cbxb

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

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

Toll Free : 1800 200 2244 | 1800 258 5555 JRPT2051014C0-1

JR) PT-02 Date: 12-10-2014

Test Syllabus Geometrical optics from lens formulae onwards. Projectile Motion, Relative Motion, Photo-electric effect, Bohrs model, de-Brogle waves, X-rays & Fluid Mechanics (Statics & dynamics), NLM, Friction, Circular Motion , WPE, SHM, Electrostatics and Gravitation (before kepplers laws)

S.No. Subject Nature of Questions No. of Questions Marks Negative Total1 to 15 MCQ 15 4 0 60

16 to 45 Subjective (Single digit) 30 3 0 9045 150

PT-2

Total Total

Maths/ Physics/

Chemistry

SECTION-1 : (One or more option correct type) [k.M1 : (,d ;k vf/kd lgh fodYi dkj)

This section contains 15 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct.

bl [k.M esa 15 cgqfodYi 'u gSA R;sd 'u esa pkj fodYi (A), (B), (C) vkSj (D) gS] ftuesa ls dsoy ,d ;k vf/kd lgh gSA

MCQ (15)

1. In the figure shown the radius of curvature of the left & right surface of the thin concave lens are 10 cm & 15 cm respectively. The radius of curvature of the mirror is 15 cm. Choose the correct options:

fp=k esa iznf'kZr ,d irys vory ySUl ds cka;s o nka;s i"`B dh ork f=kT;k e'k% 10 lseh- o 15 lseh- gSA niZ.k dh ork f=kT;k 15 lseh- gSA lgh dFkuksa dk p;u dhft,A [GO-RS]

(A*) magnitude of equivalent radius of curvature of the combination is 36 cm la;kstu dh rqY; ork f=kT;k dk ifjek.k 36 lseh- gSA (B) equivalent focal length of the combination is +36 cm la;kstu dh rqY; Qksdl nwjh +36 lseh- gSA (C*) the system behaves like a converging mirror fudk; ,d vfHklkjh niZ.k dh Hkkafr O;ogkj djrk gSA (D) the system behaves like a diverging mirror. fudk; ,d vilkjh niZ.k dh Hkkafr O;ogkj djrk gSA



Toll Free : 1800 200 2244 | 1800 258 5555 JRPT2051014C0-2

Sol.

1

1f

=3 12

1 110 15

= 1

12;

2

1f

= 4 13

215

= 245

; m

1f

= 215

1 2

1 1 1f f f

eq m

1 1 2f f f

= 118

feq = 18 cm

So, the combination behaves as a concave mirror vr% la;kstu vory niZ.k dh Hkkafr O;ogkj djrk gSA

2. A parallel beam of light is incident on a lens of focal length 10 cm. A parallel slab of refractive index 1.5 and thickness 3 cm is placed on the other side of the lens perpendicular to the principal axis as shown. Choose the correct options for position of the final image and its nature. (Assume rays to be paraxial) [GO-RP]

Qksdl nwjh 10 cm ds ysal ij ,d lekukUrj fdj.k iqat vkifrr gksrk gSA ysal ds nwljh rjQ eq[; v{k ds yEcor~ 1.5 viorZukad vkSj 3 cm eksVkbZ dh lekUrj ifdk eq[; v{k ds yEcor~ fp=kkuqlkj j[k nh tkrh gSA vfUre izfrfcEc dh fLFkfr rFkk bldh izd f`r ds fy, lgh dFkuksa dk p;u dhft,A (ekfu, fd fdj.ksa lek{kh; gS)

(A*) Final image is at distance 11 cm from lens and virtual, if X = 12 cm. (B*) Final image is at distance 11 cm from lens and virtual, if X = 14 cm. (C*) Final image is at distance 11 cm from lens and real, if X = 7 cm. (D) Final image is at distance 11 cm from lens and real, if X = 16 cm. (A*) vfUre izfrfcEc ysal ls 11 cm dh nwjh ij cusxk rFkk vkHkklh gksxk] ;fn X = 12 cm (B*) vfUre izfrfcEc ysal ls 11 cm dh nwjh ij cusxk rFkk vkHkklh gksxk] ;fn X = 14 cm (C*) vfUre izfrfcEc ysal ls 11 cm dh nwjh ij cusxk rFkk okLrfod gksxk] ;fn X = 7 cm (D) vfUre izfrfcEc ysal ls 11 cm dh nwjh ij cusxk rFkk okLrfod gksxk] ;fn X = 16 cm Sol. As rays are parallel to the principal axis, image is created by lens at the focus. By placing of glass-slab, pwafd fdj.ksa eq[; v{k ds lekUrj gS rc ySUl }kjk cuk frfcEc Qksdl ij cusxk dkp ifdk ds j[kus ij

Shift foLFkkiu = 11

.t

= 11

1.5

3 = 1 cm.

Irrespective of separation between slab and lens, image is shifted to the right by 1 cm. Total distance from lens 10 + 1 = 11 cm Ans. Image will be virtual if X is greater than 8 cm and it will be real if X is lesser than 8 cm. frfcEc 1 cm ls nka;h vksj foLFkkfir gksxk ySUl ls dqy nwjh 10 + 1 = 11 cm Ans. izfrfcEc vkHkklh gksxk] ;fn X, 8 cm ls vf/kd gks rFkk ;g okLrfod gksxkA

;fn X, 8 cm ls de gksA



Toll Free : 1800 200 2244 | 1800 258 5555 JRPT2051014C0-3

3. Let R

be the position vector of a particle performing curviliinear motion with respect to some reference point and R be its magnitude. Similarly v be its velocity vector with respect to the same reference point and v is its magnitude, then choose the correct options :

[RM-AA] ekuk R

fdlh funsZ'k fcUnq ds lkis{k o jsf[k; xfr djrs gq, d.k dk fLFkfr lfn'k gS rFkk R bldk ifjek.k

gSA blh izdkj v

leku funsZ'k fcUnq ds lkis{k d.k dk osx lfn'k gS rFkk v bldk ifjek.k gS rc lgh dFkuksa dk p;u dhft,A

(A*) v dRdt

(B) v = dRdt

(C*) v = dRdt

(D*) dR dR

Sol.

R

R' dR

dR

dR = dR cos

dRdRcos

dt dt

v = dRdt

4. A point charge q is placed at origin let AE

, BE

and CE

be the electric field intensity at three points A (1, 2, 3) , B (1, 1, 1) and C (2, 2, 2) respectively due to the charge q. Then choose the correct options :

,d fcUnq vkos'k q ewy fcUnq ij j[kk gSA ekuk vkos'k q ds dkj.k fcUnq A (1, 2, 3) , B (1, 1, 1) rFkk C (2, 2, 2) ij fo|qr {ks=k dh rhozrk,sa e'k% AE

, BE

rFkk CE

gS] rc lgh fodYi dk p;u fdft,A [ES_EF]

(A*) BC

E

E

= 4

(B) A C A CE || E , (E is parallel to E )

A C A CE || E , (E , E )

d slekUrj gSA

(C) CB

E 18E

(D*) A BE E

= 0 (Dot product AE

of and BE )

A BE E

= 0 ( AE

rFkk BE

dk vfn'k xq.kuQy 'kwU; gSA)

5. The speed of a small object undergoing uniform circular motion is 4 m/s.The magnitude of the change in the velocity during 0.5 seconds is also 4 m/s. Then choose the correct options : [CM-KN]

,d leku o`kh; xfr djrh gqbZ ,d NksVh oLrq dh pky 4 m/s gSA 0.5 lSd.M ds nkSjku blds osx esa ifjorZu dk ifjek.k] Hkh 4

m/s gSA rc lgh fodYi dk p;u fdft,A

(A*) the angular speed of object is 23

rad/sec.

(B*) the centripetal acceleration of the object is 83

m/s2



Toll Free : 1800 200 2244 | 1800 258 5555 JRPT2051014C0-4

(C*) the radius of the circle is 6

metre.

(D) the radius of circle is 3

metre.

(A*) oLrq dh dks.kh; pky 23

rad/sec gSA

(B*) oLrq dk vfHkdsUnzh; Roj.k 83

m/s2 gSA

(C*) o`k dh f=kT;k 6

ehVj gSA

(D) o`k dh f=kT;k 3

ehVj gSA

Sol. If the angular displacement is for 0.5 seconds, then the magnitude of change in velocity is 0.5 lS- esa ;fn dks.kh; foLFkkiu gS rks osx esa ifjorZu dk ifjek.k

V = 2 2 2v v 2v cos = 2v sin 2

V = 2 4 sin 2

or 2

=

6

& = 3

= 0.5

= 23

rad/s

R = v

= 4 32

= 6

m

and centripetal acceleration (vkSj vfHkdsUnzh; Roj.k) = v = 4 23

= 83

m/s2 .

6. Two particles are projected from the same point with the same speed at different angles 1 & 2 to the horizontal. They have the same range. Their times of f l ight are t1 & t2 respectively. Then choose the correct options : [PM-PH]

nks d.kksa dks ,d gh pky ls ,d gh fcUnq ls {kSfrt ls vyx&vyx dks.k 1 rFkk 2 ij iz{ksfir fd;k tkrk gSA mudh ijkl leku gSA muds mM~M;u dky e'k% t1 & t2 gSA rc lgh fodYiksa dk p;u fdft,A

(A*) 1 = 90 2 (B*) 11

tsin

= 2

2

tsin

(C*) 12

tt

= tan 1 (D) 12

tt

= tan 2

Sol. R = 2u sin2

g

is same for angles & 90 .

R = 2u sin2

g

, rFkk 90 ds fy, leku gSA

t1 = 12usin

g

= 22usin(90 )

g

t2 = 22usin

g

= 2u sin 1(90 )g

7. Figure shows two blocks A and B connected to an ideal pulley string system. In this system when bodies are released. Then choose the correct options : (neglect friction and take g = 10 m/s2) [NL-CM

fp=k esa n'kkZ;s vuqlkj nks fi.M A rFkk B ,d vkn'kZ f?kjuh&jLlh fudk; ls tqM+s gSA bl fudk; esa tc fi.Mksa dks eqDr djrs gSaA rc lgh fodYiksa dk p;u fdft,A ?k"kZ.k dks ux.; ekusa rFkk g = 10 eh-@ls-2 ysa



Toll Free : 1800 200 2244 | 1800 258 5555 JRPT2051014C0-5

(A) Acceleration of block A is 1 m/s2 fi.M A d k Roj.k 1 eh-/ls-2 gSA (B*) Acceleration of block A is 2 m/s2 fi.M A d k Roj.k 2 eh-/ls-2 gSA (C) Tension in string connected to block B is 40 N B ls t qM+h jLlh esa ruko 40 N gSA (D*) Tension in string connected to block B is 80 N B ls t qM+h jLlh esa ruko 80 N gSA

Sol. Applying NLM on 40 kg block 40 fdxzk fi.M ij U;wVu ds fu;e yxkus ij

40kg

10kg

4aT

2T2T

4T

a

10kg

4aT

2T2T

4T

a

10kg

4aT

2T2T

4T

a

400 4T = 40 a For 10 kg block T = 10.4 a Solving a = 2m/s2 10 fdxzk- fi.M ds fy, T = 10(4 a) gy djus ij a = 2m/s2 T = 80 N 8. A circular road of radius r is banked for a speed v = 40 km/hr. A car of mass m attempts to go on

the circular road. The friction coefficient between the tyre and the road is negligible. Then choose the correct options [CM-CF]

r f=kT;k dh o`kkdkj lM+d dks v = 40 km/hr dh pky ds fy;s cafdr x;k gSA m nzO;eku dh ,d dkj bl o`kkdkj iFk ij xfr djrh gSA lM+d rFkk Vk;jksa ds e/; ?k"kZ.k xq.kkad ux.; gSA rc lgh fodYiksa dk p;u fdft,A

(A*) The car can make a turn without skidding. dkj fQlys fcuk ?kwe ldrh gSA (B*) If the car turns at a speed less than 40 km/hr, it will slip downwards ;fn dkj dh eksM+ ij pky 40 km/hr, ls de gS] rks ;g uhps dh vksj fQlysxhA (C*) If the car turns at the constant speed of 40 km/hr, the force by the road on the car is equal to

222 mv(mg)

r



Toll Free : 1800 200 2244 | 1800 258 5555 JRPT2051014C0-6

;fn dkj dh eksM+ ij fu;r pky 40 km/hr gS]rks lM+d ds }kjk dkj ij yxk;k x;k cy 22

2 mv(mg)r

ds cjkcj gSA (D*) If the car turns at the correct speed of 40 km/hr, the force by the road on the car is greater

than mg as well as greater than 2mv

r

;fn dkj dh eksM+ ij pky Bhd 40 km/hr, gS] rks lM+d ds }kjk dkj ij yxk;k x;k cy mg ls vf/kd gksxk

lkFk gh ;g 2mv

rls Hkh vf/kd gksxkA

Sol. When speed of car is 40 km/hr, car can make a turn without skidding. If speed is less than 40 km/hr than tendency of slipping is downward so it will slip down. If speed is greater than 40 km/hr than tendency of slipping upward so it will slip up.

If the cars turn at correct speed 40 km/hr tc dkj dh pky 40 km/hr, gS] rks dkj fcuk fQlys ?kwe ldrh gSA ;fn pky 40 km/hr ls de gS rks blds

uhps dh vksj fQlyus dh izofk gksrh gS] vr% uhps fQly tkrh gSA ;fn pky 40 km/hr ls vf/kd gS rks bldh ij dh vksj fQlyus dh izofk gksrh gS vr% ij dh vksj fQlysxhA ;fn dkj Bhd 40 km/hr pky ij ?kwe ys

than rks N cos = mg

N sin = 2mv

r

N = 2

2 mv(mg)r

Ans.



Toll Free : 1800 200 2244 | 1800 258 5555 JRPT2051014C0-7

9. A point source of light at O at a distance x from the screen S produces light of intensity 0 at the centre of the screen. If a completely reflecting mirror M is placed at a distance x behind the source as shown in figure (neglect interference of light). Then choose the correct options [GO-PM]

insZ S ls x nwjh ij fcUnq O ij fLFkr izdk'k dk ,d fcUnq L=kksr insZ ds dsUnz ij 0 rhozrk dk izdk'k mRiUu djrk gSA ;fn iw.kZ ijkorZd niZ.k M L=kksr ds ihNs x nwjh ij fp=kkuqlkj j[kk gqvk gSA (izdk'k dk O;frdj.k ux.; ekusa) rc lgh dFku gksaxsA &

M S

x x O

(A) Total intensity of light at the centre of the screen is 1/9 times of 0 (B*) Total intensity of light at the centre of the screen nearly changes by 11%

(C*) Total intensity of light at the centre of the screen is 10/9 times of 0 (D) Total intensity of light at the centre of the screen nearly changes by 99% (A) insZ ds dsUnz ij izdk'k dh dqy rhozrk 0 dh 1/9 xquk gksxhA (B*) insZ ds dsUnz ij izdk'k dh dqy rhozrk yxHkx 11% ls ifjofrZr gksxhA

(C*) insZ ds dsUnz ij izdk'k dh dqy rhozrk 0 dh 10/9 xquk gksxhA (D) insZ ds dsUnz ij izdk'k dh dqy rhozrk yxHkx 99% ls ifjofrZr gksxhA Sol. Initially izkjEHk esa

0 = 2Kr

r d O

Screen

For the given energy : nh xbZ tkZ ds fy,

2r

d O

Screen

r

Finally

r

= 02K

99 r

Total intensity = 0109

. Hence 11% increase

dqy rhozrk = 010

9

. vr% 11% c



Toll Free : 1800 200 2244 | 1800 258 5555 JRPT2051014C0-8

10. A ball is dropped onto a pad at A and rebounds with a velocity V0 at an angle 60 with a horizontal as shown in figure. The ball will enter the opening BC, if V0 may be-

A ij fLFkr ,d xn~nh (pad) ij ,d xsan dks NksM+k tkrk gSA ;g {kSfrt ls 60 ds dks.k ij V0 osx ls fp=kkuqlkj mNyrh gSA xsan [kqys Hkkx BC esa izos'k djsxh, ;fn V0 gks ldrk gS - [PM-PH]

60

V0

3m

2

1m

B

C 0.5m

A

(A) 5 m/s (B*) 6 m/sec (C*) 7 m/sec (D*) 50 m/sec

Sol. As y = 32

tan 60 12

20

31041V4

pwafd y = 32

tan 60 12

20

31041V4

y = 20

3 152 V

As pwafd 1 y 1.5

1 1.5 20

15V

1.5

30 V0 < Possible values of V0 are 6, 7, 50 m/sec V0 ds laHko eku 6, 7, 50 m/sec gSA

11. Spring 1 has natural length of 0.5 meter and force constant K1 = 25 N/m and spring 2 has natural length of 1 meter and force constant K2 = 10 N/m. They are joined together and their free ends are stretched so that the ends are fixed to the two walls 2 meter apart as shown in figure (springs are massless) then correct options are : [CO-SP]

fLizax 1 dh lkekU; yEckbZ 0.5 ehVj rFkk cy fu;rkad k1 = 25 N/m gS rFkk fLiazx 2 dh lkekU; yEckbZ 1 ehVj rFkk cy fu;rkad k2 = 10 N/m gSA bu nksuks dks vkil esa tksM+k tkrk gS rFkk buds eqDr fljksa dks ,d nwljs ls 2 ehVj nwjh ij fLFkr nhokjksa ls tksM+k tkrk gS fLax nzO;ekughu gS rc lgh fodYi gksxsa &

(A*) Ratio of elongation in the spring-1 to spring-2 is 25

(B*) Length of the spring-1 in equilibrium is 914

m.

(C*) Ratio of potential energy stored in the spring-1 to spring-2 is 25



Toll Free : 1800 200 2244 | 1800 258 5555 JRPT2051014C0-9

(D*) Length of the spring-2 in equilibrium is 1914

m

(A*) fLiazx-1 esa foLrkj dk fLizax-2 esa foLrkj ds lkFk vuqikr 25

gSA

(B*) lkE;koLFkk esa fLiazx-1 dh yEckbZ 914

m gSA

(C*) fLiazx-1 rFkk fLizax-2 esa laxzfgr fLFkfrt tkZ dk vuqikr 25 gksxkA

(D*) lkE;koLFkk esa fLiazx-2 dh yEckbZ 914

m gSA

Sol. As pwafd F = k1 x1 = k2 x2 1 22 1

x kx k

25x1 = 10x2 x2 = 2.5 x1 and rFkk x1 + x2 = 0.5

x1 = 17

and rFkk x2 = 5

14

Hence their stretched lengths are x1 + 0.5 = 9

14 met. and x2 + 1 =

1914

met

U1 = 2

1

F2k

U2 = 2

2

F2k

1 22 1

U kU k

vr% f[kaph gqbZ voLFkk esa bldh yEckbZ;kW x1 + 0.5 = 9

14 met. rFkk x2 + 1 =

1914

met

U1 = 2

1

F2k

U2 = 2

2

F2k

1 22 1

U kU k

12. At distance of 5cm and 10cm outwards from the surface of a uniformly charged solid sphere, the potentials are 100V and 75V respectively. Then choose the correct options :

le:i vkosf'kr Bksl xksys dh lrg ls ckgj dh rjQ 5cm o 10cm nwjh ij foHko e'k% 100V rFkk 75V gSA rc lgh fodYiksa dk p;u fdft,A [ES-EF]

(A) Potential at its surface is 75V. (B*) The charge on the sphere is (5/3) 109C. (C*) The electric field on the surface is 1500 V/m. (D) The electric potential at its centre is 200V. (A) bldh lrg ij foHko 75V gSA (B*) xksys ij vkos'k (5/3) 109C gSA (C*) lrg ij fo|qr {ks=k 1500 V/m gSA (D) blds dsUnz ij foHko 200V gSA Sol. kQ(r 5cm) = 100V &

kQ(r 10cm) = 75 V

Q = 95 10 C3 , r = 10 cm

Vsurface lrg = kQr

= 150V Esurface lrg = 2kQr

= 1500 V/m

Vcentre dsUnz =32

Vsurface lrg = 32

150 = 225 V



Toll Free : 1800 200 2244 | 1800 258 5555 JRPT2051014C0-10

13. Slider block A moves to the right with constant velocity 6 m/s in the arrangement as shown in the figure. C and D are points on the string . If BV

, CV

and DV

are the velocities of B, C and D respectively then, choose correct options : [NL_CM]

CykWd A nka;h vksj fu;r osx 6 m/s ls n'kkZ;s vuqlkj xfr dj jgk gSA C o D Mksjh ij nks fcUnq gSA ;fn B,C o D ds osx e'k% BV

, CV

rFkk DV

gS] rc lgh fodYiksa dk p;u dhft,A

A

6 m/s

C D

B

(A*) C D BV V 4 V

(B) B C B D BV V V V V

(C*) B C B D BV V V V 2 V

(D*) C D BV V 2 V

Sol.

A

6 m/s

C D

B

X3 X4X2

VB

X1

Clearly Li"Vr% , VC = 6 m/s By string constraint, Mksjh ca/ku ls

1 2 3 4x x x x 0

6 + VB + VB + VB = 0 VB = 2 m/s

Also, ;g Hkh c D BV V V

2

C D

VB

VDVC



Toll Free : 1800 200 2244 | 1800 258 5555 JRPT2051014C0-11

VD = 2 m/s VD = 2 m/s

Thus vr% C DV V 6 2 8 m/ s

B CV V 4 m / s

B DV V 4 m / s

C DV V 6 2 4 m / s

14. Which of the following functions represent SHM? fuEu esa ls dkSuls Qyu ljy vkorZ xfr nf'kZr djsaxs \ [SH-EQ] (A*) sin 2t (B*) sin2 t (C*) sin t + 2 cos t (D*) sin t + 2sin (t +

3 ) + 3sin (t +

2 )+ 4sin (t + )

Sol. Equation of S.H.M x x0 = a sin (t + ) x = x0 + a sin (t + ) l-vk-x- dh lehdj.k x x0 = a sin (t + ) x = x0 + a sin (t + ) (A) x = sin2t (B) x = sin2 t = 1 cos2 t

2

= 12

12

cos2 t

x 12

= 12

cos2 t

(C) x = sin t + 2cos t = sin (t + ) { = tan1 2} So represent equation of S.H.M vr% lehdj.k l-vk-x- dh lehdj.k dks izLrqr djrh gS (D) y = A1 sint + A2 sin(t + 1) + A3 sin(t + 2) + A4 sin(t + 4) It can be represented as bls fuEu izdkj iznf'kZr dj ldrs gSA y = A sin (t + ) So it represents SHM vr% ;g SHM dks iznf'kZr djrh gSA

15. Two point charges (q1 and q2) are placed on x-axis. Figure shows graph of potential (V) on x-axis with x-co-ordinate. Then choose correct options : [ES_EP]

nks fcUnq vkos'k (q1 o q2) x-v{k ij j[ks gq, gSA fp=k esa vkjs[k x-v{k ij foHko (V) dk x-funsZ'kkad ds lkFk xzkQ n'kkZrk gSA rc lgh fodYiksa dk p;u dhft,A

(A*) q1 > 0 (B) q2 > 0 (C*) |q1| > |q2| (D) |q1| < |q2| Sol. Potential near q1 is + so q1 > 0. Potential near q2 is so, q2 < 0. Potential is zero x2 < x1 so |q1|

> |q2|.



Toll Free : 1800 200 2244 | 1800 258 5555 JRPT2051014C0-12

q1 ds lehi foHko + gS] vr% q1 > 0 gSA q2 ds lehi foHko gS] vr% q2 < 0 gSA x2 < x1 ij foHko 'kwU; gS vr% |q1| > |q2| gSA

SECTION-2 : (Integer value correct Type) [k.M2 : (iw.kkZad eku lgh dkj)

This section contains 30 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive)

bl [k.M esa 30 'u gSA R;sd 'u dk mkj 0 ls 9 rd nksauks 'kkfey ds chp dk ,dy vadh; iw.kkZad gSA

Single (Integer) (30)

16. A tube in vertical plane is shown in figure. It is filled with a liquid of density = 1000 kg/m3 and its end B is closedThen the force exerted by the fluid on the tube at end B in newton will be : [Neglect atmospheric pressure and assume the radius of the tube to be negligible in comparison to = 4 m, g = 10 m/s2] [FL-PR]

fp=k esa ,d uyh ?okZ/kj ry esa gSA bldks = 1000 kg/m3 ?kuRo okys nzo ls Hkj dj fdukjs B dks cUn

dj fn;k tkrk gS rks nzo ds }kjk uyh ds B Nksj ij yxus okyk cy U;wVu esa gksxk [ok;qnkc dks ux.; ekusa rFkk = 4 m dh rqyuk esa uyh dh f=kT;k dks ux.; ekusa, g = 10 m/s2]

A0 = 104 m2) (vuqizLFk dkV {ks=kQy = A0 = 104 m2)

Ans. 4 Sol. Pressure exerted by fluid at closed end B is can fljs B ij nzo }kjk yxk;k x;k nkc P = g force exerted by fluid at closed end B is can fljs B ij nzo }kjk yxk;k x;k cy F = PA = g A0 = 4 Ans.

17. A uniform rope lies on a table that part of it lays over. The rope begins to slide when the length of hanging part is 25 % of entire length. The co-eff icient of friction between rope and table is . Find 6. [FR-SG]

est ij j[kh gqbZ ,d le:i jLlh dk dqN Hkkx est ls uhps yVd jgk gSA tc yVdk;s x;s Hkkx dh yEckbZ] lEiw.kZ yEckbZ dh 25 % gS rc jLlh fQlyuk kjEHk dj nsrh gS] est o jLlh ds e/; ?k"kZ.k xq.kkad gS] rks 6 dk eku gksxkA



Toll Free : 1800 200 2244 | 1800 258 5555 JRPT2051014C0-13

Ans. 2 Sol. Apply system equation fudk; dh lehdj.k yxkus ij

m

4g = 3m

4g

= 13

= 0.33

Ans. 2

18. In the shown figure maximum value of M so that the blocks A, B and C move without any relative motion between them is x m kg, then x is :

fp=k esa iznf'kZr M nzO;eku dk vf/kdre~ eku rkfd CykWd A, B rFkk C muds e/; fcuk lkis{k xfr ds xfr dj lds] x m kg gS rc x gksxkA [FR_SG]

Ans. 3

Sol. fmax between A and B =

12

mg (this must be choosen)

fmax ArFkk B ds e/; =

12

mg (;g p;fur gksxk)

amax A =

g2

As, pwafd g2

= Mg

3m M

M = 3m.

19. The maximum height above point A, where the ball can reach as shown in figure is 40 L metre9X

,

then x is : [CM_VT] fp=k esa iznf'kZr xsan dh fcUnq A ls vf/kdre WpkbZ tgkW rd xsan igqWp ldrh gS] 40 L

9XehVj esa gSA rc x

gksx :

Ans. 3



Toll Free : 1800 200 2244 | 1800 258 5555 JRPT2051014C0-14

Sol. Let be the angle with the vertical where string slackes and v be the velocity at that point ekuk /okZ/kj ds lkFk cuk;k x;k dks.k gS tgk Mksjh



Toll Free : 1800 200 2244 | 1800 258 5555 JRPT2051014C0-15

Sol. Total energy d qy t kZ E = 12

m 2 A2

0.5 + 0.4 = 12

0.2 2252 A2

9 = (50)2 A2 A = 0.06 m

Ans. 6

22. A circular ring of radius a with uniform charge density is in the xy plane with centre at origin. A particle of mass m and charge q is projected from P(0, 0, a 3 ) towards origin with initial velocity u. The minimum value of the velocity so that the particle does not return to P is

0

qx m

. Find 'x'

(neglect gravity). ,dleku vkos'k ?kuRo dh a f=kT;k dh ,d o`kkdkj oy; xy ry esa fLFkr gS (ftldk dsUnz ewy fcUnq ij gS)

m nzO;eku rFkk q vkos'k ds ,d d.k dks kjfEHkd osx u ls fcUnq P(0, 0, a 3 ) ls ewy fcUnq dh vksj {ksfir fd;k tkrk gSA osx dk U;wure eku rkfd d.k P dh vksj u ykSVs]

0

qx m

gSA 'x' dks Kkr dhft,A (xq:Ro dks

ux.; ekusaA) [WE_UC] Ans. 2

Sol.

By energy equation tkZ lehdj.k }kjk

0

1 2 a4 2a

q = 12

mu2.

u = 0

q2 m

x = 2

23. A radionuclide has a halflife of 1.6 103 years. If a sample contains 3.0 1016 such nuclei at a certain instant, the activity of the sample at this time in (curie) Ci is 11 10x Ci then write the value of x. (given 1 Ci = 3.7 1010 dps)

,d jsfM;ksukfHkd dh v)Zvk;q 1.6 103 o"kZ gSA ;fn ,d uewus esa fdlh {k.k ij 3.0 1016 ,sls ukfHkd gS fd bl le; ij Ci (D;wjh) esa uewus dh lf;rk 11 10x Ci gS rks x dk eku fy[kksA (fn;k x;k gS fd 1 Ci = 3.7 1010 dps) [NP-DL]

Ans. x = 6 Sol. t1/2 = 1.6 103 3.16 107 s

Decay constant {k; fu;rkad = 2/1t

693.0

Activity lf;rk A = N

= 73 1016.3106.1693.0

= 1.4 1011 s1

A = 11 106 Ci



Toll Free : 1800 200 2244 | 1800 258 5555 JRPT2051014C0-16

24. A particle of mass m starts at t = 0 from the point A(0, 5) and moves with uniform velocity of 5i m/ s . After 5 seconds, the angular velocity of the particle about the origin is 1

13 radian/sec

then is : [CM_KN] m nzO;eku dk ,d d.k t = 0 ij fcUnq A(0, 5) ls izkjEHk gksrk gS rFkk fu;r osx 5i m/ s ls xfr djrk gSA 5

lSd.M i'pkr~ ewy fcUnq ds lkis{k d.k dk dks.kh; osx 1

13 jsfM;u/lS- gSA rc gksxk &

Ans. 2 Sol. As pwafd = about O (O ds lkis{k) =

V rr

= 5 5 1

26650 650

5 m/sect = 5

25 m

t = 0

5 m650

O

25. A swimmers speed in the direction of flow of river is 16 km h1. Swimmer's speed against the direction of flow of river is 8 km h1. Calculate the velocity of flow of the river (km h1).

,d rSjkd dh unh ds cgko dh fn'kk esa pky 16 km h1 gS rFkk cgko ds foijhr fn'kk esa rSjkd dh pky 8 km h1 gS] unh dk osx (km h1) Kkr djks \ [RL-OD]

Ans. 4 Sol. VS + Vr = 16 VS Vr = 8 VS = 12 km/hr Vr = 4 km/hr

26. The wavelength of the first line in balmer series in the hydrogen spectrum is 1. Wavelength of the

second line 2 = 1209x

, calculate x : [MP-TR]

gkbMkstu LisDVe dh ckej Js.kh dh igyh js[kk dh rjaxnS/;Z 1gSA nwljh js[kk dh rjaxnS/;Z 2 = 1209x

gS]

rks x dk eku gksxkA Ans. 3

Sol. 1

1

= R 1 14 9

1 = 4 95R

similarly blh izdkj 2

1

= R 21 14 4

2 = 163R

= 163

54 9

= 2027

Ans. 3



Toll Free : 1800 200 2244 | 1800 258 5555 JRPT2051014C0-17

27. When a centimetre thick surface is illuminated with light of wavelength , the stopping potential is V. When the same surface is illuminated by light of wavelength 2, the stopping potential is V/3. The threshold wavelength for the surface is p, find p : [MP-PE]

tc ,d lseh eksVh lrg dks rjaxnS/;Z okys izdk'k ls izdkf'kr djrs gS rks fujks/kh foHko V gSaA tc leku lrg dks 2 rjaxnS/;Z okys izdk'k ls izdkf'kr djrs gS rks fujks/kh foHko V/3 gSaA lrg dh nSgyh rjax nS/;Z p gS] rks p dk eku Kkr dhft,A

Ans. 4

Sol. hC

= + eV ....(i)

hC2

= + eV3

....(ii) 3 II I

3 12

hc

= 2 = hc4

th = 4

28. If min = 2is minimum wavelength produced in X-ray tube, k is the wavelength of k line. It is

observed that (k min) = 7. If the operating tube voltage is doubled then (k min) = x. Find x. [MP-XR]

;fn min = 2X-fdj.k ufydk esa mRiUu U;wure~ rjaxnS/;Z gS, k, k js[kk ds rjaxnS/;Z gSA ;g izsf{kr gksrk gS

fd (k min) = 7;fn ufydk ij vkjksfir oksYVst dks nqxquk fd;k tkrk gS] rc (k min) = xgS] rks x Kkr djksA

Ans. 8 Sol. Case-I :

min = 2 (k min) = 7 k= 9 Case-II : If operating voltage is doubled

min = (k min) = 8 Case-II : ;fn vkjksfir oksYVst nqxquk fd;k tkrk gS

min = (k min) = 8

29. In the shown figure the time after which the block of mass 1 kg will separate from the block of

mass 2kg is 8n15

second then n is : (g = 10 m/sec2, neglect the size of block of mass 1 kg)

og le; ftlds i'pkr~ 1 kg nzO;eku dk CykWd 2kg nzO;eku ds CykWd ls vyx gks tk;sxk] 8n15

lsd.M+ gS

rc n gksxk (g = 10 m/sec2, 1 kg CykWd dk vkdkj ux.; ekfu;sA) [FR_KF]

Ans. 1



Toll Free : 1800 200 2244 | 1800 258 5555 JRPT2051014C0-18

Sol. ablock = 51

= 5 m/s2

aplank = 30 5

2

= 252

m/s

Srel =

12

arel t2

2 = 12

152

t2

t = 2215

sec.

therefore required answer is 1 vr% vko';d mkj 1 gSA

30. A cube of wood supporting a 200 gm mass just f loats in water. When the mass is removed the cube rises by 2 cm at equilibrium. If the side of cube is a ( in cm) f ind a

2.

,d ydM+h ds ?ku ij 200 gm nzO;eku j[kk gS ;g ty esa iwjk Mwck gSA tc nzO;eku gVk fy;k tkrk gS rks

lkE;koLFkk ij ?ku 2 cm ls ij mB tkrk gSA ;fn ?ku dh Hkqtk a (cm esa) gS] rks a2 Kkr dhft,A

[FL-BY] Ans. 5 Sol. (a2) (2) = 200 g but ijUrq = 1 gm/cm 3 a = 10 cm. Ans. 5

31. If the kinetic energy of a particle decreases by 4%, its Debroglie wavelength increases by x%. Find x in nearest integer.

[MP-DB]

;fn ,d d.k dh xfrt tkZ 4% ls ?kVrh gS] rks bldh Mh&czksxyh rjaxnS/;Z x% ls c



Toll Free : 1800 200 2244 | 1800 258 5555 JRPT2051014C0-19

fp=k esa ,d ?k"kZ.kjfgr iFk iznf'kZr fd;k x;k gS] ftldk ,d Hkkx 4 m f=kT;k dk /okZ/kj o`k gSA ,d k

=3mg

4 N/m fLax fu;rkad okyh fLiazx dk ck;ka fljk n



Toll Free : 1800 200 2244 | 1800 258 5555 JRPT2051014C0-20

34. The maximum value of F for which both blocks will move together is 120x N. Find x. F dk vf/kdre~ eku ftlds fy, nksuksa CykWd ,d lkFk xfr djsaxs] 120x N gS] rks x Kkr djksA [FR_MQ]

]

Ans. 7 Sol.

fs max = 0.6 30 g = 180 fk = 0.2 35 g = 70 fs max fk = 5a a = 22 F

max fs max = 30 a F

max = 840

35. A body of mass 2 kg suspended through a vertical spring executes simple harmonic motion of period 4s. If the oscillations are stopped and the body hangs in equilibrium. The potential energy is 10 U (in J) stored in the spring then U is : [SH_SM]

,d /okZ/kj fLax ls yVdh 2 kg nzO;eku dh oLrq 4s ds vkorZ dky ls ljy vkorZ xfr djrh gSA ;fn nksyu jksd fn;s tk;sa o oLrq lkE;koLFkk esa yVdh jgrh gS] rks fLax esa laxzfgr fLFkfrt tkZ 10 U twy esa gS] rc U gksxkA

Ans. 4

Sol. K = m2 K = m 2

T2

K = 2

2

Tm4

mg = Kx equilibrium lkE;oLFkk x =

Kmg

U = 21 Kx2

=

21 K

2

22

Kgm

=

m42Tgm2

222

=

210424102 222

= 40 J

U = 4

36. A particle is projected along a rough horizontal plane, coefficient of friction varies as = 0 r2 where r is the distance from the origin in meters and 0 is a positive constant whose value is 20 m2. The initial distance of the particle is 1 m from the origin and its velocity is radially outwards. The minimum initial velocity at this point so that particle never stops is 10x m/sec then x is : (use g = 10 m/sec2)

[WE_WE] ,d d.k dks [kqjnjs {kSfrt lrg ds vuqfn'k iz{ksfir fd;k x;k gS] ftldk ?k"kZ.k xq.kkad = 0 r

2 ds vuqlkj

ifjofrZr gksrk gS tgk r ewy fcUnq ls nwjh eh- es gS rFkk 0 ,d /kukRed fLFkjkad gS ftldk eku 20 m2 gSA ewyfcUnq ls d.k dh izkjfEHkd nwjh 1 eh- gS rFkk bldk osx f=kT;h; fn'kk esa ckgj dh vksj gSA bl fcUnq ij U;wure kjfEHkd osx 10x m/sec gS ftlls d.k dHkh u :dsA rc x gksxkA (g = 10 m/sec2 )

Ans. 2 Sol. Work done against friction must equal the initial kinetic energy. ?k"kZ.k ds fo:) fd;k x;k dk;Z kjfEHkd xfrt tkZ ds cjkcj gksrk gSA



Toll Free : 1800 200 2244 | 1800 258 5555 JRPT2051014C0-21

2mv21

=

1

dxmg ; 2v2

= 0 21

1g dxx

; 2v2

= 0g

1x

1

v2 = 2g0 v = 02g v = 400 = 20 hence x = 2

37. A cubical block of copper of side 10 cm is floating in a vessel containing mercury. Water is poured into the vessel so that the copper block just gets immeresed. The height of water column in cm is :

(Hg = 13.6 g/cc , Cu= 7.3 g/cc, water =1 gm/cc [FL-BY] 10 lseh- Hkqtk dk ,d rkWcsa dk ?kukdkj CykWd ikjs ls Hkjs ik=k esa rSj jgk gSA bl ik=k esa vc ikuh bruk Hkjk

tkrk gS ftlls ;g CykWd Bhd lEiw.kZ Mwc tkrk gS] rks ikuh LrEHk dh pkbZ cm esa gksxhA (Hg = 13.6 g/cc , Cu= 7.3 g/cc, water =1 gm/cc)

Ans. 5 Sol. Let h = height to of water column ekuk h = ikuh LrEHk dh pkbZ then rks wgh + Hg g(10h) = Cu g10 h + 13.6 (10 h) = 73 63 = 12.6 h h = 5 cm

38. A block of mass 1 kg is released on wedge from position M as shown in figure (Neglect friction every where). The force exerted by vertical wall W1 on wedge, when the block is at position N is 15 P2

newton, find P : (Take g = 10 m/sec2) 1 kg nzO;eku dk CykWd ost ij fp=kkuqlkj M fLFkfr ls NksM+k tkrk gSA (izR;sd txg ?k"kZ.k ux.; gSA) /okZ/kj

nhokj W1 }kjk ost ij vkjksfir cy tc CykWd fLFkfr N ij gS] 15 P2

U;wVu gS] rc P Kkr djksaA (g = 10 m/sec2 ys) [CM_VT]

Ans. 3 Sol. v 2(10)(Rcos60 ) 10 m/s

Now, vc N mg cos60 = 2mv

R

N = 15 Newton

Force exerted by vertical wall = N sin60 = 15 3 N2

.

/okZ/kj nhokj }kjk yxk;k x;k cy = N sin60 = 15 3 N2

.

therefore vr% P = 3 Ans.

39. A small solid ball is dropped from a height h (metre) above the free surface of a liquid. It strikes the surface of the liquid at t = 0. The density of the material of the ball is 500 kg/m3 and that of

liquid is 1000 kg/m3. If the ball comes momentarily at rest at t = 2 sec, find h5

. (Neglect viscousity) [FL-BY]



Toll Free : 1800 200 2244 | 1800 258 5555 JRPT2051014C0-22

,d NksVh Bksl xsan dks h (ehVj) pkbZ ls nzo dh eqDr lrg ij fxjk;k tkrk gSA ;g nzo dh lrg ij t = 0 le; ij Vdjkrh gSA xasn ds inkFkZ dk ?kuRo 500kg / m3 rFkk nzo dk ?kuRo 1000 kg/m3 gSA vxj t = 2

lSd.M ij xsan {kf.kd :i ls :d tkrh gS] rks h5 dk eku Kkr djsA (';kurk dks ux.; ekus)

Ans. 4

Sol.

Velocity of ball when it reaches to surface of liquid tc ;g nzo dh lrg ij igqWprh gS rc xsan dk osx

a = V500

gV500gV1000 ; where V is the volume of the ball.

a = V500

gV500gV1000 ; tgkW V xasn dk vk;ru gSA

a = 10 m/sec2 apply vkjksfir djus ij v = u + at 0 = gh2 10t

gh2 = 10 (2) 2 10 h = 400 h = 20 m Ans

40. A point charge of 10 nC is at a distance of 40 cm from the surface of a sphere of radius 10 cm which was uniformly charged to a potential 15 kV. Work done by external agent to decrease the

separation between charge and surface to 30 cm is 15 Jx , then value of x is : [ES-EP]

15 kV foHko rd ,d leku :i ls vkosf'kr 10 lseh f=kT;k ds ,d xksys ds i"`B ls 40 lseh nwjh ij 10 nC dk ,d fcUnqor vkos'k fLFkr gSA vkos'k o i"`B ds chp dh nwjh 30 lseh rd ?kVkus ds fy;s ck dkjd }kjk

fd;k x;k dk;Z 15 Jx gS] rc x dk eku gksxkA

Ans. 2



Toll Free : 1800 200 2244 | 1800 258 5555 JRPT2051014C0-23

Sol. We have potential of sphere V = KQR

ge tkurs gS fd xksys dk foHko V = KQR

15000 = KQR

Q = 15000RK

=15000 10

K 100 =

1500K

initially izkjEHk esa Potential at point 1 fcUnq 1 ij foHko

V1 = KQ K 1500

5050 K100100

= 1500K 50K

100

= 3000

Finally vUr esa Potential at point 2 fcUnq 2 ij foHko

V2 = KQ 1500K40 40K100 100

= 15000

4 = 3750

So work done by external agent to decrease the separation vr% nwjh dks ?kVkus esa ckg~; dkjd }kj fd;k x;k dk;Z W = q(v2 v1) = 10 109 (3750 3000) = 750 10 109 = 7.5 J Therefore vr% x = 2

41. Masses are suspended in vertical plane as shown (the system is at rest and in equilibrium). When m1 is removed the system makes 20 oscillations in 40 sec. When m2 is also removed gently then time required for the system to make complete 30 oscillations is 8T seconds find T :

n'kkZ,uqlkj fi.Mksa dks /oZ ry esa vkyfEcr fd;k x;k gSA (fudk; fLFkj rFkk lkE;koLFkk esa) gSA m1 dks gVkus ij fudk; 40 lSd.M esa 20 nksyu djrk gSA tc m2 dks /khjs ls gVk fy;k tk, rks fudk; }kjk 30 nksyu djus esa yxk le; 8T lsd.M+ gS] rc T Kkr dhft, [SH-SM]



Toll Free : 1800 200 2244 | 1800 258 5555 JRPT2051014C0-24

Ans. 5 Sol. As, pwafd T = 2 sec.

2 = 9002k

........(i)

and rFkk T1 = 4002k

........(ii)

T1 = 43

sec.

Time required for 30 oscillations = 40 sec. 30 nksyuksa ds fy, vfHk"V le; = 40 sec.

42. A particle of mass m oscillates between P1 and P2 inside a fixed smooth spherical shell of radius 2 m. At any instant the kinetic energy of the particle is 4 J. Then the force exerted (in newton) by the particle on the shell at this instant is : [CM_VT]

m nzO;eku dk d.k 2m f=kT;k ds ?k"kZ.kjfgr fLFkj xksykdkj dks'k ds vUnj P1 rFkk P2 ds e/; nksyu djrk gSA ;fn fdlh {k.k ij d.k dh xfrt tkZ 4J gS rks bl {k.k ij dks'k ij d.k }kjk yxk;k cy U;wVu esa gksxkA

Ans. 6

Sol. N mg sin = 2mv

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

As pwafd 12

mv2 = mgh = E (say) ............(ii)

N = h 2EmgR R

= E 2E 3ER R R

therefore force is 6 N Ans. vr% cy 6 N U;wVu gSA



Toll Free : 1800 200 2244 | 1800 258 5555 JRPT2051014C0-25

43. A heavy nucleus x(A = 180) breaks into two nuclie y (A = 140) and z(A = 40). Energy released during fission reaction is 44 E Mev then E is : [NP-EF]

,d Hkkjh ukfHkd x(A = 180) nks ukfHkdksa y (A = 140) rFkk z(A = 40) esa foHkkftr gks tkrk gSA fo[k.Mu f;k ds nkSjku eqDr tkZ 44 E Mev gS] rc E gksxk :

Ans. 5 Sol. Released energy eqDr tkZ = 140 7 + 8 40 180 6 = 980 + 320 1080 = 220 MeV.

44. Two particles executing SHM of the same amplitude and same frequency. At a particular instant

their positions and direction of motion are shown by P1 and P2. The phase difference between the

two SHM is 2N

radian then N is: [SH - EQ]

nks d.k leku vk;ke rFkk leku vkof`k ls ljyvkokZ xfr dj jgs gSA fdlh fo'ks"k {k.k ij mudh fLFkfr rFkk

xfr dh fn'kk P1 rFkk P2 }kjk iznf'kZr gS] nksuksa ljvkokZ xfr;ksa ds e/; dykUrj2N jsfM;u gS rc N gksxk :

Ans. 2 Sol. for P1 x = Asin (t + ) ds fy, at t = 0 ij x = A

2 = A sin

1 =56

= 150

(Velocity is in ve direction) (osx _ .kkRed fn'kk esa gS) For P2 A sin (t + ) ds fy, t = 0 x = A

2 = A sin

2 = 116

= 330

(Velocity is in +ve direction) (osx /kukRed fn'kk esa gS) = 2 1 = 180 =



Toll Free : 1800 200 2244 | 1800 258 5555 JRPT2051014C0-26

45. A point charge Q is kept at point A the flux through the inclined surface of cone is 0

3Q.

5 Now

another charge Q is also placed at point B. If net flux through the inclined surface is 0

nQ.

5 Find n.

,d fcUnq vkos'k Q fcUnq A ij j[kk gqvk gS] 'kadq dh >qdh gqbZ lrg (inclined surface) ls xqtjus okyk yDl

0

3Q5 gSA vc vU; vkos'k Q dks Hkh fcUnq B ij j[kk tkrk gSA ;fn 'kadq dh >qdh gqbZ lrg ls xqtjus

okyk dqy yDl 0

nQ5 gSA n Kkr dhft,A [ES-GT]

H

H

B

A

Ans. 1

Sol. 0 0 0 0 0

nQ 3Q Q 3Q Q 5 5 5 5

n = 1

Page # 1

Course : JR-PT-2(eLPD) Test Date : 12.10.2014 Test Type : PT-2 Target Date :15.09.2014 Paper Time Duration : 2 Hrs.

SYLLABUS : Quantum Number, Periodic Table, Basic Inorganic Nomenclature, Gaseous

State, Chemical equiliribrium, Ionic Equilibrium-I, Chemical Bonding, Solution, Co-ordination compound , sblock, Surface Chemistry

SYLLABUS SCHEDULED SR. NO.

TOPIC NAME SYLLABUS SCHEDULED WEIGHTAGE

(BY FC)

WEIGHTAGE IN PAPER

(BY FACULTY) 1. Quantum Number

~ 5%

2. Periodic Table & Basic inorganic nomenclature

~ 5% 3. Gaseous State

~ 5%

4. Chemical equiliribrium ~ 5%

5. Ionic Equilibrium-I ~ 5%

6. Chemical Bonding ~ 15%

7. Solution ~ 20%

8. Co-ordination compound ~ 20%

9. sblock ~ 10%

10 Surface Chemistry ~ 10%

SYLLABUS : Electronic Effect, Aromaticity, Intermediate, Acid & Basic Strength,

Geometrical isomers, Optical isomers and Conformation Test Pattern :

Page # 2



PT-2

Total Total

Maths/ Physics/

Chemistry

Physical paper Organic paper MCQ(8) MCQ(7) Subjective(Single digit) 15 Subjective(Single digit) 15 Remarks (if any): Sent to pattern faculty Date : 03.09.2014

JEE (ADVANCED) CHEMISTRY PAPER SKELETON Faculty Name : PA SIR Test Name : JR (PT-2)

PAPER S.

No. TYPE (P) (I) (O) TOPIC(S) SUBTOPIC(S) DIFFICULTY LEVEL : Easy (E), Moderate (M), Tough (T)

1 MCQ (P) IEQ IEQ-CIE M

2 MCQ (P) SCP SCP-OP M

3 MCQ (P) GST GST-RG E

4 MCQ (P) SCP SCP-EBP M

5 MCQ (I) SBC SBC-CPM E

6 MCQ (P) SCH SCH-CPC M

7 MCQ (P) ATS ATS-EC M

8 MCQ (I) PTB PTB-EA M

9 MCQ (O) ART ART-BIS M

10 MCQ (O) SIM SIM-OAC M

11 MCQ (O) SIM SIM-MGI T

12 MCQ (O) SIM SIM-CIP T

Page # 3

13 MCQ (O) SIM SIM-EOS E

14 MCQ (O) GOLE GOLE-RC E

15 MCQ (O) ROHO ROHO-OXI E

16 Single digit

(Integer) (I) COR COR-OC E

17 Single digit

(Integer) (I) COR COR-BICC M

18 Single digit

(Integer) (P) SCP SCP-OP M

19 Single digit

(Integer) (I) COR COR-ACFT M

20. Single digit

(Integer) (P) SCP SCP-EBP M

21. Single digit


22 Single digit

(Integer) (I) SBC SBC-CPM M

23. Single digit

(Integer) (P) SCP SCP-VP M

24. Single digit

(Integer) (I) CBO CBO-BO M

25. Single digit

(Integer) (P) Mole Mol-BRR E

26. Single digit

(Integer) (I) CBO CBO-BABL E

27. Single digit

(Integer) (I) CBO CBO-VSEPR M

28. Single digit


29. Single digit

(Integer) (I) COR COR-IICC M

30. Single digit

(Integer) (I) CBO CBO-BABL M

31. Single digit

(Integer) (O) ART ART-TAU E

32. Single digit

(Integer) (O) SIM SIM-MGI T

Page # 4

33. Single digit

(Integer) (O) ART ART-TAU M

34. Single digit

(Integer) (O) SIM SIM-COS M

35. Single digit

(Integer) (O) SIM SIM-MGI M

36. Single digit

(Integer) (O) GOCE GOCE-RC M

37. Single digit

(Integer) (O) SIM SIM-EOS M

38. Single digit

(Integer) (O) SIM SIM-COS T

39. Single digit

(Integer) (O) ART ART-AIS M

40. Single digit

(Integer) (O) ART ART-BIS M

41. Single digit

(Integer) (O) SIM SIM-EOS M

42. Single digit

(Integer) (O) SIM SIM-CC E

43. Single digit

(Integer) (O) ART ART-TAU E

44. Single digit

(Integer) (O) GOCE GOCE-RC T

45. Single digit

(Integer) (O) ART ART-AIS T

Physical Inorganic paper

MCQ(8) 1. Which of the following statements are correct regarding the simultaneous solubility of AX2 and BX2 in water

at 25C ? [Given Ksp(AX2) = 3.888 109 , Ksp(BX2) = 1.944 108] (IEQ(P)) 25C ij ty esa AX2 rFkk BX2 dh lkFk lkFk foys;rk (simultaneous solubility) ds lanHkZ esa fuEu esa ls dkSuls dFku

lR; gSaA [fn;k gS Ksp(AX2) = 3.888 109 , Ksp(BX2) = 1.944 108] (A) The concentration of X ions in the solution is 1.296 105 M. (B*) The simultaneous solubility of AX2 is 3 104 M. (C*) The simultaneous solubility of BX2 is 1.5 103 M. (D) The concentration of X ions in the solution is 7.2 103 M. (A) foy;u esa X vk;uksa dh lkUnzrk 1.296 105 gSA (B*) AX2 dh le{kf.kd foys;rk (simultaneous solubility) 3 104 M gSA (C*) BX2 dh le{kf.kd foys;rk (simultaneous solubility) 1.5 103 M. gSA (D) foy;u esa X vk;uksa dh lkUnzrk 7.2 103 M gSA

Page # 5

Sol. sp 2sp 2

K (AX )K (BX ) =

9

83.888 101.944 10

=

15

= S5S

AX2 A2+ + 2X S 2S + 10S BX2 B2+ + 2X 5S 10S + 2S Ksp (AX2) = S(12S)2 = 3.888 109 S = 3 104 M Solubility of BX2 = 5S = 1.5 103 M BX2 dh foys;rk = 5S = 1.5 103 M [X] = 12 S = 3.6 103 M

2. Consider the following system. Three different aqueous solution each having volume 100 ml are taken and kept in contact as shown.

100 ml 0.2 M Urea

SPM SPM 100 ml 0.15 M KCl

100 ml 0.1 M AlCl3

(SCP(P)) (T) After sufficient time (Consider temp constant & 100% dissociation of strong electrolyte) (A) Volume of urea solution will be

3100

ml.

(B*) Volume of AlCl3 solution will be 3400

ml.

(C*) There will be no change in volume of KCl solution. (D) Volume of both KCl and AlCl3 solutions will increase. fuEu rU=k vuqlkj % rhu fHkUu tyh; foy;u] izR;sd dk vk;ru 100 ml ysrs gS rFkk fp=k esa fn[kk, vuqlkj lEidZ es j[krs gSa &

100 ml 0.2 M ;qfj;k

v)ZikjxE; f>Yyh

100 ml 0.15 M KCl

100 ml 0.1 M AlCl3

v)ZikjxE; f>Yyh

i;kZIr le; i'pkr~ (rki fu;r rFkk izcy oS|qr vi?kV~; dk 100% fo;kstu ekurs gq,) (A) ;wfj;k foy;u dk vk;ru

3100

ml gksxk

(B*) AlCl3 dk vk;ru 3400

ml gksxk

(C*) KCl foy;u ds vk;ru es dksbZ ifjorZu ugh gksxk (D) KCl rFkk AlCl3 foy;u nksuks dk vk;ru c

Page # 6

Sol. After sufficient time osmotic pressure of all solution will become same. as T is same i.e. molar concentration should be same, for this ratio of volume should be the same as that of ratio moles. Urea : KCl : AlCl3 20 m moles 30 m moles 40 m moles Total volume (300 ml) should be divided in 2 : 3 : 4 Vurea =

29

300 = 2003

ml

VKCl = 39

300 = 100 ml

3AlClV = 49

300 = 4003

ml

Sol. i;kZIr le; i'pkr lHkh foy;uks dk ijklj.k nkc leku gks tk;sxkA pwafd T leku gS vFkkZr~ eksyj lkUnzrk leku gksuh pkfg,] eksy vuqikr dh rjg vk;ru vuqikr Hkh leku gksuk pkfg,A ;wfj;k : KCl : AlCl3 20 feyh eksYl 30 feyh eksYl 40 feyh eksYl dqy vk;ru (300 ml) 2 : 3 : 4 esas foHkkftr gksuk pkfg,A Vurea =

29

300 = 2003

ml

VKCl = 39

300 = 100 ml

3AlClV = 49

300 = 4003

ml

3. The vander waal gas constant a is given by (GST(P)) okUMWjoky xSl fu;rkad 'a' fuEu }kjk fn;k x;k gSA

(A) 13

VC (B*) 2C C3P V (C) CC

RT18 P

(D*) 2764

2 2C

C

R TP

Sol. 2C C3P V = 3 2a

27b (3b)2 = a

Similarly for (D) option. (D) fodYi ds leku

4. 2.25 g of a Non volatile substance dissolved in 250 g of C6H6 . This solution shows depression in F.P. by 0.25K. Which of the following is /are correct : (SCP(P))

Given that : (Kb and Kf for C6H6 is 2.53 Kmolal-1 and 5.12 Kmolal-1, BP of C6H6 = 353.3 K) (A*) Molar mass of substance is = 180 (B*) B.P. of solution is = 353.42 K (C*) Relative lowering in vapour pressure of solvent = 0.0038 (D) All are not correct ,d 2.25 g vok"i'khy inkFkZ 250 g C6H6 es ?kqfyr gS] ;g foy;u 0.25K dk fgekad es voueu n'kkZrk gS& fuEu esa ls dkSulk@dkSuls lgh gS@gSa & fn;k gS : (C6H6 ds fy, Kb rFkk Kf, 2.53 K eksyy -1 rFkk 5.12 K eksyy -1 gS] C6H6 ds fy;s BP = 353.3 K) (A*) inkFkZ dk eksyj nzO;eku gS = 180 (B*) foy;u dk DoFkukad gS = 353.42 K (C*) foyk;d ds ok"inkc esa visf{kd voueu = 0.0038 gSA (D) lHkh lgh ugha gSaA Sol. Tf = m Kf =

g.wtM.M

dfoyk; solvent.)wt.g(

1000 Kf

0.256 = 2.25M.M.

1000250

5.12

Page # 7

M.M. = 180 g/mol

Tb = m Kb = g.wtM.M

d foyk; solvent.)wt.g(

1000 Kb

= 2.25180

1000250

2.53 = 0.1265

B.P. of solution (foy;u dk DoFkukad ) = 353.3 + 0.1265 = 353.425 K

solvent

V.P.P

= XB = A B

n

n n=

2.25180

2.25 250180 78

= 0.00388.

5. Which one of the following alkali metals does not give hydrated salts ? [M] (SBC(I)) fuEu esa ls dkSulh {kkj /kkrq,sa ty;ksftr yo.k ugha nsrh gSA (A) Li (B*) Na (C*) K (D*) Cs

6. Which method is suitable method for preparation (SCH(P)), Preparation of colloid) (MCQ) (T) (A*) God sol is prepared passing formaldehyde into gold (III) chloride for reduction. (B*) Silver sol can be prepared by electrical disintegration process. (C) Ferric hydroxide can be prepared by removing dissolve electrolyte by suitable membrane. (D) Ferric hydroxide can be prepared by adding large amount FeCl3 into freshly prepared precipitate. dkSulh fof/k fojpu cukus ds fy, mi;qDr fof/k gS\

(A*) xksYM (III) DyksjkbZM ds vip;u ds fy, QkWeZsfYMgkbM dks izokfgr dj xksYM lkWy cuk;k tkrk gSA (B*) oS|qr fo?kVu fof/k }kjk flYoj lkWy dks cuk;k tkrk gSA (C) mi;qDr f>Yyh }kjk ?kqfyr oS|qr vi?kV; dks fudkydj Qsfjd gkbMksDlkbM dks cuk;k tkrk gSA (D) rkts cus vo{ksi esa FeCl3 dh vf/kd ek=kk dks feykdj Qsfjd gkbMksDlkbM dks cuk;k tkrk gSA Sol. (A) 2AuCl3 + 3HCHO + 3H2O ductionRe

)sol(Au2 + 3HCOOH + 6HCl

(B) t is also called Bredrgs Arc method (NCERT) (C) It is dialysis process. (D) Peptization need small amount of electrolyte. Sol. (A) 2AuCl3 + 3HCHO + 3H2O vip;u

( )2AulkWy

+ 3HCOOH + 6HCl

(B) ;g cszx vkdZ fof/k Hkh dgykrh gSA (NCERT). (C) ;g viksgu fof/k gSA (D) isIVhdj.k ds fy, oS|qr vi?kV; dh dqN de ek=kk ek=kk vko';d gksrh gSA

7. Electrons are revolving in principal quantum number 4. Which of the following statements are incorrect for this. (ATS(P))

(A) Maximum orbital angular momentum for this electron is h3

(B*) Maximum 18 electron can have this value of principal quantum number (C*) Maximum 10 electron can have h3

angular momentum

(D*) Maximum 6 electron can have m = 0

eq[; Dok.Ve la[;k 4 esa bysDVkWu ?kw.kZu djrs gSaA fuEu esa ls dkSuls dFku xyr gSa&

(A) bysDVkWuksa ds fy, d{kdksa dk vf/kdre dks.kh; laosx h3

gS

(B*) vf/kdre 18 bysDVkWu eq[; Dok.Ve la[;k dk eku j[k ldrs gS

Page # 8

(C*) vf/kdre 10 bysDVkWu dks.kh; laosx h3

j[k ldrs gS

(D*) vf/kdre 6 bysDVkWu m = 0 j[k ldrs gSA

Sol. n = 4 (A) = 0, 1, 2, 3 L = h( 1)

2

; L = h3(3 1)

2

L = h122

, h3

(B) 2n2 2(4)2 = 32 (C) = 3, maximum electron (vf/kdre bysDVkWu) = 2(2 + 1) (D) n = 4 l = 0, 1, 2, 3 m = 0, 0, 0, 0 Total 4-orbitals can have m = 0 dqy 4 d{kdksa esa m = 0 mifLFkr gSA

8. In which of the following arrangements, the order is not correct according to the property indicated againstit: (PTB(I))

(A) increasing size : F < O < N < Ne (B) increasing E1 : B < C < O < F (C*) increasing negative value of first Heg : Se < S < O (D*) increasing E1 : Sc < Y < La fuEufyf[kr O;oLFkkvksa esa ls] e ds fo:) fy[ks x;s xq.kksa ds vuqlkj lgh e ugha gSaA (A) vkdkj dk c

[Basic strength] [{kkjh; lkeF;Z]

Page # 9

10. Which of the following statements are not true ? (OPTICAL) [SIM][O][M] (A) Molecules that are not superimposable on their mirror images are chiral. (B*) A compound with 2-chiral carbon is always chiral. (C*) All chiral molecules have at least one chiral atom (D) All the compounds having chiral molecules must be optically active fuEu esa ls dkSulk@dkSuls dFku lgh ugha gS\ (A) ,sls v.kq t ks vius niZ.k izfrfcEc ij vk/;kjksfir ugha gksrs gS] mUgsa fdjSy v.kq dgrs gSA (B*) 2-fd jSy dkcZu j[kus okyk ;kSfxd lnSo fdjSy gksrk gSA (C*) lHkh fdjSy v.kqvksa esa de ls de ,d fdjSy ijek.kq ekStwn gksrk gSA (D) fdjSy v.kq j[kus okys lHkh ;kSfxd izdkf'kd lf; gksrs gSA

11. The correct statement(s) concerning the structures E,F and G is (are) : [SIM][O][T]

(A) E,F, and G are resonating structures (B*) E,F and E, G are tautomers (C*) F and G are geometrical isomers (D*) F and G are diastereomers ;kSfxd E,F rFkk G dh lajpukvksa ls lacaf/kr lgh oDrO; gS gSa %

(A) E,F, rFkk G vuquknh (resonance) lajpuk,s gSaA (B*) E,F rFkk E, G pyko;oh (tautomers) gSA (C*) F rFkk G T;kfefr; (geometrical) leko;oh (isomers) gSA (D*) F rFkk G T;kfefr; foofje leko;oh (diastereomers) gSA Sol. (E,F) and (E,G) are Tautomers ((E,F) rFkk (E,G) pyko;oh gSaA) (F) and (G) are geometrical diastereomers. ((F) rFkk (G) T;kferh; foofje leko;oh gSaA)

12. The correct orders for physical properties of maleic acid(I) and fumaric acid (II) is/are [SIM][O][T] (A*) Dipole moment : I > II (B*) H2O Solubility : I > II (C*) Ka1 : I > II (D*)

2apK : I > II

eSysbd vEy (I) rFkk ;wesfjd vEy (II) ds HkkSfrd xq.k/keksZ ds fo"k; esa fuEu esa ls dkSulk e lgh gS\ (A*) f}/kzqo vk?kw.kZ : I > II (B*) H2O foys;rk : I > II (C*) Ka1 : I > II (D*)

2apK : I > II

13. Which of the molecules has/have all the three elements of symmetry centre, plane and axis simultaneously. [SIM][O]

fuEu esa ls dkSulk ;kSfxd lefefr dk dsUnz] lefefr dk ry rFkk lefefr dk v{k rhuksa j[krk gS\

(A) (B*)

Page # 10

(C) (D)

14. The correct statement(s) concerning the structures P,Q,R & S is/are [GOCE(O))] lajpukvksa P,Q,R rFkk S ds laca/k esa lgh dFku gksxsaA

(P) (Q) (R) (S) (A*) Q & S are not resonating structures (B*) R & S are resonating structures (C*) P & R are tautomers (D*) P & Q are resonating structures (A*) Q vkSj S vuquknh lajpuk ugha gSA (B*) R vkSj S vuquknh lajpuk gSA (C*) P vkSj R pyko;oh gSA (D*) P vkSj Q vuquknh lajpuk gSA

15. 1-Chloropropene has two diastereomers X and Y. The boiling point of X is greater than Y. Which of the following option is true about X and Y ? [SIM][O][E]

(A*) X is more stable than Y (B*) X is trans isomer (C) Melting point of Y is greater than X (D*) Water solubility of X is greater than Y 1-Dyksjksizksihu ds nks foofje:ih leko;oh X rFkk Y gSA X dk DoFkukad fcUnq Y dh vis{kk vf/kd gSA vr% fuEu esa ls

dkSulk fodYi X rFkk Y ds fo"k; esa lgh gSa \ (A*) X, Y d h vis{kk vf/kd LFkk;h gS (B*) X foi{k&leko;oh gS

(C) Y dk xyukad fcUnq X dh vis{kk vf/kd gS (D*) X dh ty esa foys;rk Y dh vis{kk vf/kd gSA

Sol.

Dipole moment : X > Y, Water solubility : X > Y, Stability : X > Y, Melting point : X > Y, Boiling point : X > Y.

Sol.

f}/kzqo vk?kw.kZ : X > Y, ty foysr;k : X > Y, fLFkkf;Ro : X > Y, xyukad fcUnq : X > Y, DoFkukad fcUnq : X > Y.

Subjective(Single Intger) 15

16. In [Ni(CO)4] complex maximum number of atoms present in same plane : (COR(I))(M))

Page # 11

ladqy [Ni(CO)4] esa] leku ry esa vf/kdre ijek.kqvksa dh la[;k fuEu gS Ans. 5 Sol. One Ni and two CO molecules are in same plane in sp3 hybrid complex. sp3 ladfjr ladqy esa ,d Ni rFkk nks CO v.kq leku ry esa gSaA

17. EAN rule is properly followed in how many of the following species. (COR(I))(M)) fuEu esa fdruh Lihf'kt EAN fu;e dk iw.kZr;k ikyu djrs gSA (i) Fe(CO)5 (ii) Ni(CO)4 (iii) Cr(CO)6 (iv) Mn(CO)5 (v) [n5C5H5)2Co]+

(vi) Co2(CO)8

(vii) [Ni(CN)4]2 (viii) [Ti(CO)6]2

(xi)

[( 5C5H5)2Fe] Ans. 7 Sol. (i) Fe(CO)5 = 26 + 10 = 36 (ii) Ni(CO)4 = 28 + 8 = 36 (iii) Cr(CO)6 = 24 + 12 = 36 (iv) Mn(CO)5 = 25 + 10 = 35 (v) [n5C5H5)2Co]+ = 273 + 12 = 36

(vi) Co (CO)8 = 27 + 1 + 8 = 36

(vii) [Ni(CN)4]2 = 28 2 + 8 = 34` (viii) [Ti(CO)6]2 = 26 + 10 = 36

(xi)

[( 5C5H5)2Fe] = 26 2 + 12 = 36

18. If urea is added in the sugar solution then how many of the following are correct ? (SCP(P)) (E) (a) Vapour pressure of solution decreases (b) Relative lowering in vapour pressure increases (c) Freezing point decreases

(d) Depression in freezing point increases (e) Elevation in boiling point as well as boiling point of solution increases (f) Osmotic pressure decreases

;fn 'kdZjk foy;u es ;wfj;k feyk;k tk, rks fuEu esa ls fdrus lgh gSa & (a) foy;u dk ok"inkc ?kVrk gSA (b) ok"inkc es vkisf{kd voueu c

Page # 12

Calculate the value of (a + b + c + d) is ..... [Fe(EDTA)] esa fdysV oy;ksa dh la[;k = a [Co (en)2(NH3)Cl]2+ esa fdysV oy;ksa dh la[;k = b [Ni(DMG)2]2+ esa fdysV oy;ksa dh la[;k = c Hkwjs oy; ladqy [Fe(H2O)5NO]SO4 esa fdysV oy;ksa dh la[;k = d (a + b + c + d) ds eku dh x.kuk djksA Ans. 9 Sol. a = 5, b = 2, c = 2, d = 0 (a + b + c + d) = 9

20. Solubility product of KAl(SO4)2, is 4x4 at 25C. If depression in freezing point when KAl(SO4)2 is dissolved in water to make its saturated solution at 25C is ax then what will be the value of a ? [Assume that there is no significant change in density of water on dissolving KAl(SO4)2, Kf = 2 K molal-1] (SCP(P))

25C rki ij] KAl(SO4)2 dk foys;rk xq.kuQy 4x4 gSA tc 25C ij KAl(SO4)2 dks ty esa foys; djds] bldk larI`r foy;u cuk;k x;k rc foy;u ds fgekad esa voueu ax gksxk] rks a dk eku D;k gksxk\ [KAl(SO4)2 dks ?kksyus ij ty ds ?kUkRo esa dksbZ lkFkZd ifjorZu ugha ekuas] Kf = 2 K molal-1]

Ans 8 Sol. For a saturated solution of KAl (SO4)2 KAl (SO4)2 ds ,d larI`r foy;u ds fy, 4x4 = 4S4 Solubility is x mol/L foys;rk x mol/L gS Tf = iKfm Tf = 4 2 x = 8 x

21. Number of chelate rings in [V(acac)2O] = a (COR(I))(M) Number of chelate rings in [Cr(gly)3] = b Number of chelate rings in [Ni(en)3]Cl3 = c Number of chelate rings in [Ni(CO)4] = d Calculate the value of (a + b + c + d) is ..... [V(acac)2O] esa fdysV oy;ks dh la[;k = a [Cr(gly)3] esa fdysV oy;ks dh la[;k = b [Ni(en)3]Cl3 esa fdysV oy;ks dh la[;k = c [Ni(CO)4] esa fdysV oy;ks dh la[;k = d (a + b + c + d) ds eku dh x.kuk djksA Ans. 8 Sol. a = 2, b = 3, c = 3, d = 0 (a + b + c + d) = 8

22. How many of the following statements are correct ? (SBC(I)) (a) BeO is amphoteric in nature. (b) LiHCO3 is not found in solid state. (c) K2O2 is diamagnetic but KO2 is paramagnetic. (d) AlCl3 is soluble in excess of NaOH and form sodium meta aluminate. (e) Anhydrous potassium nitrate on heating with potassium metal gives potassium oxide and nitrogen gas. (f) Lithium chloride is highly soluble in water. (g) Hydrated magnesium chloride on heating in dry air gives anhydrous MgCl2.

Page # 13

fuEu esa ls fdrus dFku lR; gSa \ (a) BeO dh mHk;/kehZ d f`r gSA (b) LiHCO3 Bksl voLFkk esa ugha ik;k tkrk gSA (c) K2O2 frpqEcdh; gS ijUrq KO2 vuqpqEcdh; gSA (d) AlCl3, NaOH ds vkf/kD; esa foys; gS rFkk lksfM;e esVk ,Y;qfeusV cukrk gSA (e) futZyh; ikSVsf'k;e ukbVsV dks ikSVsf'k;e /kkrq ds lkFk xeZ djus ij ikSVsf'k;e vkWDLkkbM vkSj ukbVkstu xSl curh

gSA (f) fyfFk;e DyksjkbM ty esa vR;f/kd foys; gSA (g) ty;ksftr eSXusf'k;e DyksjkbM dks 'kq"d gok esa xeZ djus ij futZyh; MgCl2 kIr gksrk gSA Ans. 5. Sol. (a), (b), (c), (d) & (e) are correct. (a), (b), (c), (d) rFkk (e) lgh gSaA (b) LiHCO3 Li2CO3 + H2O + CO2

(d) AlCl3 + 4NaOH (excess) vkf/kD; NaAlO2 + 2H2O + 3NaCl (e) KNO3 + K K2O + N2 (g) MgCl2.6H2O MgO + HCl + H2O.

23. What is the number of mole of benzene (PB0= 150 torr) per mole of toluene (PT0 = 50 torr) in vapour phase if the given solution has a vapour pressure of 120 torr in equilibrium with its vapour ?

ok"i izkoLFkk esa izfr eksy VkWywbu (PT0 = 50 torr) esa csUthu (PB0= 150 torr) ds eksyks dh la[;k D;k gS] ;fn fn;k x;k foy;u blds ok"i ds lkFk lkE; esa 120 torr ok"i nkc j[krk gks ? (SCP(P))

Ans. 7 Sol. P = PBXB + PT XT 120 = 150(XB) + 50 (1 XB) 100 XB = 70 XB = 0.7

YB = 0

B BX PP

= 0.7 x 150

120 = 0.875 YT = 1 0.875 = 0.125

B

T

Y 7Y 1

24. If Phosphorous acid, Tetrathionic acid and Pyrophosphoric acid have number of acidic hydrogen per molecule respectively as x, y and z, then find the value of x + y z.

;fn QkWLQksjl vEy , VsVkFkk;ksfud vEy rFkk ik;jksQkWLQksfjd vEy ds fy, izfr v.kq vEyh; gkbMkstu dh la[;k d s eku e'k% x, y rFkk z gS] rks (x + y z) dk eku Kkr dhft ;sA (CBO(I))

Ans. 0 Sol. Phosphorous acid (H3PO3) Dibasic x = 2 Tetrathionic acid (H2S4O6) Dibasic y = 2 Pyrophosphoric acid (H4P2O7) tetrabasic z = 4 QkWLQksjl vEy (H3PO3) f}{kkjh; x = 2 VsVkFkk;ksfud vEy (H2S4O6) f}{kkjh; y = 2 ik;jksQkWLQksfjd vEy (H4P2O7) prq"{kkjh; z = 4

25. In the disproportionation reaction of NaOH with one molecule of P4, in presence of H2O to give PH3 & sodium hypophosphite, number of molecules of NaOH reacting are .............. . (Mol-1(P))

P4 ds ,d v.kq dh] NaOH ds lkFk H2O dh mifLFkfr esa fo"kekuqikru vfHkf;k esa] ft lesa PH3 o lksfM;e gkbiksQkWLQkbV cursa gS] NaOH ds fdrus v.kq vfHkd r` gksrs gS \

Ans. 3 Sol. P4 + 3NaOH + 3H2O PH3 + 3NaH2PO2

Page # 14

26. In CI2O7, the CIO bonds showing double bond character are. (CBO(I)) CI2O7 esa, f}ca/k vfHky{k.k iznf'kZr djus okys CIO ca/kks dh la[;k gSaA Ans. 6

27. The number of vacant hybrid orbitals which participate in the formation of 3-centre 2 electron bonds i.e., banana bonds in diborane structure is : (CBO(I))

Mkbcksjsu lajpuk esa fjDr lad fjr d {kd ksa d h la[;k crkb;s t ks 3-d sUnz 2 by sDVkWu ca/k vFkkZr~ cukuk ca/k (banana bonds) fuekZ.k esa Hkkx y srs gSA

Ans. 2

Sol. No of vacant sp3 hybrid orbitals participating in the formation of banana bonds are 2. cukuk ca/k (banana bonds) d s fuekZ.k esa Hkkx y sus oky s fjDr sp3 lad fjr d {kd ksa d h l a[;k 2 gSaA

28. The C.F.S.E value of [Cr(H2O)6]Cl2 is x10 0 then what is the 'x' ? (COR(I))

[Cr(H2O)6]Cl2 ds fy, C.F.S.E dk eku x10 0 gS rks 'x' dk eku D;k gksxk \ Ans. 6. Sol. C.F.S.E = [ 0.4

2gtn + 0.6neg] 0 + np

[Cr(H2O)6]Cl2 is t2g1,1,1 eg1,0

= 0.60 = 6

100

29. Calculate total number of possible isomers (structrural and stereoisomers) of the compound [Ir(en)2(NH3)(NO2)]2+ (COR(I))

;kSfxd [Ir(en)2(NH3)(NO2)]2+ ds fy, lHkh laHko leko;oh;ksa lajpukRed rFkk f=kfoe leko;oh dh la[;k Kkr djksA Ans. 6

Sol

Optically inactive Optically active 1. trans-NH3 / NO2 3. Cis-NH3 / NO2 (d + ) 2. trans-NH3/ONO 4. Cis-NH3/ONO (d + ) gy- 6

dk'k vlf; dk'k laf; 1. foi{k(trans)-NH3 / NO2 3. lei{k (Cis)-NH3 / NO2 (d + )

Page # 15

2. foi{k (trans)-NH3/ONO 4. lei{k (Cis)-NH3/ONO (d + )

30. How many of the following statements are correct ? (CBO(I))[M] (i) Among sp3d, sp3d2 and sp3d3 hybridisation, the maximum number of 90 angles between bond pair-bond

pair of electrons is observed in sp3d2 hybridisation. (ii) Among SF4, CF4 and XeF4, interatomic bond angle of 10928 is observed only in SF4. (iii) Among N2F4, N2F2 and N2, the largest NN bond length is found in N2F4. (iv) Among NF3, NO3, BF3, H3O+ and NH3, [NF3, H3O+] and [BF3, NO3] are isostructural pairs. (v) Among NH3, PH3, AsH3 and SbH3, SbH3 has smallest bond angle. (vi) Among SbCl3, SbBr3 and SbI3, SbI3 has the largest bond angle.

fuEu dFkuksa esa ls fdrus dFku lgh gSa % (i) sp3d, sp3d2 rFkk sp3d3 esa ls sp3d2 ladj.k esa bysDVkWuksa ds cak ;qXe&cak ;qXe ds e; 90 ds cak dks.kksa dh la[;k

vfkdre izsf{kr gksrh gSA (ii) SF4, CF4 rFkk XeF4 esa ls dsoy SF4 esa varj ijek.kqd cak dks.k 10928 izsf{kr gksrk gSA (iii) N2F4, N2F2 rFkk N2 eas ls N2F4 esa ls N N cak yEckbZ vfkdre ik;h tkrh gSA (iv) NF3, NO3, BF3, H3O+ rFkk NH3 esa ls [NF3, H3O+] rFkk [BF3, NO3] lelajpukRed ;qXe gSaA (v) NH3, PH3, AsH3 rFkk SbH3, esa ls SbH3 U;wure cak dks.k j[krk gSA (vi) SbCl3, SbBr3 rFkk SbI3 esa ls SbI3 vfkdre cak dks.k j[krk gSA

Ans. 5 Sol. (i) Bond pairbond pair bond angles in sp3d2 are 12. (i) sp3d2 esa cak ;qXe&cak ;qXe ds e; cak dks.kksa dh la[;k 12 gSA

(ii) 10928 bond angle is observed only in CF4 (tetrahedral geometry) (ii) dsoy CF4 (prq"Qydh; T;kfefr) esa cak dks.k 10928 izsf{kr fd;k tkrk gSA

(iii) N N (1.10) < FN = NF(1.21) < (1.49 0.004)

(iv) BF3 and NO3 ; both trigonal planar ; NF3 and H3O+ ; both trigonal pyramidal. (iv) BF3 rFkk NO3 ; nksuks f=kdks.kh; leryh; ; NF3 rFkk H3O+ ; nksuks f=kdks.kh; fijkfeMy (v) Order is (e) NH3 > PH3 > AsH3 > SbH3 106.6 93.8 91.83 91.3

(vi) Order is (e) SbI3 > SbBr3 > SbCl3 99 98.2 97.1

Subjective(Sngle Intger) 15

31. How many of the following compound will show tautomersism ? [ART][O][E] fuEu esa ls fdrus ;kSfxd pyko;ork n'kkZ;sxsa \

(I) (II) Me3CCHO (III) MeCHO (IV) PhCHO (V) (VI)

Page # 16

(VII) (VIII) (IX) (X)

Ans. 3 Sol. (III), (VI), (IX)

32. The number of compounds which can show geometrical isomerism is : [SIM][O][T] fuEu esa ls fdrus ;kSfxd T;kferh; leko;ork n'kkZrs gS\ (I) N2H4 (II) H2N2 (III) HN3 (IV) CH2N2 (V) (CH3)2N2 (VI) (NH2)2N2 (VII) HN=NN=NH (VIII) NH2CH=NNH2 (IX) NH2CH=NH (X) CH2=NNH2 [GEOMETRICAL ISOMERISM] Ans. 6 Sol. (II), (V), (VI), (VII), (VIII), (IX)

33. How many hydrogen atoms will be exchanged by deuterium atoms when the compound is

kept in OD/D2O for a long period of time ? [ART][O][M]

tc ;kSfxd dks OD/D2O ds lkFk yEcs le; vUrjky ds fy, j[kk tkrk gS] rc fdrus gkbMkstu

ijek.kqvksa dk M~;wVsfj;e ijek.kqvksa }kjk fofue; gksxk \ Ans. 4

Sol. OD/OD 2

34. Sum of total number of geometrical isomers is : [SIM][O][M] fuEufyf[kr ;kSfxd ds T;kferh; leko;oksa dh la[;k gSa %

(Stereoisomerism) Ans. 5

Sol.

CH3

CH3

Ph

PhCH3H C3

Ph

Ph

CH3

H C3

Ph

Ph

CH3

Ph

Ph

CH3

HH

HH H H

H

H

H

HH

HHH

H

H

CH3

H

Ph

H

CH3 Ph

H H

35. How many structural isomers of C3H4Cl2 show geometrical isomerism ? [SIM][O][M] C3H4Cl2 ds fdrus lajpukRerd leko;oh T;kferh; leko;ork n'kkZrs gS\

Page # 17

Ans. 3

Sol. ClCH=CHCH2Cl , ,

36. The number of delocalised lone pairs in the following molecule is : [GOCE][O][M] uhps fn;s x;s ;kSfxd esa fdrus ,dkdh ;qXe foLFkkuhr gksrs gS\

H

O N

H H

N

N N

NNN

Ans. 3

Sol. H

O

N

N

N

H

N

N

N

N

H

37. Total number of C2-axis of symmetry in CCl4 moledulce is : CCl4 v.kq esa fdrus lefefr ds C2-v{k ik;s tkrs gS\

Ans. 3

38. Find the total number of organic products in the final product mixture? [SIM][O][T]

NH OH / excess2O3Zn / H O2

X product mixture.

fuEufyf[kr vfHkf;k ds mRikn feJ.k esa dqy fdrus dkcZfud mRikn izkIr gksrs gS\

NH OH / 2 vkf/kD;O3Zn / H O2

X mRikn feJ.k

Ans. 5

39. How many compounds will liberate CO2 gas when reacted with NaHCO3 ? [ART][O][M] fuEu esa ls fdrus ;kSfxd NaHCO3 ds lkFk vfHkf;k djus ij CO2 xSl fu"dkflr djrs gS\

(I) HCOOH, (II) O

CH COOH3 , (III) CH3SO3H, (IV) HNO3, (V)

OH

,

(VI)

OHNO2

NO2

, (VII) CH3

OHCH3

CH3

, (VIII) CH3

COOHCH3

CH3

Ans. 5

Page # 18

Sol. (I, III, IV, V, VII)

40. How many bases will liberate ammonia when reacted with NH4Cl ? [ART][O][M] fuEufyf[kr esa ls fdrus {kkj ;kSfxd NH4Cl ds lkFk vfHkf;k djds veksfu;k nsrs gS\

(I)

NH2

, (II) N

, (III) N

H

(IV) (C2H5)2NH , (V) NH

CH CNH3 2,

(VI) O

CH CNH3 2 (VII) CH2

CH CNH3 2 , (VIII) NaOH, (IX) H2O, (X) HC CNa

Ans. 3. Sol. (IV, V, VIII)

41. How many compounds have plane of symmetry ? fuEu esas ls fdrus ;kSfxd lefefr dk ry j[krs gS\ (I) CH4, (II) CH3Cl, (III) CH2BrCl, (IV) CHBrClF, (V) CFClBrI, (VI) C2FClBrI,

(VII) , (VIII) , (IX)

O

O O, (X)

F

Br Cl, (XI)

F

Br Cl

Ans. 8 Sol. (I, II, III, VI, VII, VIII, IX, X)

42. Total number of chiral carbon atoms in the following molecule is ? [SIM][O][M] uhps fn;s x;s ;kSfxsd esa dqy fdrus fdjSy dkcZu ijek.kq mifLFkr gksrs gS\

Ans. 6

43. How many of the following compounds show positive test with Br2/H2O as well as 2, 4.D.N.P. ? [ART][O][E]

fuEu esa ls fdrus ;kSfxd Br2/H2O rFkk 2, 4.D.N.P. ds lkFk /kukRed ijh{k.k nsrs gS\

(I) O

, (II) O

O, (III)

O O

, (IV) O

OC H2 5

O ,

(V)

O

, (VI) O

, (VII)

O

O

Ans. 4 Sol. (III, IV, VI, VII)

44. Maximum number of -electrons taking part in conjugation at any one instant of time in

Page # 19

CH=C=C=C=CH2

[GOCE][O][T]

CH=C=C=C=CH2 ;kSfxd esa ,d gh le; esa vf/kdre fdrus -bysDVkWu la;qXeu esa Hkkx ysrs gS\ Ans. 6

45. How many moles of propyne CH3CCH will be evolved when excess of is treated with 1 mole of following compound : [ART][O][T]

tc

ds vkf/kD; dh vfHkf;k uhps fn;s x;s ;kSfxd ds 1 eksy ds lkFk djkrs gS] rks fdrus eksy izksikbZu CH3CCH

izkIr gksrh gS\

OH

NH2

COOH

SO H3C

OH

HH

Ans. 4

[Type text] Page 1

MATHEMATICS TEST PATTERN BATCH : JR- PT-2 TEST DATE : 12-10-2014 ELPD

SYLLABUS : Quadratic Equation, Straight Line, Circle, Mathematical Reasoning, Set & Relation, Matrices & Determinant, Function and Inverse Trigonometric Function, Limits, Continuity & Derivability

NEW PATTERN (PATTERN CHANGED)



PT-2

Total Total

Maths/ Physics/

Chemistry

1. 2 tan1 ( 3) is equal to [IT-CF] 2 tan1 ( 3) cjkcj gS& (A*) 1cos 4 / 5 (B*) 1cos 4 / 5 (C*) 1/ 2 tan 4 / 3 (D) 1cot 4 / 3 Sol. Let ekuk 1tan ( 3) tan 3

2 4

i.e. < 2 < 2

2

21 tan 1 9 4

cos( 2 ) cos21 9 51 tan

1 42 cos5

1 142tan ( 3) cos5

= 1 4cos5

= + cos1 45

Further iqu% < 2 < 2

0 < 2 + < 2

tan (2 + ) = tan2 = 22tan

1 tan

=

34

tan1 34

= 2 +

2 + = cot1 43

=

2

tan1 43

2 = 2

tan1 43

= 2

+ tan1 43

[Type text] Page 2

2. The range of f(x) = cos3 x 6 cos2 x + 11 cos x 6 includes [FN-RG] f(x) = cos3 x 6 cos2 x + 11 cos x 6 ds ifjlj esa 'kkfey gS& (A*) 24 (B) 24 (C*) 0 (D) 2 Sol. Put cos x = t, j[kus ij y = t3 6t2 + 11t 6, 1 t 1 f(1) = 24 and f(1) = 0

3. Graph of which of the following curve can not represent a function [FN-RG] fuEu esa ls dkSulk o dk vkjs[k Qyu dks O;Dr ugha dj ldrk gS&

(A) y = 2x + 3 (B*) y2 = 4x 5 (C*) x2 y2 = 2014 (D*) x2 + y2 = 2014

4. If the equations x4 + ax = 1 and x5 + ax2 + 1 = 0 have a common root, then [QE-CR] (A) common roots is 2 (B*) common root is 1 (C) a = 2 (D*) a = 0 ;fn lehdj.k x4 + ax = 1 vkSj x5 + ax2 + 1 = 0 mHk;fu"B ewy j[krs gS] rc

(A) mHk;fu"B ewy 2 gSA (B*) mHk;fu"B ewy 1 gSA (C) a = 2 (D*) a = 0 Sol. 4 + a = 1 ; 5 + a2 + 1 = 0 = 1 and vkSj a = 0

5. Let f(x) = x + 2x 2x and g(x) = 2x 2x x, then [LT-IL] ekukfd f(x) = x + 2x 2x rFkk g(x) = 2x 2x x gS] rks (A*)

xlim

g(x) = 1 (B) xlim

f(x) = 1 (C*) xlim

f(x) = 1 (D) xlim

g(x) = 1

Sol. xlim

g(x) = xlim

2 22

x 2x x x 2x x

x 2x x

= 1

xlim

f(x) =xlim

2 22

x 2x x x 2x x

x 2x x

=

xlim

2x2

x 1 xx

= xlim

221 1x

= 1

6. Let fog(x) = 1 g(x) , 0 g(x) 1g(x) 1 , 1 g(x) 35 g(x) , 3 g(x) 4

and graph of g(x) is as shown. [FN-CF]

ekuk fog(x) = 1 g(x) , 0 g(x) 1g(x) 1 , 1 g(x) 35 g(x) , 3 g(x) 4

rFkk g(x) dk vkjs[k fp=kkuqlkj gS&

The graph of f(g(x)) is f(g(x)) dk vkjs[k gS&

[Type text] Page 3

(A) (B*)

(C) (D)

Sol. g(x) = 2 x 1 x 26 x 2 x 3

graph of g(x) is g(x) dk vkjs[k

fog(x) = 1 g(x) , 0 g(x) 1g(x) 1 , 1 g(x) 35 g(x) , 3 g(x) 4

=

1 g(x) , for no valueg(x) 1 , 1 x 15 g(x) , 1 x 3

=

2 x 1 , 1 x 15 (2 x) , 1 x 25 (6 x) , 2 x 3

=

x 1 , 1 x 13 x , 1 x 2x 1 , 2 x 3

the graph of f(g(x)) is as shown in option (B) f(g(x)) dk vkjs[k fodYi B esa n'kkZ;k x;k gSA

7. If f(x) = 4x2 + 12x + 10, then which of the function is always well defined is/are [IT-DG] ;fn f(x) = 4x2 + 12x + 10, rc fuEu esa ls dkSulk Qyu lnSo ifjHkkf"kr gS& (A*) sec1(f(x)) (B*) tan1(f(x)) (C) cos1(1 + f(x)) (D) sin1(1 f(x)) Sol. f(x) = (2x + 3)2 + 1 f(x) 1 1 + f(x) 2 and vkSj 1 f(x) 0

C & D are not defined C vkSj D ifjHkkf"kr ugha gSA

8. If and be real roots of the equation x2 x + 2 + 5 = 0 and f() = 2 + 2, then [QE-RC]

[Type text] Page 4

(A*) f() [2, 7] (B*) f() [2, 11] (C*) f() can be 5 (D*) f() can be 7 ;fn vkSj lehdj.k x2 x + 2 + 5 = 0 ds okLrfod ewy gS rFkk f() = 2 + 2 , rc (A*) f() [2, 7] (B*) f() [2, 11] (C*) f(), 5 gks ldrk gSA (D*) f(), 7 gks ldrk gSA Sol. f() = ( + )2 2 = 2 2(2 + 5) = 2 22 2 + 10

f() = 11 ( + 1)2 & D 0 103

2

2 f() 11

9. If log0.09(x2 + 2x) log0.3 (x 2) , then x can be [FN-DN]

;fn log0.09(x2 + 2x) log0.3 (x 2) , rc x gks ldrk gS& (A*) 1 (B) 1/2 (C*) 3/2 (D) 1/2 Sol. x2 + 2x > 0; x + 2 > 0 x > 0 2(0.3)log (x

2 + 2x) log(0.3)(x + 2)1/2

(x2 + 2x) (x + 2) (x)(x + 2 1) 0 (x)(x + 1) 0 x [1, )

10. The period of the function f(x) = sin 2 2x {x}3 3

, where {.} represents fractional part function is

Qyu f(x) = sin 2 2x {x}3 3

, tgk {.} fHkUukRed Hkkx Qyu dks O;Dr djrk gS] dk vkorZdky gS& [FN-PD]

(A*) 3 (B) 5 (C*) 6 (D) 7 Sol. sin 2

3

2x {x}

3

= sin

22 4x {x}

3 9

.

= sin 2 x3

cos 24 {x}

9

cos 2 x3

sin 24 {x}

9

Now clearly the period will be equal to 3. Li"Vr% vkorZ 3 gSA

11. Which of the following is INCORRECT? [IT-CF] (A*) sin1 2

2xx 1

= 2 tan1x if |x| 1 (B*) cos12

21 x1 x

= 2tan1x if x < 0

(C) cos1 (2x2 1) = 2 2cos1x if 1 x 0 (D*) tan1 22x

1 x = + 2tan1x if x < 1

fuEu esa ls dkSulk vlR; gS ?

(A) sin1 22x

x 1 = 2 tan1x ;fn |x| 1 (B) cos1

2

21 x1 x

= 2tan1x ;fn x < 0

(C) cos1 (2x2 1) = 2 2cos1x ;fn 1 x 0 (D) tan1 22x

1 x = + 2tan1x ;fn x < 1

[Type text] Page 5

Sol. All are standard result. lHkh ekud ifj.kke gSA

12. Which can be a point on the curve sin1x + sin1y = 2

[IT-AS]

o sin1x + sin1y = 2

ij dkSulk fcUnq gks ldrk gS&

(A*) 1 3,2 2

(B*) 3 1,2 2

(C) 1 3 ,2 2

(D) (1, 0)

Sol. sin1x = 2

sin1y x2 + y2 = 1 i.e. quadrant of circle o`k dk prqFkk'k

13. Vertices of a variable triangle are (3, 4), (5 cos, 5 sin) & (5 sin, 5 cos) where R [SL-TR] (A*) Locus of orthocenter of is a circle

(B*) Circumcentre of triangle is fixed (C*) Locus of centroid of is a circle (D) Locus of circumcentre of triangle is a circle of radius 1 unit. fdlh pj f=kHkqt ds 'kh"kZ (3, 4), (5 cos, 5 sin) vkSj (5 sin, 5 cos) gS] tgk R rc

(A*) f=kHkqt ds yEcdsUnz dk fcUnqiFk ,d o`k gSA (B*) f=kHkqt dk ifjdsUnz ,d vpj fuf'pr fcUnq gSA

(C*) f=kHkqt ds dsUnzd dk fcUnqiFk ,d o`k gSA (D) f=kHkqt ds ifjdsUnz dk fcUnqiFk] bdkbZ f=kT;k dk ,d o`k gSA Sol. Given points are equidistant from origin and centroid divides the line of join of circumcenter and

orthocenter in the ratio 1 : 2 fn, x, fcUnq] ewy fcUnq ls cjkcj nwjh ij gS rFkk ifjdsUnz vkSj yEcdsUnz dks feykus okyh js[kk dsUnzd dks 1 : 2 esa

foHkkftr djrh gSA

14. If f(x) = sin [2x 3 ]1 [2x 3 ]

, where [.] denotes the greatest integer function, then [CD-CP] (A) f(x) is continuous and differentiable in R (B) f(x) is continuous but not differentiable in R (C*) f(x) is continuous function at every point of its domain (D*) f(x) is non differentiable at some points. ;fn f(x) = sin [2x 3 ]

1 [2x 3 ]

gks] tgk [.] egke iw.kkd Qyu dks iznf'kZr djrk gS] rks

(A) vUrjky R esa f(x) lrr~ vkSj vodyuh; gSA (B) vUrjky R esa f(x) lrr~ gS ysfdu vodyuh; ugha gSA (C*) f(x) vius izkUr ds izR;sd fcUnq ij lrr~ Qyu gSA (D) f(x) dqN fcUnqvksa ij vodyuh; ugha gSA Sol. The numerator is 0 for all x R.

The denominator vanishes if 1 2x 3 < 0 i.e 3 12

x < 32

f(x) is continuous function at every point of its domain Hindi lHkh x R ds fy, va'k dk eku 0 gSA

gj dk eku 'kwU; gksxk ;fn 1 2x 3 < 0 vFkkZr~ 3 12

x < 32

izkUr ds izR;sd fcUnq ij f(x) lrr~ Qyu gSA

[Type text] Page 6

15. Consider the circles [CR-CT] x2 + y2 + 4x 4y + 4 = 0 and x2 + y2 4x 4y + 4 = 0 which of the following statement is INCORRECT ? (A*) Equation of one of the transverse common tangent is y = 4.

(B*) Equation of one of the direct common tangent is x = 0.

(C) Equaiton of one of the external common tangent is y = 0

(D*) Equation of one of the internal common tangent is x = 4. ekukfd o`k x2 + y2 + 4x 4y + 4 = 0 vkSj x2 + y2 4x 4y + 4 = 0. fn;s x;s gS] rks fuEu esa ls dkSulk dFku vlR; gS ? (A) budh ,d vuqizLFk mHk;fu"B Li'kZ js[kk dk lehdj.k y = 4 gSA

(B) budh ,d vuqLi'kZ mHk;fu"B Li'kZ js[kk dk lehdj.k x = 0 gSA

(C) budh ,d ck mHk;fu"B Li'kZ js[kk dk lehdj.k y = 0 gSA

(D) budh ,d vUr% mHk;fu"B Li'kZ js[kk dk lehdj.k x = 4 gSA

Sol.

From the figure, we can get the answer. fp=k ls gesa mkj izkIr gksrk gSA

Integer (Single digit) 16. A circle passes through the points (1, 1), (0, 6) and (5, 5). If the points on this circle the tangents at which

are parallel to the straight line joining the origin to its centre, are (a, b) and (c, d) where a > c then a b c d = [CR-LC]

,d o`k fcUnqvksa (1, 1), (0, 6) vkSj (5, 5) ls xqtjrk gSA bl o`k ij fLFkr fcUnqvksa (a, b) vkSj (c, d) ij [khaph xbZ Li'kZ js[kk,a] ewy fcUnq dks blds dsUnz ls feykus okyh ljy js[kkvksa ds lekUrj gS] tgk a > c rc a b c d =

Ans. 0 Sol. Points are A(1, 1) B(0, 6) C(5,5) Slope of AB = 5

Slope of AC = 23

Slope of BC = 15

angle B is 90 AC is the diameter Equation of circle is (x + 1)(x 5) + (y 1)(y 5) = 0 x2 + y2 4x 6y = 0

slope of OD = 32

, where D is centre

If is the inclination of EF

[Type text] Page 7

tan = 23

sin = 213

and cos = 313

coordinates of E and F are 3 22 13 , 3 1313 13

i.e. (1, 5) and (5, 1) (a,b) & (c, d) are respectively (5, 1) and (1, 5) a b c d = 0 Hindi fcUnq A(1, 1) B(0, 6) C(5,5) AB dh izo.krk = 5

AC dh izo.krk = 23

BC dh izo.krk = 15

dks.k B, 90 gS AC O;kl gSA o`k dk lehdj.k (x + 1)(x 5) + (y 1)(y 5) = 0 x2 + y2 4x 6y = 0

OD dh izo.krk = 32

;fn EF dk >qdko gks

tan = 23

sin = 213

vkSj cos =313

E rFkk F ds funsZ'kkad 3 22 13 , 3 1313 13

i.e. (1, 5) vkSj (5, 1) (a,b) vkSj (c, d) (5, 1) vkSj (1, 5) a b c d = 0

17. Consider the functions [FN-CT] (i) f(x) = (x 1)2 + log(x 1)5 (ii) f(x) = sin1x cos1x (iii) f(x) = ex + ex (iv) f(x) = x2 + 3x Find the number of onto functions. fuEufyf[kr Qyu fn;s x;s gSA

(i) f(x) = (x 1)2 + log(x 1)5 (ii) f(x) = sin1x cos1x

[Type text] Page 8

(iii) f(x) = ex + ex (iv) f(x) = x2 + 3x buesa ls vkPNknd Qyuksa dh la[;k Kkr dhft,A Ans. 1 Sol. For first option range is R izFke fodYi esa ifjlj R gSA

18. Consider the functions [FN-EO]

(i) f(x) = cos1x cot1x (ii) f(x) = x

x

2 1+

x

2 + 2014

(iii) f(x) = log2

22 tan x2 tan x

(iv) f(x) = x sin x + sec2x + 2015

Find the number of even functions. fuEufyf[kr Qyu fn;s x;s gSA

(i) f(x) = cos1x cot1x (ii) f(x) = x

x

2 1+

x

2 + 2014

(iii) f(x) = log2

22 tan x2 tan x

(iv) f(x) = x sin x + sec2x + 2015

bues

cbxb

Documents