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Parishram 2026
Physics DPP: 2
Magnetism and Matter

Q1 The SI unit of magnetization (M) is: Q6 According to Curie's law, the magnetic
(A) A/m susceptibility (χ) of a paramagnetic material is:
(B) T (Tesla) (A) Directly proportional to temperature (T)
(C) A·m² (B) Inversely proportional to temperature (T)
(D) Wb/m (C) Independent of temperature
(D) Proportional to
Q2 Magnetic susceptibility (χ) is defined as
(A) The ratio of magnetic field (B) to Q7 A material has a permeability of 0.1 H/m when
magnetization (M) the magnetic intensity is 70 A/m. What will be
(B) The ratio of magnetization (M) to magnetic the magnetic induction inside the material?
field intensity (H) (A) 7 T (B) 0.7 T
(C) The product of magnetization (M) and (C) 70 T (D) 0.07 T
magnetic field intensity (H)
Q8 A material has a magnetization of
(D) The difference between magnetic field (B)
and magnetization (M) when subjected to a
magnetic field intensity of
Q3 The SI unit of magnetic permeability (μ) is
What is the magnetic susceptibility?
(A) H/m (henry per meter)
(A) 1.01 (B) 1.21
(B) T·m/A
(C) 1.41 (D) 1.61
(C) A/m
(D) Wb/m² Q9 A solenoid has a core material with a relative
permeability of 500. If the magnetic field
Q4 If the magnetic susceptibility of a material is χ,
intensity inside the solenoid is 1000 A/m, what
its relative permeability (μᵣ) is:
is the magnetic susceptibility of the core
(A)
material?
(B)
(A) 499 (B) 500
(C)
(C) 501 (D) 1000
(D)
Q10 Which of the following is a characteristic of
Q5 Diamagnetic materials are characterized by:
soft magnetic materials?
(A) Positive susceptibility
(A) High coercivity
(B) Negative susceptibility (B) High retentivity
(C) High permeability
(C) Low hysteresis loss
(D) Permanent magnetic moments (D) Used for making permanent magnet

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Answer Key
Q1 A Q6 B
Q2 B Q7 A

Q3 A Q8 B
Q4 A Q9 A
Q5 B Q10 C

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Hints & Solutions

Note: scan the QR code to watch video solution

Q1 Text Solution:
The SI unit of magnetization (M) is ampere per
meter (A/m). Magnetization is defined as the
magnetic moment per unit volume of a
material. It's a vector quantity, meaning it has
both magnitude and direction

Video Solution: Q4 Text Solution:

Video Solution:

Q2 Text Solution:
Magnetic susceptibility (χ) is a dimensionless
quantity that describes how easily a material
Q5 Text Solution:
can be magnetized when placed in an external
Negative susceptibility
magnetic field. It is defined as the ratio of the
magnetization (M) of the material to the Video Solution:
applied magnetic field intensity

Video Solution:

Q6 Text Solution:

Q3 Text Solution:

Video Solution:

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Video Solution: Q9 Text Solution:

Using the relationship:

Video Solution:

Q7 Text Solution:
Magnetic induction (B) is given by the formula:

Video Solution:
Q10 Text Solution:
Soft magnetic materials, such as soft iron,
have low coercivity and low hysteresis loss,
making them ideal for applications like
transformer cores where rapid magnetization
and demagnetization are required
Q8 Text Solution:
Video Solution:
Magnetic susceptibility is given by:

Video Solution:

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Parishram 2026
PHYSICS DPP: 1
Magnetism and Matter

Q1 A bar magnet of pole strength q and magnetic (D) 2.5 10–5Nm

moment m is divided into two equal pieces by
Q5 A bar magnet of length 10 cm and having the
cutting it perpendicular to its length. Then
(A) q is halved and m is doubled pole strength equal to 10–3 weber is kept in a
(B) q and m both are halved magnetic field having magnetic induction (B)
(C) q is halved but m remains the same equal to 4 10–3 tesla. It makes an angle of
(D) q remains the same but m is halved 30º with the direction of magnetic induction.
The value of the torque acting on the magnet
Q2 The mid points of two small magnetic dipoles
is
of length d in end-on positions, are separated
(A) 2 10–7N-m
by a distance x, (x >> d). The force between
(B) 2 10–5N-m
them is proportional to x–n where n is (C) 0.5 N-m
(D) 0.5 10– N-m =4 10– weber/amp m)
2 (given 7

Q6 An iron bar of length L has magnetic moment

(A) 1 (B) 2
M. It is bent at the middle of its length such
(C) 3 (D) 4
that the two arms make an angle with
Q3 A bar magnet of magnetic moment 2.0 JT–1 lies each other. The magnetic moment of this new

aligned with the direction of a uniform magnet is:

magnetic field of 0.25 T. What is the amount of (A) M (B)
work required to turn the magnet so as to (C) 2M (D)
align its magnetic moment normal to the field
direction? Q7 A bar magnet of magnetic dipole moment 8
(A) 0.125 J (B) 0.25 J
Am2 has poles separated by 0.4 m. Find the
(C) 0.5 J (D) 1.0 J
pole strength of bar magnet.
Q4 The magnetic moment of a magnet is 0.1 amp (A) 20 A-m
(B) 40 A-m
m2. It is suspended in a magnetic field of
(C) 80 A-m
intensity 3 10–4 weber/m2. The torque acting
(D) 100 A-m
upon it when deflected by 30º from the
magnetic field is Q8 Two magnets of equal magnetic moments M
(A) 1 10–5Nm each are placed as shown in figure. The
(B) 1.5 10–5Nm resultant magnetic moment is M. The value of
(C) 2 10–5Nm θ is:

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(A) θ = 60º
(B) θ = 90º
(C) θ = 45º
(D) θ = 30º

Q9 A bar magnet of length ‘l’ and magnetic dipole

moment ‘M’ is bent in the form of an arc as
shown in figure. The new magnetic dipole
moment will be:

(A) (B)

(C) M (D)

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Answer Key
Q1 D Q6 B

Q2 D Q7 A

Q3 C Q8 A

Q4 B Q9 D

Q5 A

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Hints & Solutions

Note: scan the QR code to watch video solution

Q1 Text Solution:
Correct choice: D — pole strength q stays Initial :
the same, magnetic moment m becomes Final :
m/2.
Why (one-line proof):
Magnetic moment of a bar magnet is
where is the magnetic length.
Cutting perpendicular to the length leaves Video Solution:
the pole strength unchanged but halves the
length ( ), so each half gets
. NCERT’s worked Example 5.1
discusses this very division of a magnet .

Video Solution:

Q4 Text Solution:
Torque on a magnetic dipole:
(NCERT)

Q2 Text Solution:
For x ≫ d each magnet behaves like a point Video Solution:
dipole (moment m).
Axial field of dipole 1 at distance x:

Force on dipole 2 (aligned “end-on”):

.
Q5 Text Solution:
Hence , so .
Torque,
Video Solution:

Video Solution:

Q3 Text Solution:
Work needed = increase in magnetic potential
energy

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Q8 Text Solution:
(1)

Q6 Text Solution: [New NCERT Class 12th Page No. 139, 140]
Think of the two equal halves as dipoles of
Video Solution:
moment each.
After bending, the angle between their axes is
(each arm makes with the other), so
the resultant moment is

Q9 Text Solution:
(4)
Let m be strength of each pole of bar magnet

Hence the new magnet’s magnetic moment is of length l. Then M = m × l --- (i)

one-half of the original value. When the bar magnet is bent in the from of an
arc as shown in the figure
Video Solution:

Q7 Text Solution:
(1)
New magnetic dipole moment
M’ = m × 2r sin 30º

M = ml
Video Solution:

[New NCERT Class 12th Page No. 138]

Video Solution:

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PARISHRAM 2026
PHYSICS
Assignment : 1
Moving Charges and Magnetism

Q1 Q4 Find the magnitude and direction of magnetic

The magnetic field to a small current
field B, due to element of a wire lenght of 2cm
element at a distance and element having current of 2A in north direction at a
carrying current i is point 4m from the element along east.
(A) (A)
(B)
(B) (C)
(D)

(C) Q5 For having maximum magnetic field due to a

small element of current carrying conductor at
(D) a point, the angle between the element and
the line joining the element to the given point
must be
Q2 Unit of magnetic permeability is (A) 0º (B) 90º
(A) Amp/metre (C) 180º (D) 45º
(B) Amp/
(C) Henry Q6 Two long parallel wires P and Q are both
(D) Henry/metre perpendicular to the plane of the paper with
distance 5m between them. If P and Q carry
Q3 Which of the following is not correct about the current of 2.5 A and 5A, respectively, in the
Biot-Savart law? same direction, then the magnetic field at a
(A) The magnitude of the magnetic field is point half way between the wires is.
directly proportional to the current through (A)
the conductor
(B) The magnitude of the magnetic field is (B)
directly proportional to the length of the (C)
current element.
(D)
(C) The magnetic of the magnetic field is
inversely proportional to the square of the
Q7 The position of point from wire 'B', where ne
distance from the current element.
magnetic field is zero due to following current
(D) The magnetic of the magnetic field is
distribution.
directly proportional to the square of the
distance from the current element.

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Q11 A test charge is moving with

velocity in a

magnetic field .

The magnetic force on the test charge:-

(A)
(A) 4 cm (B) 2 cm
(B)
(C) 8 cm (D) 12 cm
(C)
Q8 A wire is parallel to one arm of a square
current carrying loop, which also carries (D) Zero

current. Now at any point A within the coil the Q12 A charge with and mass
magnetic field will be:
moving with a velocity of along x –
axis. A uniform static magnetic field of 0.5T is
acting along the y - axis. The mangetic force
(magnitude and direction) on charge:-
(A) Zero
(B) , along z-axis
(A) less than the magnetic field produced due (C) , along x - axis
to loop only. (D) , along y-axis
(B) more than the magnetic field produced due
Q13 An electron is moving along +x direction. To
to loop only.
get it moving along an anticlockwise circular
(C) equal to the earlier.
path in x-y plane, magnetic field applied along
(D) zero
:-
Q9 Current of 0.1 A circulates around a coil of 100 (A) + y – direction
turns and having radius equal to 5cm. The (B) + z – direction
magnetic field sets at the centre of the coil is (C) – y – direction
(A) (D) – z – direction
(B)
Q14 If cathode rays are projected at right angle to
(C)
magnetic field, their trajectory is
(D) (A) Ellipse

Q10 The magnetic field at the centre of a circular (B) Circle

(C) Parabola
coil of radius r carrying current is . The field
(D) None of these
at the centre of another coil of radius

carrying same current is then ratio of Q15 A wire of length 5 cm. is placed inside the
solenoid near its centre such that it makes an
is:-
angle of 30º with the axis of solenoid. The wire
(A) 1 : 2 (B) 2 : 1
carries a current of 5A and the magnetic field
(C) 1 : 1 (D) 4 : 1
due to solenoid is 2.5 × T. Magnetic force
on the wire is :-

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(A)
(B)
(D)
(C)
(D)
Q19 Two parallel wires are carrying electric
Q16 A wire 'LN' bent as shown in figure is placed in currents of equal magnitude and in the same
uniform perpendicular magnetic field of 5T. A direction. They exert
10A current flows through the wire. Magnetic (A) An attractive force on each other
force experienced by the wire is :- (B) A repulsive force on each other
(C) No force on each other
(D) A rotational torque on each other

Q20 Assertion: Torque on a current carrying coil is

maximum, when coil is suspended in a radial
magnetic field, such that the plane of the coil is
parallel to the external magnetic field
Reason: Torque depends upon the magnitude
of the applied magnetic field, and the
(A) 5N (B) 10N orientation of the magnetic field, with the
(C) 2.5N (D) 1.25N plane of the coil
(A) Both assertion and reason are correct and
Q17 An arbitrary shaped closed coil is made of a
reason is the correct explanation for
wire of length L and a current I ampere is
assertion.
flowing in it. If the plane of the coil is
(B) Both assertion and reason are correct but
perpendicular to magnetic field , the force reason is not the correct explanation for
on the coil is assertion.
(A) Zero (B) IBL (C) Both assertion and reason are incorrect.
(C) 2IBL (D) (D) Only assertion is correct but reason is
incorrect.
Q18
A uniform magnetic field Q21 Two parallel long wires carry current and
exists in region of space. A semi-circular wire
with . When the currents are in the
carrying a current of 1A and of radius 1m in x-y
same direction, the magnetic field midway
plane as shown in figure. The force on semi-
between the wires is 10 . When the
circular wire will be
direction of is reversed, it becomes .
The ratio is
(A) (B)
(C) (D)

Q22 Two particles and having equal charges

(A) after being accelerated through the same
potential difference, enter a region of uniform
(B)
magnetic field and describe circular path of
(C)

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radius and respectively. The ratio of . What is the net current

mass of to that of is in the conductors?
(A) (A)
(B)
(B) (C)

(C) (D)

(D)

Q23 A current flows along the length of a long thin

cylindrical shell of radius . Match the fields in
different regions given in column-I to their
respective matches in column-II.

(A) A - Q B - P C - S, T D - Q
(B) A - Q B - P, R C - S, T D - Q
(C) A - Q B - P C - S, T D - R
(D) A - Q B - P C - S, T D - Q, R

Q24 A neutron, a proton, an electron and an -

particle enter a region of uniform magnetic
field with the same velocities. The magnetic
field is perpendicular and directed into the
plane of the paper. The tracks of the particles
are labeled in the figure. The electron follows
the track

(A) (B)
(C) (D)

Q25 A closed curve encircles several conductors.

The line integral around this curve is

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Answer Key
Q1 D Q14 B

Q2 D Q15 B
Q3 D Q16 A

Q4 B Q17 A

Q5 B Q18 B
Q6 D Q19 A

Q7 A Q20 A
Q8 B Q21 D

Q9 A Q22 C

Q10 A Q23 A
Q11 D Q24 A

Q12 B Q25 C

Q13 B

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Hints & Solutions

Note: scan the QR code to watch video solution

Q1 Text Solution:

Video Solution:

Q4 Text Solution:
I = 2A
dl = 2cm = 0.02m
r = 4m

Q2 Text Solution:

or

Video Solution:

Video Solution:

Q3 Text Solution:
According to Bio-Savart Law, the magnitude of
the magnetic field is
(i) directly proportional to the current through Q5 Text Solution:
the conductor i.e. dB I The magnetic field due to small element of
(ii) directly proportional to the length of the conductor of lenght is given by
current element i.e. dB dl
(iii) directly proportional sin
The value will be maximum when
(iv) inversely proportional to the square of the
sin = 1 sin 90° or = 90º
distance from the current element i.e. dB 1/
Video Solution:

Video Solution:

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Q6 Text Solution: At point A the direction of magnetic field due

In the following figure, magnetic field at mid to wire and due to squaqre arm is same.
point m is given by

Video Solution:

Q9 Text Solution:

Video Solution:
Video Solution:

Q7 Text Solution:
Q10 Text Solution:
Current and number of turns are equal for
both coils, so

When current is in opposite direction the

magnetic field will not zero between wires A
Video Solution:
and B. Suppose it is zero at point 'P' the

Video Solution:
Q11 Text Solution:
Mangetic force on moving charge

Here:-

Q8 Text Solution:

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Hence magnetic field direction along +z-

Since º
direction.

Video Solution:

Video Solution:
Q12 Text Solution:
Magnetic force on moving charge

Q14 Text Solution:

Since will always be perpendicular to
velocity hence cathode rays will follow circular
path.

Video Solution:

Q15 Text Solution:

Video Solution: Magnetic field produced by solenoid along its
axis so magnetic force on conductor given as

Video Solution:
Q13 Text Solution:
In order to keep an electron moving along
anticlockwise circular path, it should
experience a centripetal magnetic force and at
this instant force should be along +y direction.
Now direction of external magnetic field can
be find out by right hand palm rule. Q16 Text Solution:

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and both are perpendicular to each

other, so magnetic force on wire 'LN' given as

º º

Video Solution:

Video Solution:

Q17 Text Solution:

As shown in figure, since

Q19 Text Solution:
Two straight conductors carry current in same
direction, then attractive force acts between
them.

Video Solution:

Video Solution:

Q20 Text Solution:

Assertion – correct. In a radial magnetic

field the field lines are everywhere
parallel to the plane of the coil, so the
Q18 Text Solution:
angle between the field B and the coil’s
Force on semi-circular wire will be same as the
normal is ; the torque
force on straight wire AC.
is therefore at its

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maximum value ( ) Q22 Text Solution:

throughout the motion. Result
Reason – correct and explains the
assertion. The expression above shows
that depends both on the field 2-line derivation
magnitude and on the orientation (the
sine factor); recognising this explains why 1. After the same accelerating potential :
the parallel-plane orientation gives .
maximum torque. 2. In a transverse -field the circular radius
is (NCERT XII, Eq. 4.5)
Video Solution:
.

Hence .

Video Solution:

Q21 Text Solution:

Let the magnetic field constant at the midpoint

be .

Then . Q23 Text Solution:

(A)
Same‐direction currents ⇒ fields oppose:

Reverse ⇒ fields add:

Divide (1) by (2):

Video Solution:

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Only right-curving path is labelled D,

hence the electron follows D.
B=

(C) B = (The straight path C is the neutral neutron; the

two left curves A, B are for the proton and α-
particle.)

Video Solution:

(D)

Q25 Text Solution:

Using Ampère’s circuital law

Bcentre = 0

Video Solution:

Closest choice ⇒ 0.3 A

Q24 Text Solution:
Reasoning (very short): Video Solution:

Force: .

is upward; is into the page.

Positive charges (p, α) deflect left.
Negative charge (e⁻) deflects right.

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Parishram 2026
DPP: 5
Physics
Moving Charges and Magnetism

Q1 A current carrying loop is placed in a uniform (B)

magnetic field. The torque acting on it does (C)
not depend upon the: (D)
(A) Area of loop
(B) Number of turns Q5 Current is carried in a wire of length . If the
(C) Shape of loop wire is turned into a circular coil, the
(D) Angle between normal of coil and magnetic maximum magnitude of torque in a given
field magnetic field will be
(A)
Q2 If number of turns, area and current through a
(B)
coil are given by and respectively, then
its magnetic moment is given by : (C)
(A)
(D)
(B)
(C)
Q6 A circular coil of 20 turns and radius is
(D)
placed in uniform magnetic field of
normal to the plane of the coil. If the current in
Q3 A current carrying loop is free to turn in a
coil is , then the torque acting on the coil
uniform magnetic field. The loop will then
will be
come into equilibrium when its plane is
(A)
inclined at
(B)
(A) to the direction of the field
(C)
(B) to the direction of the field
(D) Zero
(C) to the direction of the field
(D) to the direction of the field Q7 A 100 turns coil shown in figure carries a
current of in a magnetic field
Q4 A closely wound solenoid of 2000 tums and
. The torque acting on the
area of cross-section carries a
coil is
current of 2.0 A. It is suspended through its
centre and perpendicular to its length,
allowing it to turn in a horizontal plane in a
uniform magnetic field tesla making
an angle of with the axis of the solenoid.
The torque on the solenoid will be
(A)

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(A) tending to rotate out of the

the side page
(B) -m tending to rotate into the
the side page
(C) tending to rotate out of the
the side page
(A) (B)
(D) tending to rotate into the
(C) (D)
the side page

Q8 A rectangular coil, of sides and

respectively, has 10 turns in it. It carries a
current of , and is placed in a uniform
magnetic field of in such a manner that
its plane makes an angle with the field
direction. The torque on the loop is
(A)
(B)
(C)
(D)

Q9 Two circular concentric loops of radii

and are placed in
the plane as shown in the figure. A
current is flowing through them.
The magnetic moment of this loop system is

(A)

(B)

(C)

(D)

Q10 The two parts of the loop are circles of radii

and respectively and carrying same current
as shown in the given figure. What is the
magnitude of dipole moment of the current
loop ?

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Answer Key
Q1 C Q6 D

Q2 A Q7 A
Q3 C Q8 A
Q4 C Q9 C
Q5 C Q10 C

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Hints & Solutions

Note: scan the QR code to watch video solution

Q1 Text Solution: Video Solution:

Torque on a current-carrying loop in a uniform
field is

so it depends on number of turns , current

, loop area , magnetic field and the angle
Q4 Text Solution:
between the field and the loop’s normal—but
Magnetic moment of the solenoid
only the area enters, not the specific
geometric shape (square, circular, etc.).

Video Solution:

Torque in a uniform field

at angle :

Q2 Text Solution:
For a planar coil, magnetic dipole moment is
, where Video Solution:
= number of turns, = current, and
= area of one turn.
Hence .

Video Solution:

Q5 Text Solution:
Turning the wire of length into a single
circular loop gives

Q3 Text Solution:
The loop’s magnetic moment μ is
perpendicular to its plane. Magnetic moment of the loop:
Torque is τ = μ × B = μB sin θ, where θ is the
.
angle between μ and B.
Equilibrium requires τ = 0 ⇒ sin θ = 0 ⇒ θ = 0° Maximum torque in the uniform field occurs
or 180° (μ parallel or antiparallel to B). when the loop’s plane is parallel to :
Thus the plane—being perpendicular to μ—
must make 90° with the field direction.

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Using the right-hand rule for the given current

directions, the torque tries to turn the coil so
that its left side AD comes out of the page
Video Solution: (right side BC goes into the page), aligning the
coil’s normal with the field.

Video Solution:

Q6 Text Solution:
Torque on a current loop is
Q8 Text Solution:

where is the angle between the coil’s normal

and the magnetic field.
Here the field is normal to the plane of the
coil, so the normal vector is parallel to ⇒
and . Plane makes with ⇒ angle between
Hence N·m regardless of the coil’s and coil’s normal is
dimensions or current. ⇒
Video Solution:

Video Solution:

Q7 Text Solution:

Area of the rectangular coil

.
Q9 Text Solution:
Magnetic moment The outer loop and the inner loop carry the
. same current but in opposite senses (see
figure), so the net magnetic dipole moment is
the difference of their moments:
Field makes with the coil’s normal (
), so

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Direction: out of the -plane ( ).

Video Solution:

Video Solution:

Q10 Text Solution:

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Parishram 2026
Physics DPP: 4

Moving Charges and Magnetism

Q1 A wire of length 5 cm. is placed inside the Q4

A uniform magnetic field
solenoid near its centre such that it makes an
exists in region of space. A semi-circular wire
angle of 30º with the axis of solenoid. The wire
carrying a current of 1A and of radius 1m in x-y
carries a current of 5A and the magnetic field
plane as shown in figure. The force on semi-
due to solenoid is 2.5 × T. Magnetic force
circular wire will be
on the wire is :-
(A)
(B)
(C)
(D)

(A)
Q2 A wire 'LN' bent as shown in figure is placed in
uniform perpendicular magnetic field of 5T. A
(B)
10A current flows through the wire. Magnetic
force experienced by the wire is :- (C)

(D)

Q5 Two parallel wires are carrying electric

currents of equal magnitude and in the same
direction. They exert
(A) An attractive force on each other
(B) A repulsive force on each other
(C) No force on each other
(A) 5N (B) 10N
(D) A rotational torque on each other
(C) 2.5N (D) 1.25N
Q6 An infinitely long, straight conductor AB is
Q3 An arbitrary shaped closed coil is made of a
fixed and a current is passed through it.
wire of length L and a current I ampere is
Another movable straight wire CD of finite
flowing in it. If the plane of the coil is
length and carrying current is held
perpendicular to magnetic field , the force perpendicular to it and released. Neglect
on the coil is weight of the wire.
(A) Zero (B) IBL
(C) 2IBL (D)

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(A) The rod CD will move upwards parallel to (A) , Attraction

itself (B) , Repulsion
(B) The rod CD will move downwards parallel (C) , Repulsion
to itself (D) , Attraction
(C) The rod CD will move upward and turn
clockwise at the same time
(D) The rod CD will move upward and turn
anti-clockwise at the same time ANS

Q7 Two long parallel wires carrying equal current

separated by 1m, exert a force of 2 × N/m
one one another. The current flowing through
them is
(A) 2.0A
(B) 2.0×
(C) 1.0 A
(D) 1.0 ×

Q8 Three long straight wires, carrying currents are

arranged according to figure. Magnetic force
on 10cm part of the wire Q is:-

(A) , towards right

(B) , towards right
(C) , towards left
(D) , towards left

Q9 A rectangular loop ABCD is placed near on

infinite length current carrying wire. Magnetic
force on the loop is:-

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Answer Key
Q1 B Q6 C
Q2 A Q7 C
Q3 A Q8 B
Q4 B Q9 A
Q5 A

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Hints & Solutions

Note: scan the QR code to watch video solution

Q1 Text Solution:
Magnetic field produced by solenoid along its
axis so magnetic force on conductor given as

Video Solution:

Video Solution:

Q2 Text Solution:

and both are perpendicular to each Q4 Text Solution:

other, so magnetic force on wire 'LN' given as Force on semi-circular wire will be same as the
force on straight wire AC.

º º

Video Solution:

Video Solution:
Q3 Text Solution:

As shown in figure, since

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Q5 Text Solution: Q8 Text Solution:

Two straight conductors carry current in same Resultant force on 'l' lenght of wire Q
direction, then attractive force acts between
them.

Video Solution:

Video Solution:

Q6 Text Solution:
Since the force on the rod CD is non-uniform,
it will experience force and torque. From the Q9 Text Solution:
left hand side, it can be seen that the force will Force on 10cm length of wire AD :-
be upward and torque is clockwise.

...(i)
Video Solution:
...(ii)
Wire AB and DC located in non uniform
magnetic field
So force on elemental length of these wire will
be of same magnitude and opposite direction
therefore net force on wire AB and DC will be
Q7 Text Solution:
zero (always)

Video Solution:

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Video Solution:

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PARISHRAM 2026
DPP: 3
PHYSICS
Moving Charges and Magnetism

Q1 A proton and an -particle move in circular (A) (B)

paths in a transverse magnetic field. If speed (C) (D)
of proton is two times that of -particle, the
Q5 A neutron, a proton, an electron and an -
ratio of their periods of revolution is
particle enter a region of uniform magnetic
(A) 1:1 (B) 1:2
field with the same velocities. The magnetic
(C) 2:1 (D) 4:1
field is perpendicular and directed into the
Q2 An electron having mass , plane of the paper. The tracks of the particles
are labeled in the figure. The electron follows
charge and moving with a
the track
velocity of enters a region where
magnetic field exists. If it describes a circle of
radius , the intensity of magnetic field
must be
(A)
(B) (A) (B)
(C) (C) (D)
(D)
Q6 A charged particle is moving in a circular orbit
Q3 A uniform magnetic field acts at right angles to of radius with a uniform speed of
the direction of motion of electrons. As a
under the action of a uniform
result, the electron moves in a circular path of
magnetic field at right
radius . If the speed of the electrons is
angles to the plane of the orbit. The charge to
doubled, then the radius of the circular path
mass ratio of the particle is
will be
(A)
(A)
(B) (B)
(C) (C)
(D) (D)

Q4 Two ions having masses in the ratio and

Q7 An electron and a proton are moving under
charges are projected into uniform
the influence of mutual forces. In calculating
magnetic field perpendicular to the field with the change in the kinetic energy of the system
speeds in the ratio . The ratio of the radii during motion, one ignores the magnetic force
of circular paths along which the two particles of one on another. This is because,
move is

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(A) The two magnetic forces arc equal and some instant makes an acute angle with the
opposite, so they produce no net effect magnetic field. The path of the particle will be:
(B) The magnetic forces do no work on each (A) A straight line
particle (B) A circle
(C) The magnetic forces do equal and opposite (C) A helix with uniform pitch
(but non zero) work on each particle (D) A helix with non uniform pitch
(D) The magnetic forces are necessarily
Q10 In the given figure, the electron enters into the
negligible
magnetic field. It deflects in ...... direction:
Q8 When a charged particle enters a uniform
magnetic field?
(A) Speed of particle must increase
(B) Speed of particle must decrease
(C) Speed of particle will be constant
(D) Speed of particle may be decrease or (A) + ve X
increase (B) – ve X
(C) + ve Y
Q9 A charged particle moves in a uniform
(D) – ve Y
magnetic field. The velocity of the particle at

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Answer Key
Q1 B Q6 B

Q2 B Q7 B
Q3 C Q8 C
Q4 A Q9 C
Q5 A Q10 D

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Hints & Solutions

Note: scan the QR code to watch video solution

Q1 Text Solution: Substitute

Key fact (cyclotron frequency)
For any charged particle moving
perpendicularly to a uniform magnetic field ,

Particle

Proton ( )

α-particle (
Video Solution:
)

Q3 Text Solution:
(The given speed ratio is irrelevant since
For motion perpendicular to a uniform -field,
depends only on .)

Video Solution:

Doubling doubles : .

Video Solution:
Q2 Text Solution:
Formula (uniform circular motion in -
field)

Q4 Text Solution:
For motion perpendicular to a uniform -field,

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Hence

Given

So the radii are in the ratio . ,

Video Solution: ,
:

Q5 Text Solution: Video Solution:

Reasoning (very short):

Force: .
is upward; is into the page.
Positive charges (p, α) deflect left.
Negative charge (e⁻) deflects right.
Only right-curving path is labelled D, Q7 Text Solution:
hence the electron follows D. The magnetic part of the Lorentz force

is always perpendicular to the

(The straight path C is the neutral neutron; the
two left curves A, B are for the proton and α- velocity, so ; hence it cannot
particle.) change any particle’s kinetic energy, so we may
ignore it when calculating of the electron–
Video Solution:
proton system.

Video Solution:

Q6 Text Solution:
For a charged particle in uniform with
velocity ⟂ : Q8 Text Solution:
(C)
Since F is always perpendicular to velocity.
Hence speed remains constant.
[New NCERT Class 12th Page No. 138]

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Video Solution: Q10 Text Solution:

(D)
By Fleming's left hand rule.
[NCERT Class 12th, Page No. 137]

Video Solution:

Q9 Text Solution:
(3)
Conceptual question.

Video Solution:

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Parishram 2026
DPP: 2
PHYSICS
Moving Charges and Magnetism

Q1 (C)
Rank the value of for the closed
paths shown in figure from the smallest to (D)
largest.
Q5 If a long hollow copper pipe carries a direct
current, the magnetic field associated with the
current will be
(A) Only inside the pipe
(B) Only outside the pipe
(C) Both outside and outside the pipe
(A) a,b,c,d (B) a,c,d,b
(D) Neither inside nor outside the pipe
(C) a,b,c,d (D) a,c,b,d
Q6 Field inside a solenoid is
Q2 A current of 1/(4 ) ampere is flowing in a long
(A) Directly proportional to its length
straight conductor. The line integral of
(B) Directly proportional to current
magnetic induction around a closed path (C) Inversely proportional to toal number of
enclosing the current carrying conductor is. turns
(A) weber per metre
(D) Inversely proportional to current
(B) weber per metre
(C) weber per metre Q7 The expression for magnetic induction inside a
(D) zero solenoid of length L carrying current I and
having N number of turns is
Q3 A solid cylindrical wire of radius 'R' carries a (A)
current 'I'. The ratio of magnetic fields at
(B)
points which are located at R/2 and 2R
(C)
distance away from the axis of the wire:-
(A) 1 : 1 (B) 1 : 2 (D)
(C) 2 : 1 (D) 1 : 4
Q8 There are 50 turns of a wire in every cm length
Q4 A hollow cylindrical wire carries a current I, of a long solenoid. If 4 ampere current is
having inner and outer radii R and 2R flowing in the solenoid, the approximate value
respectively. Magnetic field at a point which of magnetic field along its axis at an internal
3R/2 distance away from its axis is:- point and at one end will be respectively.
(A) (A)
(B) (B)
(C)

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(D)

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Answer Key
Q1 B Q6 B
Q2 A Q7 D
Q3 A Q8 C
Q4 C
Q5 B

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Hints & Solutions

Note: scan the QR code to watch video solution

Q1 Text Solution:

We shall use where I is the Video Solution:

current enclosed by loop.
Net current enclosed by path a is zero.
Net current enclosed by path b is 5A.
Net current enclosed by path c is 1A.
Net current enclosed by path d is 3A.
Hence the order will be a < c < d < b

Video Solution: Q4 Text Solution:

Field inside cross-section of conductor

[Here, a = R, b =

2R, r 3R/2]

Q2 Text Solution:

Video Solution:

Video Solution:

Q3 Text Solution:
Field inside the conductor:-

Q5 Text Solution:
....(1) According to Ampere's law
Field outside the conductor:-

only for outside point, of line integration will

enclose the current, Hence, B = 0 inside the
....(2)
pipe .

Video Solution:
From (1) and (2)

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Video Solution:

Q6 Text Solution:

Video Solution:

Q7 Text Solution:
Video Solution:

where, number of turns per unit length n =

Video Solution:

Q8 Text Solution:
The magnetic field in the solenoid along its
axis
(i) At an internal point

(Here n = 50 truns/cm = 5000 turns/m)

(ii) At one end

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Parishram 2026
DPP: 1
PHYSICS
Moving Charges and Magnetism

Q1 Magnetic effect of current was discovered by (B) The magnitude of the magnetic field is
(A) Faraday (B) Oersted directly proportional to the length of the
(C) Ampere (D) Bohr current element.
(C) The magnetic of the magnetic field is
Q2 Magnetic field cannot be produced by
inversely proportional to the square of the
(A) Charge in uniform motion
distance from the current element.
(B) Charge at rest
(D) The magnetic of the magnetic field is
(C) Magnet
directly proportional to the square of the
(D) Current Carrying wire
distance from the current element.
Q3
The magnetic field to a small current Q6 Find the magnitude and direction of magnetic
field B, due to element of a wire lenght of 2cm
element at a distance and element having current of 2A in north direction at a
carrying current i is point 4m from the element along east.
(A) (A)
(B)
(B) (C)
(D)
(C)
Q7 For having maximum magnetic field due to a
small element of current carrying conductor at
(D)
a point, the angle between the element and
the line joining the element to the given point
must be
Q4 Unit of magnetic permeability is
(A) 0º (B) 90º
(A) Amp/metre
(C) 180º (D) 45º
(B) Amp/
(C) Henry Q8
(D) Henry/metre The magnetic field d in terms of the current

Q5 Which of the following is not correct about the density of a conducting wire is related as
Biot-Savart law? (A)
(A) The magnitude of the magnetic field is
directly proportional to the current through (B)
the conductor
(C)

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(D)

Q9 Find the direction of magnetic field at point P

and Q, if wire and point in x – y plane

(A) (B)
(C) (D)

Q10 The dimension of the magnetic field intensity

is
(A)
(B)
(C)
(D)

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Answer Key
Q1 B Q6 B

Q2 B Q7 B
Q3 D Q8 C
Q4 D Q9 D
Q5 D Q10 B

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Hints & Solutions

Note: scan the QR code to watch video solution

Q1 Text Solution:
Magnetic effect of current was discovered by
Oersted.

Video Solution:

Q5 Text Solution:
According to Bio-Savart Law, the magnitude of
the magnetic field is
(i) directly proportional to the current through

Q2 Text Solution: the conductor i.e. dB I

When a charged particle is at rest, it does not (ii) directly proportional to the length of the

produce a magnetic field. current element i.e. dB dl

(iii) directly proportional sin
Video Solution:
(iv) inversely proportional to the square of the
distance from the current element i.e. dB 1/

Video Solution:

Q3 Text Solution:

Video Solution: Q6 Text Solution:

I = 2A
dl = 2cm = 0.02m
r = 4m

Q4 Text Solution:

or

Video Solution:
Video Solution:

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direction of current. So, the magnetic field at

point P would be out from the plane of paper.
and magnetic field at a point Q would be into
the plane of paper

Q7 Text Solution:
The magnetic field due to small element of
conductor of lenght is given by

The value will be maximum when

sin = 1 sin 90° or = 90º

Video Solution:
Video Solution:

Q8 Text Solution:
We know tha current density Q10 Text Solution:
F = BIi
...(i)

Magnetic field The dimensional formula for force is [MLT–2]

Substituting (i) in (ii) The dimensional formula for current is [A]

The dimensional formula for length is [L]
Hence, the dimensional formula for B is

Therefore, we can represent the dimensional

Video Solution:
formula of magnetic field intensity is
[ML0T−2A−1]

Video Solution:

Q9 Text Solution:
By applying the Right Hand thumb rule, the
direction of the magnetic field would be
clockwise as shown in diagram around the

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PARISHRAM 2026
PHYSICS DPP: 1

Electromagnetic Waves

Q1 Which of the following is the correct Q6 A capacitor with capacitance C = 2 pF has a

expression for Ampère's Circuital Law? potential difference changing at a rate of
(A)
=1012 V/s. The displacement current is:
(B) (A) 2A (B) 3A
(C) 6A (D) 9A
(C)
Q7 Which of the following statement is correct
(D)
regarding Ampere's circuital law?
I. The law can be applied inside a conductor
Q2 In a long straight solenoid, the magnetic field or outside it but total current enclosed by the
inside is path should be known
(A) Zero II. The law is applicable for line currents,
(B) Uniform sheet currents or volume currents
(C) Varies with distance from the axis (A) Only II
(D) Varies with the square of the distance from (B) Neither I nor II
the axis (C) Both I and II
(D) Only I
Q3 A long straight wire of radius a carries a steady
current I uniformly distributed over its cross- Q8 According to Ampere's circuital Law, The Line
section. The magnetic field at a distance r from Integral of H about any closed path is
the center (where r < a) is: exactly_____to the direct current enclosed by
(A) that path.

(B) (A) equal (B) 4 times

(C) double (D) half
(C)
Q9 A long wire carrying a certain current produces
(D) Zero a magnetic field of 0.8 Tesla at a distance 0.5
cm. Then magnetic field at a distance of 1 cm
Q4 The concept of displacement current was
is:
introduced by:
(A) 0.16 Tesla (B) 0.2 Tesla
(A) Newton (B) Ampère
(C) 0.8 Tesla (D) 0.4 Tesla
(C) Maxwell (D) Faraday
Q10 A current of 5 A passes along the axis of a
Q5 The SI unit of displacement current is:
cylinder of 5 cm radius. The flux density at the
(A) Henry (B) Coulomb
surface of the cylinder is
(C) Ampere (D) Farad
(A) 2 μT (B) 20 μT
(C) 200 μT (D) 2000 μT

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Answer Key
Q1 A Q6 A

Q2 B Q7 C
Q3 B Q8 A
Q4 C Q9 D
Q5 C Q10 B

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Hints & Solutions

Note: scan the QR code to watch video solution

Q1 Text Solution: accounting for time-varying electric fields

Ampere's Circuital Law states that the line Video Solution:
integral of the magnetic field around a
closed loop is equal to μ0 times the total
current I enclosed by the loop.

Video Solution:

Q5 Text Solution:
Displacement current has the same unit as
conduction current, which is Ampere (A).

Video Solution:

Q2 Text Solution:
Inside a long straight solenoid, the magnetic
field is uniform and parallel to the axis of the
solenoid

Video Solution:
Q6 Text Solution:
Displacement current Id = C = 2 × 10–12 X

1012 = 2 A.

Video Solution:

Q3 Text Solution:
Inside the wire (r < a), the magnetic field
increases linearly with distance r from the
center, given by

B=
Q7 Text Solution:
Video Solution:
• Ampere’s Circuital Law states that the line
integral of the magnetic field around any
closed path is equal to the permeability of free
space times the total current enclosed by the
path.
• The law is mathematically expressed as
, where is the line
Q4 Text Solution:
integral of the magnetic field B around a
James Clerk Maxwell introduced the concept of
closed loop, μ0 is the permeability of free
displacement current to modify Ampère's Law,

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space, and lenc is the total current enclosed by

the loop.
• The law can be applied in various scenarios
including inside a conductor or outside it,
provided the total current enclosed by the
path is known.
o This is crucial for determining the magnetic
field in different regions around current-
carrying conductors. Where dl is a small element, μ0 is the
o Hence, statement I is correct.
permeability of free space and I is the electric
• The law is applicable for different types of
current.
currents such as line currents, sheet currents,
or volume currents.
o Line currents refer to currents flowing Video Solution:
through thin wires or conductors.
o sheet currents are currents distributed over
a surface.
o Volume currents are currents distributed
throughout a volume.
o The versatility of Ampere's Circuital Law in
handling these different types of current Q9 Text Solution:
distributions makes it a powerful tool in Given:
electromagnetism. B1 = 0.8 T, d1 = 0.5 cm
o Hence, statement II is correct. The intensity of the magnetic field due to wire
of infinite length at a distance d from it is given
Video Solution:
by

Where μ0 = permeability of free space, I =

current in a wire, d = distance
As current is constant in the wire, then the
magnetic field varies with the distance 'd' as
Q8 Text Solution:
Ampere's Circuital Law:
• It gives the relationship between the
current and the magnetic field created by it.
• This law says that the integral of magnetic
field density (B) along an imaginary closed
Video Solution:
path is equal to the product of current
enclosed by the path and permeability of the
medium.

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At the surface of the cylinder, r = R, and the

flux density becomes:

With I = 5 A, and R = 5 x 10–2 m, the flux

density at the surface will be:
Q10 Text Solution:
Concept:
B = 20 × 10–6T
The magnetic flux density inside a cylinder
B = 20 μT
with uniform current density is given by:
Video Solution:

r = distance from the center of the cylinder (r <

R)
R = Radius of the cylinder
I = Current flowing across the cylinder
Application:

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PARISHRAM 2026
PHYSICS Assignment : 1
Alternating Current

Q1 'In' an ideal transformer, the turns ratio is (B)

(C)
. The ratio is equal to (the
(D)
symbols carry their usual meaning) :
(A) 1 : 2 (B) 2 : 1 Q5 Assertion: Electrical power through
(C) 1 : 1 (D) 1 : 4 transmission lines is transmitted at high
voltage.
Q2 A 12 V, 60 W lamp is connected to the
Reason: Due to high voltage, high current will
secondary of a step down transformer, whose
be produced and heat loss through wires will
primary is connected to ac mains of 220 V.
be more.
Assuming the transformer to be ideal, what is
(A) Both Assertion and Reason are true and
the current in the primary winding ?
the Reason is the correct explanation of the
(A) 2.7 A (B) 3.7 A
Assertion.
(C) 0.37 A (D) 0.27 A
(B) Both Assertion and Reason are true but the
Q3 In step-up transformer, the turn ratio is . Reason is not the correct explanation of the
A leclanche cell is connected Assertion.

across the primary, the voltage developed in (C) Assertion is true but Reason is false.

the secondary would be: (D) Both Assertion and Reason are false.

(A) Q6 Assertion: The power output of a practical

(B) transformer is always smaller than the power
(C) input.
(D) Zero Reason: A transformer works on the principle

Q4 Match column I with column II, and choose the of mutual induction.

correct combination from the options given (A) Both Assertion and Reason are true and

below: the Reason is the correct explanation of the

Assertion.
(B) Both Assertion and Reason are true but the
Reason is not the correct explanation of the
Assertion.
(C) Assertion is true but Reason is false.
(D) Both Assertion and Reason are false.

Q7 The transformation ratio in the step-up

transformer is
(A)
(A) One

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(B) Greater than one

(C) Less than one
(D) The ratio greater or less than one depends
on the other factor

Q8 The core of any transformer is laminated so as

to:
(A) Make it light weight
(B) Make it robust and strong
(C) Increase the secondary voltage
(D) Reduce the energy loss due to eddy current

Q9 In an ideal transformer, the voltage and the

current in the primary coil are and ,
respectively. If the voltage in the secondary
coil is , then the value of current in the
secondary coil will be:
(A)
(B)
(C)
(D)

Q10 Alternating current is transmitted at far off

places:
(A) at high voltage and low current
(B) at high voltage and high current
(C) at low voltage and low current
(D) at low voltage and high current

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Answer Key
Q1 B Q6 B

Q2 D Q7 B
Q3 D Q8 D
Q4 B Q9 A
Q5 C Q10 A

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Hints & Solutions

Note: scan the QR code to watch video solution

Q1 Text Solution:

For an ideal transformer, Vp/Vs = Np/Ns.

Given Np/Ns = 1/2. The ratio Vs : Vp is
Vs/Vp = Ns/Np = (Np/Ns)⁻¹ = (1/2)⁻¹ = 2/1.
(NEW NCERT 12th Page No. 194, 195,
196) Q4 Text Solution:
Match the columns:
Video Solution:
(A) Peak voltage in a 220 V AC source =
→

(R)
(B) RMS current of constant current of
2.8 A = RMS of DC = 2.8 A → (P)
Q2 Text Solution: (C) Transformers work only in AC
circuits → (Q)
Ps = 60 W (given for the lamp).
(D) Capacitors act as infinite resistance
Pp = Ps = 60 W. in DC circuits → (S)
Vp = 220 V (primary voltage).
Matching:
Pp = Vp * Ip 60 = 220 * Ip.
Ip = 60 / 220 = 6 / 22 = 3 / 11 A ≈ 0.27 A. A–R
B–P
(NEW NCERT 12th Page No. 194) C–Q
D–S
Video Solution:
Video Solution:

Q3 Text Solution:
A step-up transformer requires alternating Q5 Text Solution:
current (AC) to function. A Leclanché cell Assertion is true:
provides direct current (DC). Electrical power is transmitted at high voltage
Transformers do not work with DC — there’s to reduce power loss during transmission.
no changing magnetic flux, so no EMF is Reason is false:
induced in the secondary. At high voltage, for same power , the

Video Solution:

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current decreases, and heat loss is

minimized — not increased.
Laminating the core increases resistance to
Video Solution: eddy current paths, minimizing energy loss.

Video Solution:

Q6 Text Solution:
Assertion is true:
In a practical transformer, some power is Q9 Text Solution:
lost due to heat, eddy currents, hysteresis, Given (ideal transformer):
etc., so output < input.
Reason is true: Primary voltage , current
A transformer does work on the principle of
mutual induction. Secondary voltage
But — mutual induction is not the reason
In an ideal transformer:
for power loss; it's just the working principle.

Video Solution:

Video Solution:

Q7 Text Solution:
In a step-up transformer, the secondary
voltage is greater than the primary voltage.
So:
Q10 Text Solution:
AC is transmitted at high voltage to minimize
current for a given power . Lower
current reduces power loss due to resistance
Video Solution:
in transmission lines, since power loss .

Video Solution:

Q8 Text Solution:
The core of a transformer is laminated to:

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PARISHRAM 2026
PHYSICS DPP: 2
Alternating Current

Q1 Match List-I with List-II (D) Voltage leads the current by 60º
List-I List-II
Q4 For a series RLC circuit R = = . The
impedance of the circuit and phase difference
Purely capacitive
A I. between V and i will be
circuit
(A)

Purely inductive (B)

B II.
circuit
(C)
LCR series at (D)
C III.
resonance
Q5 For the series LCR circuit shown in the figure,
what is the resonance frequency and the
D LCR series circuit IV.
current at the resonating frequency?

Choose the correct answer from the options

given below:
(A) A-IV, B-I, C-II, D-III
(A) 2500 rad s–1and A
(B) A-I, B-IV, C-III, D-II
(B) 2500 rad s–1and 5 A
(C) A-I, B-IV, C-II, D-III
(C) 2500 rad s–1and A
(D) A-IV, B-I, C-III, D-II
(D) 25 rad s–1and A
Q2 A bulb and a capacitor are connected in series
across an AC supply. A dielectric is then placed Q6 In a series circuit R = 300 W, L = 0.9 H, C = 2.0
between the plates of the capacitor. The glow mF and w = 1000 rad/sec. The impedance of
of the bulb: the circuit is:
(A) remains same (B) decrease (A) 1300W (B) 900W
(C) increase (D) becomes zero (C) 500W (D) 400W

Q3 Voltage and current in an ac circuit are given Q7 The reading of the A.C. voltmeter in the
by and network shown in figure is (where V is in volt)

(A) Voltage leads the current by 30º

(B) Current leads the voltage by 30º
(C) Current leads the voltage by 60º

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(A) 400 V (B) 220 V

(C) 200 V (D) Zero

Q8 The phase difference between current and

voltage in an circuit is radian. If the
frequency of is , then the phase
difference is equivalent to the time difference
of:
(A)
(B)
(C)
(D)

Q9 A coil of resistance and inductance

is connected across an alternating voltage of
frequency 300/2 . Calculate the phase
difference between voltage and current in the
circuit.
(A)
(B)
(C)
(D)

Q10 The quality factor of series L-C-R circuit is Q,

then the value capacitance in terms of
resistance R and inductance (L) is:
(A) (B)

(C) (D)

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Answer Key
Q1 C Q6 C

Q2 C Q7 C
Q3 C Q8 C
Q4 B Q9 C
Q5 B Q10 B

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Hints & Solutions

Note: scan the QR code to watch video solution

Q1 Text Solution:
(3)
In purely capacitive, current leads voltage by
90° In purely inductive, current lags voltage by
90°
At resonance, XC = XL
i.e., Z = R Q4 Text Solution:
(2)
Video Solution:

Also

Q2 Text Solution:
(3)
As dielectric is placed, capacitance will [New NCERT Class 12th Page No. 186-189]
increase.
Video Solution:

So capacitive Reactance decreases. So

impedance of circuit decreases. Hence current
increases. So power increases. So glow of the
bulb increases.

Video Solution:
Q5 Text Solution:
(2)
Resonance frequency

rad/sec
Resonance current =
Q3 Text Solution:
(3) Video Solution:
Current leading the voltage by 60°.

Video Solution:

Q6 Text Solution:

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(3)
For series R-L-C circuit,

Q9 Text Solution:
rad/s
[New NCERT Class 12th Page No. 187] Using phasor analysis,

Video Solution:

Q7 Text Solution:
(3)
At resonance

Video Solution:

[NCERT Page No. 246]

Video Solution:

Q10 Text Solution:

(2)
We know that quality factor (Q),
Q8 Text Solution:
(3)
Frequency of
Time [New NCERT Class 12th Page No. 187 to 189]

Video Solution:

Video Solution:

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PARISHRAM 2026
PHYSICS DPP: 1

Alternating Current

Q1 The peak value of an alternating e.m.f. which is reaching from zero to maximum value and the
given by E = cos is 10 volts and its peak value of current will be

frequency is 50 Hz. At time t = s, the (A) and 14.14 amp

(B) and 7.07 amp
instantaneous e.m.f. is
(A) 10V (B) 5 (C) and 7.07 amp
(D) and 14.14 amp
(C) 5V (D) 1V
Q7 The frequency of an alternating voltage is 50
Q2 A current in circuit is given by i = 3 + 4 sin .
cycles/sec and its amplitude is 120V. Then the
Then the effective value of current is:
r.m.s. value of voltage is
(A) 5 (B)
(A) 101.3V (B) 84.8V
(C) (D) (C) 70.7V (D) 56.5V

Q3 The peak value of 220 volts of ac mains is Q8 The maximum value of a.c. voltage in a circuit
(A) 155.6 volts is 707V. Its rms value is approximately:-
(B) 220.0 volts (A) 70.7V (B) 100V
(C) 311.0 volts (C) 500V (D) 707V
(D) 440 volts
Q9 The r.m.s. voltage of domestic electricity
Q4 If an A.C. main supply is given to be 220 V. supply is 220 volt. Electrical appliances should
What would be the average e.m.f. during a be designed to withstand an instantaneous
positive half cycle:- voltage of
(A) 198 V (A) 220V (B) 310V
(B) 386 V (C) 330V (D) 440V
(C) 256 V
(D) None of these Q10 Assertion (A): The average value of alternating
current over a complete cycle is zero.
Q5 If represents the peak value of the voltage Reason (R): In a complete cycle, the positive
in an ac circuit, the r.m.s. value of the voltage and negative halves of the current cancel each
will be other.
(A) (B) (A) Both A and R are true and R is the correct
explanation of A.
(C) (D)
(B) Both A and R are true but R is not the
correct explanation of A.
Q6 The r.m.s. value of an ac of 50Hz is 10 amp.
(C) A is true but R is false.
The time taken by the alternating current in
(D) A is false but R is true.

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Answer Key
Q1 B Q6 D
Q2 C Q7 B

Q3 C Q8 C
Q4 A Q9 B
Q5 D Q10 A

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Hints & Solutions

Note: scan the QR code to watch video solution

Q1 Text Solution: Q4 Text Solution:

For a sinusoidal voltage:

Vavg​=π2×311​≈3.1416622​≈. 198V

Video Solution:

Video Solution:

Q5 Text Solution:

Q2 Text Solution:

Video Solution:

So, effective value of current (r.m.s.)

Video Solution: Q6 Text Solution:

Time taken by the current to reach the
maximum value

Video Solution:

Q3 Text Solution:

Video Solution:

Q7 Text Solution:

Video Solution:

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Video Solution:

Q8 Text Solution:
Q10 Text Solution:
The alternating current reverses its direction
Video Solution: periodically. Over one complete cycle, the area
under the curve above the time axis is equal
and opposite to the area below the axis.
Hence, their average value is zero.
The reason correctly explains this cancellation.

Video Solution:

Q9 Text Solution:

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PARISHRAM 2026
PHYSICS DPP: 3

Electromagnetic Induction

Q1 Consider a pair of coils arranged coaxially solenoid is 4 H, then the mutual inductance
parallel to each other, in a vertical plane. When of the solenoid due to solenoid will be:
the current in one coil increases from 0 to 10 A (A) 4 H (B) 8 H
in 0.5 s, the emf induced in the other coil is 20 (C) 2 H (D) None of these
V. The mutual inductance of the coil is:
(A) 4.0 H (B) 0.5 H Q5 The mutual inductance between a pair of a
(C) 2.0 H (D) 1.0 H coils A and B placed close to each other
depends upon
Q2 The maximum possible mutual inductance (A) the rate of change of current in A
between the two coils of self inductance (B) the rate of change of current in A and B
is (C) the material of the wire of the coils
(A) (B) (D) the relative position and orientation of A
and B
(C) (D)
Q6 A 100 turn coil of area 0.1 m2 rotates at half a
Q3 Consider two coaxial cylinder of same Length revolution per second. It is placed in a uniform
L, with vaccum inside them. The radii of the magnetic field of 0.01 T perpendicular to the
inner solenoid and the outer solenoid are axis of rotation of the coil. Calculate the
respectively. The number maximum voltage generated in the coil?
of turns per unit length are for the (A) 256.33 V (B) 89.12V
inner solenoid and the outer solenoid, (C) 0.314 V (D) 3.1455 V
respectively. The mutual inductance is:
Q7 Name the current induced in solid metallic
(A)
masses when the magnetic flux threading
(B) through them changes.
(C) (A) Ampere currents
(D) (B) Faraday currents
(C) Eddy currents
Q4 Two long solenoids have equal (D) Solenoidal currents
lengths and the solenoid is placed co-axially
Q8 An a.c. generator consists of a coil of 50 turns
inside the solenoid . The ratio of the radius
and an area 2.5 m2 rotating at an angular
of the solenoid is 1 : 2. If the
speed of 60 rad s–1 in a uniform magnetic field
mutual inductance of the solenoid due to
of 0.3 T between two fixed pole pieces. What is

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the flux through the coil, when the current is Q10 Assertion: The mutual inductance between
zero? two coils is maximum when the coils are
(A) Maximum wound on each other.
(B) Minimum Reason: The flux linkage between two coils is
(C) Zero maximum when they are wound on each
(D) Independent of current other.
(A) Both Assertion and Reason are true and
Q9 An a.c. generator consists of a coil of 1000
Reason is the correct explanation of the
turns and cross sectional area of 3m2, rotating Assertion .
at a constant angular speed of 60 rad s–1 in a (B) Both Assertion and Reason are true, but
uniform magnetic field 0.04 T. The resistance Reason is not the correct explanation of the
of the coil is 500Ω. Calculate the maximum Assertion .
current drawn from the generator. (C) Assertion is true, but Reason is false.
(A) 2500 A (B) 14.4 A (D) Assertion is false and Reason is also false.
(C) 6.25 A (D) 0.55 A

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Answer Key
Q1 D Q6 C

Q2 A Q7 C
Q3 B Q8 A
Q4 A Q9 B
Q5 D Q10 A

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Hints & Solutions

Note: scan the QR code to watch video solution

Q1 Text Solution: 1. When two coils are brought in proximity

with each other the magnetic field in one of
When two coils are arranged in such a
the coils tends to link with the other. If this
way that a change of current in one coil
magnetic field of the first coil is changed then
causes an emf to be induced in the other,
the magnetic. flux associated with the second
the coils are said to have mutual
coil changes and this leads to the generation
inductance.
of voltage in the second coil.
The mutual inductance in denoted letter
2. This property of a coil that affects or
M and measured in Henry.
changes the current and voltage in a
Induced emf in the coil is secondary coil is called mutual inductance.
Where, di = change in current, dt =
change in time, M = mutual inductance
Let us first consider a case when the total flux
Calculation: associated with one coil links with the other.
Given, change in current di = 10 A, change in i.e., a case of maximum flux linkage. consider
time dt = 0.5 s emf induced in the other coil, e two coils placed adjacent to each other. Thus,
= 20 V
so, from the equation

The self -inductance of the coil can be written

as,
Video Solution:

By multiplying equation 1 and equation 2, we

get

Q2 Text Solution:
Self-induction:
1. Self - induction is the property of the current
carrying coil that resists or opposses the
change of current flowing through it.
2. This occurs mainly due to the self induced Hence, option A is correct.
emf producecd in the coil itself. Video Solution:

Where N = number of turns in the coil, L = self-

induction of the coil, = flux associated with
the coil and I = current in the coil.
Mutual -inductance

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Q4 Text Solution:
Mutual - inductance

When two coils are brought in proximity

with each other the magnetic field in one
of the coils tends to link with the other. If
this magnetic field of the first coil is
Q3 Text Solution:
changed then the magnetic flux
Mutual Inductance:
associated with the seconds coil changes
when two coils are brought in proximity to
and this leads to the generation of
each other the magnetic field in one of the
voltage in the second coil.
coils tends to link with the other.
This property of a coil that effects or
changes the current and voltage in a
secondary coil is called mutual
inductance.
The SI unit of the mutual inductance is
henry.

Let two coils of the number of turns

are placed close to each other.
Then the mutual inductance of coil 1 with
respect to coil 2 is given as,

This further leads to the generation of voltage

Where = number of turns in coil 1, =flux
in the second coil.
The property of a coil that affects or changes linked with coil 1, and current in coil 2.

the current and voltage in a secondary coil is Mutual inductance for two co-axial solenoids:

mutual inductance
Let two long solenoids of
The mutual inductance, equal length are placed co-axially as
Explanation: shown in the figure.
Lets conside L is the length of the coaxial
The solenoid is placed inside the
cylinder.
solenoid .
are the numbers of turns of the
The mutual inductance of both the
inner solenoid and the outer solenoid
solenoids will be equal and it is given as,
respectively.
The mutual inductance is .
For medium of relative permeability
Video Solution:

Where = number of turns per unit

length of solenoid 1, number of
turns per unit length of solenoid 2, =
radius of the inner solenoid, and l =
length of both solenoids

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Factors Affecting Mutual Inductance:

Relative position and orientation of coils:

Coils placed closer and aligned properly
will have higher mutual inductance.
Number of turns in the coils: More turns
in a coil increase the linkage of magnetic
Explanation: flux.
Given Permeability of the core material: A
ferromagnetic core increases mutual
inductance.

Where = number of turns per unit Explanation of Options:

length of solenoid 1, = number of
turns per unit length of solenoid 2, = Rate of change of current in A (Incorrect):
radius of the inner solenoid, and l = Mutual inductance is a geometric
length of both the solenoids property and does not depend on how
fast the current changes.
Rate of change of current in A and B
(Incorrect): While emf depends on the
Hence, option A is correct. rate of change of current, mutual
inductance does not.
Video Solution: Material of the wire (Incorrect): The
material of the wire does not significantly
affect mutual inductance.
Relative position and orientation of coils
(Correct Answer): The distance and
alignment of the coils directly influence
their mutual inductance.
Q5 Text Solution:
The mutual inductance between A and B
Mutual inductance (M) between two coils depends on the relative position and
quantifies the ability of one coil to induce orientaton of A and B.
an electromotive force (emf) in the other Video Solution:
due to a changing current.
It depends on the geometric
arrangement of the coils, not on the rate
of change of current.

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On the other hand, the current is maximum,

when the coil is horizontal. So, when the
current is zero, the flux through the coil is
maximum.

Video Solution:

Q6 Text Solution:
The maximum voltage generated in the coil,
e0 = nBAω
e0 = nBA × (2πv)
e0= 100 × 0.01 × 0.1 × 2π × 0.5
e0 = 0.314 V Q9 Text Solution:
e0 = nBAω
Video Solution:
e0 = 1000 × 0.04 × 3 × 60 e0 = 7200 V.

The maximum current drawn

I = 14.4 A.

Video Solution:

Q7 Text Solution:
Eddy currents are the currents induced in solid
metallic masses when the magnetic flux
threading through them changes. These
currents look like eddies or whirlpools in water Q10 Text Solution:
and so they are known as eddy currents. Both Assertion and Reason are true and
Video Solution: Reason is the correct explanation of the
Assertion.

Video Solution:

Q8 Text Solution:
The current is zero when the coil is vertical. In
this position, flux through the coil is maximum.

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PARISHRAM 2026
Physics DPP: 2
Electromagnetic Induction

Q1 A rod of length I rotated with a uniform (C) anticlockwise

angular velocity omega about its
(D) zero
perpendicular bisector. A uniform magnetic
field B exists parallel to the axis of rotation. Q4 Consider the situation shown in the figure. The
The potential diference between the two ends wire AB is sliding on the fixed rails with a
of the rod is constant velocity. If the wire AB is replaced by
(A) zero (B) semicircular wire, the magnitude of the
induced current will
(C) (D)

Q2 A conductor of 3 m in length is moving

perpendicularly to magnetic field of 10–3 tesla
with the speed of 102 m/s, then the e.m.f.
produced across the ends of conductor will be
(A) 0.03 volt
(B) 0.3 volt
(C) 3 × 10–3volt
(A) Increase
(D) 3 volt
(B) same
Q3 A conducting square loop of side l and (C) Decrease
resistance R moves in its plane with a uniform (D) Depends upon bulging of curvature
velocity v perpendicular to one of its sides. A
Q5 The magnetic potential energy stored in a
uniform and constant magnetic field B exists
certain inductor is 25 mJ, when the current in
along the perpendicular to the plane of the
the inductor is 60 mA. This inductor is of
loop as shown in figure. The current induced
inductance
in the loop is
(A) 1.389 H (B) 138.88 H
(C) 0.138 H (D) 13.89 H

Q6 What will be self inductance of a coil of 200

turns in which a current of 2A produces a
magnetic flux of 4 mWb?
(A) 0.1 H (B) 0.2 H
(C) 0.3 H (D) 0.4 H
(A) clockwise
Q7 The mutual inductance between a pair of a
(B) antickockwise
coils A and B placed close to each other

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depends upon
(A) the rate of change of current in A
(B) the rate of change of current in A and B
(C) the material of the wire of the coils
(D) the relative position and orientation of A
and B

Q8 Consider two coaxial cylinder of same Length

L, with vaccum inside them. The radii of the
inner solenoid and the outer solenoid are
respectively. The number
of turns per unit length are for the
inner solenoid and the outer solenoid,
respectively. The mutual inductance is:
(A)
(B)
(C)
(D)

Q9 Two coils, A and B are arranged parallel to

each other. When the current in coil A
increases at rate of 20 A/s and current in coil B
is 5 A/s, the induced emf in coil B is 60 mV. The
mutual inductance of the two coils is:
(A) 5 mH (B) 3 mH
(C) 6 mH (D) 4 mH

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Answer Key
Q1 A Q6 D

Q2 B Q7 D
Q3 D Q8 B
Q4 B Q9 B
Q5 D

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Hints & Solutions

Note: scan the QR code to watch video solution

Q1 Text Solution:

flux is constnat i=0

Video Solution:

Q4 Text Solution:

Video Solution:

Q2 Text Solution:
volt
l = effective length
Video Solution:

Video Solution:

Q3 Text Solution:

Q5 Text Solution:
Energy stored in inductor coil is given by

B = constant
A = l2 = constant

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Hence option (D) is correct.

Video Solution:

Factors Affecting Mutual Inductance:

Relative position and orientation of coils:

Coils placed closer and aligned properly
Q6 Text Solution:
will have higher mutual inductance.
Self-inductance: The property of the current-
Number of turns in the coils: More turns
carrying coil that opposes or resists the
in a coil increase the linkage of magnetic
change of current flowing through itself is
flux.
called self-inductance.
Permeability of the core material: A
ferromagnetic core increases mutual
When N isthe number of turns, is the inductance.
magnetic flux, i is current, and L is the self-
inductance. Explanation of Options:

Calculation:
Rate of change of current in A (Incorrect):
Given that N = 200 turns; i = 2A; = 4mWb = 4 Mutual inductance is a geometric
× 10–3 Wb; property and does not depend on how
fast the current changes.
Rate of change of current in A and B
(Incorrect): While emf depends on the
L = 0.4 H rate of change of current, mutual
So the correct answer is option D. inductance does not.
Video Solution: Material of the wire (Incorrect): The
material of the wire does not significantly
affect mutual inductance.
Relative position and orientation of coils
(Correct Answer): The distance and
alignment of the coils directly influence
their mutual inductance.
Q7 Text Solution:
The mutual inductance between A and B
Mutual inductance (M) between two coils depends on the relative position and
quantifies the ability of one coil to induce orientaton of A and B.
an electromotive force (emf) in the other Video Solution:
due to a changing current.
It depends on the geometric
arrangement of the coils, not on the rate
of change of current.

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Video Solution:

Q8 Text Solution:
Mutual Inductance: Q9 Text Solution:
when two coils are brought in proximity to When two coils are arranged in such a way
each other the magnetic field in one of the that a change of current in one coil causes an
coils tends to link with the other. emf to be induced in the other, the coils are
said to have mutual inductance.
The mutual inductance is denoted letter M and
measured in Henry.
Induced emf in coil A due to the change in
current in coil B is
Where, di = change in current , dt = change in
time, M = mutual inductance
calculation:
Given,

This further leads to the generation of voltage

From coil B, from the reciporicity theorem
in the second coil.
The property of a coil that affects or changes
the current and voltage in a secondary coil is
mutual inductance
The mutual inductance, Video Solution:

Explanation:
Lets conside L is the length of the coaxial
cylinder.
are the numbers of turns of the
inner solenoid and the outer solenoid
respectively.
The mutual inductance is .

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PARISHRAM 2026
PHYSICS DPP: 1
Electromagnetic Induction

Q1 We have a coil as shown in the figure. Firstly Q3 In the given figure when the tapping key K is
the north pole of a bar magnet is moved pressed:
towards the coil on the left side and then the
south pole of the bar magnet is moved away
from the coil on the right side, then the
direction of the deflection in the galvanometer
will be:

(A) Galvanometer will show continuous

deflection until the key K is released
(B) Galvanometer will show momentary
deflection and returns to zero immediately
(C) The deflection of the galvanometer will
increase continuously until the key K is
released
(D) None of these

Q4 A coil and magnet are moved in the same

direction and with same speed. What will
(A) Same in both the case happen?
(B) Different in both the cases (A) The coil will expeprience a force
(C) No deflection will occur in both the case (B) The magnet will experience a force
(D) Can't say (C) Electric current will be induced in the coil
(D) Electric current will not be induced in the
Q2 In the second experiment of Faraday and coil
Henry, the primary coil is connected to the
galvanometer and the secondary coil is Q5 During Faraday's electromagnetic induction
connected to a battery. If the primary coil is experiment the mechanical effort of
rotated about its axis, then: movement of magnet near a coil produces
(A) The current will induce in the primary coil electric energy within the coil. This
(B) No current will induce in the primary coil phenomenon can be best explained on the
(C) Momentary current will induce in the basis of:
primary coil (A) Lenz's law and conservation of energy
(D) Can't say (B) Lenz's law and conservation of charge
(C) Faraday's law and conservation of energy

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(D) Faraday's law and conservation of charge T, where is constant.The magnitude of flux
passing through the saqure is:
Q6 A bar magnet is dropped and it passes
(A) (B)
through a copper ring as shown in the
(C) (D)
diagram. The acceleration of the falling
magnet while passing through the ring is

(A) equal to acceleration due to gravity

(B) More than the acceleration due to gravity
(C) Less than the acceleration due to gravity
(D) Zero

Q7 The flux linked with a coil at any instant t is

given by = – 50t + 250. The induced
emf at t = 3s is
(A) 190 V (B) –190 V
(C) –10V (D) 10V

Q8 A circular disc of radius 0.2m is placed in a

uniform magnetic field of induction

in such a way that it axis make an angle of

sixty degree with B. The magnetic flux linked
with the disc is :
(A) 0.02Wb (B) 0.06Wb
(C) 0.08Wb (D) 0.01 Wb

Q9 The net magnetic flux through any closed

surface is:
(A) Positive (B) Infinity
(C) Negative (D) Zero

Q10 A square of side L meters lies in the x – y plane

in a region, where the magnetic feld is give by

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Answer Key
Q1 B Q6 C

Q2 B Q7 C
Q3 B Q8 A
Q4 D Q9 D
Q5 A Q10 C

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Hints & Solutions

Note: scan the QR code to watch video solution

Q1 Text Solution: According to the second experiment of

Faraday and Henry, we can say that when
According to the first experiment of
a current-carrying coil is moved towards
Faraday and Henry, we can say that when
or away from another coil then an emf
a bar magnet is moved towards or away
gets induced in the coil.
from a coil then an emf gets induced in
If the circuit is closed in which the coil is
the coil.
connected then a current will also get
If the circuit is closed in which the coil is
induced in the coil.
connected then a current will also get
The relative motion between the two
induced in the coil.
coils is required to induce a current in
The relative motion between the magnet
the coil
and the coil is required to induce a
In the given case when the primary coil is
current in the coil.
rotated about its axis there will be no
When the North-pole of a bar magnet is
relative motion between the primary and
pushed towards the coil, the pointer in
the secondary coil.
the galvanometer deflects indicating the
Since there is no relative motion between
presence of electric current in the coil.
the primary and the secondary coil so no
When the magnet is pulled away from
current will induce in the primary coil.
the coil, the galvanometer shows
Hence, option B is correct.
deflection in the opposite direction,
which indicates the reversal of the
Video Solution:
current's direction.
Moreover, when the South-pole of the
bar magnet is moved towards or away
from the coil, the deflections in the
galvanometer are opposite to that
observed with the North-pole for similar
movements.
So we can conclude that for the following Q3 Text Solution:
motion of the magnet the direction of the
In this experiment two coils, and
deflection in the galvanometer will be the
held stationary. Coil is connected to
different. Hence, option B is correct.
galvanometer G while the second coil
Video Solution: is connected to a battery through a
tapping key K
It is observed that the galvanometer
shows a momentary deflection when the
tapping key K is pressed and the pointer
in the galvanometer returns to zero
immediately.
Q2 Text Solution:

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If the key is held pressed continuously, Video Solution:

there is no deflection in the
galvanometer. Hence, option B is correct.

Video Solution:

Q5 Text Solution:

During Faraday's electromagnetic

induction experiment the mechanical

Q4 Text Solution: effort of movement of a magnet near a

Faraday's Laws of Electromagnetic Induction: coil produces electric energy within the
coil.
Whenever the number of magnetic lines of
force (magnetic flux) passing through a This phenomenon can be best explained

circuit/coil changes an emf is produced in the on the basis of Lenz's law.

circuit called induced emf. The induced emf is Lenz's law depends on the principle of

given by the rate of change of magnetic flux conservation of energy and Newton's

linked with the circuit i.e. third law.

Video Solution:

When a bar magnet is pushed toward the

coil, the magnetic field linked with the
coil increases, hence induced
current/induced e.m.f. is set up in the
coil. Hence galvanometer deflects right.
When the bar magnet is pulled away
Q6 Text Solution:
from the coil, the magnetic field linked
Magnetic flux is a measurement of the total
with the coil decreases, hence induced
magnetic field which passes through a given
current/ induced e.m.f. is set up in the
area.
coil. Hence galvanometer deflects left.
When the bar magnet held stationary
inside the coil there is no change in the
magnetic field, hence no induced current.
Such that galvanometer doesn't show
any deflection.
When the coil and a magnet are moved
in the same direction and at the same
speed, then there is no change in the
magnetic field across the coil and hence
no induced current. Therefore option D is
correct.

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Lenz law:

Lenz's law depends on the principle of

conservation of energy and Newton's
third law.
It is the most convenient method to
determine the direction of the induced
current. Video Solution:
It states that the direction of an induced
current is always such to oppose the
change in the circuit or the magnetic field
that produces it.

Explanation:
Q8 Text Solution:
As the magnet falls the flux through the
Magnetic flux is given by the formula:
ring changes.
Regardless of whether the flux increases
Where :
or decreases, by Lenz's law a current is
induced in the ring that opposes the – is the magnetic flux.
change in flux. – B is the magnetic field induction (in Wb/ ).
This induced current induced a magnetic – A is the area of the surface
field that interacts with the falling – is the angle between the magnetic field and
magnet and decreases the acceleration the normal (perpendicular) to the surface.
of fall below g. Hence, the acceleration of The area A of a circular disc is given by the
the falling magnet while passing through formula:
the ring is less than the acceleration due
to gravity. Where r is the radius of the disc. Given r =
0.2m:
Video Solution:

The magnetic field induction B is given as:

The angle between the magnetic field and

then normal to the disc is given as 60º.
Now, substituting the values into the magnetic
Q7 Text Solution:
flux formula:
As we lerant in
º
Rate of change of magnetic Flux.
º
Now, simplifying the expression:
Where in

= change in flux Answer:

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The magnetic flux linked with the disc is: Since the net magnetic flux through any closed
surface is zero. We conclude that:
Net magnetic flux = 0
Video Solution: The net magnetic flux through any closed
surface is zero.

Video Solution:

Q9 Text Solution:
To solve the question regarding the net
magnetic flux through any closed surface, we Q10 Text Solution:
can follow these steps: The magnetic flux linked with uniform surface
Magnetic flux through a surface is defined of area of A in uniform magnetic field given by
as the integral of the magnetic field (B) over
that surface. Mathematically, it is expressed
as:

Where is a differential area vector on the

surface.
According to Gauss's Law for magnetism, the Video Solution:
net magnetic flux through any closed surface
is always zero. This is expressed as:

This law states that there are no magnetic

monopoles; hence, magnetic field lines that
enter a closed surface must also exit it.

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PARISHRAM 2026
PHYSICS Assignment : 1
Current Electricity

Q1 There are 8.4 × free electrons per in

copper. The current in the wire is 0.21 A (e =
1.6 × C). Then the drift velocity of
electron in a copper wire of cross
section will be:
(A)
(B) (A) More if the experiment is performed at
(C) higher temperature
(B) More if a wire of steel of same dimension is
(D) None of these
used
Q2 A conductor of lenght 40cm where a potential (C) More if the lenght of the wire is increased
difference of 10V is maintained between the (D) Less if the lenght of the wire is increased
ends of the conductor. Find the mobility of the
electrons provided the drift velocity of the
Q5 The temperature (T) dependence of resistivity
electrons is 5 ×
(A) of a semiconductor is represented by:
(B) (A)
(C)
(D)

Q3 All of the following statements are true except

(A) Conductance is the reciprocal of resistance
and is measured in Siemens
(B) ohm's law is not applicable at very low and
very high temperatures
(C) ohm's law is applicable to semiconductors
(D) ohm's law is not applicable to electron
tubes, discharge tubes and electrolytes

Q4 I-V characteristic of a copper wire of length L

and area of cross-section A is shown in figure.
The slope of the curve becomes

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(B) (C) 5V (D)

Q9 Consider the circuit shown in the figure. The

current is equal to

(C)

(D) (A) 5 A
(B) 3 A
(C) –3A
(D) –5/6 A

Q10 In the following figure what is the resultant

Q6 If the resistance of conductor is 5 at 50ºC resistance between A and C?
and 7 at 100ºC, then the mean temperature
coefficient of resistance of the meterial is.
(A) 0.008/ºC (B) 0.006/ºC
(C) 0.004/ºC (D) 0.001/ºC

Q7 A metal wire of lenght l has resistance R. Half

of its lenght is streched uniformly such that
length of wire becomes 2l. The new resistance
of wire will be? (A) (B)

(C) (D)

Q11 In a mixed grouping of identical cells, five rows

(A) 2R (B) 3R are connected in parallel and each row
(C) 4R (D) 5R contains 10 cell. This combination sends a
current I through an external resistance of 20
Q8 A battery of emf 10V is connected to resistance
. If the emf and internal resistance of each
as shown in figure. The potential difference
cell is 2.2V and 1 , respectively, then find the
between the points A and B is
value of i.
(A) 1 (B) 0.5
(C) 2 (D) 2.5

Q12 Four identical cell each of emf 2 V are joined in

parallel providing current to an external circuit
consisting of two 15 resistors joined in
parallel. The terminal voltage of the equivalent
cell as read by an ideal voltmeter is 1.6V.
(A) –2V (B) 2V Calculate the internal resistance of each cell.

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(A) 3.5 (B) 9.5 (C) (2) and (3)

(C) 5.5 (D) 7.5 (D) (1), (2) and (3)

Q13 n identical cell each of emf and internal Q16 A high-voltage transmission line has an
resistance r, are joined in series to from a aluminium cable of diameter 3.0 cm, 200 km
closed circuit. The potential difference across long. The resistivity is . The
any one cell is- resistance of this cable is;
(A) Zero (A) (B)
(B) (C) (D)
(C)
(D) Q17 Two copper wires have their masses in the
ratio 2 : 3 and the lengths in the ratio 3 : 4, The
Q14 Three 60 W, 120 V light bulbs are connected ratio of their resistances is;
across a 120V power source. If resistance of (A) 4 : 9 (B) 27 : 32
each bulb does not change with current, the (C) 16 : 9 (D) 27 : 128
find out total power delivered to the three
Q18 Statement I : A uniform wire of resistance 80
bulbs.
is cut into four equal parts. These parts are
now connected in parallel. The equivalent
resistance of the combination will be 5 .
Statement II : Two resistance 2R and 3R are
connected in parallel in a electric circuit. The
value of thermal energy developed in 3R and
(A) 180 W (B) 20 W 2R will be in the ratio 3 : 2.
(C) 40 W (D) 60 W Among the above given statements, choose
the most appropriate answer from the options
Q15 Consider the following statements regarding
given below:
the network shown in the figure :
(A) Both statement I and statement II are
correct
(B) Both statement I and statement II are
incorrect
(C) Statement I is correct but statement II is
incorrect
(D) Statement I is incorrect but statement II is
correct.
1. The equivalent resistance of the network
between points A and B is independent of Q19 The charge flowing through a resistance
value of G.
varies with time as , where
2. The equivalent resistance of the network
and are positive constants. The total heat
between points A and B is
produced in is
3. The current through G is zero.
(A) (B)
Which of the above statements is/are true?
(A) (1) alone (C) (D)
(B) (2) alone

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Answer Key
Q1 C Q10 C
Q2 B Q11 A
Q3 C Q12 D
Q4 D Q13 A
Q5 C Q14 C
Q6 A Q15 D
Q7 D Q16 C
Q8 B Q17 B
Q9 D Q18 C

Q19 A

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Hints & Solutions

Note: scan the QR code to watch video solution

Q1 Text Solution: Video Solution:

Video Solution:

Q2 Text Solution:

Video Solution:

Q3 Text Solution:
Q5 Text Solution:
Ohm's Law is not obeyed by semiconductors.
Temperature dependence of resistivity for a
Video Solution: typical semiconductor.

Q4 Text Solution: Temperature T

Slope of V-I curve = R . But in given T. .= const

graph is rectangular hyperbola
curve axis of I and V are interchanged. So
Video Solution:
slope of given curve i.e. with the

increase in lenght of the wire. Slope of the

curve will decrease.

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Q6 Text Solution: Potential difference between C and A,

Using .....(i)
Potential difference between C and B,
.....(ii)
On solving (i) and (iii)

Video Solution:
Video Solution:

Q9 Text Solution:
Q7 Text Solution: Suppose current through different paths of
the circuit is as follows.

After applying KVL for loop (1) and loop (2)

We get
Video Solution:
And

Hence,

Video Solution:

Q8 Text Solution:
2A and
current in each branch = 1A
Q10 Text Solution:

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Video Solution:

Video Solution:

Q13 Text Solution:

Current in circuit i
Q11 Text Solution:
The number of cells in a row is n = 10 and the
number of such rows is m = 5.

The equivalent circuit of one cell is shown in

Video Solution: figure.
The potential difference across the cell is

Video Solution:

Q12 Text Solution:

Total internal resistance of the cell
combination
Total e.m.f. Q14 Text Solution:
Total external resistance

Current drawn from the equivalent cell

Total power supplied

Video Solution:

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Q15 Text Solution: Since current through G is zero, it does not

affect the equivalent resistance between A
Analyzing the circuit: and B.
✔️Statement 1 is TRUE
We are given a symmetrical resistive network:
Video Solution:
Top arms: each
Bottom arms: each
Middle branch: resistor between
midpoints of top and bottom branches

Step 1: Check symmetry

Q16 Text Solution:
The circuit is symmetrical with respect to a
Given:
vertical line through the middle. Therefore,
potentials at points on either side of are Diameter
equal.
Length
⇒ No potential difference across G ⇒ current
Resistivity
through G is zero.
✔️Statement 3 is TRUE
Use resistance formula:
Step 2: Use of symmetry (since
current through G = 0)

We can remove G when calculating resistance

between A and B. Now,
So, treat the circuit as:

Two branches from A to B Video Solution:

Upper path:
Lower path:

Total resistance of each path:

Upper:
Lower:
Q17 Text Solution:
These two paths are in parallel, so:
Given:

Mass ratio:

✔️Statement 2 is TRUE Length ratio:

For same material (copper), resistance:

Step 3: Independence from G

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Energy in :
So,

Video Solution:

Ratio:

⛔ Statement II is incorrect (It says ratio is 3:2

for 3R : 2R, which is reversed)
Q18 Text Solution:
Video Solution:
Statement I:

Original resistance =
Cut into 4 equal parts ⇒ each part =

Connected in parallel:
Q19 Text Solution:
We are given:

✅ Statement I is correct
Current is:
Statement II:

Resistors: and in parallel

Current divides inversely with resistance
Heat produced in a resistor:
So power/energy ∝

Let total current = , and current through

= , =
Using current division: Find the time when current becomes zero:

Energy in : Now compute:

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Expand:

Simplify each term:

Now integrate:

Integrate term by term:

Now:

Video Solution:
Now evaluate from to :

Plug in :

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PARISHRAM 2026
DPP: 4
PHYSICS
Current Electricity

Q1 An electric heater is rated 1500 watt. If electric Q5 A filament bulb is to be used

power costs Rs. 2 per kilo-watt hour, then the in a main supply. When a resistance R
cost of power for 10 hours running of the is connected in series, it works perfectly and
heater is; the bulb consumes . The value of is
(A) Rs. 30 (B) Rs. 15 (A)
(C) Rs. 150 (D) Rs. 25 (B)
(C)
Q2 A resistor develops 300 J of thermal energy in
(D)
15 s, when a current of 2 A is passed through
it. If the current increases to 3A, the energy Q6 A resistor has power dissipation with cell
developed in 10 s is;
voltage . The resistor is cut in equal parts
(A) 450 J (B) 410 J
and all parts are connected in parallel with
(C) 540 J (D) 45 J
same cell. The new power dissipation is
Q3 An electric bulb is marked 100 W, 230 V. If the (A) (B)
supply voltage drops to 115 V, what is the total (C) (D)
energy produced by the bulb in 10 min?
Q7 A 100 watt bulb working on 200 volt and a 200
(A) 30 kJ (B) 20 kJ
watt bulb working on 100 volt have:
(C) 15 kJ (D) 10 kJ
(A) Resistances in the ratio of 4 : 1.
Q4 Three identical resistance A,B and C are (B) Maximum current ratings in the ratio of 1 :
connected as shown in figure. 4.
(C) Resistances in the ratio of 2 : 1.
(D) Maximum current ratings in the ratio of 1 :
2.

Q8 A light bulb is rated at 44 W, 220 V, and a table

fan is rated at 60 W, 110 V.
Which statement is correct if each of the two
devices is connected to a power supply of 220
The heat produced will be maximum V separately?
(A) in B (1) The light bulb has a greater resistance and
(B) in B and C draws a greater current than the table fan.
(C) in A (2) The light bulb has a greater resistance and
(D) same for A, B and C draws a smaller current than the table fan.

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(3) The light bulb has a smaller resistance and

draws a greater current than the table fan.
(4) The light bulb has a smaller resistance and
draws a smaller current than the table fan.

Q9 Rated power of a bulb at V voltage is P. Now,

same voltage V is applied in all conditions
mentioned in List-I. Match this List-I with List-
II in which actual total power consumed is
given.
List-I List-II
Two bulbs
are
(A) (I) P
connected in
parallel.
Two bulbs
are
(B) (II) 2P
connected in
series.
Two bulbs
are
connected in
parallel and
(C) (III)
one bulb in
series with
this
combination.
A group of
two-two
bulbs in
(D) parallel are (IV) None
mutually
connected in
series.
Choose the correct answer from the options
given below:
(A) A-II, B-III, C-IV, D-I
(B) A-IV, B-I, C-II, D-III
(C) A-I, B-III, C-IV, D-II
(D) A-III, B-II, C-I, D-IV

Q10 The maximum power drawn from a cell which

is connected across an external variable
resistance R, (where r is internal resistance)
(A) (B)
(C) (D)

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Answer Key
Q1 A Q6 C

Q2 A Q7 B
Q3 C Q8 (2)
Q4 C Q9 A
Q5 A Q10 B

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Hints & Solutions

Note: scan the QR code to watch video solution

Q1 Text Solution:
(A)
Given:

Power rating
Time
Cost per kWh = Rs. 2 Q3 Text Solution:
Given:
Energy consumed:
Power rating at

Total cost: Supply voltage

Time

First, calculate the resistance of the bulb:

Video Solution:

Now, power at 115 V:

Q2 Text Solution:
Given (Case 1):
Energy produced in 600 s:
Energy
Time
Current
Video Solution:
Using:

Now (Case 2):

, Q4 Text Solution:
Given: Three identical resistors A, B, and C,
where B and C are in parallel, and this
combination is in series with A.
Video Solution: Let each resistance be , and total voltage
supplied be .

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Step 1: Calculate equivalent Bulb rating:

resistance Mains supply:
Bulb works perfectly when resistance
Resistors B and C in parallel:
is added in series.

Total resistance:
Step 1: Resistance of the bulb

Step 2: Current in circuit

Step 2: Current required (same in

series circuit)
Step 3: Heat produced =

Heat in A:

Step 3: Voltage drop across total

Heat in B and C: resistance
Current splits equally in B and C (since both Since total supply is 230 V and current is 5 A:
are identical), so each gets:

So, Step 4: Value of series resistance

Video Solution:

Clearly:

Video Solution:

Q6 Text Solution:
Given:

A resistor dissipates power with cell

voltage :

Q5 Text Solution:
Given:

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Q8 Text Solution:
The resistor is cut into equal parts:
Calculate the current for the light bulb using its
each part has resistance
power rating:

Step 2
Calculate the resistance for the light bulb:
All parts are connected in parallel.

Step 1: Equivalent resistance of Step 3

resistors of in parallel: Calculate the current for the table fan using its
power rating:

Video Solution:
Step 2: New power dissipation:

Video Solution:
Q9 Text Solution:
(1)

In series
In parallel

Video Solution:
Q7 Text Solution:
(2)

Q10 Text Solution:

Maximum current rating
(B)
So,

Power is maximum when r = R Þ

Video Solution: [New NCERT Class 12th Page No. 92]

Video Solution:

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PARISHRAM 2026
DPP: 3
PHYSICS
Current Electricity

Q1 A wire of length 'l' and resistance is (A) (B)

divided into 10 equal parts. The fiirst 5 parts (C) (D)
are connected in series while the next 5 parts
Q4 The current in the resistor shown in the
are connected in parallel. The two
circuit is;
combinations are again connected in
series.The resistance of this final combination
is:
(A) (B)
(C) (D)

Q2 From the graph between current I and voltage

V shown in figure, identify the portion
(A) (B) 3A
corresponding to negative resistance.
(C) 6A (D) 2A

Q5 Four cells of identical emf E and internal

resistance r are connected in series to a
variable resistor. The following graph shows
the variation of terminal voltage of the
combination with current. The emf of each cell
used is;

(A) DE (B) CD
(C) BC (D) AB

Q3 In the given circuit, if the ammeter reads 5.0 A

and voltmeter reads 20 V, the value of
resistance R is.....;

(A) 1.4 V (B) 5.6 V

(C) 2 V (D) 1 V

Q6 Two resistors of resistances and are

connected in parallel. This combination is then

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connected to a battery of emf 2V and internal

resistance . What is the current flowing (B)
through the battery ?
(C)
(A) 4A (B)
(D)
(C) (D) 1A
Q10 For the circuit given below, the Kirchhoff's loop
Q7 If the galvanometer does not show any rule for the loop BCDEB is given by the
deflection in the circuit shown, the value of R is equation
given by

(A) (B) (A)

(C) (D) (B)
Q8 Assertion (A): In the circuit shown, (C)
or
(D)
, if
Q11 10 resistors, each of resistance are
connected in series to a battery of emf and
negligible internal resistance. Then those are
connected in parallel to the same battery, the
Reason (R): Potential difference across the current is increased times. The value of is
terminals of a non ideal battery is less than its (A) 100 (B) 1
emf when a current flows through it. (C) 1000 (D) 10
(A) If both assertion and reason are true and
reason is the correct explanation of the Q12 An electric heater is rated 1500 watt. If electric
assertion. power costs Rs. 2 per kilo-watt hour, then the
(B) If both assertion and reason are true but cost of power for 10 hours running of the
reason is not correct explanation of the heater is;
assertion. (A) Rs. 30 (B) Rs. 15
(C) If assertion is true, but reason is false. (C) Rs. 150 (D) Rs. 25
(D) Assertion is false, reason is true
Q13 A resistor develops 300 J of thermal energy in
Q9 In the electrical circuit shown in the figure, the 15 s, when a current of 2 A is passed through

current through the side is it. If the current increases to 3A, the energy
developed in 10 s is;
(A) 450 J (B) 410 J
(C) 540 J (D) 45 J

Q14 An electric bulb rated 100 W at 220 V is

operating at 110 V. What is the power
(A)

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consumed?
(A) 50 W (B) 75 W
(C) 100 W (D) 25 W

Q15 The power dissipated in the circuit shown in

the figure is 30 Watts. The value of R is; (A)
(B)
(C)
(D)

Q19 Kirchhoff's first and second laws for electrical

circuits are consequences of
(A) Conservation of energy
(B) Conservation of electric charge and energy

(A) (B) respectively

(C) (D) (C) Conservation of electric charge

(D) Conservation of energy and electric charge
Q16 The magnitude and direction of the current in respectively
the following circuit is
Q20 100 cells each of e.m.f. 5 V and internal
resistance 1 ohm are to be arranged so as to
produce maximum current in a 25 ohms
resistance. Each row is to contain equal
number of cells. The number of rows should
be
(A) from to through (A) 2 (B) 4
(B) from A to B through E (C) 5 (D) 10

(C) 1.5A from B to A through E

(D) from B to A through E

Q17 A filament bulb is to be used

in a main supply. When a resistance R
is connected in series, it works perfectly and
the bulb consumes . The value of is
(A)
(B)
(C)
(D)

Q18 A part of a circuit is shown in figure.

is equal to is equal to

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Answer Key
Q1 B Q11 A

Q2 B Q12 A

Q3 A Q13 A

Q4 D Q14 D

Q5 A Q15 C

Q6 D Q16 A

Q7 B Q17 A

Q8 B Q18 C

Q9 A Q19 B,D

Q10 A Q20 A

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Hints & Solutions

Note: scan the QR code to watch video solution

Q1 Text Solution:

Given:

Total resistance of wire =

Divided into 10 equal parts ⇒ each part
has resistance:
Q3 Text Solution:

Given:

Current
First 5 parts in series: Voltage across resistor

Using Ohm’s Law:

Next 5 parts in parallel:

Each , so: Video Solution:

Q4 Text Solution:
Final combination (series of and
): Step-by-step solution:

🔹 Step 1: Simplify the circuit

Video Solution:
The 4Ω resistors on top and bottom are
in parallel.

Now the circuit reduces to:

Q2 Text Solution:
The slope of CD is negative, and offer a 2Ω (from parallel 4Ω)

negative resistance. In series with 1Ω

In series with 6V battery
Video Solution:

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🔹 Step 2: Total resistance: Two resistors: , in

parallel
Battery EMF , internal
resistance
🔹 Step 3: Total current in the circuit:

Step 1: Find equivalent resistance of

parallel combination

🔹 Step 4: Current in the 1Ω resistor

(series):

Since it's in series, current through 1Ω

Step 2: Total resistance in circuit
resistor = total current =

Video Solution:

Step 3: Apply Ohm’s Law

Video Solution:

Q5 Text Solution:

From the graph:

At , terminal voltage

This is the maximum voltage, i.e., when no Q7 Text Solution:

current is drawn, so: For no reading galvanometer. Potential across
same.

Video Solution:

Q6 Text Solution:

Given:
Video Solution:

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less than emf, but equal to IR.

So while both statements are true, the
reason is not the correct explanation for
the assertion (which is a specific numerical
match).

Video Solution:
Q8 Text Solution:

Let's analyze both the Assertion (A)

and Reason (R):

Assertion (A):
Q9 Text Solution:
"In the circuit shown, if
"

EMF of battery = 4 V
Resistor
Current

Use Ohm’s Law:

Video Solution:
This is the voltage drop across the resistor,
and direction of current matches the
polarity of battery.
So,

Q10 Text Solution:

Apply Kirchhoff's Loop Rule to loop

Reason (R):
BCDEB
“Potential difference across terminals of a non-
Loop: B → C → D → E → B (clockwise)
ideal battery is less than its emf when current
flows through it.”
Yes — due to internal resistance, terminal Step-by-step:
voltage ,
1. B → C:
so terminal voltage < emf.
Current through resistor
So Reason is also correct.
Drop =
But in this case, the assertion is about a 2. C → D:
terminal voltage becoming zero — not just Battery , going from – to +

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Gain =
3. D → E:
Battery , going from + to –
Drop =
4. E → B:
Current through resistor
Q12 Text Solution:
Drop =
(A)
Given:
Total sum of voltages:
Power rating
Time
Cost per kWh = Rs. 2
Video Solution:
Energy consumed:

Total cost:

Q11 Text Solution:

Video Solution:
Case 1: Series connection

Total resistance
Current

Case 2: Parallel connection Q13 Text Solution:

Given (Case 1):
Total resistance

Current Energy
Time
Current
Ratio of currents:
Using:

Video Solution: Now (Case 2):

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Power dissipated:

Video Solution:

Given , so:

Q14 Text Solution:

Given:
Video Solution:
Rated power at

Operating voltage

Step 1: Calculate resistance of the

bulb

Q16 Text Solution:

Step 2: Power consumed at 110 V Step-by-step solution:

🔹 Step 1: Simplify the circuit

Video Solution: From the diagram:

Battery 1: 10 V (left, positive at top)

Battery 2: 5 V (right, positive at bottom)
Resistors:
from A to E
from E to B
Q15 Text Solution: from D to C
Given:
Path: A → E → B → C → D → A (full loop)
Voltage
Power dissipated 🔹 Step 2: Net EMF in the loop
One resistor is , and another is ,
connected in parallel Battery 1: +10 V (from A to E)
Battery 2: -5 V (from E to B)
Let the equivalent resistance be:

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🔹 Step 3: Total resistance in loop

Step 3: Voltage drop across total

🔹 Step 4: Current in loop resistance

Since total supply is 230 V and current is 5 A:

🔹 Step 5: Direction
Step 4: Value of series resistance
Since net EMF is from A to B via E (10 V
dominates),
👉 Current flows from A to B through E
Video Solution:
Explanation: Net EMF is 5 V, total resistance is
10 Ω, giving 0.5 A in the direction of the
stronger battery (10 V).

Video Solution:

Q18 Text Solution:

Given:

Resistor , current flows

Q17 Text Solution: upward
Given: Current from A to C = 2 A

Bulb rating:
Mains supply: Step:
Bulb works perfectly when resistance
From point A to C:
is added in series.
Voltage drop across 2 Ω =
Battery = 2 V (drop)
Step 1: Resistance of the bulb
→

Assume
Then,

Step 2: Current required (same in

series circuit)

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Video Solution: Q20 Text Solution:

Let n be the number of cells in a row and m be
the number or rows.
Now, maximum current is obtained for r = R in
Mixed combination total resistance of the
circuit is given by.

Q19 Text Solution:

Kirchhoff's laws for electrical circuits, both the
In mixed combination, we have,
first and second, are consequences
nm= 1000 ...(ii)
of fundamental conservation
From equation (i) and (ii) we have,
principles. Specifically, the first law (Kirchhoff's
Current Law or KCL) is a direct result of the
conservation of electric charge, while the
second law (Kirchhoff's Voltage Law or KVL)
stems from the conservation of energy Video Solution:

Video Solution:

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PARISHRAM 2026
DPP: 2
PHYSICS
Current Electricity

Q1 The resistance of a wire (its temperature Q5 Which of the following graphs represents the
coefficient ) is at . If its variation of resistivity ( ) with temperature (T)
temperature is increased twice the initial for copper
value, the final resistance of the wire is; (A)
(A) (B)
(C) (D)

Q2 A high-voltage transmission line has an (B)

aluminium cable of diameter 3.0 cm, 200 km
long. The resistivity is . The
resistance of this cable is;
(A) (B) (C)
(C) (D)

Q3 The current —voltage graph for a given

metallic wire at two different temperatures
(D)
and is shown in the figure. The
temperatures and are related as;

Q6 Two solid conductors are made up of same

material have same length and same
resistance. One of them has a circular cross
section of area and the other one has a
square cross section of area . The ratio
(A) (B) is
(C) (D) (A) 1 (B) 0.8
(C) 2 (D) 1.5
Q4 Two copper wires have their masses in the
ratio 2 : 3 and the lengths in the ratio 3 : 4, The Q7 Column-1 gives certain physical terms
ratio of their resistances is; associated with flow of current through a
(A) 4 : 9 (B) 27 : 32 metallic conductor. Column-II gives some
(C) 16 : 9 (D) 27 : 128 mathematical relations involving electrical
quantities. Match Column-I and Column-II with
appropriate relations.

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(B) 3 : 2 : 1
(C) 6 : 3 : 2
(D) 2 : 3 : 6

(A) A-(R), B-(S), C-(Q), D-(P)

(B) A-(R), B-(P), C-(S), D-(Q)
(C) A-(R), B-(Q), C-(S), D-(P)
(D) A-(R), B-(S), C-(P), D-(Q)

Q8 The resistance of platinum wire at is

and at . The temperature
coefficient of resistance of the wire is
(A)
(B)
(C)
(D)

Q9 Assertion (A): The drift velocity of electrons in

a metallic wire will decrease, if the
temperature of the wire is increased.
Reason (R): On increasing temperature,
conductivity of metallic wire decreases.
(A) Both Assertion (A) and Reason (R) are the
true, and Reason (R) is a correct
explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are the
true, but Reason (R) is not a correct
explanation of Assertion (A).
(C) Assertion (A) is true, and Reason (R) is false.
(D) Assertion (A) is false, and Reason (R) is true.

Q10 An electric current passes through non-

uniform cross-section wire. If j1, j2 and j3 are
the current densities, than j1 : j2 : j3 is:

(A) 1 : 2 : 3

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Answer Key
Q1 D Q6 A

Q2 C Q7 D
Q3 B Q8 B
Q4 B Q9 B
Q5 B Q10 C

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Hints & Solutions

Note: scan the QR code to watch video solution

Q1 Text Solution: Observation from graph:

For the same voltage, current is higher

at temperature than at
So, resistance is lower at

For metallic wires:

Video Solution: Resistance increases with temperature

So,

Video Solution:
Q2 Text Solution:

Given:

Diameter
Length
Resistivity
Q4 Text Solution:

Use resistance formula: Given:

Mass ratio:
Length ratio:

For same material (copper), resistance:

Now,

Video Solution:

So,

Video Solution:
Q3 Text Solution:

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Q5 Text Solution: Q7 Text Solution:

For some metals like copper, resistivity is
nearly proportional to temperature although a
non linear region always exists at very low
temperature

Video Solution:

Video Solution:

Q6 Text Solution:

Given:

Same material, length, and resistance Q8 Text Solution:

One wire: circular cross-section of area Step 1: Calculate the change in resistance

Other wire: square cross-section of area Given:

- Resistance at
-Resistance at
Resistance formula: The change in resistance is:

Step 2: Calculate the change in temperature

The change in temperature is:

Since both have same resistance, their cross-
sectional areas must be equal:
Step 3: Substitute values into the formula for a
Now, substituting the values into the formula:

Step 4: Simplify the expression

Video Solution:
Calculating the denominator:

Now substituting back:

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Step 5: Calculate a

Video Solution:

Q10 Text Solution:

Q9 Text Solution:
On increasing temperature of wire the kinetic
energy of free electrons increase and so they Video Solution:

collide more rapidly with each other and

hence their drift velocity decreases. Also when
temperature increases, resistivity increase and
resistivity is inversely proportional to
conductivity fo material.

Video Solution:

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PARISHRAM 2026
DPP: 1
PHYSICS
Current Electricity

Q1 10,000 alpha particles pass per minute section of the conductor during t = 2s to t = 3s
through a straight tube of radius r. What is the is :-
resulting electric current? (A) 10 C (B) 24 C
(A) (C) 33 C (D) 44 C
(B)
Q6 In a Neon discharge tube
(C)
(D) ions move to the right each second, while
electrons move to the left per
Q2 The current through a wire depends on time
second; electron charge is . The
as i = (2 + 3t) A. Calculate the charge crossed
current in the discharge tube is :-
through a cross section of the wire in first 10
(A) 1 A towards right
seconds.
(B) 0.66 A towards right
(A) 150 C (B) 165 C
(C) 0.66 A towards left
(C) 170 C (D) 175 C
(D) zero
Q3 When no current flows through a conductor
Q7 The plot represents the flow of current
(A) the free electrons do not move
through a wire for different time intervals. The
(B) the average speed of a free electron over a
ratio of charges flowing through the wire
large period of time is zero.
corresponding to these time intervals is (see
(C) the average velocity of a free electron over
figure) :-
a large period of time is zero.
(D) the average of square of velocities of all the
free electrons at an instant is zero.

Q4 In an atom, electron revolve around the

nucleus along a path of radius 0.72 A making
revolutions per second. The (A) 2 : 1 : 2 (B) 1 : 3 : 3
(C) 1 : 1 : 1 (D) 2 : 3 : 4
equivalent current is [Given e = 1.6 × 10-19C]
(A) 1.4 A (B) 1.8 A Q8 Given below are two statement: one is labelled
(C) 1.2 A (D) 1.5 A as Assertion A and the other is labelled as
Reason R:
Q5 The current in a conductor varies with time t
Assertion : There is no current in the metals in
as A where I i amperes and t in the absence of electric field.
seconds. Electric charge flowing through a Reason : Motion of free electron are random.

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In the light of the above statements, choose Q9 Statement I: There is no current in the metals
the correct answer from the options given in the absence of electric field.
below: Statement II: Motion of free electrons is
(A) Both A and R are true but R is NOT the random.
correct explanation of A. (A) If both statements are true and statement
(B) Both A and R are true and R is the correct II is the correct explanation of statement I.
explanation of A. (B) If both statements are true but statement II
(C) A is false but R is true. is not the correct explanation of statement
(D) A is true but R is false. I.
(C) If statement I is true but statement II is
false.
2
(D) If both statements are false.

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Answer Key
Q1 C Q6 B

Q2 C Q7 C

Q3 C Q8 B

Q4 D Q9 A

Q5 B

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Hints & Solutions

Note: scan the QR code to watch video solution

Q1 Text Solution: Electron revolutions per second

Charge of electron

Equivalent current:
Video Solution:

Video Solution:

Q2 Text Solution:

Q5 Text Solution:

Video Solution:

Video Solution:

Q3 Text Solution:
Due to random motion net velocity of all the
free e– is zero, so no current flows through the
Q6 Text Solution:
conductor at room temperature.
i = 4.1 × 1018 × 1.6 × 10–19
Video Solution:
= 0.66 A towards right

Video Solution:

Q4 Text Solution:

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Q7 Text Solution:

Video Solution:

Q9 Text Solution:
Q8 Text Solution:
"There is no current in the metals in the absence
True – In absence of an electric field, the net
of electric field."
current is zero in metals because there's no
Without an electric field, free electrons move
directed movement of electrons.
randomly, so net current = 0
True – The free electrons in metals move
randomly when no external field is applied, so "Motion of free electrons is random."
their net motion cancels out. Even without a field, electrons move randomly
The random motion of electrons explains due to thermal energy
why net current is zero without an electric
field. The random motion of electrons results in no
net flow, hence no current.
Video Solution:
So, Statement II is the correct explanation
of Statement I.

Video Solution:

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PARISHRAM 2026
PHYSICS ASSIGNMENT: 1

Electrostatic Potential and Capacitance

Q1 The electric potential inside a conducting Q5 Which of the following about potential
sphere difference between any two points is true?
(A) increases from centre to surface I. It depends only on the initial and final
(B) decreases from centre to surface position.
(C) remains constant from centre to surface II. It is the work done per unit positive charge
(D) is zero at every point inside in moving from one point to other.
III. It is more for a positive charge of two units
Q2 Which of the following about potential at a
as compared to a positive charge of one unit.
point due to a given point charge is true?
(A) I only (B) II only
The potential at a point P due to a given point
(C) I and II (D) I, II and III
charge
(A) is a function of distance from the point Q6 Consider the following statements and select
charge. the correct option
(B) varies inversely as the square of distance I. In an external electric field, the positive and
from the point charge. negative charges of a non-polar molecule are
(C) is a vector quantity displaced in opposite directions.
(D) is directly proportional to the square of II. In non -polar molecules displacement stops
distance from the point charge. when the external force on the constituent
charges of the molecule is balanced by the
Q3 To obtain capacity from three capacitors
restoring force.
of each, they will be arranged. III. The non-polar molecule develops an
(A) all the three in series induced dipole moment.
(B) all the three in parallel (A) I and II (B) II and III
(C) two capacitors in series and the third in (C) I and III (D) I, II and III
parallel with the combinatioin of first two
(D) two capacitors in parallel and the third in Q7 In which of the following cases is the electric
series with the combination of first two field zero but potential is not zero at a point on
-axis ?
Q4 The work done in placing a charge of (A)
coulomb on a condenser of
capacity 100 micro-farad is
(A) joule
(B) joule
(C) joule
(D) joule

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(B)

(C)
(A) changes, V remains unchanged

(B) remains unchanged, V changes

(C) both and V change

(D) and V remain unchanged

(D)

Q10 Which of the following figure shows the

correct equipotential surfaces of a system of
two positive charges?
(A)

Q8 The following figures show an electric dipole in

four orientations in uniform electric field.
(B)
Arrange them in increasing order of potential
energy.

(C)

(A) (i), (ii), (iii), (iv)

(D)
(B) (iv), (iii), (ii), (i)
(C) (iv), (ii), (i), (iii)
(D) (iv), (ii), (iii), (i)

Q9 Charges are placed on the vertices of a square

as shown. Let be the electric field and V the

potential at the centre. If the charges on A and
Q11 When a dielectric slab is inserted between the
B are interchanged with those on D and C
plates of one of the two identical capacitors
respectively, then
shown in the figure then match the following:

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assertion
(C) Assertion is correct, reason is incorrect
(D) Assertion is incorrect, reason is correct.

Q14 Two equally charged spheres of radii a and b

Column I Column II are connected together. What will be the ratio
Charge on of electric field intensity on their surfaces?
(A) (1) Increases
A (A) (B)
Potential
(C) (D)
(B) difference (2) Decreases
across A
Potential Q15 In the given circuit diagram, both capacitors
(C) difference (3) Remains constant are initially uncharged. The capacitance C1 = 2
across B F and C2 = 4F emf of battery A and B are 2 V
Charge on
(D) (4) Cannot say and 4 V respectively.
B
(A)
(B)
(C)
(D)

Q12 What is the effective capacitance between

points X and Y ?
Column II
Column I
(Magnitude only)
On closing switch
S1 with S2 open
(A) (1)
work done by
battery A is
(A) Switch S1 is open
(B) and S2 is closed,
(B) (2) 4
(C) work done by
(D) battery B is
Charge on
Q13 Assertion: The potential difference between capacitor C2 is
any two points in an electric field depends only (C) (3) 8
(after S1 open and
on initial and final position. S2 closed)
Reason: Electric field is a conservative field so Charge on C1
the work done per unit positive charge does (D) when (4)
not depend on path followed. both are closed
(A) Assertion is correct, reason is correct;
reason is a correct explanation for (A)
assertion. (B)
(B) Assertion is correct, reason is correct; (C)
reason is not a correct explanation for (D)

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Q16 An electric charge is placed at the disconnected from it. A dielectric slab of

origin (0,0) of co-ordinate system. Two dielectric constant K, which can just fill the air
gap of the capacitor, is now inserted in it.
points A and B are situated at and
Which of the following is incorrect?
(2,0) respectively. The potential difference
(A) The energy stored in the capacitor
between the points A and B will be
decreases K times.
(A) 4.5 volt (B) 9 volt
(B) The chance in energy
(C) zero (D) 2 volt
stored is
Q17 The expression (C) The charge on the capacitor is not
implies, that electric
conserved.
field is in that direction in which
(D) The potential difference between the plates
(A) increase in potential is steepest.
decreases K times.
(B) decrease in potential is steepest.
(C) change in potential is minimum.
(D) None of these

Q18 Assertion : Electric energy resides out of the

spherical isolated conductor.
Reason : The electric field at any point inside
the conductor is zero.
(A) Assertion is correct, reason is correct;
reason is a correct explanation for
assertion.
(B) Assertion is correct, reason is correct;
reason is not a correct explanation for
assertion
(C) Assertion is correct, reason is incorrect
(D) Assertion is incorrect, reason is correct.

Q19 A foil of aluminium of negligible thickness is

inserted in between the space of a parallel
plate condenser. If the foil is electrically
insulated, the capacity of the condenser will

(A) increase
(B) decrease
(C) remain unchanged
(D) become zero

Q20 A parallel plate air capacitor of capacitance C is

connected to a cell of emf V and then

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Answer Key
Q1 C Q11 B

Q2 A Q12 D
Q3 C Q13 A

Q4 C Q14 C

Q5 C Q15 D
Q6 D Q16 C

Q7 C Q17 B
Q8 D Q18 A

Q9 A Q19 C

Q10 C Q20 C

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Hints & Solutions

Note: scan the QR code to watch video solution

Q1 Text Solution:
Electric potential inside a conductor is
constant and it is equal to that on the surface
of the conductor.

Video Solution:

Q5 Text Solution:
Since , more work will be done for a
positive charge of two units as compared to
positive charge of one unit, but the ratio is
same. Therefore potential difference is same.
Q2 Text Solution:
Video Solution:
Since , for a given point charge, q
is constant, therefore V depends only on r .
Hence V is a function of distance.

Video Solution:

Q6 Text Solution:
In an external electric field, the positive and
negative charges of a non-polar molecule are
displaced in opposite directions. The
Q3 Text Solution: displacement stops when the external force on
the constituent charges of the molecule is
balanced by the restoring force (due to
Video Solution:
internal fields in the molecule). The non-polar
molecule thus develops an induced dipole
moment. The dielectric is said to be polarised
by the external field.

Video Solution:

Q4 Text Solution:
Work done

Video Solution:
Q7 Text Solution:
In this case electric fields due to the two
charges at origin are just equal and opposite

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and thus cancel each other whereas potential

due to the two charges add up and is not zero.

Video Solution:

Video Solution:

Q8 Text Solution:
The potential energy of a dipole in uniform
electric field is given by

Q10 Text Solution:

(i) For . Equipotential surfaces are normal to the
This is maximum value. electric field lines. The following figure shows
(ii) For the equipotential surfaces along with electric
(iii) For , field lines for a system of two positive charges.
is negative and hence U is positive.
(iv) For , is positive
and hence U is negative.
Therefore the correct increasing order is (iv),
(ii), (iii), (i)

Video Solution:

Video Solution:

Q9 Text Solution:
As shown in the figure, the resultant electric
Q11 Text Solution:
fields before and after interchanging the
If V is the potential applied across the
charges will have the same magnitude, but
capacitor then p.d. across each capacitor will
opposite directions.
be . When
Also, the potential will be same in both cases
as it is a scalar quantity.S A-1: dielectric is inserted in capacitor B, then
B-1:
C-2 : and
D-2 : On solving above equations, we get

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Clearly potential of A increases and that of B are in parallel hence net

decreases. Initial charges on the capacitors are
capacitance
:
Video Solution:

charges:

Charge on capacitor A will increase, and on B

will decrease.

Video Solution:
Q13 Text Solution:
Assertion is correct, reason is correct; reason
is a correct explanation for assertion.

Video Solution:

Q12 Text Solution:

Q14 Text Solution:

Equivalent circuit

Let charge on each sphere = q when they are

connected together their potential will be
equal.
Now let charge on a = q1 and on b = 2 q - q1

Hence no charge will flow through

b:a

Video Solution:
C1 and C2 are in series, also C3 and C4 are in
series.
Hence

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and B is given by

Q15 Text Solution: Video Solution:

W.d. by battery A,

W.d. by battery

Q17 Text Solution:

As we move towards a positive charge

Video Solution: distribution V increases i.e., is positive. The

increase in potential is steepest when we
move exactly towards charge distribution. But
E is in a direction exactly away from charge
distribution, therefore E is in exactly opposite
direction in which increase in potential is
steepest. Hence .
Q16 Text Solution:
Video Solution:

Q18 Text Solution:

The distance of point from the As these is no electric field inside the
origin, conductor, and so no energy inside it.

Video Solution:

The distance of point B(2,0) from the origin,

Now, potential at
Q19 Text Solution:
remain unchanged
Potential difference between the points A

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Video Solution:

As the capacitor is isolated, so change will

remain conserved p.d. between two plates of
the capacitor

Q20 Text Solution:

Video Solution:
Capacitance of the capacitor,
After inserting the dielectric, new capacitance

New potential difference

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 PARISHRAM 2026
PHYSICS
ELECTROSTATIC POTENTIAL AND CAPACITANCE DPP: 7

Q1 A capacitor has a capacitance of 50 pF , which (D) depends on the nature of the material
increases to 175 pF with a dielectric material inserted
between its plates. What is the dielectric
Q5 Can a metal be used as a medium for
constant of the material?
dielectric?
(A) 2.5 (B) 3.5
(A) Yes
(C) 4.5 (D) 5.5
(B) No
Q2 The capacity and the energy stored in a (C) Depends on its shape
parallel plate condenser with air between its (D) Depends on dielectric
plates are respectively C0 and W0. If the air
Q6 Assertion (A): Inserting a dielectric into a
between the plates is replaced by glass
charged capacitor connected to a battery
(dielectric constant = 5) find the capacitance of
increases the stored energy.
the condenser and the energy stored in it.
Reason (R): Capacitance increases and voltage
(A) (B)
remains constant, so energy increases.
(C) (D) (A) Both A and R are true, and R is the correct
explanation of A.
Q3 A parallel plate capacitor is to be designed (B) Both A and R are true, but R is not the
with a voltage rating 1 kV using a material of correct explanation of A.
dielectric constant 10 and dielectric strength (C) A is true but R is false.

. What minimum area of the plates (D) A is false but R is true.

is required to have a capacitance of 88 . 5 p F? Q7 A parallel plate air capacitor has a capacitance

(A) C. When it is half filled with a dielectric of
(B) dielectric constant 5, the percentage increase
(C) in the capacitance will be :-
(D)

Q4 When a slab of dielectric medium is placed

between the plates of a parallel plate capacitor
which is connected with a battery, then the
charge on plates in comparison with earlier
charge : (A) 400% (B) 66.6%
(A) is less (C) 33.3% (D) 200%
(B) is same
(C) is more

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Q8 Capacitor Capacitance
Select the correct statements about (A) connected to a (P) increases, charge
the effect of inserting a dielectric in a battery increases
capacitor: Capacitance
Capacitor isolated
(B) (Q) increases, voltage
after charging
1. The capacitance of a parallel plate increases
capacitor increases by a factor of K. Voltage is kept
(C) (R) Energy increases
constant
2. The electric field between the plates Charge is kept
(D) (S) Energy decreases
decreases if the battery is disconnected. constant
(A) Code (A) (B) (C) (D) P Q R S
3. The charge on the capacitor remains (B) P Q S R
constant if the battery is disconnected. (C) Q P R S
(D) P Q R R
4. The energy stored always increases upon
Q10 The distance between the plates of a parallel
inserting a dielectric.
plate capacitor is 'd'. Another thick metal plate
(A) 1, 2 and 3 only of thickness d/2 and area same as that of
(B) 1, 2 and 4 only plates is so placed between the plates, that it
(C) 2 and 4 only does not touch them. The capacity of the
(D) All of the above resulting capacitor :-
(A) remains the same
Q9 Column II (Effect (B) becomes double
Column I
of inserting (C) becomes half
(Situation)
dielectric) (D) becomes one fourth

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 Answer Key
Q1 B Q6 A

Q2 C Q7 D
Q3 A Q8 A

Q4 C Q9 A

Q5 B Q10 A

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 Hints & Solutions
Note: scan the QR code to watch video solution

Q1 Text Solution: also increased

Video Solution:

Video Solution:

Q5 Text Solution:
No

Q2 Text Solution: Video Solution:

Video Solution:

Q6 Text Solution:
Both A and R are true, and R is the correct
explanation of A.

Q3 Text Solution:

Video Solution:

Video Solution:

Q7 Text Solution:

Q4 Text Solution:
Q = CV; Video Solution:
when a slab of dielectric medium is placed
capacitance is increased so charge is

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Q8 Text Solution:
Video Solution:
Statement 4 is incorrect because energy
increases only if voltage remains constant
(battery connected). In isolated cases, energy
decreases.

Video Solution:

Q10 Text Solution:

1

Video Solution:

Q9 Text Solution:
(A)

(B)

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PARISHRAM 2026
DPP: 6
PHYSICS
Electrostatic Potential and Capacitance

Q1 The equivalent capacitance between the (D) 16 component capacitors

points A and B in the given diagram is :
Q4 The effective capacity of the network between
terminals A and B is :

(A) 8 mF (B) 6 mF
(C) (D)
(A)
Q2 In an adjoining figure three capacitors C1, C2 (B)
(C)
and C3 are joined to a battery. The correct
(D)
condition will be : (Symbols have their usual
meanings) Q5 The effective capacitance between the points P
and Q of the arrangement shown in the figure
is :

(A)
(B) (A)

(C) (B)

(D) (C)
(D)
Q3 A number of capacitors, each of capacitance 1
and each one of which gets punctured if a Q6 Two capacitors of capacity C1 and C2 are

potential difference just exceeding 500 volt is connected as shown in figure.

applied, are provided. Then an arrangement Now the switch is closed. Calculate the charge
suitable for giving a capacitor of capacitance on each capacitor.
across which 2000 volt may be applied
requires at least :
(A) 4 component capacitors
(B) 12 component capacitors (A)
(C) 48 component capacitors (B)

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(C)
(D)

Q7 Five identical plates each of area A are joined

as shown in the figure. The distance between
successive plates is d. The plates are
connected to potential difference of V volt.
(A) (B)
Find the charges of plates 1 and 4
(C) (D)

(A)

(B)

(C)

(D)

Q8 Two spheres of radii R1 and R2 having equal

charges are joined together with a copper
wire. If V is the potential of each sphere after
they are separated from each other, then the
initial charge on both spheres was:
(A) (B)

(C) (D)

Q9 Two spheres of radii 1 cm and 2 cm have been

charged with
coulombs of positive charge. When they are
connected with a wire, charge:
(A) will flow from the first to the second
(B) will flow from the second to the first
(C) will not flow at all
(D) may flow either from first to second, or
from the second to first, depending upon
the length of the connecting wire

Q10 In the following circuit the resultant

capacitance between A & B is . Find the
value of C:

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Answer Key
Q1 C Q6 C

Q2 C Q7 A
Q3 C Q8 B
Q4 A Q9 C
Q5 B Q10 B

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Hints & Solutions

Note: scan the QR code to watch video solution

Q1 Text Solution:

Video Solution:

total capacitors

Video Solution:

Q2 Text Solution:
V2 = V3 {Because C2 and C3 in parallel}
Q1 = Q2 + Q3
Q4 Text Solution:
V = V1 + V2
Given circuit is balanced wheat stone bridge.
Hence effective capacity is
Video Solution: Video Solution:

Q3 Text Solution: Q5 Text Solution:

Identify Wheatstone bridge

Suppose m rows of given capacitors are CPQ = 1

connected in parallel and each row now
Video Solution:
contains n capacitors then potential difference

across each capacitor and equivalent

capacitance of network on putting

values,

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 Boards

Q6 Text Solution:

Common potential

Charge on
Video Solution:

Charge on

Video Solution:
Q9 Text Solution:
As the potential of two spheres is same hence
there will be no flow of charge.

Video Solution:

Q7 Text Solution:

Charge on plate

Q10 Text Solution:

Charge on plate

Video Solution:

Video Solution:

Q8 Text Solution:
Let Initial charge = q
then from charge conservation

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 PARISHRAM 2026
PHYSICS DPP: 5
Electrostatic Potential and Capacitance

Q1 A capacitor of capacitance C has a charge Q. volt. Calculate the charge spent for
The net charge on the capacitor is always each flash.
(A) zero (B) infinite (A) 0.07 C (B) 0.09 C
(C) positive (D) Negative (C) 0.08 C (D) 0.06 C

Q2 A capacitor gets a charge of when it is Q6 Two capacitors C1 and C2 have equal amount
connected to a battery of emf 5 V . Calculate
of energy stored in them. What is the ratio of
the capacity of the capacitor.
potential differences across their plates?
(A)
(A)
(B)
(C) (B)
(D)
(C)
Q3 When 2 × 1016 electrons are transferred from
(D)
one conductor to another, a potential
difference of 10 V appears between the
Q7 To increase the charge on the plate of a
conductors. Calculate the capacitance of the
capacitor implies to :-
two conductors system.
(A) decrease the potential difference between
(A)
the plates.
(B)
(B) decrease the capacitance of the capacitor.
(C)
(C) increase the capacitance of the capacitor.
(D)
(D) increase the potential difference between
Q4 The graph shows the variation of voltage V the plates.
across the plates of two capacitors A & B with
Q8 Capacitors are used in electrical circuits where
charge Q . Which of the two capacitors has
appliances need rapid :
larger capacitance?
(A) Current (B) Voltage
(C) Watt (D) Resistance

Q9 Assertion (A): The capacitance of a capacitor

depends only on the geometry of the plates
(A) CA= CB (B) CA< CB
and the nature of dielectric.
(C) CA> CB (D) None of these
Reason (R): Capacitance is given by
Q5 For flash pictures, a photographer uses a (A) Both Assertion and Reason are true, and
capacitor and a charger that supplies Reason is the correct explanation of the

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 Assertion.
(B) Both Assertion and Reason are true, but
Reason is not the correct explanation of the
Assertion.
(C) Assertion is true, but Reason is false.
(D) Assertion is false, but Reason is true.

Q10 The capacitance of a parallel plate capacitor is

. If the distance between the plates is
tripled and area doubled then new
capacitance will be
(A)
(B)
(C)
(D)

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 Answer Key
Q1 A Q6 C

Q2 B Q7 D
Q3 D Q8 A
Q4 C Q9 A
Q5 B Q10 B

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 Hints & Solutions
Note: scan the QR code to watch video solution

Q1 Text Solution:
Zero

Video Solution:

Q5 Text Solution:
Q = CV = 0.09 C

Video Solution:
Q2 Text Solution:
Capacity of the capacitor

Video Solution:

Q6 Text Solution:
According to question

Q3 Text Solution:
Video Solution:

Video Solution:

Q7 Text Solution:
Q = CV
C does not depen on Q and V
Q4 Text Solution: Q V

A Because C = Video Solution:

Video Solution:

Q8 Text Solution:

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 Current Q10 Text Solution:

Video Solution: ...(1)

...(2)
on deviding (1) / (2)
we get

Q9 Text Solution:
Video Solution:
Capacitance is independent of voltage and
charge–depends only on A, d, . Both
statements are true and related.

Video Solution:

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 PARISHRAM 2026
PHYSICS DPP: 4
Electrostatic Potential and Capacitance

Q1 Certain positive charge is given to a conductor.

Then its potential: -
(A) is maximum at the surface
(B) is maximum at the centre
(C) remains same throughout the conductor
(D) is maximum somewhere between the
surface and the centre
(A) VA> VC, VB=VD
Q2 A solid conducting sphere having a charge Q is
(B) VA< VC, VB=VD
surrounded by an uncharged concentric
(C) VA= VC, VB< VD
conducting spherical shell. Let the potential
(D) VA= VC, VB> VD
difference between the surface of the solid
sphere and the outer surface of the shell be V. Q5 The electric field is constant in both magnitude
If the shell is now given a charge -3 Q the new and direction. Consider a path of length d at
potential difference between the same two an angle with respect to field lines as shown in
surfaces is :- figure. The potential difference between points
(A) V (B) 2V 1 and 2 is :-
(C) 4V (D) –2V

Q3 Electric field at a distance x from the origin is

given as . Then potential

difference between the points situated at x =
10 m and x = 20 is :- (A) (B)
(A) 5 V (B) 10 V
(C) (D)
(C) 15 V (D) 4 V

Q4 A circle of radius R is drawn in a uniform Q6 If a unit positive charge is taken from one
electric field E as shown in the fig. VA, VB, VC point to another over an equipotential surface,
and VD are respectively the potentials of points then
(A) Work is done on the charge
A, B, C and D on the circle then :-
(B) Work is done by the charge
(C) Work done is constant
(D) No work is done

Q7 There are two equipotential surface as shown

in figure. The distance between them is r. The

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 charge of q coulomb is taken from the surface (A) Both Assertion and Reason are true, and
A to B, the resultant work done will be Reason is the correct explanation of the
Assertion.
(B) Both Assertion and Reason are true, but
Reason is not the correct explanation of the
Assertion.
(C) Assertion is true, but Reason is false.
(A) (D) Assertion is false, but Reason is true.
(B)

(C)

(D) W = zero

Q8 The electric potential V at any point O(x, y, z all

in metres) in space is given by V = 4x2 volt The
electric field at the point ( 1m, 0,2m) in
volt/metre is
(A) 8 along negative X-axis
(B) 8 along positive X-axis
(C) 16 along negative X-axis
(D) 16 along positive Z-axis

Q9 Assertion (A): The equipotential surfaces due

to a single point charge are concentric
spheres.
Reason (R): The electric field of a point charge
is radial and spherically symmetric.
(A) Both Assertion and Reason are true, and
Reason is the correct explanation of the
Assertion.
(B) Both Assertion and Reason are true, but
Reason is not the correct explanation of the
Assertion.
(C) Assertion is true, but Reason is false.
(D) Assertion is false, but Reason is true.

Q10 Assertion (A): Equipotential surfaces are

closer together where the electric field is
stronger.
Reason (R): Electric field is the negative
gradient of electric potential.

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 Answer Key
Q1 C Q6 D

Q2 A Q7 D
Q3 A Q8 A
Q4 C Q9 A
Q5 B Q10 A

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 Hints & Solutions
Note: scan the QR code to watch video solution

Q1 Text Solution:
A and C lie on perpendicular to
inside conduction They are equipotential.
V remains same throughout the conductor.
Video Solution:
Video Solution:

Q5 Text Solution:
Q2 Text Solution: V1 –V2 = Ed cos60°
Potential Difference between inner and outer
Video Solution:
sphere will remain unchanged of the charges
kept on outer shell.

Video Solution:

Q6 Text Solution:

for points on equipotential

Q3 Text Solution:
surface

Video Solution:

Potential Difference

Video Solution:

Q7 Text Solution:
Surfaces are equipotential
VA = VB
W = q(VB – VA) = 0
Q4 Text Solution:
Video Solution:
VD > VB as V in direction of

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 distance r. So, equipotentials are spheres.
The electric field is radial, which is
perpendicular to these spherical surfaces.

Video Solution:

Q8 Text Solution:

Q10 Text Solution:

The electric field , which means the
Video Solution:
greater the rate of change of potential with
distance (closer equipotentials), the stronger
the field.

Video Solution:

Q9 Text Solution:
For a point charge, potential at a distance r is

, which is same at all points at

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 PARISHRAM 2026
PHYSICS DPP: 3
Electrostatic Potential and Capacitance

Q1 The distance between H+ and Cl– ions in HCl Q5 If an electric dipole is placed in an electric field
generated by a point charge then: -
molecule is 1.28Å What will be the potential
(A) the net electric force on the dipole must be
due to this dipole at a distance of 12Å on the
zero
axis of dipole
(B) the net electric force on the dipole may be
(A) 0.13 V (B) 1.3 V
zero
(C) 13V (D) 130V
(C) the torque on the dipole due to the field
Q2 must be zero
An electric dipole of dipole moment is
(D) the torque on the dipole due to the field
placed at the origin along the x -axis. The angle
may be zero
made by electric field with x-axis at a point P,
whose position vector makes an angle with Q6 An electric dipole is placed in non-uniform

x-axis, is: (where, ) electric field, then it experiences -

(A) a force equal to zero
(A) (B)
(B) A non - zero torque
(C) (D)
(C) a force which must be non-zero
Q3 At any point on the perpendicular bisector of (D) can't predict behaviour
the line joining two equal and opposite
Q7 Two charges
charges :-
(A) the electric field is zero Å
(B) the electric potential is zero apart forms a dipole. If it is kept in uniform
(C) the electric potential decreases with electric field of intensity then
increasing distance from their mid-point what will be its electrical energy in equilibrium
(D) the electric field is perpendicular to the line (A)
joining the charges (B)
(C)
Q4 The electric potential at a point due to an
(D)
electric dipole will be :-
(A)
Q8
An electric dipole of dipole moment is lying
(B)
along a uniform electric field . The work
(C) done in rotating the dipole by 90° is: -
(A) 2pE (B) pE
(D) (C) (D)

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Q9 An electric dipole of moment P is placed in the
position of stable equilibrium in uniform
electric field of intensity E. It is rotated through
an angle from the initial position. The
potential energy of electric dipole in the final
position is
(A)
(B)
(C)
(D)

Q10 The electric field due to an electric dipole at a

distance r from its centre in axial position is E.
If the dipole is rotated through an angle of 90°
about its perpendicular axis, the electric field
at the same point will be
(A) E (B) E/4
(C) E/2 (D) 2E

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 Answer Key
Q1 A Q6 C

Q2 C Q7 B
Q3 B Q8 B
Q4 A Q9 D
Q5 D Q10 C

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 Hints & Solutions
Note: scan the QR code to watch video solution

Q1 Text Solution: Q3 Text Solution:

Electric potential due to an electric dipole

Video Solution:

According to given data

Video Solution:
Q4 Text Solution:

Q2 Text Solution:

Video Solution:

In shown diagram, Net electric field

vector due to dipole. (by derivation) Q5 Text Solution:
The torque on the dipole to the field may be
Angle made by
zero.
Torque acting on a dipole placed in an electric
field.
Video Solution:
So, for the torque will be zero.
Hence, the amount of torque acting on the
dipole depends on the orientation of the
dipole in the given electric field.

Video Solution:

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 Q8 Text Solution:

Video Solution:

Q6 Text Solution:
a force which must be non-zero

Video Solution:

Q9 Text Solution:

Potential Energy

Video Solution:

Q7 Text Solution:
Given

Q10 Text Solution:

Video Solution:

Video Solution:

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 PARISHRAM 2026
DPP: 2
PHYSICS
Electrostatic Potential and Capacitance

Q1 1 Volt is equivalent to: Q6 Two unlike charges of magnitude q are

(A) (B) separated by a distance 2d The potential at a
point midway between them is
(C) (D)
(A) Zero
(B)
Q2 Two unlike charges of magnitude q are
separated by a distance 5d. The potential at a (C)
point midway between them:
(D)
(A) Zero
(B)
Q7 An electric charge of 20 Q is situated at the
(C) origin of X - Y co-ordinate system. The
(D) potential difference between the points. (5 a,
0) and ( -3 a, 4 a ) will be
Q3 Two points A and B are maintained at the (A) a (B) 2a
potentials of 18 V and –5V respectively. What (C) Zero (D)
will be the magnitude of work done in moving
150 α-particles from A to B? Q8 Four charges 2C,-3 C,-4C and 5C respectively
(A) 4 × 10–12J are placed at the four corners of a square.
(B) 8 × 10–8J Which of the following statements is true for
(C) 2 × 10–15J the point of intersection of the diagonals?
(D) 1.1 × 10–15J (A)
(B)
Q4 The electric potential and electric field at a (C)
point due to a point charge are 600 V and 200
(D)
N/C respectively. Then magnitude of the point
charge should be :- Q9 Point charge are
(A) (B) kept at points x = 0 and x = 6 respectively.
(C) (D) Electrical potential will be zero at points
(A) x = 2 and x = 9
Q5 The potential at a point, due to a positive
(B) x = 1 and x = 5
charge of at a distance of 9 m, is (C) x = 4 and x = 12
(A) (B) (D) x = -2 and x = 2
(C) (D)
Q10 Two charges of each are placed at the
corners A and B of an equilateral triangle ABC

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of side length 0.2m in air. The electric potential

at C is

(A)
(B)
(C)
(D)

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 Answer Key
Q1 C Q6 A

Q2 A Q7 C
Q3 D Q8 B
Q4 C Q9 C
Q5 B Q10 C

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 Hints & Solutions
Note: scan the QR code to watch video solution

Q1 Text Solution:
One volt (1 V) is defined as the potential
difference between two points in a circuit
Video Solution:
when one joule (1 J) of work is done to move
one coulomb (1 C) of charge between those
points.

Video Solution:

Q4 Text Solution:

Q2 Text Solution:
The electric potential at a point due to a
charge q at a distance r is given by:

Video Solution:
Since potential is a scalar quantity, we can
directly add the contributions from both

=0 Q5 Text Solution:
Video Solution: By using

Video Solution:

Q3 Text Solution:

Q6 Text Solution:
Potential at mid-point

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 Video Solution: Q9 Text Solution:
Case-I

Q7 Text Solution:
Case-II

Video Solution:

Video Solution:

Q10 Text Solution:

Q8 Text Solution:

Video Solution: Video Solution:

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Android App | iOS App | PW Website
 PARISHRAM 2026
DPP: 1
PHYSICS
Electrostatic Potential and Capacitance

Q1 Pick out the expression of electric potential Q4 Three point charges q, and q are placed
energy from the following. at the corners of an equilateral triangle of side
(A) a. The electrostatic potential energy of the
system is:
(B)
(A) (B)
(C)
(C) (D) Zero
(D) U

Q5 A system consists of four identical point

Q2 Two isolated metallic spheres, one with a charges +q placed at the corners of a square
radius R and another with a radius 5R, each of side a. The electrostatic potential energy of
carries a charge 'q' uniformly distributed over this configuration is:
the entire surface. Which sphere stores more (A) (B)
electric potential energy?
(A) The sphere with radius 5R (C) (D)
(B) Both of the spheres will have the same
energy Q6 Two equal charges q are placed at a distance
(C) The sphere with radius R of 2a and a third charge is placed at the
(D) Initially it will be the sphere with radius 5R midpoint. The potential energy of the system
then it will shift to the sphere with radius R is
(A)
Q3 What should be the value of Q so that the
potential energy of the system is zero? (B)

(C)

(D) None of the above

Q7 A positively charged particle is released from

rest in a uniform electric field. The electric
potential energy of the charge:
(A) Remains constant because the electric field

(A) 2q (B) is uniform

(B) Increases because the charge moves along
(C) (D) the electric field
(C) Decreases because the charge moves along
the electric field

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 (D) Decreases because the charge moves
opposite to the electric field

Q8 Choose the incorrect statement :-

(A) the potential energy per unit positive
charge in an electric field at some point is
called the electric potential.
(B) the work required to be done to move a
point charge from one point to another in
an electric field depends on the position of
the points
(C) the potential energy of the system will
increase if a positive charge is moved
against the Coulombian force
(D) the value of fundamental charge is not
equivalent to the electronic charge.

Q9 When one electron is taken towards the other

electron, then the electric potential energy of
the
(A) decreases
(B) increases
(C) remains unchanged
(D) will become zero

Q10 Three charges are placed along x-axis at x = –a,

x = 0 and x = a as shown in the figure. The
potential energy of the system is:

(A)

(B)

(C)

(D)

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 Answer Key
Q1 A Q6 C

Q2 C Q7 C
Q3 D Q8 D
Q4 B Q9 B
Q5 C Q10 B

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 Hints & Solutions
Note: scan the QR code to watch video solution

Q1 Text Solution: energy is zero, i.e.

Explanation: electric potential energy of a
.
system of charges is the total amount of work
done in bringing the various charges to their Therefore, simplifying the equation, we get Q
respective positions from infinitely large = .
mutual separations. The expression for electric
potential energy is given by: Video Solution:

Video Solution:

Q4 Text Solution:
Total potential energy = sum of interaction
energies:
Q2 Text Solution:
Explanation: The sphere with radius R stores
more electric potential energy. According to Video Solution:
the electric potential energy equation
Potential energy is

inversely proportional to radius. Therefore, the

sphere with lesser radius will store more
energy. So, the smaller sphere will store more
energy.
Q5 Text Solution:
Video Solution: There are 6 unique pairs: 4 edges contribute

each, and 2 diagonals contribute

U=

Q3 Text Solution: Video Solution:

Explanation: The distance between the

charges Q and + q is a according to
Pythagoras theorem. Therefore, the total
potential energy of the system becomes

. But

according to the questions, the total potential Q6 Text Solution:

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 1. Between charge q at A and charge -2q at 0:
Distance = a

Potential energy
2. Between charge q at B and charge -2q at 0:
Distance = a

Potential energy Q9 Text Solution:

Potential energy of system will be given by
Potential energy
Substituting the values we calculated:
decrease then potential energy will be
increases.

Video Solution:

Video Solution:

Q10 Text Solution:

(2)

Q7 Text Solution:
The electric potential energy of the positively
charged particle decreases as it moves in the
direction of the electric field.

Video Solution:

Utotal = U1 + U2 + U3

Q8 Text Solution: Video Solution:

the value of fundamental charge is not

equivalent to the electronic charge.

Video Solution:

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 Boards

PARISHRAM 2026
PHYSICS Assignment - 01

ELECTRIC CHARGES AND FIELDS

Q1 Assertion (A): Equal amount of positive and electric field of magnitude, will be at a
negative charges are distributed uniformly on
distance of
two halves of a thin circular ring as shown in
(A) 1 m (B) 2 m
figure. The resultant electric field at the centre
(C) 3 m (D) 6 m
O of the ring is along OC.
Reason (R): It is so because the net potential at Q4 A point P lies at a distance x from the mid
O is not zero. point of an electric dipole on its axis. The
electric field at point P is proportional to
(A) (B)

(C) (D)

Q5 Beams of electrons and protons move parallel

to each other in the same direction. They
(A) attract each other.
(A) both Assertion (A) and Reason (R) are true
(B) repel each other.
and Reason (R) is correct explanation of
(C) neither attract nor repel.
Assertion (A).
(D) force of attraction or repulsion depends
(B) both Assertion (A) and Reason (R) are true
upon speed of beams.
and Reason (R) is not the correct
explanation of Assertion (A). Q6 Assertion (A) : Work done in moving a charge
(C) Assertion (A) is true but Reason ( R ) is false. around a closed path, in an electric field is
(D) both Assertion (A) and Reason (R) are false. always zero.
Reason (R) : Electrostatic force is a
Q2 An electric dipole of length 2 cm is placed at an
conservative force.
angle of with an electric field
(A) Both Assertion (A) and Reason (R) are true
. If the dipole experiences a and Reason (R) is the correct explanation of
torque of , the magnitude of the Assertion (A).
either charge of the dipole, is (B) Both Assertion (A) and Reason (R) and true,
(A) (B) but Reason (R) is not the correct
(C) (D) explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
Q3 The magnitude of the electric field due to a (D) Assertion (A) is false, but Reason (R) is true.
point charge object at a distance of 4.0 m is
Q7 If a body is charged by rubbing it, its weight
. From the same charged object the
(A) Always decrease slightly

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 Boards

(B) Always increase slightly (C) directed radially away from the point
(C) May increase or decrease slightly charge
(D) No change in weight (D) directed radially towards the point charge

Q8 In figure, two positive charges q2 and q3 fixed Q11 A positive charge Q is uniformly distributed
along the y axis, exert a net electric force in along a circular ring of radius R. A small test
the +x direction on a charge q1 fixed along the charge q is placed at the centre of the ring.
Then
x axis. If a positive charge Q is added at (x, 0)
the force on q1

(A) shall increase along the positive x-axis (A) if q > 0 and is displaced away from the
(B) shall decrease along the positive x-axis centre in the plane of the ring., it will be
(C) shall point along the negative x-axis pushed back towards the centre
(D) shall increase but the direction 2and3 (B) if q < 0 and is displaced away from the
changes because of the intersection q centre in the plane of the ring, it will never
of Q with q return to the centre and will continue
moving till it hits the ring
Q9 The electric flux through the surface
(C) if q < 0, it will perform SHM for small
displacement along the axis
(D) All of the above

Q12 What will be the total flux through the faces of

the cube with side of length a if a charge q is
placed at corner A of the cube.
(A) fig (iv) is the largest
(B) fig (iii) is the least
(C) fig (ii) is same as fig (iii) but is smaller than
fig (iv)
(D) is the same for all the figures

Q10 A point charge +q is placed at a distance d

from an isolated conducting plane. The field at
a point P the other side of the plane is
(A) directed perpendicular to the plane and
away from the plane
(B) directed perpendicularly to the plane (A)
towards the plane (B)
(C)

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 Boards

(D) None of the above

Q13 Two small balls having equal charges Q, are

suspended from a hook with two insulating
threads each of length L. This arrangement is
carried in the space, where there is no
gravitation. The tension in each string will be-

Correct order of charging is

(A) (B) 0
(A)
(C) (D) (B)
(C)
(D)
Q14 A copper atom consists of copper nucleus
surrounded by 29 electrons. The atomic Q17 Assertion : The property that the force with
weight of copper is 63.5 g/ mole. Let us now which two charges attract or repel each other
take two pieces of copper each wieghing 10 g . are not affected by the presence of a third
Let us transfer one electron from one piece to charge.
another for every 1000 atoms in that piece. Reason : Force on any charge due to a number
What will be the coulomb force between the of other charge is the vector sum of all the
two pieces after the transfer of electron if they forces on that charge due to other charges,
are 1 cm . apart. [Avogadro number N = 6 × taken one at a time.
1023/g. mole charge on an electron (=-1.6 × 10- (A) Assertion is correct, reason is correct;
19 reason is a correct explanation for
coulomb]
assertion.
(A) 2.057 × 1016N
(B) Assertion is correct, reason is correct;
(B) 2.05 × 1020N
reason is not a correct explanation for
(C) 3.2 × 106N
assertion
(D) 3.02 × 106N
(C) Assertion is correct, reason is incorrect
Q15 The bob of a pendulum carries an electric (D) Assertion is incorrect, reason is correct.
charge of coulomb in an
Q18 The surface density on the copper sphere is .
electric field of and it is at rest. The electric field strength on the surface of the
The angle made by the pendulum with the sphere is
vertical will be, if the mass of pendulum is (A) (B)
- (C) (D)
(A) 27° (B) 45°
(C) 87° (D) 127° Q19 The electric intensity due to a dipole of length
10 cm and having a charge of , at a
Q16 A metal sphere is being charged by induction point on the axis at a distance 20 cm from one
using a charged rod, but the sequence of of the charges in air, is
diagrams showing the process misplaced. (A)
(B)

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 Boards

(C)
(D)

Q20 An electric dipole has the magnitude of its

charge as q and its dipole moment is p. It is
placed in uniform electric field E . If its dipole
moment is along the direction of the field, the
force on it and its potential energy are
respectively.
(A) q.E and max.
(B) 2q. E and min
(C) q.E and min
(D) zero and min.

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 Boards

Answer Key
Q1 C Q11 D

Q2 A Q12 A
Q3 C Q13 A

Q4 A Q14 A

Q5 A Q15 B
Q6 A Q16 C

Q7 C Q17 B
Q8 A Q18 D

Q9 D Q19 A

Q10 A Q20 D

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 Boards

Hints & Solutions

Note: scan the QR code to watch video solution

Q1 Text Solution:
Electric field is from positive charge to negative
charge, hence the direction of electric field will which simplifies to:
be along OC. Also, the net potential at O is
zero. Hence, the assertion is true and the
reason is false. By cross-multiplying and solving for :

Video Solution:

Finally, taking the square root of both sides:

Thus, the distance at which the electric field of

magnitude will be:

Q2 Text Solution: Video Solution:

Q4 Text Solution:
Expression for electric field due to dipole at
Or,
geueral point in given as
Or,

So,
Video Solution:
Video Solution:

Q3 Text Solution:
Q5 Text Solution:
Using the inverse square law relationship for
Since, electron and proton has opposite
electric fields from a point charge:
charge and moving in the same direction, So
there must be an attractive force between
them.
Substituting the known values: Video Solution:

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 Boards

is the same for all the figures as charge is

same

Video Solution:

Q6 Text Solution:
Electric field is said to be conservative since,
work done in this field is path independent. It
depends on the end points only. So, in an
Q10 Text Solution:
electric field, total work done in moving a
directed perpendicular to the plane and away
charge from a point to the next point and
from the plane
again back to the first point is zero.
Video Solution:
Hence, assertion and reason both are true and
reason explains the assertion.

Video Solution:

Q11 Text Solution:

The correct answer is
a) if q > 0 and is displaced away from the
Q7 Text Solution:
centre in the plane of the ring., it will be
May increase or decrease slightly
pushed back towards the centre
Video Solution: b) if q < 0 and is displaced away from the
centre in the plane of the ring, it will never
return to the centre and will continue moving
till it hits the ring
d) q at the centre of the ring is in an unstable
equilibrium within the plane of the ring for q >
0
Q8 Text Solution:
Video Solution:
shall increase along the positive x-axis.

Video Solution:

Q12 Text Solution:

When we consider the charged particle to be
Q9 Text Solution: placed at the centre of the cube whose side is
2 a , then the charge is equally distributed

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 Boards

among 8 cubes. Therefore, the total flux Video Solution:

through the faces of the cube

Video Solution:

Q15 Text Solution:

Q13 Text Solution:

(electrostatic force acting between

two charged balls)

Video Solution:

Video Solution:

Q14 Text Solution:

Sol. 63.5 g copper contains N = 6 × 1023 copper
atoms. Therefore number of copper atoms in
10 g copper = 6 × 1023 / 63.56 × 10
As only one electron is transferred for every
1000 atoms, therefore the number of electron Q16 Text Solution:
transferred When charged rod is brought near uncharged
conductor near end of conductor has opposite
Magnitude of charge charge. When for end of this conductor is
connected is ground (i.e., earthed), charge of
Coul.
far end flows down to ground when for end
connection and rod are removed charge on
Separation between pieces conductor spreads uniformly on surface.
One piece of copper has Video Solution:
positive charge and the other negative charge,
therefore force of
attraction between the pieces

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 Boards

Q17 Text Solution: of the point on the axis from the mid-point of
Force on any charge due to a number of other the dipole
charges is the vector sum of all the forces on We know that the electric field intensity due to
that charge due to the other charges, taken dipole on the given point (E)
one at a time. The individual force are
unaffected due to the presence of other
charges. This is the principle of superposition
of charges.

Video Solution:

Video Solution:

Q18 Text Solution:

According to Gauss's theorem,

Q20 Text Solution:

When the dipole is in the direction of field then
net force is
Video Solution:

and its potential energy is minimum = - p.E.= –

qaE.

Video Solution:
Q19 Text Solution:
Given : Length of the dipole
0.05 m
Charge on the dipole
and distance

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 PARISHRAM 2026
PHYSICS DPP: 5

ELECTRIC CHARGES AND FIELDS

Q1 The electric field due to a continuous charge Q5 The electric field at a point on the axis of a
distribution depends on: uniformly charged ring of radius R and total
(A) The geometry of the charge distribution charge Q is:
and the location of the point of interest (A) (B)
(B) Only the total charge
(C) Only the distance from the charge (C) (D)
(D) None of the above

Q2 For a charge distribution with a volume charge Q6 An electron and a proton are at distance of 1Å .
density , the total charge in a spherical The moment of this dipole will be (C × m)
volume of radius R is given by: (A) 1.6 ×
(A) (B) 1.6 ×
(B) (C) 3.2 ×
(D) 3.2 ×
(C)
(D) Q7 A given charge is situated at a certain distance
from an electric dipole in the end-on position
Q3 The surface charge density is defined as:
experiences a force F. If the distance of the
(A) Charge per unit volume
charge is doubled, the force acting on the
(B) Charge per unit length
charge will be
(C) Charge per unit area
(A) 2F (B) F/2
(D) Charge per unit time
(C) F/4 (D) F/8
Q4 A uniformly charged infinite line of charge with
Q8 The spatial distribution of the electric field
linear charge density creates an electric field
lines due to charges (A, B) is shown in figure.
at a point a distance r from the line. If the
Which one of the following statements is
electric field at this point is E, what will happen correct?
to the electric field if the linear charge density
is doubled?
(A) The electric field will increase by a factor of
2.
(B) The electric field will increase by a factor of
4.
(C) The electric field will decrease by a factor of
(A) A is +ve and B is -ve and |A| > |B|
2.
(B) A is – ve and B is + ve; |A| = |B|
(D) The electric field will remain unchanged.

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 (C) Both are + ve but A > B (A) A net force, but no torque
(D) Both are – ve A > B (B) No net force, but a torque
(C) A net force and a torque
Q9 When an electric dipole is placed in a uniform
(D) Neither a force nor a torque
electric field, it experiences:

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 Answer Key
Q1 A Q6 B

Q2 B Q7 D
Q3 C Q8 A
Q4 A Q9 B
Q5 A

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 Hints & Solutions
Note: scan the QR code to watch video solution

Q1 Text Solution: The electric field due to a uniformly charged

The electric field at a point due to a continuous infinite line is given by . If is
charge distribution depends on the geometry
doubled, the electric field will also double.
of the distribution and the location of the
point. Video Solution:

Video Solution:

Q5 Text Solution:
The electric field at a point on the axis of a
Q2 Text Solution:
uniformly charged ring at a distance x from the
The total charge in a spherical volume with
centre of the ring is given by the equation:
radius R and volume charge density is given
by the volume of the sphere multiplied
by . Video Solution:
Thus,

Video Solution:

Q6 Text Solution:

Q3 Text Solution:
Surface charge density is the charge per unit
Video Solution:
area, typically expressed as , where Q
is the charge and A is the surface area.

Video Solution:

Q7 Text Solution:

Q4 Text Solution:

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 Video Solution: In a uniform electric field, there is no net force
acting on the dipole because the forces on the
positive and negative charges cancel each
other out. However, the dipole experiences a
torque, which tends to align the dipole with
the direction of the electric field. The torque is
given by:
Q8 Text Solution: T = pE sin
More field lines are comming out of A than B Where p is the dipole moment, E is the electric
Hence, |A| > |B|. field streghth, and is the angle between the
Video Solution: dipole moment and the electric field.

Video Solution:

Q9 Text Solution:

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 PARISHRAM 2026
PHYSICS DPP: 6

ELECTRIC CHARGES AND FIELDS

Q1 If the electric field in a region is uniform and (D) +q1and –q1

makes an angle of 30° with the surface, and
Q4 Assertion: The electric flux through a closed
the magnitude of the electric field is 5 N/C and
surface surrounding a positive charge is
the area is 10 m2 then the electric flux passing
proportional to the charge enclosed inside the
through the surface is:
surface.
(A) 50 Nm2/C (B) 43.3 Nm2/C
Reason: According to Gauss's Law, the flux
(C) 100 Nm2/C (D) 25 Nm2/C
through any closed surface is proportional to
Q2 Which of the following is NOT a valid the total charge enclosed by the surface and is
application of Gauss's Law? given by
(A) Calculation of the electric field due to a
(A) Both assertion and reason are true, and the
uniformly charged infinite plane sheet.
reason is the correct explanation of the
(B) Calculation of the electric field due to a
assertion.
uniformly charged spherical shell at points
(B) Both assertion and reason are true, but the
inside and outside the shell.
reason is not the correct explanation of the
(C) Calculation of the electric field due to a
assertion.
point charge.
(C) Assertion is true, but reason is false.
(D) Calculation of the electric field due to a
(D) Assertion is false, but reason is true.
charged conductor using Coulomb’s law.
Q5 An electric charge q is placed at the centre of a
Q3 Consider the charge configuration and a
cube of side a. The electric flux through one of
spherical Gaussian surface as shown in the
its faces will be
figure. When calculating the flux of the electric
(A)
field over the spherical surface, the electric
field will be due to (B)

(C)

(D)

Q6 Total electric flux coming out of a unit positive

charge put in air is
(A)
(B)
(A) q2
(B) Only the positive charges (C)
(C) All the charges (D)

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Q7 The S.I. unit of electric flux is
(A) Weber
(B) Newton per coulomb
(C) Volt × metre
(D) Joule per coulomb

Q8 A charge q is placed at the centre of the open

end of cylindrical vessel. The flux of the electric
field through the surface of the vessel is
(A) Zero (B)

(C) (D)

Q9 Assertion: The electric flux through a

spherical surface surrounding a point charge is
independent of the size of the sphere.
Reason: According to Gauss’s Law, the electric
flux through a spherical surface only depends
on the charge enclosed and not on the radius
or size of the sphere.
(A) Both assertion and reason are true, and the
reason is the correct explanation of the
assertion.
(B) Both assertion and reason are true, but the
reason is not the correct explanation of the
assertion.
(C) Assertion is true, but reason is false.
(D) Assertion is false, but reason is true.

Q10 The electric flux through a closed surface area

S enclosing charge Q is . If the surface area is
doubled, then the flux is
(A) (B)

(C) (D)

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 Answer Key
Q1 B Q6 B

Q2 D Q7 C
Q3 C Q8 C
Q4 A Q9 A
Q5 A Q10 D

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 Hints & Solutions
Note: scan the QR code to watch video solution

Q1 Text Solution:
The electric flux is
Substituting the values:

Video Solution: Q4 Text Solution:

The assertion is true because the flux through
a closed surface surrounding a charge is
proportional to the charge enclosed. The
reason is also true, as it is the mathematical
statement of Gauss's Law.

Video Solution:
Q2 Text Solution:
Gauss’s Law can be used to derive the electric
field due to a point charge, spherical shell, and
infinite plane sheet. However, it cannot
directly be applied to charged conductors
unless symmetry is considered. Coulomb’s Law
is typically used to calculate the electric field Q5 Text Solution:
for such conductors.
By Gauss's theorem. Total flux
Video Solution:
So the flux through one face

Video Solution:

Q3 Text Solution:
At any point over the spherical Gaussian
surface, net electric field is the vector sum of Q6 Text Solution:
electric fields due to +q2, q1 and q2. Don't Total flux coming out form unit charge
confuse with the electric flux which is zero
(net) passing over the Gaussian surface as the
Video Solution:
net charge enclosing the surface is zero.

Video Solution:

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 Video Solution:

Q7 Text Solution:
S.I. unit of electric flux is Q9 Text Solution:
The assertion is true because the flux through
a spherical surface surrounding a point charge
Video Solution: is the same regardless of the size of the
sphere. The reason is also true because
Gauss’s Law states that the flux depends only
on the charge enclosed, not on the radius of
the sphere.

Video Solution:

Q8 Text Solution:
To apply Gauss's theorem it is essential that
charge should be placed inside a closed
surface. So imagine another similar cylindrical
vessel above it as shown in figure (dotted).

Q10 Text Solution:

Flux depends on charge enclosed by a close
surface.

Video Solution:

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 Parishram 2026
PHYSICS DPP: 2

ELECTRIC CHARGES AND FIELDS

Q1 When a glass rod is rubbed with silk, the glass (B) Induced charge and electrostatic force
rod becomes: (C) Gravitational attraction
(A) Negatively charged (D) Surface tension
(B) Positively charged
Q7 Assertion (A): The electrostatic force between
(C) Neutral
two charges increases if the distance between
(D) Electrically inert
them is decreased.
Q2 Which of the following is true about charging Reason (R): According to Coulomb's Law, force
by friction? is inversely proportional to the square of the
(A) Both bodies always lose charge distance between charges.
(B) Electrons are transferred from one body to (A) Both A and R are true, and R is the correct
another explanation of A
(C) Protons are transferred (B) Both A and R are true, but R is not the
(D) Neutrons are transferred correct explanation of A
(C) A is true but R is false
Q3 During charging by friction, the total charge of
(D) A is false but R is true
the system:
(A) Increases Q8 Assertion (A): Electrostatic force between
(B) Decreases charges is greater in vacuum than in water.
(C) Remains conserved Reason (R): Water has a high dielectric
(D) Becomes zero constant which reduces the force between
charges.
Q4 The electrostatic force is a:
(A) Both A and R are true, and R is the correct
(A) Non-conservative and contact force
explanation of A
(B) Conservative and long-range force
(B) Both A and R are true, but R is not the
(C) Short-range and non-conservative force
correct explanation of A
(D) Gravitational type short-range force
(C) A is true but R is false
Q5 If the distance between two charges is tripled, (D) A is false but R is true
the force becomes:
Q9 There are two charges +1 microcoulomb and
(A) (B)
+5 microcoulomb. The ratio of the forces
(C) 3 times (D) 9 times acting on them will be
(A) 1: 5 (B) 1: 1
Q6 When a balloon is rubbed with hair, it sticks to
(C) 5: 1 (D) 1 : 25
the wall. This is due to:
(A) Magnetic attraction

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Q10 Two point charges repel is added to each of them, then the
each other with a force of 40 N . If a charge of force between them will become
(A) -10 N (B) +10 N
(C) +20 N (D) -20 N

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 Answer Key
Q1 B Q6 B

Q2 B Q7 A
Q3 C Q8 A
Q4 B Q9 B
Q5 A Q10 A

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 Hints & Solutions
Note: scan the QR code to watch video solution

Q1 Text Solution: Conservative and long-range force

Positively charged Electrostatic forces are conservative (work is
Electrons move from the glass rod to silk, so path-independent) and long-range (act over
the glass loses electrons and becomes distance).
positively charged.
Video Solution:
Video Solution:

Q5 Text Solution:
Q2 Text Solution:
Electrons are transferred from one body to
another
Only electrons are mobile. Friction causes Video Solution:
electrons to be transferred from one body to
another.

Video Solution:

Q6 Text Solution:
Induced charge and electrostatic force
The charged balloon induces opposite charges
Q3 Text Solution: on the wall, causing electrostatic attraction.
Remains conserved Video Solution:
Law of conservation of charge – the total
charge before and after friction remains
constant.

Video Solution:

Q7 Text Solution:
Coulomb's Law: , so decreasing
distance increases the force. The reason
directly explains the assertion.
Q4 Text Solution:
Video Solution:

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 Video Solution:

Q8 Text Solution:
In Coulomb's law with a medium: Q10 Text Solution:
In second case, charges will be

Water's dielectric constant reduces the

force significantly. Since

Video Solution:

Video Solution:

Q9 Text Solution:
The same force will act on both bodies
although their directions will be different.

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 Parishram 2026
PHYSICS DPP: 1

ELECTRIC CHARGES AND FIELDS

Q1 The minimum possible charge on an object is (C) Charged negatively

(A) 1 coulomb (D) An insulator
(B) 1 stat coulomb
Q6 When 1014 electrons are removed from a
(C) 1.6 × 10–19coulomb
(D) 3.2 × 10–19coulomb neutral metal sphere, the charge on the
sphere becomes
Q2 Which of the following charges can not be (A) 16µC
present on an oil drop in Millikan's (B) –16µC
experiment: (C) 32µC
(A) 4.0 × 10–19C (D) –32µC
(B) 6.0 × 10–19C
(C) 10.0 × 10–19C Q7 Which of the following is a characteristic
(D) none of them property of a conductor?
(A) Allows free movement of electrons
Q3 Consider a neutral conducting sphere. A (B) Prevents the flow of electrons
positive point charge is placed outside the (C) Contains tightly bound electrons
sphere. The net charge on the sphere is then: (D) Does not allow charge transfer
(A) negative and distributed uniformly over its
surface Q8 When a negatively charged rod is brought near
(B) negative and appears only at the point on a neutral conducting sphere without touching
the sphere closest to the point charge it, the sphere will:
(C) negative and distributed non-uniformly (A) Gain a negative charge overall
over its entire surface of the sphere (B) Gain a positive charge overall
(D) zero (C) Have a negative charge on the near side
and positive on the far side
Q4 When a glass rod is rubbed with silk, it (D) Have a positive charge on the near side
(A) Gains electrons from silk and negative on the far side
(B) Gives electrons to silk
(C) Gains protons from silk Q9 What is the SI unit of electric charge?
(D) Gives protons to silk (A) Ampere (B) Coulomb
(C) Volt (D) Farad
Q5 When a body is earth connected, electrons
from the earth flow into the body. This means Q10 Which of the following is a property of electric
the body is charge?
(A) Uncharged (A) Charges exist only in pairs
(B) Charged positively (B) Like charges attract

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(C) Unlike charges repel (D) Charge is quantized

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 Answer Key
Q1 C Q6 A

Q2 D Q7 A
Q3 D Q8 D
Q4 B Q9 B
Q5 B Q10 D

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 Hints & Solutions
Note: scan the QR code to watch video solution

Q1 Text Solution: Q4 Text Solution:

The minimum possible charge on an object is When a glass rod is rubbed with silk:
the charge of an electron or a proton, which is • Electrons are transferred from the glass rod
approximately 1.6 × 10–19 coulomb. to the silk.
Thus, the correct answer is: Video Solution:
1.6 × 10–19 coulomb

Video Solution:

Q5 Text Solution:
When a body is earth-connected and electrons
flow from the earth into the body, it means:
Q2 Text Solution:
• The body is positively charged initially.
In Millikan's oil drop experiment, the charge
• To neutralize this positive charge, electrons
on an oil drop is always a multiple of the
from the earth move into the body.
elementary charge, which is approximately e =
Video Solution:
1.6 × 10–19 C.
Since none of the given charges are integer
multiples of the elementary charge, none of
them can be present on an oil drop.

Video Solution:

Q6 Text Solution:
Q=n×e
Q = 1014 × 1.6 × 10–19
Q = 1.6 × 10–5 C
Q3 Text Solution: Q = 16µC
The induced charge distribution is non- When electrons are removed, the sphere
uniform but the net charge of the sphere itself becomes positively charged.
remains zero. Video Solution:
Video Solution:

Q7 Text Solution:

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 In conductors, electrons are free to move, Q9 Text Solution:
allowing easy charge transfer. Metals like The SI unit of charge is the coulomb (C).
copper and aluminum are good conductors.
Video Solution:
Video Solution:

Q10 Text Solution:

Q8 Text Solution: Electric charge is quantized and exists in
In charging by induction, the near side of the discrete packets of ±e, where e = 1.6 × 10–19 C.
sphere attracts positive charges, while
Video Solution:
negative charges are repelled to the far side.

Video Solution:

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 PARISHRAM 2026
PHYSICS DPP: 4

ELECTRIC CHARGES AND FIELDS

Q1 The unit of electric field intensity is....... They represent the path a positive test charge
(A) N/C would follow in an electric field.
(B) V/m (D) The density of lines of force increases with
(C) Both (A) and (B) the strength of the electric field.
(D) None of these
Q6 Which of the following is true about the electric
Q2 The field due to a charge at a distance 'x' from it lines of force?
is 'E'. When the distance is doubled, the intensity (A) They never intersect each other.
of the field is (B) They form closed loops.
(A) E/8 (B) E (C) They always point towards the positive
(C) 2E (D) E/4 charge.
(D) They do not depend on the magnitude of the
Q3 The magnitude of electric field intensity E is such
charge.
that, an electron placed in it would experience an
electrical force equal to its weight is given by Q7 The electric field lines due to a positive point
mg
(A) mge (B) e
charge:
(C) (D) (A) Point radially outward from the charge.
2
e e
g
mg 2
m
(B) Point radially inward towards the charge.
Q4 A drop of 10 −6
kg water carries 10 −6
C charge. (C) Form closed loops around the charge.
What electric field should be applied to balance (D) Are parallel to each other.
its weight (assume g = 10 m/s
2
)
(A) 10 V/m upward Q8 Some electric lines of force are shown in figure,
(B) 10 V/m downward for point A and B.
(C) 0.1 V/m downward
(D) 0.1 V/m upward

Q5 Which of the following statements about electric

lines of force is false?
(A) They always start at positive charges and end
at negative charges.
(B) They form closed loops.
(C)

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 (B) Both assertion and reason are true, but the
reason is not the correct explanation of the
assertion
(C) Assertion is true, but reason is false.
(D) Assertion is false, but reason is true.

(A) EA > EB

(B) EA < EB

(C) EA = EB

(D) Can't determine from the given information

Q9 The given figure gives electric lines of force due

to two charge q 1 and q 2 . What are the signs of
the two charges?

(A) Both are negative

(B) Both are positive
(C) q1 is positive but q2 is negative
(D) q1 is negative but q2 is positive

Q10 Assertion: The electric field lines between two

point charges, one positive and one negative, are
straight lines that form a direct path from the
positive charge to the negative charge.
Reason: The electric field lines are always
directed from positive to negative charges, and
their symmetry depends on the relative
magnitude and distance between the charges.
(A) Both assertion and reason are true, and the
reason is the correct explanation of the
assertion.

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 Answer Key
Q1 (C) Q6 (A)

Q2 (D) Q7 (A)

Q3 (B) Q8 (A)

Q4 (A) Q9 (A)

Q5 (B) Q10 (D)

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 Hints & Solutions
Note: scan the QR code to watch video solution

Q1 Text Solution: Video Solution:

Both (A) and (B)

Video Solution:

Q5 Text Solution:
Electric lines of force do not form closed loops.

Q2 Text Solution: Video Solution:

The intensity of the field when the distance is
doubled is .
E

Video Solution:

Q6 Text Solution:
Electric lines of force do not intersect each other,
as this would imply that the electric field has two
directions at a point, which is impossible.
Q3 Text Solution:
According to the question, Video Solution:
mg
eE = mg ⇒ E =
e

Video Solution:

Q7 Text Solution:
The electric field lines due to a positive point
charge radiate outward from the charge.
Q4 Text Solution:
By using QE = mg Video Solution:
mg −6

; upward
10 ×10
⇒ E = = = 10 V /m
−6
Q 10

because charge is positive.

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Q8 Text Solution:
Lines are denser at A. so EA > EB

Video Solution:

Q9 Text Solution:
Both are negative

Video Solution:

Q10 Text Solution:

The assertion is false because the electric field
lines between two point charges are not always
straight, especially when the charges are of
unequal magnitudes or the distance between
them is large. They are straight only in the case
of equal magnitude charges at a large distance
apart. The reason, however, is true: electric field
lines always point from positive to negative
charges.

Video Solution:

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 CHEMISTRY 112 WORKSHEET
NOMENCLATURE OF COORDINATION COMPOUNDS AND IONS

A. Name the following complex compounds or ions.

1. [Al (H2O)6] Br3 Hexaaquaaluminum (III) bromide

2. [Cr (NH3)6] Cl3 Hexaamminechromium (III) chloride

3. K3 [FeF6] Potassium hexafluoroferrate (III)

4. [Zn (OH)4]-2 Tetrahydroxozincate (II) ion

5. [Co (H2O)4Cl2] Cl Tetraaquadichlorocobalt (III) chloride

6. [Cu (NH3)4]+2 Tetraamminecopper (II) ion

7. K2 [SnCl6] Potassium hexachlorostannate (IV)

8. [Pt (NH3)4Cl2] [PtCl6] Tetraamminedichloroplatinum (IV) hexachloroplatinate (IV)

B. Write the formula for each of the following complex compounds or ions.

1. Hexaamminecobalt (III) chloride [Co (NH3)6] Cl3

2. Diamminetetrabromoplatinum (VI) bromide [Pt (NH3)2 Br4] Br2

3. Tetraaquacadmium (II) nitrate [Cd (H2O)4] (NO3)2

4. Diamminesilver (I) ion [Ag (NH3)2]+

5. Sodium tetracyanocuprate (I) Na3[Cu (CN)4]

6. Silver hexacyanoferrate (II) Ag4[Fe (CN)6]

7. Tetraammineoxalatonickel (II) [Ni (NH3)4 C2O4]