Mains Level Problems with Solutions JEE Notes | EduRev

JEE : Mains Level Problems with Solutions JEE Notes | EduRev

 Page 1


 1 
 
PHYSICS, CHEMISTRY AND MATHEMATICS 
Class II IIT-JEE Achiever 2017-2018   Max. Marks              360     
 
 
Solution to Test - 01   (Integrated) 
Main    
Duration 
 Date 
           3 Hours 
       15-04-2017 
 
 
PART I - PHYSICS 
Multiple choice questions with one correct alternative    
1. The figure indicates the electric field lines corresponding to an electric field E. Then  
(A) E
P
 = E
Q
 = E
R
 
(B) E
P 
> E
Q 
> E
R
 
(C) E
P
 = E
R 
> E
Q
 
(D) E
P 
< E
Q
< E
R
 
Ans (C) 
Count the number of lines of force 
2. There are n electrons of charge e on a drop of oil of density ?. It is in equilibrium (at rest) in an electric 
field E. The radius of the drop is 
(A) 
1
2
2neE
4 g
? ?
? ?
? ?
? ?
  (B) 
1
2
neE
g
? ?
? ?
?
? ?
  (C) 
1
3
3neE
4 g
? ?
? ?
? ?
? ?
  (D) 
1
3
2neE
g
? ?
? ?
? ?
? ?
 
Ans (C) 
1
3
3
4 3neE
mg qE r g neE r
3 4 g
? ?
? ? ? ? ? ? ?
? ?
? ?
? ?
 
3. The electric force experienced by a 1 ?C charge is 1.5 ? 10
–3
 N. the magnitude of the electric field at the 
position of the charge is 
(A) 1.5 ? 10
2
 NC
–1
  (B) 1.5 ? 10
3
 NC
–1
 (C) 1.5 ? 10
4
 NC
–1
 (D) 1.5 ? 10
5
 NC
–1
 
Ans (B) 
3
3 1
6
F 1.5 10
F qE E 1.5 10 NC
q 1 10
?
?
?
?
? ? ? ? ? ?
?
 
4. A water droplet of mass 10 mg and having charge 1.0 ?C stays suspended in a room. The electric field in 
the room is  
(A) 10
5
 NC
–1
 upwards    (B) 10
5
 NC
–1
 downwards 
(C) 10
6
 NC
–1
 upwards    (D) 10
6
 NC
–1
 downwards 
Ans (A) 
3
5 1
6
mg 10 10 10
mg qE E 10 NC
q 1.0 10
?
?
?
? ?
? ? ? ? ?
?
 
5. A pendulum bob of mass 80 mg, carrying a charge 2 ? 10
–8
 C is at rest in a uniform horizontal electric 
field of 20 ? 10
3
 NC
–1
. The tension in the thread is nearly  
(A) 8 ? 10
–4
 N  (B) 9 ? 10
–4
 N  (C) 8 ? 10
–8
 N  (D) 9 ? 10
–8
 N 
Ans (B) 
2 2 6 2 8 3 2 4
T (mg) (qE) (80 10 10) (2 10 20 10 ) 9 10 N
? ? ?
? ? ? ? ? ? ? ? ? ? ? 
P R 
Q 
Page 2


 1 
 
PHYSICS, CHEMISTRY AND MATHEMATICS 
Class II IIT-JEE Achiever 2017-2018   Max. Marks              360     
 
 
Solution to Test - 01   (Integrated) 
Main    
Duration 
 Date 
           3 Hours 
       15-04-2017 
 
 
PART I - PHYSICS 
Multiple choice questions with one correct alternative    
1. The figure indicates the electric field lines corresponding to an electric field E. Then  
(A) E
P
 = E
Q
 = E
R
 
(B) E
P 
> E
Q 
> E
R
 
(C) E
P
 = E
R 
> E
Q
 
(D) E
P 
< E
Q
< E
R
 
Ans (C) 
Count the number of lines of force 
2. There are n electrons of charge e on a drop of oil of density ?. It is in equilibrium (at rest) in an electric 
field E. The radius of the drop is 
(A) 
1
2
2neE
4 g
? ?
? ?
? ?
? ?
  (B) 
1
2
neE
g
? ?
? ?
?
? ?
  (C) 
1
3
3neE
4 g
? ?
? ?
? ?
? ?
  (D) 
1
3
2neE
g
? ?
? ?
? ?
? ?
 
Ans (C) 
1
3
3
4 3neE
mg qE r g neE r
3 4 g
? ?
? ? ? ? ? ? ?
? ?
? ?
? ?
 
3. The electric force experienced by a 1 ?C charge is 1.5 ? 10
–3
 N. the magnitude of the electric field at the 
position of the charge is 
(A) 1.5 ? 10
2
 NC
–1
  (B) 1.5 ? 10
3
 NC
–1
 (C) 1.5 ? 10
4
 NC
–1
 (D) 1.5 ? 10
5
 NC
–1
 
Ans (B) 
3
3 1
6
F 1.5 10
F qE E 1.5 10 NC
q 1 10
?
?
?
?
? ? ? ? ? ?
?
 
4. A water droplet of mass 10 mg and having charge 1.0 ?C stays suspended in a room. The electric field in 
the room is  
(A) 10
5
 NC
–1
 upwards    (B) 10
5
 NC
–1
 downwards 
(C) 10
6
 NC
–1
 upwards    (D) 10
6
 NC
–1
 downwards 
Ans (A) 
3
5 1
6
mg 10 10 10
mg qE E 10 NC
q 1.0 10
?
?
?
? ?
? ? ? ? ?
?
 
5. A pendulum bob of mass 80 mg, carrying a charge 2 ? 10
–8
 C is at rest in a uniform horizontal electric 
field of 20 ? 10
3
 NC
–1
. The tension in the thread is nearly  
(A) 8 ? 10
–4
 N  (B) 9 ? 10
–4
 N  (C) 8 ? 10
–8
 N  (D) 9 ? 10
–8
 N 
Ans (B) 
2 2 6 2 8 3 2 4
T (mg) (qE) (80 10 10) (2 10 20 10 ) 9 10 N
? ? ?
? ? ? ? ? ? ? ? ? ? ? 
P R 
Q 
2IIT1718PCMT1SM(Intg) 2 
 
6. Consider a uniformly charged ring of radius R. The point on the axis of the ring where the electric field 
is maximum is at a distance x from the centre of the ring. The value of x is 
(A) 
R
2
   (B) 
R
3
   (C) 
R
3
  (D) 
R
2
 
Ans (D) 
7. A circular wire loop of radius R carries a total charge Q distributed uniformly over its length. A small 
length dl of the wire is cut off. The electric field at the centre due to the remaining wire is 
(A) 
2 3
0
Q(d )
8 R ? ?
l
  (B) 
3
0
Q(d )
4 R ? ?
l
  (C) 
2 3
0
Q(d )
4 R ? ?
l
  (D) 
3
0
Q(d )
8 R ? ?
l
 
Ans (A) 
The electric field due to the remaining wire is equal and opposite to the field produced by the charge 
element dl. 
? 
2 2 2 3
0 0 0
1 d 1 Q d Q(d )
E
4 R 4 2 R R 8 R
?
? ? ?
? ? ? ? ? ? ?
l l l
 
8. A wire of length l with a uniform charge density ? is bent in the form of a regular hexagon. The electric 
field at the centre of the hexagon is 
(A) 
0
4
?
? ? l
   (B) 
0
2
?
? ? l
  (C) 
0
8
?
? ? l
  (D) zero 
Ans (D) 
9. A charged particle of mass m = 2 kg and charge 1 ?C is projected from a horizontal ground at an angle  
? = 45 ? with speed 10 ms
–1
. In space, a horizontal electric field toward the direction of projection  
E = 2 ? 10
7
 NC
–1
 exists. The range of the projectile is 
(A) 20 m   (B) 60 m  (C) 200 m  (D) 180 m 
Ans (A) 
6 7
2
x
qE 10 2 10
a 10 ms
m 2
?
?
? ?
? ? ? 
Time of flight 
1
2 10
2usin
2
T 2 s
g 10
? ?
?
? ? ? 
Hence, 
2
x x
1
R u T a T
2
? ? 
 
2
x
1
10cos45 T a T
2
? ? ? ?
10 1
2 10 2
2 2
? ? ? ? ? = 20 m 
10. Three positive point charges q
1
, q
2
 and q
3
 form an isolated system. Suppose the 
charges have generated a property due to which like charges also attract. The charges 
are moving along a circle with same speed maintaining angles as shown in the figure. 
The charge q
1
 experiences a force f
1
 due to other two charges. Similarly, q
2
 
experiences a force f
2
 and q
3
, a force f
3
. The ratio f
1
 : f
2
 : f
3
 is 
(A) 1 : 1 : 1   
(B) q
1
 : q
2
 : q
3
 
(C) 1 : 3 : 2 
(D) this ratio cannot be calculated 
Ans (C) 
Page 3


 1 
 
PHYSICS, CHEMISTRY AND MATHEMATICS 
Class II IIT-JEE Achiever 2017-2018   Max. Marks              360     
 
 
Solution to Test - 01   (Integrated) 
Main    
Duration 
 Date 
           3 Hours 
       15-04-2017 
 
 
PART I - PHYSICS 
Multiple choice questions with one correct alternative    
1. The figure indicates the electric field lines corresponding to an electric field E. Then  
(A) E
P
 = E
Q
 = E
R
 
(B) E
P 
> E
Q 
> E
R
 
(C) E
P
 = E
R 
> E
Q
 
(D) E
P 
< E
Q
< E
R
 
Ans (C) 
Count the number of lines of force 
2. There are n electrons of charge e on a drop of oil of density ?. It is in equilibrium (at rest) in an electric 
field E. The radius of the drop is 
(A) 
1
2
2neE
4 g
? ?
? ?
? ?
? ?
  (B) 
1
2
neE
g
? ?
? ?
?
? ?
  (C) 
1
3
3neE
4 g
? ?
? ?
? ?
? ?
  (D) 
1
3
2neE
g
? ?
? ?
? ?
? ?
 
Ans (C) 
1
3
3
4 3neE
mg qE r g neE r
3 4 g
? ?
? ? ? ? ? ? ?
? ?
? ?
? ?
 
3. The electric force experienced by a 1 ?C charge is 1.5 ? 10
–3
 N. the magnitude of the electric field at the 
position of the charge is 
(A) 1.5 ? 10
2
 NC
–1
  (B) 1.5 ? 10
3
 NC
–1
 (C) 1.5 ? 10
4
 NC
–1
 (D) 1.5 ? 10
5
 NC
–1
 
Ans (B) 
3
3 1
6
F 1.5 10
F qE E 1.5 10 NC
q 1 10
?
?
?
?
? ? ? ? ? ?
?
 
4. A water droplet of mass 10 mg and having charge 1.0 ?C stays suspended in a room. The electric field in 
the room is  
(A) 10
5
 NC
–1
 upwards    (B) 10
5
 NC
–1
 downwards 
(C) 10
6
 NC
–1
 upwards    (D) 10
6
 NC
–1
 downwards 
Ans (A) 
3
5 1
6
mg 10 10 10
mg qE E 10 NC
q 1.0 10
?
?
?
? ?
? ? ? ? ?
?
 
5. A pendulum bob of mass 80 mg, carrying a charge 2 ? 10
–8
 C is at rest in a uniform horizontal electric 
field of 20 ? 10
3
 NC
–1
. The tension in the thread is nearly  
(A) 8 ? 10
–4
 N  (B) 9 ? 10
–4
 N  (C) 8 ? 10
–8
 N  (D) 9 ? 10
–8
 N 
Ans (B) 
2 2 6 2 8 3 2 4
T (mg) (qE) (80 10 10) (2 10 20 10 ) 9 10 N
? ? ?
? ? ? ? ? ? ? ? ? ? ? 
P R 
Q 
2IIT1718PCMT1SM(Intg) 2 
 
6. Consider a uniformly charged ring of radius R. The point on the axis of the ring where the electric field 
is maximum is at a distance x from the centre of the ring. The value of x is 
(A) 
R
2
   (B) 
R
3
   (C) 
R
3
  (D) 
R
2
 
Ans (D) 
7. A circular wire loop of radius R carries a total charge Q distributed uniformly over its length. A small 
length dl of the wire is cut off. The electric field at the centre due to the remaining wire is 
(A) 
2 3
0
Q(d )
8 R ? ?
l
  (B) 
3
0
Q(d )
4 R ? ?
l
  (C) 
2 3
0
Q(d )
4 R ? ?
l
  (D) 
3
0
Q(d )
8 R ? ?
l
 
Ans (A) 
The electric field due to the remaining wire is equal and opposite to the field produced by the charge 
element dl. 
? 
2 2 2 3
0 0 0
1 d 1 Q d Q(d )
E
4 R 4 2 R R 8 R
?
? ? ?
? ? ? ? ? ? ?
l l l
 
8. A wire of length l with a uniform charge density ? is bent in the form of a regular hexagon. The electric 
field at the centre of the hexagon is 
(A) 
0
4
?
? ? l
   (B) 
0
2
?
? ? l
  (C) 
0
8
?
? ? l
  (D) zero 
Ans (D) 
9. A charged particle of mass m = 2 kg and charge 1 ?C is projected from a horizontal ground at an angle  
? = 45 ? with speed 10 ms
–1
. In space, a horizontal electric field toward the direction of projection  
E = 2 ? 10
7
 NC
–1
 exists. The range of the projectile is 
(A) 20 m   (B) 60 m  (C) 200 m  (D) 180 m 
Ans (A) 
6 7
2
x
qE 10 2 10
a 10 ms
m 2
?
?
? ?
? ? ? 
Time of flight 
1
2 10
2usin
2
T 2 s
g 10
? ?
?
? ? ? 
Hence, 
2
x x
1
R u T a T
2
? ? 
 
2
x
1
10cos45 T a T
2
? ? ? ?
10 1
2 10 2
2 2
? ? ? ? ? = 20 m 
10. Three positive point charges q
1
, q
2
 and q
3
 form an isolated system. Suppose the 
charges have generated a property due to which like charges also attract. The charges 
are moving along a circle with same speed maintaining angles as shown in the figure. 
The charge q
1
 experiences a force f
1
 due to other two charges. Similarly, q
2
 
experiences a force f
2
 and q
3
, a force f
3
. The ratio f
1
 : f
2
 : f
3
 is 
(A) 1 : 1 : 1   
(B) q
1
 : q
2
 : q
3
 
(C) 1 : 3 : 2 
(D) this ratio cannot be calculated 
Ans (C) 
 3 
 
Since the charges exert force on one another that are in action-reaction pair, sum of forces 
1 2 3
f , f and f
? ? ?
 
is zero. So force vectors form a triangle, as shown in the figure.  
Applying sine rule, we have 
3 1 2
f f f
sin30 sin 60 sin90
? ? 
or 
2
1 3
2f
2f f K (sup pose)
3
? ? ? 
 
1 2 3
k 3
f , f K, f K
2 2
? ? ? 
So,  
1 2 3
1 3
f : f : f : : 1 1 : 3 : 2
2 2
? ? 
11. Two mutually perpendicular wire carry charge densities ?
1
 and ?
2
. The electric 
lines of force makes angle ? with second wire, then 
1
2
?
?
 is 
(A) tan
2 
? 
(B) cot
2
? 
(C) sin
2
? 
(D) cos
2
? 
Ans (B) 
1
1
0
E
2 x
?
?
? ?
 and 
2
2
0
E
2 y
?
?
? ?
 
1 1
2 2
E y
E x
? ? ? ? ?
?
? ? ? ?
?
? ?
? ?
 
1
2
1
tan
tan
? ? ?
? ?
? ?
? ?
? ?
 or 
2 1
2
cot
?
? ?
?
 
12. A non-conducting ring of mass m and radius R is charged as shown. The 
magnitude of charge density, (i.e., charge per unit length) is ?. It is then placed 
on a smooth non-conducting horizontal plane. At time t = 0, a uniform electric 
field 
0
ˆ
E E i ?
?
 is switched on.  The angular acceleration of the ring at that instant 
is:   
(A) 
0
2 E
m
?
   (B) 
0
E
m
?
  
(C) 
0
2 E
3m
?
  (D) none of these 
Ans (B) 
Consider an infinitesimal element subtending an angle d ? at the centre and at angle ? as shown in figure.  
dF = ?Rd ?E
0
 
A force of same magnitude but in opposite direction acts on a corresponding element in the region of 
negative charge. Net torque due to both types of charges is  d ? = ( ?Rd ?E
0
) (2Rsin ?) 
? 
2
2
0
0
2 R E sin d
?
? ? ? ? ?
?
? 
2
0
2 R E ? ? ? 
Equation for pure rolling motion is ? ? = I ? 
? 2 ?R
2
E
0
 = mR
2
?   
y 
x 
E
0 
+ 
+ 
+ 
? 
? 
? 
? 
Page 4


 1 
 
PHYSICS, CHEMISTRY AND MATHEMATICS 
Class II IIT-JEE Achiever 2017-2018   Max. Marks              360     
 
 
Solution to Test - 01   (Integrated) 
Main    
Duration 
 Date 
           3 Hours 
       15-04-2017 
 
 
PART I - PHYSICS 
Multiple choice questions with one correct alternative    
1. The figure indicates the electric field lines corresponding to an electric field E. Then  
(A) E
P
 = E
Q
 = E
R
 
(B) E
P 
> E
Q 
> E
R
 
(C) E
P
 = E
R 
> E
Q
 
(D) E
P 
< E
Q
< E
R
 
Ans (C) 
Count the number of lines of force 
2. There are n electrons of charge e on a drop of oil of density ?. It is in equilibrium (at rest) in an electric 
field E. The radius of the drop is 
(A) 
1
2
2neE
4 g
? ?
? ?
? ?
? ?
  (B) 
1
2
neE
g
? ?
? ?
?
? ?
  (C) 
1
3
3neE
4 g
? ?
? ?
? ?
? ?
  (D) 
1
3
2neE
g
? ?
? ?
? ?
? ?
 
Ans (C) 
1
3
3
4 3neE
mg qE r g neE r
3 4 g
? ?
? ? ? ? ? ? ?
? ?
? ?
? ?
 
3. The electric force experienced by a 1 ?C charge is 1.5 ? 10
–3
 N. the magnitude of the electric field at the 
position of the charge is 
(A) 1.5 ? 10
2
 NC
–1
  (B) 1.5 ? 10
3
 NC
–1
 (C) 1.5 ? 10
4
 NC
–1
 (D) 1.5 ? 10
5
 NC
–1
 
Ans (B) 
3
3 1
6
F 1.5 10
F qE E 1.5 10 NC
q 1 10
?
?
?
?
? ? ? ? ? ?
?
 
4. A water droplet of mass 10 mg and having charge 1.0 ?C stays suspended in a room. The electric field in 
the room is  
(A) 10
5
 NC
–1
 upwards    (B) 10
5
 NC
–1
 downwards 
(C) 10
6
 NC
–1
 upwards    (D) 10
6
 NC
–1
 downwards 
Ans (A) 
3
5 1
6
mg 10 10 10
mg qE E 10 NC
q 1.0 10
?
?
?
? ?
? ? ? ? ?
?
 
5. A pendulum bob of mass 80 mg, carrying a charge 2 ? 10
–8
 C is at rest in a uniform horizontal electric 
field of 20 ? 10
3
 NC
–1
. The tension in the thread is nearly  
(A) 8 ? 10
–4
 N  (B) 9 ? 10
–4
 N  (C) 8 ? 10
–8
 N  (D) 9 ? 10
–8
 N 
Ans (B) 
2 2 6 2 8 3 2 4
T (mg) (qE) (80 10 10) (2 10 20 10 ) 9 10 N
? ? ?
? ? ? ? ? ? ? ? ? ? ? 
P R 
Q 
2IIT1718PCMT1SM(Intg) 2 
 
6. Consider a uniformly charged ring of radius R. The point on the axis of the ring where the electric field 
is maximum is at a distance x from the centre of the ring. The value of x is 
(A) 
R
2
   (B) 
R
3
   (C) 
R
3
  (D) 
R
2
 
Ans (D) 
7. A circular wire loop of radius R carries a total charge Q distributed uniformly over its length. A small 
length dl of the wire is cut off. The electric field at the centre due to the remaining wire is 
(A) 
2 3
0
Q(d )
8 R ? ?
l
  (B) 
3
0
Q(d )
4 R ? ?
l
  (C) 
2 3
0
Q(d )
4 R ? ?
l
  (D) 
3
0
Q(d )
8 R ? ?
l
 
Ans (A) 
The electric field due to the remaining wire is equal and opposite to the field produced by the charge 
element dl. 
? 
2 2 2 3
0 0 0
1 d 1 Q d Q(d )
E
4 R 4 2 R R 8 R
?
? ? ?
? ? ? ? ? ? ?
l l l
 
8. A wire of length l with a uniform charge density ? is bent in the form of a regular hexagon. The electric 
field at the centre of the hexagon is 
(A) 
0
4
?
? ? l
   (B) 
0
2
?
? ? l
  (C) 
0
8
?
? ? l
  (D) zero 
Ans (D) 
9. A charged particle of mass m = 2 kg and charge 1 ?C is projected from a horizontal ground at an angle  
? = 45 ? with speed 10 ms
–1
. In space, a horizontal electric field toward the direction of projection  
E = 2 ? 10
7
 NC
–1
 exists. The range of the projectile is 
(A) 20 m   (B) 60 m  (C) 200 m  (D) 180 m 
Ans (A) 
6 7
2
x
qE 10 2 10
a 10 ms
m 2
?
?
? ?
? ? ? 
Time of flight 
1
2 10
2usin
2
T 2 s
g 10
? ?
?
? ? ? 
Hence, 
2
x x
1
R u T a T
2
? ? 
 
2
x
1
10cos45 T a T
2
? ? ? ?
10 1
2 10 2
2 2
? ? ? ? ? = 20 m 
10. Three positive point charges q
1
, q
2
 and q
3
 form an isolated system. Suppose the 
charges have generated a property due to which like charges also attract. The charges 
are moving along a circle with same speed maintaining angles as shown in the figure. 
The charge q
1
 experiences a force f
1
 due to other two charges. Similarly, q
2
 
experiences a force f
2
 and q
3
, a force f
3
. The ratio f
1
 : f
2
 : f
3
 is 
(A) 1 : 1 : 1   
(B) q
1
 : q
2
 : q
3
 
(C) 1 : 3 : 2 
(D) this ratio cannot be calculated 
Ans (C) 
 3 
 
Since the charges exert force on one another that are in action-reaction pair, sum of forces 
1 2 3
f , f and f
? ? ?
 
is zero. So force vectors form a triangle, as shown in the figure.  
Applying sine rule, we have 
3 1 2
f f f
sin30 sin 60 sin90
? ? 
or 
2
1 3
2f
2f f K (sup pose)
3
? ? ? 
 
1 2 3
k 3
f , f K, f K
2 2
? ? ? 
So,  
1 2 3
1 3
f : f : f : : 1 1 : 3 : 2
2 2
? ? 
11. Two mutually perpendicular wire carry charge densities ?
1
 and ?
2
. The electric 
lines of force makes angle ? with second wire, then 
1
2
?
?
 is 
(A) tan
2 
? 
(B) cot
2
? 
(C) sin
2
? 
(D) cos
2
? 
Ans (B) 
1
1
0
E
2 x
?
?
? ?
 and 
2
2
0
E
2 y
?
?
? ?
 
1 1
2 2
E y
E x
? ? ? ? ?
?
? ? ? ?
?
? ?
? ?
 
1
2
1
tan
tan
? ? ?
? ?
? ?
? ?
? ?
 or 
2 1
2
cot
?
? ?
?
 
12. A non-conducting ring of mass m and radius R is charged as shown. The 
magnitude of charge density, (i.e., charge per unit length) is ?. It is then placed 
on a smooth non-conducting horizontal plane. At time t = 0, a uniform electric 
field 
0
ˆ
E E i ?
?
 is switched on.  The angular acceleration of the ring at that instant 
is:   
(A) 
0
2 E
m
?
   (B) 
0
E
m
?
  
(C) 
0
2 E
3m
?
  (D) none of these 
Ans (B) 
Consider an infinitesimal element subtending an angle d ? at the centre and at angle ? as shown in figure.  
dF = ?Rd ?E
0
 
A force of same magnitude but in opposite direction acts on a corresponding element in the region of 
negative charge. Net torque due to both types of charges is  d ? = ( ?Rd ?E
0
) (2Rsin ?) 
? 
2
2
0
0
2 R E sin d
?
? ? ? ? ?
?
? 
2
0
2 R E ? ? ? 
Equation for pure rolling motion is ? ? = I ? 
? 2 ?R
2
E
0
 = mR
2
?   
y 
x 
E
0 
+ 
+ 
+ 
? 
? 
? 
? 
2IIT1718PCMT1SM(Intg) 4 
 
13. A positively charged thin metal ring of radius R is fixed in the x-y plane with its centre at the origin O. A 
negatively charged particle is released from rest at the point (0, 0, z
0
) where z
0
> 0. Then the motion of P is 
(A) periodic, for all values of z
0
 satisfying 0 < z
0 
< ?. 
(B) simple harmonic, for all the values of z
0
 satisfying 0 <  z
0 
? R. 
(C) approximately simple harmonic for all values  of z
0
. 
(D) such that crosses O and continues to move along the negative z-axis. 
Ans (A) 
Here electric field; E = 
? ?
0
3 2
2 2
0 0
Qz
4 R z ? ? ?
 where Q is charge on the ring and z
0
 is the axial point distance 
from O. Then F = qE = 
? ?
0
3 2
2 2
0 0
Qqz
4 R z ? ? ?
 
When charge ( ?q) crosses origin, force is again towards the centre; i.e. motion is periodic. Now for  
z
0 
<< R; F = 
0
3
0
Qqz
4 R ? ?
 
where clearly F is proportional to the displacement z
0
 from the mean position. 
? Motion is SHM only for z
0 
<< R . 
14. Two point charges 
a b
q and q whose strengths are equal in magnitude are positioned 
at a certain distance from each other. Assuming the field strength is positive in the 
direction coinciding with the positive direction of x-axis, determine the signs of 
charges for the distribution of the field strength between the charges shown in figure. 
(A) q
a
? + ve, q
b
? ?ve    (B) q
a
? ?ve, q
b
? + ve 
(C) q
a
? + ve, q
b
? + ve    (D) q
a
? ?ve, q
b
? ?ve 
Ans (A) 
This problem is related to the graphical representation of electric field due to a system of charges. 
To the right of q
a
 field is + ve.  So q
a
 is + ve. To the left of q
b
 field is + ve so q
b
 is ?ve. 
15. Two positive point charges q are placed on the x-axis, one at x = a  and another at x = -a on the x-axis. 
The graph of the x-component of the electric field as a function of x, for values of x between  
-4a and +4a is correctly shown in 
 
 
(A)       (B)  
 
 
 
 
 
(C)        (D) 
 
 
 
Ans (A) 
Check the direction of electric field at a point and then trace the graph 
E
x 
a
 
 
a
 
x
 
 
E
x 
a
 
 
a
 
 
 
 
E
x
a
 
 
a
 
 
 
x
 
 
a
 
 
a
 
 
 
E
x 
Page 5


 1 
 
PHYSICS, CHEMISTRY AND MATHEMATICS 
Class II IIT-JEE Achiever 2017-2018   Max. Marks              360     
 
 
Solution to Test - 01   (Integrated) 
Main    
Duration 
 Date 
           3 Hours 
       15-04-2017 
 
 
PART I - PHYSICS 
Multiple choice questions with one correct alternative    
1. The figure indicates the electric field lines corresponding to an electric field E. Then  
(A) E
P
 = E
Q
 = E
R
 
(B) E
P 
> E
Q 
> E
R
 
(C) E
P
 = E
R 
> E
Q
 
(D) E
P 
< E
Q
< E
R
 
Ans (C) 
Count the number of lines of force 
2. There are n electrons of charge e on a drop of oil of density ?. It is in equilibrium (at rest) in an electric 
field E. The radius of the drop is 
(A) 
1
2
2neE
4 g
? ?
? ?
? ?
? ?
  (B) 
1
2
neE
g
? ?
? ?
?
? ?
  (C) 
1
3
3neE
4 g
? ?
? ?
? ?
? ?
  (D) 
1
3
2neE
g
? ?
? ?
? ?
? ?
 
Ans (C) 
1
3
3
4 3neE
mg qE r g neE r
3 4 g
? ?
? ? ? ? ? ? ?
? ?
? ?
? ?
 
3. The electric force experienced by a 1 ?C charge is 1.5 ? 10
–3
 N. the magnitude of the electric field at the 
position of the charge is 
(A) 1.5 ? 10
2
 NC
–1
  (B) 1.5 ? 10
3
 NC
–1
 (C) 1.5 ? 10
4
 NC
–1
 (D) 1.5 ? 10
5
 NC
–1
 
Ans (B) 
3
3 1
6
F 1.5 10
F qE E 1.5 10 NC
q 1 10
?
?
?
?
? ? ? ? ? ?
?
 
4. A water droplet of mass 10 mg and having charge 1.0 ?C stays suspended in a room. The electric field in 
the room is  
(A) 10
5
 NC
–1
 upwards    (B) 10
5
 NC
–1
 downwards 
(C) 10
6
 NC
–1
 upwards    (D) 10
6
 NC
–1
 downwards 
Ans (A) 
3
5 1
6
mg 10 10 10
mg qE E 10 NC
q 1.0 10
?
?
?
? ?
? ? ? ? ?
?
 
5. A pendulum bob of mass 80 mg, carrying a charge 2 ? 10
–8
 C is at rest in a uniform horizontal electric 
field of 20 ? 10
3
 NC
–1
. The tension in the thread is nearly  
(A) 8 ? 10
–4
 N  (B) 9 ? 10
–4
 N  (C) 8 ? 10
–8
 N  (D) 9 ? 10
–8
 N 
Ans (B) 
2 2 6 2 8 3 2 4
T (mg) (qE) (80 10 10) (2 10 20 10 ) 9 10 N
? ? ?
? ? ? ? ? ? ? ? ? ? ? 
P R 
Q 
2IIT1718PCMT1SM(Intg) 2 
 
6. Consider a uniformly charged ring of radius R. The point on the axis of the ring where the electric field 
is maximum is at a distance x from the centre of the ring. The value of x is 
(A) 
R
2
   (B) 
R
3
   (C) 
R
3
  (D) 
R
2
 
Ans (D) 
7. A circular wire loop of radius R carries a total charge Q distributed uniformly over its length. A small 
length dl of the wire is cut off. The electric field at the centre due to the remaining wire is 
(A) 
2 3
0
Q(d )
8 R ? ?
l
  (B) 
3
0
Q(d )
4 R ? ?
l
  (C) 
2 3
0
Q(d )
4 R ? ?
l
  (D) 
3
0
Q(d )
8 R ? ?
l
 
Ans (A) 
The electric field due to the remaining wire is equal and opposite to the field produced by the charge 
element dl. 
? 
2 2 2 3
0 0 0
1 d 1 Q d Q(d )
E
4 R 4 2 R R 8 R
?
? ? ?
? ? ? ? ? ? ?
l l l
 
8. A wire of length l with a uniform charge density ? is bent in the form of a regular hexagon. The electric 
field at the centre of the hexagon is 
(A) 
0
4
?
? ? l
   (B) 
0
2
?
? ? l
  (C) 
0
8
?
? ? l
  (D) zero 
Ans (D) 
9. A charged particle of mass m = 2 kg and charge 1 ?C is projected from a horizontal ground at an angle  
? = 45 ? with speed 10 ms
–1
. In space, a horizontal electric field toward the direction of projection  
E = 2 ? 10
7
 NC
–1
 exists. The range of the projectile is 
(A) 20 m   (B) 60 m  (C) 200 m  (D) 180 m 
Ans (A) 
6 7
2
x
qE 10 2 10
a 10 ms
m 2
?
?
? ?
? ? ? 
Time of flight 
1
2 10
2usin
2
T 2 s
g 10
? ?
?
? ? ? 
Hence, 
2
x x
1
R u T a T
2
? ? 
 
2
x
1
10cos45 T a T
2
? ? ? ?
10 1
2 10 2
2 2
? ? ? ? ? = 20 m 
10. Three positive point charges q
1
, q
2
 and q
3
 form an isolated system. Suppose the 
charges have generated a property due to which like charges also attract. The charges 
are moving along a circle with same speed maintaining angles as shown in the figure. 
The charge q
1
 experiences a force f
1
 due to other two charges. Similarly, q
2
 
experiences a force f
2
 and q
3
, a force f
3
. The ratio f
1
 : f
2
 : f
3
 is 
(A) 1 : 1 : 1   
(B) q
1
 : q
2
 : q
3
 
(C) 1 : 3 : 2 
(D) this ratio cannot be calculated 
Ans (C) 
 3 
 
Since the charges exert force on one another that are in action-reaction pair, sum of forces 
1 2 3
f , f and f
? ? ?
 
is zero. So force vectors form a triangle, as shown in the figure.  
Applying sine rule, we have 
3 1 2
f f f
sin30 sin 60 sin90
? ? 
or 
2
1 3
2f
2f f K (sup pose)
3
? ? ? 
 
1 2 3
k 3
f , f K, f K
2 2
? ? ? 
So,  
1 2 3
1 3
f : f : f : : 1 1 : 3 : 2
2 2
? ? 
11. Two mutually perpendicular wire carry charge densities ?
1
 and ?
2
. The electric 
lines of force makes angle ? with second wire, then 
1
2
?
?
 is 
(A) tan
2 
? 
(B) cot
2
? 
(C) sin
2
? 
(D) cos
2
? 
Ans (B) 
1
1
0
E
2 x
?
?
? ?
 and 
2
2
0
E
2 y
?
?
? ?
 
1 1
2 2
E y
E x
? ? ? ? ?
?
? ? ? ?
?
? ?
? ?
 
1
2
1
tan
tan
? ? ?
? ?
? ?
? ?
? ?
 or 
2 1
2
cot
?
? ?
?
 
12. A non-conducting ring of mass m and radius R is charged as shown. The 
magnitude of charge density, (i.e., charge per unit length) is ?. It is then placed 
on a smooth non-conducting horizontal plane. At time t = 0, a uniform electric 
field 
0
ˆ
E E i ?
?
 is switched on.  The angular acceleration of the ring at that instant 
is:   
(A) 
0
2 E
m
?
   (B) 
0
E
m
?
  
(C) 
0
2 E
3m
?
  (D) none of these 
Ans (B) 
Consider an infinitesimal element subtending an angle d ? at the centre and at angle ? as shown in figure.  
dF = ?Rd ?E
0
 
A force of same magnitude but in opposite direction acts on a corresponding element in the region of 
negative charge. Net torque due to both types of charges is  d ? = ( ?Rd ?E
0
) (2Rsin ?) 
? 
2
2
0
0
2 R E sin d
?
? ? ? ? ?
?
? 
2
0
2 R E ? ? ? 
Equation for pure rolling motion is ? ? = I ? 
? 2 ?R
2
E
0
 = mR
2
?   
y 
x 
E
0 
+ 
+ 
+ 
? 
? 
? 
? 
2IIT1718PCMT1SM(Intg) 4 
 
13. A positively charged thin metal ring of radius R is fixed in the x-y plane with its centre at the origin O. A 
negatively charged particle is released from rest at the point (0, 0, z
0
) where z
0
> 0. Then the motion of P is 
(A) periodic, for all values of z
0
 satisfying 0 < z
0 
< ?. 
(B) simple harmonic, for all the values of z
0
 satisfying 0 <  z
0 
? R. 
(C) approximately simple harmonic for all values  of z
0
. 
(D) such that crosses O and continues to move along the negative z-axis. 
Ans (A) 
Here electric field; E = 
? ?
0
3 2
2 2
0 0
Qz
4 R z ? ? ?
 where Q is charge on the ring and z
0
 is the axial point distance 
from O. Then F = qE = 
? ?
0
3 2
2 2
0 0
Qqz
4 R z ? ? ?
 
When charge ( ?q) crosses origin, force is again towards the centre; i.e. motion is periodic. Now for  
z
0 
<< R; F = 
0
3
0
Qqz
4 R ? ?
 
where clearly F is proportional to the displacement z
0
 from the mean position. 
? Motion is SHM only for z
0 
<< R . 
14. Two point charges 
a b
q and q whose strengths are equal in magnitude are positioned 
at a certain distance from each other. Assuming the field strength is positive in the 
direction coinciding with the positive direction of x-axis, determine the signs of 
charges for the distribution of the field strength between the charges shown in figure. 
(A) q
a
? + ve, q
b
? ?ve    (B) q
a
? ?ve, q
b
? + ve 
(C) q
a
? + ve, q
b
? + ve    (D) q
a
? ?ve, q
b
? ?ve 
Ans (A) 
This problem is related to the graphical representation of electric field due to a system of charges. 
To the right of q
a
 field is + ve.  So q
a
 is + ve. To the left of q
b
 field is + ve so q
b
 is ?ve. 
15. Two positive point charges q are placed on the x-axis, one at x = a  and another at x = -a on the x-axis. 
The graph of the x-component of the electric field as a function of x, for values of x between  
-4a and +4a is correctly shown in 
 
 
(A)       (B)  
 
 
 
 
 
(C)        (D) 
 
 
 
Ans (A) 
Check the direction of electric field at a point and then trace the graph 
E
x 
a
 
 
a
 
x
 
 
E
x 
a
 
 
a
 
 
 
 
E
x
a
 
 
a
 
 
 
x
 
 
a
 
 
a
 
 
 
E
x 
 5 
 
16. The dielectric constant K of an insulator cannot be  
(A) 4   (B) 6   (C) ?   (D) 9  
Ans (C) 
K can have all values lying between zero (for perfect insulator) to infinity (for perfect conductor) option 
(C) is correct. 
17. A force F acts between sodium and chlorine ions of salt [sodium chloride] when put 1 cm apart in air. 
The permittivity of air and dielectric constant of water are ?
0
 and K, respectively. When a piece of salt is 
put in water, electrical force acting between sodium and chlorine ions 1 cm apart is  
(A) F   (B) 
0
FK
?
  (C) 
F
K
   (D) 
0
F
K
?
 
Ans (C) 
2
0
e e
F ,in air
4 r
?
?
? ?
 
2
0
e e
F , in medium
4 Kr
?
? ?
? ?
 
F
F
K
? ? 
18. The charge contained in 500 cc of water due to protons will be  
(A) 6.0 ? 10
27
C   (B) 2.67 ? 10
7
 C (C) 6 ? 10
23
 C  (D) 1.67 ? 10
23
 C 
Ans (B) 
No. of H
2
O molecules in 500 cc.  
23 Av
mass N 500
6 10
Mol.wt 18
?
? ? ? ? 
Each molecule contains 10 protons  
23 19 7
10 500
q 6 10 1.6 10 coul 2.67 10 C
18
?
?
? ? ? ? ? ? ? ? 
19. Two identical copper spheres are separated by 1 m in vacuum. The number of electrons that have to be 
removed from one sphere and added to the other so that they attract each other with a force of 0.9 N is  
(A) 6.25 ? 10
15
  (B) 42.5 ? 10
15
  (C) 6.25 ? 10
13
  (D) 42.5 ? 10
13
 
Ans (C)  
1 2
1 2 2
0
1 q q
F , q q ne
4 r
? ? ?
? ?
 
(n ? No. of electrons removed/added)  
20. Two charges 2C and 6C are separated by a finite distance. The force between them is 12 ? 10
3
 N. If a 
charge ?4C is added to each of them, the force now becomes  
(A) 4 ? 10
3
 N repulsion     (B) 4 ? 10
2
 N repulsion  
(C) 6 ? 10
3
 N attraction     (D) 4 ? 10
3
 N attraction 
Ans (D)  
1 2
1 2 1
1 2
F q q
F q q , , q 2C 4C 2C
F q q
? ? ?
? ? ? ? ? ? ? 
      
2
q 6C 4C 2C ? ? ? ? 
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