d.k dk osx gS) dk le; ds lkFk ifjorZu çnf'kZr gSA t = 3 sec ij d.k dk rkR{kf.kd

Roj.k D;k gksxk

3. All surfaces are smooth. Find the acceleration (in m/s2) of sphere in the given figure. (g = 10 m/s2)

lHkh lrg fpduh gS fn;s x;s fp=k esa xksys dk Roj.k (m/s2 esa) esa Kkr djksA

¼Å/okZ/kj

nhokj

½

¼{kSfrt lrg½

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4. A block of mass m is placed on a weighing machine and it is also attached with a spring balance.

The whole system is placed in a lift as shown in figure. Lift is moving with constant acceleration

10 m/s2 in upward direction. The reading of the weighing machine is 500 N and the reading of

spring balance is 300 N. If the actual mass of the block is m kg then fill value ofm

10. (Take g = 10 m/s2)

m nzO;eku dk ,d CykWd ,d Hkkj rkSyus okyh e'khu (weighing machine) ij j[kk gS rFkk ;g ,d

fLizax&rqyk ls Hkh tqM+k gSA iwjk fudk; fp=kkuqlkj ,d fy¶V esa j[kk gqvk gSA fy¶V 10 m/s2 ds fu;r Roj.k ls

Åij dh vksj xfr dj jgh gSA Hkkj&rkSyus okyh e'khu (weighing machine) dk ikB~;kad 500 N gS rFkk

fLizax&rqyk dk ikB~;kad 300 N gSA CykWd dk okLrfod nzO;eku m ¼fdxzk- esa½ gks rks m

10dk eku Kkr dhft,A

(g = 10 m/s2 yhft,)

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5. Two towns A and B are connected by a regular bus service with a bus leaving in either direction

every T minutes with same speed. A man cycling with speed of 20km/h in the direction A to B,

notices that a bus goes past him every t1 = 18minutes in the direction of motion, and every

t2 = 6 minutes in the opposite direction. What is the period T (in min.) of the bus service? Assume

that velocity of cyclist is less than velocity of bus–

nks 'kgj A rFkk B fu;fer cl lsok }kjk tqM+s gq, gSA tgk¡ fdlh Hkh fn'kk esa çR;sd T feuV ds ckn cl

fudyrh gSA ,d vkneh lkbZfdy ij 20km/h dh pky ls A ls B dh rjQ tk jgk gSA og xfr djrs le; ;g

ns[krk gS fd cl çR;sd t1 = 18 feuV esa mldh xfr dh fn'kk ls o çR;sd t2 = 6 feuV esa mldh xfr dh

foijhr fn'kk ls xqtjrh gSA rks cl&lsok dk le;kUrjky T D;k gksxk ;fn ;g ekuk fd lkbZfdy dk osx cl

ds osx ls de gSA

6. Due to explosion under water, a gas bubble is formed which oscillates with a Time Period T.

Experimentally, it was found that T is proportional to Pa db Ec, where P is the static pressure, d is

the density and E is the total energy of explosion. Find value of a + b + c.

ikuh ds vUnj foLQksV ls cuk ,d xSl dk cqycqyk vkorZdky T ls nksyu djrk gSA iz;ksfxd :i ls ;g ik;k

x;k fd T dk eku Pa db Ec ds lekuqikrh gS] tgkW P fLFkj nkc] d ?kuRo rFkk E foLQksV dh dqy ÅtkZ gSA

a + b + c dk eku gksxk :

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7. Hail stones falling vertically with a speed of 12 3 m/s hits the wind screen which makes an angle

30º with the horizontal. If car is running at velocity v(in m/s) so that the driver find the hailstones

striking perpendicular to the wind screen. Find the value ofv

2?

12 3 m/s dh pky ls Å/okZ/kj fxj jgs vksys {kSfrt ls 30º dks.k cuk jgs ,d dkj ds vkxs okys dkWp(wind screen) ij Vdjkrs gSA ;fn dkj v(m/s) osx ls bl izdkj xfr'khy gS rkfd pkyd dks ,slk izrhr gksrk

gS fd vksys dkj ds dkWp ij yEcor~ Vdjkrs gSA v

2dk eku Kkr djksA

8. A boy and a block, both of same mass, are suspended at the same horizontal level, from each end

of a light string that moves over a frictionless pulley as shown. The boy starts moving upwards with

an acceleration 2.5 m/s2 relative to the rope. If the block is to travel a total distance 10 m before

reaching at the pulley, find time taken by the block in doing so.

?k"kZ.k jfgr f?kjuh ls ikfjr ,d gYdh jLlh ds izR;sd fljs ij] leku {kSfrt Lrj ij] ,d yM+dk ,oa ,dfi.M nksuksa leku nzO;eku ds yVds gSA yM+dk Åij dh vksj jLlh ds lkis{k 2.5 eh-/ls-2 ds Roj.k ls pyukizkjEHk djrk gSA ;fn fi.M] f?kjuh rd igqapus esa dqy nwjh 10 eh- r; djrk gS] rks fi.M dks ,slk djus esa yxkle; Kkr djksaA

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9. A particle moves along a straight line, net force acting on it is given by F ma

where m = mass of

particle a

= acceleration of particle. If particle has mass 2 gm and initially it's speed was zero thenfind final speed of particle. Find your answer in terms of v0 km/s and fill the value of v0 in answer

key. (consider given curve as half circle)

,d js[kk ds vuqfn'k xfr'khy d.k ij dk;Zjr dqy cy F ma }kjk fn;k tkrk gSA tgk¡ m = d.k dk

nzO;eku] a

= d.k dk Roj.k gSA ;fn d.k dk nzO;eku 2 gm rFkk izkjfEHkd pky 'kwU; gks rks d.k dh vfUrepky Kkr djksaA viuk mÙkj v0 km/s ds inksa esa fnft, rFkk mÙkj iqfLrdk esa v0 dk eku Hkfj;s (fn;s x;s oØdks vnZo`r ekfu;s)

10. A block of mass m is released on a smooth inclined wedge of inclination 30º and mass M. Heightof the block (from ground) varies with time as h = 1.5 – 1.5t2. (h in metre and t = time in second).

The acceleration of M is3

xm/s2. Find the value of x :

M nzO;eku rFkk 30º dks.k dh fpduh ur lrg okys ost ij m nzO;eku ds CykWd dks NksM+k tkrk gSA CykWd dhtehu ls Å¡pkbZ le; ds lkFk h = 1.5 – 1.5t2 }kjk nh tkrh gSA (h ehVj esa gS rFkk t = lsd.M esa gSA) M dk

Roj.k3

xm/s2gks rks x dk eku Kkr djks :

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SECTION – 2 : (Maximum Marks : 40) This section contains TEN questions Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these

four option(s) is(are) correct For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS Marking scheme :

+4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened 0 If none of the bubbles is darkened–1 In all other cases

[kaM 2 : (vf/kdre vad : 40) bl [kaM esa nl iz'u gSaA izR;sd iz'u esa pkj fodYi (A), (B), (C) rFkk (D) gSaA bu pkj fodYiksa esa ls ,d ;k ,d ls vf/kd fodYi lgh gSaA izR;sd iz'u esa] lHkh lgh fodYi ¼fodYiksa½ ds vuq:i cqycqys ¼cqycqykas½ dks vks- vkj- ,l- esa dkyk djsaA vadu ;kstuk :

+4 ;fn flQZ lHkh lgh fodYi ¼fodYiksa½ ds vuq:i cqycqys ¼cqycqykas½ dks dkyk fd;k tk,A 0 ;fn dksbZ Hkh cqycqyk dkyk u fd;k gksA–1 vU; lHkh voLFkkvksa esa

11. A particles of mass 25 kg, moving at 6 m/s, is acted upon by a force in the opposite direction to thevelocity. The variation of force with time is shown in the graph. Then :(A) Its speed will be 2 m/s when the force stops acting.

(B) its magnitude of average acceleration for the whole time interval in which force acts is2

5m/s2

(C) its magnitude of average acceleration for the whole time interval in which force acts is4

5m/s2

(D) Its direction of motion will be reversed atleast once during the time interval.

25 kg nzO;eku dk ,d d.k 6 m/s ds osx ls xfr'khy gSA bl ij osx dh foijhr fn'kk esa ,d cy vjksfir gSAcy rFkk le; ds e/; xzkQ fp=k esa iznf'kZr gS rks :(A) tc cy vkjksfir gksuk can gks tkrk gS rc d.k dh pky 2 m/s gksxhA(B) tc rc cy vjksfir gksrk gSA bl nkSjku d.k ds vkSlr Roj.k dk ifjek.k 2

5 m/s2gSA

(C) tc rc cy vjksfir gksrk gSA bl nkSjku d.k ds vkSlr Roj.k dk ifjek.k 4

5m/s2gSA

(D) fn;s x;s le; vUrjky esa d.k de ls de ,d ckj viuh fn'kk ifjofrZr djrk gSA

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12. The acceleration v/s time graph for the motion of a particle starting from rest from origin and

moving towards positive x-axis is shown in figure. Pick out correct option.

(A) Particle keeps on moving along positive x-axis throughout the motion

(B) Velocity of particle at point B is lesser than velocity at C.

(C) Maximum velocity of the particle is at point A.

(D) The velocities of particle at point C and D are equal.

ewy fcUnq ls ¼fojke voLFkk ls½ izkjEHk gq, rFkk /kukRed x-v{k ds vuqfn'k d.k dh xfr ds fy, Roj.k≤

xzkQ fp=k esa iznf'kZr gSA lgh fodYiksa dk p;u dhft;sa

(A) lEiw.kZ xfr ds nkSjku d.k /kukRed x-v{k ds vuqfn'k xfr djrk jgrk gSA

(B) fcUnq B ij d.k dk osx C ij osx dh rqyuk esa de gksxkA

(C) fcUnq A ij d.k dk osx vf/kdre gSA

(D) fcUnq C rFkk fcUnq D ij d.k ds osx cjkcj gSA

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13. Lines 1 and 2 represent velocity time graphs of two particles A and B having initial velocities

VA and VB respectively. Both particles move along a straight line starting from different positions

such that right side of motion is taken positive and left side is taken negative. Select correct

alternatives.

(A) Both the particles are approaching each other during motion.

(B) At the point P, both particles have same velocity as well as same position.

(C) Particle A is retarding whereas B is speeding up.

(D) Till time instant t0 (from t = 0), the displacement of particle A is more than that of B.

js[kk 1 rFkk 2 nks d.k A rFkk B ds osx le; xzkQ dks iznf'kZr djrh gS] rFkk d.kksa ds izkjfEHkd osx Øe'k%

VA rFkk VB gSA nksuksa d.k vyx&vyx fLFkfr;ksa ls izkjEHk gksrs gq, ljy js[kk ds vuqfn'k bl izdkj xfr djrs

gS] fd xfr ds nka;h rjQ /kukRed rFkk cka;h rjQ _.kkRed ysrs gSA lR; fodYiksa dk p;u dhft,A

(A) nksuksa d.k xfr ds nkSjku ,d nwljs dh vksj xfr djrs gSA

(B) fcUnq P ij nksuksa d.kksa ds osx rFkk fLFkfr leku gSaA

(C) d.k A eafnr gksrk gS tcfd B Rofjr gksrk gSA

(D) t0 le; rd (t = 0 ls) d.k A dk foLFkkiu B ds foLFkkiu ls vf/kd gksrk gSA

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14. In a situation, a board is moving with a velocity v with respect to earth, while a man A and man B

are running with a velocity 2v with respect to earth and both men are running from the opposite

ends of the board at the same time, as shown. Length of the board is L. If they meet after time T,

then

(A) value of T is L/4v

(B) value of T is L/2v

(C) Displacement of man B with respect to board in time T is 3L/4

(D) Displacement of man A with respect to board in time T is L/4

fp=k esa iznf'kZr O;oLFkk esa ,d r[rk /kjkry ds lkis{k v osx ls xfr dj jgk gSA tcfd O;fDr A rFkk O;fDr B

/kjkry ds lkis{k 2v pky ls nkSM+ jgs gS rFkk nksuksa O;fDr leku le; ij r[rs ds foifjr fljksa ls fp=kkuqlkj

nkSM+ jgs gSA r[rs dh yEckbZ L gSA ;fn os T le; i'pkr feyrs gS rc&

(A) T dk eku L/4v gksxkA

(B) T dk eku L/2v gksxkA

(C) T le; esa O;fDr B dk r[rs ds lkis{k foLFkkiu 3L/4 gksxkA

(D) T le; esa O;fDr A dk r[rs ds lkis{k foLFkkiu L/4 gksxkA

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15. A ball is thrown vertically upward from ground level with certain speed v0. Which of the following

graph(s) may be correct (for one time of flight).v – speed of ball h = height of ball from ground level.

(A) (B)

(C) (D)

,d xsan dks tehu ls v0 pky }kjk m/okZ/kj Åij dh rjQ Qsadk tkrk gSA ,d mM~M;u dky ds le; fuEu esals dkSu ls xzkQ lgh gks ldrs gSA ¼,d mM~M;u dky ds fy,½v – xsan dh pky h = tehu ls xsan }kjk izkIrdh xbZ Å¡pkbZ

(A) (B)

(C) (D)

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16. A particle is dropped from a finite height H above the ground level under gravity. Due to airresistance net acceleration of particle become a = g – kv in the direction of velocity. Wherek = positive constant & v = speed of particle. Then which of the following graph(s) is/are possible(t0 = time when particle touches the ground):

(A) (B)

(C) (D)

,d d.k dks /kjkry ls fuf'pr Å¡pkbZ H ls xq:Ro ds v/khu NksM+k tkrk gSA gok ds izfrjks/k ds dkj.k d.k dkifj.kkeh Roj.k a = g – kv osx dh fn'kk esa gS tgkW k = /kukRed fu;rkad rFkk v = d.k dh pky gS fuEu esa lsdkSuls xzkQ lEHko gSA (t0 = og le; gS] tc d.k /kjkry dks Nqrk gS)

(A) (B)

(C) (D)

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17. In the figure shown a particle of mass m is tied to a string and the other end of string is tied to theside wall of a carriage, which is accelerating horizontally with constant acceleration a. Assumingthe string remains taut and the particle is not in contact with the base of carriage. Then : (Particleis at rest with respect to carriage)

(A) The angle that string makes with the horizontal is tan–1g

a

(B) The angle that string makes with the horizontal is tan–1a

g

(C) Tension in the string is 2 2m g a

(D) Tension in the string is 2 2m g a

fp=kkuqlkj m nzO;eku dk ,d d.k ,d jLlh }kjk ca/kk gqvk gS rFkk jLlh dk nwljk fljk xkM+h dh nhokj lsca/kk gSA xkM+h {kSfrt] fu;r Roj.k a ls Rofjr gS ekuuk gS fd jLlh ges'kk ruh gqbZ jgrh gS rFkk d.k] xkM+h dsvk/kkj ls lEidZ esa ugha gS] rc: (xkM+h ds lkis{k d.k fojkekoLFkk esa gSA)

(A) jLlh }kjk] {kSfrt ls cuk;k x;k dks.k , tan– 1g

a

gSA

(B) jLlh }kjk] {kSfrt ls cuk;k x;k dks.k tan– 1a

g

gSA

(C) jLlh esa ruko 2 2m g a gSA

(D) jLlh esa ruko 2 2m g a gSA

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18. All pulleys are light & smooth, strings are light & inextensible. Consider the case when M (see fig.)

(A) acceleration of m is g upward (B) Net force on m is mg/3 upward

(C) Tension in the string connecting M is4mg

9(D) Tension in the string connecting m is 2 mg

lHkh f?kjfu;kW gYdh o fpduh gS] Mksfj;kW gYdh o vforkU; gSA Mdh fLFkfr ij fopkj fdft,A (fp=k ns[ks)

(A) m dk Roj.k Åij dh vksj g gSA (B) m ij ifj.kkeh cy mg/3 Åij dh vksj gSA

(C) M ls tqMh Mksjh es ruko 4mg

9gSA (D) m ls tqMh Mksjh esa ruko 2 mg gSA

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19. A spring connects two particles of masses m1 and m2. A horizontal force F acts on m1 . Ignore

friction, When the elongation of the spring is x and accelerations of m1 and m2 are a1 and a2

respectively then choose the correct alternative(s) –

(A) a2=2

kx

m

(B) a1 =1

F kx

m

(C) F = m1a1 + m2a2

(D) a1=a2=1 2

F

m m, at the instant of natural length of the spring

0 0 0 0 0 00 0 0 0 0 0m2 m1

a2a1

F

nzO;eku m1 rFkk m2 ds nks d.k dekuh }kjk fp=kkuqlkj ;qfXer gSA m1 ij {kSfrt cy F fØ;k'khy gSA ?k"kZ.kux.; gSA tc dekuh dk çlkj x gS rFkk m1 o m2 ds Roj.k Øe'k% a1 o a2 gSA rc lgh fodYi pqfu, –

(A) a2=2

kx

m

(B) a1 =1

F kx

m

(C) F = m1a1 + m2a2

(D) a1=a2=1 2

F

m m, dekuh dh çkdf̀rd yEckbZ ds {k.k ij

PHYSICS

PH

YS

ICS





20. In the figure shown, two blocks one of mass 5 kg and the other of mass 2 kg are connected by

light and inextensible string. Pulleys are light and frictionless. Choose the correct statement.

(A) The acceleration of 5 kg mass is5g

11m/s2 (B) The acceleration of 2 kg mass is

5g

11m/s2

(C) Tension in the string is12g

11N (D) Tension in the string is

10g

11N

5 kg

2 kg

iznf'kZr fp=k esa 5 kg rFkk 2 kg nzO;eku ds nks CykWd gYdh rFkk vfoRkkU; Mksjh ls tqM+s gq, gSA f?kjfu;k gYdh

rFkk ?k"kZ.k jfgr gSA lgh oDrO; dk p;u dhft, &

(A) 5 kg nzO;eku dk Roj.k 5g

11m/s2 gSA (B) 2 kg nzO;eku dk Roj.k 5g

11 m/s2 gSA

(C) Mksjh esa ruko 12g

11 N gSA (D) Mksjh esa ruko 10g

11N gSA

Code

0ADVANCED PATTERN ONLINE TEST-1

(AONT-1) XITARGET : JEE (MAIN+ADVANCED) 2017

DATE : 05-01-2017 | COURSE : VIJETA (JP), VISHWAAS (JF), ANOOP (EP), AKHIL (EF)11. Since it is not possible to erase and correct pen filled

bubble, you are advised to be extremely careful while

darken the bubble corresponding to your answer.

11. pawfd isu ls Hkjs x, xksys feVkuk vkSj lq/kkjuk laHko ugha gSblfy, vki lko/kkuh iwoZd vius mÙkj ds xksyksa dks HkjsaA

12. Neither try to erase / rub / scratch the option nor

make the Cross (X) mark on the option once filled.

Do not scribble, smudge, cut, tear, or wrinkle the

ORS. Do not put any stray marks or whitener

anywhere on the ORS.

12. fodYi dks u feVk,a@u LØsp djsa vkSj u gh xyr (X) fpUgdks HkjsaA ORS dks dkVs u gh QkMs u gh xUnk ugha djsa rFkkdksbZ Hkh fu'kku ;k lQsnh ORS ij ugha yxk;sA

13. If there is any discrepancy between the written data

and the bubbled data in your ORS, the bubbled data

will be taken as final.

13. ;fn ORS esa fdlh izdkj dh fy[ks x, vkadMksa rFkk xksysfd, vkadMksa esa fojks/kkHkkl gS] rks xksys fd, vkadMksa dks ghlgh ekuk tkosxkA

C. Question Paper Format

The question paper consists of one part. This part

consists are two section.

C. iz'u&i=k dk izk:ibl iz'u&i=k esa ,d Hkkx gSaA bl Hkkx esa nks [kaM gSaA

14. Section 1 contains 10 questions. The answer to

each question is a single-digit integer, ranging from

0 to 9 (both inclusive).

14. [kaM 1 esa 10 iz'u gSaA izR;sd iz'u dk mÙkj 0 ls 9 rd ¼nksuksa'kkfey½ ds chp ,d ,dy vadh; iw.kkZad gSA

15. Section 2 contains 10 multiple choice questions.

Each question has Four choices (A), (B), (C) and (D)

out of which ONE OR MORE THAN ONE is correct.

15. [kaM 2 esa 10 cgqfodYi iz'u gaSA gj iz'u esa pkj fodYi (A),

(B), (C) vkSj (D) gSa ftuesa ls ,d ;k ,d ls vf/kd fodYilgh gSaA

D. Marking Scheme

16. For each question in Section 1, you will be awarded

4 marks if you darken the bubble corresponding to

only the correct answer and zero mark if no bubbles

are darkened. In all other cases, minus one (–1) mark

will be awarded.

D. vadu ;kstuk16. [kaM 1 esa gj iz'u esa dsoy lgh mÙkj okys cqycqys(BUBBLES)

dks dkyk djus ij 4 vad vkSj dksbZ Hkh cqycwyk dkyk ughaa djus ij(0) vad iznku fd, tk;saxsA vU; lHkh fLFkfr;ksa esa _.kkRed ,d(–1) vad iznku fd;k tk;sxkA

17. For each question in Section 2, you will be awarded

4 marks if you darken the bubble corresponding to

only the correct answer and zero mark if no bubbles

are darkened. In all other cases, minus one (–1) mark

will be awarded.

17. [kaM 2 esa gj iz'u esa dsoy lgh mÙkj okys cqycqys(BUBBLES)

dks dkyk djus ij 4 vad vkSj dksbZ Hkh cqycwyk dkyk ughaa djus ij(0) vad iznku fd, tk;saxsA vU; lHkh fLFkfr;ksa esa _.kkRed ,d(–1) vad iznku fd;k tk;sxkA

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AONT-XI TEST-1TARGET : (JEE MAIN+ADVANCED) 2017

DATE : 05-01-2017 COURSE : JP,EP, JF, EF

HINTS & SOLUTIONS ¼l ad sr ,o a gy½

(PHYSICS)1. From the slopes on the graph, we can obtain the

rat io of accelerat ions of the two cycl ists:

xzkQ esa izo.krkvksa ls] ge nksuksa lkbfdy pkydksa ds Roj.kksa dkvuqikr izkIr dj ldrs gSA

2

1

v ' 0a 4 3

4v ' 0a

4

The distances travel led by the two cycl ists mustbe equal:

nksuksa lkbfdy pkydksa }kjk pyh xbZ nwfj;k¡ cjkcj gksxhA2 2

1 2a t a (t 3)

2 2

;

2

2

1

a t

a t 3

= 2 ;

t

t 3 t = 6 min.

2.1

v= – t + c

–2

1

v.dv

dt = – 1

dv

dt = v2 at t = 3 ij, v = 3

dv

dt= 3.

3. Suppose of acceleration of wedge a1 towards right and

acceleration of sphere a2 downward

ekuk ost dk Roj.k a1 nka;h rjQ gS rFkk xksys dk Roj.k a2 uhpsdh rjQ gSA

mg – N cos = ma2 ...........(i)N sin = ma1 ...........(ii)

From wedge certain c a f / kr xfr lsa2 cos = a1sin ...........(iii)

From (i),(ii) & (iii) lsa2 = g sin2= 5 m/s2 .

4. m = mass of the blockApplying Newton's Law on the block in vertical directionT + N – mg = ma ...............(1)T = 300 N (given reading of spring balance)

N = 500 N (given reading of W.M.)

a = 10 m/s2

m =T N

g a

=500 300

20

m = 40 kg

Ans. 4

gy% m = CykWd dk nzO;ekuCykWd ij Å/okZ/kj fn'kk esa U;wVu dk fu;e yxkus ijT + N – mg = ma ...............(1)

T = 300 N (fLizax rqyk dk fn;k x;k ikB~;kad)

N = 500 N (Hkkj rqyk e'khu dk fn;k x;k ikB~;kad)

a = 10 m/s2

m =T N

g a

=500 300

20

m = 40 kg

Ans. 4

5. Let velocity of bus is V. Then separation between twonearest buses is VT

vxj cl dk osx V gS rks nks utnhdh clksa ds chp dh nwjh VT

gksxhAA bus moving in same direction travels the distance VT in 18minutes to meet the cyclist.

leku fn'kk esa xqtjus okyh clsa VT nwjh 18 feuV esa r; djdslkbfdy lokj dks feyxhA




so t = rel

rel

s

v

18 VT

hrs60 (V 20)

...(1)

(Vrel = V – 20)

A bus moving in opposite direction travels the distance VT in6 minutes tomeet the cyclist.

foijhr fn'kk esa xqtjus okyh clsa VT nwjh 6 ehuV esa r; djdslkbfdy lokj dks feyxhA

So vr% t = rel

rel

S

V

6 VT

60 V 20

...(2)

(here Vrel = V + 20)

from (1) & (2)

(1) rFkk (2) ls

3 =V 20

V 20

V = 40 Km/hr.

from (1)

(1) ls18 40T

60 (40 20)

18 40

T60 20

T =9

hrs60

=9

6060 minutes

= 9 minutes

6. T Pa db Ec

T = K Pa db Ec

where k is a dimensionless constant.

tgkW k foekghu fu;rakd gSA[Mº Lº T’] = [ML–1 T–2]a [ML–3]b [ML2 T–2]c

[Mº Lº T1] = [Ma + b + c L–a – 3b + 2c T–2a – 2C]Comparing dimensions of both sides

nksuksa vksj foekvksa dh rqyuk djus ij

a =5

6

, b =

1

2 , c =

1

3

7. tan30º =v

12 3 v = 12 m/s

v

2 = 6 m/s Ans.

8. Acceleration of boy and block will be same equal to

1.25 m/s2 w.r.t. ground. Hence

yM+ds ,oa CykWd nksuksa dk tehu ds lkis{k Roj.k cjkcj ,oa1.25 eh-/ls-2 ds cjkcj gksxkA vr%

10 =1

2 (1.25) t2 t = 4 sec.

9.

v = adt =F

dtm

v =3

1

2 10F dt

=3

1

2 10 ×

(2)(2)

2

= × 103 m/s = km/s

v0 = 1 Ans.

10. ax and ay component of acceleration of m and a

acceleration of M.

ax rFkk ay, m nzO;eku ds Roj.k ds ?kVd gS rFkk a, nzO;eku M

dk Roj.k gSA

y

x

a

a a=

1

3

a + ax = 3 ay = 3 3

mg – N cos30º = mayN sin 30º = max

ay =

2

23

d h

dton solving gy djus ij

a =2

3x = 2




12. (i) Since the acceleration of particle is always positive, soparticle always moves along positive x-axis & its velocitygoes on increasing with increase in time.(ii) VB > VD > VA > VC

(i) pwafd d.k dk Roj.k lnSo /kukRed gS] vr% d.k /kukRed x–

v{k dh fn'kk esa gh xfr djrk jgsxk rFkk mldk osx le; dslkFk c<+rk tk;sxkA(ii) vr% VB > VD > VA > VC

13. Acceleration is slope of v-t curveAt point P, both particles may have different positions.Since displacement equals area under v-t graph,displacement of particle A is more than that of B.

Roj.k v-t oØ dh <ky gSAfcUnq P ij nksuks d.kksa dh fLFkfr;k¡ vyx&vyx gks ldrh gSaApq¡fd foLFkkiu v-t xzkQ ds vUrZxr leku gSA d.k A dkfoLFkkiu] d.k B ds foLFkkiu ls vf/kd gSaA

14. Velocity of A w.r.t plank = V

r[rs ds lkis{k A dk osx = V

Velocity of B w.r.t plank = 3V

r[rs ds lkis{k B dk osx = 3V

T =L

V 3V =

L

4V

15.dv

dh= –ve therough the motion xfr ds nkSjku _.kkRed gSA

a =dv

vdh

= costant fu;r dv 1

dh v

16.dv

dt= g–kv

v

0

dv

g – kv =

t

t 0

dt

–1 g – kv

lnk g

= t

g – kv = ge–ktv =–ktg

1– ek

a =–ktg

0 – e (–k)k

= g e–kt

V =g a a g

– –k k k k

V –g

k= –

a

kkv –g = –aa = g – kv= –kv + g

17. tan =g

a

T = m2 2g a

18. a =g – a

2a =

g

3

3T – mg = maT =4mg

9

& 3T =4mg

3

Net force on m ma = mg/3 upward

m ij ifj.kkeh cy ma = mg/3 Åij dh vksj gSA

19. For m2, ds fy, a2 =2

kx

m

For m1, ds fy, a1 =1

F kx

m

For (m1 + m2), ds fy, F = m1a1 + m2a2

At natural length (lkekU; yEckbZ ij) a2 = 0, a1 =1

F

m.




20.5 kg

2 kg

TT T T T

T

a1

5ga2

2g

5T – 5g = 5a1 ....(i)

2g – T = 2a2 ...(ii)

5a1 = a2 ...(iii)

by (i), (ii) & (iii) ls a1 =g

11 m/s2

a2 =5g

11 m/s2

T =12g

11 N




AONT-XI TEST-1TARGET : (JEE MAIN+ADVANCED) 2017

DATE : 05-01-2017 COURSE : JP,EP, JF, EF

ANSWER KEY

(PHYSICS)

1. (6) 2. (3) 3. (5) 4. (4) 5. (9) 6. (0) 7. (6)

8. (4) 9. (1) 10. (2) 11. (AB) 12. (A) 13. (CD) 14. (ACD)

15. (BD) 16. (AB) 17. (AC) 18. (BC) 19. (ABC) 20. (BC)

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