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TESTPATTERN2015-16 | 1

COURSE NAME : VIJAY (JR) (E-LPD) TEST : PT-1 (JEE ADV. PATTERN) TEST DATE : 16-08-2015 Syllabus: Rectilinear Motion+ Projectile Motion + Relative Motion + Geometrical Optics

TEST LEVEL : PHYSICS

S.No. Subject Nature of Questions No. of Questions Marks Negative Total

1 to 22 MCQ 22 4 �2 88

23 to 40 Integer type (Single Digit) 18 4 �1 72

40 160

(P5) Pattern-5 (2 HRS. EACH SUBJECT)

Total Total

Maths/ Physics/

Chemistry

SECTION � 1 : (Maximum Marks : 88)

This section contains 22 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 �2 In all other cases

[kaM 1 : (vf/kdre vad : 88) bl [kaM esa 22 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 �2 vU; lHkh voLFkkvksa esa

http://www.resonance.ac.in

mailto:[email protected]




MCQ 1. Two particles are fired from the same point O. T1 and T2 are their time of flight. Choose the correct

options (shaded portion is a step of height h above the ground) :

nks d.k leku fcUnq O ls Qsds tkrs gSA T1 rFkk T2 buds mM~M;u dky gSA lgh dFku NkVks (nka;h rjQ

Nka;kfdr Hkkx tehu ls h Å WpkbZ dh lh<h gS) :

(A*) T2 < T1 for small values of h

(B*) V2 > V1

(C*) If height h tends to a limiting value of zero then (T1 � T2) will tend to a limiting value of zero

(D*) If h tends to 2 2

1 1V sin2g

or h tends to

2 22 2V sin

2g

, then in both cases 2T2 will tends to T1

(A*) h ds NksVs eku ds fy, T2 < T1 gSA

(B*) V2 > V1

(C*) ;fn h dk eku 'kwU; dh rjQ vxzlj gks rks (T1 � T2) dk eku Hkh 'kwU; dh rjQ vxzlj gksxkA

(D*) ;fn h dk eku 2 2

1 1V sin2g

;k

2 22 2V sin

2g

dh rjQ vxzlj gks rks nksuks fLFkfr;ks esa 2T2 dk eku T1 dh

rjQ vxzlj gksxkA

Sol. H1 = H2

V1sin1 = V2sin2

but (ii) reaches little above ground,

ijUrq (ii) /kjkry ls Å ij fxjsxh

T2 < T1

If ;fn h = 0

T1 = T2

h = H1 = H2






T2 = 12

T1 .

2. A particle is projected up an incline (inclination angle = 30º) with 15 3 m/s at an angle of 30º with

the incline (as shown in figure) (g = 10 m/s2)

,d d.k urry (urdks.k = 30º) ij Å ij dh vksj 15 3 m/s dh pky ls urry ls 30º dks.k ij iz{ksfir

fd;k tkrk gSA n'kkZ;s vuqlkj (g = 10 m/s2)

(A*) 1.5 sec later, angle between acceleration & velocity is 120º.

(B) 1.5 sec later, angle between acceleration & velocity is 60º.

(C*) Range on the incline is 45 m

(D*) Time of flight on the incline is 3 sec.

(A*) 1.5 lsd.M ckn Roj.k o osx ds e/; dks.k 120º gSA

(B) 1.5 lsd.M ckn Roj.k o osx ds e/; dks.k 60º gSA

(C*) urry ij ijkl 45 m gSA

(D*) urry ij mM~M;u dky 3 lsd.M gSA

Sol. For motion to incline 0 = ( u sin 30º) t � 12

(g cos 30º)t2

t = 2usin30º

gcos30º = 3 sec = Time of flight

at 1.5 sec, velocity is parallel to incline

angle between velocity & acceleration is 120º

30º

for motion parallel to incline

R = (u cos 30º) t � 12

(g cos 30º) t2 & t = 3 sec

R = 45 m

gy . urry ds yEcor~ xfr ds fy, 0 = ( u sin 30º) t � 12

(g cos 30º)t2

t = 2usin30º

gcos30º = 3 sec = mM~M;u dky

1.5 lsd.M ij osx urry ds lekUrj gks tkrk gSA






osx o Roj.k ds e/; dks.k 120º gSA

urry ds lekUrj xfr ds fy,

R = (u cos 30º) t �12

(g cos 30º) t2 & t = 3 sec

R = 45 m

3. A particle is projected from ground with velocity 40 2 m/s at 45º. At time t = 2s : (g = 10 m/s2)

,d d.k dks tehu ls 40 2 m/s osx rFkk 45º dks.k ij ç{ksfir fd;k tkrk gSA t = 2s ij : (g = 10 m/s2)

(A*) displacement of particle has magnitude 100 m (B*) vertical component of velocity is 20

m/s

(C*) velocity makes an angle of tan�1(2) with vertical (D*) particle is at height of 60m from

ground

(A*) d.k ds foLFkkiu dk ifjek.k 100 m gSA (B*) osx dk Å /okZ/kj ?kVd 20 m/s gSA

(C*) Å /okZ/kj ls osx }kjk cuk;k x;k dks.k tan�1(2) gSA (D*) tehu ls d.k dh Å ¡pkbZ 60m gSA

Sol. ux = 40 m/s, uy = 40 m/s

x = 4x.t = 80 m

y = 4yt � 12

8t2 = 60 m

4y = 4y � gt = 20 m/s

tan = x

y

v

v = 2

= tan�1(2) with vertical.

= tan�1(2) Å /okZ/kj lsA

4. At an instant particle-A is at origin and moving with constant velocity � �(3i 4 j) m / s and particle-B

is at (4,4)m and moving with constant velocity � �(4i 3 j) m / s . Then at this instant :

fdlh {k.k ij d.k A ewy fcUnq ij rFkk fu;r osx � �(3i 4 j) m / s ls xfr'khy gS] rFkk d.k B fcUnq (4,4)m

ij gS] rFkk fu;r osx � �(4i 3 j) m / s ls xfr'khy gS] rks bl le; :

(A*) relative velocity of B w.r.t. A is � �( i 7 j) m / s A ds lkis{k B dk lkis{k osx � �( i 7 j) m / s gSA

(B*) approach velocity of A and B is 3 2 m/s A rFkk B dk lkehI; osx 3 2 m/s gSA

(C*) relative velocity of B w.r.t. A remains constant A ds lkis{k B dk lkis{k osx fu;r jgrk gSA

(D) approach velocity of A and B remains constant A rFkk B dk lkehI; osx fu;r jgrk gSA

Sol.






vBA = vB � vA

= � � � �[4i 3 j] [3i 4 j] = � �i 7 j

vapp = 4 cos45º + 3 cos45º + 3 cos45º � 4 cos45º

= 6 cos45º

= 3 2 m/s .

5. Two particle A and B are located in x-y plane at points (0, 0) and (0, 4 m). They simultaneously

start moving with velocities. A�v 2jm/ s

and B�v 2i m / s

. Select the correct alternative(s).

(A) The distance between them is constant

(B*) The distance between them first decreases and then increases

(C*) The shortest distance between them is 2 2 m

(D*) Time after which they are at minimum distance is 1s

nks d.k A o B , x-y ry esa fcUnq (0, 0) rFkk (0, 4 m) ij O;ofLFkr gSA os A�v 2jm/ s

rFkk B�v 2i m / s

ds osx ls ,d lkFk xfr djuk izkjEHk djrs gSA lgh fodYiksa dk p;u djksA

(A) buds e/; nwjh fu;r gS

(B*) buds e/; nwjh igys ?kVrh gS rFkk fQj c<+rh gS

(C*) buds e/; y?kwÙke nwjh 2 2 m gSA

(D*) og le; ftlds i'pkr os U;wure nwjh ij gksrs gS] 1s gS

Sol. AB A Bv v v

= � �2 ( j i) m/s or ABv

= 2 2 m/s

Assuming B to be at rest, A will move with velocity ABv

in the direction shown in figure. The

distance between them will first decrease from A to C and then increase beyond C.

45°

B

A

4mC

vAB






Minimum distance between them is BC which is equal to 4

2 or 2 2 and the time after which

they are at closest distance is :

t = AB

AC 2 2v 2 2

= 1s.

Hindi. AB A Bv v v

= � �2 ( j i) m/s ;k ABv

= 2 2 m/s

ekfu, B fojkekoLFkk ij gS A, ABv

osx ls fp=k esa iznf'kZr fn'kk esa xfr djsxkA buds e/; nwjh igys A ls C

rd ?kVrh gS rFkk fQj C ds uhps tkus ij c<+rh gSA

45°

B

A

4mC

vAB

buds e/; U;wure nwjh BC gS tks 4

2 ;k 2 2 ds cjkcj gS rFkk og le; ftlds i'pkr~ os lehiorhZ nwjh ij

gksrs gSa] gS:

t = AB

AC 2 2v 2 2

= 1s.

6. An open elevator is ascending with zero acceleration and speed 10 m/s. A ball is thrown vertically

up by a boy when he is at a height 10 m from the ground, the velocity of projection is 30m/s with

respect to elevator. Choose correct option, assuming height of the boy very small : (g = 10 m/s2)

,d [kqyh fy¶V 'kwU; Roj.k rFkk 10 m/s pky ds lkFk Å ij dh rjQ xfr'khy gSA tc fy¶V tehu ls 10 m

Å ¡pkbZ ij gS rc yM+dk ,d xsan Å ij dh rjQ QSadrk gSA fy¶V ds lkis{k xsan dk ç{ksi.k osx 30m/s gSA lgh

fodYi Nk¡VksA ekuk yM+ds dh Å ¡pkbZ cgqr de gSA (g = 10 m/s2) (A*) Maximum height attained by the ball from ground is 90 m.

(B*) Maximum height attained by the ball with respect to lift from the point of projection is 45 m.

(C*) Time taken by the ball to meet the elevator again is 6 sec

(D*) The speed of the ball when it comes back to the boy is 20 m/s with respect to ground.

(A*) tehu ds lkis{k xsan }kjk çkIr vf/kdre Å ¡pkbZ 90 m gSA

(B*) fy¶V ds lkis{k xsan }kjk çkIr vf/kdre Å ¡pkbZ 45 m ¼ç{ksi.k fcUnq ls½ gSA

(C*) xsan }kjk nqckjk fy¶V ls Vdjkus esa yxk le; 6 sec gSA

(D*) tehu ds lkis{k tc xsan okil yM+ds ds ikl igq¡prh gS rc xsan dh pky 20 m/s gSA






Sol. (A) Absolute velocity of ball = 40 m/s (upwards)

hmax = hi = ff

= 10 + 2(40)

2 10

h = 90 m

(B) Maximum height from left = 2(30)

2 10

= 45 m

(C) The ball unless meet the elevator again when displacement of ball = displacement of lift

40 t � 12

× 10 × t2 = 10 × t

t = 6s.

(D) Let t0 be the total time taken by the ball to reach the ground then � 10 = 40 × to � 12

× 10 × to2

t0 = 8.24 s.

time taken by the ball for each the ground after crossing the elevator = t0 � t = 2.24 s.

Sol. (A) xsan dk okLrfod osx = 40 m/s (Å ij dh vksj)

hmax = hi = ff

= 10 + 2(40)

2 10

h = 90 m

(B) fy¶V ls vf/kdre Å ¡pkbZ = 2(30)

2 10

= 45 m

(C) xsan nqckjk fy¶V ls Vdjk;sxh tc xsan dk foLFkkiu = fy¶V dk foLFkkiu

40 t � 12

× 10 × t2 = 10 × t

t = 6s.

(D) ekuk t0 xsan }kjk nqckjk tehu ij igq¡pus esa yxk le; gS � 10 = 40 × to � 12

× 10 × to2

t0 = 8.24 s.

fy¶V ls xqtjus ds ckn xsan }kjk fy;k le; = t0 � t = 2.24 s.

7. A man is standing on a road and observes that rain is falling at angle 45º with the vertical. The

man starts running on the road with constant acceleration 0.5 m/s2. After a certain time from the

start of the motion, it appears to him that rain is still falling at angle 45º with the vertical, with speed

2 2 m/s . Motion of the man is in the same vertical plane in which the rain is falling. Then which

of the following statement(s) are true.






(A) It is not possible

(B) Speed of the rain relative to the ground is 2 m/s.

(C*) Speed of the man when he finds rain to be falling at angle 45º with the vertical, is 4m/s.

(D*) The man has travelled a distance 16m on the road by the time he again finds rain to be falling

at angle 45°.

,d vkneh ,d lM+d ij [kM+k gS rFkk m/okZ/kj ds lkFk 45º ds dks.k ij ckfj'k dks fxjrh gqbZ ikrk gSA vc

vkneh lM+d ij fu;r Roj.k 0.5 m/s2 ls nkSM+uk izkjEHk djrk gSA vkneh dh xfr izkjEHk djus ds dqN nsj ckn

og vkneh ckfj'k dks fQj ls m/okZ/kj ds lkFk 45º dk dks.k cukrs gq, 2 2 m/s dh pky ls fxjrh gqbZ ikrk gS

ckfj'k ds fxjus dk m/okZ/kj ry gh vkneh dh xfr dk ry gSA rks fuEu esa ls dkSuls dFku lR; gS &

(A) ;g laHko ugha gSA

(B) tehu ds lkis{k ckfj'k dh pky 2 m/s gSA

(C*) vkneh dh pky tc og ckfj'k dks m/okZ/kj ls 45º dks.k ij ikrk gS rc 4m/s gSA

(D*) vkneh dks tc nqckjk ckfj'k 45° dks.k ij feyrh gS rks bl le; vUrjky esa r; nwjh 16m gSA

Sol. r g r m m gV V V

r g rg m gV V V

45°45°

Initial Final

VVmg

Vrg Vrm

Vrm cos 45° = Vrg cos45°

Vrm = 2 2 m/s = Vrg

Vrm cos 45° = Vmg � Vrg cos45°

m g

1V 2 2

2 +

12 2

2 = 4 m/s

VrgVrm

�Vmg

45° 45°

using v2 = u2 + 2as for the motion of man,

v2 = u2 + 2as vkneh dh xfr ds fy,]

s = 16 m.






8. A balloon rises vertically upwards from the ground, starting from rest, such that its acceleration a

varies with the height x according to the equation a = 0

xa 1

H , 0 x H. Choose the correct

statements

lrg ls ,d xqCckjk Å /okZ/kj Å ij dh vksj fojkekoLFkk ls bl izdkj mBuk izkjEHk djrk gS] fd bldk Roj.k 'a'

Å WpkbZ x ds lkFk lehdj.k a = 0

xa 1

H , 0 x H ds vuqlkj ifjofrZr gksrk gSA lgh dFku dk p;u

dhft,A

(A) the speed of the balloon when it reaches the height H is 0a H

3

tc xqCckjk H Å WpkbZ ij igqWprk gS] bldh pky 0a H

3 gSA

(B*) The speed of the balloon when it reaches the height H is 0a H2

3

tc xqCckjk H Å WpkbZ ij igqWprk gS] bldh pky 0a H2

3 gSA

(C*) The maximum acceleration of the balloon is a0

xqCckjs dk vf/kdre~ Roj.k a0 gSA

(D) The variation of velocity v with x is given by v2 = 04a H x1 1

3 H

x ds lkFk osx v esa ifjorZu v2 = 04a H x1 1

3 H

}kjk fn;k x;k gSA

Sol. The graph should be vkjs[k gksuk pkfg,

a = 0

dv xv a 1

dx H

3 / 22

0

v 2H xa 1 c

2 3 H

H

a0

a

x

x = 0, v = 0 c = 02a H

3






v2 = 3 / 2

04a H x1 1

3 H

At x = H ij, v2 = 04a H

3, v = 0a H

23

9. If the position of a particle moving along a straight line as x =3

2t2t 4t 6

3

, (x is in m & t is in

second):

;fn ljy js[kk ds vuqfn'k xfr'khy ,d d.k dh fLFkfr x =3

2t2t 4t 6

3

ds vuqlkj gS] (x ehVj esa rFkk

t lSd.M esa gS) :

(A) then particle is moving with constant acceleration

(B*) At t = 2 second, particle has zero velocity

(C) At t = 2 second, particle has acceleration 1 m/s2

(D*) At t = 3 second, particle has acceleration 2 m/s2

(A) rc d.k vpj Roj.k ls xfr'khy gksxkA

(B*) t = 2 lSd.M ij] d.k 'kwU; osx j[krk gSA

(C) t = 2 lSd.M ij] d.k dk Roj.k 1 m/s2 gSA

(D*) t = 3 lSd.M ij, d.k dk Roj.k 2 m/s2 gSA

Sol. V = dxdt

= t2 � 4t + 4

V = (t � 2)2

a = 2(t � 2)

At t = 2 ij , V = 0, a = 0

At t = 3 ij , a = 2 m/s2 .

10. A particle moves along a straight line and its velocity depends on time 't' as v = 4t � t2. Here v is in

m/sec. and t is in second. Then for the first 5 seconds :

(A*) Magnitude of average velocity is 53

m/s (B*) Average speed is 135

m/s

(C) Average speed is 115

m/s (D*) Average acceleration is � 1m/s2|

,d d.k ljy js[kk ds vuqfn'k xfr'khy gS vkSj bldk osx le; 't' ij v = 4t � t2. ds vuqlkj fuHkZj djrk gSA

;gka v m/s esa gS rFkk t lsd.M esa gSA rc izFke 5 lsd.M ds fy,






(A*) vkSlr osx dk ifjek.k 53

m/s gS (B*) vkSlr pky 135

m/s gS

(C) vkSlr pky 115

m/s gS (D*) vkSlr Roj.k � 1m/s2| gS

Sol. Average velocity vkSlr osx = st

= vavg

S = 5

0

vdt = 5

2

0

(4t t )dt =253

m

vavg = 25 / 3 m5sec.

= 53

msec

Average speed = dis tance covered

time taken =

dis tancet

vkSlr pky = r; dh xb Znwjh fy;k x;k le;

= t

nwjh

Distance nwjh = 4 5

0 4

v dt (�v)dt

= 32 73 3 =

393

m = 13 m

Average speed vkSlr pky = 13m5sec

Average acceleration vkSlr Roj.k (aavg) = f iv v

t

vf = 4 × 5 � 52 = 20 � 25 = �5

vi = 0

aavg = 5 05

= � 1 m/s2 .

11. A particle is moving along x-axis such that its position is given by x = 4 � 9t + 3t3

where t is time in

seconds, x is in meters. Mark the correct statement(s) : (A) Direction of motion is not changing at any of the instants

(B*) Direction of the motion is changing at t = 3 seconds

(C*) For 0 < t < 3 sec. the particle slowing down

(D*) For 3 < t < 6 sec. the particle is speeding up

,d d.k x-v{k ds vuqfn'k bl izdkj xfr'khy gS fd bldh fLFkfr x = 4 � 9t + 3t3

}kjk nh xbZ gS tgka t

lsd.M esa gS] x ehVj esa gSA lgh fodYiksa dk p;u dhft,A






(A) xfr dh fn'kk fdlh Hkh {k.k ifjofrZr ugh gksrh gSA

(B*) xfr dh fn'kk t = 3 sec ij ifjofrZr gksrh gSA

(C*) 0 < t < 3 sec ds fy,] d.k dh pky ?kVsxh

(D*) 3 < t < 6 sec ds fy,] d.k dh pky c<sxh

Sol. v = dxdt

= 0 � 9 +23t

3

v = t2 � 9

v = 0 gS at t = 3 ij

Also, rFkk a = 2t gS

The particle's velocity will be zero at t = 3 sec. where it changes its direction of motion. For 0 < t <

3 sec. v is �ve and a is +ve so particle is slowing down.

d.k dh pky t = 3 sec ij 'kwU; gSA tgka bldh xfr dh fn'kk ifjofrZr gskrh gSA 0 < t < 3 sec ds fy, v �

ve gS o a +ve gS vr% d.k dh pky ?kVsxh

12. A point object is placed at a distance of 5R3

from the pole of concave mirror of small aperature

and radius of curvature R. Point object oscillates with amplitude 1mm perpendicular to the optical

axis. Then

NksVs }kjd rFkk oØrk f=kT;k R ds vory niZ.k ds lkeus çdkf'kd v{k ij ,d fcUnq oLrq /kzqo ls 5R3

nwjh

ij fLFkr gSA fcUnq oLrq izdkf'kr v{k ds yEcor~ 1mm vk;ke ls nksyu dj jgh gSA rc

(A*) amplitude of image is 3

mm.7

(B) phase difference between motion of object and its image when object crosses optical axis is .

(C*) position of image of object from pole is 5R

� ,07

, when object at �O�.

(D*) Image of object is real.

(A) çfrfcEc dk vk;ke 3

mm7

gSA

(B) oLrq dh xfr rFkk blds çfrfcEc ds e/; dykUrj gS ] tc oLrq izdkf'kd v{k ls xqtjrh gSA

(C) oLrq ds çfrfcEc dh /kzqo ls fLFkfr 5R

� ,07

gS tc oLrq �O� ij gSA






(D) oLrq dk çfrfcEc okLrfod gSA

Sol. 1 1 1v u f

1 3 �2�

v 5R R

1 �2 3 7

�v R 5R 5R

5Rv �

7

m = v �5R / 7

� �0 u �5R / 3

= � 37

amplitude vk;ke = 3

mm.7

phase difference is .

dykUrj gSA

Ans. (A), (C) & (D).

13. Which of the following statements are incorrect for spherical mirrors.

fuEu esa ls dkSulk dFku xksyh; niZ.k ds fy, xyr gS &

(A*) a concave mirror forms only virtual images for any position of real object

,d vory niZ.k okLrfod oLrq dh fdlh Hkh fLFkfr ds fy, dsoy vkHkklh çfrfcEc cukrk gSA

(B) a convex mirror forms only virtual images for any position of a real object

,d mÙky niZ.k okLrfod oLrq dh fdlh Hkh fLFkfr ds fy, dsoy vkHkklh çfrfcEc cukrk gSA

(C*) a concave mirror forms only a virtual diminished image of an object placed between its pole

and the focus

,d vory niZ.k] /kzqo o Qksdl ds chp j[kh oLrq dk dsoy vkHkklh o NksVk çfrfcEc cukrk gSA

(D*) a convex mirror forms a virtual enlarged image of an object if it lies between its pole and the

focus.

;fn oLrq /kzqo rFkk Qksdl ds e/; gks rks mÙky niZ.k vkHkklh rFkk vkof/kZr çfrfcEc cukrk gSA

Sol. (A) No, when object is between infinite and focus ,image is real.

ugh, tc oLrq vuUr rFkk Qksdl ds e/; gS] izfrfcEc okLrfod gS

(C) when object is between pole and focus, image is magnified.

tc oLrq /kzqo rFkk Qksdl ds e/; gS] izfrfcEc vkof/kZr gksxk

(D) when object is between pole and focus image formed by convex mirror is real.

tc oLrq /kzqo rFkk Qksdl ds e/; gS] mÙky niZ.k }kjk cuk izfrfcEc okLrfod gksxk






14. In the shown figure two long plane mirror are kept parallel to each other with their reflecting side

facing one another. A point object 'O' is situated at the distance of 20 cm from A then :

iznf'kZr fp=k esa nks yEcs lery niZ.k ,d nwljs ds lekUrj j[ks gq, gS] rFkk muds ijkorZd i"B ,d nwljs ds

vkeus lkeus gSA ,d fcUnq oLrq 'O'] A ls 20 cm dh nwjh ij fLFkr gS] rc %

(A*) distance of first image formed in A is 50 cm from B.

(B*) distance between fifth image formed in the two mirrors is 3 m

(C) distance between fifth image formed in the two mirrors is 240 cm

(D*) total of infinite images are formed in the two mirrors

(A*) A esa cus izFke izfrfcEc dh B ls nwjh 50 cm gksxhA

(B*) nksuksa niZ.kksa esa cus ik¡posa izfrfcEc ds e/; nwjh 3 m gksxhA

(C) nksuksa niZ.kksa esa cus ik¡posa izfrfcEc ds e/; nwjh 240 cm gksxhA

(D*) nksuksa niZ.kksa ds e/; vuUr izfrfcEc cusxsaA

Sol.






15. Two plane mirrors are inclined to each other with their reflecting faces making acute angle. A light

ray is incident on one plane mirror. The total deviation after two successive reflections is:

nks lery niZ.k vius ijkorhZ i"Bksa ds lkFk U;wu dks.k (acute angle) cukrs gq, ,d nwljs ds lkFk >qds gq,

gSA ,d çdk'k fdj.k fdlh ,d lery niZ.k ij vkifrr gksrh gSA nks mÙkjksÙkj ijkorZuksa ds ckn dqy fopyu&

(A*) independent of the initial angle of incidence çkjfEHkd vkiru dks.k ij fuHkZj ugha djrk gSA

(B) independent of the angle between the mirrors niZ.kksa ds chp dks.k ij fuHkZj ugha djrk gSA

(C) dependent on the initial angle of incidence çkjfEHkd vkiru dks.k ij fuHkZj djrk gSA

(D*) dependent on the angle between the mirrors. niZ.kksa ds chp dks.k ij fuHkZj djrk gSA

Sol.

i r2 2

i + r = .........(i)

= 2i + 2r

= 2 (Anticlockwise okekorZ)

16. Position time graph for a particle moving on straight line is shown in figure. Assume same slope of

x�t graph at t = 0 and t = 20 s. Select correct alternative/s :

ljy js[kk esa xfr dj jgs ,d d.k ds fy, fLFkfr le;&xzkQ fp=kkuqlkj gSA ;g ekfu, fd t = 0 rFkk t = 20 s

ij x�t xzkQ dk <ky leku gSA lgh fodYi@fodYiksa dk p;u dhft, :

(A*) Average velocity of particle from t = 0 to t = 20 sec. is zero.

(B*) Acceleration of particle from t = 10 sec. to t = 15 sec. is positive.

(C*) Average acceleration of particle from t = 0 to t = 20 is zero.

(D) Average velocity & instantaneous velocity becomes same in magnitude and direction more

than once from t = 0 to t = 20 sec.






(A*) t = 0 ls t = 20 sec esa d.k dk vkSlr osx 'kwU; gksxkA

(B*) t = 10 sec ls t = 15 sec esa d.k dk Roj.k /kukRed gksxkA

(C*) t = 0 ls t = 20 sec esa d.k dk vkSlr Roj.k 'kwU; gksxkA

(D) t = 0 ls t = 20 sec ds nkSjku d.k dk vkSlr osx rFkk rkR{kf.kd osx ifjek.k rFkk fn'kk esa ,d ls vf/kd

ckj leku gks tk,xkA

17. Particle thrown from O, parallel to the incline as shown hits the incline perpendicular to it. Choose

the correct options.

urry ds lekUrj fcUnq O ls, QSadk x;k d.k urry ij fp=kkuqlkj yEcor~ Vdjkrk gSA lgh fodYiksa dk p;u

dhft,A

(A*) d is equal to 9375

m16

(B) Particle hits the incline after the time 10 sec

(C) d is equal to 7225

m16

(D*) Particle hits the incline after time 12.5 sec

(A*) d, 9375

m16

ds cjkcj gS (B) 10 sec i'pkr~ d.k urry ij Vdjk;sxk

(C) d, 7225

m16

ds cjkcj gS (D*) 12.5 sec i'pkr~ d.k urry ij Vdjk;sxk

Sol. yV

60= tan 37 = 3/4 vy = 45 m/s

� 45 = 80 � 10t

10t = 125

t = 252

sec

37º

X0

60

y

d + x0 = 25

602

= 750 m.

(45)2 = (80)2 + 2(�10)y

y = 875

4m.






= tan 53 = 4/3 x0 = 3/4 y =

262516

m.

d = 750 � 262516

= 937516

m

18. A ray of light is incident normally on one face of a prism as shown in figure. The refractive index of

the material of the prism is 53

and the prism is immersed in water of refractive index 4

,3

then

iznf'kZr fp=k esa fizTe ds ,d Qyd ij fp=kkuqlkj ,d izdk'k fdj.k yEcor~ vkifrr gksrh gSA fizTe ds inkFkZ

dk viorZukad 53

gS rFkk fizTe dks 4

,3

viorZukd okys ty esa j[kk x;k gS] rc

60°

30°

P

(A*) The angle of emergence 2 of the ray is �1 5sin .

8

fuxZr izdk'k fdj.k dk dks.k 2,�1 5

sin .8

gksxkA

(B) The angle of emergence 2 of the ray is �1 5sin .

4 3

fuxZr izdk'k fdj.k dk dks.k 2, �1 5

sin .4 3

gksxkA

(C) The angle of emergence 2 of the ray is �1 7sin .

3 4

fuxZr izdk'k fdj.k dk dks.k 2, �1 7

sin .3 4

gksxkA

(D*) Total internal reflection will not occur at P if the refractive index of water increases to a value

greater than 5

2 3 by dissolving some substance

dqN v'kqf) ?kksy dj ;fn ty dk viorZukad 5

2 3 ds eku ls] vf/kd c<+k ns rks fcUnq P ij iw.kZ vkUrfjd

ijkorZu ?kfVr ugha gksxkA






Sol. Initially at P, there is TIR, Therefore at Q (snell�s law)

izkjaHk esa P ij, iw.kZ vkarfjd ijkorZu gS] vr% Q ij (Lusy ds fu;e ls)

2

5 4sin30 sin

3 3

�1

2 2

5 5sin sin

8 8

If w >5

2 3

5 / 2 32 / 3

sinC > C > 60°

Therefore, now TIR will not occur at P.

vr%] P ij iw.kZ vkUrfjd ijkorZu ugha gksxkA

Ans. (A, D)

19. In the figure the light is incident at an angle (slightly greater than the critical angle, for refraction

from n2 to n1) as shown in figure (a). Now keeping the incident ray fixed a parallel slab of refractive

index n3 is introduced between medium n1 and n2 as shown in figure (b). Choose the correct

options

fp=k (a esa) esa izdk'k dks.k (ØkfUrd dks.k ls FkksM+k cM+k) ij n2 ls n1 esa viorZu ds fy, vkifrr gksrk gSA

vc vkifrr fdj.k dks fLFkj j[krs gq;s n3 viorZukad dh ,d lekUrj ifV~Vdk ek/;e n1 rFkk n2 ds e/; fp=k

(b) ds vuqlkj j[kh tkrh gSA lgh fodYiksa dk p;u dhft,A

(A*) total internal reflection occurs at AB for n3 < n1

(B) total internal reflection occurs at AB for n3 > n1






(C*) the ray will finally return back to the medium n2 for all values of n3

(D) total internal reflection occurs at CD for n3 < n1.

(A*) n3 < n1 ds fy;s AB ij iw.kZ vkUrfjd ijkorZu gksrk gSA

(B) n3 > n1 ds fy;s AB ij iw.kZ vkUrfjd ijkorZu gksrk gSA

(C*) n3 ds lHkh ekuksa ds fy;s fdj.k vUr esa mlh ek/;e n2 esa ykSV tkrh gSA

(D) n3 < n1 ds fy;s CD ij iw.kZ vkUrfjd ijkorZu gksrk gSA

20. A small air bubble is trapped inside a transparent cube of size 12 cm. When viewed from one of the

vertical faces, the bubble appears to be at 5 cm from it. When viewed from opposite face, it appears

at 3 cm from it.

(A*) The distance of the air bubble from the first face is 7.5 cm.

(B) The distance of the air bubble from the first face is 9 cm.

(C) Refractive index of the material of the cube is 2.0.

(D*) Refractive index of the material of the cube is 1.5.

,d NksVk ok;q dk cqycqyk 12 cm Hkqtk ds ikjn'khZ ?ku ds vUnj fLFkr gSA tc bls ?ku dh Å /okZ/kj lrg ls

ns[kk tkrk gSA rc ;g cqycqyk lrg ls 5 cm dh nwjh ij izfrr gksrk gSA tc bls foifjr lrg ls ns[kk tkrk

gSA rc ;g lrg ls 3 cm dh nwjh ij izfrr gksrk gSA

(A*) izFke lrg ls ok;q ds cqycqys dh nwjh 7.5 cm gskxhA

(B) izFke lrg ls ok;q ds cqycqys dh nwjh 9 cm gskxhA

(C) ?ku ds inkFkZ dk viorZukad 2.0 gSA

(D*) ?ku ds inkFkZ dk viorZukad 1.5 gSA

Sol.

Let the bubble B is at distance H from the face F1 of the cube.

ekuk cqycqyk B ?ku dh lrg F1 ls H nwjh ij gSA

h1 = a

C

n

nH = 5 cm

Similarly when looking from opposite face F2,

blh izdkj tc foifjr lrg F2 ls ns[krs gS] rc






h2 = a

C

n

n (12 � H) = 3 cm

Solving H = 7.5 cm and nC = 1.5

gy djus ij H = 7.5 cm rFkk nC = 1.5

21. Which of the following statements is/are correct about the refraction of light from a plane surface

when light ray is incident in denser medium. [C is critical angle]

tc fdj.k l?ku ek/;e esa izos'k djrh gS rks izdk'k fdj.k ds lery lrg ls viorZu ds fy, dkSuls dFku

lR; gSA [C ØkfUrd dks.k gS]

(A*) The maximum angle of deviation during refraction is 2

� C, it will be at angle of incidence is

C.

(B*) The maximum angle of deviation for all angle of incidences is � 2C, when angle of

incidence is slightly greater than C.

(C*) If angle of incidence is less than C then deviation increases if angle of incidence is also

increased.

(D*) If angle of incidence is greater than C then angle of deviation decreases if angle of

incidence is increased.

(A*) viorZu ds nkSjku vf/kdre fopyu dks.k 2

� C gksxk, ;g vkiru dks.k C ij gksxkA

(B*) lHkh vkiru dks.kksa ds fy, vf/kdre fopyu dks.k � 2C gS] tc vkiru dks.k C ls FkksM+k lk T;knk

gSA

(C*) ;fn vkiru dks.k C ls de gks rks vkiru dks.k c<+kus ij fopyu dks.k c<+sxkA

(D*) ;fn vkiru dks.k C ls T;knk gks rks vkiru dks.k c<+kus ij fopyu dks.k ?kVsxkA

Sol.

22. A parachutist jumps from height 100 m. He is to reach at ground with zero velocity. For this

purpose he switches on a parachute propeller after falling freely for certain height. Given that after

the parachute propeller is switched on total acceleration of the man varies with velocity as a = �

2v, where v is the instantaneous velocity of the man. Choose the correct options for this situation.

(use g = 10m/s2).

(A*) Man switches on parachute propeller after falling freely for time 4 second.






(B) Man swithces on parachute propeller after falling freely for time 2 second.

(C*) Distance covered by the man with parachute propeller switched on is 20 m.

(D) Distance covered by the man during free fall is 60 m.

,d iSjk'kqV /kkjh O;fDr 100 m Å pkbZ ls dqnrk gSAog tehu ij 'kwU; osx ls igqpuk pkgrk gSA blds fy, og

dqN Å pkbZ fxjus ds i'pkr~ iSjk'kqV [kksyrk gSA ;g fn;k gS fd iSjk'kqV [kksyus ds i'pkr~ O;fDr dk dqy Roj.k

osx ds lkFk a = �2v ds vuqlkj ifjofrZr gksrk gSA, tgk¡ v O;fDr dk rkR{kf.kd osx gSA bl fLFkfr ds fy, lgh

fodYi pqfu,&(g = 10m/s2).

(A*) O;fDr dks 4 lSd.M+ i'pkr~ isjk'kqV [kksyuk pkfg,A

(B) O;fDr dks 2 lSd.M+ i'pkr~ isjk'kqV [kksyuk pkfg,A

(C*) iSjk'kqV [kqyus ds i'pkr~ O;fDr }kjk r; nwjh 20 eh- gSA

(D) O;fDr }kjk eqDr :i ls fxjus dh nwjh 60 eh gSA

Sol. After switching on parachuite propeller

iSjk'kwV [kksyus ds i'pkr~

dv

vdy

= �2V

0

0

2gx

dv =

0

100

x

2 dy

02gx = 2(100�x0)

x02 � 205x0 + 10000 = 0

x0 = 80m

time of free fall eqDr :i ls fxjus dk le; t = 2(80)10

= 4 sec

SECTION � 2 : (Maximum Marks : 72)

This section contains 18 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

�1 In all other cases

[kaM 2 : (vf/kdre vad : 72)

bl [kaM esa 18 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

�1 vU; lHkh voLFkkvksa esa

Integer 23. A bird is at a point P (4m, �1m, 5m) and sees two points P1 (�1m, � 1m, 0m) and P2 (3m, � 1m, �

3m). At time t = 0, it starts flying in a plane of the three positions, with a constant speed of 1 m/s in

a direction perpendicular to the straight line P1 P2 till it sees P1 and P2 collinear at time t. Find the

time t. ( in s)

,d i{kh fcUnq P (4m, �1m, 5m) ij gS rFkk ;g nks fcUnqvksa P1 (�1m, � 1m, 0m) rFkk P2 (3m, � 1m, � 3m)

dks ns[krk gSA t = 0 le; ij ;g rhuksa fLFkfr;ksa ds ry esa 1 m/s dh fu;r pky ls ljy js[kk P1 P2 ds

yEcor~ fn'kk es rc rd mM+rk gS tc rd fd ;g t le; ij P1 rFkk P2 dks ljSf[kd ugha ns[k ysrk gSA t le;

¼lSd.M esa½ Kkr djksA

Ans. 7

Sol. (Tough) Let bird flying along PQR

ekuk i{kh PQR ds vuqfn'k mM+ jgk gS

1� �PP 5i 5k

2 1 1 2� �P P P P 4i 3k

component of PP1 along P1P2 (PP1x)






P1P2 ds vuqfn'k PP1 (PP1x) dk ?kVd

1 2 1

2 1

PP .P P

| P P |

=

20 155

= 1

and rFkk 1PP

= 50

From triangle PP1 Q

f=kHkqt PP1 Q esa

50 = (1)2 + (PQ)2

PQ = 7m

time le; = 71

= 7 s

Ans. : 7 s

24. A particle is moving along a straight line. Its velocity varies as v = 6 � 2t where v is in m/s and t in

seconds. Find the difference between distance covered and magnitude of displacement in first 4

seconds.

,d d.k ljy js[kk ds vuqfn'k xfr'khy gSA bldk osx le; ds lkFk v = 6 � 2t }kjk fn;k tkrk gSA ;gk¡ v,

m/s esa rFkk t le; lSd.M esa gSA izFke 4 lSd.M esa d.k }kjk r; nwjh rFkk foLFkkiu ds ifjek.k es vUrj Kkr

djksA

Ans. 2

Sol. Displacement foLFkkiu = t

0

vdt = 4

0

(6 2t)dt = 2m

Distance nwjh = 4

0

| v | dt = 3 4

0 3

(6 2t)dt (2t 6)dt = 10 m

Distance � displacements nwjh&foLFkkiu = 10 � 8 = � 2

25. A particle moves in a straight line along x-axis such that its acceleration along x-axis at any instant

is proportional to square of its velocity at that instant, porportionality constant being C = 12

. If at t =

0, particle is at x = 0 and velocity at that instant is v0 = 1 � 1e

. Then find the position of particle at t

= 2sec in meters.

[x is in meters and t is in seconds] [Given : n e = 1]






,d d.k x-v{k ds vuqfn'k ljy js[kk esa bl izdkj xfr djrk gS fd x-v{k ds vuqfn'k fdlh {k.k ij bldk

Roj.k ml {k.k ij blds osx ds oxZ ds lekuqikrh gksrk gSA lekuqikrh fu;rkad C = 12

gSA ;fn t = 0 ij d.k

x = 0 ij gS rFkk ml {k.k osx v0 = 1 �1e

gS rks t = 2sec ij d.k dh fLFkfr ehVj esa Kkr djksA

[x ehsVj esa rFkk t lSd.M esa gS] [fn;k x;k gS : n e = 1]

Ans. x = 2

Sol. a v2

(1) dvdt

= CV2

0

v t

2v 0

1 dvdt

c v

1c

. 0

v

v

1v

= t

1c 0

1 1v v

= t �

0

1 1v v

= ct 1v

=0

1v

� ct

v = 0

0

v

(1 v ct)

(2) from here, we obtain ;gk¡ ls ge izkIr djrs gS

dxdt

= 0

0

v

1 v ct

0

1v

x

0

dx = t

00

dt1 v ct

0

xv

= � 0

1cv

t

0 0n 1 v ct

x = � 1c 0[ n [1 v ct] n 1] x = �

1c 0n [1 v ct]

(3) Putting values ekuksa dks izfrLFkkfir djus ij

x = � 1 1 1

n 1 1 21 e 22

= � 2 1

ne

so vr%, x = 2 n e

x = 2 m






26. Two particles A and B move in x-y plane such that both have constant acceleration

A�a 10 j

m/s2 and B�a 5 j

m/s2 respectively. The velocities of particles at t = 0 are

A� �u 5 i 20 j

m/s and B� �u 2.5 i 10 j

m/s. At time t = 0, particle A is at origin and particle B

is at point having coordinates (5 meters, 0). Find the instant of time in seconds at which angle

between velocity of A and velocity of B is 180°.

nks d.k A o B, x-y ry esa bl izdkj xfr djrs gS fd muds fu;r Roj.k Øe'k% A�a 10 j

m/s2 rFkk

B�a 5 j

m/s2 gSA t = 0 ij d.k dk osx A� �u 5 i 20 j

m/s rFkk B� �u 2.5 i 10 j

m/s gSA t = 0

ij ,d d.k A ewy fcUnq ij gS rFkk d.k B ml fcUnq ij gS ftlds funsZ'kkad (5 meters, 0) gSA og le;

¼lSd.M esa½ Kkr dhft;s tc A rFkk B ds osxksa ds e/; dks.k 180° gksxkA

Ans. 2

Sol. At t = 2 sec. y component of velocity of A and B is zero and x-components are in opposite

direction.

t = 2 lS- ij A rFkk B ds osx ds y ?kVd 'kwU; gS ,oa osx ds x ?kVd ijLij foijhr fn'kk esa gksaxsA

27. A particle moves in a straight line such that it moves in the same direction but its speed decreases

uniformly with time to zero in time interval T = 10s. It travels a total distance 'S' = 12.5 m. Find the

velocity (in m/s) at any intermediate time t = 6.s.

,d d.k ljy js[kk esa bl izdkj xfr djrk gS fd ;g leku fn'kk esa xfr djrk gS fdUrq bldh pky

le;kUrjky T = 10s esa ,d leku :i ls 'kwU; rd ?kVrh gSA ;g dqy 'S' = 12.5 m nwjh r; djrk gSA buds

e/; ds fdlh le; t = 6.s ij osx ¼m/s esa½ Kkr djksaA

Ans. 1

Sol. Speed decreasing to zero uniformly. Let a be a positive constant (acceleration) then

pky ,d leku :i ls ?kVrh gqbZ 'kwU; gks tkrh gSA ekuk a /kukRed fu;rkad ¼Roj.k½ gSA rksa

v = a (T � t)

s = T

0

v dt = T

2

0

1aTt at

2

= aT2 � 2aT

2 = 21

aT2

a = 2

2s

T

v = 2

2S

T (T � t)

Ans. v = 2

2S

T (T � t) = 1 m/s






28. 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 of v2

?

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 rc v2

dk eku Kkr djksA

Ans. 6

Sol. tan30º = v

12 3

v = 12 m/s

v2

= 6 m/s Ans.

29. Particle A starts from rest at t = 0 from x = 0 with constant acceleration to reach x = 1m at t = 1

second. Particle B starts with uniform velocity at t = 0 from x = 1m to reach x = 2m at t = 1 sec. The

distance covered, by particle B in the frame of reference attached to particle A from t = 0 to t = 1

sec. is x ×10�1 m. Find �x�.

d.k A, t = 0 ij x = 0 ls t = 1 sec. ij x = 1m rd igq¡pus ds fy, fu;r Roj.k ls fojke ls pyuk çkjEHk

djrk gSA d.k B , t = 0 ij x = 1m ls t = 1sec ij x = 2m rd igq¡pus ds fy, ,dleku osx ls pyuk

çkjEHk djrk gSA d.k A ls tqM+s funsZ'k rU=k esa le; t = 0 ls t = 1 sec. ds e/; d.k B }kjk r; dh xbZ nwjh x

×10�1 m gSA �x� dk eku Kkr djksA

Ans. 5

Sol. For A (A ds fy,)

s = ut + 21at

2

1 = (0) (1) + 12

(a)(1)2

aA = 2m/s2

vA = A + aAt

vA = 2t

For B. (B ds fy,)






uB = 1m/s

1B Av

= 1 � 2t

1B Av

= | 1 � 2t |

Distance nwjh = 0

| 1� 2t | dt 0.5

30. Two points A and B are moving with the same speed u = 4 2 m/s in the positive direction of x�

axis and

y�axis respectively. Find the magnitude of relative velocity of C w.r.t. A (where C is the mid point of

AB).

nks fcUnq A rFkk B leku pky u = 4 2 m/s ls Øe'k% /kukRed x�v{k ,oa /kukRed y�v{k esa xfreku gSA ;fn

C, AB dk e/; fcUnq gks] rks C dk A ds lkis{k osx dk ifjek.k Kkr djksA

Ans. 4 m/s

Sol. A�V ui

B�V uj

Let at any time t, coordinate of point A and B be (x,0) and (0, y) respectively.

ekuk fdlh le; t ij fcUnq A rFkk B ds funsZ'kkad (x,0) rFkk (0, y) gSA

Then, coordinate of point C (a,b) will be

fcUnq C ds funsZ'kkad (a,b) gksxsa

a = x2

, dadt

= 12

dxdt

or Vx = u2

and rFkk b = y2

, dbdt

= 1 dy2 dt

or Vy = u2

Hence velocity of C would be vr% C dk osx

C

u u� �V i j2 2

Velocity of C with respect to A is : vr% C dk A ds lkis{k osx

CA C AV V V

= � u u� �i j2 2

CA

uV

2

= 4 m/s






31. A man can swim with a velocity Vmw in still water. When the water moves with a velocity Vw , the

man crosses the river to and fro in minimum time T1. If the man intends to cross the river

perpendicularly, he takes time T2 for to and fro journey. Now he swims in the downstream and

comes back to his initial position by swiming upstream along the shore. For to and fro journey

(same distance) along the shore the man takes a time T3. If T1 = 3 sec and T3 = 12 sec. Find out

the value of T2 .

,d O;fDr fLFkj ikuh esa Vmw osx ls rSj ldrk gSA tc ikuh Vwosx ls xfr dj jgk gS] O;fDr unh dks U;wure

lEHko le; esa ikj djds okfil vkus esa le; T1 ysrk gSA ;fn O;fDr unh ds izokg ds yEcor~ ikj djds

okfil vkus esa T2 le; ysrk gSA vc O;fDr unh ds cgko dh fn'kk esa unh dh pkSM+kbZ ds cjkcj nwjh tkdj

okfil izkjfEHkd fLFkfr rd vkus esa T3 le; ysrk gSA ;fn T1 = 3 sec rFkk T3 = 12 sec gSA T2 dk eku Kkr

dhft,A

Ans. 6

Sol. Ist case fLFkfr esa: T1 = m

2dv

IInd case fLFkfr esa : T2 = 2 2

m

2d

v v

IIIrd case fLFkfr esa : T3 = m m

d dv v v v

= m2 2

m

2dv

v v

Then rc, T22 = T1T3

= 3 × 12 T2 = 6 sec.

32. A person is standing on a plank which starts slipping over a fixed smooth wedge from rest position

hence person also moves with plank. After 12

second of motion begins he observed that rain is

falling vertically down ward. After 73

sec. he again observed that rains is coming horizontally.

From the given information find the speed of rain w.r.t. ground. Find your answer in form of x m/s

and fill value of x in OMR sheet.

(use g = 10 m/s2)

,d O;fDr xqVds ij [kMk gqvk gSA xqVdk fp=kkuqlkj tMor~ fpdus ost ij fLFkjkoLFkk ls fQlyuk izkjEHk

djrk gSA O;fDr Hkh xqVds ds lkFk xfr djuk izkjEHk djrk gSA 12

lSd.M i'pkr~ og izsf{kr djrk gS fd ckfj'k

m/okZ/kj fxj jgh gSA 73

sec. i'pkr~ og iqu% izsf{kr djrk gS fd ckfj'k {kSfrt fn'kk ls vk jgh gSA nh xbZ






lwpukvksa ds vk/kkj ij ckfj'k dh tehu ds lkis{k pky Kkr djksaA viuk mÙkj x m/s ds :i esa Kkr djksaA rFkk

x dk eku OMR 'khV esa HkjksA

(g = 10 m/s2)

Ans. 6

Sol. xr

V = V1 cos 37º = (g sin 37º t1) cos 37º

yr

V = V2 sin 37º = g sin2 37º t2

Vr = x y

2 2r rV V

= 10 3

25 2 2

1 216t 9t

= 65

4 21 6m / s

33. A ball is thrown from the roof of a building of height 44m with speed v0 at an angle below the

horizontal. It lands 2 seconds later at a point 30m from the base of the building. If tan = X

10 then,

find the value of X. (Take g = 10 m/s2)

,d xsan dks 44m Å ph ehukj dh Nr ls v0 pky ls {kSfrt ds uhps dks.k ij Qsadk tkrk gSA ;g ehukj ds

vk/kkj ls 30m nqjh ij fLFkr ,d fcUnq ij 2 lSd.M esa igqprh gSA ;fn tan = X

10 rks X dk eku Kkr djks.

(g = 10 m/s2)






Ans. 8

Sol. With the origin at the base

vk/kkj ij ewy fcUnq ds lkFk

x = v0cos . t, y = 44 � v0sint �12

(10) t2

putting t = 2 sec and y = 0 , x = 30

t = 2 sec rFkk y = 0 ij j[ksu ij , x = 30

v0cos = 15 and rFkk v0sin = 242

= 12

tan = 1215

= X

10 =

810

Hence vr% X = 8

34. A particle is projected up an inclined plane of inclination to the horizontal as shown. The angle of

projection of particle with horizontal is . If the particle strikes the plane horizontally, then tan =

tanx

. Find the value of x.

{kSfrt ls >qdko okys ur ry ij ,d d.k dks Å ij dh vksj fp=kkuqlkj iz{ksfir fd;k tkrk gSA {kSfrt ds

lkFk d.k dk iz{ksI; dks.k gSA ;fn d.k ry ij {ksfrtr% Vdjkrk gS] rks tan = tan

x

, x dk eku Kkr

djksA

Ans. x = 2

Sol. As the particle strikes the plane horizontally so its velocity is parallel to horizontal axis and vertical

velocity is zero.

D;ksafd d.k ry ij {kSfrtr% Vdjkrk gS vr% bldk osx {kSfrt i"B ds lekUrj gksxk rFkk m/okZ/kj osx 'kwU;

gksxkA

In vertical direction, 0 = u sin � gt

m/okZ/kj fn'kk esa 0 = u sin � gt

or, ;k] u sin = g 2usin( � )

gcos

or, ;k] sin cos = 2 [sin cos � cos sin ]






2 cos sin = sin cos

2 tan = tan or ;k [tan = 1/2 tan ].

35. If the projectile hits the inclined plane perpendicularly when thrown horizontally with v0 from a

tower of height H as shown then the value of 2

0

2gH

(v ) is

;fn n'kkZ;s vuqlkj H Å ¡pkbZ dh ehukj ls ,d ç{ksi dks {kSfrt osx v0 ls Qsadrs gS rFkk ;g ç{ksi urry ij

yEcor~ Vdjkrk gSA rc 2

0

2gH

(v ) dk eku gS &

Ans. 5

Sol.

ax = � g sin 30º

ay = g cos 30º

ux = V0 cos 30º

uy = V0 sin 30º

Vx = ux + axt

0 = 0V 3 gt

2 2

t = 0V 3

g ............(i)

Sy = uyt + ayt2






(H cos 30º � 0) = V0 sin 30º t + 12

g cos 30º t2 ............(ii)

From (i) & rFkk (ii) ls ,

2

0

2gH

V = 5.

36. One particle each projected from point A and B. If at time t = x

10, both particles collide in air (both

particles collide in air before any of them strike on ground or inclined). Than value of x will be :

nks d.kksa dks A o B ls ç{ksfir fd;k tkrk gSA ;fn nksauks d.k t = x

10 le; ij gok esa Vdjkrs gSA ¼nksauks d.k

gok esa Vdjkrs gS vkSj blls igys dksbZ Hkh d.k lrg ;k urry ls ugha Vdjkrs gSA ½ rc x dk eku gksxk &

Ans. x = 8

Sol. w.r.t particle B, tan = 1020

= 4y / 5

3y / 5 10

d.k B ds lkis{k tan = 1020

= 4y / 5

3y / 5 10

3y + 50 = 8y ; y = 10m

time le; = 10 6

20

= 45

sec.

x = 8






37. A particle is projected up with 15 m/s perpendicular to the incline. (see fig.) . If the particle hits the

incline at a distance R (in meter) from the point of projection then find the value of R5

? (g = 10

m/s2)

,d d.k (fp=k esa) urry ds yEcor~ 15 m/s osx ls Å ij dh vksj iz{ksfir fd;k tkrk gSA ;fn d.k urry ij

iz{ksi.k fcUnq ls R (ehVj dks) nwjh ij Vdjkrk gS rks R5

dk eku Kkr djksaA (g = 10 m/s2)

Ans. 6

Sol. For motion to incline urry ds yEcor~ xfr ds fy,

0 = 15 t � 12

(g cos 30º) t2 ..................(i)

for motion along the incline urry ds vuqfn'k xfr ds fy,

R = 0 + 12

(g sin 30º) t2 ..................(ii)

Solving (i) & (ii) (i) o (ii) dks gy djus ij

we get ge izkIr djrs gSA R = 30 m R5

= 6 m

38. A man of mass m starts moving w.r.t. a platform of mass 2m with a velocity 9

u m / s13

as shown

in the figure. The platform is fitted with a concave mirror of focal length f. The velocity of image (in

m/s) at the initial moment is :

m nzO;eku dk ,d O;fDr 2m nzO;eku ds IysVQkeZ ds lkis{k 9

u m / s13

ls fp=kkuqlkj xfr dj jgk gSA

IysVQkWeZ f Qksdl nwjh ds vory niZ.k ls tqM+k gqvk gSA çkjfEHkd {k.k ij çfrfcEc dk osx (m/s esa) gksxkA

Ans. 3m/s






Sol. Magnification vko/kZu f

m3f

f2

m = �2

Taking the direction in the right hand side 2im omV m V

nka;h vksj fn'kk ysus ij 2im omV m V

imV velocity of image w.r.t. mirror niZ.k ds lkis{k çfrfcEc dk ossx

omV velocity of object w.r.t mirror niZ.k ds lkis{k oLrq dk osx

imV 4u

m

mu uV

m 2m 3

velocity of platform (minor) IysVQkeZ (niZ.k) dk osx

im i mV V V

Vi velocity of image w.r.t ground tehu ds lkis{k çfrfcEc dk osx

Vi = �44

u3

i

13V u

3

Vi = 3m/s

39. When an object is placed at a distance of 25 cm from a concave mirror, the magnification is m1.

The object is moved 15 cm further away from the mirror with respect to the earlier position, and the

magnification becomes m2. If m1/m2 = 4 then find the focal length (in decimeter) of the mirror

(Assume image is real and m1, m2 are numerical values of transverse magnification of the image)

tc ,d oLrq dks ,d vory niZ.k ls 25 cm nwj j[kk tkrk gS rks vko/kZu m1 gSA oLrq dks iwoZ fLFkfr ls niZ.k

ds lkis{k 15 cm nwj f[kldk;k tkrk gS rc vko/kZu m2 gks tkrk gSA ;fn m1/m2 = 4 rks niZ.k dh Qksdl nwjh

(MslhehVj esa) Kkr djksA ekuuk gS fd çfrfcEc okLrfod gS rFkk m1 vkSj m2 çfrfcEc ds vuqçLFk vko/kZu dk

xf.krh; eku gSA

Ans 2

Sol. 1f

= 1

1 125 m 25

............(i)

1f

= 1

25 15 �

2

1m .40






=1

1 1m40 .404

= 1 1 25

140 10 f

f = 40 1

1.53 15 = 20 = 2 decimeter

40. A man is sitting in a room at 2 m from a wall W1 wants to see the full height of the wall W2 behind

him 4 m high and 6 m away from the facing wall W1. What is the minimum vertical length in meter

of mirror on the facing wall required for the purpose ?

,d dejs esa ,d O;fDr nhokj W1 ls 2 m nwjh ij cSBk gSA ;g vius ihNs 4m Å ¡ph ,oa lkeus dh nhokj W1

ls 6m nwjh ij fLFkr W2 nhokj dh iwjh Å ¡pkbZ ns[kuk pkgrk gSA bl m)s'; ds fy, lkeus dh nhokj W1 ij

yxs niZ.k dh Å /okZ/kj yEckbZ de ls de fdruh gksuh pkfg,

Ans. 1

Sol.

Let the min. length of mirror be L

ekuk niZ.k dh U;wure yEckbZ L gSA

L4

= 28

L = 1 m



Page # 1

Course : JR-PT-1 (E-LPD)

Test Date : 16-08-2015

Test Type : (JEE ADVANCED PATTERN)

Paper

Time Duration : 2 Hrs.

SYLLABUS: Mole Concept, QMM and Periodic Table

IUPAC NOMENCLATURE & STRUCTURAL ISOMERISM

Test Pattern :


1 to 22 MCQ 22 4 �2 88


40 160


Total Total

Maths/ Physics/

Chemistry

Physical Inorganic paper Organic paper

MCQ (14) MCQ (8)

Integer type (Single) 11 Integer type (Single) 7

CHEMISTRY

MCQ (22) 1. Which of the following statement is correct for 3dxy orbitals?

(A*) It has two nodal planes, xz and yz. (B) The minimum probability point lie along = 45º. (C) + ve and � ve signs represent charge. (D*) It is a non-axial orbital. 3dxy d{kd ds lUnHkZ esa dkSulk@dkSuls dFku lgh gS@gSa ?

(A*) n'kkZ;k x;k d{kd nks uksMy ry xz o yz j[krk gSA (B) = 45º ds vuqfn'k U;wure izkf;drk fcUnq gksrk gSA

Page # 2

(C) + ve o � ve fpUg vkos'k dks iznf'kZr djrs gSA (D*) ;g ,d fuj{kh; d{kd gSA Sol. These are the fact. ;g rF; gSA 2. Which of the following statements is/are correct regarding 6.4g of 216

8 O ion :

(A*) Total number of proton present is about 1.92 × 10

24. (B*) Total number of electrons present is 4NA (C) Magnitude of net charge on sample is 0.4 NAC (Coulomb.) (D*) Total number of neutrons present is 3.2 NA

2168 O vk;u ds 6.4 xzke ds lEcU/k esa fuEu esa ls dkSulk@dkSuls dFku lgh gS@gSa ?

(A*) mifLFkr izksVkWuksa dh dqy la[;k 1.92 × 1024 gSA

(B*) mifLFkr bysDVªkWuksa dh dqy la[;k 4NA gSA (C) uewus ij dqy vkos'k dk ifjek.k 0.4 NAC (dwyke) gSA (D*) mifLFkr U;wVªkWuksa dh dqy la[;k 3.2 NA gSA

Sol. Total number of proton present = 8N16

4.6A = 3.2 NA = 1.926 × 10

24

mifLFkr izksVkWuksa dh dqy la[;k= 8N16

4.6A = 3.2 NA = 1.926 × 10

24

Total number of electrons present = 16

4.6 × NA × 10 = 4NA

mifLFkr bysDVªkWuksa dh dqy la[;k= 16

4.6 × NA × 10 = 4NA

Magnitude of net charge on sample = 16

4.6 × NA × 2e = 0.8 NAeC

uewus ij dqy vkos'k dk ifjek.k = 16

4.6 × NA × 2e = 0.8 NAeC

Total number of neutrons present = 16

4.6 × NA × 8 = 3.2 NA

mifLFkr U;wVªkWuksa dh dqy la[;k = 16

4.6 × NA × 8 = 3.2 NA

3. Ammonia (NH3) gas combines with oxygen gas over Pt catalyst to produce Nitric oxide (NO) and water. If

13.6 g of NH3 gas is taken initially, then :

(A*) Volume of oxygen gas required at NTP is 22.4 L

(B*) Volume of H2O() produced at 4° C (assuming density of water as 1000 Kg/m3) is 21.6 mL. (C*) Total mass of products obtained is 45.6 g. (D*) Number of moles of NO produced is 0.8. Pt mRiszjd dh mifLFkfr esa veksfu;k xSl (NH3), vkWDlhtu xSl (O2) ls tqM+dj ukbfVªd vkWDlkbM (NO) ,oa ty

mRikfnr djrh gSA ;fn izkjEHk esa 13.6 g NH3 xSl yh tk;s] rkss % (A*) NTP ij vkWDlhtu xSl ds 22.4 L vk;ru dh vko';drk gksxhA

(B*) 4°C rki ij mRiUu H2O() dk vk;ru 21.6 mL gSA ¼ekuk ty dk ?kuRo 1000 Kg/m3 gS½ (C*) izkIr mRiknksa dk dqy Hkkj 45.6 g gSA (D*) NO ds mRikfnr eksyksa dh la[;k 0.8 gSA

Page # 3

Sol. 4NH3 + 5O2 4NO + 6H2O

Volume of oxygen required at NTP = 13.617

54

22.4 = 22.4 L

Mass of water produced = 13.617

64

18 = 21.6 g

volume of H2O () = 21.6 mL ( dH2O = 1000 Kg/m3 = 1 g/mL)

moles of NO produced =13.617

=45

= 0.8

Mass of NO produced = 0.8 30 = 24 g Total mass of products = mNO +

2H Om = 24 + 21.6 = 45.6 g

Sol. 4NH3 + 5O2 4NO + 6H2O

NTP ij O2 dk vko';d vk;ru = 13.617

54

22.4 = 22.4 L

ty dk mRikfnr Hkkj = 13.617

64

18 = 21.6 g

H2O () dk vk;ru = 21.6 mL ( dH2O = 1000 Kg/m3 = 1 g/mL)

NO ds mRikfnr eksy = 13.617

=45

= 0.8

NO dk mRikfnr Hkkj = 0.8 30 = 24 g

mRiknksa dk dqy Hkkj = mNO + 2H Om = 24 + 21.6 = 45.6 g

4. Which is/are correct about the K+ ion

(A*) For the last electron n = 3, = 1 (B) Eight electrons have m = 0

(C*) Six electron have (n +) = 4 (D*) It's spin magnetic moment () is Zero. K+ vk;u ds ckjs esa lgh gS@gSa %

(A*) vfUre bysDVªkWu ds fy;s n = 3, = 1 (B) vkB bysDVªkWu m = 0 eku j[krs gSaA

(C*) N% bysDVªkWu (n +) = 3 eku j[krs gSa (D*) bldk pØ.k pqEcdh; vk?kw.kZ () 'kwU; gSA Sol. 19K

+ = 1s2, 2s22p6, 3s23p6 5. In which of the following arrangements, the order is correct according to the property indicated against it:

(A*) increasing size : Cu2+ < Cu+ < Cu (B) increasing EA1 (magnitude): Te < Se < O < S

(C*) increasing EA1 (magnitude) : < Br < F < Cl (D) increasing E1 : Li < Na < K < Rb

fuEufyf[kr O;oLFkkvksa esa ls] Øe ds vkxs fy[ks x;s xq.kksa ds vuqlkj lgh Øe esa gS@gSa% (A*) c<+rk gqvk vkdkj : Cu2+ < Cu+ < Cu (B) c<+rh gqbZ EA1 (ifjek.k) : Te < Se < O < S

(C*) c<+rh gqbZ EA1 (ifjek.k) : < Br < F < Cl (D) c<+rh gqbZ E1 : Li < Na < K < Rb Sol. Increasing EA1 (magnitude): O < Te < Se < S

c<+rh gqbZ EA1 (ifjek.k) : O < Te < Se < S 6. Which of the following has/have no unit ? (A*) Electronegativity (B) Electron gain enthalpy (C) Ionisation enthalpy (D*) Metallic character

\\server-1\Session 2015-16\JEE ADVANCED\CHEMISTRY\Sheets\Class-XI\Inorganic Chemistry\Periodic Table & Properties\With Answer\MCQ Q.13)

fuEufyf[kr esa ls fdldh@fdudh dksbZ bdkbZ ugha gksrh gS \ (A*) fo|qr_ .krk (B) bysDVªkWu xzg.k ,UFkSYih (C) vk;uu ,UFkSYih (D*) /kkfRod xq.k

Page # 4

7. Which of the following statements is/are true ?

(A*) Among Sc3+, Cl� & Ca2+, the Sc3+ ion is smallest. (B*) Among Al, Ar & Mg, the E1 is smallest for Al.

(C*) Among Na, Mg & Al, the E2 of Mg is the smallest.

(D) Oxidising power of halogens decreases with decrease in their atomic number. fuEu esa ls dkSulk dFku lR; gS@gSa ?

(A*) Sc3+, Cl o Ca2+ esa ls] Sc3+ vk;u lcls NksVk gksrk gSA (B*) Al, Ar o Mg esa ls] Al dk E1 lcls de gksrk gSA

(C*) Na, Mg o Al esa ls] Mg dk E2 lcls de gksrk gSA

(D) gSykstuksa dh vkWDlhdkjh {kerk] ijek.kq Øekad esa deh ds lkFk de gksrh tkrh gSA 8. 1.2 Kg of iron pyrites (FeS2) is roasted to convert sulphur in (FeS2) to SO2 with 80% efficiency. It is then

oxidised to SO3 which is then absorbed in water to give H2SO4. If % yield of each of 2nd reactions is either 60% or 30% then amount of H2SO4 can be :

1.2 Kg vk;ju ikbjkbV (FeS2) 80% n{krk ds lkFk HkftZr gksdj (FeS2) esa mifLFkr lYQj dks SO2 esa ifjofrZr djrs

gSaA ;g fQj SO3 esa vkWDlhd r gks tkrk gS tks ty esa vo'kksf"kr gksdj H2SO4 nsrk gSA ;fn izR;sd f}rh; vfHkfØ;kvksa dh

izfr'kr yfC/k ;k rks 60% ;k 30% gks] rks H2SO4 dh ek=kk fuEu gks ldrh gS % (A) 6.82 mol (B*) 1.44 mol (C) 3.62 mol (D*) 5.76 mol

Sol. FeS2 2SO2

1200120

16 mol (80%)

= 10 mol

SO2 60%or 30%

60% 30%

;k SO3

16 16 × 0.6 or 16 × 0.3

SO3 + H2O 60%or 30% 60% 30%

;k H2SO4

16 × 0.6 16 × 0.6 × 0.6 = 5.76 mol or or 16 × 0.3 16 × 0.3 × 0.3 = 1.44 mol

9. If value of n + + m = 4 then which options follow this : (A) No electron in F (B*) 3 or 4 electrons in Cl (C) 5 electrons in Ca (D*) 7 electrons in Mn

;fn n + + m = 4 gS rc dkSulk dFku lgh gSa \ (A) F esa dksbZ bysDVªkWu ugha gksrkA (B*) Cl esa 3 ;k 4 bysDVªkWu gksrs gSaA (C) Ca esa 5 bysDVªkWu gkssrs gSaA (D*) Mn esa 7 bysDVªkWu gksrs gSaA

Sol. n + + m = 4 4 0 0 4s 3 1 0 one 3p orbital 3 2 �1 one 3d orbital 2 1 1 one 2p orbital only these orbitals are satisfying given condition and now write the configuration of each and count the

electrons.

gy- n + + m = 4 4 0 0 4s 3 1 0 ,d 3p d{kd

3 2 �1 ,d 3d d{kd 2 1 1 ,d 2p d{kd

Page # 5

nh xbZ ifjfLFkfr ds vuqlkj dsoy mijksDr d{kd izkIr gksrs gSA vr% izR;sd dk bysDVªkWfu; foU;kl fy[kdj bysDVªkWuksa dks fxu ldrs gSA

10. 64g of sulphur and 128 g of Oxygen are made to combine according to the following unbalanced reaction : S + O2 SO3

Then, select the correct option(s) : (A*) Sulphur is the limiting reagent.

(B*) 2 moles of SO3 will be produced. (C*) 1 mole of excess reagent is left behind. (D*) Sum of number of moles of the reactants and products present finally is 3.

64g lYQj rFkk 128 g vkWDlhtu fuEu vlUrqfyr vfHkfØ;k ds vuqlkj tqM+rs gS %

S + O2 SO3

rc] lgh fodYi@fodYiksa dk p;u dhft,: (A*) lYQj lhekUr vfHkdeZd gSA

(B*) 2 eksy SO3 mRiUu gksxsaA

(C*) 1 eksy vkf/kD; vfHkdeZd 'ks"k cprk gSA (D*) vUr esa mifLFkr vfHkdkjd rFkk mRiknksa ds eksyksa dh la[;k dk ;ksx 3 gSA

Sol. 2S + 3O2 2SO3

Mole 6432

= 2 12832

= 4 = 0

(LR) Moles of O2 (ER) left = 4 � 3 = 1 no. of moles of SO3 produced = 2 no. of moles of O2 left + no. of moles of SO3 produced = 1 + 2 = 3

Sol. 2S + 3O2 2SO3

eksy 6432

= 2 12832

= 4 = 0

(LR) O2 (ER) ds 'ks"k eksy = 4 � 3 = 1

mRiUu SO3 ds eksyksa dh la[;k = 2 'ks"k O2 ds eksyksa dh la[;k + mRiUu SO3 ds eksyksa dh la[;k = 1 + 2 = 3 11. Which of the following will be same for Glucose and Acetic acid ? (A) Molar mass (B*) Empirical formula mass (C*) % of carbon by mass (D*) Empirical formula Xywdkst rFkk ,lhfVd vEy ds fy, fuEu esa ls dkSu leku gksxsa \ (A) eksyj nzO;eku (B*) ewykuqikrh lw=k nzO;eku

(C*) dkcZu dk % nzO;eku ds vuqlkj (D*) ewykuqikrh lw=k Sol. Refer notes. uksV~l nsa[ksA 12. Which of the following is/are correct order regarding radius ? fuEu esa ls dkSulk@dkSuls f=kT;k ds lEcU/k esa lgh Øe gS@gSa\

(A) Be+2 < B+3 < Li+ (B*) B+3 < Be+2 < Li+

(C*) 3�2�� NOF (D*) B < Ga < Al

Page # 6

Sol. Fact. rF; 13. Dissolving 120 g of urea (molar mass 60) in 990 g of water gave a solution of density 1.11 g/ml. Select the

correct statements: (A*) Molarity of solution is 2 M. (B*) Molality of solution is 2.02 m. (C) Molarity of solution is 1.78 M (D*) Mole fraction of urea is 0.035.

990 xzke ty esa ;wfj;k (eksyj nzO;eku 60) ds 120 xzke ?kksys tkrs gS] rks 1.11 g/ml ?kuRo okyk ,d foy;u izkIr gksrk gSA lgh dFkuksa dk pquko dhft,& (A*) foy;u dh eksyjrk 2 M gSA (B*) foy;u dh eksyyrk 2.02 m gSA

(C) foy;u dh eksyjrk 1.78 M gSA (D*) ;wfj;k dk eksy izHkkt 0.035 gSA Sol. Mass of solute = 120 g Mass of water = 990 g Mass of solution = 1110 g

Volume of Solution md

= 11101.11

mL = 1000 mL

Now Milli mole = M × Vin mL

12060

× 1000 = M 1110

1.11

M = 2 molality of solution

m = urea

M 10001000 d �M M

m = 2.02

mole fraction of urea =

12060

120 99060 18

= 2

2 55= 0.035

Sol. foys; dk nzO;eku = 120 g

ty dk nzO;eku = 990 g foy;u dk nzO;eku = 1110 xzke

foy;u dk vk;ru md

= 11101.11

mL = 1000 mL

vc] feyh eksy = M × VmL esa

12060

× 1000 = M 1110

1.11

M = 2 foy;u dh eksyyrk

m = M 1000

1000 d �M M

;wfj;k

m = 2.02

;wfj;k dk eksy izHkkt =

12060

120 99060 18

= 2

2 55

= 0.035

Page # 7

14. The process(es) requiring the absorption of energy is/are : fuEu esa ls dkSuls izØe@izØeksa esa Å tkZ vo'kksf"kr gksrh gS@gSa \

\\server-1\Session 2014-15\TEST PAPERS\JEE ADVANCED\VISHESH (JD)\2. 15-06-2014 (CT-1)\Chemistry\1. JD1 TO JD5 & JH1 to JH3\MCQ(52)

(A) Cl Cl� (B*) S S2� (C) H H� (D*) Ar Ar � Sol. (B, D) (B) Addition of second electron to S� is opposed by electrostatic repulsion due to same charge. Hence

energy is given for the addition of IInd electron. (D) Ar has stable electronic confiugration (ns2 np6). Hence energy has to be given to add an extra electron

to form Ar � gy& (B) S� esa f}rh; bysDVªkWu tksM+us ij leku vkos'k ds dkj.k fLFkj fo|qr izfrd"kZ.k gksrk gSA vr% f}rh; bysDVªkWu tksM+us

esa Å tkZ nh tkrh gSA (D) Ar esa LFkk;h bysDVªkWfud foU;kl (ns2 np6) gksrk gSA vr% vfrfjDr bysDVªkWu tksM+dj Ar � cukus esa Å tkZ nsuh iM+rh

gSA

15.

Which alkyl group is attached directly to the cyclohexane ring in this compound? (A) Isopropyl (B*) t-Butyl (C*) Isobutyl (D*) Neopentyl

mijksDr ;kSfxd esa dkSuls ,fYdy lewg lkbDyksgSDlsu oy; ls lh/ks tqM+s gq;s gSA

(A) vkblksizksfiy (B*) t-C;wfVy (C*) vkblksC;qfVy (D*) fu;ksisfUVy

Sol. In the given compound isopropyl group is not attached directely to the cyclohexane ring.

fn;s ;kSfxd esa vkblksizksfiy lewg lkbDyksgSDlsu oy; ls lh/ks tqM+k gqvk ugha gSA 16. Professor I.L. Finar instructed his student to convert the -Amino acid (X) into a new compound (Y) in

which the �NH2 group is replaced by �COOH group. The student was confused in writing the correct IUPAC names of these two compounds. The correct IUPAC names of X and Y will be respectively.

(X) =

(A) 2-Ethyl-2-amino ethanoic acid, 2-Ethyl-2-carboxy ethanoic acid (B*) 2-Aminobutanoic acid, Ethylpropanedioic acid (C) 2-Aminobutanoic acid, 2-Carboxybutanoic acid (D) 2-Carboxypropan-1-amine, Propane-1, 1-dicarboxylic acid.

izksQslj I.L. Finar us vius fo|kFkhZ dks ,d mnkgj.k fn;k ftlesa -vehuks vEy (X) dks ,d u;s ;kSfxd (Y) ftlesa

�NH2 lewg dks �COOH lewg }kjk izfrLFkkfir fd;k x;k Fkk] esa ifjofrZr djus ds fy, fn;kA bu nksuksa ;kSfxd ds

IUPAC uke fy[kus esa fo|kFkhZ dks my>u (confused) gqvkA X vkSj Y dk lgh IUPAC uke gksxkA

Page # 8

(X) =

(A) 2-,fFky -2-,ehuks ,FksukWbd vEy , 2-,fFky -2-dkcksZDlh ,FksukWbd vEy

(B*) 2-,ehuksC;wVsukWbd vEy , ,fFkyizksisuMkbZvkWbd vEy

(C) 2-,ehuksC;wVsukWbd vEy , 2-dkcksZDlhC;wVsukWbd vEy

(D) 2-dkcksZDlhizksisu-1-,sehu, izksisu-1, 1-MkbZdkWcksZfDlfyd vEy

Sol. X = Y =

2-Aminobutanoic acid Ethylpropanedioic acid

2-,ehuksC;wVsukWbd vEy ,fFkyizksisuMkbvkWbd vEy

17. How many of them are phenolic compounds (have phenolic group) ? (A*) Vanillin (B*)

Salicylaldehyde (C*) Glycerol (D*) Catechol

fuEu esa ls fdrus fQukWfyd ;kSfxd gS ¼fQukWfyd lewg j[krs gS½ \

(A*) osfuyhu (B*) lsfyflysfYMgkbM (C*) fXyljkWy (D*) dsVsdkWy

18. Aspirin is a pain relieving drug. It is an 2-Alkanoyloxybenzoic acid, having molecular formula (C9H8O4). Its

structure is :

,Lizhu ,d nnZ fuokjd vkS"k/kh gS] ;g 2-,YdsukW;yvkWDlhcsUtkWbd vEy gS ftldk v.kqlw=k (C9H8O4) gSA bldh lajpuk

gksxh %

(A) (B*)

(C) (D)

Sol.

Aspirin (2-Ethanoyloxybenzoic acid)

,Lizhu (2-,sFksukW;yvkWDlhcsUtkWbd vEy )

19. Which of the following IUPAC name is/are correct ?

(A*) Benzene carbonyl chloride

Page # 9

(B) 3-Cyano benzenamide

(C*) Phenyl ethanoate

(D*)

Ethanoic propanoic anhydride

fuEu esa ls dkSulk@dkSuls IUPAC uke lgh gSa &

(A*) csUthudkcksZfuyDyksjkbM (B) 3-lk;ukscsUthus,ekbM

(C*) Qsfuy ,sFksukW,V (D*)

,FksukWbd izksisukWbd ,ugkbMªkbM

Sol. 3-Cyano benzenecarboxamide ¼3-lk;uks csUthudkcksZDlsekbM½

20. Which of the following are functional isomers of methyl ethanoate ?

fuEu esa ls dkSuls esfFky ,FksukW,V ds fØ;kRed leko;oh gksaxs \

(A*) CH3�CH2�COOH (B*) CH|OH

CCH3

H

||O

(C*) CH � O � CH � C � H3 2

||O

(D*) CH � C � CH2 3

||O

|OH

Sol. All have different functional groups than ester.

lHkh esa ,LVj ls fHkUu fØ;kRed lewg gSA

21. Select the correct relationship.

(A*) and are fuctional isomers

Page # 10

(B*) and are chain isomers

(C) and are chain isomers

(D*) and are metamers

lgh lEcU/k dk p;u dhft, &

(A*)

OH

o

OH

fØ;kRed leko;oh gSaA

(B*) o Ja[kyk leko;oh gSaA

(C) o Ja[kyk leko;oh gSaA

(D*) o e/;ko;oh gSaA

Sol. (C) and are homologs.

(C) o letkr gSaA

22. Select the structures with correct numbering for IUPAC name of the compound.

fuEu ;kSfxdksa ds IUPAC ukedj.k ds fy;s lgh Øekadu okyh lajpukvksa dk p;u dhft;s &

(A) (B*) (C) (D*)

Sol.

&

represent correct numbering

Sol.

rFkk

esa lgh Øekadu fd;k x;k gSA

Integer type (Single) 18

Page # 11

23. Find the weight of a sample(in g) which contains 0.5 mole of Ne gas and about 1.505 ×1023 molecules of

an unknown gas X. If the sample has average molar mass 100 g/mole. Report your answer dividing the

weight of sample by 15.

,d izkn'kZ dk Hkkj (xzke esa) Kkr dhft,] ftlesa Ne xSl ds 0.5 eksy rFkk ,d vKkr xSl X ds yxHkx 1.505 ×1023

v.kq gSA ;fn izkn'kZ dk vkSlr eksyj nzO;eku 100 g/mol gS rks viuk mÙkj] izkn'kZ ds Hkkj dks 15 ls Hkkx nsus ds i'pkr~

nhft,A

Ans. 5

Sol. Average Molar mass = total weighttotal moles

100 = 23

23

total weight

1.505 100.5

6.023 10

= total weight0.5 0.25

100 = total weight

0.75 total weight = 100 × 0.75 = 75

1575

= 5

vkSlr eksyj nzO;eku = eksydqy Hkkjdqy

100 =

23

23

10023.6

10505.15.0

Hkkjdqy =

25.05.0

Hkkjdqy

100 = 75.0

Hkkjdqy dqy Hkkj = 100 × 0.75 = 75

1575

= 5

24. Mole fraction of solute in solution is 0.25. What is the moles of solvent in solution having 1 mole of solute?

,d foy;u esa foys; dh eksy fHkUu 0.25 gSA 1 eksy foys; j[kus okys foy;u esa foyk;d ds eksyksa dh la[;k fdruh gksxh \

Ans. 3

Sol. nsolvent = 0.750.25

× 1 = 3

nfoyk;d = 0.750.25

× 1 = 3

25. 27 Kg of SO2Cl2 is reacted with excess of NaOH completely. If the difference between masses of Na2SO4

and NaCl produced is x Kg, then x is : (SO2Cl2 + NaOH Na2SO4 + NaCl + H2O)

27 Kg SO2Cl2, NaOH ds vkf/kD; ds lkFk iw.kZr% vfHkfØ;k djrk gSA ;fn mRIkkfnr Na2SO4 o NaCl ds nzO;ekuksa ds e/; vUrj x Kg gS, rc x dk eku Kkr dhft, :

(SO2Cl2 + NaOH Na2SO4 + NaCl + H2O)

Page # 12

Ans. 5 Kg Sol. SO2Cl2 + 4NaOH Na2SO4 + 2NaCl + 2H2O

moles ¼eksy½ = 27000135

=200 200 moles ¼eksy½ 400 moles ¼eksy½

m = 200 × 142 g m = 400 × 58.5 g = 28.400 Kg = 23.4 Kg x = 28.4 � 23.4 = 5 Kg. 26. If H2SO4 is formed from it�s elements by taking 6.023 × 1023 atom of �O� 5.6 litre of H2 gas at STP and 8 g S.

. How many of the following conclusions are correct based on this information ?

(i) 0.125 moles of H2SO4 are formed (ii) 0.25 moles of H2SO4 are formed (iii) no moles of �S� are left (iv) 1/4 mole of O2 is left (v) 1/2 mole of H2 is left ;fn H2SO4 vius la?kVd rRoksa 'O' ds 6.023 × 1023 ijek.kq] STP ij H2 xSl ds 5.6 yhVj vkSj S ds 8 xzke ls cuk gS]

rks fuEu esa ls fdrus fu"d"kZ bl lwpuk ds vk/kkj ij lgh gS \ (i) H2SO4 ds 0.125 eksy curs gSaA (ii) H2SO4 ds 0.25 eksy curs gSaA (iii) �S� ds dksbZ eksy 'ks"k ugha jgrs gSA (iv) O2 ds 1/4 eksy 'ks"k jgrs gSaA (v) gkbMªkstu ds 1/2 eksy 'ks"k jgrs gSA Ans. 2 Sol. (ii + iii)

H2 + S + 2O2 H2SO4

2Hn =

5.6

22.4 =

14

ns = 832

= 14

2On =

12

As all reactants are in stoichiometric ratios, none will be left behind. Hence 1

4 mole of H2SO4 is formed.

gy- H2 + S + 2O2 H2SO4

2Hn =

5.6

22.4 =

14

ns = 832

= 14

2On =

12

tSlk fd lHkh vfHkdkjd jllehdj.kferh; vuqikr esa gS] vr% dksbZ Hkh 'ks"k ugha cpsxkA vr% 1

4 eksy H2SO4 cusxkA

27. An element 'E' has exceptional valence shell electron configuration as given 4d10. What is the period of 'E'

in the modern periodic table ? ,d rRo 'E' ds la;ksth dks'k dk bysDVªkWfud vfHkfoU;kl vioknLo:i 4d10 gSA vk/kqfud vkorZ lkj.kh esa 'E' dk vkorZ

D;k gksxk?

Ans. 5. Sol. Expected electron configuration should be [Kr] 4d10 5s0. The principal quantum number (n) of valence shell

is thus equal to 5. So the period of the element to which it belongs is 5. lEHkkfor bysDVªkWfud vfHkfoU;kl [Kr] 4d10 5s0 gksuk pkfg,A vr% la;ksth dks'k dh eq[; Dok.Ve la[;k (n) dk eku 5 ds

cjkcj gSA blfy, bl rRo dk vkorZ 5 gksxkA

Page # 13

28. What is the screening constant for outer electron of H?

H ds ckg~; bysDVªkWu ds fy, ifjj{k.k fu;rkad D;k gS\

Ans. 0 29. Determine the sum of fully filled and half filled orbitals, which have number of radial nodes less than the

number of angular nodes, in ground state of Fe3+ ion. Fe3+ vk;u dh ewy voLFkk esa v)Z iwfjr o iw.kZ iwfjr d{kdks dh la[;k dk ;ksx Kkr dhft;s ftuds fy;s f=kT;h; uksMks

dh la[;k dks.kh; uksMks dh la[;k ls de gksA Ans. 8 Sol. Fe3+ : 1s2 2s2 2p6 3s2 3p6 3d5

(R=A=0) (R=1,A=0) (R=0,A=1) (R=2,A=0) (R=1,A=1) (R=0,A=2) So, condition in the question corresponds to orbitals of 2p and 3d subshells. Total number of half filled and fully filled orbitals = 3 + 5 = 8. vr% iz'u esa nh x;h fLFkfr 2p o 4p midks'kksa ds laxr gSA

vr% v)Ziwfjr o iw.kZiwfjr d{kdks dh dqy la[;k = 3 + 5 = 8. 30. XeF6 fluorinates 2 to F7 and liberates Xenon(g). 3.5 mmol of XeF6 can yield a maximum of_____ mmol of

IF7. XeF6, 2 dks F7 esa ¶yksjhuhd r dj nsrk gS rFkk thukWu ¼xSl½ eqDr gksrh gSA XeF6 ds 3.5 feyh eksy ls IF7 ds vf/kdre

------------------ feyh eksy izkIr fd;s tk ldrs gSaA Ans. 3

Sol. XeF6 + I2 IF7 + Xe POAC on �F� : 6 (m.mole of XeF6) = 7 (m.mole of IF7)

3.5 6

7

= 3 m.moles of IF7

gy % XeF6 + I2 IF7 + Xe

�F� ijek.kq ij POAC yxkus ij : 6 (XeF6 ds feyh eksy ) = 7 (IF7 ds feyh eksy )

3.5 6

7

= IF7 ds 3 feyh eksy

31. Balance the following equation and write the number which is the sum of the coefficients of all species.

................. CS2 + ................. Cl2 CCl4 + ................. S2Cl2

fuEu lehdj.k dks lUrqfyr dhft, o og la[;k fyf[k, tks Lih'kht ds xq.kkadks dk ;ksx gks &

................. CS2 + ................. Cl2 CCl4 + ................. S2Cl2 Ans. 6

Sol. CS2 + 3Cl2 CCl4 + S2Cl2

1 + 3 + 1 + 1 = 6 32. 200 g impure CaCO3 on heating gives 5.6 litre CO2 gas at STP. Find the percentage of calcium in the lime

stone sample. STP ij v'kq) CaCO3 ds 200 g dks xeZ djus ij ;g CO2 xSl ds 5.6 yhVj nsrk gSA pwus ds iRFkj ds uewus esa

dSfY'k;e dk izfr'kr gksxkA

Page # 14

Ans. 5

Sol. CaCO3 CaO + CO2

5.622.4

= 14

mole

mole of CaO = mole of Ca = 14

mass of Ca = 14

× 40 = 10

% of Ca in sample = 10200

× 100 = 5%

gy . CaCO3 CaO + CO2

5.622.4

= 14

eksy

CaO ds eksy = Ca ds eksy = 14

Ca ds eksy =14

× 40 = 10

uewus esa Ca dk izfr'kr = 10200

× 100 = 5%

33. Total number of elements which have more ionization energy as compared to their next higher atomic

number elements. Li, Be, C, N, O, F, Ne [Reff. SGF_2015] Li, Be, C, N, O, F, Ne fuEu esa ls ,sls rRoksa dh la[;k crkb;s ftudh vk;uu Å tkZ vxys ijek.kq Øekad okys rRo ls

vf/kd gS \

Ans. 3 Sol. Be, N, Ne 34. How many parent chains have root word hex or hexa in the following structures ?

fuEu lajpukvksa esa fdruh tud Ja[kykvksa essa ewy'kCn gsDl ;k gsDlk iz;qDr gksrk gS \

, , ,

, , , Ans. 3 Sol. (iii), (vi), (viii) 35. What is the possible number of isomers of the aromatic compounds of molecular formula C7H7Cl.

C7H7Cl v.kqlw=k okys ,jkseSfVd ;kSfxd ds lHkh lEHko leko;fo;ksa dh la[;k crkb;sA Ans. 4

Sol.

Page # 15

36. How many structural isomers are possible having molecular formula C6H14. v.kqlw=k C6H14 ds fdrus lajpukRed leko;oh lEHko gSA Ans. 5

Sol. , , , ,

37. The number of structural isomers for the compound with molecular formula C2BrClFI. v.kqlw=k C2BrClFI ds dqy lajpukRed leko;fo;ksa dh la[;k gksxhA Ans. 3

Sol. F

C=C Cl

Br

;

F C=C

Br

Cl

;

FC=C

Br

Cl

38. How many position isomer are possible for Dibromobutane. MkbZczkseksC;wVsu ds fy, fdrus fLFkfr leko;oh lEHko gS \ Ans. 6

Sol. , ,

, ,

39. How many carbon atoms are present in parent chain of the following compound ? fuEu ;kSfxd dh tud Ja[kyk esa fdrus dkcZu ijek.kq mifLFkr gS \

Ans. 6

Sol.

40. Find the total number of cyclic structural isomers having molecular formula C5H10. v.kqlw=k C5H10 ds dqy pfØ; lajpukRed leko;fo;ksa dh la[;k Kkr dhft,A Ans. 5

Sol. , , ,

,




TEST PATTERN

COURSE NAME : VIJAY (JR)

TEST TYPE : PT-1 (JEE ADV. PATTERN)

TEST DATE : 16-08-2015

TEST SYLLABUS : FUNDAMENTALS OF MATHEMATICS, QUADRATIC EQUATION (UPTO

NATURE OF ROOTS)


1 to 22 MCQ 22 4 �2 88


40 160


Total Total

Maths/ Physics/

Chemistry

MCQ

1. Possible value(s) of �� for which the equation x2 � (sin � 2)x � (1 + sin) = 0 has roots whose

sum of squares is least is

�� ds laHkkfor eku ftlds fy, lehdj.k x2 � (sin � 2)x � (1 + sin) = 0 ewy j[krk gS vkSj mu ewyksa ds

oxksZ dk ;ksxQy U;wure gS&

(A) 4

(B) 3

(C*) 2

(D*) 52

Sol. (C), (D)

2 +2 = (sin � 2)2 + 2(1 + sin)

= sin2

+ 4 � 4sin + 2 + 2sin

= sin2 � 2sin + 6 = (sin � 1)2 + 5

= 2

, 52

2. The equation |x + 2| � |x + 1| + |x � 1| = K, x R has two solutions if K is equal to

lehdj.k |x + 2| � |x + 1| + |x � 1| = K, x R ds nks gy j[krk gS ;fn K cjkcj gS&

(A*) 4 (B*) 3/2 (C) 5/2 (D) 3 Sol. Graph of y = |x + 2| � |x + 1| + |x � 1|

y = |x + 2| � |x + 1| + |x � 1| dk vkjs[k






for two solutions nks gy ds fy, K (1, 2) (3, )

3. Solution set of the inequality 22cos x � 1

0x

sin2

, can be

(A*) 2n � 1 2n 1

x4 4

, where n = 1, 3, 6

(B*) n n 1

x4 4

, where n = 8

(C*) n � 1 n

x4 4

, where n = 16

(D) 2n � 1 2n 1

x4 4

, where n = 5

vlfedk 22cos x � 1

0x

sin2

dk gy leqPp; gks ldrk gS&

(A*) 2n � 1 2n 1

x4 4

, tgk¡ n = 1, 3, 6

(B*) n n 1

x4 4

, tgk¡ n = 8

(C*) n � 1 n

x4 4

, tgk¡ n = 16

(D) 2n � 1 2n 1

x4 4

, tgk¡ n = 5

Sol. Case fLFkfr-1 �1 cosx < �1

2

1

2< cosx 1 and vkSj sin

x2

< 0

x (2, 4)

2< x <94

114

< x < 13

4

15

4

< x < 4

CasefLFkfr -2 : � 1

2 < cosx <

1

2 and vkSj sin

x2

> 0

sinx2

> 0 x2

(0, )






x (0, 2)

4

< x < 34

54

< x < 74

Alter : oSdfYid

cos2x

0x

sin2

Let us consider for x (0, 4)

cos2x

0x

sin2

ekukfd x (0, 4) ds fy,

CasefLFkfr-I sin x2

< 0 , cos 2x > 0

sin x2

< 0 x2

(2) x(2, 4)

2x (4, 8)

graph of cos 2x is dk vkjs[k

cos 2x > 0 4< 2x 92

112

< 2x < 13

2

15

2

< 2x < 8

2< x <94

114

< x < 13

4

15

4

< x < 4

CasefLFkfr-II sinx2

> 0, cos 2x < 0

sinx2

> 0 x2

(0, ) x (0, 2)

2x (0, 4)

cos 2x < 0

2

< 2x < 32

52

< 2x < 72

4

< x < 34

54

< x < 74






4. If |sin2x + 17 � x2| = |16 � x2| + 2sin2x + cos2x then subsets of solution are

;fn |sin2x + 17 � x2| = |16 � x2| + 2sin2x + cos2x rc gy dk mileqPp; gS&

(A*) {0} (B*) [� 4, 4] (C) [� 8, 8] (D) 17, 17

Sol. |sin2x + 17 � x2| = |16 � x2| + |sin2x + 1|

|x + y| = |x| + |y|

x y 0

(16 � x2) (sin2x + 1) 0

x [� 4, 4]

5. Let f(x) = cos ([2] x) + cos( [�2 ] x) + [e] � [� ] , where [ . ] is the greatest integer function, then

ekuk f(x) = cos ([2 ] x) + cos( [�2] x) + [e] � [� ], tgk¡ [ . ] egÙke iw.kk±d Qyu gS] rc

(A*) f2

= 5 (B) f() = 7 (C*) f(�) = 6 (D) f4

= 8

Sol. f(x) = cos 9x + cos 10x + 2 � (� 4).

6. If a, b respectively be the numbers of solutions and sum of solutions of 2x

x 1� |x| =

2x| x 1|

, then

;fn a vkSj b Øe'k% lehdj.k 2x

x 1� |x| =

2x| x 1|

ds gyksa dh la[;k vkSj gyksa dk ;ksxQy gS] rc

(A*) a = 3 (B*) b = 1 (C) b = 2 (D) a = 2

Sol. 2 | x |

| x 1|� |x| =

2| x || x 1|

2 � |x � 1| = |x|, x = 0 I. x < 0 2 + x � 1 = � x x = �1/2 II. 0 x < 1 2 + x � 1 = x 2 1 III. x > 1 2 � x + 1 = x x = 3/2 7. Let a and c be odd prime numbers and b be an integer. If the quadratic equation

ax2 + bx + c = 0 has rational roots, then roots of equation are

ekuk a vkSj c fo"ke vHkkT; la[;k,a gS rFkk b ,d iw.kk±d gSA ;fn f}?kkr lehdj.k ax2 + bx + c = 0 ds ewy

ifjes; gS] rc lehdj.k ds ewy gS&

(A) 1 (B*) �1 (C*) c

�a

(D) ca

Sol. Since ax2 + bx + c = 0 has rational roots

pawfd ax2 + bx + c = 0 ifjes; ewy gSA

b2 � 4ac is square of an integer

b2 � 4ac ,d iw.kk±d dk oxZ gSA i.e. b2 � 4ac = k2 for some integer k

vFkkZr~ b2 � 4ac = k2 fdlh iw.kk±d k ds fy,

b2 � k2 = 4ac (b � k) (b + k) = 4ac ... (A) Since b and k are integer, therefore, (b � k) and (b + k) are either both even.






Since b and k are integer, therefore, (b � k) and (b + k) are either both even. since 4ac is even. b � k = 2a and b + k = 2c or { b � k = 2c and b + k = 2a} on adding, 2b = 2a + 2c i.e. a � b + c = 0 x = � 1 is a root of ax2 + bx + c = 0

Product of roots = ca

� 1 × other root = ca

other root = � ca

Roots are � 1 & c

�a

Hindi pawfd ax2 + bx + c = 0 ifjes; ewy gSA

b2 � 4ac ,d iw.kk±d dk oxZ gSA

vFkkZr~ b2 � 4ac = k2 fdlh iw.kk±d k ds fy,

b2 � k2 = 4ac (b � k) (b + k) = 4ac ... (A)

pwfd b vkSj k iw.kkZad gS blfy, (b � k) vkSj (b + k) ;k nksuksa le gS

pwfd 4ac le gSA

b � k = 2a vkSj b + k = 2c or { b � k = 2c vkSjb + k = 2a}

2b = 2a + 2c tksMus ij

vFkkZr~ a � b + c = 0

x = � 1 lehdj.k ax2 + bx + c = 0 ds ewy gSA

ewyks dk xq.kuQy = ca

� 1 × vU; ewy = ca

vU; ewy = � ca

ewy � 1 vkSj c

�a

8. The equation 2xlog 16 + log2 x64 = 3 has :

(A*) one irrational solution (B*) no prime solution (C*) two real solutions (D*) one integral solution lehdj.k 2x

log 16 + log2 x64 = 3 ds fy;s &

(A) ,d vifjes; gy fo|eku gSaA (B) dksbZ vHkkT; gy fo|eku ughaA (C) nks okLrfod gy fo|eku gSaA (D) ,d iw.kk±d gy fo|eku gSaA Sol. 2x

log 16 + log2x 64 = 3 4 2xlog 2 + 6 log2x 2 = 3

2

2

4

log x +

2

6log 2x

= 3 2

2log x

+ 2

61 log x

= 3

but ysfdu log2 x = t

2t

+ 6

1 t = 3 2 + 2t + 6t = 3t + 3t2

3t2 � 5t � 2 = 0 3t2 � 5t � 2 = 0 (3t + 1)(t � 2) = 0






t = � 13

, t = 2

log2 x = � 13

log2 x = 2

x = 2�1/3 x = 4

= 1/ 3

1

2.

9. The domain of f(x) = x

1[log 2]

contains the set (where [] is greatest integer function)

f(x) = x

1[log 2]

dk izkUr leqPp; dks j[krk gS (tgk¡ [] egÙke iw.kkZad Qyu gS)

(A) (1, 3) (B*) (0,1) (C*) (1, 2] (D) [2, )

Sol. f(x) = x

1[log 2]

for logx 2 to be defined, x > 0, x 1

for f(x) to be defined [logx 2] 0

logx 2 [0, 1)

logx2 < 0 or logx2 1

Hence domain is (0, 1) (1, 2]

Hindi f(x) = x

1[log 2]

logx 2 ds fy, ifjHkkf"kr, x > 0, x 1

f(x) dks ifjHkkf"kr ds fy, [logx 2] 0

logx 2 [0, 1)

logx2 < 0 or logx2 1

vr% izkUr (0, 1) (1, 2] gSA

10. The set of all solution of the inequality

2x �3x13

> 9 is contained by the set

vlfedk 2x �3x

13

> 9 ds lHkh gyksa dk leqPp;] leqPp; dks j[krk gS -

(A) (�, 0) (B*) (0, 2) (C) (0, 1) (D*) 1

, 32

Sol. x2 � 3x < � 2

x2 � 3x + 2 < 0

(x � 2)(x � 1) < 0

x (1, 2)






11. Solution set of the inequality 7 � x 2x � 3 , is contained by the set

vlfedk 7 � x 2x � 3 dk gy leqPp;] fdl leqPp; j[krk gS -

(A) 11 89

,8

(B*) (� ,7) (C*) 11 89

� ,8

(D*) (� ,3)

Sol. 7 � x 0 x 7 ........(i)

case fLFkfr -1 2x�3 > 0 x > 32

7 � x > 4x2 + 9 � 12x

4x2 � 11 x + 2 < 0

x 11� 89 11 89

,8 8

x 3 11 89

,2 8

....(ii)

case fLFkfr -2 2x � 3 0 x 32

positive > negative

/kukRed > _ .kkRed

x 3

� ,2

....(iii)

(i) (iii) 3

� ,2

....(iv)

(ii) (iv)

x 11 89

� ,8

12. Let A = Minimum (x2 � 2x + 7), x R and B = Minimum (x2 � 2x + 7), x [2, ), then : (A*) log(B�A)(A + B) is not defined (B*) A + B = 13 (C*) log(2B�A) A < 1 (D*) log(2A�B) A > 1

ekuk A = U;wure (x2 � 2x + 7), x R rFkk B = U;wure (x2 � 2x + 7), x [2, ) gks] rks (A*) log(B�A)(A + B) ifjHkkf"kr ugha gSA (B*) A + B = 13 (C*) log(2B�A) A < 1 (D*) log(2A�B) A > 1

13. Consider the equation log5 [(2 + 5 )x + ( 5 � 2 )x ] = 12

� log1/52. Which of the following is TRUE?

(A*) The equation has integral roots only

(B*) The absolute value of the difference of the roots of the equation is 2.

(C) The sum of the roots of the equation is 2.

(D*) The roots are equal in magnitude.






ekuk fd lehdj.k log5 [(2 + 5 )x + ( 5 � 2 )x ] = 12

� log1/52 gS] rc fuEu esa ls dkSuls lR; gS ?

(A*) lehdj.k dsoy iw.kkZad ewy j[krk gSA

(B*) lehdj.k ds ewyksa ds vUrj dk fujis{k eku 2 gSA

(C) lehdj.k ds ewyksa dk ;ksxQy 2 gSA

(D*) ekikad esa ewy cjkcj gSA

Sol. 1

5 25 2

log5 x x

5 2 5 2

=

12

+ log52

x x( 5 2) ( 5 2)

2

= 5

Let ekuk x

5 2 t ..........(i)

t + 1t

= 2 5

t2 � 2 5 t + 1 = 0 t = 5 2 , 5 2

Putting these values of t in (i) we get

lehdj.k (i) esa t dk eku j[kus ij

x = 1, � 1

14. The solution set of the system of inequations

2 sin2x � 3 sinx + 1 0 and x2 + x � 12 0 has

(A*) three integers (B*) one prime numbers

(C*) two natural numbers (D*) no composite number

vlfedk 2 sin2x � 3 sinx + 1 0 vkSj x2 + x � 12 0 ds fudk; dk gy leqPp; j[krk gS �

(A*) rhu iw.kkZad (B*) ,d vHkkT; la[;k,a

(C*) nks izkd r la[;k,a (D*) dksbZ la;qDr la[;k ugh

Sol. (A,B,C,D)

2 sin2x � 3 sinx + 1 0

(2sinx � 1)(sinx � 1) 0






12 sinx 1

2n + 6

x 2n + 56

, n ....(A)

x2 + x � 12 0 (x + 4)(x � 3) 0 x [� 4, 3] .........(B)

(A) and vkSj (B)

x �7

�4,6

5

,6 6

Integers are � 4, 1, 2 iw.kkZad � 4, 1, 2 gSA 15. If the equation tan |x| = |tanx| , where x (�2, 2), then

(A) solution is 3

�2 , � � , 0 ,2 2 2

(B*) solution is 3

� , �2

� ,2 2

3

,2

(C) number of natural number in solution of given equation is one (D*) number of natural number in solution of given equation is two

;fn lehdj.k tan |x| = |tanx| tgk¡ x (�2, 2) rc -

(A) 3

�2 , � � , 0 ,2 2 2

gy gSA

(B*) 3

� , �2

� ,2 2

3

,2

gy gSA

(C) nh xbZ lehdj.k ds gyksa esa izkd r la[;kvksa dh la[;k 1 gSA

(D*) nh xbZ lehdj.k ds gyksa esa izkd r la[;kvksa dh la[;k 2 gSA Sol. tan |x| = |tanx|

Case fLFkfr-I x 0 and vkSj tanx 0 n x < n2

(n w) ...........(A)

Case fLFkfr-II x 0 and vkSj tanx < 0 tanx = � tanx tanx = 0 x

Case fLFkfr-III x < 0 and vkSj tanx 0 � tanx = tanx

tanx = 0 x = n, n I � ...........(B)

Case fLFkfr-IV x < 0 and vkSj tanx < 0

� tanx = � tanx n �2

< x < n n I � {0} ...........(C)

(A) (B) (C) gives ls

x �(2n 1) , � n2

n , (2n 1)2

, n N

In the interval (�2, 2)

(�2, 2) vUrjky esa

x 3

� , �2

3

,2






16. Which of the following equations have no real solutions ?

(A*) x2 � x + 6 + ex = 0

(B) x4 � 2x2 sin2 x2

+ 1 = 0

(C*) log{x}12

= � 2 (where {.} denotes fractional part function

(D) cosx = |x|

fuEu esa ls dkSulh lehdj.k okLrfod gy ugh j[krh gS ?

(A*) x2 � x + 6 + ex = 0

(B) x4 � 2x2 sin2 x2

+ 1 = 0

(C*) log{x}12

= � 2 (tgk¡ {.} fHkUukRed Hkkx Qyu dks O;Dr djrk gSA)

(D) cosx = |x|

Sol. (A) ex = � x2 + x � 6

ex > 0 and � x2 + x � 6 < 0 x R (A) has no solution dksbZ gy ugh j[krk gSA

(B) 2 sin2 x2

= x2 + 2

1x

The equality holds when x = 1, �1

valfedk larq"B gksrh gS tc x = 1, �1

(C) 0 < {x} < 1

log{x}12

> 0 (C) has no real solution. dksbZ okLrfod gy ugha j[krh gSA

(D) cosx = |x|

graph of cosx and |x| interesect at two points

cosx vkSj |x| dk vkjs[k nks fcUnqvksa ij izfrPNsn djrk gS

17. Solution(s) of the equation, 3x2 � 2x3 = log2 (x2 + 1) � log2x is/are

lehdj.k 3x2 � 2x3 = log2 (x2 + 1) � log2x ds gy gS -

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

Sol. Here ;gk¡ x > 0

log2

2x 1x

= 3x2 � 2x3

2x 1x

= 2 33x �2x2

x + 1x

= 2 33x �2x2 21 (

1x

x 2 if x > 0)

3x2 � 2x3 1

or ;k 2x3 � 3x2 + 1 0

or ;k (x � 1)2 (2x + 1) 0

x 1

� , �2

{1}






But ijUrq x > 0

x = 1

18. Integral values of 'a' such that quadratic equation x2 + ax + a + 1 = 0 has integral roots is equal to

'a' ds iw.kkZad eku gksxsa tcfd f}?kkr lehdj.k x2 + ax + a + 1 = 0 iw.kkZad ewy j[krk gS] cjkcj gS -

(A*) �1 (B) 2 (C) 1 (D*) 5

Sol. a I and vkSj D = a2 � 4 (a + 1) = (a � 2)2 �8 = 2 (let I)

(a�2)2 = 8 + 2 = ± 1

(a � 2)2 = 9

a = 5, � 1

19. Which of the following is/are correct ?

(A*) The graph of y = 2

2

| log x |log x

is

(B) The graph of y = 2

2

| log x |log x

is

(C*) The graph of y = |log x| is

(D*) The graph of y = | x |x

is

�1

1

O x

y

fuEu esa ls dkSulk lgh gS ?

(A*) y = 2

2

| log x |log x

dk vkjs[k gSA

(B) y = 2

2

| log x |log x

dk vkjs[k gSA






(C*) y = |log x| dk vkjs[k gSA

(D*) y = | x |x

dk vkjs[k

�1

1

O x

y

gSA

Sol. y = 2

2

| log x |log x

= 2

2

1 log x 0 x 1

1 log x 0 0 x 1

20. If 4 cos2 + 3 = 2 ( 3 + 1) cos , then =

;fn 4 cos2 + 3 = 2 ( 3 + 1) cos , rc =

(A) 2n,n (B) 2n ± /4 ,n

(C*) 2n ± /6 ,n (D*) 2n ± /3 ,n

Sol. 4 cos2 + 3 = 2 ( 3 + 1) cos

4 cos2 � 2 cos = 2 3 cos � 3

2 cos (2 cos � 1) = 3 (2 cos � 1)

cos = 3

2 or ;k cos =

12

= 2n ± /6, n or ;k 2n ± /3, n

21. Which of the following is/are CORRECT ?

fuEu esa ls dkSulk lgh gS?

(A*) sin 38

= 2 2

2

(B*) cos 25

= 5 � 1

4

(C*) tan 98

= 2 �1 (D*) tan 512

= 2 3

Sol. Standard results ekud ifj.kke

22. The graph of y = f(x) is as shown

Which of the following is/are CORRECT ?






(A*) The graph of y = f(�|x|) is

(B) The graph of y = f(�|x|) is

(C*) The graph of y = f(|x|) is

(D) The graph of y = f(|x|) is

y = f(x) dk vkjs[k n'kkZ;k gS -

fuEu esa dkSulk lgh gS ?

(A*) y = f(�|x|) dk vkjs[k s

gS

(B) y = f(�|x|) dk vkjs[k

gSA

(C*) y = f(|x|) dk vkjs[k gSA






(D) y = f(|x|) dk vkjs[k

gSA

Sol.

Integer Que. 23. Solution of the inequation � 2 ||x| + 2| < 5 is (�a, b). Find the value of a + b

vlfedk � 2 ||x| + 2| < 5 dk gy (�a, b) gS rks a + b dk eku Kkr dhft,A

Ans. 6

Sol. |x| + 2 is always positive

|x| + 2 lnSo /kukRed gSA

0 |x| + 2 < 5

or ;k �2 |x| < 3

�3 < x < 3

24. Complete solution set of the inequality |x2 � x � 2| + |x + 1| 0 contain 'n' integer(s). Then 'n' is : vlfedk |x2 � x � 2| + |x + 1| 0 lEiw.kZ gy leqPp; esa 'n' iw.kkZad gS rc 'n' gS - Ans. 1

Sol. |x2 � x � 2| + |x + 1| 0 |x2 � x � 2| + |x + 1| = 0 It is possible only when ;g dsoy laHko gS tc

x2 � x � 2 = 0 and vkSj x + 1 = 0

x = � 1

25. If 2

5

5

log (x � 5x 7)

log (0.001)

> 0 , then x (a, b). Find 2b � 3a.

;fn 2

5

5

log (x � 5x 7)

log (0.001)

> 0 rc x (a, b) gks] rks 2b � 3a dk eku gS&

Ans. 0

Sol. log5(0.001) < 0

log5 (x2 � 5x + 7) < 0 and vkSj x2 � 5x + 7 > 0 x R

and vkSj x2 � 5x + 7 < 1

or;k x2 � 5x + 6 < 0 x (2, 3)

26. The polynomials P(x) = 3kx2 + 2x + 1 and Q(x) = 2x2 � k when divided by x � 1

leaves the same remainder, then the value of k is� 1m

. Find m.

cgqin P(x) = 3kx2 + 2x + 1 vkSj Q(x) = 2x2 � k dks tc x � 1 ls foHkkftr djrs gS rc 'ks"kQy

leku jgrk gS rc k dk eku �1m

gS rc m dk eku Kkr dhft,A






Ans. 4 Sol. P(1) = 3k + 2 + 1 = Q(1) = 2 � k 4k = � 1

k = � 14

27. The value of 28 cos 15

cos 215

cos 315

cos 415

cos 515

cos 615

cos 715

is

28 cos 15

cos 215

cos 315

cos 415

cos 515

cos 615

cos 715

dk eku gS -

Ans. 2

Sol. E = � 28 (cos 15

cos 215

cos 415

cos 815

) (cos 315

cos515

cos 615

)

= �28

16sin

15

16sin15

×

12

× 5 �1 5 1

4 4

= 2

28. Solution of inequality [x]2 � 9[x] � 52 < 0 is [a, b). Find b + 2a (where [.] denotes greatest integer

function)

vlfedk [x]2 � 9[x] � 52 < 0 dk gy [a, b) gS rc b + 2a dk eku Kkr dhft, (tgk¡ [.] egÙke iw.kk±d Qyu

gS) Ans. 7 Sol. [x]2 � 9[x] � 52 < 0 ([x] � 13)([x] + 4) < 0 � 4 < [x] < 13 � 3 [x] 12 x [�3, 13) 29. Number of values of x satisfying |x � 3| + |x + 3| = |2x| + 6 is/are.

lehdj.k |x � 3| + |x + 3| = |2x| + 6 dks larq"B djus okys x ds ekuksa dh la[;k gS - Ans. 1 Sol.

30. If 2

2x 1x 12x 5x 2

, then complete solution set is (a, �1) (b, c) . Find |a + 3b + 2c|.

;fn 2

2x 1x 12x 5x 2

gks rks lEiw.kZ gy leqPp; (a, �1) (b, c) gS rc |a + 3b + 2c| Kkr dhft,A

Ans. 5

Sol. Given tgk¡ 2

2x

2x 5x 2 >

1x 1

2x

(2x 1)(x 2) >

1(x 1)






2x

(2x 1)(x 2) �

1(x 1)

> 0

2x(x 1) (2x 1)(x 2)

(x 1)(2x 1)(x 2)

> 0

2 22x 2x 2x 4x x 2

(x 1) (x 2) (2x 1)

> 0

3x 2

(x 1)(x 2)(2x 1)

> 0

3x 2

0(x 1)(x 2)(2x 1)

x (�2, �1) 2 1

� , �3 2

31. If log1/3 3x �1

x 2

< 1 then x must lie in the interval (�, �a)

5,

. Find � a.

;fn log1/3 3x �1

x 2

< 1 gks rks x dk vUrjky (�, �a)

5,

esa fLFkr gS rc � a dk eku Kkr

dhft,A

Ans. 6

Sol. 3x �1

x 2>

13

8x � 5

x 2 > 0

x (�, � 2) 5

,8

32. The number of solution of the equation tan x + sec x = 2 cos x lying in the interval [0, 2] is

vUrjky [0, 2] esa fLFkr lehdj.k tan x + sec x = 2 cos x ds gyksa dh la[;k gSA

Ans. 2

Sol. tan x + sec x = 2 cos x ; x (2n ± 1) /2

sin x + 1 = 2 cos2 x

sin x + 1 = 2 (1 � sin2 x)

sin x = � 1, 1/2

sin x = 1/2 ( sin x � 1)

x = /6, 5/6

33. The number of solutions of the equation sgn(x2) = |x| is equal to

(where sgn(.) denotes the signum function)

lehdj.k sgn(x2) = |x| ds gyksa dh la[;k cjkcj gS -

(tgk¡ sgn(.) flXue Qyu dks O;Dr djrk gS)

Ans. 3






Sol.

Number of solutions = 3

gyksa dh la[;k = 3

34. Number of real solutions of the equation 2x �6x 8| x � 2 | 1

is

lehdj.k 2x �6x 8| x � 2 | 1

ds okLrfod gyksa dh la[;k gSA

Ans. 3

Sol. 2x �6x 8| x � 2 | 1

Either |x � 2| = 1 or ;k x2 � 6x + 8 = 0

x = 3, 1 x = 2, 4

x = 2 is not possible laHko ugh gSA x = 1, 3, 4

35. If the solution set of the inequations 6{x}2 � 5{x} + 1 0

(where { .} denotes fractional part function) and x

(5x �12)(sinx � 2)0

(e 2)(x � 4)

is [a, b] [c, d] where

a < b < c < d then find the value of 5a �2b �3c + 2d.

;fn vlfedk 6{x}2 � 5{x} + 1 0 vkSj x

(5x �12)(sinx � 2)0

(e 2)(x � 4)

dk gy leqPp; [a, b] [c, d] gS (tgk¡ { .}

fHkUukRed Hkkx Qyu gS) vkSj a < b < c < d rc 5a �2b �3c + 2d dk eku Kkr dhft,A Ans. 4 Sol. 6{x}2 � 5{x} + 1 0

13 {x}

12

n + 13 x n +

12

...(1) (n )

x

(5x �12)(sinx � 2)0

(e 2)(x � 4)

5x �12

0x � 4

125 x < 4 .......(2)

Intersection of (1) and (2) (1) vkSj (2) dk izfrPNsnu

4/3 3/2 7/3 5/2 10/3 7/2

12/5 4






x 12 5

,5 2

10 7

,3 2

5a �2b �3c + 2d= 4

36. The number of solutions of the equation

1 � 2x = 3[x] + 2{x}, (where [.] denotes the greatest integer function and {.} denotes the fractional part function) is

lehdj.k 1 � 2x = 3[x] + 2{x} ds gyksa dh la[;k gS (tgk¡ [.] egÙke iw.kkZad Qyu vkSj {.} fHkUukRed Hkkx

Qyu gS) Ans. 1 Sol. 1 � 2x = 3[x] + 2{x}

put x = + f j[kus ij 1 � 2 � 2f = 3 + 2f

4f = 1 � 5 f = 1 5

4

0 1 5

4

< 1 �1 < 5 1

4

0

�4 < 5 � 1 0

3

5

< 15 = 0 f =

14

, x = 14

37. Find the value of 213 13

sin sin � 4sin sin10 10 10 10

.

213 13

sin sin � 4sin sin10 10 10 10

dk eku Kkr dhft,A

Ans. 0

Sol. 2 7 3 13

2sin cos � 4sin sin10 5 10 10

= 3 3 3

4sin cos 4sin sin10 5 10 10

= 3 3

4sin cos sin10 5 10

= 3

4sin � sin sin10 10 10

= 0

38. If a, b, c are real and distinct numbers, then the value of 2 2 3 2 2 3 2 2 3(a � b ) (b � c ) (c � a )

(a � b)(b � c)(c � a)(a b)(b c)(c a)

is

;fn a, b, c okLrfod vkSj fofHkUu la[;k,¡ gS rc 2 2 3 2 2 3 2 2 3(a � b ) (b � c ) (c � a )

(a � b)(b � c)(c � a)(a b)(b c)(c a)

dk eku gS -

Ans. 3

Sol. 2 2 3 2 2 3 2 2 3(a � b ) (b � c ) (c � a )

(a � b)(b � c)(c � a)(a b)(b c)(c a)

= 2 2 2 2 2 2

2 2 2 2 2 2

3(a � b )(b � c )(c � a )

(a � b )(b � c )(c � a ) = 3

39. The number of values of x in the interval 11

0,2

satisfying the equation

6sin2x + sin x � 2 = 0 is . Find � 3.






vUrjky 11

0,2

esa fLFkr lehdj.k 6sin2x + sin x � 2 = 0 dks lrq"B djus okys x ds ekuksa dh la[;k

gS rc � 3 Kkr dhft,A Ans. 8

Sol. sinx = 1 1 48

12

= 1 712

= �

23

, 12

There is one solution in each of

izR;sd esa ,d gy gS

0,2

, ,2

,

3,

2

, .. .. 11

5 ,2

,

40. If 4x � 1

x�23 =

1x

23

� 2x�12 then find the value of 4x.

;fn 4x � 1

x�23 =

1x

23

� 2x�12 gks rks 4x dk eku Kkr dhft,A Ans. 6

Sol. 4x + 2x�12 = 1

x23

+ 1

x�23

22x + 12

22x = x3.3 + x3

3

2x32

2= x4

.33

x

43

= 4

3×

23

=8

3 3

2x

2

3

= 3

2

3

2x = 3 4x = 6



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