JEE Main 2020 Physics January 7 Shift 2 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

 Page 1


       
 
      
    
    
      
 
                       
 
 
  
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. If weight of an object at pole is 196 N then weight at equator is [g = 10 m/s
2
 ; radius of earth = 6400 Km] 
 ;fn ,d oLrq dk Hkkj /kzqo ij 196N gS] rks Hkwe/; js[kk ij bldk Hkkj gksxkA [g = 10 m/s
2
 ; i`Foh dh f=kT;k = 6400 Km] 
 (1) 194.32 N  (2) 194.66 N  (3) 195.32 N  (4) 195.66 N 
Ans. (3) 
Sol. at pole, weight /kzqo ij Hkkj = mg = 196 
 m = 19.6 kg 
 at equator, weight Hkwe/; js[kk ij Hkkj = mg – m ?
2
R 
 = 196 – (19.6) 
2
3
2
6400 10
24 3600
? ? ?
? ?
? ?
?
? ?
 
 = 195.33 N 
 
2. In a house 15 Bulbs of 45 W, 15 bulbs of 100 W, 15 bulbs of 10 W and Two heaters of 1 KW each is 
connected to 220 V Mains supply then find minimum fuse current  
 ,d ?kj esa 45W ds 15 cYc 100W ds 15 cYc] 10W ds 15 cYc rFkk 1KW ds nks ghVj 220V dh eq[; /kkjk ls 
tqMs+ gq,s gS] rks ¶;wt+ rkj ls izokfgr U;wure /kkjk dk eku gksxkA 
 (1) 5 A   (2) 20 A   (3) 25 A   (4) 15 A 
Ans. (2) 
Sol. Total power is dqy 'kfDr (15 × 45) + (15 × 100) + (15 × 10) + (2 × 1000)  
 = 4325 W 
 So current is vr% /kkjk = 
4325
19.66
220
? A 
 Ans is 20 Amp. 
 
3. In an adiabatic process, volume is doubled then find the ratio of final average relaxation time & initial 
relaxation time. Given ? ?
V
P
C
C
 
 ,d :ðks"e izØe esa vk;ru dks nksxquk dj fn;k tkrk gS] rks vfUre rFkk izkjfEHkd vkSlr foJkafr dky dk vuqikr 
Kkr djsaA fn;k gS ? ?
V
P
C
C
 
 (1) 
2
1
   (2) 2   (3) 
?
?
?
?
?
?
?
2
1
  (4) 
1
2
2
1
?
?
?
?
?
?
?
?
  
Ans. (Bonus) 
Sol. relaxation time foJkafr dky ( ?) ? 
T
V
 
Page 2


       
 
      
    
    
      
 
                       
 
 
  
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. If weight of an object at pole is 196 N then weight at equator is [g = 10 m/s
2
 ; radius of earth = 6400 Km] 
 ;fn ,d oLrq dk Hkkj /kzqo ij 196N gS] rks Hkwe/; js[kk ij bldk Hkkj gksxkA [g = 10 m/s
2
 ; i`Foh dh f=kT;k = 6400 Km] 
 (1) 194.32 N  (2) 194.66 N  (3) 195.32 N  (4) 195.66 N 
Ans. (3) 
Sol. at pole, weight /kzqo ij Hkkj = mg = 196 
 m = 19.6 kg 
 at equator, weight Hkwe/; js[kk ij Hkkj = mg – m ?
2
R 
 = 196 – (19.6) 
2
3
2
6400 10
24 3600
? ? ?
? ?
? ?
?
? ?
 
 = 195.33 N 
 
2. In a house 15 Bulbs of 45 W, 15 bulbs of 100 W, 15 bulbs of 10 W and Two heaters of 1 KW each is 
connected to 220 V Mains supply then find minimum fuse current  
 ,d ?kj esa 45W ds 15 cYc 100W ds 15 cYc] 10W ds 15 cYc rFkk 1KW ds nks ghVj 220V dh eq[; /kkjk ls 
tqMs+ gq,s gS] rks ¶;wt+ rkj ls izokfgr U;wure /kkjk dk eku gksxkA 
 (1) 5 A   (2) 20 A   (3) 25 A   (4) 15 A 
Ans. (2) 
Sol. Total power is dqy 'kfDr (15 × 45) + (15 × 100) + (15 × 10) + (2 × 1000)  
 = 4325 W 
 So current is vr% /kkjk = 
4325
19.66
220
? A 
 Ans is 20 Amp. 
 
3. In an adiabatic process, volume is doubled then find the ratio of final average relaxation time & initial 
relaxation time. Given ? ?
V
P
C
C
 
 ,d :ðks"e izØe esa vk;ru dks nksxquk dj fn;k tkrk gS] rks vfUre rFkk izkjfEHkd vkSlr foJkafr dky dk vuqikr 
Kkr djsaA fn;k gS ? ?
V
P
C
C
 
 (1) 
2
1
   (2) 2   (3) 
?
?
?
?
?
?
?
2
1
  (4) 
1
2
2
1
?
?
?
?
?
?
?
?
  
Ans. (Bonus) 
Sol. relaxation time foJkafr dky ( ?) ? 
T
V
 
       
 
      
    
    
      
 
                       
 
 
  
PAGE # 2 
 
 and rFkk T ? 
1
1
V
? ?
 
 ? ? 
2
1
1
V
? ?
?
 
 ? ? 
2
1
V
? ?
 
 
2
1
i
f
V
V 2
? ?
?
?
?
?
?
?
?
?
?
 
 ? ? 2
1
i
f
2
? ?
?
?
?
 
 
4. A block of mass 10kg is suspended from string of length 4m. When pulled by a force F along horizontal 
 from midpoint. Upper half of string makes 45° with vertical, value of F is    
 ,d 10kg æO;eku dk CykWd 4m yEckbZ dh jLlh ls tqM+k gqvk gSA ;fn jLlh ds e/; fcUnq ls ,d {kSfrt+ cy F 
yxk;k tkrk gS] rks Åij dk vk/kk Hkkx Å/okZ/kj ls 45° dksa.k cukrk gS] rks F dk eku gksxkµ 
 (1) 100N  (2) 90N   (3) 75N   (4) 70N  
Ans. (1) 
Sol.  
 
 
10kg 
T 
100N 
100N 
F 
45° 
 
2
T
 = 100 
 
2
T
 = F 
 F = 100N 
 
5. The surface mass density of a disc of radius a varies with radial distance as ? = A + Br where A & B are 
positive constants then moment of inertia of the disc about an axis passing through its centre and 
perpendicular to the plane   
 ,d pdrh dk lrgh æO;eku ?kuRo f=kfTt; nwjh ds vuqlkj ? = A + Br ls ifjofrZr gksrk gS] rFkk pdrh dh f=kT;k 
a gS] rFkk A  vkSj B /kukRed fu;rkad gSa] rks pdrh ds dsUæ ls ry ds yEcor~ xqt+jus okyh v{k ds lkis{k tM+Ro 
vk?kw.kZ gksxk 
 (1) ?
?
?
?
?
?
? ?
5
Ba
4
A
a 2
4
     (2) ?
?
?
?
?
?
? ?
5
B
4
Aa
a 2
4
  
(3) ?
?
?
?
?
?
? ?
5
Ba
4
A
a
4
     (4) ?
?
?
?
?
?
? ?
4
Ba
5
A
a 2
4
  
Ans. (1) 
Page 3


       
 
      
    
    
      
 
                       
 
 
  
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. If weight of an object at pole is 196 N then weight at equator is [g = 10 m/s
2
 ; radius of earth = 6400 Km] 
 ;fn ,d oLrq dk Hkkj /kzqo ij 196N gS] rks Hkwe/; js[kk ij bldk Hkkj gksxkA [g = 10 m/s
2
 ; i`Foh dh f=kT;k = 6400 Km] 
 (1) 194.32 N  (2) 194.66 N  (3) 195.32 N  (4) 195.66 N 
Ans. (3) 
Sol. at pole, weight /kzqo ij Hkkj = mg = 196 
 m = 19.6 kg 
 at equator, weight Hkwe/; js[kk ij Hkkj = mg – m ?
2
R 
 = 196 – (19.6) 
2
3
2
6400 10
24 3600
? ? ?
? ?
? ?
?
? ?
 
 = 195.33 N 
 
2. In a house 15 Bulbs of 45 W, 15 bulbs of 100 W, 15 bulbs of 10 W and Two heaters of 1 KW each is 
connected to 220 V Mains supply then find minimum fuse current  
 ,d ?kj esa 45W ds 15 cYc 100W ds 15 cYc] 10W ds 15 cYc rFkk 1KW ds nks ghVj 220V dh eq[; /kkjk ls 
tqMs+ gq,s gS] rks ¶;wt+ rkj ls izokfgr U;wure /kkjk dk eku gksxkA 
 (1) 5 A   (2) 20 A   (3) 25 A   (4) 15 A 
Ans. (2) 
Sol. Total power is dqy 'kfDr (15 × 45) + (15 × 100) + (15 × 10) + (2 × 1000)  
 = 4325 W 
 So current is vr% /kkjk = 
4325
19.66
220
? A 
 Ans is 20 Amp. 
 
3. In an adiabatic process, volume is doubled then find the ratio of final average relaxation time & initial 
relaxation time. Given ? ?
V
P
C
C
 
 ,d :ðks"e izØe esa vk;ru dks nksxquk dj fn;k tkrk gS] rks vfUre rFkk izkjfEHkd vkSlr foJkafr dky dk vuqikr 
Kkr djsaA fn;k gS ? ?
V
P
C
C
 
 (1) 
2
1
   (2) 2   (3) 
?
?
?
?
?
?
?
2
1
  (4) 
1
2
2
1
?
?
?
?
?
?
?
?
  
Ans. (Bonus) 
Sol. relaxation time foJkafr dky ( ?) ? 
T
V
 
       
 
      
    
    
      
 
                       
 
 
  
PAGE # 2 
 
 and rFkk T ? 
1
1
V
? ?
 
 ? ? 
2
1
1
V
? ?
?
 
 ? ? 
2
1
V
? ?
 
 
2
1
i
f
V
V 2
? ?
?
?
?
?
?
?
?
?
?
 
 ? ? 2
1
i
f
2
? ?
?
?
?
 
 
4. A block of mass 10kg is suspended from string of length 4m. When pulled by a force F along horizontal 
 from midpoint. Upper half of string makes 45° with vertical, value of F is    
 ,d 10kg æO;eku dk CykWd 4m yEckbZ dh jLlh ls tqM+k gqvk gSA ;fn jLlh ds e/; fcUnq ls ,d {kSfrt+ cy F 
yxk;k tkrk gS] rks Åij dk vk/kk Hkkx Å/okZ/kj ls 45° dksa.k cukrk gS] rks F dk eku gksxkµ 
 (1) 100N  (2) 90N   (3) 75N   (4) 70N  
Ans. (1) 
Sol.  
 
 
10kg 
T 
100N 
100N 
F 
45° 
 
2
T
 = 100 
 
2
T
 = F 
 F = 100N 
 
5. The surface mass density of a disc of radius a varies with radial distance as ? = A + Br where A & B are 
positive constants then moment of inertia of the disc about an axis passing through its centre and 
perpendicular to the plane   
 ,d pdrh dk lrgh æO;eku ?kuRo f=kfTt; nwjh ds vuqlkj ? = A + Br ls ifjofrZr gksrk gS] rFkk pdrh dh f=kT;k 
a gS] rFkk A  vkSj B /kukRed fu;rkad gSa] rks pdrh ds dsUæ ls ry ds yEcor~ xqt+jus okyh v{k ds lkis{k tM+Ro 
vk?kw.kZ gksxk 
 (1) ?
?
?
?
?
?
? ?
5
Ba
4
A
a 2
4
     (2) ?
?
?
?
?
?
? ?
5
B
4
Aa
a 2
4
  
(3) ?
?
?
?
?
?
? ?
5
Ba
4
A
a
4
     (4) ?
?
?
?
?
?
? ?
4
Ba
5
A
a 2
4
  
Ans. (1) 
       
 
      
    
    
      
 
                       
 
 
  
PAGE # 3 
 
Sol. 
 
 
a 
 
 ? = A + Br 
 ? ?
? ?
? ? ? rdr 2 Br A dm 
 I = 
2
r dm
?
 
 ? ? dr r 2 Br A
3
a
0
? ? ?
?
 
 = 
?
?
?
?
?
?
?
?
? ?
5
a
B
4
a
A 2
5 4
 
 = 2 ?a
4
 ?
?
?
?
?
?
?
5
Ba
4
A
   
 
6. Cascaded Carnot engine is an arrangement in which heat sink of one engine is source for other. If high 
 temperature for one engine is T 1, low temperature for other engine is T 2 (Assume work done by both 
 engine is same) Calculate lower temperature of first engine. 
 la;qfXed dkuksZ±V btau ,d O;oLFkk gS ftlesa ,d btau dk Å"eh; flad nwljs btau ds fy, L=kksr dh rjg dk;Z 
djrk gSA ;fn ,d btau dk mPp rki T 1 gS rFkk nwljs btau dk fuEu rki T 2 gSA (ekfu;s fd nksuksa btauks }kjk fd;k 
x;k dk;Z leku gS) rc izFke btau dk fuEu rki Kkr djksA 
 (1) 
2 1
2 1
T T
T T 2
?
   (2) 
2
T T
2 1
?
   (3) 0   (4) 
2 1
T T  
Ans. (2) 
Sol. Let, Q H : Heat input to I
st
 engine 
  Q L : Heat rejected from I
st
 engine 
  Q L
'
: Heat rejected from II
nd
 engine   
 Work done by I
st
 engine = work done by II
nd
 engine 
 QH – QL = QL – QL
'
  
 2QL = QH + QL
'
  
 2 = 
T
T
T
T
2 1
?  
 T = 
2
T T
2 1
?
 
 ekuk, Q H : I
st 
btau dks nh xbZ Å"ek  
  Q L : I
st
 btau ls fudkyh xbZ Å"ek 
  Q L
'
 : II
nd 
btau ls fudkyh xbZ Å"ek 
 I
st 
btau }kjk fd;k x;k dk;Z = II
nd
 btau }kjk fd;k x;k dk;Z 
 Q H – Q L = Q L – Q L
'
  
 2Q L = Q H + Q L
'
  
 2 = 
T
T
T
T
2 1
?  
 T = 
2
T T
2 1
?
 
 
 
Page 4


       
 
      
    
    
      
 
                       
 
 
  
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. If weight of an object at pole is 196 N then weight at equator is [g = 10 m/s
2
 ; radius of earth = 6400 Km] 
 ;fn ,d oLrq dk Hkkj /kzqo ij 196N gS] rks Hkwe/; js[kk ij bldk Hkkj gksxkA [g = 10 m/s
2
 ; i`Foh dh f=kT;k = 6400 Km] 
 (1) 194.32 N  (2) 194.66 N  (3) 195.32 N  (4) 195.66 N 
Ans. (3) 
Sol. at pole, weight /kzqo ij Hkkj = mg = 196 
 m = 19.6 kg 
 at equator, weight Hkwe/; js[kk ij Hkkj = mg – m ?
2
R 
 = 196 – (19.6) 
2
3
2
6400 10
24 3600
? ? ?
? ?
? ?
?
? ?
 
 = 195.33 N 
 
2. In a house 15 Bulbs of 45 W, 15 bulbs of 100 W, 15 bulbs of 10 W and Two heaters of 1 KW each is 
connected to 220 V Mains supply then find minimum fuse current  
 ,d ?kj esa 45W ds 15 cYc 100W ds 15 cYc] 10W ds 15 cYc rFkk 1KW ds nks ghVj 220V dh eq[; /kkjk ls 
tqMs+ gq,s gS] rks ¶;wt+ rkj ls izokfgr U;wure /kkjk dk eku gksxkA 
 (1) 5 A   (2) 20 A   (3) 25 A   (4) 15 A 
Ans. (2) 
Sol. Total power is dqy 'kfDr (15 × 45) + (15 × 100) + (15 × 10) + (2 × 1000)  
 = 4325 W 
 So current is vr% /kkjk = 
4325
19.66
220
? A 
 Ans is 20 Amp. 
 
3. In an adiabatic process, volume is doubled then find the ratio of final average relaxation time & initial 
relaxation time. Given ? ?
V
P
C
C
 
 ,d :ðks"e izØe esa vk;ru dks nksxquk dj fn;k tkrk gS] rks vfUre rFkk izkjfEHkd vkSlr foJkafr dky dk vuqikr 
Kkr djsaA fn;k gS ? ?
V
P
C
C
 
 (1) 
2
1
   (2) 2   (3) 
?
?
?
?
?
?
?
2
1
  (4) 
1
2
2
1
?
?
?
?
?
?
?
?
  
Ans. (Bonus) 
Sol. relaxation time foJkafr dky ( ?) ? 
T
V
 
       
 
      
    
    
      
 
                       
 
 
  
PAGE # 2 
 
 and rFkk T ? 
1
1
V
? ?
 
 ? ? 
2
1
1
V
? ?
?
 
 ? ? 
2
1
V
? ?
 
 
2
1
i
f
V
V 2
? ?
?
?
?
?
?
?
?
?
?
 
 ? ? 2
1
i
f
2
? ?
?
?
?
 
 
4. A block of mass 10kg is suspended from string of length 4m. When pulled by a force F along horizontal 
 from midpoint. Upper half of string makes 45° with vertical, value of F is    
 ,d 10kg æO;eku dk CykWd 4m yEckbZ dh jLlh ls tqM+k gqvk gSA ;fn jLlh ds e/; fcUnq ls ,d {kSfrt+ cy F 
yxk;k tkrk gS] rks Åij dk vk/kk Hkkx Å/okZ/kj ls 45° dksa.k cukrk gS] rks F dk eku gksxkµ 
 (1) 100N  (2) 90N   (3) 75N   (4) 70N  
Ans. (1) 
Sol.  
 
 
10kg 
T 
100N 
100N 
F 
45° 
 
2
T
 = 100 
 
2
T
 = F 
 F = 100N 
 
5. The surface mass density of a disc of radius a varies with radial distance as ? = A + Br where A & B are 
positive constants then moment of inertia of the disc about an axis passing through its centre and 
perpendicular to the plane   
 ,d pdrh dk lrgh æO;eku ?kuRo f=kfTt; nwjh ds vuqlkj ? = A + Br ls ifjofrZr gksrk gS] rFkk pdrh dh f=kT;k 
a gS] rFkk A  vkSj B /kukRed fu;rkad gSa] rks pdrh ds dsUæ ls ry ds yEcor~ xqt+jus okyh v{k ds lkis{k tM+Ro 
vk?kw.kZ gksxk 
 (1) ?
?
?
?
?
?
? ?
5
Ba
4
A
a 2
4
     (2) ?
?
?
?
?
?
? ?
5
B
4
Aa
a 2
4
  
(3) ?
?
?
?
?
?
? ?
5
Ba
4
A
a
4
     (4) ?
?
?
?
?
?
? ?
4
Ba
5
A
a 2
4
  
Ans. (1) 
       
 
      
    
    
      
 
                       
 
 
  
PAGE # 3 
 
Sol. 
 
 
a 
 
 ? = A + Br 
 ? ?
? ?
? ? ? rdr 2 Br A dm 
 I = 
2
r dm
?
 
 ? ? dr r 2 Br A
3
a
0
? ? ?
?
 
 = 
?
?
?
?
?
?
?
?
? ?
5
a
B
4
a
A 2
5 4
 
 = 2 ?a
4
 ?
?
?
?
?
?
?
5
Ba
4
A
   
 
6. Cascaded Carnot engine is an arrangement in which heat sink of one engine is source for other. If high 
 temperature for one engine is T 1, low temperature for other engine is T 2 (Assume work done by both 
 engine is same) Calculate lower temperature of first engine. 
 la;qfXed dkuksZ±V btau ,d O;oLFkk gS ftlesa ,d btau dk Å"eh; flad nwljs btau ds fy, L=kksr dh rjg dk;Z 
djrk gSA ;fn ,d btau dk mPp rki T 1 gS rFkk nwljs btau dk fuEu rki T 2 gSA (ekfu;s fd nksuksa btauks }kjk fd;k 
x;k dk;Z leku gS) rc izFke btau dk fuEu rki Kkr djksA 
 (1) 
2 1
2 1
T T
T T 2
?
   (2) 
2
T T
2 1
?
   (3) 0   (4) 
2 1
T T  
Ans. (2) 
Sol. Let, Q H : Heat input to I
st
 engine 
  Q L : Heat rejected from I
st
 engine 
  Q L
'
: Heat rejected from II
nd
 engine   
 Work done by I
st
 engine = work done by II
nd
 engine 
 QH – QL = QL – QL
'
  
 2QL = QH + QL
'
  
 2 = 
T
T
T
T
2 1
?  
 T = 
2
T T
2 1
?
 
 ekuk, Q H : I
st 
btau dks nh xbZ Å"ek  
  Q L : I
st
 btau ls fudkyh xbZ Å"ek 
  Q L
'
 : II
nd 
btau ls fudkyh xbZ Å"ek 
 I
st 
btau }kjk fd;k x;k dk;Z = II
nd
 btau }kjk fd;k x;k dk;Z 
 Q H – Q L = Q L – Q L
'
  
 2Q L = Q H + Q L
'
  
 2 = 
T
T
T
T
2 1
?  
 T = 
2
T T
2 1
?
 
 
 
       
 
      
    
    
      
 
                       
 
 
  
PAGE # 4 
 
7. Activity of a substance changes from 700 s
–1 
to 500 s
–1
 in 30 minute. Find its half-life in minutes 
 ,d inkFkZ dh lfØ;rk 700 s
–1 
ls 500 s
–1 
rd 30 feuV esa cnyrh gS] rc bldk v/kZvk;qdky feuV esa Kkr djksA 
 (1) 66   (2) 62   (3) 56   (4) 50 
Ans. (2) 
Sol. t
A
A
n
t
0
? ?
?
?
?
?
?
?
? 
 ? ?n2 = ?t 1/2  …(i) 
? ? ?n
700
500
? ?
? ?
? ?
 = ?(30 min) …(ii) 
? (i)/(ii) 
 ? 
1/2
t n2
n(7 / 5) (30min)
?
?
?
  
 ?  (2.06004) 30 = t 1/2 = 61.8 min. 
 
8. In YDSE, separation between slits is 0.15 mm, distance between slits and screen is 1.5 m and 
wavelength of light is 589 nm, then fringe width is    
 YDSE esa fNæksa ds e/; nwjh 0.15mm gS] fNæksa rFkk inZs ds e/; nwjh 1.5m gS rFkk izdk'k dh rjaxnS/;Z 589 nm gS] 
rks fÝat pkSM+kbZ gksxhA 
 (1) 5.9 mm  (2) 3.9 mm  (3) 1.9 mm  (4) 2.3 mm 
Ans. (1) 
Sol. mm 9 . 5
10 15 . 0
5 . 1 10 589
d
D
3 –
9 –
?
?
? ?
?
?
? ? 
 
9. An ideal fluid is flowing in a pipe in streamline flow. Pipe has maximum and minimum diameter of  
6.4 cm and 4.8 cm respectively. Find out the ratio of minimum to maximum velocity.  
 ,d vkn'kZ æo fdlh ufydk esa /kkjk izokg :i ls cg jgk gSA ufydk dk vf/kdre rFkk U;wure O;kl Øe'k% 6.4cm 
rFkk 4.8cm gS] rks U;wure ls vf/kdre osx dk vuqikr Kkr djksA 
 (1) 
256
81
   (2) 
16
9
   (3) 
4
3
   (4) 
16
3
 
Ans. (2) 
Sol. Using equation of continuity 
 lkarR;rk lehdj.k yxkus ij 
A1V1 = A2V2  
16
9
4 . 6
8 . 4
A
A
V
V
2
1
2
2
1
? ?
?
?
?
?
?
? ?  
 
 
 
Page 5


       
 
      
    
    
      
 
                       
 
 
  
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. If weight of an object at pole is 196 N then weight at equator is [g = 10 m/s
2
 ; radius of earth = 6400 Km] 
 ;fn ,d oLrq dk Hkkj /kzqo ij 196N gS] rks Hkwe/; js[kk ij bldk Hkkj gksxkA [g = 10 m/s
2
 ; i`Foh dh f=kT;k = 6400 Km] 
 (1) 194.32 N  (2) 194.66 N  (3) 195.32 N  (4) 195.66 N 
Ans. (3) 
Sol. at pole, weight /kzqo ij Hkkj = mg = 196 
 m = 19.6 kg 
 at equator, weight Hkwe/; js[kk ij Hkkj = mg – m ?
2
R 
 = 196 – (19.6) 
2
3
2
6400 10
24 3600
? ? ?
? ?
? ?
?
? ?
 
 = 195.33 N 
 
2. In a house 15 Bulbs of 45 W, 15 bulbs of 100 W, 15 bulbs of 10 W and Two heaters of 1 KW each is 
connected to 220 V Mains supply then find minimum fuse current  
 ,d ?kj esa 45W ds 15 cYc 100W ds 15 cYc] 10W ds 15 cYc rFkk 1KW ds nks ghVj 220V dh eq[; /kkjk ls 
tqMs+ gq,s gS] rks ¶;wt+ rkj ls izokfgr U;wure /kkjk dk eku gksxkA 
 (1) 5 A   (2) 20 A   (3) 25 A   (4) 15 A 
Ans. (2) 
Sol. Total power is dqy 'kfDr (15 × 45) + (15 × 100) + (15 × 10) + (2 × 1000)  
 = 4325 W 
 So current is vr% /kkjk = 
4325
19.66
220
? A 
 Ans is 20 Amp. 
 
3. In an adiabatic process, volume is doubled then find the ratio of final average relaxation time & initial 
relaxation time. Given ? ?
V
P
C
C
 
 ,d :ðks"e izØe esa vk;ru dks nksxquk dj fn;k tkrk gS] rks vfUre rFkk izkjfEHkd vkSlr foJkafr dky dk vuqikr 
Kkr djsaA fn;k gS ? ?
V
P
C
C
 
 (1) 
2
1
   (2) 2   (3) 
?
?
?
?
?
?
?
2
1
  (4) 
1
2
2
1
?
?
?
?
?
?
?
?
  
Ans. (Bonus) 
Sol. relaxation time foJkafr dky ( ?) ? 
T
V
 
       
 
      
    
    
      
 
                       
 
 
  
PAGE # 2 
 
 and rFkk T ? 
1
1
V
? ?
 
 ? ? 
2
1
1
V
? ?
?
 
 ? ? 
2
1
V
? ?
 
 
2
1
i
f
V
V 2
? ?
?
?
?
?
?
?
?
?
?
 
 ? ? 2
1
i
f
2
? ?
?
?
?
 
 
4. A block of mass 10kg is suspended from string of length 4m. When pulled by a force F along horizontal 
 from midpoint. Upper half of string makes 45° with vertical, value of F is    
 ,d 10kg æO;eku dk CykWd 4m yEckbZ dh jLlh ls tqM+k gqvk gSA ;fn jLlh ds e/; fcUnq ls ,d {kSfrt+ cy F 
yxk;k tkrk gS] rks Åij dk vk/kk Hkkx Å/okZ/kj ls 45° dksa.k cukrk gS] rks F dk eku gksxkµ 
 (1) 100N  (2) 90N   (3) 75N   (4) 70N  
Ans. (1) 
Sol.  
 
 
10kg 
T 
100N 
100N 
F 
45° 
 
2
T
 = 100 
 
2
T
 = F 
 F = 100N 
 
5. The surface mass density of a disc of radius a varies with radial distance as ? = A + Br where A & B are 
positive constants then moment of inertia of the disc about an axis passing through its centre and 
perpendicular to the plane   
 ,d pdrh dk lrgh æO;eku ?kuRo f=kfTt; nwjh ds vuqlkj ? = A + Br ls ifjofrZr gksrk gS] rFkk pdrh dh f=kT;k 
a gS] rFkk A  vkSj B /kukRed fu;rkad gSa] rks pdrh ds dsUæ ls ry ds yEcor~ xqt+jus okyh v{k ds lkis{k tM+Ro 
vk?kw.kZ gksxk 
 (1) ?
?
?
?
?
?
? ?
5
Ba
4
A
a 2
4
     (2) ?
?
?
?
?
?
? ?
5
B
4
Aa
a 2
4
  
(3) ?
?
?
?
?
?
? ?
5
Ba
4
A
a
4
     (4) ?
?
?
?
?
?
? ?
4
Ba
5
A
a 2
4
  
Ans. (1) 
       
 
      
    
    
      
 
                       
 
 
  
PAGE # 3 
 
Sol. 
 
 
a 
 
 ? = A + Br 
 ? ?
? ?
? ? ? rdr 2 Br A dm 
 I = 
2
r dm
?
 
 ? ? dr r 2 Br A
3
a
0
? ? ?
?
 
 = 
?
?
?
?
?
?
?
?
? ?
5
a
B
4
a
A 2
5 4
 
 = 2 ?a
4
 ?
?
?
?
?
?
?
5
Ba
4
A
   
 
6. Cascaded Carnot engine is an arrangement in which heat sink of one engine is source for other. If high 
 temperature for one engine is T 1, low temperature for other engine is T 2 (Assume work done by both 
 engine is same) Calculate lower temperature of first engine. 
 la;qfXed dkuksZ±V btau ,d O;oLFkk gS ftlesa ,d btau dk Å"eh; flad nwljs btau ds fy, L=kksr dh rjg dk;Z 
djrk gSA ;fn ,d btau dk mPp rki T 1 gS rFkk nwljs btau dk fuEu rki T 2 gSA (ekfu;s fd nksuksa btauks }kjk fd;k 
x;k dk;Z leku gS) rc izFke btau dk fuEu rki Kkr djksA 
 (1) 
2 1
2 1
T T
T T 2
?
   (2) 
2
T T
2 1
?
   (3) 0   (4) 
2 1
T T  
Ans. (2) 
Sol. Let, Q H : Heat input to I
st
 engine 
  Q L : Heat rejected from I
st
 engine 
  Q L
'
: Heat rejected from II
nd
 engine   
 Work done by I
st
 engine = work done by II
nd
 engine 
 QH – QL = QL – QL
'
  
 2QL = QH + QL
'
  
 2 = 
T
T
T
T
2 1
?  
 T = 
2
T T
2 1
?
 
 ekuk, Q H : I
st 
btau dks nh xbZ Å"ek  
  Q L : I
st
 btau ls fudkyh xbZ Å"ek 
  Q L
'
 : II
nd 
btau ls fudkyh xbZ Å"ek 
 I
st 
btau }kjk fd;k x;k dk;Z = II
nd
 btau }kjk fd;k x;k dk;Z 
 Q H – Q L = Q L – Q L
'
  
 2Q L = Q H + Q L
'
  
 2 = 
T
T
T
T
2 1
?  
 T = 
2
T T
2 1
?
 
 
 
       
 
      
    
    
      
 
                       
 
 
  
PAGE # 4 
 
7. Activity of a substance changes from 700 s
–1 
to 500 s
–1
 in 30 minute. Find its half-life in minutes 
 ,d inkFkZ dh lfØ;rk 700 s
–1 
ls 500 s
–1 
rd 30 feuV esa cnyrh gS] rc bldk v/kZvk;qdky feuV esa Kkr djksA 
 (1) 66   (2) 62   (3) 56   (4) 50 
Ans. (2) 
Sol. t
A
A
n
t
0
? ?
?
?
?
?
?
?
? 
 ? ?n2 = ?t 1/2  …(i) 
? ? ?n
700
500
? ?
? ?
? ?
 = ?(30 min) …(ii) 
? (i)/(ii) 
 ? 
1/2
t n2
n(7 / 5) (30min)
?
?
?
  
 ?  (2.06004) 30 = t 1/2 = 61.8 min. 
 
8. In YDSE, separation between slits is 0.15 mm, distance between slits and screen is 1.5 m and 
wavelength of light is 589 nm, then fringe width is    
 YDSE esa fNæksa ds e/; nwjh 0.15mm gS] fNæksa rFkk inZs ds e/; nwjh 1.5m gS rFkk izdk'k dh rjaxnS/;Z 589 nm gS] 
rks fÝat pkSM+kbZ gksxhA 
 (1) 5.9 mm  (2) 3.9 mm  (3) 1.9 mm  (4) 2.3 mm 
Ans. (1) 
Sol. mm 9 . 5
10 15 . 0
5 . 1 10 589
d
D
3 –
9 –
?
?
? ?
?
?
? ? 
 
9. An ideal fluid is flowing in a pipe in streamline flow. Pipe has maximum and minimum diameter of  
6.4 cm and 4.8 cm respectively. Find out the ratio of minimum to maximum velocity.  
 ,d vkn'kZ æo fdlh ufydk esa /kkjk izokg :i ls cg jgk gSA ufydk dk vf/kdre rFkk U;wure O;kl Øe'k% 6.4cm 
rFkk 4.8cm gS] rks U;wure ls vf/kdre osx dk vuqikr Kkr djksA 
 (1) 
256
81
   (2) 
16
9
   (3) 
4
3
   (4) 
16
3
 
Ans. (2) 
Sol. Using equation of continuity 
 lkarR;rk lehdj.k yxkus ij 
A1V1 = A2V2  
16
9
4 . 6
8 . 4
A
A
V
V
2
1
2
2
1
? ?
?
?
?
?
?
? ?  
 
 
 
       
 
      
    
    
      
 
                       
 
 
  
PAGE # 5 
 
 
10. There is a electric circuit as shown in the figure. Find potential difference between points a and b.  
 fp=kkuqlkj ifjiFk esa ,d fo|qr /kkjk gS] rks fcUnqvksa a rFkk b ds e/; foHkokUrj Kkr djksA 
 
30V
b
a
10k ?
10k ?
10k ?
 
(1) 0V    (2) 15V   (3) 10V   (4) 5V 
Ans. (3) 
Sol. Diode is in forward bias, so it will behave as simple wire so,  
 Mk;ksM vxz ck;l voLFkk esa gS] vr% ;g lk/kkj.k rkj dh rjg dk;Z djsxkA 
 
30V 
b
a
5k ?
10k ?
b a
?
 
 So vr%, V ab = V 10 5
10 5
30
? ?
?
 
   
11. A particle of mass m and positive charge q is projected with a speed of v 0 in y–direction in the presence 
 of electric and magnetic field are in x–direction. Find the instant of time at which the speed of particle 
 becomes double the initial speed.  
 x-fn'kk esa mifLFkr fo|qr rFkk pqEcdh; {ks=k esa ,d m nzO;eku rFkk /kukRed vkos'k q dk d.k v 0 pky ls y–fn'kk esa 
iz{ksfir fd;k tkrk gSA og le; Kkr dhft, tc d.k dh pky izkjfEHkd pky dh nksxquk gks tk,sxh 
 (1) t = 
qE
3 mv
0
   (2) t = 
qE
2 mv
0
  (3) t = 
qE
mv
0
   (4) t = 
qE 2
mv
0
  
Ans. (1) 
Sol.  
 
 
m,q 
x 
y 
B = i
ˆ
B
0
 
E = i
ˆ
E
0
 
V0 j
ˆ
 
 
As j
ˆ
v v
0
?
?
 (magnitude of velocity does not change in y–z plane) 
 (2v0)
2
 = 
2
x
2
0
v v ?  ;    vx = 
0
v 3 
 ? t
m
qE
0 v 3
0
? ?   ;   t = 
qE
3 mv
0