test type : main pattern cumulative test-4 (mct-4)

®

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®

TEST TYPE : MAIN PATTERN CUMULATIVE TEST-4 (MCT-4) TARGET : JEE (MAIN+ADVANCED) 2021

COURSE NAME : VIJETA (01JP, TCHP, 02JP)

TEST DATE : 10-10-2021 TARGET DATE : 00-00-20

Test Syllabus : Geometrical Optics, Electrostatics, Gravitation, Current electricity, Heat transfer, Capacitance, EMF, Centre of mass, RBD, SHM, Fluid mechanics,

Elasticity and Viscosity, Surface tension, Units and Dimensions

PART-A (Hkkx– A)

SECTION – 1 : (Maximum Marks : 80)

This section contains TWENTY (20) questions. Each question has FOUR options (1), (2), (3) and (4) ONLY ONE of these four option is correct Marking scheme :

Full Marks : +4 If ONLY the correct option is chosen. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : –1 In all other cases

[kaM 1 : (vf/kdre vad : 80)

bl [kaM esa chl (20) iz'u gSaA

izR;sd iz'u esa pkj fodYi (1), (2), (3) rFkk (4) gSaA bu pkjksa fodYiksa esa ls dsoy ,d fodYi lgh gSaA

vadu ;kstuk : iw.kZ vad % +4 ;fn flQZ lgh fodYi gh pquk x;k gSA

'kwU; vad % 0 ;fn dksbZ Hkh fodYi ugha pquk x;k gS ¼vFkkZr~ iz'u vuqÙkfjr gS½A _.k vad % –1 vU; lHkh ifjfLFkfr;ksa esaA

1. If an electron and a proton having same momentum enter perpendicularly to a magnetic field, then :

;fn ,d bysDVªkWu o çksVkWu leku laosx ls ,d pqEcdh; {ks=k esa yEcor~ ços'k djrs gS rks

(1*) curved path of electron and proton will be same (ignoring the sense of revolution) (2) they will move undeflected (3) curved path of electron is more curved than that of proton (4) path of proton is more curved

(1*) bysDVªkWu o çksVkWu dk oØh; iFk leku gksxk (?kw.kZu dh fn'kk fopkj.kh; ugha gS) (2) os vfo{ksfir xfr djsaxsA

(3) bysDVªkWu dk oØh; iFk çksVkWu dh rqyuk esa vf/kd oØh; gksxkA

mailto:[email protected]

http://www.resonance.ac.in/reso/results/jee-main-2014.aspx

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(4) çksVkWu dk iFk vf/kd oØh; gksxkA

Sol. Since, electron and proton have same momenta so, the same force will act on them by the magnetic field.

pqafd bysDVªkWu ,oa izksVkWu dk laosx leku gS vr% nksuks ij pqEcdh; {ks=k }kjk yxus okyk cy leku gksxkA F = qvB sin = qvB ( =90º) Hence, both will move on same trajectory (curved path)

vr% nksuksa leku oØh; iFk ij xfr djsaxs

2. A bar magnet of length ‘I’ and magnetic dipole moment ‘M’ is bent in the form of an arc as shown in

figure. The new magnetic dipole moment will be :

,d NM+ ¼n.M½ pqEcd dh yEckbZ '' vkSj bldk pqEcdh; f}/kzqo cy&vk?kw.kZ 'M' gSA ;fn bls vkjs[k ¼fp=k½ eas n'kkZ;s x;s vuqlkj ,d pki ds vkdkj eas eksM+ fn;k tk;s rks bldk u;k pqEcdh; f}/kzqo cy&vk?kw.kZ gksxk &

(1*) 3

M

(2) 2

M

(3) M

2 (4) M

Sol. M = m × M' = m × r

= r

3

So vr%, m' = 3

M

3. A network of four capacitors of capacity equal to C1 = C, C

2 =

2C, C

3 = 3C and C

4 = 4C are connected to

a battery as shown in the figure. The ratio of the charges on C2 an C

4 is :

/kkfjrk C1 = C, C

2 =

2C, C

3 = 3C rFkk C

4 = 4C ds pkj la/kkfj=k ,d cSVjh ls fp=kkuqlkj tqM+s gSA C

2 o C

4 ij

vkos'kksa dk vuqikr gksxk :

(1) 22

3 (2*)

3

22 (3)

7

4 (4)

4

7

Sol. Key Idea : Charge on a capacitor is the product of capacitance and potential difference across it. The charge flowing through C

4 is

q4 = C

4 × V = 4 CV

The series combination of C1, C

2 and C

3 gives

1 1 1 1

C' C 2C 3C



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= 6 3 2 11

6C 6C

C’ = 6C

11

Now, C’ and C4 form parallel combination giving

C’’ = C’ + C4

= 6C 50C

4C11 11

Net charge q = C’’ V

= 50

11 CV

Total charge flowing through C1, C

2, C

3 will be

q’ = q – q4

= 50 6CVCV – 4CV11 11

Since, C1, C

2 and C

3 are in series combination hence, charge flowing through these will be same.

q2 = q

1 = q

3 = q’ = 6CV

11

Thus, 2

4

q 6CV / 11 3

q 4CV 22

la/kkfj=k ij vkos'k bldh /kkfjrk ,ao mlds vuqfn'k foHkorkUrj ds xq.kuQy ds cjkcj gksrk gSA C

4 ls izkokfgr gksus oky vkos'k

q4 = C

4 × V = 4 CV

C1, C

2 ,ao C

3 dk Js.kh la;kstu ls izkIr gksrk gSA

1 1 1 1

C' C 2C 3C =

6 3 2 11

6C 6C

C’ = 6C

11

C’ ,oa C4 ds lekUrj la;kstu ls izkIr gksrk gSA

C’’ = C’ + C4 =

6C 50C4C

11 11

dqy vkos'k q = C’’ V = 50

11 CV

C1, C

2, C

3 ls izkokfgr gksus okyk dqy vkos'k

q’ = q – q4 =

50 6CVCV – 4CV11 11

pwafd C1, C

2 ,oa C

3 Js.kh Øe esa tqM+s gq, gS vr% izR;sd ls cgus okyk vkos'k leku gksxk

q2 = q

1 = q

3 = q’ = 6CV

11

bl izdkj, 2

4

q 6CV / 11 3

q 4CV 22



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4. Two thin dielectric slabs of dielectric constants K1

and K2

( K1

< K2) are inserted between plates of a

parallel plate capacitor, as shown in the figure. The variation of electric field 'E' between the plates with

distance 'd' as measured from plate P is correctly shown by:

,d lekUrj iV~fVdk ¼IysV½ la/kkfj=k dh nks IysVksa ds chp esa] K1 rFkk K

2 ( K

1 < K

2) ijkoS|qrkad ds nks irys LySac

¼iV~fVdk½ fp=k esa n'kkZ;s x;s vuqlkj j[kh xbZ gSaaA la/kkfj=k dh nks iV~fVdkvksa ds chp fo|qr {ks=k dk eku 'E' esa]

iV~fVdk P ls nwjh 'd' ds lkFk ifjorZu dks dkSulk xzkQ lgh :i ls n'kkZrk gS\

(1) (2)

(3*) (4) Ans. (3) Sol.



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5. A ring is made of a wire having a resistance R0 = 12 . Find the points A and B as shown in the figure,

at which a current carrying conductor should be connected so that the resistance R of the sub circuit

between these points is equal to 8

3 .

rkj ls cus ,d ,d oy; dk izfrjks/k R0 = 12 gSA bl oy; esa ,sls fdu nks fcUnqvksa A vkSj B ij /kkjkokgh

pkyd dks tksM+k tk; rkfd] bu nks fcUnqvksa ds chp mi ifjiFk dk izfrjks/k R = 8

3 gksA

(1) 1

2

5

8 (2) 1

2

1

3 (3) 1

2

3

8 (4*) 1

2

1

2

Sol. Let x is the resistance per unit legth then ekuk x ,dkad yEckbZ dk izfrjks/k gS vr%

equivalent resistance dqy izfrjks/k

1 2

1 2

R RR

R R

1 1 2 2

1 2

x x

x x

8

3 = 1 2

1 2

x

1

1

2

x

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

also vr% R0 = x

1 + x

2

12 = x (

1+

2)

= x2

1

2

1

(i)

(ii)

8

312

1

= 1

1

2

12

2

x

1

X 1

= 1

2

12

2

1

2

1

2

1

× 8

36 = 1

2

(y2 + 1+2y) × 8

36 = y (where tgk y = 1

2

)

8y2 + 8 +16y = 36 y

8y2 – 20y + 8 = 0 2y2 – 5y + 2 = 0

2y2 – 4y – y + 2 = 0 2y [y – 2] – 1(y–2) = 0

(2y – 1) (y–2) = 0 y = 1

2

=1

2 or 2



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6. A potentiometer wire has length 4m and resistance 8. The resistance that must be connected in series with the wire and an accumulator of e.m.f 2V, so as to get a potential gradient 1mV per cm on the wire is :

fdlh iksVsfU'k;ksehVj ¼foHkoekih½ ds rkj dh yEckbZ 4m gS vkSj bldk izfrjks/k 8 gSA bl esa Js.kh Øe esa] 2V fo|qr okgd cy dk ,d lapk;d lsy rFkk ,d izfrjks/kd tksM+k x;k gS rkfd bl rkj ij foHko izo.krk 1mV izfr lasVhehVj gks tk;A rks] bl izfrjks/kd dk izfrjks/k gksxk%

(1) 40 (2) 44 (3) 48 (4*) 32 Sol. Total potential difference across potentio meter wier iksVsfU'k;ksehVj ¼foHkoekih½ ds vuqfn'k foHkokUrjj = 10–3

× 400 volt = 0.4 volt

so vr%, 2 1

R 8 20

R + 8 = 40 R = 32 7. If two spheres of same masses and radius are brought in contact, then the force of attraction between

them will be proportional to

;fn leku nzO;eku ,oa f=kT;k ds nks xksys Li'kZ djk, tk;sa rks muds e/; vkd"kZ.k cy lekuqikrh gksxkA

(1) r2 (2) r3 (3) r6 (4*) r4

Sol. Gravitational force acting between the spheres,

nks xksyksa ds chp cy

F = G2

.

(2 )

m m

r =

3 3

2

4 4.3 3

4

G r r

r

=

2 2 44

9G r

F r4

8. A body of mass ‘m’ is taken from the earth’s surface to the height equal to twice the radius (R) of the earth. The change in potential energy of body will be :

‘m’ æO;eku dh ,d oLrq dks i`Foh dh lrg ¼i`"B½ ls mldh f=kT;k (R) ls nks xquk Å¡pkbZ rd ys tk;k tkrk gSA oLrq dh fLFkfrt ÅtkZ esa ifjorZu gksxk :

(1*) 2

3mgR (2) 3mgR (3)

1

3mgR (4) mg2R

Sol. i

GMmU

R

3

f

GMmU

R

1 2 2

13 3 3 3

GMm GMm GMm GMmU MgR

R R R R

Alternate: U =

1

mgh

h

R

h = 2R.



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9. If potential (in volts) in a region is expressed as V(x, y, z) = 6 xy – y + 2yz, the electric field (in N/C) at point (1, 1, 0) is :

;fn fdlh {ks=k esa foHko (oksYV esa) dks V(x, y, z) = 6 xy – y + 2yz, ls fufnZ"V fd;k tk;s rks fcUnq (1, 1, 0) ij fo|qr {ks=k (N/C esa) gS %

(1*) ˆ ˆ ˆ–(6i 5 j 2k) (2) ˆ ˆ ˆ–(2i 3 j k) (3) ˆ ˆ ˆ–(6i 9 j k) (4) ˆ ˆ ˆ–(3i 5 j 3k)

Sol. (1) V = 6xy – y + 24z

V V V

E I j kx y z

E 6y I 6x 1 2z j 2y k

E

(1,1,0) = ˆ ˆ ˆ–(6i 5 j 2k)

10. Two pith balls carrying equal charges are suspended from a common point by strings of equal length,

the equilibrium separation between them is r. Now the strings are rigidly clamped at half the height. The equilibrium separation r' between the balls now become :

ljdaMs (fiFk) dh nks ckWyksa (xksfy;ksa) ij leku (cjkcj) vkos'k gSaA bUgsa leku yEckbZ dh Mksfj;ksa (/kkxs) ls ,d fcUnq ls yVdk;k x;k gSA larqyu dh voLFkk esa buds chp dh nwjh r gSA nksuksa Mksfj;ksa dks mudh vk/kh yEckbZ ij dl dj ck¡/k fn;k tkrk gSA vc lUrqyu dh fLFkfr esa nksuksa ckWyksa ds chp dh nwjh gksxh :

r

y

r'

y/2

(1*) 3

r

2

(2) 2r

3

(3) 2r

3

(4)

2r

2

Ans. (1) Sol.

r

y

r

y/2

tan = eF

mg

r / 2

y =

2

2

kq

r

mg

r3 y

r’3 y

2

r '

r =

1/3

1

2



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11. In an astronomical telescope in normal adjustment a straight black line of lenght L is drawn on just right side of objective lens. The eye-piece forms a real image of this line. The length of this image is I. The angular magnification of the telescope is :

(1) I

L –1 (2) I

I

L

L –

(3*) I

L (4)

I

L1

lkekU; lek;kstu dh fLFkfr eas] fdlh [kxksyh; nwjn'kZd ds vfHkn`';d ysal ds Bhd ckgj nkfguh vkSj ij L yEckbZ ds ,d dkyh ljy js[kk f[kph xbZ gSA usf=kdk bl ljy js[kk dk okLrfod izfrfcEc cukrh gSA izfrfcEc dh yEckbZ I

gS rks nwjn'kZd dk dks.kh; vko/kZu gS :

(1) I

L –1 (2) I

I

L

L –

(3*) I

L (4)

I

L1

Sol. (3)

Magnification by eyepiece usf=kdk }kjk vko/kZu

f

mf u

e

e 0 e

f–L f (–(f f )

e

0

f

L f

0

e

f Lm.p.

f

12. A ball moving with velocity 2 m/s collides head on with another stationary ball of double the mass. If the

coefficient of restitution is 0.5, then their velocities (in m/s) after collision will be 2 eh/ls dh pky ls xfr djrh gqbZ xsan vius ls nksxqus nzO;eku okyh ,d vU; xsan (tks 'kq:okr esa fLFkj gS) ls

Vdjkrh gSA ;fn izR;koLFku xq.kkad 0.5 gS rc Vdjkus ds ckn muds osx (eh/ls esa) gksaxs

(1*) 0, 1 (2) 1, 1 (3) 1, 0.5 (4) 0, 2

Sol. Here ;gkWa, m1 = m, m

2 = 2m

u1 = 2 m/s, u

2 = 0

coefficient of restitution izR;koLFku xq.kkad, e = 0.5

Let v1 and v

2 be their respective velocities after collision.

Applying the law of conservation of linear momentum, we get

ekuk v1 vkSj v

2 Øe'k% m1 vkSj m2 ds VDdj ds ckn ds osx gS

js[kh; laosx ds laj{k.k dk fu;e yxkus ij gesa feysxkA

m1u

1 + m

2u

2 = m

1v

1 + m

2v

2

m × 2 + 2m × 0 = m × v1 + 2m × v

2

or 2m = mv1 + 2mv

2

or 2 = (v1 + 2v

2) ...(i)

By definition of coefficient of restitution izR;koLFku xq.kkad dh ifjHkk"kk ds vk/kkj ij,

e = 2 1

1 2

v v

u u

or e(u1 – u

2) = v

2 – v

1

0.5(2 – 0) = v2 – v

1 ...(ii)

1 = v2 – v

1

Solving equations (i) and (ii), we get

lehdj.k (i) vkSj (ii) gy djus ij



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v1 = 0 m/s, v

2 = 1 m/s

13. Consider the following two statements :

fuEu nks dFkuks ij fopkj dhft;s&

A. Linear momentum of a system of particles is zero d.kks ds fudk; dk js[kh; laosx 'kwU; gSA

B. Kinetic energy of a system of particles is zero, d.kks ds fudk; dh xfrt ÅtkZ 'kwU; gSA

Then, rc

(1) A does not imply B and B does not imply A (2) A implies B but B does not imply A (3*) A does not imply B but B implies A (4) A implies B and B implies A

(1) A dk fu"d"kZ B ugh gS rFkk B dk fu"d"kZ A ugh gSA (2) A dk fu"d"kZ B gS fdUrq B dk fu"d"kZ A ugh gS A

(3*) A dk fu"d"kZ B ugh gS fdUrq B dk fu"d"kZ A gSA

(4) A dk fu"d"kZ B rFkk B dk fu"d"kZ A gSA Sol. If kinetic energy of system is zero, then momentum of system is necessarily zero.

;fn fudk; dh xfrt ÅtkZ 'kwU; gS rks fudk; dk laosx vko';d :i ls 'kwU; gksxkA

14. If a current is passed through a spring then the spring will :

;fn ,d fLçax ls /kkjk çokfgr dh tk;s rks fLçax

(1) expand (2*) compress (3) remain same (4) none of these

(1) çlkfjr gksxh (2*) ladqfpr gksxh (3) ;Fkkor jgsxh (4) buesa ls dksbZ ugha Sol. Due to flow of current in same direction in two adjacent sides, an attractive magnetic force will be

produced due to which spring will get compressed. '

nks lehiorhZ ywiksa esa leku fn'kk esa /kkjk izokfgr gksus ds dkj.k pqEcdh; cy vkd"kZ.k izd`fr dk gksxk ftlls fLizax lEihfM+r gksxh

15. In a mass spectrometer used for measuring the masses of ions, the ions are initially accelerated by an

electric potential V and then made to describe semicircular paths of radius R using a describe semicircular paths of radius R using a magnetic field B. If V and B are kept constant, the ratio

charge on the ion

mass of the ion

will be proportional to :

vk;uksa ds æO;eku ekius ds fy;s ,d æO;eku ekih LiSDVªksehVj esa vk;uksa dks igys oS|qr foHko V }kjk Rofjr dj fQj pqEcdh; {ks=k B dk iz;ksx dj R f=kT;k ds v)Zo`Ùkh; iFk ij pyk;k tkrk gSA ;fn V vkSj B dks fLFkjekuh j[kk

tk;s rks vuqikr

vk;u dk vkos'k

vk;u dk æO;eku vuqikrh gksxk :

(1) 1

R (2*)

2

1

R (3) R2 (4) R

Sol. Centripetal force is provided by the magnetic force qvB. The radius of the orbit in which ions moving is determined by the relation as given below.

2mv

qvBR

The angular frequency of rotation of the ions about the vertical field B is given by

v qB

2R m

where is frequency.



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Energy of ion is given by

E = 21mv

2 = 21

m(R )2

= 2

2 2

2

1 qmR B

2 m

or E = 2 2 21 R B q

2 m ...(i)

If ions are accelerated by electric potential V, then energy attained by ions E = qV From Eqs. (i) and (ii), we get

qV = 2 2 21 R B q

2 m

2 2

q 2V

m R B

If V and B are kept constant, then

2

q 1

m R

gy % vfHkdsUnzh; cy pqEcdh; cy qvB }kjk iznku fd;k tkrk gSA ftl d{k esa vk;u py jgs gS mldh f=kT;k fuEufyf[kr lq=k ls fu/kkfjr gksrh gSA

2mv

qvBR

vk;u dh yEcor pqEcdh; {kS=k B ds pkjks vkSj ?kq.kZu dh dks.kh; vko`rh fuEufyf[kr ls fu/kkfjr gksrh gSA

v qB

2R m

tgkWa vko`fr gS vk;u dh ÅtkZ gS

E = 21mv

2 = 21

m(R )2

= 2

2 2

2

1 qmR B

2 m

or ;k E = 2 2 21 R B q

2 m ...(i)

vxj vk;u dks fo|qr foHko V }kjk Rofjr fd;k tk;s rc vk;u fd ÅtkZ gksxh

E = qV

lehdj.k (i) vkSj (ii) ls ges feysxk

qV = 2 2 21 R B q

2 m

2 2

q 2V

m R B

vxj V vkSj B dks fu;r j[kk tk;s rc

2

q 1

m R



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16. A series combination of n1 capacitors, each of value C

1 is charged by a source of potential difference

4V. When another parallel combination of n2 capacitors, each of value C

2, is charged by a source of

potential difference V, it has the same (total) energy stored in it, as the first combination has. The value of C

2, in terms of C

1, is then

Js.kh Øe es tqM+s n1 la/kkfj=k ftles ls izR;sd ftles ls izR;sd dh /kkfjrk C

1 gS] 4V oksYV ds foHkokUrj ls vkosf'kr

gS ,d vU; Js.kh Øe es tqM+s n2 la/kkfj=k ftlesa izR;sd dh /kkfjrk C

2 gSA V oksYV ds foHkokUrj }kjk vkosf'kr gSA

nksuks leku ÅtkZ ,d=k djrs gS rc C2 dk eku C

1 ds inks es gksxkA

(1) 1

1 2

2C

n n (2) 2

11

n16 C

n (3) 2

11

n2 C

n (4*) 1

1 2

16C

n n

Sol. A series combination of n1 capacitors each of capacitance C

1 are connected to 4V source as shown in

the figure.

Total capacitance of the series combination of the capacitors is

s 1 1 1

1 1 1 1

C C C C ......... upto n

1 terms = 1

1

n

C

or Cs = 1

1

C

n

Total energy stored in a series combination of the capacitors is

us = 2

s

1C (4V)

2

= 21

1

C1(4V)

2 n

(using (i))...(ii)

A parallel combination of n2 capacitors each of capacitance C

2 are connected to V source as shown in

the figure.

Total capacitance of the parallel combination of capacitors is C

p = C

2 + C

2 + ..... + upto n

2 terms = n

2C

2

or Cp = n

2C

2 ...(iii)

Total energy stored in a parallel combination of capacitors is

up = 2

p

1C V

2

= 22 2

1(n C )(V)

2 (Using (iii))....(iv)

According to the given problem, U

s = U

p

Substituting the values of u

s and u

p from equations (ii) and (iv), we get



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2 212 2

1

C1 1(4V) (n C )(V)

2 n 2

or 12 2

1

C 16n C

n or C

2 = 1

1 2

16C

n n

C1 /kkfjrk ds n

1 la/kkfj=kksa dk Js.kh la;kstu 4 oksYV dh cSVjh ls tqM+k gS tSlk dh fp=k esa fn[kk;k x;k gS

Js.kh la;kstu dh dqy /kkfjrk

s 1 1 1

1 1 1 1

C C C C ......... upto n

1 terms = 1

1

n

C or C

s = 1

1

C

n

la/kkfj=kksa ds Js.kh la;kstu dh dqy lafpr ÅtkZ

us = 2

s

1C (4V)

2 = 21

1

C1(4V)

2 n

(using (i))...(ii)

C2 /kkfjrk okys n

2 la/kkfj=kksa dk lekUrj Øe la;kstu V oksYV dh cSVjh ls tqM+k gS tSlk dh fp=k esa fn[kk;k x;k

gSA

la/kkfj=kksa ds lekUrj Øe la;kstu dh dqy /kkfjrk

Cp = C

2 + C

2 + ..... + upto n

2 terms = n

2C

2

or Cp = n

2C

2 ...(iii)

lekUrj Øe la;kstu dh dqYk lafpr ÅtkZ

up = 2

p

1C V

2

= 22 2

1(n C )(V)

2 (Using (iii))....(iv)

fn, iz'u ds vuqlkj U

s = U

p

lehdj.k (ii) ,oa (iv) ls u

s ,ao u

p ds eku j[kus ij gesa izkIr gksrk gS

2 212 2

1

C1 1(4V) (n C )(V)

2 n 2 or 1

2 21

C 16n C

n or C

2 = 1

1 2

16C

n n



®




17. A capacitor of 2F is charged as shown in the diagram. When the switch S is turned to position 2, the percentage of its stored energy dissipated is :

vkjs[k esa n’kkZ;s vuqlkj 2F /kkfjrk ds fdlh la/kkfj=k dk vkos’ku fd;k x;k gSA tc fLop S dks fLFkfr 2 ij ?kqek;k tkrk gS] rks blesa lafpr ÅtkZ dk izfr’kr {k; gksxkA

1 2

8F 2F

V

(1*) 80% (2) 0% (3) 20% (4) 75%

Sol. Initial energy stored in the 2 F capacitor is = 1

2 (2 )V2 = V2 J

Energy loss =

1 2

1 2

C C

2 C C (V1 - V2)2 =

2 8

2 2 8

(V - 0)2

Eloss = 5

4 V2 J

% loss = 2

2

5 4 V

V × 100 = 80%

izkjaHk esa 2F la/kkfj=k esa lafpr ÅtkZ gS = 1

2 (2 )V2 = V2 J

ÅtkZ izfrkr =

1 2

1 2

C C

2 C C (V1 - V2)2 =

2 8

2 2 8

(V - 0)2

ÅtkZ izfrkr = 5

4 V2 J

% ÅtkZ izfrkr = 2

2

5 4 V

V × 100 = 80%

18. Two metal wires of identical dimension are connected in series. If 1 and

2 are the conductivities of

the metal wires respectively, the effective conductivity of the combination is :

loZle foLrkj ¼eki½ ds /kkrq ds nks rkj Js.kh Øe esa tqM+s gSaA ;fn bu rkjksa dh pkydrk Øe'k% 1 rFkk

2 gS

rks] buds bl la;kstu dh pkydrk gksxh

(1) 1 2

1 22

(2) 1 2

1 2

(3) 1 2

1 2

(4) 1 2

1 2

2

Sol. (4)

Rec

= 1 2A A

= eq

eq eqA

eq

2

A =

A1 2

1 2

eq = 1 2

1 2

2

Ans. (4)



®




19. In a region the potential is represented by V(x, y, z) = 6x – 8xy –8y + 6yz, where V is in volts and x, y, z, are in meters. The electric force experienced by a charge of 2 coulomb situated at point (1, 1,1) is :

fdlh {ks=k esa foHko dks] V(x, y, z) = 6x – 8xy –8y + 6yz ls] fu:fir fd;k tkrk gS] tgk¡ V] oksYV esa rFkk x, y, z,

ehVj esa gSaA rks fcUnq (1, 1,1) ij fLFkr 2 dwyk¡e vkos'k ij yxus okyk fo|qr cy gksxk :

(1) 6 5 N (2) 30N (3) 24N (4*) 4 35 N

Sol. V(x, y, z)

= 6x – 8xy – 6y + 6yz

Ex =

V

x

= – 6 + 8y

Ey =

V

y

= 8 x + 8 – 6z

Ez =

V

z

= – 6y

E = (–6 + 8y), i + (8x + 8 – 6z) j – 6y k

(1, 1, 1)ˆ ˆ ˆE 2i 10j 6k

| E | = 2 35 NC–1

F = qE = 2 × 2 35 = 4 35 N

20. Charges +q and –q are placed at points A and B respectively which are a distance 2 L apart, C is the

midpoint between A and B. The work done in moving a charge +Q along the semicircle CRD is :

vkos'k +q o –q Øe'k% fcUnqvksa A o B ij 2 L nwjh ij j[ks gSa] C, A o B dk e/; fcUnq gSA vkos'k +Q dks v)Z o`Ùk CRD ij pykus esa fd;k x;k dk;Z gS

(1) 0

qQ

4 L (2)

0

qQ

2 L (3)

0

qQ

6 L (4*)

0

qQ–6 L

Sol. Work done is equal to change in potential energy. In Ist case, when charge + Q is situated at C.

fd;k x;k dk;Z fLFkfrt ÅtkZ esa ifjorZu ds cjkcj gksxk

izFke fLFfr esa tc vkos'k + Q ,C ij fLFkr gS Electric potential energy of system in that case

fudk; dh fo|qr fLFkfrt ÅtkZ

U1 =

0

1

4(q)(–q)

2L +

0

1

4(–q)Q

L +

0

1

4 .

qQ

L

In IInd case, when charge +Q is moved from C to D. Electric potential energy of system in that case

f}rh; fLFkft esa tc vkos'k +Q, C ls D rd ys tk;k tkrk gSA

bl fLFkfr esa fudk; dh fo|qr fLFkfrt ÅtkZ

U2 =

0

1

4(q)(–q)

2L +

0

1

4 .

qQ

3L

+ 0

1

4(–q)(Q)

L



®




Work done fd;k x;k dk;Z = U = U2 – U

1

= 2

0 0 0

1 q 1 qQ 1 qQ– –4 2L `4 3L 4 L

– 2

0 0 0

1 q 1 qQ 1 qQ– – . .4 2L `4 L 4 L

= 0

qQ

4 .

1 1–3L L

= 0

qQ

4 1– 3

3L c

= 0

–2qQ12 L

= – 0

qQ

6 L


This section contains TEN (10) questions. The answer to each question is NUMERICAL VALUE with two digit integer and decimal upto two digit.

If the numerical value has more than two decimal places truncate/round-off the value to TWO decimal placed.

There are 10 Questions & you have attempt any 5 Questions. If a student attempts more than 5 questions, then only first 5 questions which he has attempted will be checked.

Marking scheme : Full Marks : +4 If ONLY the correct option is chosen. Zero Marks : 0 In all other cases

[kaM 2 ¼vf/kdre vad% 20)

bl [kaM esa nl (10) iz'u gSA izR;sd iz'u dk mÙkj la[;kRed eku (NUMERICAL VALUE) gSa] tks f}&vadh; iw.kkZad rFkk n'keyo f)&vadu eas gSA

;fn la[;kRed eku esa nks ls vf/kd n’'keyo LFkku gS ] rks la[;kRed eku dks n'keyo ds nks LFkkuksa rd VªadsV@jkmaM vkWQ (truncate/round-off) djsaA

bl [kaM esa 10 iz'u gaS ftuesa ls vkidks dsoy fdUgh 5 iz'uksa dk mÙkj nsuk gS ;fn vki 5 ls vf/kd iz'uksa dk mÙkj nsrs gS] rks mÙkj fn;s x;s izFke 5 iz'uksa dh gh tk¡p dh tk;sxhA


'kwU; vad % 0 vU; lHkh ifjfLFkfr;ksa esaA

21. The length of an elastic string is 5 metre when the longitudinal tension is 4 N and 6 metre when the

tension is 5 N. If the length of the string (in metre) is "2X" when the longitudinal tension is 9 N

is(assume Hooke’s law is valid) then the value of X will be :

tc ,d izR;kLFk Mksjh esa vuqnS/;Z ruko 4 N gS] rc bldh yEckbZ 5 ehVj gS rFkk tc vuqnS/;Z ruko 5N rc bldh

yEckbZ 6 ehVj gSA ;fn blesa vuqnS/;Z ruko 9N gksusa ij bldh yEckbZ "2X" (ehVj esa) gS] rks X dk eku gksxk ¼gqd ds

fu;e dks oS| eku dj gy djs½

Ans. 05.00

Sol. Let the original length of the string be L.

ekuk L Mksjh dh okLrfod yEckbZ gSA

Applying F = kx mi;ksx djus ij, we have 4 = k (5 – L) izkIr gksxkA

5 = k (6 – L)

9 = k (2X – L). From these equations vr% x = 5



®




22. The cylindrical tube of a spray pump has radius, R, one end of which has n fine holes, each of radius r. If the speed of the liquid in the tube is V, the speed of the ejection of the liquid through the holes is

nr

VR, then + =

fdlh Lizs iEi dh csyukdkj uyh dh f=kT;k R gSA bl uyh ds fljs ij n lw{e fNnz gS] ftuesa izR;sd dh f=kT;k r gSA

;fn uyh esa nzo dh pky V gS rks bu fNnzksa ls ckgj fudyrs gq, nzo dh pky

nr

VR gS] rc + = :

Ans. 04.00 Sol. Volume inflow rate = volume anflow rate

vk;ru izokg nj = vk;kru cfgokZg nj

R2V = nr2 (v) 2 2

2 2

R V VRv

n r nr

23. A force ˆ ˆ ˆF i 3 j 6k is acting at a point ˆ ˆ ˆr 2i – 6 j – 12k . The value of || for which angular

momentum about origin is conserved is :

fdlh fcUnq ˆ ˆ ˆr 2i – 6 j – 12k ij ,d cy ˆ ˆ ˆF i 3 j 6k yx jgk gSA rks || ds fdl eku ds fy, ewy fcUnq ds

ifjr% dks.kh; laosx ljf{kr jgsxk :

Ans. 01.00

Sol. If vxj L = constant then fu;r gS rc = 0

blfy, so r F 0

F should be parallel to r so coefficient should be in same ratio.So 3 6

2 –6 –12

F r ds lekUrj gS] blfy, xq.kkad ,d gh vuqikr esa gksus pkfg,A 3 6

2 –6 –12

So blfy, –Ans (4)

24. The moment of inertia of an equilateral triangular plate about the axis passing through its centre of

mass and lying in the plane is . The moment of inertia of a hexagonal plate of side 'a' and made of same material and same thickness, about an axis passing through the centre of mass and lying in its

plane will be ‘n’ times , n =

fp=k esa iznf'kZr leckgq f=kHkqtkdkj IysV dk blds nzO;eku dsUnz ls ikfjr rFkk blds ry esa fLFkr v{k ds lkis{k tM+Ro vk?kw.kZ gSA 'a' Hkqtk dh le"kV~ Hkqtkdkj IysV tks leku inkFkZ rFkk leku eksVkbZ dh cuh gS] dk tM+Ro vk?kw.kZ blds nzO;eku dsUnz ls xqtjus okyh rFkk blds ry esa fLFkr v{k ds lkis{k ‘n’ xquk gS, rc n =

a

a a

a a

Ans. 30.00

Sol. mr2

M 4 m

R 2r

Moment of inertia of upper half = 16 – = 15 Åijh vk/kh IysV dk tM+Ro vk?kw.kZ = 16 – = 15 Total moment of inertia dqy tM+Ro vk?kw.kZ = 30



®




16

2a

25. A wall has two layer A and B each made of different material, both the layers have the same thickness.

The thermal conductivity of material A is twice that of B. Under thermal equilibrium the temperature difference across the wall B is 36°C. The temperature difference across the wall A (in °C) is –

,d nhokj dh nks ijr A ,oa B fHkUu /kkrq dh cuh gqbZ gSA nksuksa ijr leku eksVkbZ j[krh gSA /kkrq A Å"eh; pkydrk /kkrq B ls nksxquk gSA Å"eh; lkE;koLFkk esa nhokj B ds vUnj rkikUrj 36°C gSA rc nhokj A vuqfn'k rkikUrj gksxk

(°C esa) –

Ans. 18.00

Sol. Since dT 1

dx k

(area & thickness is same) ({ks=kQy ,oa eksVkbZ leku gSA) A B

dT 1 dT

dx 2 dx

= 1/2 × 36 = 18°C 26. For a normal eye, the cornea of eye provides a converging power of 40 D and the least converging

power of the eye lens behind the cornea is 20 D. Using this information, the distance between the retina and the cornea - eye lens can be estimated to be :

lkekU; us=k esa dkWfuZ;k (LoPN eaMy) dh vfHklkjh 'kfDr 40 D gS rFkk dkWfuZ;k ds ihNs us=k ysal dh U;wure vfHklkjh 'kfDr 20 D gSA bl lwpuk ls us=k ds jsfVuk (n`f"ViVy) rFkk ysUl ds chp dh vuqekfur nwjh gksxh :

Ans. 01.67 Sol. Given P

1 = 40 D

P2 = 20 D

P = P1 + P

2 = 60 D

P = 100

f (in cm) So, f =

100

60

f = 5

3 = 1.67 cm.

27. The Young's modulus of steel is twice that of brass. Two wires of same length and of same area of

cross section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weights added to the steel and brass wires respectively must be in the ratio of n : 1, then n =

LVhy dk ;ax izR;kLFkrk xq.kkad] ihry ls nks xquk gSA ,d gh yEckbZ rFkk ,d gh vuqizLFk dkV ds nks rkjksa] ,d LVhy dk rFkk ,d ihry dk dks ,d gh Nr ls yVdk;k tkrk gSA ;fn Hkkj yVdkus ij nksuks rkjks ds fupys fljs ,d gh ry ij gS rks LVhy rFkk ihry ds rkjks ls yVdk;s Hkkjksa dk vuqikr Øe'k% n : 1 gks, rks n =

Ans. 02.00 Sol.



®




W

Y .A

so blfy, w

AY

1 2e e

1 2

1 2

w w

AY AY

1 1

2 2

w Y2

w Y

28. A particle moving in a straight line has magnitude of velocity v given by v2 = 12 – 3x2, where v is in m/s

and x is in m. Find the amplitude of the oscillation of the particle in m.

lh/kh js[kk esa xfr dj jgs ,d d.k ds osx v dk ifjek.k v2 = 12 – 3x2 }kjk fn;k x;k gS] tgk¡ v osx m/s esa] o x

ehVj esa gSA d.k ds nksyu dk vk;ke ehVj esa Kkr dhft;sA

Ans. 02.00 Sol. (Moderate) v2 = 12 – 3x2

= 3(4 – x) 2

Also, v2 = 2 (A2 – x2)

On comparing A = 2m. (Moderate) v2 = 12 – 3x2

= 3(4 – x) 2

v2 = 2 (A2 – x2)

rqyuk djus ij A = 2m

29. A particle executes simple harmonic motion under the restoring force provided by a spring. The time

period is T. If the spring is divided in two equal parts and one part is used to continue the simple

harmonic motion, the time period become n

T, then value of n is :

fLçax }kjk vkjksfir çR;ku;u cy ds dkj.k ,d d.k ljy vkorZ xfr djrk gSA vkorZ dky T gSA ;fn fLçax dks nks

Hkkxksa esa foHkkftr dj fn;k tk;s rFkk blds vk/ks ,d Hkkx ls ljy vkorZ xfr djkbZ tk; rks vkorZ dky n

Tgks

tkrk gS rc n dk eku gksxk &

Ans. 02.00

Sol. Time perisd vkorZdky = T = 2m

K

Spring dirided into two equal parts Lengh reduced to half

fLizax nks cjkcj Hkkxks esa foHkkftr dh x;h gS vr% yEckbZ vk/kh gks tk;sxh

We know ge tkurs gS K 1

K become twice K nqxuk gks tk;sxkA

Tnew = 2new

m

K = 2

m

2K =

1

2

m2

K

=

T

2



®




30. The resistances of the four arms P, Q, R and S in a Wheatstone’s bridge are 10 ohm, 30 ohm, 30 ohm and 90 ohm, respectively. The e.m.f. and internal resistance of the cell are 7 Volt and 5 ohm respectively. If the galvanometer resistance is 50 ohm, the current (in Amp.) drawn from the cell will be :

fdlh OghVLVksu fct dh pkj Hkqtkvksa P, Q, R rFkk S ds izfrjks/k Øe'k% 10, 30 , 30 rFkk 90 gSaA blesa tqM+s lsy dk bZ-,e-,Q (e.m.f.) 7 oksYV rFkk vkUrfjd izfrjks/k 5 gSA ;fn xSYosuksehVj dk izfrjks/k 50 gS rks lsy }kjk izokfgr fo|qr /kkjk dk eku (Amp. esa) gksxk :

Ans. 00.02

Sol. i = 7

40 1205

40 120

= 7

5 30 =

1

5 = 0.2 amp.



1

®

Course : 01JP MCT-4) Test Date : 10-10-2021 Test Type : (JEE MAIN PATTERN) Target Date : 00.00.2021 Paper 1 Time Duration :

01JP & 02JP & TCHP (PAPER LEVEL)

SYLLABUS : Solution & Colligative Properties, Coordination Compounds, Solid State,

Electrochemistry, Metallurgy, Qualitative Analysis (anion & Cation), p-Block (15 to 16 groups), Atomic Structure

SYLLABUS :

ORM-I, ORM-II, Redox Reaction. Q.No. Subject Nature of Questions No. of Questions Marks Negative Total

1 to 20 PHYSICS SCQ 20 4 –1 80

21 to 30 PHYSICSNumerical type (2 digit, 2 decimal)

(Do Any 5) (Optional Type)10 4 0 20

31 to 50 CHEMISTRY SCQ 20 4 –1 80

51 to 60 CHEMISTRYNumerical type (2 digit, 2 decimal)


61 to 80 MATHS SCQ 20 4 –1 80

81 to 90 MATHSNumerical type (2 digit, 2 decimal)


90 300

For All Main Pattern Test | Pattern : P1-21 | NEW PATTERN | w.e.f. on 21-12-2020

Total Total

PAPER


This section contains TWENTY (20) questions. Each question has FOUR options (1), (2), (3) and (4) ONLY ONE of these four option is correct Marking scheme : Full Marks : +4 If ONLY the correct option is chosen. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : –1 In all other cases






31. Total number of geometrical isomers of [Co(gly)Br2Cl2]2– are : [Co(gly)Br2Cl2]2– ds dqy T;kfefr leko;oh;ksa dh la[;k gS %

(1) 5 (2*) 4 (3) 3 (4) 2 Sol. M(AB)c2d2

(Ac)(Bc)(dd) (Ad)(Bd)(cc) (Ac)(Bd)(cd) (Ad)(Bc)(cd)

Four geometrical isomers (pkj T;kfefr leko;oh)

2

32. The number of de-Broglie waves made by a Bohr electron in an orbit of maximum magnetic quantum number 3 is –

,d cksgj bysDVªkWu }kjk cuk;h x;h Mh&czksXyh rjaxks dh la[;k D;k gksxh] ftldh cksgj d{kk esa pqEcdh; DokaVe la[;k dk vf/kdre eku 3 gS&

(1) 5 (2*) 4

(3) 6 (4) None of these (buesa ls dksbZ ugha) Sol. No. of de-Broglie waves = n (Orbit number)

Mh&czksXyh rjaxksa dh la[;k = n (d{k la[;k) If ;fn mmax = 3

max = 3

n = 4

33. CaC2 + N2 1100ºC A + B

ionic element compound

Which of the following statement is incorrect ? (1) Anion of A is isoelectronic with CO2. (2) On hydrolysis of A, NH3 is formed. (3*) B has only two allotropes. (4) Shape of anion of A is linear.

CaC2 + N2 1100ºC A + B

vk;fud rRo ;kSfxd fuEu esa ls dkSulk dFku xyr gS ?

(1) A dk _.kk;u CO2 ds lkFk lebysDVªkWfud gSA (2) A ds ty vi?kVu ij NH3 curk gSA

(3*) B dsoy nks vij:i j[krk gSA (4) A ds _.kk;u dh vkdfr js[kh; gSA

Sol. 2 2CaC N CaNCN C

OHNH2CaCOOH5CaCN 4322

34. Which of the following property increases down the group for nitrogen family? (1) Acidic character of oxide (2*) Reducing power of hydride (3) Basic character of hydride (4) Boiling point of hydride

ukbVªkstu ifjokj ds fy, fuEu esa ls dkSulk xq.k lewg esa uhps tkus ij c<+rk gS \

(1) vkWDlkbM dh vEyh; izÑfr (2*) gkbMªkbM dh vipk;d {kerk (3) gkbMªkbM dk {kkjh; vfHky{k.k (4) gkbMªkbM dk DoFkukad Sol. (1) Acidic character of oxide decreases down the group (3) Basic character of hydride decreases down the group (4) PH3 < AsH3 < NH3 < SbH3 < SiH3 (Boiling point)

Sol. (1) lewg esa uhps tkus ij vkWDlkbM dk vEyh; y{k.k ?kVrk gSA

(3) lewg esa uhps tkus ij gkbMªkbM dk {kkjh; y{k.k de gksrk gSA

(4) PH3 < AsH3 < NH3 < SbH3 < SiH3 (DoFkukad)

35. Two solutions of non-volatile and non-electrolyte solute A and B are prepared separately. The molar

mass ratio A

B

M 1

M 3 . Both are prepared as 5% by weight solution in water. Then what is the ratio of

freezing point depressions, ƒ A

ƒ B

( T )

( T )

of the solutions?

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

vok"i'khy rFkk oS|qr vuvi?kV~; (non-electrolyte) foys; A rFkk B ds nks foy;u i`Fkd&i`Fkd cuk;s x;sA buds

eksyj nzO;ekuksa dk vuqikr A

B

M 1

M 3 gSA nksuksa dk foy;u] ty esa Hkkj ds vuqlkj 5% cuk;k x;k] rc bu foy;uksa

ds fgekad esa voueu dk vuqikr ƒ A

ƒ B

( T )

( T )

D;k gS \

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

3

Sol. Tƒ = kƒ × m

(Tƒ)A = kƒ × A

5 1000

M 95

(Tƒ)B = kƒ × B

5 1000

M 95

ƒ A

ƒ B

( T )

( T )

= B

A

M

M

= 3

1 = 3.

36. ZnCO3AZnO

BImpure

ZnC

pure

Zn correct coding of A, B & C is

(A) (B) (C) (1*) Calcination Coke Fractional distillation (2) Calcination Carbon monoxide Fractional distillation (3) Calcination Electrolysis Liquation (4) Roasting Coke Poling

ZnCO3AZnO

B Znv'kq)

C Zn'kq)

A, B rFkk C dk lgh feyku gS&

(A) (B) (C)

(1*) fuLrkiu dksd izHkkth vklou

(2) fuLrkiu dkcZu eksuksvkWDlkbM izHkkth vklou

(3) fuLrkiu oS/kqrvi?kVu nzohdj.k (4) HktZu dksd n.M foyksMu 37. MnO

2(s) and conc. H

2SO

4 is added to NaCl(s). The greenish yellow gas liberated is :

(1*) Cl2 (2) HCl (3) SO

3 (4) ClO

2

MnO2(s) ,oa lkUnz H

2SO

4 dks NaCl(s) esa feyk;k tkrk gSA mRlftZr gjh ihyh xSl gksxh %

(1*) Cl2 (2) HCl (3) SO

3 (4) ClO

2

Sol. Yellowish-green gas (chlorine) with suffocating odour is evolved when the solid chloride mixed with manganese dioxide is heated with concentrated H

2SO

4.

NaCl + H2SO

4 NaHSO

4 + HCl

MnO2 + 4HCl MnCl

2 + 2H

2O + Cl

2

gy.

tc lksfM;e DyksjkbM dks eSXuht MkbZvkWDlkbM ds lkFk fefJr djds lkUnz H2SO

4 ds lkFk xeZ djrs gS rks ne

?kksVus okyh xa?k ds lkFk ihyh gjh xSl ¼Dyksjhu½ mRlftZr gksrh gSA

NaCl + H2SO

4 NaHSO

4 + HCl

MnO2 + 4HCl MnCl

2 + 2H

2O + Cl

2

38. H2O2 can be prepared by successive reactions : 2NH4HSO4 H2 + (NH4)2S2O8

(NH4)2S2O8 + 2H2O 2NH4HSO4 + H2O2

The first reaction is an electrolytic reaction and second is steam distillation. What amount of current would have to be used in first reaction to produce enough intermediate to yield 10.2 g pure H2O2 per hour. Assume current efficiency 50%.

(1) 42.17 A (2) 12.17 A (3) 22.17 A (4*) 32.17 A

Øekxr vfHkfØ;kvksa ds }kjk H2O2 dks cuk;k tk ldrk gSA 2NH4HSO4 H2 + (NH4)2S2O8

(NH4)2S2O8 + 2H2O 2NH4HSO4 + H2O2

izFke vfHkfØ;k oS|qr vi?kVuh vfHkfØ;k gS o f}rh; eki vklou gSA 10.2 g 'kq) H2O2 izfr ?k.Vs cukus ds fy;s izFke vfHkfØ;k esa fdruhs ek=kk esa /kkjk iz;qDr gksuh pkfg, rkfd i;kZIr e/;orhZ cusA /kkjk n{krk 50% ekusA

(1) 42.17 A (2) 12.17 A (3) 22.17 A (4*) 32.17 A

Sol. 22OnH formed = nOS)NH(n 8224

and neq(NH4)2S2O8 = F

Q

96500

5.03600i2

34

2.102n

4

]e2OSSO2[ 282

24

i = 32.17 A

Sol. fufeZr 22OnH = nOS)NH(n 8224

o neq(NH4)2S2O8 = F

Q

96500

5.03600i2

34

2.102n

]e2OSSO2[ 282

24

i = 32.17 A

39. Which of the following statement are incorrect with respect to Fe3+ ion ?

(I) Maximum number of electrons with ( + m) = 0 is 11.

(II) Total number of electrons with one radial node and ms = 1

2 is 8.

(III) Modulus of total spin is 2. (IV) Spin multiplicity = 6 (1) II, IV (2*) II, III (3) II, III, IV (4) I, III

fuEu esa ls dkSuls dFku Fe3+ vk;u ds lanHkZ esa xyr gSa \

(I) ( + m) = 0 eku okys bysDVªkWuksa dh vf/kdre la[;k 11 gSA

(II) ,d f=kT;h; uksM+ rFkk ms = 1

2 eku okys bysDVªkWuksa dh dqy la[;k 8 gSA

(III) dqy pØ.k dk ekikad (Modulus) 2 gSA

(IV) pØ.k cgqxq.kdrk = 6

(1) II, IV (2*) II, III (3) II, III, IV (4) I, III

Sol. (1) 1s2, 2s2, 2x2p , 3s2, 2

x3p and 1xy3d orbitals have ( + m) = 0. So no. of electrons = 11

(3) |Total spin| = 5 × 1

2 = 2.5

(4) Spin states = spin multiplicity = n + 1 = 5 + 1 = 6.

Sol. (1) 1s2, 2s2, 2x2p , 3s2, 2

x3p rFkk 1xy3d d{kd ( + m) = 0 bysDVªkWuksa dh la[;k = 11 j[krs gSA

(3) |dqy pØ.k | = 5 × 1

2 = 2.5

(4) pØ.k voLFkk = pØ.k cgqxq.kdrk = n + 1 = 5 + 1 = 6.

40. Identify the correct order of wavelength of light absorbed by the following complex ions.

fuEu ladqy vk;uksa ds }kjk vo'kksf"kr çdk'k dh rjaxnS/;Z dk lgh Øe igpkfu,A

[Co(H2O)

6]3+ ; Co(CN)

6]3– ; [Co(F)

6]3– ; [Co(en)

3]3+

I II III IV

(1*) III > I > IV > II (2) II > IV > I > III

(3) III > I > II > IV (4) None of these buesa ls dksbZ ugha

Sol. corresponds to the wave length of the visible light and = hc

The order of ability to produce d-orbital splitting of various ligands :

CO > CN– > NO2

– > en > NH3 > H

2O > OH– > F– > Cl– > Br– > I–

strong field weak field ligands ligands

(larger ) (smaller )

gy- n`'; çdk'k dh rjaxnS/;Z ls lacaf/kr gS rFkk = hc

d-d{kd ds foikVu ds fy, fofHkUu fyxs.Mksa dh {kerk dk Øe gS&

CO > CN– > NO2

– > en > NH3 > H

2O > OH– > F– > Cl– > Br– > I–

çcy {kS=k nqcZy {kS=k fyxs.M fyxs.M

(vf/kdre) (U;wure)

41. Chromyl chloride test is employed for the detection of chloride ions. A similar test can be employed for

(1) Br– (2) – (3) for both (4*) None of these

5

Øksfey DyksjkbM ijh{k.k] DyksjkbM vk;uksa ds fu/kkZj.k ds fy, iz;ksx fd;k tkrk gSA blh izdkj dk ijh{k.k fuEu ds fy, fd;k tk ldrk gSA

(1) Br– (2) – (3) nksuksa ds fy, (4*) buesa ls dksbZ ugha Sol. K

2Cr

2O

7 will oxidise Br– & – into Br

2 and

2 which give colorless solutions with NaOH.

Sol. K2Cr

2O

7 }kjk Br

2 o

2 dks Br– vkSj – esa vkWDlhÑr fd;k tkrk gS tks NaOH ds lkFk jaxghu foy;u nsrk gSA

42. In which of the following cases metal obtained by carbon reduction is in liquid state?

fuEu es ls dkSulh fLFkfr esa dkcZu vip;u }kjk izkIr /kkrq nzo voLFkk esa gksrh gS\

(1)

oG

C CO

M MO

T

(2)

oG

C CO

M MO

T

(3*)

oGC CO

M MO

T

(4) None of these buesa ls dksbZ ugha

Sol. When state of metal changes from solid to liquid and then gas, there is steep increase in value of oG .

In case of (1), (2) metal obtained is in gaseous state. In case of (3) it is in liquid state.

tc /kkrq dh voLFkk Bksl ls nzo esa rFkk ckn esa xSl esa cnyrh gS] rks ;gk¡ oG ds eku esa vf/kd o`f) gksrh gSA fLFkfr (1), (2) esa izkIr /kkrq xSlh; voLFkk esa gksrh gSA fLFkfr (3) esa ;g nzo voLFkk esa gksrh gSA

43. Product of given reaction is :

O=C–NH2

C=O

H

H / Catalyst2

Excess

mijksDr vfHkfØ;k dk mRikn gS %

O=C–NH2

C=O

H

(1) (2*)

O=C–NH2

CH OH2

(3)

CH –NH2 2

CH OH2

(4)

CH –NH2 2

CH –OH2

6

44. Product

Product is -

mRikn

mRikn gS -

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

45. 2Br

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

46. The product of reaction is I

4

Br

CCl

I

4

Br

CCl vfHkfØ;k dk mRikn gS %

(1)

Br

I (2)

Br

I (3) Br

I

(4*) Br

I

47. Which of the following reagent give markovnikov product with propene :

fuEu esa ls dkSulk vfHkdeZd izksihu ds lkFk ekjdksZfudkWo mRikn nsxk \

(1) HCl (2) CH3COOH (3) dil. (ruq) H2SO4 (4*) All of these ;s lHkh 48. Identify final product in the following :

fuEufyf[kr vfHkfØ;k esa vafre mRikn dkSulk gS\

HOBr /H (2eq.)

(1*)

CCHBr2

||O

(2)

CCH Br2

||O

(3) (4)

CHCHO

Br

Sol. + HO Br+-

M.A.

C = CHBr|OH

HO Br

- +

C – CHBr2

|OH

|OH

C – CHBr2

||O

– H O2

7

49. Choose the incorrect statements : (1) Benzene reacts with electrophile to form reaction intermediate known as sigma complex (arenium

ion) (2) Formation of carbon- electrophile bond is rate-determining step in ArSE reaction. (except reversible

reaction) (3*) Breaking of C–H bond is rate-determining step in nitration of benzene. (4) Sulphonation of benzene is a reversible reaction

xyr dFkuksa dk p;u dhft, % (1) csUthu] bysDVªkWuLusgh ds lkFk fØ;k djds vfHkfØ;k e/;orhZ cukrk gS] ftls flXek ladqy ¼,jsfu;e vk;u½ dgrs

gSA (2) ArSE vfHkfØ;k esa dkcZu bysDVªkWuLusgh cU/k dk fuekZ.k] nj fu/kkZjd in gksrk gSA ¼mRØe.kh; vfHkfØ;k dks

NksM+dj½ (3*) csUthu ds ukbVªhdj.k esa C–H cU/k dk VwVuk] nj fu/kkZjd in gksrk gSA

(4) csUthu dk lYQksuhdj.k mRØe.kh; vfHkfØ;k gSA

50. Br2

H2O Final product is : vfUre mRikn gS %

(1*)

Br

OH

(2)

OH

Br

(3)

O

(4)

O

Sol. Br2

H2O

Br

OH












51. Maximum of how many quantum numbers can be different for any two electrons in the ground state configuration of Na?

Na ds vk| voLFkk foU;kl esa dksbZ Hkh nks bysDVªkWuksa ds fy, vf/kdre fdruh Dok.Ve la[;k;sa fHkUu gks ldrh gSa\ Ans. 04.00 Sol. Theory Based

8

52. What is the value of pKb(CH3COO–), if m = 390 mho cm2/mol and m = 7.8 mho cm2/mol for 0.04 M of

a CH3COOH solution at 25ºC? (log2 = 0.30)

pKb(CH3COO–) dk eku D;k gS] ;fn 25ºC ij CH3COOH foy;u ds 0.04 M ds fy, m = 390 mho cm2/mol

rFkk m = 7.8 mho cm2/mol gS\ (log2 = 0.30)

Ans. 09.20

Sol. = 7.8

390= 2×10–2

Ka = c2 = 16×10–6

or ;k pKa = 4.8

53. In a sample of hydrogen atoms, all the electron jump from n = 5 to ground level finally (directly or

indirectly) without emitting any line in Balmer series. The number of possible different lines is :

gkbMªkstu ijek.kqvks ds izkn'kZ es] lHkh bysDVªkWu ckej Js.kh es fdlh Hkh js[kk ds mRltZu ds fcuk vUr es ¼izR;{k ;k vizR;{k :i ls½ n = 5 ls vk| voLFkk esa dwnrs gSaA lEHkkfor fHkUu js[kkvksa dh la[;k gS&

Ans. 06.00

Sol.

5

3

2

1

5 2, 4 2, 3 2, 2 1 lines will not occur.

5 2, 4 2, 3 2, 2 1 js[kk;sa ugha gksxhA 54. Indium (III) oxide is used for making electrically conducting transparent glass of touch-screens. In its

structure indium ions occupy FCC positions and oxide ions occupy positions wholly inside the unit cell. How many oxide ions are present per FCC unit cell?

bf.M;e(III) vkWDlkbM Vp LØhu ds fo|qr :i ls pkfyr ikjn'khZ dk¡p cukus ds fy, iz;qä gksrk gSSA blh lajpuk esa bf.M;e vk;u FCC fLFkfr;ksa dks ?ksjrs gSa rFkk vkWDlkbM vk;u ,dd dksf"Bdk esa lEiw.kZ Hkhrj fLFkfr;ksa dks ?ksjrs gSaA fdrus vkWDlkbM vk;u izfr FCC ,dd dksf"Bdk esa mifLFkr gSa\

Ans. 06.00 Sol. I2O3: There are 4 indium ions per FCC unit cell. Thus, 6 oxide ions must be present per FCC unit cell as

per stoichiometry of the compound.

I2O3: ;gk¡ izfr FCC ,dd dksf"Bdk esa 4 bafM;e vk;u gSA blfy, ;kSfxd dh jllehdj.kferh ds vuqlkj izfr FCC ,dd dksf"Bdk esa 6 vkWDlkbM vk;u mifLFkr gksuk pkfg,A

55. How many Zn+2 ions are found on one body-diagonal of FCC unit cell of ZnS crystal ?

ZnS fØLVy dh FCC ,dd dksf"Bdk ds ,d dk; fod.kZ ij ik;s tkus okys Zn+2 vk;uksa dh la[;k fdruh gksxh \

Ans. 01.00 Sol. Zn+2 occupy alternate tetrahedral voids

Zn+2 ,dkUrfjr prq"Qydh; fjfDr;k¡ ?ksjrk gSA 56. 250 mL sample of a 0.20 M Cr3+ is electrolysed with a current of 96.5 A. If the remaining [Cr3+] is 0.1M

then the duration of process is (in sec) : (Assume volume remain constant during process)

0.20 M Cr3+ ds 250 mL izkn'kZ dks 96.5 A /kkjk ds lkFk oS|qr vi?kfVr fd;k tkrk gSA ;fn [Cr3+] 0.1M 'ks"k jgrk gS] rks izØe dh vof/k gS ¼lSd.M esa½ & (ekuk izØe ds nkSjku vk;ru fu;r jgrk gS)

Ans. 75.00 Sol. Initial moles of Cr3+ = 0.25 × 0.2 = 0.05 mol Final moles of Cr3+ = 0.25 × 0.1 = 0.025 mol Therefore moles of Cr3+ reduced is : 0.05 – 0.025 = 0.025 mol

9

or eq. of Cr3+ reduced 0.025×3 = t 96.5

96500

t = 75 sec

Sol. Cr3+ ds izkjfEHkd eksy = 0.25×0.2 = 0.05 mol

Cr3+ ds vfUre eksy = 0.25 × 0.1 = 0.025 mol

blfy, Cr3+ ds vipf;r eksy gS&

0.05 – 0.025 = 0.025 mol

;k vipf;r Cr3+ ds rqY;kad 0.025×3 = t 96.5

96500

t = 75 sec 57. How many of the following reaction(s) can evolve phosphine ?

(I) White P + Ca(OH)2 (II) AlP + H2O

(III) H3PO4 (IV) PH4 + NaOH

(V) H3PO3 (VI) H3PO2

fuEu es ls dkSulh vfHkfØ;k@vfHkfØ;k,sa QkWLQhu mRlftZr dj ldrh gS@gSa \

(I) lQsn P + Ca(OH)2 (II) AlP + H2O

(III) H3PO4 (IV) PH4 + NaOH

(V) H3PO3 (VI) H3PO2

Ans. 05.00

Sol. (I) 2P4 + 3Ca(OH)2 + 6H2O 3Ca(HPO2)2 + 2PH3

(II) AlP + 3H2O Al(OH)3 + PH3

(III) 2H3PO4 o220 C

H4P2O7 o316 C2HPO3

P4O10

(IV) PH4 + NaOH Na + H2O + PH3

(V) H3PO3 H3PO4 + PH3

(VI) H3PO2 H3PO3 + PH3

58. How many of the following are free radical (substitution or addition) reaction ?

fuEu esa ls fdruh eqDr eqyd ¼izfrLFkkiu ;k ;ksxkRed½ vfHkfØ;k,as gSa \

(a)

2Cl /h (b)

2 3Cl / AlCl

(c)

2 4Cl /CCl (d)

HCl

(e)

HBr

Peroxide

ijkWDl kb M HBr

(f)

HCl

Peroxide

ijkWDl kb M HCl

Ans. 02.00 Sol. 2 (a, e) (a) is free radical substitution reaction (b) is electrophilic substitution reaction (c, d, f) are electrophilic addition reaction (e) is free radical addition reaction Sol. 2 (a, e)

(a) eqDr ewyd izfrLFkkiu vfHkfØ;k gSA

(b) bysDVªkWuLusgh izfrLFkkiu vfHkfØ;k gSA

(c, d, f) bysDVªkWuLusgh ;ksxkRed vfHkfØ;k gSA

(e) eqDr ewyd ;ksxkRed vfHkfØ;k gSA

59. How many alkene/s react faster than propene with dil.H

2SO

4?

fuEu esa ls fdruh ,Ydhu] ruq H2SO

4 ds lkFk izksihu dh rqyuk esa rhozrk ls fØ;k djrh gSa\

(a)

(b) Ph

(c)

Ph

Ph

(d) CH3O

10

(e)

(f)

O

(g)

(h) Cl

Ans. 06.00 Sol. (a, b , c ,d , e , g)

rate of addition of E stability of cation

E ds ;ksx dh nj èkuk;u dk LFkkf;Ro

60. C4H

8 (unsaturated hydrocarbons) Total number of products formed :

C4H

8 (vlar`Ir gkbMªksdkcZu) dqy fdrus mRikn curs gS \

Ans. 06.00

Sol. CH2=CH–C

2H

5

Meso

CH3–CH=CH–CH

3

trans

Sol. CH2=CH–C

2H

5

felks

CH3–CH=CH–CH

3

foi{k

®

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

Website : www.resonance.ac.in | E-mail : [email protected] M02EAMCT2101021-1


®

TEST PATTERN

COURSE NAME : 01JP-02JP-TCHP TEST TYPE : MCT-4

Test Date : 10-10-2021 Test Syllabus : Relation, Function & ITF, Probability, LCD, MOD, AOD, Matrix &

Determinant, Definite Integration, Indefinite integration, Set, Mathematical reasoning, Statistics, FOM-I, FOM-II, Trigonometry, SOT, Quadratic Equation, Sequence and Series

PART-C (Hkkx– C) (MATHEMATICS)


This section contains TWENTY (20) questions. Each question has FOUR options (1), (2), (3) and (4) ONLY ONE of these four option is correct Marking scheme :

Full Marks : +4 If ONLY the correct option is chosen. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : –1 In all other cases






61. If a relation S is defined by S = {(x, y)| x, y N and x2 – 4xy + 3y2 = 0} where N = the set of natural numbers, then S is

(1*) reflexive (2) symmetric

(3) transitive (4) reflexive and transitive

;fn lEcU/k S, S = {(x, y)| x, y N ls ifjHkkf"kr gS rFkk x2 – 4xy + 3y2 = 0} tgk¡ N = izkd`r la[;kvksa dk leqPp; gS] rc S gS&

(1*) LorqY; (2) lefer

(3) laØked (4) LorqY; vkSj laØked

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

1 to 20 PHYSICS SCQ 20 4 –1 80

21 to 30 PHYSICSNumerical type (2 digit, 2 decimal)


31 to 50 CHEMISTRY SCQ 20 4 –1 80

51 to 60 CHEMISTRYNumerical type (2 digit, 2 decimal)


61 to 80 MATHS SCQ 20 4 –1 80

81 to 90 MATHSNumerical type (2 digit, 2 decimal)


90 300

For All Main Pattern Test | Pattern : P1-21 | NEW PATTERN | w.e.f. on 21-12-2020

Total Total



®




Sol. For S = {(x, y) | x, y N and x2 – 4xy + 3y2 = 0}; x2 – 4xy + 3y2 = 0

S = {(x, y) | x, y N ds fy, rFkk x2 – 4xy + 3y2 = 0}; x2 – 4xy + 3y2 = 0

(x – y)(x – 3y) = 0. So blfy, , x –y = 0 or ;k x – 3y = 0

x = y or ;k x = 3y

if x = y, then (x, x) S

;fn x = y, rc (x, x) S

S is reflexive S LorqY; gSA

Taking x = 3 and y = 1, x = 3y will be true.

x = 3 rFkk y = 1 ysus ij, x = 3y lgh gSA

(3, 1) S, but (1, 3) S because x2 – 4xy + 3y2 = 1 – 12 + 27 0

(3, 1) S, ijUrq (1, 3) S D;kasfd x2 – 4xy + 3y2 = 1 – 12 + 27 0

S is not symmetric.

S lefer ugha gSA

Similarly, if x = 9, y = 3, then (9, 3) S. Also (3, 1) S

blh izdkj ;fn x = 9, y = 3, rc (9, 3) S. rFkk (3, 1) S

But (9, 1) S. So, S is not transitive (81 – 36 + 3 0)

ijUrq (9, 1) S. blfy, S laØked ugha gS (81 – 36 + 3 0)

S is reflexive but not symmetric and transitive

S LorqY; gS ijUrq lefer ugha gS rFkk laØked

62. Let ]5,0[]5,0[:f be an invertible function defined by f(x) = ax2 + bx + c, Rc,b,a & 0abc then

one of the root of the equation cx2 + bx + a = 0 is

ekuk ]5,0[]5,0[:f esa ifjHkkf”"kr Qyu f(x) = ax2 + bx + c, Rc,b,a vkSj 0abc izfryksfe; Qyu gS rc

lehdj.k cx2 + bx + a = 0 dk ,d ewy gS -

(1*) a (2) b (3) c (4) a + b + c

Sol. f invertible function

f is one. one function

Also 5c5)0(f0)0(f & f(5) = 0

ax2 + bx + c = 0 has 2 roots 5,

a

c.5

a

5.5

a

1



®




equation cx2 + bx + a = 0 has 2 roots 1

,5

1

so a1

Hindi f izfryksfe; Qyu gSA

f ,dSdh Qyu gSA

rFkk 5c5)0(f0)0(f vkSj f(5) = 0

ax2 + bx + c = 0 ds nks ewy 5, gSA

a

c.5

a

5.5

a

1

lehdj.k cx2 + bx + a = 0 ds nks ewy 1

,5

1 gS

blfy, a1

63. One hundred identical coins, each with probability p of showing heads are tossed once. If 0 < p < 1 and the probability of heads showing on 50 coins is equal to that of heads showing on 51 coins, the value of p is

,d lkS loZle flDds] izR;sd 'kh"kZ n'kkZus dh izkf;drk p gSA ftudks ,d ckj mNkyk tkrk gSA ;fn 0 < p < 1 rFkk 50

flDdksa ij 'kh"kZ n'kkZus dh izkf;drk] 51 flDdksa ij 'kh"kZ n'kkZus ds cjkcj gS] rc p dk eku gS&

(1) 1

2 (2*)

51

101 (3)

49

101 (4)

3

101

Sol. Let x be the number of coins showing heads. Then x follows a binomial distribution with parameters n = 100 and p. Since p(x = 50) = p(x = 51), we get

ekuk x, 'kh"kZ n'kkZus ds flDdksa dh la[;k gS] rc x izkpy esa f}in forj.k dks n'kkZrk gS n = 100 vkSj p. pawfd

p(x = 50) = p(x = 51), ;gk¡

100C50p50(1 – p)50 = 100C51p51(1 – p)49

100!

50! 50! ×

51! 49!

100! =

p1

p

p =

51

101

64. The probability that a rectangle picked up from a chessboard has the area 6 cm2 where the distance between consecutive parallel lines on the board is 1 cm, is

'karjt ds cksMZ ls 6 lsehs2 ds {ks=kQy ds vk;r ds pqus tkus dh izkf;drk gS] tgk¡ cksMZ esa Øekxr lekUrj js[kkvksa ds e/; dh nwjh 1 lseh gS -

(1) 56

3 (2)

28

3 (3)

56

9 (4*)

108

11



®




Sol. Rectangle of dimension 8 6 18

1 11 6 C C 24

Rectangle of dimension 8 2 1 8 3 1

1 12 3 C C 42

Total number of rectangle of 65cm 2 66 = 132

Total number of rectangle 9 9

2 2C C

Required probability 132 4 11

9 8 9 8 108

Hindi vk;r (foek;sa (1 × 6) gS) = 8 6 18

1 1C C 24

vk;r (foek;sa (2 × 3) gS) = 8 2 1 8 3 1

1 1C C 42

vk;rksa dh la[;k 65cm 2 66 = 132

vk;rksa dh la[;k 9 9

2 2C C

vHkh"V izkf;drk 132 4 11

9 8 9 8 108

65. If and are the roots of the quadratic equation ax2 + bx + c = 0, then

/1x

Lim2

2

)x1(2

)abxcxcos(1

=

(1*)

1

2

c (2)

1

2

c (3)

1c

(4) None of these

;fn vkSj f}?kkr lehdj.k ax2 + bx + c = 0 ds ewy gS rc -

/1x

Lim2

2

)x1(2

)abxcxcos(1

=

(1*)

1

2

c (2)

1

2

c (3)

1c

(4) buesa ls dksbZ ugha

Sol. ax2 + bx + c = 0 roots ,

ax2 + bx + c = 0 ewy, gSA

So blfy,, cx2 + bx + a = c (x – 1/) (x – 1/)

/1x

Lim22 )/1x(2

))/1x)(/1x(ccos(1

/1xLim

22

2

)/1x(2

)/1x)(/1x(2

csin2

/1x

Lim4

c)/1x(

2

c)/1x)(/1x(

2

c)1x)(/1x(sin

1 22

2

2

2

/1xLim 2

2

2)/1/1(

4

c1

=

11

2

c



®




66. Let f(x) =

2

4 2

n

n

x

x

x e

x e

then lim

xf(x) is:

ekuk f(x) =

2

4 2

n

n

x

x

x e

x e

rc lim

xf(x) gS&

(1) 1 (2*) 1/2 (3) 2 (4) None of these buesa ls dksbZ ugha

Sol.

2

42

2

1

lim

1

x

x

xx

x

xn e

e

xn e

e

2

4

2

11

lim2

2 1

x

x

x

xx n

e

xx n

e

67. If two curves C1 : y = 1 – cotx, x (0, ) and C2 : y = 2|x| + touch each other, then the value of can

be

;fn nks oØ C1 : y = 1 – cotx, x (0, ) vkSj C2 : y = 2|x| + ,d nwljs dks Li'kZ djrs gS rc dk eku gS -

(1*) 2

(2)

2

(3)

2

32

(4) 2

2

Sol : 1C : y 1 cot x x 0,

2C : y 2x (for interval of C1 dk vUrjky)

For 1C and 2C to touch :

1 2c c

dy dy

dx dx

1C o 2C ds fy, Li'kZ djrs gS :

1 2c c

dy dy

dx dx

2 3cosec x 2 x ,

4 4

For touch Li'kZ djus fy, : 2.

1 cot4 4 2

and vkSj3 3

1 cot 2. 24 4

2

32

68. If the function 3 2 22 9 12 1,f x x ax a x where a > 0 , attains its maximum and minimum at p and q

respectively such that p2 = q, then a equals

;fn Qyu 3 2 22 9 12 1,f x x ax a x tgk¡ a > 0 bldk mfPp"B o fufEu"B Øe'k% p vkSj q ij j[krk gS

tcfd p2 = q rc a cjkcj gS&

(1) 1

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



®




Sol. 3 2 22 9 12 1f x x ax a x

2 2' 6 18 12f x x ax a

2 26 3 2x ax a

6 2x a x a

x = a, 2a

'' 12 18 f x x a

'' 6 0 f a a x a then in maximum rc mfPp"B gSA

p = a & rFkk q = 2a

p2 = q a2 = 2a a=2

69. Differentiation of 2

1

2

1 xcos

1 x

with respect to 1 2sin 2x 1 x can be equal to (where x

10,

2

21

2

1 xcos

1 x

dk 1 2sin 2x 1 x ds lkis{k vodyu fdlds cjkcj gS (tgk¡ x

10,

2

)

(1) 2

2

1 x

1 x

(2) 21 x

x

(3) 21 x (4*)

2

2

1 x

1 x

Sol. Let ekuk y = sin–1 22x 1– x

and vkSj z = cos–12

2

1– x1 x

Put x = sinj[kus ij = sin–1x

y = sin–1 22sin 1– sin = sin–1 (sin2)

y =2

= 2 sin–1 x

dy

dx =

2

12

1– x =

2

2

1– x

z = cos–12

2

1– x1 x

= 2 tan–1x

dz

dx=

2

12

1 x

dz dz / dx

dy dy / dx

2

2

1

1

x

x



®




70. The value of 1 2

2

–1

x – 1x 1 dx + 2

1

0

1– x1 x dx is equal to

1 2

2

–1

x – 1x 1 dx + 2

1

0

1– x1 x dx dk eku cjkcj gS&

(1) –2 (2) –1 (3) 2 (4*) 0

Sol. I = 1 2

2

0

x – 1x 1 dx + 2

1

0

1– x1– x dx

= 21 2

2

0

x 1– 2x 1

dx + 2

1

20

1– x1– x dx

= 21

0

1 dx – 41

2

0

1

1 x dx + 21

20

1

1– x dx + 1

2 –1/ 2

0

(1– x ) (–2x) dx

= 2 – 4× 4

+ 2×

2

+ 2

2 1

0[ 1– x ]

= 2 – + + 2[0 – 1] = 0

71. Let f(x) be a function defined by f(x) = x

2

1

t(t – 3t 2) dt, 1 x 3 then the maximum value of f(x) is

ekuk f(x) ,d Qyu bl izdkj ifjHkkf"kr gS fd f(x) = x

2

1

t(t – 3t 2) dt, 1 x 3 rc f(x) dk vf/kdre eku gS&

(1) 0 (2) 1 (3*) 2 (4) 3

Sol. f (x) = x(x2 – 3x + 2) = x(x – 1)(x – 2) the sign scheme for f (x) is as below

f (x) = x(x2 – 3x + 2) = x(x – 1)(x – 2), f (x) ds fpUgksa dh fLFkfr ls

+ + – –

0 1 2

f (x) 0 in 1 x 2 and f (x) 0 in 2 x 3

f (x) 0 esa 1 x 2 vkSj f (x) 0 esa 2 x 3

f(x) is decreasing in [1, 2] and increasing in [2, 3]

[1, 2] esa f(x) áleku gS rFkk [2, 3] esa o/kZeku gSA

maximum f(x) = the greatest among {f(1), f(3)}

vf/kdre f(x) = {f(1), f(3)} esa ls vf/kdre

f(1) = 1

2

1

x(x – 3x 2) dx = 0, f(3) = 3

2

1

x(x – 3x 2) dx = 2

maximum f(x) = 2 f(x) dk vf/kdre eku = 2



®




72. Let f(x) be a non constant continuous function defined in R such that f2(x) = x

0

2sec ²tf(t). dt

4 tant and

f(0) = 0, If m

f n4 n

, where m and n are relatively prime numbers, then 10(m + n) is equal to

ekuk f(x), R esa lrr~ Qyu tks vpj ugha gS] bl izdkj gS fd f2(x) = x

0

2sec ²tf(t). dt

4 tant rFkk f(0) = 0, ;fn

mf n

4 n

, tgk¡ m vkSj n lgvHkkT; la[;k,¡ gS rc 10(m + n) cjkcj gS -

(1*) 90 (2) 90 log4 (3) 9 (4) –19

Sol. We have, x

0

2sec ²t(f(x))² f(t). dt

4 tant

Differentiating both sides w.r. to x we get

2sec ²x sec ²x

2f(x).f '(x) f(x) f '(x)dx dx4 tanx 4tanx

f(x) = log (4 + tan x) + C

Given, f(0) = 0 = log 4 + C C= – log4

f(x) = log (4 + tan x) – log4

5 mf log5 – log4 log log (given)

4 4 n

m = 5, n = 4 m + n = 9

Hindi ;gk¡, x

0

2sec ²t(f(x))² f(t). dt

4 tant

x ds lkis{k nksuks rjQ vodyu ls

2sec ²x sec ²x

2f(x).f '(x) f(x) f '(x)dx dx4 tanx 4tanx

f(x) = log (4 + tan x) + C

fn;k x;k gS, f(0) = 0 = log 4 + C C= – log4

f(x) = log (4 + tan x) – log4

5 mf log5 – log4 log log (given)

4 4 n

m = 5, n = 4 m + n = 9

73. The converse of p (q r) is

p (q r) dk izfrykse gS -

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

Sol. Converse of p q is q p

p q dk izfrykse q p gSA

(q r) p

= (~ q r) p = (q ~ r) p



®




74. The mean of 5 observations is 4.4 and their variance is 8.24. If three of the observations are 1, 2 and 6. Find the other two observation:

5 izs{k.kksa dk ek/; 4.4 gS rFkk mudk izlj.k 8.24 gSA ;fn mu izs{k.kksa esa ls rhu 1, 2 vkSj 6 gS] rc vU; nks izs{k.k Kkr dhft,&

(1*) 4, 9 (2) 8, 4 (3) 4, 4 (4) 9, 9

Sol. 13ba4.45

621ba

2

222 )4.4(

5

3641ba24.8

a2 + b2 = 97

75. If 2r

r1

r

r1

3

1

2

1

, then the triangle is (where r1, r2, r3 are the radii of ex-circles of the given triangle is

(1) Equilateral (2) Isosceles

(3*) Right angled (4) Right angled isosceles

;fn 2r

r1

r

r1

3

1

2

1

, rc f=kHkqt gS& (tgk¡ r1, r2, r3 fn, x, f=kHkqt ds cfg"oÙkksa dh f=kT;k,a gS)

(1) leckgq (2) lef}ckgq

(3*) ledks.k f=kHkqt (4) ledks.k lef}ckgq

Sol. L.H.S. = 2as

cs1

as

bs1

or ;k (b – a) (c – a) = 2 (s – a)2

or ;k 2(bc – ac – ab + a2) = (b + c – a)2

a2 = b2 + c2

A = 90

76. The most general solution of simultaneous equation 02sin5sin22 and

0113tantan32 is

;qxir lehdj.k 02sin5sin22 vkSj 0113tantan3

2 dk O;kid gy gS -

(1) 6

5 n,

n I (2) 6

52

n,

n I

(3*) 6

52

n , n I (4) 2 n , n I

Sol. 02sin5sin22 2,

2

1sin



®




3 tan tan 1 1 tan 1 0 3

1 ,1tan

'' lies in II quadrant

'' f}rh; prqFkkZa'k esa gSA

6

52

n

n I

77. Possible values of a such that the equation x2 + 2ax + a =2 1 1

a + x - - ,x -a16 16

, has two distinct real

roots can be equal to

lehdj.k x2 + 2ax + a =2 1 1

a + x - - ,x -a16 16

ds nks fHkUu fHkUUk okLrfod ewy gksus ds fy, a ds lEHko eku gksxsa&

(1*) 0 (2) 3

1 (3)

2

1 (4*) 1

Sol. The equation is x2 + 2ax + 1

16 = a +

2 1a x

16 , f(x) = f1(x)

nh xbZ lehdj.k x2 + 2ax + 1

16 = a +

2 1a x

16 , f(x) = f1(x)

which has the solution if x2 + 2ax + 1

16= x

ftldk gy gksxk ;fn x2 + 2ax + 1

16= x

2 1x (2a 1)x 0

16

For real and distinct roots (2a 1)2 1

016

okLrfod ,oa fHkUu fHkUu ewy gkssus ds fy, (2a 1)2 1

016

1 1 1 3

2a 1 or 2a 1 a or a2 2 4 4

Also rFkk a2 – a – 16

1< 0

4

52,

4

3

4

1,

4

5–2

78. Let 12

128.......

325

24

65

16

5

8S

18 , then

ekuk 12

128.......

325

24

65

16

5

8S

18 , rc

(1*) 545

1088S (2)

1088

545S (3)

545

1056S (4)

1056

545S



®




Sol.

16

1r4 )1r4(

r8S

16

1r 22 )1r2r2()1r2r2(

r8

=

16

1r22 1r2r2

1

1r2r2

12

=

545

1

481

1......

13

1

13

1

5

1

5

1

1

12

=

545

112 =

545

1088

79. Let f(x) =

2

2 2 2

2 2

sec x cosx sec x cot xcosecx

cos x cos x cosec x

1 cos x cos x

. If

/2

0

f(x)

dx = –

q

p

4. Where p, q are co-prime

numbers, then find the value of (3p – q)

ekuk f(x) =

2

2 2 2

2 2

sec x cosx sec x cot xcosecx

cos x cos x cosec x

1 cos x cos x

;fn

/2

0

f(x)

dx = –

q

p

4tgk¡ p, q lgvHkkT; la[;k,¡ gS rc

(3p – q) dk eku Kkr dhft, -

(1*) 9 (2) 6 (3) 0 (4) 13

Sol. taking sec x and cos x common from c1 and c2 respectively we get

sec x vkSj cos x dks Øe'k% c1 vkSj c2 ls ysus ij

f(x) =

2

3 2

2

1 1 sec x cot x.cosecx

cos x cosx cosec x

cosx cosx cos x

c1 c1 – c2 and then simplify, we get

c1 c1 – c2 rFkk rc ljy djus ij

f(x) = – sin2x – cos5x

/2

0

f(x)

dx = /2

2 5

0

(– sin x – cos x)

dx

= – 8

4 15

3p – q = 9



®




80. If 2

2

xsec x – tanxtan x dx = f(x) + c such that

x 0lim

f(x) = – 1, then x 0lim

n

f(x)

, (n N) is equal to

(where [.] denotes greatest integer function and c is constant of integration)

;fn 2

2

xsec x – tanxtan x dx = f(x) + c bl izdkj gS fd

x 0lim

f(x) = – 1 rc x 0lim

n

f(x)

, (n N) dk eku cjkcj gS -

(tgk¡ [.] egÙke iw.kkZd Qyu gS rFkk c lekdy fu;rkd gS)

(1) n (2*) –n – 1 (3) n – 1 (4) 1 – n

Sol. 2

2

2

xsec x – tanxx

tanx

x

dx = 2

dt

t = – 1

t + c = – x

tan x + c

f(x) = – x

tan x

x 0lim

tanx–nx

= – n – 1












81. The value of the expression 9

cosec6

log 3 2 2 3 2 2

can be expressed in the lowest form as p

q

where p and q are relatively prime, then p + q equals.

;fn O;atd 9

cosec6

log 3 2 2 3 2 2

dk eku p

q ds :i esa O;Dr fd;k tk, tgk¡ p vkSj q lgvHkkT; gS] rks

p + q dk eku gS &



®




Ans: 07.00

Sol: As 3 + 2 2 = 2

2 1 and 3 – 2 2 = 22 1

Expression = 92log ( 2 1) ( 2 1) = 9 9

3/2

2 2log (2 2) log 2 =

3

2 9 =

1

6 =

p

q

p + q = 7 Ans.

82. Let 2 i 4, 1 j 5 and 2 k 4 where i, j, kN. If S9 = {(ai, bj, ck)| i + j + k = 9}, then the total number of elements of S9 will be

ekuk 2 i 4, 1 j 5 vkSj 2 k 4 tgk¡ i, j, kN. ;fn S9 = {(ai, bj, ck)| i + j + k = 9}, rc S9 ds vo;oksa dh dqy la[;k gksxh&

Ans. 09.00

Sol. i A and 2 i 4 so that, i 1

i A vkSj 2 i 4 blfy, i 1

if i = 2 then j + k = 7

;fn i = 2 rc j + k = 7

j = 3 and k = 4, j = 4 and k = 3 and j = 5 and k = 2

j = 3 vkSj k = 4, j = 4 vkSj k = 3 vkSj j = 5 vkSj k = 2

but 2 k 4 so that, j = 2 and k = 5 and j = 1, k = 6 are impossible

ijUrq 2 k 4 blfy, j = 2 vkSj k = 5 vkSj j = 1, k = 6 vlaHko gSA

(2, 3, 4),(2, 4, 3) and (2, 5, 2) are possible elements of S9.

(2, 3, 4),(2, 4, 3) vkSj (2, 5, 2), S9 ds laHkkfor vo;o gSA

if i = 3, then j + k = 6. ;fn i = 3, rc j + k = 6.

j = k = 3 and j = 2, k = 4 and j = 4, k = 2 are possible but j = 5 and k = 1 are impossible.

j = k = 3 vkSj j = 2, k = 4 vkSj j = 4, k = 2 laHko gS ijUrq j = 5 vkSj k = 1 vlaHko gSA

j = 1 and k = 5 are impossible.

j = 1 vkSj k = 5 vlaHko gSA

(3, 3, 3), (3, 2, 4),(3, 4, 2) are in S9

(3, 3, 3), (3, 2, 4),(3, 4, 2), S9 esa gSA

If i = 4, then j + k = 5

;fn i = 4, rc j + k = 5

j = 1, k = 4; j = 2, k = 3 and j = 3, k = 2 are possible, but j = 4, k = 1 are impossible.

j = 1, k = 4; j = 2, k = 3 vkSj j = 3, k = 2 laHko gS ijUrq j = 4, k = 1 vlaHko gSA

(4, 1, 4), (4, 2, 3) and (4, 3, 2) are in S9.

(4, 1, 4), (4, 2, 3) vkSj (4, 3, 2), S9 esa gSA

Thus, S9 has total number of 9 elements which are (4, 1, 4), (2, 5, 2), (3, 2, 4), (3, 3, 3),(3, 4, 2), (4, 2, 3), (4, 3, 2),(2, 4, 3), (2, 3, 4)

vr% S9 esa vo;oksa dh dqy la[;k 9 gS tks fd (4, 1, 4), (2, 5, 2), (3, 2, 4), (3, 3, 3),(3, 4, 2),

(4, 2, 3), (4, 3, 2),(2, 4, 3), (2, 3, 4)

83. Consider a function f(x) = 1xsec2

24x10xcoslog 1

21

5

where [.] and {.} denotes greatest

integer function and fractional part function respectively. Find the number of integers lying in the domain of f(x).

ekukfd f(x) = 1xsec2

24x10xcoslog 1

21

5

tgk¡ [.] rFkk {.} Øe'k% egÙke iw.kk±d Qyu rFkk

fHkUukRed Hkkx Qyu dks O;Dr djrk gSA f(x) ds izkUr esa fLFkr iw.kk±dksa dh la[;k gS&



®




Ans. 04.00

Sol. sec–1[{x} + 1]

= sec–1([{x}] + 1) ([x + n] = [x] + n if n I)

= sec–1 (0 + 1)

= sec–1 (1) defined for all x

= sec–1 (1) lHkh x ds fy, ifjHkkf"kr

domain of izkUr sec–1 [{x} + 1] is x R

Now for vc

2

24x10xcoslog

21

5 ds fy,

02

24x10x2

(x + 6) (x + 4) 0

x (–, –6] [– 4, )

And rFkk 12

24x10x2

x2 + 10x + 24 < 4

x2 + 10x + 20 < 0

x (– 5 – 5 , –5 + 5 )

domain of

2

24x10xcoslog

21

5 is x 55,46,55

2

24x10xcoslog

21

5 dk izkUr gS blfy, x 55,46,55

Integer in the domain are –7, –6, –4, –3

izkUr esa iw.kk±d gS –7, –6, –4, –3

–7, –6, –4, –3

Hencevr% n = 4

84. How many roots of the equation (x – 1)(x – 3)(x – 5) + (x – 1)(x – 3)(x – 7) + (x – 1)(x – 5)(x – 7) + (x – 3)(x – 5) (x – 7) = 0 are positive :

lehdj.k (x – 1)(x – 3)(x – 5) + (x – 1)(x – 3)(x – 7) + (x – 1)(x – 5)(x – 7) + (x – 3)(x – 5) (x – 7) = 0 ds fdrus ewy /kukRed gS&

Ans. 03.00

Sol. Let ekuk f ’(x) = (x – 1) (x – 3) (x – 5) + (x – 1) (x – 3) (x – 7)

+ (x – 1) (x – 5) (x – 7) + (x – 3) (x – 5) (x – 7)

f(x) = (x – 1) (x – 3) (x – 5) (x – 7) + d



®




f(1) = f(3) = f(5) = f(7) = d

so by Rolle’s theorem jksy izes; ls

f ’(x) = 0 has atleast one root each in (1, 3), (3, 5), (5, 7) respectively

f ’(x) = 0 dk de ls de ,d ewy (1, 3), (3, 5), (5, 7) esa gSA

but it is cubic equation so it has 3 positive roots.

ijUrq ?kuh; lehdj.k blfy, rhu /kukRed ewy gSA

85. A = a b

c d

, AAT = 2, bc > 0. If P(, ) divides the line joining Q(1, 2) and R(2, 5) in the ratio c : b

then | + | is equal to

A = a b

c d

, AAT = 2, bc > 0. ;fn js[kk[k.M dks tksM+us okys fcUnq Q(1, 2) vkSj R(2, 5) dks P(, )

c : b esa foHkkftr djrk gS] rks | + | cjkcj gS &

Ans. 05.00

Sol. A = a b

c d

and AAT = 2

= a b

c d

a c

b d

= 2 0

0 2

a2 + b2 = 2, ac + bd = 0, c2 + d2 = 2

ac + bd = 0 a

d = – b

c = k

a = dk, b = – ck

a2 + b2 = 2 k2(d2...................)..............

k2 = 1 k = ± 1

b–c

= ±1 b

c = – 1 or

b

c = 1

P(, ) divides Q(1, 2)and R(2, 5) in the ratio k : 1

= 1 2

2

, =

2 5

2

( k = 1)

= 3

2, =

2

7

| + | = 2

7

2

3 = 5



®




86. If x

1

dt)t(fx

11)x(f , then the value of )e(f 1–

is

;fn x

1

dt)t(fx

11)x(f , rc )e(f 1–

dk eku gS -

Ans. 00.00

Sol. Given fn;k x;k gS

dt)t(fx)x(xf

x

1

)x(f1)x(fx)x(f

c|x|log)x(f

f(1) = 1

1|x|log)x(f

0)e(f 1

87. If in a triangle ABC, 2cosA cosB 2cosC a b

a b c bc ac , then angle A equals to ( in degree)–

;fn ABC esa 2cosA cosB 2cosC a b

a b c bc ac , rc dks.k A cjkcj gS –

Ans. 90.00

Sol. We have 2cosA cosB 2cosC a b

a b c bc ac

Multiplying both sides by abc

2bc cosA + ac cosB + 2ab cosC = a2 + b2

2 2 2

2 2 2 2 2 2 2 2c a b

(b c a ) a b c a b2

2 2 2 2 2(c a b ) 2a 2b

2 2 2b c a A 90

Hence (A) is the correct answer.

;gk¡ 2cosA cosB 2cosC a b

a b c bc ac

abc ls xq.kk djus ij

2bc cosA + ac cosB + 2ab cosC = a2 + b2

2 2 2

2 2 2 2 2 2 2 2c a b

(b c a ) a b c a b2



®




2 2 2 2 2(c a b ) 2a 2b

2 2 2b c a A 90

vr % (A) lgh mÙkj gSA

88. If A = {x |2sinx – 3 = 0; 0 < x < } and B = {x|2cosx + 1 = 0; 0 < x < 2}, then find n(A B)

;fn A = {x |2sinx – 3 = 0; 0 < x < } vkSj B = {x|2cosx + 1 = 0; 0 < x < 2}, rc n(A B) dk eku Kkr dhft,A

Ans. 01.00

Sol. A = {x|2sinx – 3 = 0}

2sinx = 3

sinx = 3

2 > 0

Now vc x (0, )

Soblfy, , x = 3

or

2

3

A = 2

,3 3

B = {x|2cos x + 1 = 0}

2 cos x = – 1

cos x = – 1

2 < 0

Now, x (0, 2) is given.

vc x (0, 2) fn;k x;k gSA

But ijUrq cos x < 0. Soblfy, , x 3

,2 2

x = – 3

or;k +

3

x = 2

3

or ;k

4

3

B = 2 4

,3 3

A B = 2

3



®




89. It is given that the polynomial cbxaxx)x(P 23 and P(x) = 0 has three distinct positive integral roots

and P(14) = 231. Let 232x2x)x(Q 2 . It is also given that the polynomial equation 0)x(QP has no

real roots. Then |a| =

;g fn;k x;k gS fd cbxaxx)x(P 23 rFkk P(x) = 0 ds rhu fHkUu fHkUu /kukRed iw.kk±d ewy gS rFkk

P(14) = 231 ekuk 232x2x)x(Q 2 ;g fn;k x;k gS fd cgqin lehdj.k 0)x(QP dk dksbZ okLrfod ewy ugha gS]

rc |a| =

Ans. 21.00

Sol. 231)1x()x(Q 2

),231[)x(Q

)xx)(xx)(xx()x(P 321

)x14)(x14)(x14(231)14(P 321

= 3 × 7 × 11

11x14,7x14,3x14 321

)xxx(a 321 = –(11 + 7 + 3)

|a| = 21

90. The least integral value of x satisfying the inequation 1

5

log (x2 – 6x + 18) + 2 log5 (x – 4) < 0 is then

value of is

vlfedk 1

5

log (x2 – 6x + 18) + 2 log5 (x – 4) < 0 dks larq"V djus okyk x dk U;wure iw.kk±d eku gS] rc dk eku gS&

Ans. 05.00

Sol. 21

5

log (x – 6x 18) + 2 log5 (x – 4) < 0

– log5(x2 – 6x + 18) + 2 log5(x – 4) < 0

log5 2

2

(x – 4)(x – 6x 18)

< 0

2

2

(x – 4)x – 6x 18

< 1

2

2

(x – 4) – 1 0x – 6x 18

2

–2x – 2x – 6x 18

< 0

– 2 (x + 1) < 0

(x2 – 6x + 18 > 0 x R)

x > – 1 ... (i)

alsorFkk x2 – 6x + 18 > 0, x R and vkSj x – 4 > 0

x > 4 ... (ii) by (i) andvkSj (ii) x > 4 (i) andvkSj (ii) x > 4

least integral value of x is 5.

x ds de ls de eku 5 gSA



test type : main pattern cumulative test-4 (mct-4)

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