JEE Main 2020 Physics January 7 Shift 1 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. A block of mass m is suspended from a pulley in form of a circular disc of mass m & radius R. The system 
is released from rest, find the angular velocity of disc when block has dropped by height h. (there is no 
slipping between string & pulley)      
,d m nzO;eku dk CykWd f?kjuh ls jLlh ds }kjk tqM+k gqvk gSA f?kjuh pdrh ds :i esa gSA ftldk nzO;eku m rFkk 
f=kT;k R gSA ;fn fudk; dks fojke ls NksM+k tkrk gSA rc CykWd }kjk h Å¡pkbZ uhps vkus ij pdrh dk dks.kh; osx 
gksxk (f?kjuh rFkk pdrh esa dksbZ fQlyu ugha gS)  
 
m
m
 
 (1) 
3
gh 4
R
1
    (2) 
3
gh 2
R
1
    (3) 
3
gh 2
R    (4) 
3
gh 4
R  
Ans. (1) 
Sol. mgh = 
2 2
2
1
mv
2
1
? ? ? 
 v = ?R (no slipping dksbZ fQlyu ugha) 
 mgh = 
2
2
2 2
2
mR
2
1
R m
2
1
? ? ? 
 mgh = 
2 2
R m
4
3
? 
 
3
gh 4
R
1
R 3
gh 4
2
? ? ? 
 
2. Three point masses 1kg, 1.5 kg, 2.5 kg are placed at the vertices of a triangle with sides 3cm, 4cm and 
5cm as shown in the figure. The location of centre of mass with respect to 1kg mass is : 
2.5 kg 
 
5cm 
 
4 cm 
 
1 kg 
 
3 cm 
 
1.5 kg 
 
   
 
 (1) 0.6 cm to the right of 1 kg and 2 cm above 1 kg mass 
  (2) 0.9 cm to the right of 1kg and 2 cm above 1 kg mass 
 (3) 0.9 cm to the left of 1kg and 2 cm above 1kg mass 
 (4) 0.9 cm to the right of 1 kg and 1.5 cm above 1kg mass  
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. A block of mass m is suspended from a pulley in form of a circular disc of mass m & radius R. The system 
is released from rest, find the angular velocity of disc when block has dropped by height h. (there is no 
slipping between string & pulley)      
,d m nzO;eku dk CykWd f?kjuh ls jLlh ds }kjk tqM+k gqvk gSA f?kjuh pdrh ds :i esa gSA ftldk nzO;eku m rFkk 
f=kT;k R gSA ;fn fudk; dks fojke ls NksM+k tkrk gSA rc CykWd }kjk h Å¡pkbZ uhps vkus ij pdrh dk dks.kh; osx 
gksxk (f?kjuh rFkk pdrh esa dksbZ fQlyu ugha gS)  
 
m
m
 
 (1) 
3
gh 4
R
1
    (2) 
3
gh 2
R
1
    (3) 
3
gh 2
R    (4) 
3
gh 4
R  
Ans. (1) 
Sol. mgh = 
2 2
2
1
mv
2
1
? ? ? 
 v = ?R (no slipping dksbZ fQlyu ugha) 
 mgh = 
2
2
2 2
2
mR
2
1
R m
2
1
? ? ? 
 mgh = 
2 2
R m
4
3
? 
 
3
gh 4
R
1
R 3
gh 4
2
? ? ? 
 
2. Three point masses 1kg, 1.5 kg, 2.5 kg are placed at the vertices of a triangle with sides 3cm, 4cm and 
5cm as shown in the figure. The location of centre of mass with respect to 1kg mass is : 
2.5 kg 
 
5cm 
 
4 cm 
 
1 kg 
 
3 cm 
 
1.5 kg 
 
   
 
 (1) 0.6 cm to the right of 1 kg and 2 cm above 1 kg mass 
  (2) 0.9 cm to the right of 1kg and 2 cm above 1 kg mass 
 (3) 0.9 cm to the left of 1kg and 2 cm above 1kg mass 
 (4) 0.9 cm to the right of 1 kg and 1.5 cm above 1kg mass  
        
 
 
    
    
     
 
   
  
PAGE # 2 
 
fp=k esa fn[kk, vuqlkj rhu fcUnq nzO;eku ftuds nzO;eku Øe'k% 1kg, 1.5kg rFkk 2.5kg gS] ,d f=kHkqt ds dksuksa ij 
j[ks gSA f=kHkqt dh Hkqtk,sa Øe'k% 3 cm, 4cm rFkk 5cm gSA rc nzO;eku dsUnz dh fLFkfr 1kg nzO;eku ds lkis{k gksxh : 
2.5 kg 
 
5cm 
 
4 cm 
 
1 kg 
 
3 cm 
 
1.5 kg 
 
 
 (1) 1kg ds nka;h vksj 0.6 cm nwj rFkk 1kg nzO;eku ds 2 cm Åij  
  (2) 1kg ds nka;h vksj 0.9 cm nwj rFkk 1kg nzO;eku ds 2 cm Åij 
 (3) 1kg ds cka;h vksj 0.9 cm nwj rFkk 1kg nzO;eku ds 2 cm Åij 
 (4) 1 kg ds nka;h vksj 0.9 cm nwj rFkk 1kg nzO;eku ds 1.5 cm Åij 
Ans. (2) 
Sol. Take 1kg mass at origin  
 1 kg nzO;eku dks ewy fcUnq ysus ij  
 
2.5 kg 
 
5cm 
4 cm 
1 kg 
 
3 cm 
 
1.5 kg 
 
 
Y 
 
X 
 
 
 
cm
1 0 1.5 3 2.5 0
X 0.9cm
5
? ? ? ? ?
? ? 
 
cm
1 0 1.5 0 2.5 4
Y 2cm
5
? ? ? ? ?
? ? 
 
3. In a single slit diffraction set up, second minima is observed at an angle of 60°. The expected position of 
first minima is    
 ,dy fNnz foorZu izfØ;k esa] 2
nd
 fufEu"B (minima) 60° ds dks.k ij fn[kkbZ nsrk gS] rks 1
st
 fufEu"B (minima) dh 
visf{kr fLFkfr gksxh &  
 (1) 25°     (2) 20°     (3) 30°     (4) 45°  
Ans. (1) 
Sol. For 2
nd
 minima   f}rh; fufEu"B ds fy, 
 d sin ? = 2 ?    
 sin ? = 
2
3
 (given) 
 ? 
4
3
d
?
?
 … (i) 
 So for 1
st
 minima is  
 blfy, izFke fufEu"B ds fy, 
d sin ? = ?    
 sin ? = 
4
3
d
?
?
 (from equation (i) (lehdj.k (i) ls 
 ? = 25.65° (from sin table) (sin rkfydk ls) 
 ? ? ? 25 
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. A block of mass m is suspended from a pulley in form of a circular disc of mass m & radius R. The system 
is released from rest, find the angular velocity of disc when block has dropped by height h. (there is no 
slipping between string & pulley)      
,d m nzO;eku dk CykWd f?kjuh ls jLlh ds }kjk tqM+k gqvk gSA f?kjuh pdrh ds :i esa gSA ftldk nzO;eku m rFkk 
f=kT;k R gSA ;fn fudk; dks fojke ls NksM+k tkrk gSA rc CykWd }kjk h Å¡pkbZ uhps vkus ij pdrh dk dks.kh; osx 
gksxk (f?kjuh rFkk pdrh esa dksbZ fQlyu ugha gS)  
 
m
m
 
 (1) 
3
gh 4
R
1
    (2) 
3
gh 2
R
1
    (3) 
3
gh 2
R    (4) 
3
gh 4
R  
Ans. (1) 
Sol. mgh = 
2 2
2
1
mv
2
1
? ? ? 
 v = ?R (no slipping dksbZ fQlyu ugha) 
 mgh = 
2
2
2 2
2
mR
2
1
R m
2
1
? ? ? 
 mgh = 
2 2
R m
4
3
? 
 
3
gh 4
R
1
R 3
gh 4
2
? ? ? 
 
2. Three point masses 1kg, 1.5 kg, 2.5 kg are placed at the vertices of a triangle with sides 3cm, 4cm and 
5cm as shown in the figure. The location of centre of mass with respect to 1kg mass is : 
2.5 kg 
 
5cm 
 
4 cm 
 
1 kg 
 
3 cm 
 
1.5 kg 
 
   
 
 (1) 0.6 cm to the right of 1 kg and 2 cm above 1 kg mass 
  (2) 0.9 cm to the right of 1kg and 2 cm above 1 kg mass 
 (3) 0.9 cm to the left of 1kg and 2 cm above 1kg mass 
 (4) 0.9 cm to the right of 1 kg and 1.5 cm above 1kg mass  
        
 
 
    
    
     
 
   
  
PAGE # 2 
 
fp=k esa fn[kk, vuqlkj rhu fcUnq nzO;eku ftuds nzO;eku Øe'k% 1kg, 1.5kg rFkk 2.5kg gS] ,d f=kHkqt ds dksuksa ij 
j[ks gSA f=kHkqt dh Hkqtk,sa Øe'k% 3 cm, 4cm rFkk 5cm gSA rc nzO;eku dsUnz dh fLFkfr 1kg nzO;eku ds lkis{k gksxh : 
2.5 kg 
 
5cm 
 
4 cm 
 
1 kg 
 
3 cm 
 
1.5 kg 
 
 
 (1) 1kg ds nka;h vksj 0.6 cm nwj rFkk 1kg nzO;eku ds 2 cm Åij  
  (2) 1kg ds nka;h vksj 0.9 cm nwj rFkk 1kg nzO;eku ds 2 cm Åij 
 (3) 1kg ds cka;h vksj 0.9 cm nwj rFkk 1kg nzO;eku ds 2 cm Åij 
 (4) 1 kg ds nka;h vksj 0.9 cm nwj rFkk 1kg nzO;eku ds 1.5 cm Åij 
Ans. (2) 
Sol. Take 1kg mass at origin  
 1 kg nzO;eku dks ewy fcUnq ysus ij  
 
2.5 kg 
 
5cm 
4 cm 
1 kg 
 
3 cm 
 
1.5 kg 
 
 
Y 
 
X 
 
 
 
cm
1 0 1.5 3 2.5 0
X 0.9cm
5
? ? ? ? ?
? ? 
 
cm
1 0 1.5 0 2.5 4
Y 2cm
5
? ? ? ? ?
? ? 
 
3. In a single slit diffraction set up, second minima is observed at an angle of 60°. The expected position of 
first minima is    
 ,dy fNnz foorZu izfØ;k esa] 2
nd
 fufEu"B (minima) 60° ds dks.k ij fn[kkbZ nsrk gS] rks 1
st
 fufEu"B (minima) dh 
visf{kr fLFkfr gksxh &  
 (1) 25°     (2) 20°     (3) 30°     (4) 45°  
Ans. (1) 
Sol. For 2
nd
 minima   f}rh; fufEu"B ds fy, 
 d sin ? = 2 ?    
 sin ? = 
2
3
 (given) 
 ? 
4
3
d
?
?
 … (i) 
 So for 1
st
 minima is  
 blfy, izFke fufEu"B ds fy, 
d sin ? = ?    
 sin ? = 
4
3
d
?
?
 (from equation (i) (lehdj.k (i) ls 
 ? = 25.65° (from sin table) (sin rkfydk ls) 
 ? ? ? 25 
        
 
 
    
    
     
 
   
  
PAGE # 3 
 
4. There are two infinite plane sheets each having uniform surface charge density + ? C/m
2
. They are 
inclined to each other at an angle 30° as shown in the figure. Electric field at any arbitrary point P is:  
 fp=k esa nks vuUr yEckbZ dh lery ifêdk gS ftudk le:i i`"Bh; vkos'k ?kuRo + ? C/m
2 
gSA nksuksa ifêdk ,d nwljs 
ls 30° dks.k ij >qdh gSA rc fp=kkuqlkj fdlh Hkh ;kn`fPNd fcUnq P ij fo|qr {ks=k gksxk: 
 
30°
+ ?
+ ?
P
Y 
 
X 
 
 
 (1) 
0
3 1
ˆ ˆ 1– y – x
2 2 2
? ? ? ?
?
? ?
? ?
?
? ?
? ?
? ?
    (2) 
0
3 1
ˆ ˆ 1 y – x
2 2 2
? ? ? ?
?
? ? ?
? ?
?
? ?
? ?
? ?
      
(3) 
0
3 1
ˆ ˆ 1– y x
2 2 2
? ? ? ?
?
? ? ?
? ?
?
? ?
? ?
? ?
   (4)  
0
3 1
ˆ ˆ 1 y x
2 2 2
? ? ? ?
?
? ? ? ?
? ?
?
? ?
? ?
? ?
 
Ans. (1) 
Sol.  
 
 
30° 
+ ? 
+ ? 
P
60°
 
? ? ? ?
0 0 0
ˆ ˆ E cos60 –x – sin60 y
2 2 2
? ? ? ? ?
? ? ? ?
? ?
? ? ?
? ?
?
 
 E
?
= 
0
3 1
ˆ ˆ 1– y – x
2 2 2
? ? ? ?
?
? ?
? ?
?
? ?
? ?
? ?
 
 
5. A parallel plate capacitor with plate area A & plate separation d is filled with a dielectric material of 
dielectric constant given by k = k 0(1 + ?x). Calculate capacitance of system: (given ?d << 1)  
 ,d lekUrj iV~V la/kkfj=k ftldh lrg dk {ks=kQy A gS rFkk ifV~Vdkvksa ds e/; dh nwjh d gSA buds e/; ,d 
ijkoS|qrkad inkFkZ Hkjk gSA inkFkZ dk ijkoS|qrkad fu;rkad k = k 0(1 + ?x) gSA rc fudk; dh /kkfjrk gksxh:  
(fn;k gS ?d << 1) 
 
x 
 
 (1)  ? ?
2 2 0 0
d 1
d
A k
? ?
?
  (2) ?
?
?
?
?
? ?
?
?
2
d
1
d
A k
0 0
   (3) ? ? d 1
d 2
A k
0 0
? ?
?
  (4) ?
?
?
?
?
? ?
?
?
2
d
1
d 2
A k
0 0
  
Ans. (2) 
 
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. A block of mass m is suspended from a pulley in form of a circular disc of mass m & radius R. The system 
is released from rest, find the angular velocity of disc when block has dropped by height h. (there is no 
slipping between string & pulley)      
,d m nzO;eku dk CykWd f?kjuh ls jLlh ds }kjk tqM+k gqvk gSA f?kjuh pdrh ds :i esa gSA ftldk nzO;eku m rFkk 
f=kT;k R gSA ;fn fudk; dks fojke ls NksM+k tkrk gSA rc CykWd }kjk h Å¡pkbZ uhps vkus ij pdrh dk dks.kh; osx 
gksxk (f?kjuh rFkk pdrh esa dksbZ fQlyu ugha gS)  
 
m
m
 
 (1) 
3
gh 4
R
1
    (2) 
3
gh 2
R
1
    (3) 
3
gh 2
R    (4) 
3
gh 4
R  
Ans. (1) 
Sol. mgh = 
2 2
2
1
mv
2
1
? ? ? 
 v = ?R (no slipping dksbZ fQlyu ugha) 
 mgh = 
2
2
2 2
2
mR
2
1
R m
2
1
? ? ? 
 mgh = 
2 2
R m
4
3
? 
 
3
gh 4
R
1
R 3
gh 4
2
? ? ? 
 
2. Three point masses 1kg, 1.5 kg, 2.5 kg are placed at the vertices of a triangle with sides 3cm, 4cm and 
5cm as shown in the figure. The location of centre of mass with respect to 1kg mass is : 
2.5 kg 
 
5cm 
 
4 cm 
 
1 kg 
 
3 cm 
 
1.5 kg 
 
   
 
 (1) 0.6 cm to the right of 1 kg and 2 cm above 1 kg mass 
  (2) 0.9 cm to the right of 1kg and 2 cm above 1 kg mass 
 (3) 0.9 cm to the left of 1kg and 2 cm above 1kg mass 
 (4) 0.9 cm to the right of 1 kg and 1.5 cm above 1kg mass  
        
 
 
    
    
     
 
   
  
PAGE # 2 
 
fp=k esa fn[kk, vuqlkj rhu fcUnq nzO;eku ftuds nzO;eku Øe'k% 1kg, 1.5kg rFkk 2.5kg gS] ,d f=kHkqt ds dksuksa ij 
j[ks gSA f=kHkqt dh Hkqtk,sa Øe'k% 3 cm, 4cm rFkk 5cm gSA rc nzO;eku dsUnz dh fLFkfr 1kg nzO;eku ds lkis{k gksxh : 
2.5 kg 
 
5cm 
 
4 cm 
 
1 kg 
 
3 cm 
 
1.5 kg 
 
 
 (1) 1kg ds nka;h vksj 0.6 cm nwj rFkk 1kg nzO;eku ds 2 cm Åij  
  (2) 1kg ds nka;h vksj 0.9 cm nwj rFkk 1kg nzO;eku ds 2 cm Åij 
 (3) 1kg ds cka;h vksj 0.9 cm nwj rFkk 1kg nzO;eku ds 2 cm Åij 
 (4) 1 kg ds nka;h vksj 0.9 cm nwj rFkk 1kg nzO;eku ds 1.5 cm Åij 
Ans. (2) 
Sol. Take 1kg mass at origin  
 1 kg nzO;eku dks ewy fcUnq ysus ij  
 
2.5 kg 
 
5cm 
4 cm 
1 kg 
 
3 cm 
 
1.5 kg 
 
 
Y 
 
X 
 
 
 
cm
1 0 1.5 3 2.5 0
X 0.9cm
5
? ? ? ? ?
? ? 
 
cm
1 0 1.5 0 2.5 4
Y 2cm
5
? ? ? ? ?
? ? 
 
3. In a single slit diffraction set up, second minima is observed at an angle of 60°. The expected position of 
first minima is    
 ,dy fNnz foorZu izfØ;k esa] 2
nd
 fufEu"B (minima) 60° ds dks.k ij fn[kkbZ nsrk gS] rks 1
st
 fufEu"B (minima) dh 
visf{kr fLFkfr gksxh &  
 (1) 25°     (2) 20°     (3) 30°     (4) 45°  
Ans. (1) 
Sol. For 2
nd
 minima   f}rh; fufEu"B ds fy, 
 d sin ? = 2 ?    
 sin ? = 
2
3
 (given) 
 ? 
4
3
d
?
?
 … (i) 
 So for 1
st
 minima is  
 blfy, izFke fufEu"B ds fy, 
d sin ? = ?    
 sin ? = 
4
3
d
?
?
 (from equation (i) (lehdj.k (i) ls 
 ? = 25.65° (from sin table) (sin rkfydk ls) 
 ? ? ? 25 
        
 
 
    
    
     
 
   
  
PAGE # 3 
 
4. There are two infinite plane sheets each having uniform surface charge density + ? C/m
2
. They are 
inclined to each other at an angle 30° as shown in the figure. Electric field at any arbitrary point P is:  
 fp=k esa nks vuUr yEckbZ dh lery ifêdk gS ftudk le:i i`"Bh; vkos'k ?kuRo + ? C/m
2 
gSA nksuksa ifêdk ,d nwljs 
ls 30° dks.k ij >qdh gSA rc fp=kkuqlkj fdlh Hkh ;kn`fPNd fcUnq P ij fo|qr {ks=k gksxk: 
 
30°
+ ?
+ ?
P
Y 
 
X 
 
 
 (1) 
0
3 1
ˆ ˆ 1– y – x
2 2 2
? ? ? ?
?
? ?
? ?
?
? ?
? ?
? ?
    (2) 
0
3 1
ˆ ˆ 1 y – x
2 2 2
? ? ? ?
?
? ? ?
? ?
?
? ?
? ?
? ?
      
(3) 
0
3 1
ˆ ˆ 1– y x
2 2 2
? ? ? ?
?
? ? ?
? ?
?
? ?
? ?
? ?
   (4)  
0
3 1
ˆ ˆ 1 y x
2 2 2
? ? ? ?
?
? ? ? ?
? ?
?
? ?
? ?
? ?
 
Ans. (1) 
Sol.  
 
 
30° 
+ ? 
+ ? 
P
60°
 
? ? ? ?
0 0 0
ˆ ˆ E cos60 –x – sin60 y
2 2 2
? ? ? ? ?
? ? ? ?
? ?
? ? ?
? ?
?
 
 E
?
= 
0
3 1
ˆ ˆ 1– y – x
2 2 2
? ? ? ?
?
? ?
? ?
?
? ?
? ?
? ?
 
 
5. A parallel plate capacitor with plate area A & plate separation d is filled with a dielectric material of 
dielectric constant given by k = k 0(1 + ?x). Calculate capacitance of system: (given ?d << 1)  
 ,d lekUrj iV~V la/kkfj=k ftldh lrg dk {ks=kQy A gS rFkk ifV~Vdkvksa ds e/; dh nwjh d gSA buds e/; ,d 
ijkoS|qrkad inkFkZ Hkjk gSA inkFkZ dk ijkoS|qrkad fu;rkad k = k 0(1 + ?x) gSA rc fudk; dh /kkfjrk gksxh:  
(fn;k gS ?d << 1) 
 
x 
 
 (1)  ? ?
2 2 0 0
d 1
d
A k
? ?
?
  (2) ?
?
?
?
?
? ?
?
?
2
d
1
d
A k
0 0
   (3) ? ? d 1
d 2
A k
0 0
? ?
?
  (4) ?
?
?
?
?
? ?
?
?
2
d
1
d 2
A k
0 0
  
Ans. (2) 
 
        
 
 
    
    
     
 
   
  
PAGE # 4 
 
Sol. Capacitance of element ?kVd dh /kkfjrk  = 
dx
A k
0
?
 
 
 
dx
x
 
 Capacitance of element ?kVd dh /kkfjrk, C' = 
dx
A ) x 1 ( k
0 0
? ? ?
 
 
?
?
? ? ?
?
d
0
0 0
) x 1 ( A k
dx
' C
1
 
 ) d 1 ( n
A k
1
C
1
0 0
? ?
? ?
? ? 
 Given fn;k gS& ?d << 1 
 
?
?
?
?
?
?
?
?
?
? ?
? ?
?
2
d
d
A k
1
C
1
2 2
0 0
 
 ?
?
?
?
?
? ?
?
?
?
2
d
1
A k
d
C
1
0 0
 
 ?
?
?
?
?
? ?
?
?
?
2
d
1
d
A k
C
0 0
 
 
6. A long solenoid of radius R carries a time dependent current I = I
0 
t(1 – t). A ring of radius 2R is placed 
coaxially near its centre. During the time interval 0 ? t ? 1, the induced current ?R and the induced emf VR 
in the ring vary as: 
 (1) current will change its direction and its emf will be zero at t = 0.25sec. 
   (2) current will not change its direction & emf will be maximum at t = 0.5sec 
 (3) current will not change direction and emf will be zero at 0.25sec. 
 (4) current will change its direction and its emf will be zero at t = 0.5sec. 
 ,d R f=kT;k dh ijhuyhdk ftlesa ,d le; fuHkZj /kkjk I = I
0 
t(1 – t) izokfgr gks jgh gSA ,d 2R f=kT;k dh oy; 
ifjuyhdk ds lek{kh; :i ls mlds dsUnz ds ikl fLFkr gSA le; vUrjky 0 ? t ? 1, ds nkSjku oy; esa izsjhr /kkjk ? R 
vkSj izsfjr fo+|qr okgd cy V R  ds laxr lgh fodYi gksxk 
 (1) /kkjk viuh fn'kk cnysxh vkSj fo|qr okgd cy le; t = 0.25sec ij 'kwU; gksxkA 
   (2) /kkjk viuh fn'kk ugha cnysxh vkSj fo|qr okgd cy le; t = 0.5sec ij vf/kdre gksxkA 
 (3) /kkjk viuh fn'kk ugha cnysxh vkSj fo|qr okgd cy le; t = 0.25sec ij 'kwU; gksxkA  
 (4) /kkjk viuh fn'kk cnysxh vkSj fo|qr okgd cy t = 0.5sec ij 'kwU; gksxkA  
Ans. (4) 
Sol. I = I 0t – I 0t
2
  
 ? = BA 
 ? = ? 0nIA 
 V R = 
dt
d
–
?
 = – ? 0nAI 0 (1 – 2t)  
 VR = 0 at   t = 
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. A block of mass m is suspended from a pulley in form of a circular disc of mass m & radius R. The system 
is released from rest, find the angular velocity of disc when block has dropped by height h. (there is no 
slipping between string & pulley)      
,d m nzO;eku dk CykWd f?kjuh ls jLlh ds }kjk tqM+k gqvk gSA f?kjuh pdrh ds :i esa gSA ftldk nzO;eku m rFkk 
f=kT;k R gSA ;fn fudk; dks fojke ls NksM+k tkrk gSA rc CykWd }kjk h Å¡pkbZ uhps vkus ij pdrh dk dks.kh; osx 
gksxk (f?kjuh rFkk pdrh esa dksbZ fQlyu ugha gS)  
 
m
m
 
 (1) 
3
gh 4
R
1
    (2) 
3
gh 2
R
1
    (3) 
3
gh 2
R    (4) 
3
gh 4
R  
Ans. (1) 
Sol. mgh = 
2 2
2
1
mv
2
1
? ? ? 
 v = ?R (no slipping dksbZ fQlyu ugha) 
 mgh = 
2
2
2 2
2
mR
2
1
R m
2
1
? ? ? 
 mgh = 
2 2
R m
4
3
? 
 
3
gh 4
R
1
R 3
gh 4
2
? ? ? 
 
2. Three point masses 1kg, 1.5 kg, 2.5 kg are placed at the vertices of a triangle with sides 3cm, 4cm and 
5cm as shown in the figure. The location of centre of mass with respect to 1kg mass is : 
2.5 kg 
 
5cm 
 
4 cm 
 
1 kg 
 
3 cm 
 
1.5 kg 
 
   
 
 (1) 0.6 cm to the right of 1 kg and 2 cm above 1 kg mass 
  (2) 0.9 cm to the right of 1kg and 2 cm above 1 kg mass 
 (3) 0.9 cm to the left of 1kg and 2 cm above 1kg mass 
 (4) 0.9 cm to the right of 1 kg and 1.5 cm above 1kg mass  
        
 
 
    
    
     
 
   
  
PAGE # 2 
 
fp=k esa fn[kk, vuqlkj rhu fcUnq nzO;eku ftuds nzO;eku Øe'k% 1kg, 1.5kg rFkk 2.5kg gS] ,d f=kHkqt ds dksuksa ij 
j[ks gSA f=kHkqt dh Hkqtk,sa Øe'k% 3 cm, 4cm rFkk 5cm gSA rc nzO;eku dsUnz dh fLFkfr 1kg nzO;eku ds lkis{k gksxh : 
2.5 kg 
 
5cm 
 
4 cm 
 
1 kg 
 
3 cm 
 
1.5 kg 
 
 
 (1) 1kg ds nka;h vksj 0.6 cm nwj rFkk 1kg nzO;eku ds 2 cm Åij  
  (2) 1kg ds nka;h vksj 0.9 cm nwj rFkk 1kg nzO;eku ds 2 cm Åij 
 (3) 1kg ds cka;h vksj 0.9 cm nwj rFkk 1kg nzO;eku ds 2 cm Åij 
 (4) 1 kg ds nka;h vksj 0.9 cm nwj rFkk 1kg nzO;eku ds 1.5 cm Åij 
Ans. (2) 
Sol. Take 1kg mass at origin  
 1 kg nzO;eku dks ewy fcUnq ysus ij  
 
2.5 kg 
 
5cm 
4 cm 
1 kg 
 
3 cm 
 
1.5 kg 
 
 
Y 
 
X 
 
 
 
cm
1 0 1.5 3 2.5 0
X 0.9cm
5
? ? ? ? ?
? ? 
 
cm
1 0 1.5 0 2.5 4
Y 2cm
5
? ? ? ? ?
? ? 
 
3. In a single slit diffraction set up, second minima is observed at an angle of 60°. The expected position of 
first minima is    
 ,dy fNnz foorZu izfØ;k esa] 2
nd
 fufEu"B (minima) 60° ds dks.k ij fn[kkbZ nsrk gS] rks 1
st
 fufEu"B (minima) dh 
visf{kr fLFkfr gksxh &  
 (1) 25°     (2) 20°     (3) 30°     (4) 45°  
Ans. (1) 
Sol. For 2
nd
 minima   f}rh; fufEu"B ds fy, 
 d sin ? = 2 ?    
 sin ? = 
2
3
 (given) 
 ? 
4
3
d
?
?
 … (i) 
 So for 1
st
 minima is  
 blfy, izFke fufEu"B ds fy, 
d sin ? = ?    
 sin ? = 
4
3
d
?
?
 (from equation (i) (lehdj.k (i) ls 
 ? = 25.65° (from sin table) (sin rkfydk ls) 
 ? ? ? 25 
        
 
 
    
    
     
 
   
  
PAGE # 3 
 
4. There are two infinite plane sheets each having uniform surface charge density + ? C/m
2
. They are 
inclined to each other at an angle 30° as shown in the figure. Electric field at any arbitrary point P is:  
 fp=k esa nks vuUr yEckbZ dh lery ifêdk gS ftudk le:i i`"Bh; vkos'k ?kuRo + ? C/m
2 
gSA nksuksa ifêdk ,d nwljs 
ls 30° dks.k ij >qdh gSA rc fp=kkuqlkj fdlh Hkh ;kn`fPNd fcUnq P ij fo|qr {ks=k gksxk: 
 
30°
+ ?
+ ?
P
Y 
 
X 
 
 
 (1) 
0
3 1
ˆ ˆ 1– y – x
2 2 2
? ? ? ?
?
? ?
? ?
?
? ?
? ?
? ?
    (2) 
0
3 1
ˆ ˆ 1 y – x
2 2 2
? ? ? ?
?
? ? ?
? ?
?
? ?
? ?
? ?
      
(3) 
0
3 1
ˆ ˆ 1– y x
2 2 2
? ? ? ?
?
? ? ?
? ?
?
? ?
? ?
? ?
   (4)  
0
3 1
ˆ ˆ 1 y x
2 2 2
? ? ? ?
?
? ? ? ?
? ?
?
? ?
? ?
? ?
 
Ans. (1) 
Sol.  
 
 
30° 
+ ? 
+ ? 
P
60°
 
? ? ? ?
0 0 0
ˆ ˆ E cos60 –x – sin60 y
2 2 2
? ? ? ? ?
? ? ? ?
? ?
? ? ?
? ?
?
 
 E
?
= 
0
3 1
ˆ ˆ 1– y – x
2 2 2
? ? ? ?
?
? ?
? ?
?
? ?
? ?
? ?
 
 
5. A parallel plate capacitor with plate area A & plate separation d is filled with a dielectric material of 
dielectric constant given by k = k 0(1 + ?x). Calculate capacitance of system: (given ?d << 1)  
 ,d lekUrj iV~V la/kkfj=k ftldh lrg dk {ks=kQy A gS rFkk ifV~Vdkvksa ds e/; dh nwjh d gSA buds e/; ,d 
ijkoS|qrkad inkFkZ Hkjk gSA inkFkZ dk ijkoS|qrkad fu;rkad k = k 0(1 + ?x) gSA rc fudk; dh /kkfjrk gksxh:  
(fn;k gS ?d << 1) 
 
x 
 
 (1)  ? ?
2 2 0 0
d 1
d
A k
? ?
?
  (2) ?
?
?
?
?
? ?
?
?
2
d
1
d
A k
0 0
   (3) ? ? d 1
d 2
A k
0 0
? ?
?
  (4) ?
?
?
?
?
? ?
?
?
2
d
1
d 2
A k
0 0
  
Ans. (2) 
 
        
 
 
    
    
     
 
   
  
PAGE # 4 
 
Sol. Capacitance of element ?kVd dh /kkfjrk  = 
dx
A k
0
?
 
 
 
dx
x
 
 Capacitance of element ?kVd dh /kkfjrk, C' = 
dx
A ) x 1 ( k
0 0
? ? ?
 
 
?
?
? ? ?
?
d
0
0 0
) x 1 ( A k
dx
' C
1
 
 ) d 1 ( n
A k
1
C
1
0 0
? ?
? ?
? ? 
 Given fn;k gS& ?d << 1 
 
?
?
?
?
?
?
?
?
?
? ?
? ?
?
2
d
d
A k
1
C
1
2 2
0 0
 
 ?
?
?
?
?
? ?
?
?
?
2
d
1
A k
d
C
1
0 0
 
 ?
?
?
?
?
? ?
?
?
?
2
d
1
d
A k
C
0 0
 
 
6. A long solenoid of radius R carries a time dependent current I = I
0 
t(1 – t). A ring of radius 2R is placed 
coaxially near its centre. During the time interval 0 ? t ? 1, the induced current ?R and the induced emf VR 
in the ring vary as: 
 (1) current will change its direction and its emf will be zero at t = 0.25sec. 
   (2) current will not change its direction & emf will be maximum at t = 0.5sec 
 (3) current will not change direction and emf will be zero at 0.25sec. 
 (4) current will change its direction and its emf will be zero at t = 0.5sec. 
 ,d R f=kT;k dh ijhuyhdk ftlesa ,d le; fuHkZj /kkjk I = I
0 
t(1 – t) izokfgr gks jgh gSA ,d 2R f=kT;k dh oy; 
ifjuyhdk ds lek{kh; :i ls mlds dsUnz ds ikl fLFkr gSA le; vUrjky 0 ? t ? 1, ds nkSjku oy; esa izsjhr /kkjk ? R 
vkSj izsfjr fo+|qr okgd cy V R  ds laxr lgh fodYi gksxk 
 (1) /kkjk viuh fn'kk cnysxh vkSj fo|qr okgd cy le; t = 0.25sec ij 'kwU; gksxkA 
   (2) /kkjk viuh fn'kk ugha cnysxh vkSj fo|qr okgd cy le; t = 0.5sec ij vf/kdre gksxkA 
 (3) /kkjk viuh fn'kk ugha cnysxh vkSj fo|qr okgd cy le; t = 0.25sec ij 'kwU; gksxkA  
 (4) /kkjk viuh fn'kk cnysxh vkSj fo|qr okgd cy t = 0.5sec ij 'kwU; gksxkA  
Ans. (4) 
Sol. I = I 0t – I 0t
2
  
 ? = BA 
 ? = ? 0nIA 
 V R = 
dt
d
–
?
 = – ? 0nAI 0 (1 – 2t)  
 VR = 0 at   t = 
2
1
  
        
 
 
    
    
     
 
   
  
PAGE # 5 
 
and rFkk ? R = 
loop of ce tan sis Re
V
R
 
sèk dk izfrjk yqi
R
V
 
 
 
e 
1/2 
t 
 
 
7. If 10% of intensity is passed from analyser, then, the angle by which analyser should be rotated such 
that transmitted intensity becomes zero. (Assume no absorption by analyser and polarizer). 
 ;fn ,d èkqzod ls 10% rhozrk xqtjrh gks rks èkqzod dks fdrus dks.k ls ?kqek;s tk;s fd fuxZr rhozrk 'kwU; gks tk;sA 
(eku fyft,s dh /kzqod ls vo'kks"k.k 'kwU; gS) 
(1) 60°   (2) 18.4°  (3) 45°   (4) 71.6° 
Ans. (B) 
Sol. I = I 0 cos
2
? ?
 ? ?
2
0
0
cos
10
I
I
  
 cos ? = 
10
1
 = 0.31 < 
2
1
 which is 0.707 
 So ? > 45° and rFkk 90 – ? < 45º so only one option is correct blfy, dsoy ,d fodYi lgh gksxkA 
i.e. vFkkZr 18.4º 
 angle rotated should be ?kqek;k gqvk dks.k = 90° – 71.6° = 18.4° 
 
8. Three moles of ideal gas A with 
3
4
C
C
V
P
? is mixed with two moles of another ideal gas B with 
3
5
C
C
V
P
? .   
The 
v
P
C
C
 of mixture is (Assuming temperature is constant)    
vkn'kZ xSl A ds 3 eksy ftlds fy, fd 
3
4
C
C
V
P
? gS] dks nwljh vkn'kZ xSl B ds 2 eksy ftlds fy, 
3
5
C
C
V
P
? gS] ds 
lkFk feyk;k tkrk gS rks feJ.k dh 
v
P
C
C
 dk eku crkb, (rkieku dks ,d&leku eku ekfu;sA)  
(1) 1.5   (2) 1.42   (3) 1.7      (4) 1.3  
Ans. (2) 
Sol. 
1 –
R n
1 –
R n
1 –
R
n
1 –
R
n
C n C n
C n C n
2
2
1
1
2
2
2
1
1
1
V 2 V 1
P 2 P 1
mixture
2 1
2 1
?
?
?
?
?
?
?
?
?
?
?
? ? 
 on rearranging we get iqu O;ofLFkr djus ij, 
1 –
n
1 –
n
1 –
n n
2
2
1
1
mix
2 1
?
?
?
?
?
?
;  
3 / 2
2
3 / 1
3
1 –
5
mix
? ?
?
 
 
1 –
5
mix
?
 = 9 + 3 = 12  ? 
12
5
1
12
17
mixure
? ? ? ?  ;  ? mix = 1.42