JEE Main 2020 Physics January 9 Shift 2 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

 Page 1


     
 
      
    
    
     
 
                       
 
  
 
 
 
PART : PHYSICS 
 
Single Choice Type   (,dy fodYih; izdkj) 
This section contains 20 Single choice questions. Each question has 4 choices (1), (2), (3) and (4) for its 
answer, out of which Only One is correct. 
bl [k.M esa 20 ,dy fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA 
 
1. A mass m attached to spring of natural length ? 0 and spring constant k. One end of string is attached to 
centre of disc in horizontal plane which is being rotated by constant angular speed ?. Find extension 
per unit length in spring (given k >> m ?
2
) :  
 ,d nzO;eku m, ? 0 yEckbZ o fLizax fu;rkad k dh fLizax ls tqM+k gSA fLizax dk ,d fljk {kSfrt ry esa j[kh pdrh ds 
dsUnz ls tqM+k gS rFkk pdrh fu;r dks.kh; osx ? ls ?kwwe jgh gSA fLizax dh bdkbZ yEckbZ esa izlkj Kkr dhft,&  
 (fn;k gS k >> m ?
2
) 
 (1) 
k
m
2
?
    (2) 
k
m
3
2
2
?
    (3) 
k 2
m
2
?
    (4) 
k
m 3
2
?
  
Ans. (1) 
Sol.  
? ?
k, ?0
m
 
m ?
2 
( ?0 + x) = kx 
 
2
0
m
k
1
x ?
? ?
?
?
?
?
?
?
?
 
 x = 
2
2
0
m k
m
? ?
? ?
 
 k >> m ?
2
 
 So, 
0
x
?
 is equal to 
k
m
2
?
.  
 vr% 
0
x
?
, 
k
m
2
?
ds cjkcj gS  
 
Page 2


     
 
      
    
    
     
 
                       
 
  
 
 
 
PART : PHYSICS 
 
Single Choice Type   (,dy fodYih; izdkj) 
This section contains 20 Single choice questions. Each question has 4 choices (1), (2), (3) and (4) for its 
answer, out of which Only One is correct. 
bl [k.M esa 20 ,dy fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA 
 
1. A mass m attached to spring of natural length ? 0 and spring constant k. One end of string is attached to 
centre of disc in horizontal plane which is being rotated by constant angular speed ?. Find extension 
per unit length in spring (given k >> m ?
2
) :  
 ,d nzO;eku m, ? 0 yEckbZ o fLizax fu;rkad k dh fLizax ls tqM+k gSA fLizax dk ,d fljk {kSfrt ry esa j[kh pdrh ds 
dsUnz ls tqM+k gS rFkk pdrh fu;r dks.kh; osx ? ls ?kwwe jgh gSA fLizax dh bdkbZ yEckbZ esa izlkj Kkr dhft,&  
 (fn;k gS k >> m ?
2
) 
 (1) 
k
m
2
?
    (2) 
k
m
3
2
2
?
    (3) 
k 2
m
2
?
    (4) 
k
m 3
2
?
  
Ans. (1) 
Sol.  
? ?
k, ?0
m
 
m ?
2 
( ?0 + x) = kx 
 
2
0
m
k
1
x ?
? ?
?
?
?
?
?
?
?
 
 x = 
2
2
0
m k
m
? ?
? ?
 
 k >> m ?
2
 
 So, 
0
x
?
 is equal to 
k
m
2
?
.  
 vr% 
0
x
?
, 
k
m
2
?
ds cjkcj gS  
 
     
 
      
    
    
     
 
                       
 
  
 
 
 
2. A loop of radius R and mass m is placed in a uniform magnetic field B with its plane perpendicular to 
the field. Current I is flowing in it. Now loop is slightly rotated about its diameter and released. Find time 
period of oscillation.      
 R f=kT;k rFkk m nzO;eku dk ywi fu;r pqEcdh; {ks=k B ds YkEcor~ j[kk gSA blesa /kkjk I cg jgh gSA vc ywi dks 
mlds O;kl ds vuqfn'k vYi dks.k ls ?kqekrs gS rks NksM+us ds i'pkr~ vkorZdky Kkr djks& 
 (1) 
IB
m
2
?
  (2) 
IB
m 2 ?
     (3) 
IB
m
2
?
    (4) 
IB
m
?
  
Ans. (2) 
Sol. ? = MB sin ? = I ?   
 ?R
2
I B ? = 
2
mR
2
?   
 ? = 
m
B 2 ? ?
 = 
T
2 ?
 
 T = 
IB
m 2 ?
  
 
3. A string of mass per unit length ? = 6 × 10
–3
 kg/m is fixed at both ends under the tension 540 N. If the 
string is in resonance with consecutive frequencies 420 Hz and 490 Hz. Then find the length of the 
string?   
 ,d jLlh ftldk js[kh; nzO;eku ?kuRo ? = 6 × 10
–3
 kg/m gS] rFkk ruko 540 N gS] nksuks fljks ij ca/kh gqbZ gSA ;fn 
jLlh dh nks Øekxr vko`fÙk;k¡ 420 Hz rFkk 490 Hz ds lkFk vuqukn esa gS rks jLlh dh yEckbZ Kkr djks& 
 (1) 2.1 m     (2) 1.1 m    (3) 4.8 m   (4) 4.2 m    
Ans. (1) 
Sol. Fundamental frequency ewy vko`fÙk = 490 – 420 = 70 Hz 
 70 = 
? 2
1
?
T
 
? ?70 = 
? 2
1
3
10 6
540
?
?
 
? ? = 
70 2
1
?
 
3
10 90 ? = 
140
300
 
? ? ? ? 2.14 m     
 
4. Ratio of energy density of two steel rods is 1 : 4 when same mass is suspended from the rods. If length 
of both rods is same then ratio of diameter of rods will be.  
 nks LVhy dh NM+ksa esa ÅtkZ ?kuRo dk vuqikr 1 : 4 gS] tc buls leku nzO;eku yVdk;k tkrk gSA ;fn budh 
yEckbZ;k¡ leku gS rks buds O;klksa dk vuqikr Kkr djks& 
 (1) 1 : 2    (2) 2 : 1    (3) 1 : 2    (4) 2 : 1  
Ans. (1) 
Page 3


     
 
      
    
    
     
 
                       
 
  
 
 
 
PART : PHYSICS 
 
Single Choice Type   (,dy fodYih; izdkj) 
This section contains 20 Single choice questions. Each question has 4 choices (1), (2), (3) and (4) for its 
answer, out of which Only One is correct. 
bl [k.M esa 20 ,dy fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA 
 
1. A mass m attached to spring of natural length ? 0 and spring constant k. One end of string is attached to 
centre of disc in horizontal plane which is being rotated by constant angular speed ?. Find extension 
per unit length in spring (given k >> m ?
2
) :  
 ,d nzO;eku m, ? 0 yEckbZ o fLizax fu;rkad k dh fLizax ls tqM+k gSA fLizax dk ,d fljk {kSfrt ry esa j[kh pdrh ds 
dsUnz ls tqM+k gS rFkk pdrh fu;r dks.kh; osx ? ls ?kwwe jgh gSA fLizax dh bdkbZ yEckbZ esa izlkj Kkr dhft,&  
 (fn;k gS k >> m ?
2
) 
 (1) 
k
m
2
?
    (2) 
k
m
3
2
2
?
    (3) 
k 2
m
2
?
    (4) 
k
m 3
2
?
  
Ans. (1) 
Sol.  
? ?
k, ?0
m
 
m ?
2 
( ?0 + x) = kx 
 
2
0
m
k
1
x ?
? ?
?
?
?
?
?
?
?
 
 x = 
2
2
0
m k
m
? ?
? ?
 
 k >> m ?
2
 
 So, 
0
x
?
 is equal to 
k
m
2
?
.  
 vr% 
0
x
?
, 
k
m
2
?
ds cjkcj gS  
 
     
 
      
    
    
     
 
                       
 
  
 
 
 
2. A loop of radius R and mass m is placed in a uniform magnetic field B with its plane perpendicular to 
the field. Current I is flowing in it. Now loop is slightly rotated about its diameter and released. Find time 
period of oscillation.      
 R f=kT;k rFkk m nzO;eku dk ywi fu;r pqEcdh; {ks=k B ds YkEcor~ j[kk gSA blesa /kkjk I cg jgh gSA vc ywi dks 
mlds O;kl ds vuqfn'k vYi dks.k ls ?kqekrs gS rks NksM+us ds i'pkr~ vkorZdky Kkr djks& 
 (1) 
IB
m
2
?
  (2) 
IB
m 2 ?
     (3) 
IB
m
2
?
    (4) 
IB
m
?
  
Ans. (2) 
Sol. ? = MB sin ? = I ?   
 ?R
2
I B ? = 
2
mR
2
?   
 ? = 
m
B 2 ? ?
 = 
T
2 ?
 
 T = 
IB
m 2 ?
  
 
3. A string of mass per unit length ? = 6 × 10
–3
 kg/m is fixed at both ends under the tension 540 N. If the 
string is in resonance with consecutive frequencies 420 Hz and 490 Hz. Then find the length of the 
string?   
 ,d jLlh ftldk js[kh; nzO;eku ?kuRo ? = 6 × 10
–3
 kg/m gS] rFkk ruko 540 N gS] nksuks fljks ij ca/kh gqbZ gSA ;fn 
jLlh dh nks Øekxr vko`fÙk;k¡ 420 Hz rFkk 490 Hz ds lkFk vuqukn esa gS rks jLlh dh yEckbZ Kkr djks& 
 (1) 2.1 m     (2) 1.1 m    (3) 4.8 m   (4) 4.2 m    
Ans. (1) 
Sol. Fundamental frequency ewy vko`fÙk = 490 – 420 = 70 Hz 
 70 = 
? 2
1
?
T
 
? ?70 = 
? 2
1
3
10 6
540
?
?
 
? ? = 
70 2
1
?
 
3
10 90 ? = 
140
300
 
? ? ? ? 2.14 m     
 
4. Ratio of energy density of two steel rods is 1 : 4 when same mass is suspended from the rods. If length 
of both rods is same then ratio of diameter of rods will be.  
 nks LVhy dh NM+ksa esa ÅtkZ ?kuRo dk vuqikr 1 : 4 gS] tc buls leku nzO;eku yVdk;k tkrk gSA ;fn budh 
yEckbZ;k¡ leku gS rks buds O;klksa dk vuqikr Kkr djks& 
 (1) 1 : 2    (2) 2 : 1    (3) 1 : 2    (4) 2 : 1  
Ans. (1) 
     
 
      
    
    
     
 
                       
 
  
 
 
Sol. 
dv
du
 = 
2
1
 stress¼izfrcy½ × 
y
stress
 
 = 
2
1
 
y A
F
2
2
 
 
dv
du
?
4
d
1
 
 
2
1
dv
du
dv
du
?
?
?
?
?
?
?
?
?
?
?
?
 = 
4
1
4
2
d
d
 = 
4
1
 
 
2
1
d
d
 = (4)
1/4
   
 
2
1
d
d
 = 1 : 2 
 
5. A particle is projected from the ground with speed u at angle 60° from horizontal. It collides with a 
second particle of same mass moving with horizontal speed u in same direction at highest point of its 
trajectory. If collision is perfectly inelastic then find horizontal distance travelled by them after collision 
when they reached at ground  
 ,d d.k dks /kjkry ls u pkky ls 60° ds dks.k ij isz{ksfir fd;k tkrk gSA ;g vius mPpre fcUnq ij {kSfrt fn'kk 
esa u pky ls leku fn'kk esas xfr'khy leku nzO;eku ds d.k ls iw.kZr% vizR;kLFk VDdj djrk gS] rks buds }kjk 
VDdj ds i'pkr~  /kjkry rd igqpusa esa pyh x;h {kSfrt nwjh Kkr djks& 
 (1) 
2
3 6u
8g
    (2) 
2
3 3u
8g
    (3) 
2
u
8g
    (4)  
2
3u
g
 
Ans. (2) 
Sol. 
 
?=60° 
ucos ? 
u 
m 
m 
2m 
v 
 
 p i = p f 
mu + mucos ? ?= 2mv  
? 
? ? u 1 cos60 3
v u
2 4
? ?
? ? 
so horizontal range after collision VDdj ds i'pkr~ {kSfrt ijkl = vt 
     = 
max
2H
v
g
 
    = 
? ?
2 2
2
2u sin 60 3
u
4
2g
?
 
    = 
2
2
3
3 3 3u
4
u
4 g 8g
?  
 
Page 4


     
 
      
    
    
     
 
                       
 
  
 
 
 
PART : PHYSICS 
 
Single Choice Type   (,dy fodYih; izdkj) 
This section contains 20 Single choice questions. Each question has 4 choices (1), (2), (3) and (4) for its 
answer, out of which Only One is correct. 
bl [k.M esa 20 ,dy fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA 
 
1. A mass m attached to spring of natural length ? 0 and spring constant k. One end of string is attached to 
centre of disc in horizontal plane which is being rotated by constant angular speed ?. Find extension 
per unit length in spring (given k >> m ?
2
) :  
 ,d nzO;eku m, ? 0 yEckbZ o fLizax fu;rkad k dh fLizax ls tqM+k gSA fLizax dk ,d fljk {kSfrt ry esa j[kh pdrh ds 
dsUnz ls tqM+k gS rFkk pdrh fu;r dks.kh; osx ? ls ?kwwe jgh gSA fLizax dh bdkbZ yEckbZ esa izlkj Kkr dhft,&  
 (fn;k gS k >> m ?
2
) 
 (1) 
k
m
2
?
    (2) 
k
m
3
2
2
?
    (3) 
k 2
m
2
?
    (4) 
k
m 3
2
?
  
Ans. (1) 
Sol.  
? ?
k, ?0
m
 
m ?
2 
( ?0 + x) = kx 
 
2
0
m
k
1
x ?
? ?
?
?
?
?
?
?
?
 
 x = 
2
2
0
m k
m
? ?
? ?
 
 k >> m ?
2
 
 So, 
0
x
?
 is equal to 
k
m
2
?
.  
 vr% 
0
x
?
, 
k
m
2
?
ds cjkcj gS  
 
     
 
      
    
    
     
 
                       
 
  
 
 
 
2. A loop of radius R and mass m is placed in a uniform magnetic field B with its plane perpendicular to 
the field. Current I is flowing in it. Now loop is slightly rotated about its diameter and released. Find time 
period of oscillation.      
 R f=kT;k rFkk m nzO;eku dk ywi fu;r pqEcdh; {ks=k B ds YkEcor~ j[kk gSA blesa /kkjk I cg jgh gSA vc ywi dks 
mlds O;kl ds vuqfn'k vYi dks.k ls ?kqekrs gS rks NksM+us ds i'pkr~ vkorZdky Kkr djks& 
 (1) 
IB
m
2
?
  (2) 
IB
m 2 ?
     (3) 
IB
m
2
?
    (4) 
IB
m
?
  
Ans. (2) 
Sol. ? = MB sin ? = I ?   
 ?R
2
I B ? = 
2
mR
2
?   
 ? = 
m
B 2 ? ?
 = 
T
2 ?
 
 T = 
IB
m 2 ?
  
 
3. A string of mass per unit length ? = 6 × 10
–3
 kg/m is fixed at both ends under the tension 540 N. If the 
string is in resonance with consecutive frequencies 420 Hz and 490 Hz. Then find the length of the 
string?   
 ,d jLlh ftldk js[kh; nzO;eku ?kuRo ? = 6 × 10
–3
 kg/m gS] rFkk ruko 540 N gS] nksuks fljks ij ca/kh gqbZ gSA ;fn 
jLlh dh nks Øekxr vko`fÙk;k¡ 420 Hz rFkk 490 Hz ds lkFk vuqukn esa gS rks jLlh dh yEckbZ Kkr djks& 
 (1) 2.1 m     (2) 1.1 m    (3) 4.8 m   (4) 4.2 m    
Ans. (1) 
Sol. Fundamental frequency ewy vko`fÙk = 490 – 420 = 70 Hz 
 70 = 
? 2
1
?
T
 
? ?70 = 
? 2
1
3
10 6
540
?
?
 
? ? = 
70 2
1
?
 
3
10 90 ? = 
140
300
 
? ? ? ? 2.14 m     
 
4. Ratio of energy density of two steel rods is 1 : 4 when same mass is suspended from the rods. If length 
of both rods is same then ratio of diameter of rods will be.  
 nks LVhy dh NM+ksa esa ÅtkZ ?kuRo dk vuqikr 1 : 4 gS] tc buls leku nzO;eku yVdk;k tkrk gSA ;fn budh 
yEckbZ;k¡ leku gS rks buds O;klksa dk vuqikr Kkr djks& 
 (1) 1 : 2    (2) 2 : 1    (3) 1 : 2    (4) 2 : 1  
Ans. (1) 
     
 
      
    
    
     
 
                       
 
  
 
 
Sol. 
dv
du
 = 
2
1
 stress¼izfrcy½ × 
y
stress
 
 = 
2
1
 
y A
F
2
2
 
 
dv
du
?
4
d
1
 
 
2
1
dv
du
dv
du
?
?
?
?
?
?
?
?
?
?
?
?
 = 
4
1
4
2
d
d
 = 
4
1
 
 
2
1
d
d
 = (4)
1/4
   
 
2
1
d
d
 = 1 : 2 
 
5. A particle is projected from the ground with speed u at angle 60° from horizontal. It collides with a 
second particle of same mass moving with horizontal speed u in same direction at highest point of its 
trajectory. If collision is perfectly inelastic then find horizontal distance travelled by them after collision 
when they reached at ground  
 ,d d.k dks /kjkry ls u pkky ls 60° ds dks.k ij isz{ksfir fd;k tkrk gSA ;g vius mPpre fcUnq ij {kSfrt fn'kk 
esa u pky ls leku fn'kk esas xfr'khy leku nzO;eku ds d.k ls iw.kZr% vizR;kLFk VDdj djrk gS] rks buds }kjk 
VDdj ds i'pkr~  /kjkry rd igqpusa esa pyh x;h {kSfrt nwjh Kkr djks& 
 (1) 
2
3 6u
8g
    (2) 
2
3 3u
8g
    (3) 
2
u
8g
    (4)  
2
3u
g
 
Ans. (2) 
Sol. 
 
?=60° 
ucos ? 
u 
m 
m 
2m 
v 
 
 p i = p f 
mu + mucos ? ?= 2mv  
? 
? ? u 1 cos60 3
v u
2 4
? ?
? ? 
so horizontal range after collision VDdj ds i'pkr~ {kSfrt ijkl = vt 
     = 
max
2H
v
g
 
    = 
? ?
2 2
2
2u sin 60 3
u
4
2g
?
 
    = 
2
2
3
3 3 3u
4
u
4 g 8g
?  
 
     
 
      
    
    
     
 
                       
 
  
 
 
 
6. H-like atom with ionization energy of 9R. Find the wavelength of light emitted (in nm) when electron 
jumps from second excited state to ground state. (R is Rydberg constant)  
 ,d gkbMªkstu tSls ijek.kq dh vk;uu ÅtkZ 9R gS] rks mRlftZr izdk'k dh rjaxnS/;Z (nm esa) Kkr djks tc bysDVªkWu 
f}rh; mRrsftr voLFkk ls ewy voLFkk esa dwnrk gSA (R fjMcxZ fu;rkad gS) 
 (1) 11.39   (2) 12.86    (3) 8.02   (4) 2.19 
Ans. (1) 
Sol. ? ?
2
2 2
1 2
hc 1 1
13.6eV z –
n n
? ?
?
? ?
?
? ?
? ?
 
 n 1 = 1 
 n 2 = 3 
 ? ?
? ?
2
2 2
hc 1 1
13.6eV 3 –
1 3
? ?
?
? ?
?
? ?
 
 ? ? ? ? ?
hc 8
13.6eV 9
9
? ?
?
? ?
? ? ?
 
 wavelength rjaxnS/;Z = 
1240
nm
8 13.6 ?
 
 ? = 11.39nm 
 
7. Two planets of masses M and 
2
M
 have radii R and 
2
R
 respectively. If ratio of escape velocities from 
their surfaces 
2
1
v
v
 is 
4
n
, then find n :   
 nks xzg ftuds nzO;eku M rFkk 
2
M
 rFkk f=kT;k;sa Øe'k% R o 
2
R
 gSA ;fn buds fy;s lrg ls iyk;u osxks dk vuqikr 
2
1
v
v
 dk eku 
4
n
 gS rks n  Kkr djs &   
 (1) 3    (2) 1     (3) 2     (4) 4 
Ans. (4) 
Sol. v e = 
R
GM 2
 
 ? 
2 / R
2 / GM 2
R
GM 2
v
v
2
1
? = 1 = 
4
n
 
? ? ? ? n = 4 
8. Find centre of mass of given rod of linear mass density ? = 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
2
x
b a
?
, x is distance from one of its 
end. Length of the rod is ?.  
 js[kh; nzO;eku ?kuRo ? = 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
2
x
b a
?
, tgka x ,d fljs ls nwjh gS] okyh NM+ dk nzO;eku dsUnz Kkr djksA ? NM+ 
dh yEckbZ gSA 
 (1) ?
?
?
?
?
?
?
?
b a 2
b a 3
4
3 ?
  (2) ?
?
?
?
?
?
?
?
b a 3
b a 2
4
3 ?
  (3) ?
?
?
?
?
?
?
?
b a 3
b a 2
4
?
  (4) ?
?
?
?
?
?
?
?
b a 3
b a 2
? 
Ans. (2) 
Page 5


     
 
      
    
    
     
 
                       
 
  
 
 
 
PART : PHYSICS 
 
Single Choice Type   (,dy fodYih; izdkj) 
This section contains 20 Single choice questions. Each question has 4 choices (1), (2), (3) and (4) for its 
answer, out of which Only One is correct. 
bl [k.M esa 20 ,dy fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA 
 
1. A mass m attached to spring of natural length ? 0 and spring constant k. One end of string is attached to 
centre of disc in horizontal plane which is being rotated by constant angular speed ?. Find extension 
per unit length in spring (given k >> m ?
2
) :  
 ,d nzO;eku m, ? 0 yEckbZ o fLizax fu;rkad k dh fLizax ls tqM+k gSA fLizax dk ,d fljk {kSfrt ry esa j[kh pdrh ds 
dsUnz ls tqM+k gS rFkk pdrh fu;r dks.kh; osx ? ls ?kwwe jgh gSA fLizax dh bdkbZ yEckbZ esa izlkj Kkr dhft,&  
 (fn;k gS k >> m ?
2
) 
 (1) 
k
m
2
?
    (2) 
k
m
3
2
2
?
    (3) 
k 2
m
2
?
    (4) 
k
m 3
2
?
  
Ans. (1) 
Sol.  
? ?
k, ?0
m
 
m ?
2 
( ?0 + x) = kx 
 
2
0
m
k
1
x ?
? ?
?
?
?
?
?
?
?
 
 x = 
2
2
0
m k
m
? ?
? ?
 
 k >> m ?
2
 
 So, 
0
x
?
 is equal to 
k
m
2
?
.  
 vr% 
0
x
?
, 
k
m
2
?
ds cjkcj gS  
 
     
 
      
    
    
     
 
                       
 
  
 
 
 
2. A loop of radius R and mass m is placed in a uniform magnetic field B with its plane perpendicular to 
the field. Current I is flowing in it. Now loop is slightly rotated about its diameter and released. Find time 
period of oscillation.      
 R f=kT;k rFkk m nzO;eku dk ywi fu;r pqEcdh; {ks=k B ds YkEcor~ j[kk gSA blesa /kkjk I cg jgh gSA vc ywi dks 
mlds O;kl ds vuqfn'k vYi dks.k ls ?kqekrs gS rks NksM+us ds i'pkr~ vkorZdky Kkr djks& 
 (1) 
IB
m
2
?
  (2) 
IB
m 2 ?
     (3) 
IB
m
2
?
    (4) 
IB
m
?
  
Ans. (2) 
Sol. ? = MB sin ? = I ?   
 ?R
2
I B ? = 
2
mR
2
?   
 ? = 
m
B 2 ? ?
 = 
T
2 ?
 
 T = 
IB
m 2 ?
  
 
3. A string of mass per unit length ? = 6 × 10
–3
 kg/m is fixed at both ends under the tension 540 N. If the 
string is in resonance with consecutive frequencies 420 Hz and 490 Hz. Then find the length of the 
string?   
 ,d jLlh ftldk js[kh; nzO;eku ?kuRo ? = 6 × 10
–3
 kg/m gS] rFkk ruko 540 N gS] nksuks fljks ij ca/kh gqbZ gSA ;fn 
jLlh dh nks Øekxr vko`fÙk;k¡ 420 Hz rFkk 490 Hz ds lkFk vuqukn esa gS rks jLlh dh yEckbZ Kkr djks& 
 (1) 2.1 m     (2) 1.1 m    (3) 4.8 m   (4) 4.2 m    
Ans. (1) 
Sol. Fundamental frequency ewy vko`fÙk = 490 – 420 = 70 Hz 
 70 = 
? 2
1
?
T
 
? ?70 = 
? 2
1
3
10 6
540
?
?
 
? ? = 
70 2
1
?
 
3
10 90 ? = 
140
300
 
? ? ? ? 2.14 m     
 
4. Ratio of energy density of two steel rods is 1 : 4 when same mass is suspended from the rods. If length 
of both rods is same then ratio of diameter of rods will be.  
 nks LVhy dh NM+ksa esa ÅtkZ ?kuRo dk vuqikr 1 : 4 gS] tc buls leku nzO;eku yVdk;k tkrk gSA ;fn budh 
yEckbZ;k¡ leku gS rks buds O;klksa dk vuqikr Kkr djks& 
 (1) 1 : 2    (2) 2 : 1    (3) 1 : 2    (4) 2 : 1  
Ans. (1) 
     
 
      
    
    
     
 
                       
 
  
 
 
Sol. 
dv
du
 = 
2
1
 stress¼izfrcy½ × 
y
stress
 
 = 
2
1
 
y A
F
2
2
 
 
dv
du
?
4
d
1
 
 
2
1
dv
du
dv
du
?
?
?
?
?
?
?
?
?
?
?
?
 = 
4
1
4
2
d
d
 = 
4
1
 
 
2
1
d
d
 = (4)
1/4
   
 
2
1
d
d
 = 1 : 2 
 
5. A particle is projected from the ground with speed u at angle 60° from horizontal. It collides with a 
second particle of same mass moving with horizontal speed u in same direction at highest point of its 
trajectory. If collision is perfectly inelastic then find horizontal distance travelled by them after collision 
when they reached at ground  
 ,d d.k dks /kjkry ls u pkky ls 60° ds dks.k ij isz{ksfir fd;k tkrk gSA ;g vius mPpre fcUnq ij {kSfrt fn'kk 
esa u pky ls leku fn'kk esas xfr'khy leku nzO;eku ds d.k ls iw.kZr% vizR;kLFk VDdj djrk gS] rks buds }kjk 
VDdj ds i'pkr~  /kjkry rd igqpusa esa pyh x;h {kSfrt nwjh Kkr djks& 
 (1) 
2
3 6u
8g
    (2) 
2
3 3u
8g
    (3) 
2
u
8g
    (4)  
2
3u
g
 
Ans. (2) 
Sol. 
 
?=60° 
ucos ? 
u 
m 
m 
2m 
v 
 
 p i = p f 
mu + mucos ? ?= 2mv  
? 
? ? u 1 cos60 3
v u
2 4
? ?
? ? 
so horizontal range after collision VDdj ds i'pkr~ {kSfrt ijkl = vt 
     = 
max
2H
v
g
 
    = 
? ?
2 2
2
2u sin 60 3
u
4
2g
?
 
    = 
2
2
3
3 3 3u
4
u
4 g 8g
?  
 
     
 
      
    
    
     
 
                       
 
  
 
 
 
6. H-like atom with ionization energy of 9R. Find the wavelength of light emitted (in nm) when electron 
jumps from second excited state to ground state. (R is Rydberg constant)  
 ,d gkbMªkstu tSls ijek.kq dh vk;uu ÅtkZ 9R gS] rks mRlftZr izdk'k dh rjaxnS/;Z (nm esa) Kkr djks tc bysDVªkWu 
f}rh; mRrsftr voLFkk ls ewy voLFkk esa dwnrk gSA (R fjMcxZ fu;rkad gS) 
 (1) 11.39   (2) 12.86    (3) 8.02   (4) 2.19 
Ans. (1) 
Sol. ? ?
2
2 2
1 2
hc 1 1
13.6eV z –
n n
? ?
?
? ?
?
? ?
? ?
 
 n 1 = 1 
 n 2 = 3 
 ? ?
? ?
2
2 2
hc 1 1
13.6eV 3 –
1 3
? ?
?
? ?
?
? ?
 
 ? ? ? ? ?
hc 8
13.6eV 9
9
? ?
?
? ?
? ? ?
 
 wavelength rjaxnS/;Z = 
1240
nm
8 13.6 ?
 
 ? = 11.39nm 
 
7. Two planets of masses M and 
2
M
 have radii R and 
2
R
 respectively. If ratio of escape velocities from 
their surfaces 
2
1
v
v
 is 
4
n
, then find n :   
 nks xzg ftuds nzO;eku M rFkk 
2
M
 rFkk f=kT;k;sa Øe'k% R o 
2
R
 gSA ;fn buds fy;s lrg ls iyk;u osxks dk vuqikr 
2
1
v
v
 dk eku 
4
n
 gS rks n  Kkr djs &   
 (1) 3    (2) 1     (3) 2     (4) 4 
Ans. (4) 
Sol. v e = 
R
GM 2
 
 ? 
2 / R
2 / GM 2
R
GM 2
v
v
2
1
? = 1 = 
4
n
 
? ? ? ? n = 4 
8. Find centre of mass of given rod of linear mass density ? = 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
2
x
b a
?
, x is distance from one of its 
end. Length of the rod is ?.  
 js[kh; nzO;eku ?kuRo ? = 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
2
x
b a
?
, tgka x ,d fljs ls nwjh gS] okyh NM+ dk nzO;eku dsUnz Kkr djksA ? NM+ 
dh yEckbZ gSA 
 (1) ?
?
?
?
?
?
?
?
b a 2
b a 3
4
3 ?
  (2) ?
?
?
?
?
?
?
?
b a 3
b a 2
4
3 ?
  (3) ?
?
?
?
?
?
?
?
b a 3
b a 2
4
?
  (4) ?
?
?
?
?
?
?
?
b a 3
b a 2
? 
Ans. (2) 
     
 
      
    
    
     
 
                       
 
  
 
 
Sol.  
  
 xcm = 
?
?
0
dM . x
M
1
 
 dM = ? .dx = 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
2
x
b a
?
 . dx 
 x
cm
 = 
?
?
dM
xdM
 = 
?
?
?
?
dx
dx x
 = 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
0
2
2
0
2
2
dx
bx
a
dx
bx
a x
 
        = 
?
?
? ?
?
?
0
3
2
0
0
4
2
0
2
3
x b
) x ( a
4
x b
2
x
a
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
 
       = 
3
b
a
4
b
2
a
2 2
?
?
? ?
?
?
 = 3
4 ) b a 3 (
) b a 2 (
?
?
? ?
 
       = ?
?
?
?
?
?
?
?
b a 3
b a 2
4
3 ?
 
 
9. If a point source is placed at a depth h in a liquid of refractive index 
3
4
. Find percentage of energy of 
light that escapes from liquid.  (assuming 100% transmission of emerging light) 
 fdlh nzo esa h xgjkbZ ij ,d fcUnq L=kksr j[kk gS rFkk nzo dk vioZrukad 
3
4
 gSa] rks nzo ls fudyus okyh ÅtkZ dk 
izfr'kr eku Kkr djks ¼ckgj fudyus okys izdk'k dk izfjxeu 100% gS½& 
 (1) 15%   (2) 17%   (3) 21 %  (4) 34%  
Ans. (2) 
dx 
x