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# 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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