Rts-main & Advance-(t-1)-smj-pcm 4-03-18.pdf

CAREER POINT

Revision Test Series For JEE Main-2018 Test-1 [T-1] Physics, Chemistry & Mathematics Time : 3 Hours

Maximum Marks : 360

SYLLABUS Physics

: Unit dimension, Motion in one dimension, Projectile Motion, Laws of motion, Friction.

Chemistry

: Gaseous State, Mole concept, Redox, Equivalent concept, Solution, Atomic Structure.

Mathematics

:

Point, Straight line, Circle, Parabola, Ellipse, Hyperbola.

IMPORTANT INSTRUCTIONS

1.

This paper contains 90 Qs. in all. All questions are compulsory.

2.

There is Negative Marking. Guessing of answer is harmful.

3.

The question paper contains blank space for your rough work. No additional sheet will be provided for rough work.

4.

The answer sheet, machine readable Optical Mark Recognition (OMR) is provided separately.

5.

Do not break the seals of the question paper booklet before being instructed to do so by the invigilator.

6.

Blank papers, Clipboards, Log tables, Slide Rule, Calculators, Cellular Phones, Pagers & Electronic Gadgets in any form are not allowed to be carried inside the examination hall.

MARKING SCHEME: 1.

Each Question has four options, only one option is correct. For each incorrect response, one-fourth of the weightage marks allotted to the question would be deducted.

2.

In Physics

:

Q.

In Chemistry

:

In Mathematics

:

1

-

30

carry

4

marks each,

Q. 31 -

60

carry

4

marks each,

Q. 61 -

90

carry

4

marks each,

RTS/Main-Adv./18/T-1/PCM CAREER POINT: CP Tower, Road No. 1, IPIA, Kota (Raj.), Ph: 0744-5151200 | www.careerpoint.ac.in

SEAL

GENERAL :

PHYSICS Q.1

Q.2

Q.3

Q.4

The distance-time graph of a particle at time t makes angle 45º with respect to time axis. After one second, it makes angle 60º with respect to the time axis. What is the acceleration of the particle? (1)

3

(2)

(3)

3 –1

(4) 1

Q.1

v{k ds lkFk 45º dk dks.k cukrk gSA ,d lsd.M ds ckn] og le; v{k ds lkFk 60º dks.k cukrk gSA d.k dk Roj.k D;k gS &

3 +1

A stone is thrown upwards which rises to a height of 100 m. The relative velocity of the stone with respect to the earth will be maximum at (1) height of 50 m (2) the highest point (3) the ground (4) height of 5 m

Q.2

A parachutist drops feely from an aeroplane for 10 s before the parachute opens out. Then he descends with net retardation of 2.5 ms–2. If he bails out of the plane at a height of 2495 m his velocity on reaching the ground will be (take g = 10 ms–2) (1) 20 ms–1 (2) 10 ms–1 (3) 15 ms–1 (4) 5 ms–1

Q.3

A body starts from the origin and moves along the x-axis such that the velocity at any instant is given by 4t3 – 2t where t is in sec and velocity in ms–1. What is the acceleration of particle when it is 2 m from the origin? (1) 28 ms–2 (2) 22 ms–2 (3) 12 ms–2 (4) 10 ms–2

Q.4

CAREER POINT

t le; ij ,d d.k dk foLFkkiu≤ xzkQ le;

(1)

3

(2)

(3)

3 –1

(4) 1

3 +1

,d iRFkj dks Åij dh vksj QSadus ij og 100 m dh Å¡pkbZ rd igq¡prk gSA i`Foh ds lkis{k iRFkj dk vkisf{kd osx vf/kdre gksxk & (1) 50 dh Å¡pkbZ ij

(2) mPpre fcUnq ij

(3) /kjkry ij

(4) 5 m dh Å¡pkbZ ij

gokbZ tgkt ls iSjk'kwV ;qDr O;fDr iSjk'kwV [kksyus ds igys 10 s ds fy;s ds Lora=k :Ik ls fxjrk gSA rc og 2.5 ms–2 ds dqy eanu ls uhps fxjrk gSA ;fn og tgkt ls dqnus ij 2495 m dh Å¡pkbZ ij Fkk] rks /kjkry ij igq¡pus ij mldk osx gksxk & (g = 10 ms–2 ysa) (1) 20 ms–1 (3) 15 ms–1

(2) 10 ms–1 (4) 5 ms–1

,d fi.M ewy fcUnq ls izkjEHk dj x-v{k ds vuqfn'k xfr djrk gS rkfd fdlh {k.k ij osx 4t3 – 2t }kjk fn;k tkrk gS tgk¡ t lsd.M esa rFkk osx ms–1 esa gSaA d.k dk Roj.k D;k gSa] tc og ewy fcUnw ls 2 m ij gS & (1) 28 ms–2 (3) 12 ms–2

(2) 22 ms–2 (4) 10 ms–2

SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg

2

Q.5

Q.6

In an experiment to determine acceleration due to gravity by simple pendulum, a student commits 1% positive error in the measurement of length and 3% negative error in the time then % error in g will be (1) 3% (2) 4% (3) 7% (4) 10%

Q.5

The equation of motion of a projectile is

Q.6

y = 12 x –

iz;ksx esa ,d Nk=k yEckbZ ds ekiu esa 1% /kukRed =kqfV rFkk le; esa 3% _.kkRed =kqfV ikrk gS] rc g esa % =kqfV gksxh & (1) 3%

3 2 x . The horizontal component of 4

A particle is projected up an inclined plane

Q.7

tan α = … . Given angle of projection with the horizontal is β.

(3) 2 tan β Q.8

(1)

(4) 2 tan β

angle θ with the horizontal, from a distance 23 m from a 10 m high wall, such that it just clears the wall and falls 36 m away from the wall on the other side of the wall. Find v and θ.

3 2 x 4

Q.8

(2) 30.6 m (4) 12.4 m

,d d.k dks {kSfrt ls α dks.k ij vkufrr ,d vkur ry ij Åij dh vksj iz{ksfir fd;k x;k gSA ;fn d.k ry ij {kSfrt :Ik ls Vdjkrk gS] rks tan α = …A {kSfrt ds lkFk iz{ksI; dks.k β fn;k x;k gS & tan β 2 (3) 2 tan β

(2) tan β

A projectile is fired with a velocity v making an

(4) 10%

,d iz{ksI; dh xfr dh lehdj.k y = 12 x –

(1) 36 m (3) 21.6 m

particle strikes the plane horizontally then

tanβ 2

(3) 7%

g = 10 ms–2, iz{ksI; dh ijkl D;k gS &

of inclination α to the horizontal. If the

(1)

(2) 4%

gSA osx dk {kSfrt ?kVd 3 ms–1 gSA fn;k x;k gS

velocity is 3 ms–1. Given that g = 10 ms–2. What is the range of the projectile? (1) 36 m (2) 30.6 m (3) 21.6 m (4) 12.4 m Q.7

ljy yksyd }kjk xq:Roh; Roj.k Kkr djus ds

(2) tan β (4) 2 tan β

,d iz{ksI; dks {kSfrt ls θ dks.k cukus gq, 10 m Å¡ph nhokj ls 23 m nwjh ls v osx ls nkxk x;k gS rkfd og Bhd nhokj dks ikj dj lds rFkk nhokj dh nwljh vksj nhokj ls 36 m nwj fxjsA v o θ Kkr dhft, &

(1) 49.1 ms–1, tan–1

5 5 (2) 59.1 ms–1, tan–1 4 4

(1) 49.1 ms–1, tan–1

5 5 (2) 59.1 ms–1, tan–1 4 4

(3) 39.1 ms–1, tan–1

4 4 (4) 29.1 ms–1, tan–1 5 5

(3) 39.1 ms–1, tan–1

4 4 (4) 29.1 ms–1, tan–1 5 5

CAREER POINT

SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg

3

Q.9

A food packet is released from a helicopter flying at a height of 1 km with a velocity 80 ms– 1 (when on a flood relief mission), the distance at which the packet falls from the point of release is nearly (1) 1500 m (3) 1000 m

Q.10

Q.9

iSdsV fdruh nwjh ij fxjsxk & (1) 1500 m (3) 1000 m

(2) 800 2 m (4) None of these

A projectile is fired at an angle θ with the horizontal. Find the condition under which it lands perpendicular on an inclined plane inclination α as shown in figure.

1 km dh Å¡pkbZ ij mM+ jgs gSfydkWIVj ls [kkus dk iSdsV 80 ms–1 osx ls fxjk;k tkrk gS (pØokr lgk;rk fe'ku ij), iSdsV dks NksM+us ds fcUnq ls

Q.10

,d iz{ksI; dks {kSfrt ls θ dks.k ij nkxk x;k gSA og 'krZ Kkr dhft, ftlds v/khu og fp=kkuqlkj α vkur dks.k ds vkur ry ij yEcor~ fxjs &

θ α

θ α

(1) sin α = cos (θ – α) (2) cos α = sin (θ – α) (3) 2 tan α = cot (θ – α) (4) None of these Q.11

A balloon of mass M is under a drag force F and upthrust T. It is moving down with a uniform velocity ν. What amount of mass m be removed that it starts rising up with same velocity ν? T 2T (2) (1) M – g g (3)

Q.12

(1) sin α = cos (θ – α) (2) cos α = sin (θ – α) (3) 2 tan α = cot (θ – α) (4)) buesa ls dksbZ ugha Q.11

Consider the situation shown in figure. The acceleration of block of mass m is

30º

(1) g/3 up the plane (3) g/2 up the plane CAREER POINT

2m

(2) g/3 down the plane (4) g/2 down the plane

M nzO;eku dk ,d xqCckjk d"kZ.k (drag) cy F rFkk mRIykou T ds v/khu gSA og uhps dh vksj ,dleku osx ν ls xfr'khy gSA nzO;eku dh fdruh ek=kk m mlls gVkuh gksxh rkfd og leku osx ν ls Åij

mBuk izkjEHk dj nsa (1) M –

⎛ T⎞ (4) 2 ⎜⎜ M – ⎟⎟ g⎠ ⎝

T g

(2) 800 2 m (4) buesa ls dksbZ ugha

(3)

Q.12

T g

(2)

2T g

⎛ T⎞ (4) 2 ⎜⎜ M – ⎟⎟ g⎠ ⎝

T g

fp=k ls n'kkZ;h fLFkfr ij fopkj dhft,A m nzO;eku ds CykWd dk Roj.k gS &

30º

2m

(1) g/3 ry ds Åij dh vksj (2) g/3 ry ds uhps dh vksj (3) g/2 ry ds Åij dh vksj (4) g/2 ry ds uhps dh vksj

SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg

4

Q.13

Q.14

Q.15

When the spring is loaded with 5 N its length is α and when loaded with 4 N its length is β. When loaded with 9 N, its length will be (1) α + β (2) 5α – 4β (4) None of these (3) 4α – 5β

Q.13

The motion of a particle is given by dυ( t ) = 6 – 3υ(t) where υ(t) is speed in ms–1 dt and t is time in s. If the body is at rest at t = 0, then (1) the terminal speed is 2 ms–1 (2) the speed varies with time as υ(t) = 2 (1 – e–3t) ms–1 (3) speed is 0.1 ms–1 when the acceleration is half the initial value (4) the magnitude of the initial acceleration is 0.6 ms–2

Q.14

The acceleration of a particle is increasing linearly with time t as βt. If the particle starts from origin with initial velocity u, the distance traveled by it in t second is 1 3 1 2 (1) ut + βt (2) ut + βt 2 2

Q.15

(3) ut +

Q.16

1 3 βt 3

(4) ut +

CAREER POINT

(1) α + β (3) 4α – 5β

(2) 5α – 4β (4) buesa ls dksbZ ugha

,d d.k dh xfr

dυ( t ) = 6 – 3υ(t) }kjk nh xbZ gS dt

tgk¡ υ(t)] ms–1 esa pky rFkk t lsd.M esa gSA ;fn t = 0 ij fi.M fojke esas gS] rks (1) lhekUr pky 2 ms–1 gS (2) le; ds lkFk pky υ(t) = 2 (1 – e–3t) ms–1 ds

vuqlkj ifjofrZr gksrh gS (3) pky 0.1 ms–1 gS] tc Roj.k izkjfEHkd eku dk

vk/kk gS (4) izkjfEHkd Roj.k dk ifjek.k 0.6 ms–2 gS

βt 3 6

Engine of a vehicle can give it an acceleration of 1 ms–2 and its brakes can retard it at 3 ms–2. The minimum time in which the vehicle can make a journey between stations A and B having a distance of 1200 m is (1) 55.6 s (2) 65.5 s (3) 50.6 s (4) 56.5 s

tc fLizax dks 5 N ls Hkkfjr fd;k tkrk gS] rks mldh yEckbZ α gS rFkk tc 4 N ls Hkkfjr fd;k tkrk gS mldh yEckbZ β gSA tc 9 N ls Hkkfjr fd;k tkrk gS] rks mldh yEckbZ gksxh &

Q.16

le; t ds lkFk ,d d.k dk Roj.k js[kh; :Ik ls βt ds vuqlkj c<+rk gSA ;fn d.k ewy fcUnq ls izkjfEHkd osx u ls xfr izkjEHk djrk gS] rks t lsd.M esa mlds }kjk r; dh xbZ nwjh gS & (1) ut +

1 3 βt 2

(2) ut +

1 2 βt 2

(3) ut +

1 3 βt 3

(4) ut +

βt 3 6

,d okgu dk rFkk czsd mls U;wure le; nwjh ij fLFkr ysrk gS &

batu mls 1 ms–2 dk Roj.k nsrk gS 3 ms–2 ij eafnr dj ldrk gSA og Kkr dhft, tks okgu 1200 m dh nks LVs'kuksa A o B dh ;k=kk djus esa

(1) 55.6 s

(2) 65.5 s

(3) 50.6 s

(4) 56.5 s

SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg

5

Q.17

Q.18

A particle moving in straight line covers half the distance with speed 3 ms –1 . The other half is covered in two equal time intervals with speed 4.5 ms –1 and 7.5 ms–1 respectively. The average speed of the particle during motion is (in ms–1 ) (1) 4 ms–1 (2) 5 ms–1 (3) 5.5 ms–1 (4) 4.8 ms–1

Q.17

From the top of a tower a stone is thrown up which reaches the ground in a time t1. A second stone thrown down with the same speed reaches the ground in time t2. A third stone released from rest from same location reaches the ground in time t3, then 1 1 1 + (1) t32 = t12 – t22 (2) 3 = t2 t1 t3

Q.18

(3) t3 = Q.19

Q.20

t1t 2

(4) t3 =

t1 + t 2 2

Two motorcycles M1 and M2 are heading towards each other with a speed of 30 km h–1 each. A bird flies off M1 at 60 km h–1 when distance between the motorcycles is 60 km. It heads towards M2 and then back to M1 and so on. Time for second trip of bird is 2 1 1 (1) hr (2) hr (3) 2 hr (4) hr 9 9 3

Q.19

A van with a constant speed υ is after a motorcycle on a straight road. The motorcycle starts when the van is at a distance X away. If the motorcycle has constant acceleration a, the motorcycle will be overtook if υ

Q.20

(1) ≥ 2aX

(2) ≥

(3) ≤ (2aX)

1/3

CAREER POINT

(4) ≤

ljy js [ kk es a xfr'khy ,d d.k vk/kh nw j h 3 ms –1 pky ls r; djrk gS A nw l jh vk/kh nw j h Øe'k% 4.5 ms –1 o 7.5 ms –1 pky ls nks leku le; va r jky es a r; djrk gS A xfr ds nkS j ku d.k dh vkSl r pky gS & (1) 4 ms–1

(2) 5 ms–1

(3) 5.5 ms–1

(4) 4.8 ms–1

,d VkWoj ds 'kh"kZ ls ,d iRFkj dks Åij dh vksj QSadus ij og /kjkry ij t1] le; esa igq¡prk gSA nwljs iRFkj dks leku pky ls uhps dh vksj QSadus ij og t2 le; esa /kjkry ij igq¡prk gSA rhljs iRFkj dks leku fLFkfr ls fojke ls eqä djus ij og /kjkry ij t3 le; esa igq¡prk gS] rc & 1

(1) t32 = t12 – t22

(2)

(3) t3 =

(4) t3 =

t1t 2

t 33

=

1 1 + t2 t1

t1 + t 2 2

nks eksVjlkbZfdy M1 o M2 ,d&nwljs dh vksj izR;sd 30 km h–1 dh pky ls igq¡p jgh gSA M1 ls ,d i{kh 60 km h–1 dh pky ij mM+rk gS tc nks eksVj lkbZfdyksa ds e/; nwjh 60 km gSA og M2 dh vksj tkrk gS rFkk fQj M1 ij iqu% ykSVrk rFkk blh izdkj xfr djrk gSA i{kh dh nwljh ;k=kk ds fy;s le; gS & (1)

2 hr 9

(2)

1 hr 9

(3) 2 hr

(4)

1 hr 3

,d osu lh/kh lM+d ij ,d okgu ds ihNs fu;r pky υ ls xfr djrh gSA okgu rc xfr izkjEHk djrk gS tc osu X nwjh ij gSA ;fn okgu fu;r Roj.k a j[krk gS] rks osu] okgu dks idM+ ysxh ;fn υ gks &

2aX

(1) ≥ 2aX

2aX

(3) ≤ (2aX)

1/3

(2) ≥

2aX

(4) ≤

2aX

SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg

6

Q.21

A smooth track of incline of length l is joined smoothly with circular track of radius R. A mass of m kg is projected up from the bottom of the inclined plane. The minimum speed of the mass to reach the top of the track is given by, υ =

Q.21

l yEckbZ ds vkufrr ?k"kZ.kghu iFk dks R f=kT;k ds o`Ùkh; iFk ls fcuk ?k"kZ.k ds tksM+k x;k gSA m kg ds

,d nzO;eku dks vkur ry ds isans ls Åij dh vksj iz{ksfir fd;k tkrk gSA iFk ds 'kh"kZ rd igq¡pus esa nzO;eku dh U;wure pky υ = gS & R

R L

L θ

θ

(1) [2g (l cos θ + R) (1 + cos θ)]1/2 (2) (2g l sin θ + R)1/2 (3) [2g{l sin θ + R (1 – cos θ)}]1/2 (4) (2g l cos θ + R)1/2

(1) [2g (l cos θ + R) (1 + cos θ)]1/2 (2) (2g l sin θ + R)1/2 (3) [2g{l sin θ + R (1 – cos θ)}]1/2 (4) (2g l cos θ + R)1/2 Q.22

Q.23

A mass of 4 kg is suspended by a rope of length 3.0 m from the ceiling. A force of 30 N in the horizontal direction is applied at the mid point of the rope. The angle the rope makes with the vertical in equilibrium is (Take g = 10 ms–2 and neglect the mass of the rope.) (1) 37º (2) 47º (3) 57º (4) 27º

Q.22

A uniform rod of length 1 m having mass 1 kg rests against a smooth wall at an angle of 30º with the ground. Calculate the force exerted by the ground on the rod in vertical direction is

Q.23

4 kg ds ,d nzO;eku dks Nr ls 3.0 m yEckbZ dh

,d jLlh }kjk yVdk;k x;k gSA jLlh ds ek/; fcUnq ij {kSfrt fn'kk esa 30 N dk cy yxk;k x;k gSA lkE;koLFkk esa jLlh Å/oZ ds lkFk fdruk dks.k cukrh gS (g = 10 ms–2 ys rFkk jLlh ds nzO;eku dks ux.; ekusa) (1) 37º

CAREER POINT

(3) 57º

(4) 27º

1 kg nzO;eku dh 1 m yEckbZ dh ,dleku NM+ /kjkry ls 30º ds dks.k ij ?k"kZ.kghu nhokj ds lkis{k fojke esa gSaA Å/okZ/kj fn'kk esa NM+ ij /kjkry }kjk vkjksfir cy Kkr dhft, &

30º

(1) 10 N (3) 23.1 N

(2) 47º

30º

(2) 2.31 N (4) 12.3 N

(1) 10 N (3) 23.1 N

(2) 2.31 N (4) 12.3 N

SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg

7

Q.24

Q.25

A plumb line is hanging from the ceiling of

Q.24

,d dkj dh Nr ls Iyac js[kk yVd jgh gSA tc

a car. The plumb line gets inclined at angle

dkj a Roj.k ls xfr djrh gS rks Iyac js[kk α dks.k

α when the car moves with acceleration a, then-

ij vkufrr gksrh gS] rc &

(1) a = g/tanα

(2) a = g tanα

(1) a = g/tanα

(2) a = g tanα

(3) a = g cos α

(4) a = g sin α

(3) a = g cos α

(4) a = g sin α

A body of mass m starting from rest slides down a frictionless inclined surface of gradient α fixed on the floor of a lift

Q.25

m nzO;eku dk ,d fi.M fojke ls a Roj.k ls Åij

dh vksj Rofjr fy¶V ds Q'kZ ij α dks.k ij vkufrr ?k"kZ.kghu vkur lrg ij

uhps dh vksj

accelerating upward with acceleration a. Taking width of inclined plane as W, the

fQlyrk gSA vkur ry dh pkSM+kbZ W ysa] ry ds

time taken by body to slide from top to

'kh"kZ ls iSansa rd fi.M dks fQlyus esa fy;k x;k

bottom of the plane is

le; gS &

a

a

α

α W 1

1

1

⎞2 ⎛ 4W ⎟⎟ (2) ⎜⎜ ⎝ (g – a ) sin α ⎠

⎛ ⎞2 2W ⎟⎟ (1) ⎜⎜ ⎝ (g + a ) sin α ⎠ 1

⎛ ⎞2 4W ⎟⎟ (3) ⎜⎜ ⎝ (g + a ) sin 2α ⎠ CAREER POINT

W 1

⎞2 ⎛ 4W ⎟⎟ (2) ⎜⎜ ⎝ (g – a ) sin α ⎠

⎛ ⎞2 2W ⎟⎟ (1) ⎜⎜ ⎝ (g + a ) sin α ⎠ 1

⎛ ⎞2 W ⎟⎟ (4) ⎜⎜ ⎝ (g + a ) sin 2α ⎠

1

⎛ ⎞2 4W ⎟⎟ (3) ⎜⎜ ⎝ (g + a ) sin 2α ⎠

1

⎛ ⎞2 W ⎟⎟ (4) ⎜⎜ ⎝ (g + a ) sin 2α ⎠

SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg

8

Q.26

A block of mass M is situated on a smooth horizontal table. A thread tied to the block passes through a hole in the table and carries a mass m at its other end. If the length of thread above the table is l and M is revolving in horizontal circle with angular speed ω on the table, then value of m so that it remains suspended at a constant height h is ω M l

h

(1) Mghω2 M lω 2 (3) g Q.27

Q.28

Q.26

fLFkr gSA est esa ,d fNnz ls xqtj jgh jLlh CykWd dks dlrh gS rFkk nwljs fljs ij m nzO;eku j[krh gSA ;fn est ds Åij jLlh dh yEckbZ l gS rFkk M est ij dks.kh; pky ω ls {kSfrt o`Ùk esa ?kwerk gS] rks M dk eku rkfd og fu;r Å¡pkbZ h ij yVdk jgs] gS ω

h m

(4) Mlω2 Q.27

A board is balanced on a rough horizontal semicircular log. Equilibrium is obtained with the help of addition of a weight to one of the ends of the board when the board makes an angle θ with the horizontal. Coefficient of friction between the log and the board is

Q.28

CAREER POINT

m

(2) Mglω2

(1) Mghω2 M lω 2 (3) g

(2) Mglω2

(2) cos θ (4) sin θ

M

l

A parachute of mass m starts coming down with a constant acceleration a. Determine the ballast mass to be released for the parachute to have an upward acceleration of same magnitude. Neglect air drag. 2ma ma ma 2ma (3) (4) (1) (2) a+g a–g a+g a–g

(1) tan θ (3) cot θ

M nzO;eku dk ,d CykWd ?k"kZ.kghu {kSfrt est ij

(4) Mlω2

m nzO;eku dk ,d iSjk'kwV fu;r Roj.k a ls uhps dh vksj vkuk izkjEHk djrk gSA leku ifjek.k ds Å/okZ/kj Roj.k ls xfr djus ds fy;s iSjk'kwV ls fdruk nzO;eku fudkyuk gksxkA ok;q ?k"kZ.k dks ux.; ekusa & (1)

2ma a+g

(2)

ma a–g

(3)

ma 2ma (4) a+g a–g

,d cksMZ ,d [kqjnjsa {kSfrt v)Zo`Ùkh; yês ij larqfyr gSA lkE;koLFkk cksMZ ds fdlh ,d fljs ij vfrfjDr Hkkj dh lgk;rk ls izkIr gksrk gS] tc cksMZ {kSfrt ls θ dks.k cukrk gSA yês rFkk cksMZ ds e/; ?k"kZ.k xq.kkad gSa &

(1) tan θ (3) cot θ

(2) cos θ (4) sin θ

SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg

9

Q.29

A large free mass M and a small mass m are connected to a string such that m moves in

Q.29

horizontal circle. Length of string is l and θ is the angle this length makes with vertical. The frequency of rotation of mass m so that M remains at rest is M

,d cM+s Lora=k nzO;eku M rFkk ,d NksVs nzO;eku m dks Mksjh }kjk tksM+k x;k gS rkfd m {kSfrt o`Ùk esa xfr djsaA Mksjh dh yEckbZ l gS rFkk θ og dks.k gS] tks Mksjh Å/oZ ds lkFk cukrh gSA m nzO;eku dh ?kw.kZu vko`fÙk rkfd M fojke esa cuk jgsa] gS & M θ

θ

l

l

m

m

(1) 2π

(3)

Q.30

ml Mg

1 2π

ml Mg

(2)

1 2π

mg Ml

(1) 2π

(4)

1 2π

Mg ml

(3)

A mass m kg is subjected to a constant force F kgf which cause it move in t second to a distance x m. The velocity acquired is v ms–1. Then distance covered x is given by (1)

v2m 2Fg

(2)

(3)

2 Fgt 2 3 m

(4) None of these

CAREER POINT

2 v2m 3 Fg

Q.30

ml Mg

1 2π

ml Mg

(2)

1 2π

mg Ml

(4)

1 2π

Mg ml

m kg nzO;eku fu;r cy F kgf ls lEcaf/kr gS]

ftlds dkj.k og t lsd.M esa x m nwjh r; djrk gSA izkIr osx v ms–1 gSA rc r; dh xbZ nwjh x gS& (1)

v2m 2Fg

(2)

(3)

2 Fgt 2 3 m

(4) buesa ls dksbZ ugha

SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg

2 v2m 3 Fg

10

CHEMISTRY Q.31

Q.32

For the reaction, N2 + 3H2 → 2NH3, if molecular mass of NH3 and N2 are M1 and M2, their equivalent masses are E1 and E2. Then (E1 – E2) is 2M1 – M 2 (1) (2) M1 – M2 6 (3) 3M1 – M2 (4) M1 – 3M2

Q.31

In a reaction FeS2 + KMnO4+ H+ → Fe+3 + SO2 + Mn+2 + H2O

Q.32

2M1 – M 2 6 (3) 3M1 – M2 (1)

(3)

Q.33

Q.34

Q.35

molar mass 11

,d vfHkfØ;k

(1) eksyj nzO;eku

molar mass 13

(3)

The density of gas A is twice that to B. At the same temperature the molecular weight of gas B is twice that of A. The ratio of pressure of gas A and B will be : (1) 1 : 6 (2) 1 : 1 (3) 4 : 1 (4) 1 : 4

Q.33

2.4g of pure Mg (at. mass = 24) is dropped in 100 mL of 1M HCl. Which of the following statement is wrong ? (1) 1.12 L of hydrogen is produced at S.T.P. (2) 0.05 mol of magnesium is left behind (3) HCl is the limiting reagent. (4) None of these

Q.34

In the equation Cr2O72– + Fe2+ + H+ → Cr3+ + Fe3+ + H2O the coefficients of Fe2+ & H+ are respectively (1) 6, 7 (2) 6, 14 (3) 5, 7 (4) 5, 14

Q.35

CAREER POINT

(4) M1 – 3M2

esa FeS2 dk rqY;kadh Hkkj fuEu ds cjkcj gksxk -

molar mass (2) 10

(4)

(2) M1 – M2

FeS2 + KMnO4+ H+ → Fe+3 + SO2 + Mn+2 + H2O

the equivalent mass of FeS2 would be equal to (1) molar mass

vfHkfØ;k N2 + 3H2 → 2NH3 ds fy,, ;fn NH3 rFkk N2 ds vkf.od nzO;eku M1 rFkk M2 gS rFkk rqY;kadh Hkkj E1 rFkk E2 gSA rc (E1 – E2) gksxk-

eksyj nzO;eku 11

(2)

eksyj nzO;eku 10

(4)

eksyj nzO;eku 13

xSl A dk ?kuRo B dk nksxquk gSA mlh rki ij xSl B dk vkf.od Hkkj xSl A dk nksxquk gSA rks xSl A rFkk xSl B ds nkcksa dk vuqikr gksxk : (1) 1 : 6 (3) 4 : 1

(2) 1 : 1 (4) 1 : 4

2.4g 'kq) Mg (ijek.kq Hkkj = 24) dks 100 mL/1M HCl esa Mkyk tkrk gS rks fuEu esa ls dkSulk xyr gS? (1) 1.12 L gkbMªkstu S.T.P. ij fufeZr gksrh gS (2) 0.05 mol esXusf'k;e var esa cprk gS (3) HCl lhekar vfHkdeZd gS (4) buesa ls dksbZ ugha

lehdj.k Cr2O72– + Fe2+ + H+ → Cr3+ + Fe3+ + H2O

esa Fe2+ rFkk H+ ds xq.kkad Øe'k% gS (1) 6, 7

(2) 6, 14 (3) 5, 7

SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg

(4) 5, 14 11

Q.36

Q.37

Q.38

If 20 mL of 0.2 M K3 [Fe(CN)6] is reduced by some equivalents of N2H4, then the number of moles of N2H4 required are – N2H4 + K3 [Fe(CN)6] + KOH K4[Fe(CN)6] + N2 + H2O –5 –3 (1) 10 (2) 10 (3) 10–6 (4) 10–2

Q.36

The approximate atomic mass of a metal having Cal is – specific heat 0.16 gm Kelvin (1) 32 (2) 40 (3) 64 (4) 128

Q.37

1 mole of equimolar and ferrous oxalate KMnO4 in acidic oxidation, x is (1) 0.5 mole (3) 1.2 mole

Q.38

mixture of ferric oxalate will require x mole of medium for complete

;fn 20 mL, 0.2 M K3 [Fe(CN)6] dks N2H4 ds dqN rqY;kadks }kjk vipf;r fd;k tk, rks vko';d N2H4 ds eksyksa dh la[;k Kkr djks – N2H4 + K3 [Fe(CN)6] + KOH K4[Fe(CN)6] + N2 + H2O (1) 10–5 (2) 10–3 (3) 10–6 (4) 10–2

,d /kkrq ftldh fof'k"V Å"ek 0.16

Cal gS] gm Kelvin

dk ijek.kq nzO;eku yxHkx gS (1) 32 (3) 64

(2) 0.9 mole (4) 4.5 mole

(2) 40 (4) 128

Qsfjd vkWDlsysV rFkk Qsjl vkWDlsysV ds 1 eksy le eksyj feJ.k ds vEyh; ek/;e esa laiw.kZ vkWDlhdj.k ds fy, KMnO4 ds x eksyksa dh vko';drk gksrh gS] rks x gS (1) 0.5 mole (3) 1.2 mole

(2) 0.9 mole (4) 4.5 mole

Q.39

5 g sample contain only Na2CO3 and Na2SO4. This sample is dissolved and the volume made up to 250 mL, 25 mL of this solution neutralizes 20 mL of 0.1 M H2SO4. Calculate the % of Na2SO4 in the sample (1) 42.4 (2) 57.6 (3) 36.2 (4) None of these

Q.39

5 g ds ,d uewus esa dsoy Na2CO3 rFkk Na2SO4 gSA bl uewus dks ?kksydj 250 mL rd dk vk;ru cuk;k tkrk gSA bl foy;u dk 25 mL, 20 mL 0.1 M H2SO4 dks mnklhu djrk gSA rks uewus esa Na2SO4 dk % Kkr djks (1) 42.4 (2) 57.6 (3) 36.2 (4) buesa ls dksbZ ugha

Q.40

The equivalent mass of H3PO4 in the reaction given below is H3PO4+ NaOH → NaH2PO4+ H2O (1) 49 (2) 98 (3) 32.6 (4) 40

Q.40

uhps nh xbZ vfHkfØ;k esa H3PO4 dk rqY;kadh Hkkj gksxk -

What is the energy content per photon (J) for light of frequency 4.2 × 1014 Hz ? (1) 2.8 × 10–21 (2) 2.5 × 10–19 –19 (3) 2.8 × 10 (4) 2.5 × 10–18

Q.41

Q.41

CAREER POINT

H3PO4+ NaOH → NaH2PO4+ H2O (1) 49 (2) 98 (3) 32.6 (4) 40

vko`fÙk 4.2 × 1014 gV~Zt ds izdk'k ds fy, izfr QksVksu (J) ÅtkZ vo;o D;k gksaxs ? (1) 2.8 × 10–21 (3) 2.8 × 10–19

(2) 2.5 × 10–19 (4) 2.5 × 10–18

SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg

12

Q.42

The energy of an electron moving in nth Bohr's orbit of an element is given by –13.6 2 En = Z eV/atom (Z = atomic number). n2 The graph of E Vs Z2 (keeping "n" constant) will be Z2 (1) E

(2)

Q.42

fdlh rRo dh nth cksj d{kk esa xfr'khy bysDVªkWu dh ÅtkZ –13.6 2 Z eV/ijek.kq (Z = ijek.qk Øekad) n2 }kjk nh xbZA E Vs Z2 dk oØ ("n" fu;r j[krs gq,) gksxk Z2 En =

(1) E

E

Z2

(3) E

(2)

E

Z2 (4) E

Z2

(3) E

Z2

(4) E

Z2

Z2

Q.43

What is the shortest wavelength line in the Paschen series of Li2+ ion ? R 9 1 9R (2) (3) (4) (1) 9 R R 4

Q.43

Li2+ vk;u dh ik'pu Js.kh eas lcls NksVh rjaxnS/;Z dh js[kk D;k gS ? R 9 1 9R (1) (2) (3) (4) 9 R R 4

Q.44

Be3+ and a proton are accelerated by the same

Q.44

Be3+ rFkk ,d izksVksu leku foHko ls Rofjr gksrs gS]

ratio (assume mass of proton = mass of neutron) -

rks mudh Mh-czksXyh rjaxnS/;ksZ dk vuqikr gksxk (ekuk izksVksu dk nzO;eku = U;qVªkWu dk nzO;eku) -

(1) 1 : 2

(1) 1 : 2

potential, their de-Broglie wavelengths have the

Q.45

(2) 1 : 4

(3) 1 : 1 (4) 1 : 3 3

A dye absorbs a photon of wavelength λ and re-emits two photon of wavelengths λ1 and λ2 respectively. The wavelength λ is related with λ1 and λ2 is λ + λ2 λ1λ 2 (1) λ = 1 (2) λ = λ1λ 2 λ1 + λ 2 (3) λ =

λ21λ22 λ1 + λ 2

CAREER POINT

(4) λ =

λ1λ 2 ( λ1 + λ 2 ) 2

Q.45

(2) 1 : 4

(3) 1 : 1 (4) 1 : 3 3

,d o.kZd rjaxnS/;Z λ ds ,d QksVksu dks vo'kksf"kr djrk gS rFkk nks QksVksu ftudh rjaxnS/;Z Øe'k% λ1 rFkk λ2 gS] dks iqu% mRlftZr djrk gSA rks rjaxnS/;Z λ, λ1 rFkk λ2 ls fdl izdkj lacaf/kr gS (1) λ =

λ1 + λ 2 λ1λ 2

(2) λ =

λ1λ 2 λ1 + λ 2

(3) λ =

λ21λ22 λ1 + λ 2

(4) λ =

λ1λ 2 ( λ1 + λ 2 ) 2

SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg

13

Q.46

Which of the following statements about an

Q.46

electron with ml = + 2 is incorrect ? (1) The electron could be in the third shell (2) The electron is in a non-spherical orbital 1 (3) The electron may have ms = 2 (4) The electron can not be in a d-orbital Q.47

For which of the following sets of quantum numbers, an electron will have the highest energy ? (1) (2) (3) (4)

Q.48

Q.49

Q.50

n 3 4 4 5

l 2 3 1 0

m 1 –1 –1 0

dFku xyr gS ? (1) bysDVªkWu rhljs dks'k esa gks ldrk gS (2) bysDVªkWu xksyh; d{kd esa ugha gksrk gS (3) bysDVªkWu ds fy, ms =

1 gks ldrk gS 2

(4) bysDVªkWu d-d{kd esa ugha gks ldrk Q.47

fuEu esa ls fdl DokaVe la[;kvksa ds leqPp; fy, ,d bysDVªkWu mPpre mtkZ j[ksxk ?

s –1/2 +1/2 +1/2 –1/2

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

Radiation corresponding to the transition n = 4 to n = 2 in hydrogen atoms falls on a certain metal (work function = 2.5 eV). The maximum kinetic energy of the photo-electrons will be (1) 0.55 eV (2) 2.55 eV (3) 4.45 eV (4) None of these

Q.48

The freezing point (in ºC) of a solution containing 0.1 g of K3[Fe(CN)6] (Mol. Wt. 329) in 100 g of water (Kf = 1.86 K kg mol–1) is(1) – 2.3 × 10–2 (2) – 5.7 × 10–2 (3) – 5.7 × 10–3 (4) –1.2 × 10–2

Q.49

Dissolving 120 g of urea (Mol. Wt. 60) in 1000 g of water gave a solution of density 1.15 g/mL. The molarity of the solution is (1) 1.78 M (2) 2.00 M (3) 2.05 M (4) 2.22 M

Q.50

CAREER POINT

ml = + 2 ds ,d bysDVªkWu ds fy, fuEu esa ls dkSulk

n

l

m

s

3 4 4 5

2 3 1 0

1 –1 –1 0

–1/2 +1/2 +1/2 –1/2

gkbMªkstu ijek.kq esa laØe.k n = 4 ls n = 2 ds lkis{k fofdj.k ,d fuf'pr /kkrq (dk;Z Qyu = 2.5 eV) ij fxjrh gSA QksVks-bysDVªkWu dh vf/kdre xfrt ÅtkZ gksxh (1) 0.55 eV (3) 4.45 eV

(2) 2.55 eV (4) buesa ls dksbZ ugha

100 g ty esa mifLFkr 0.1 g K3[Fe(CN)6] (v.kq Hkkj 329) ds ,d foy;u dk fgekad (ºC esa) gksxk (Kf = 1.86 K kg mol–1) (1) – 2.3 × 10–2 (2) – 5.7 × 10–2 (3) – 5.7 × 10–3 (4) –1.2 × 10–2 1000 g ty esa 120 g ;wfj;k dks ?kksyus ij 1.15 g/mL ?kuRo dk ,d foy;u curk gSA foy;u

dh eksyjrk gS (1) 1.78 M (3) 2.05 M

(2) 2.00 M (4) 2.22 M

SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg

14

Q.51

Q.52

Q.53

Which of the following solutions has maximum freezing point depression at equimolal concentration ? (1) [Co(H2O)6]Cl3 (2) [Co(H2O)5Cl]Cl2.H2O (3) [Co(H2O)4Cl2]Cl.2H2O (4) [Co(H2O)3Cl3].3H2O

Q.51

A solution containing 4 g of polyvinyl chloride in 1 litre of dioxane was found to have an osmotic pressure of 6 × 10–4 atm at 300 K. The molecular mass of polymer is (1) 3 × 103 (2) 1.6 × 105 4 (3) 5 × 10 (4) 6.4 × 102

Q.52

Two solutions of H2SO4 of molarities x and y are mixed in the ratio of V1 mL : V2 mL to form a solution of molarity M1. If they are mixed in the ratio of V2 mL : V1 mL, they form a solution of molarity M2. Then x : y is M x 5 (Given : V1/V2 = > 1 and 1 = ) 4 y M2

Q.53

H2SO4 ds nks foy;uksa dks ftudh eksyjrk,¡ x rFkk y gS] dks eksyjrk M1 dk ,d foy;u cukus ds fy, V1 mL : V2 mL ds vuqikr esa feyk;k tkrk gSA ;fn bUgsa V2 mL : V1 mL ds vuqikr esa feykrs gS rks eksyjrk M2 dk ,d foy;u curk gSA rc x : y gS M1 x 5 (fn;k x;k gS : V1/V2 = > 1 rFkk = ) 4 y M2 (1) 2 : 1 (2) 4 : 1 (3) 1 : 2 (4) 3 : 1

The ratio of the value of any colligative property for K4[Fe(CN)6] to that of Fe4[Fe(CN)6]3 solution is nearly (1) 0.62 (2) 0.71 (3) 1.4 (4) 1.2

Q.54

K4[Fe(CN)6] foy;u ds rFkk Fe4[Fe(CN)6]3 foy;u

An aqueous solution is 1.00 molal in KI. Which change will cause the vapour pressure of the solution to increase ? (1) Addition of NaCl (2) Addition of Na2SO4 (3) Addition of 1.00 molal HgI2 (4) Addition of water

Q.55

(1) 2 : 1 (3) 1 : 2 Q.54

Q.55

CAREER POINT

fuEu esa ls dkSulk foy;u leku eksyy lkUnz.k ij vf/kdre fgekad voueu j[krk gS ? (1) [Co(H2O)6]Cl3 (2) [Co(H2O)5Cl]Cl2.H2O (3) [Co(H2O)4Cl2]Cl.2H2O (4) [Co(H2O)3Cl3].3H2O

,d foy;u ftlesa 1 litre MkbvkWDlsu esa 4g ikWyhokbfuy DyksjkbM gS] esa 300 K ij 6 × 10–4 atm dk ijklj.k nkc ik;k x;kA cgqyd dk vkf.od nzO;eku gksxk (1) 3 × 103 (3) 5 × 104

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

(2) 1.6 × 105 (4) 6.4 × 102

ds fdlh v.kqla[;d xq.k ds eku dk vuqikr yxHkx gS (1) 0.62 (3) 1.4

(2) 0.71 (4) 1.2

KI dk tyh; foy;u 1.00 eksyy gSA fdl ifjroZu

ds gksus ij foy;u dk ok"i nkc c<+sxk ? (1) NaCl dk la;kstu (2) Na2SO4 dk la;kstu (3) 1.00 eksyy HgI2 dk la;kstu (4) ty dk la;kstu

SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg

15

Q.56

Q.57

Q.58

Q.59

Q.60

Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be added to 4 kg of water to prevent it from freezing at –6ºC will be : (Kf for water = 1.86 K kg mol–1 and Molar mass of ethylene glycol = 62 g mol–1) (1) 304.60 g (2) 800.00 g (3) 204.30 g (4) 400.00 g

Q.56

Phosphine (PH3) decomposes to produce vapours of phosphorus and H2 gas. What will be the change in volume when 100 mL of phosphine is decomposed ? (1) +50 mL (2) 500 mL (3) +75 mL (4) –500 mL

Q.57

In the reaction 1 CO + O2 → CO2; N2 + O2 → 2NO 2 10 mL of mixture containing carbon monoxide and nitrogen required 7 mL oxygen to form CO2 and NO, on combustion. The volume of N2 in the mixture will be (1) 7/2 mL (2) 17/2 mL (3) 4 mL (4) 7 mL

Q.58

0.54 gm of metal “M” yields 1.02 gm of its oxide M2O3. The at. wt. of metal “M” is(1) 9 (2) 18 (3) 27 (4) 54

Q.59

26.8 gm of Na2SO4.nH2O contains 12.6 gm of

Q.60

CAREER POINT

(2) 10

(3) 6

(1) 304.60 g (3) 204.30 g

(2) 800.00 g (4) 400.00 g

QkWLQhu (PH3) vi?kfVr gksdj QkWLQksjl dh ok"i rFkk H2 xSl mRikfnr djrh gSA 100 mL QkWLQhu ds vi?kfVr gksus ij vk;ru esa D;k ifjorZu gksxk ? (1) +50 mL (3) +75 mL

(2) 500 mL (4) –500 mL

bl vfHkfØ;k esa] 1 O2 → CO2; N2 + O2 → 2NO 2 10 mL ds feJ.k esa dkcZu eksuks vkWDlkbM rFkk

CO +

ukbVªkstu gS tks ngu djus ij 7 mL vkWDlhtu dh lgk;rk ls CO2 rFkk NO cukrh gSA rks feJ.k esa N2 dk vk;ru gksxk (1) 7/2 mL (3) 4 mL

(2) 17/2 mL (4) 7 mL

0.54 gm /kkrq “M”, blds vkWDlkbM M2O3 dk 1.02 gm nsrh gSA /kkrq “M” dk ijek.kq Hkkj gS (1) 9 (2) 18 (3) 27 (4) 54 26.8 gm Na2SO4.nH2O esa 12.6 gm ty mifLFkr

gS rks 'n' dk eku gS -

water. The value of 'n' is (1) 1

,fFkyhu XYkkbdksy dks BaMh tyok;q esa cQZjks/kh ds :i esa iz;qDr fd;k tkrk gSA 4 kg ty dks –6ºC ij teus ls jksdus ds fy, blesa feyk, x, ,fFkyhu Xykbdksy dk nzO;eku gksxk : (ty ds fy, Kf = 1.86 K kg mol–1 rFkk ,fFkyhu Xykbdksy dk eksyj nzO;eku = 62 g mol–1)

(4) 7

(1) 1

(2) 10

(3) 6

SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg

(4) 7

16

MATHEMATICS Q.61

If in triangle ABC, A ≡ (1, 10), circumcentre ≡ (–1/3, 2/3) and orthocenter ≡ (11/3, 4/3), then the co-ordinates of mid point of side opposite to A is ⎛ − 11 ⎞ (1) ⎜1, ⎟ ⎝ 3 ⎠ (3) (1, –3)

Q.62

Q.63

Q.64

Q.61

rFkk yEcdsUnz ≡ (11/3, 4/3) gS] rks A ds lEeq[k Hkqtk ds e/; fcUnq ds funs'Z kkad gksxas -

(2) (1, 5)

⎛ − 11 ⎞ (1) ⎜1, ⎟ ⎝ 3 ⎠

(2) (1, 5)

(4) (1, 6)

(3) (1, –3)

(4) (1, 6)

Let A ≡ (3, – 4), B ≡ (1, 2). Let P ≡ (2k – 1, 2k + 1) be a variable point such that PA + PB is minimum then k is 7 7 7 (2) 0 (3) (4) (1) 9 8 10

Q.62

If two tangents to the parabola y2 = 4ax make angle θ1 and θ2 with x-axis where tan2θ1+ tan2θ2 = C, then the locus of their point of intersection is (1) y2 – ax = 2cx2 (2) y2 – 2ax = cx2 2 2 (3) y + 2ax = cx (4) y2 + ax = cx2

Q.63

If (b2 – b1) (b3 – b1) + (a2 – a1) (a3 – a1) = 0 then circumcentre of the triangle having vertices (a1, b1), (a2, b2) and (a3, b3) is-

Q.64

⎛a +a b +b ⎞ (1) ⎜ 1 2 , 1 2 ⎟ 2 ⎠ ⎝ 2

⎛a +a +a b +b +b ⎞ (2) ⎜ 1 2 3 , 1 2 3 ⎟ 3 3 ⎝ ⎠ ⎛a +a b +b ⎞ (3) ⎜ 2 3 , 2 3 ⎟ 2 ⎠ ⎝ 2 ⎛a +a b +b ⎞ (4) ⎜ 1 3 , 1 3 ⎟ 2 ⎠ ⎝ 2

CAREER POINT

;fn f=kHkqt ABC eas, A ≡ (1, 10), ifjdsUnz ≡ (–1/3, 2/3)

ekuk A ≡ (3, – 4), B ≡ (1, 2) gSA ekuk P ≡ (2k – 1, 2k + 1) ,d pj fcUnq bl çdkj gS fd PA + PB U;wure gS] rc k dk eku gS (1)

7 9

(2) 0

(3)

7 8

(4)

7 10

;fn ijoy; y2 = 4ax dh nks Li'kZ js[kk,sa x-v{k ds lkFk θ1 rFkk θ2 dks.k cukrh gS tgk¡ tan2θ1+ tan2θ2 = C gS] rc buds çfrPNsnu fcUnq dk fcUnqiFk gS (1) y2 – ax = 2cx2 (3) y2 + 2ax = cx2

(2) y2 – 2ax = cx2 (4) y2 + ax = cx2

;fn (b2 – b1) (b3 – b1) + (a2 – a1) (a3 – a1) = 0 gS] rc 'kh"kksZ (a1, b1), (a2, b2) rFkk (a3, b3) okys f=kHkqt dk ifjdsUæ gS ⎛a +a b +b ⎞ (1) ⎜ 1 2 , 1 2 ⎟ 2 ⎠ ⎝ 2 ⎛a +a +a b +b +b ⎞ (2) ⎜ 1 2 3 , 1 2 3 ⎟ 3 3 ⎠ ⎝

⎛a +a b +b ⎞ (3) ⎜ 2 3 , 2 3 ⎟ 2 ⎠ ⎝ 2 ⎛a +a b +b ⎞ (4) ⎜ 1 3 , 1 3 ⎟ 2 ⎠ ⎝ 2

SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg

17

Q.65

Q.66

Q.67

Q.68

Q.69

The set of values of 'b' for which the origin and the point (1, 1) lie on the same side of the line a2 x + aby + 1 = 0 (∀ a ∈ R, b > 0) is (1) (2, 4) (2) (0, 2) (3) [0, 2] (4) (–2, 0)

Q.65

If it is possible to draw a line which belongs to all the given family of lines y – 2x + 1 + λ1 (2y – x – 1) = 0, 3y – x – 6 + λ2(y – 3x + 6) = 0, ax + y – 2 + λ3(6x + ay – a) = 0, then (1) a = 4 (2) a = 3 (3) a = – 2 (4) a = 2

Q.66

If P is a point on the line y = x such that PA + PB is minimum where A (3, 4), B(7, 13). Then, coordinates of P are ⎛ 13 13 ⎞ ⎛ 23 23 ⎞ (2) ⎜ , ⎟ (1) ⎜ , ⎟ 7 7 ⎝ ⎠ ⎝ 7 7 ⎠ ⎛ 26 26 ⎞ ⎛ 31 31 ⎞ (3) ⎜ , ⎟ (4) ⎜ , ⎟ ⎝7 7⎠ ⎝ 7 7 ⎠

Q.67

Through the point P(α, β), where αβ > 0, the x y straight line + = 1 is drawn so as to form a b with coordinate axes a triangle of area S. If ab > 0, then least value of S is -

Q.68

(2)

(3) αβ

(4) 3αβ

Lines x + 2y – 1 = 0, ax + y + 3 = 0 and bx – y + 2 = 0 are concurrent and let s be the curve denoting locus of (a, b). Then the least distance of s from the origin is (1)

5 57

CAREER POINT

(2)

5 51

(3)

(1) (2, 4) (3) [0, 2]

5 58

(4)

;fn ,d js[kk [khapuk laHko gks tks fd lHkh js[kk fudk; y – 2x + 1 + λ1 (2y – x – 1) = 0,

rc& (1) a = 4 (3) a = – 2

(2) a = 3 (4) a = 2

;fn fcUnq P js[kk y = x ij bl izdkj fo|eku gS fd PA + PB U;wure gS tgk¡ A (3, 4), B(7, 13) gS] rc P ds funsZ'kkad gS ⎛ 13 13 ⎞ (1) ⎜ , ⎟ ⎝7 7⎠ ⎛ 31 31 ⎞ (3) ⎜ , ⎟ ⎝7 7⎠

⎛ 23 23 ⎞ (2) ⎜ , ⎟ ⎝ 7 7 ⎠ ⎛ 26 26 ⎞ (4) ⎜ , ⎟ ⎝ 7 7 ⎠

fcUnq P(α, β), tgk¡ αβ > 0, ls xqtjrh gqbZ ljy js[kk

x y + = 1 a b

bl izdkj [khaph tkrh gS fd ;g

funsZ'kkad v{kksa ds lkFk S {ks=kQy dk f=kHkqt cukrh gS ;fn ab > 0 rc S dk U;wure eku gS -

Q.69

1 αβ 2

(1) 2αβ

(2)

(3) αβ

(4) 3αβ

js[kk,¡ x + 2y – 1 = 0, ax + y + 3 = 0 rFkk bx – y + 2 = 0 laxkeh gS rFkk ekuk s ,d oØ gS] tks (a, b) ds fcUnqiFk dks O;Dr dj jgk gS] rc s dh ewy fcUnq ls U;wure nwjh gS

5 59

(2) (0, 2) (4) (–2, 0)

3y – x – 6 + λ2(y – 3x + 6) = 0, ax + y – 2 + λ3(6x + ay – a) = 0 ls lacaf/kr gS]

1 αβ 2

(1) 2αβ

'b' ds mu ekuksa dk leqPp; ftuds fy, ewyfcUnq o fcUnq (1, 1), js[kk a2 x + aby + 1 = 0 (∀ a ∈ R, b > 0) ds ,d gh vksj fLFkr gksa] gSa-

(1)

5 57

(2)

5

(3)

51

SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg

5 58

(4)

5 59

18

Q.70

A line L passes through the points (1, 1) and (2, 0) and another line L' passes through

Q.70

⎛1 ⎝2

L ds yEcor~ gSA rc js[kkvksa L, L' rFkk y-v{k ds

of the triangle formed by the lines L, L' and y-axis, is (1) 15/8 (2) 25/4 (3) 25/8 (4) 25/16

Q.72

Q.73

}kjk cus f=kHkqt dk {ks=kQy gS (1) 15/8

A variable line x/a + y/b = 1 moves in such a way that harmonic mean of a and b is 8. Then the least area of triangle made by the line with the coordinate axes is (1) 8 sq. unit (2) 16 sq. unit (3) 32 sq. unit (4) 64 sq. unit

Q.71

If tangent at (1, 2) to the first circle x2 + y2 = 5 intersects the second circle x2 + y2 = 9 at P and Q and tangents at P and Q to the second circle meet at point R, then the coordinates of R are-

Q.72

(2) 25/4

,d pj js[kk x/a + y/b fd a rFkk b dk gjkRed lkFk js[kk }kjk fufeZr gksxk(1) 8 oxZ bdkbZ (3) 32 oxZ bdkbZ

⎛9 8⎞ (2) ⎜ , ⎟ ⎝ 15 5 ⎠

(1) (2, 3)

⎛ 9 18 ⎞ (3) ⎜ , ⎟ ⎝5 5 ⎠

⎛9 8 ⎞ (4) ⎜ , ⎟ ⎝ 5 15 ⎠

⎛ 9 18 ⎞ (3) ⎜ , ⎟ ⎝5 5 ⎠

x + y – (λ + 6) x + (8 – 2λ)y – 3 = 0, λ being 2

Q.73

(3) 25/8

(4) 25/16

= 1 bl izdkj xfr djrh gS

ek/; 8 gS] rc funsZ'kh v{kksa ds f=kHkqt dk U;wure {ks=kQy (2) 16 oxZ bdkbZ (4) 64 oxZ bdkbZ

;fn izFke o`Ùk x2 + y2 = 5 ds fcUnq (1, 2) ij [khaph xbZ Li'kZ js[kk] f}rh; o`Ùk x2 + y2 = 9 dks P o Q ij çfrPNsfnr djrh gS rFkk nwljs o`Ùk ds P o Q ij [khaph xbZ Li'kZ js[kk,sa] fcUnq R ij feyrh gS] rks R ds funsZ'kkad gksaxs -

(1) (2, 3)

Tangents are drawn to the circle x2 + y2 = 1 at the points where it is met by the circles,

⎞ ⎠

rFkk nwljh js[kk L' fcUnq ⎜ , 0 ⎟ ls xqtjrh gS rFkk

⎛1 ⎞ ⎜ , 0 ⎟ and perpendicular to L. Then the area ⎝2 ⎠

Q.71

,d js[kk L fcUnqvksa (1, 1) rFkk (2, 0) ls xqtjrh gS

⎛9 8⎞ (2) ⎜ , ⎟ ⎝ 15 5 ⎠ ⎛9 8 ⎞ (4) ⎜ , ⎟ ⎝ 5 15 ⎠

o`Ùk x2 + y2 = 1 ds mu fcUnqvksa ij Li'kZ js[kk,as [khaph tkrh gS tgk¡ ;s o`Ùkksa x2 + y2 – (λ + 6) x + (8 – 2λ)y – 3 = 0 ls

2

the variable. The locus of the point of intersection of these tangents is :(1) 2x – y +10 = 0 (2) x + 2y – 10 = 0 (3) x – 2y + 10 = 0 (4) 2x + y – 10 = 0

CAREER POINT

feyrk gS, λ pj gSAs bu Li'kZ js[kkvksa ds izfrPNsn fcUnq dk fcUnqiFk gksxk(1) 2x – y +10 = 0

(2) x + 2y – 10 = 0

(3) x – 2y + 10 = 0

(4) 2x + y – 10 = 0

SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg

19

Q.74

Q.75

Q.76

The point on the circle (x – 3)2 + (y – 4)2 = 4 which is at least distance from the circle x2 + y2 = 1 is ⎛ 9 12 ⎞ ⎛3 4⎞ (2) ⎜ , ⎟ (1) ⎜ , ⎟ ⎝5 5 ⎠ ⎝5 5⎠ ⎛ 9 12 ⎞ (3) (9, 12) (4) ⎜ , ⎟ ⎝7 7 ⎠

Q.74

Tangents are drawn to the circle x2 + y2 = 12 at the points where it is met by the circle x2 + y2 – 5x + 3y – 2 = 0; the point of intersection of these tangents is (1) (6, –18/5) (2) (6, 18/5) (3) (18/5, 6) (4) (– 6, – 18/5)

Q.75

The range of values of m for which the line y = mx + 2 cuts the circle x2 + y2 = 1 at distinct or coincident points is -

Q.76

⎛ 9 12 ⎞ (2) ⎜ , ⎟ ⎝5 5 ⎠

(3) (9, 12)

⎛ 9 12 ⎞ (4) ⎜ , ⎟ ⎝7 7 ⎠

o`Ùk x2 + y2 = 12 ds mu fcUnqvksa tgk¡ ;g o`Ùk x2 + y2 – 5x + 3y – 2 = 0 dks feyrk gS ij Li'kZ js[kk,sa [khpha tkrh gSA bu Li'kZ js[kkvksa dk çfrPNsn fcUnq gS (2) (6, 18/5) (4) (– 6, – 18/5)

m ds mu ekuksa dk ifjlj ftuds fy, js[kk y = mx + 2 o`Ùk x2 + y2 = 1 dks fofHkUu ;k lEikrh fcUnqvksa ij dkVrh gS] gS (1) (– ∞, – 3 ] ∪ [ 3 , ∞)

(2) [– 3 , 3 ]

(2) [– 3 , 3 ]

(3) [ 3 , ∞) (4) None of these

(3) [ 3 , ∞) (4) buesa ls dksbZ ugha

If the line x cosθ + y sinθ = 2 is the equation of a transverse common tangent to the circles

Q.77

x2 + y2 = 4 and x2 + y2 – 6 3x – 6y + 20 = 0, then the value of θ is (1) 5π/6 (2) 2π/3 Q.78

⎛3 4⎞ (1) ⎜ , ⎟ ⎝5 5⎠

(1) (6, –18/5) (3) (18/5, 6)

(1) (– ∞, – 3 ] ∪ [ 3 , ∞)

Q.77

o`Ùk (x – 3)2 + (y – 4)2 = 4 ij og fcUnq tks o`Ùk x2 + y2 = 1 ls y?kqÙke nwjh ij gS, gS

(3) π/3

(1) 1 CAREER POINT

(2)

2

(3) 2

(1) 5π/6

(4) π/6

Two co-centric circles are such that the smaller circle divides the larger circle into two equal areas. If radius of smaller circle is 2, then length of tangent drawn from any point P on the larger circle to smaller circle is -

(4) 4

;fn js[kk x cosθ + y sinθ = 2, o`Ùkksa x2 + y2 = 4 rFkk x2 + y2 – 6 3x – 6y + 20 = 0, dh fr;Zd mHk;fu"B Li'kZ js[kk dk lehdj.k gS rc θ dk eku gksxk-

Q.78

(2) 2π/3

(3) π/3

(4) π/6

nks ladsUnzh; o`Ùk bl izdkj gS fd NksVk o`Ùk cM+s o`Ùk dks nks leku {ks=kQyksa esa foHkkftr djrk gSA ;fn NksVs o`Ùk dh f=kT;k 2 gS] rks cM+s o`Ùk ij fLFkr fdlh fcUnq P ls NksVs o`Ùk ij [khaph xbZ Li'kZ js[kk dh yEckbZ gksxh(1) 1

(2)

2

(3) 2

SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg

(4) 4 20

Q.79

An equilateral triangle SAB is inscribed in the parabola y2 = 4ax having its focus at S. If the chord AB lies towards the left of S, then the side length of this triangle is (1) 2a(2 –

3)

(3) a(2– 3 ) Q.80

Q.81

Q.82

Q.83

Q.79

(2) 4a(2– 3 )

(1) 2a(2 –

(4) 8a(2 – 3 )

(3) a(2– 3 )

A variable parabola y2 = 4ax, a (where a ≠ –1/4) being the parameter meets the curve y2 + x – y – 2 = 0 at two points. The locus of the point of intersection of tangents at these point is (1) x – 2y – 4 = 0 (2) x – 4y + 2 = 0 (3) x – 4y – 1 = 0 (4) 2x – y + 1 = 0

Q.80

Two straight lines y – b = m1 (x + a) and y – b = m2 (x + a) are tangents of y2 = 4 ax. Then (1) m1 + m2 = 0 (2) m1m2 = 1 (3) m1 = m2 (4) m1m2 = – 1

Q.81

Parabolas y2 = 4a(x – c1) and x2 = 4a(y – c2), where c1 and c2 are variables, touch each other. The locus of their point of contact is (1) xy = 2a2 (2) xy = 4a2 2 (3) xy = a (4) xy = 8a2

Q.82

In the adjacent figure a parabola is drawn to pass through the vertices B, C and D of the square ABCD. If A(2, 1), C(2, 3), then the focus of this parabola is C

Q.83

(2) 4a(2– 3 ) (4) 8a(2 – 3 )

(1) x – 2y – 4 = 0 (3) x – 4y – 1 = 0

(2) x – 4y + 2 = 0 (4) 2x – y + 1 = 0

nks ljy js[kk,sa y – b = m1 (x + a) rFkk y – b = m2 (x + a) ijoy; y2 = 4 ax dh Li'kZ js[kk,sa gS] rc (1) m1 + m2 = 0 (3) m1 = m2

(2) m1m2 = 1 (4) m1m2 = – 1

ijoy; y2 = 4a(x – c1) rFkk x2 = 4a(y – c2) tgk¡ c1 rFkk c2 pj gS ,d nwljs dks Li'kZ djrs gSA buds Li'kZ fcUnq dk fcUnqiFk gS (1) xy = 2a2 (3) xy = a2

(2) xy = 4a2 (4) xy = 8a2a

layXu fp=k esa] oxZ ABCD ds 'kh"kksZ B, C rFkk D ls xqtjrk gqvk ,d ijoy; [khapk x;k gSA ;fn A(2, 1), C(2, 3), rc bl ijoy; dh ukfHk gS C B

D A

A

CAREER POINT

3)

,d pj ijoy; y2 = 4ax, a (tgk¡ a ≠ –1/4) çkpy gS] oØ y2 + x – y – 2 = 0 dks nks fcUnqvksa ij feyrk gSA bu fcUnqvksa ij [khaph xbZ Li'kZ js[kkvksa ds çfrPNsnu fcUnq dk fcUnqiFk gksxk -

B

D

⎛ 11 ⎞ (1) ⎜1, ⎟ ⎝ 4⎠ ⎛ 13 ⎞ (3) ⎜ 3, ⎟ ⎝ 4⎠

,d leckgq f=kHkqt SAB, ijoy; y2 = 4ax eas vUrfufgr gS ftldh ukfHk S ij gSA ;fn thok AB, S ds ckbZ vksj fLFkr gS] rc bl f=kHkqt dh Hkqtk dh yEckbZ gS -

⎛ 11 ⎞ (2) ⎜ 2, ⎟ ⎝ 4⎠ ⎛ 13 ⎞ (4) ⎜ 2, ⎟ ⎝ 4⎠

⎛ 11 ⎞ (1) ⎜1, ⎟ ⎝ 4⎠ ⎛ 13 ⎞ (3) ⎜ 3, ⎟ ⎝ 4⎠

⎛ 11 ⎞ (2) ⎜ 2, ⎟ ⎝ 4⎠ ⎛ 13 ⎞ (4) ⎜ 2, ⎟ ⎝ 4⎠

SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg

21

Q.84

If a and c are the lengths of segments of any focal chord of the parabola y2 = 2bx (b > 0), then the roots of the equation ax2 + bx + c = 0 are (1) real and distinct (2) real and equal (3) imaginary (4) None of these

Q.84

;fn ijoy; y2 = 2bx (b > 0) dh fdlh ukHkh; thok ds [k.Mksa dh yEckbZ;k¡ a rFkk c gS] rc lehdj.k ax2 + bx + c = 0 ds ewy gSa (1) okLrfod rFkk fHkUu (2) okLrfod rFkk cjkcj (3) dkYifud (4) buesa ls dksbZ ugha

Q.85

x – 2y + 4 = 0 is a common tangent to y2 = 4x x 2 y2 + = 1. Then the value of b and the and 4 b2 other common tangent are given by (1) b = 3 ; x + 2y + 4 = 0

Q.85

y2 = 4x rFkk

Q.86

x – 2y + 4 = 0 gSA rc b dk eku rFkk vU; mHk;fu"B

Li'kZ js[kk gksxh -

(2) b =

3 ; x + 2y + 4 = 0

(1) b = 3 ; x + 2y + 4 = 0 (2) b = 3 ; x + 2y + 4 = 0

(3) b =

3 ; x + 2y – 4 = 0

(3) b =

3 ; x + 2y – 4 = 0

(4) b =

3 ; x – 2y – 4 = 0

(4) b =

3 ; x – 2y – 4 = 0

Let the major axis of a standard ellipse, equals the transverse axis of a standard hyperbola and their director circles have radius equal to 2R and R respectively. If e1 and e2 are the eccentricities of the ellipse and hyperbola then the correct relation is (1) 4e12 − e22 = 6 (3)

Q.87

x 2 y2 + = 1 dh mHk;fu"B Li'kZ js[kk 4 b2

4e 22

− e12

=6

(2) e12 − 4e22 = 2 (4)

2e13

− e 22

=4

If p is the length of the perpendicular from a focus upon the tangent at any point P of the ellipse

Q.86

x 2 y2 + = 1 and r is the distance of P a 2 b2

Q.87

ekuk ,d ekud nh?kZo`Ùk dk nh?kZ v{k] ekud vfrijoy; ds vuqizLFk v{k ds cjkcj gS rFkk muds fu;ked o`Ùkks dh f=k;k,s Øe'k% 2R rFkk R ds cjkcj gSA ;fn e1 rFkk e2 nh?kZo`Ùk rFkk vfrijoy; dh mRdsUnzrk gS] rks lgh lEcU/k gS (1) 4e12 − e22 = 6

(2) e12 − 4e22 = 2

(3) 4e 22 − e12 = 6

(4) 2e13 − e 22 = 4

;fn nh?kZo`Ùk

[khaph xbZ Li'kZ js[kk ij ,d ukfHk ls Mkys x;s yEc dh yEckbZ p gS rFkk r, fcUnq P dh ukfHk ls nwjh gS]

2

from the focus, then (1) –1 CAREER POINT

(2) 0

b 2a – 2 is equal to r p (3) 1

(4) 2

x 2 y2 + = 1 ds fdlh fcUnq P ij a 2 b2

rc

b2 2a – 2 dk eku gS r p

(1) –1

(2) 0

(3) 1

SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg

(4) 2 22

Q.88

Q.89

If equation (10x – 5)2 + (10y – 4)2 = λ2 (3x + 4y – 1)2 represent a hyperbola, then (1) – 2 < λ < 2 (2) λ > 2 (4) 0 < λ < 2 (3) λ < – 2 or λ > 2 A tangent drawn to hyperbola

x 2 y2 − = 1 at a 2 b2

Q.88

(1) – 2 < λ < 2 (3) λ < – 2 or λ > 2

Q.89

⎛π⎞ P ⎜ ⎟ forms a triangle of area 3a2 square units, ⎝6⎠ with coordinate axes, then the square of its eccentricity is (1) 15 (2) 24 (3) 17 (4) 14 Q.90

Tangents drawn from the point (c, d) to the hyperbola

x

2

a

2

–

y

2

b2

= 1 make angles α and β

;fn lehdj.k (10x – 5)2 + (10y – 4)2 = λ2 (3x + 4y – 1)2 , vfrijoy; dks fu:fir djrh gS] rc -

vfrijoy;

(2) λ > 2 (4) 0 < λ < 2

x 2 y2 ⎛π⎞ − 2 = 1, ds fcUnq P ⎜ ⎟ ij 2 a b ⎝6⎠

[khaph xbZ ,d Li'kZ js[kk funsZ'kkad v{kksa ds lkFk 3a2 oxZ bdkbZ {ks=kQy dk f=kHkqt cukrh gS rc bldh mRdsUnzrk dk oxZ gS(1) 15 Q.90

(2) 24

(3) 17

fcUnq (c, d) ls vfrijoy;

x2 a2

gSA ;fn tanα. tanβ = 1 rc &

(1) a2 + b2 = c2 + d2 (3) a2 – b2 = c2 + d2

(1) a2 + b2 = c2 + d2 (3) a2 – b2 = c2 + d2

CAREER POINT

–

y2 b2

= 1 ij [khph

xbZ Li'kZ js[kk,W x-v{k ds lkFk α rFkk β dks.k cukrh

with the x-axis. If tanα. tanβ = 1, then (2) a2 + b2 = c2 – d2 (4) a2 – b2 = c2 – d2

(4) 14

(2) a2 + b2 = c2 – d2 (4) a2 – b2 = c2 – d2

SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg

23

Time : 3 Hours

Maximum Marks : 360

SYLLABUS HkkSfrd foKku

:

bdkbZ ,oa foek] ,d fofe; xfr] iz{ksI; xfr] xfr ds fu;e] ?k"kZ.k

jlk;u foKku

:

xSlh; voLFkk] eksy vfHk/kkj.kk] jsMkWDl] lkE; vfHk/kkj.kk] foy;u] ijek.kq lajpuk

xf.kr

:

fcUnq] ljy js[kk] o`Ùk] ijoy;] nh?kZo`Ùk] vfrijoy;

IMPORTANT INSTRUCTIONS

SEAL

lkekU; % 1.

bl iz'u i=k esa dqy 90 iz'u gSaA lHkh iz'u gy djus vfuok;Z gSaA

2.

blesa _.kkRed vadu gS vr% mÙkj vuqekfur djuk gkfudkjd gks ldrk gSA

3.

bl iz'u i=k esa gh jQ odZ ds fy, [kkyh LFkku fn;k x;k gSA jQ odZ ds fy, dksbZ vfrfjDr 'khV ugha nh tk,sxhA

4.

mÙkj O.M.R.(Optical Marks Recognisation) 'khV esa vafdr djus gSaA ;g vyx ls nh xbZ gSA

5.

iz'u i=k dh lhy rc rd u [kksysa tc rd ,slk djus ds fy, ifjoh{kd }kjk dgk u tk,sA

6.

[kkyh dkx+t] fDyi cksMZ] ykWx lkj.kh] LykbM :y] dsYdqysVj] lsY;qyj Qksu] istj ;k fdlh Hkh izdkj dk vU; bysDVªkWfud midj.k fdlh Hkh :i esa ijh{kk gkWy ds vUnj ys tk;s tkus dh vuqefr ugha gSA

vadu i)fr 1.

:

izR;sd iz'u esa pkj fodYi fn;s x;s gSa] dsoy ,d fodYi lgh gSA izR;sd xyr mÙkj ds fy, ml iz'u ds fy, fu/kkZfjr vadksa esa ls ,d&pkSFkkbZ vad dkV fy, tk,saxsA

2.

HkkSfrd foKku esa

:

Q. 1

-

30

izR;sd ds fy, 4 vad,

jlk;u foKku esa

:

Q. 31

-

60

izR;sd ds fy, 4 vad,

Xkf.kr esa

:

Q. 61

-

90

izR;sd ds fy, 4 vad,

TYPE-6 CAREER POINT: CP Tower, Road No. 1, IPIA, Kota (Raj.), Ph: 0744-5151200 | www.careerpoint.ac.in