Question Set B

cbse 12th Standard Physics Electrostatic Potential and Capacitance Model Questions Papers Study Material - QB365

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Questions:

Multiple Choice Question

10 x 1 = 10

The equivalent resistance between A and B of the network shown in figure is

(a)

3RΩ

(b)

(3/2)RΩ

(c)

2RΩ

(d)

(2/3)RΩ

Three charges q₁ -q and q₀ are placed as shown in figure. The rnagnitude of the net force on the 1 charge q₀at point O is \(\left(\text { Take, } K=\frac{1}{4 \pi \varepsilon_0}\right)\)

(a)

(b)

\(\frac{2 K q q_0}{a^2}\)

(c)

\(\frac{\sqrt{2} K q q_0}{a^2}\)

(d)

\(\frac{1}{\sqrt{2}} \frac{K q q_0}{a^2}\)

A conducting circular loop of radius r carries a constant current i. It is placed in a uniform magnetic field B, such that B is perpendicular to the plane of the loop. The magnetic force acting on the loop is

(a)

irB

(b)

2πriB

(c)

zero

(d)

πrib

The radii of two metallic spheres A and Bare r₁ and r₂ respectively (r₁ > r₂). They are connected by a thin wire and the system is given a certain charge. The charge will be greater

(a)

on the surface of the sphere B.

(b)

on the surface of the sphere A.

(c)

equal on both.

(d)

zero on both.

Four equal charges q are placed at the four corners A, B, C, D of a square of length a. The magnitude of the force on the charge at B will be

(a)

\(\frac{3 q^{2}}{4 \pi \varepsilon_{0} a^{2}}\)

(b)

\(\frac{4 q^{2}}{4 \pi \varepsilon_{0} a^{2}}\)

(c)

\(\frac{(1+2 \sqrt{2}) q^{2}}{2 \times 4 \pi \varepsilon_{0} a^{2}}\)

(d)

\(\frac{\left(\frac{2+1}{\sqrt{2}}\right)^{q^{2}}}{4 \pi \varepsilon_{0} a^{2}}\)

The electrostatic potential of a uniformly charged thin spherical shell of charge Q and radius R at a distance r from the centre is

(a)

\(\frac{Q}{4 \pi \varepsilon_{0} r}\) for points outside and \(\frac{Q}{4 \pi \varepsilon_{0} R}\) for points inside the shell.

(b)

\(\frac{Q}{4 \pi \varepsilon_{0} R}\) for both points inside and outside the shell

(c)

zero for points outside and \(\frac{Q}{4 \pi \varepsilon_{0} R}\) for points inside the shell.

(d)

zero for both points inside and outside the shell

In a system, 'n' electric dipole are placed in a closed surface. The value of emergent electric flux from enclosed surface is

(a)

\(\frac{q}{\varepsilon_{0}}\)

(b)

\(\frac{2 q}{\varepsilon_{0}}\)

(c)

\(-\frac{2 q}{\varepsilon_{0}}\)

(d)

zero

According to the Kirchhoff's law the sum of the products of current and resistance as well as emfs in a closed loop is:

(a)

greater than zero

(b)

zero

(c)

less than zero

(d)

determained by the emf

Two resistance when connected in parallel have equivalent resistance of 3\(\Omega\). When one of the resistance is burnt and broken, the net resistance is 12\(\Omega\). What is the resistance of the burnt resistor?

(a)

4\(\Omega\)

(b)

8\(\Omega\)

(c)

12\(\Omega\)

(d)

16\(\Omega\)

10.

A current of 5 A is flowing through a circular coil of diameter 14 cm having 100 turns. The magnetic dipole moment associated with this coil is :

(a)

\(0.077{ Am }^{ 2 }\)

(b)

\(0.77{ Am }^{ 2 }\)

(c)

\(7.7{ Am }^{ 2 }\)

(d)

\(77{ Am }^{ 2 }\)

Case Study Questions

3 x 4 = 12

11.

Whenever an electric current is passed through a conductor, it becomes hot after some time. The phenomenon of the production of heat in a resistor by the flow of an electric current through it is called heating effect of current or Joule heating. Thus, the electrical energy supplied by the source of emf is converted into heat. In purely resistive circuit, the energy expended by the source entirely appears as heat. But if the circuit has an active element like a motor, then a part of the energy supplied by the source goes to do useful work and the rest appears as heat. Joule's law of heating form the basis of various electrical appliances such as electric bulb, electric furnace,
electric press etc.
(i) Which of the following is a correct statement?

(a) Heat produced in a conductor is independent of the current flowing

(b) Heat produced in a conductor varies inversely as the current flowing

(c) Heat produced in a conductor varies directly as the square of the current flowing

(d) Heat produced in a conductor varies inversely as the square of the current flowing

(ii) If the coil of a heater is cut to half, what would happen to heat produced?

(a) Doubled

(b) Halved

(c) Remains same

(d) Becomes four times

(iii) A 25 Wand 100 Ware joined in series and connected to the mains. Which bulbs will glow brighter?

(a) 100W	(b) 25 W
(c) both bulbs will glow brighter	(d) none will glow brighter

(iv) A rigid container with thermally insulated wall contains a coil of resistance \(100 \Omega\) carrying current 1A. Change in its internal' energy after 5 min will be

(a) 0 kJ

(b) 10 kJ

(c) 20 kJ

(d) 30 kJ

(v) The heat emitted by a bulb of 1.90W in 1 min is

(a) 100 J

(b) 1000 J

(c) 600 J

(d) 6000 J

12.

Potential difference (\(\Delta\)V) between two points A and B separated by a distance x, in a uniform electric field E is given by \(\Delta V=-E x\),where x is measured parallel to the field lines. If a charge qo moves from P to Q, the changein potential energy \((\Delta U)\) is given as \(\Delta U=q_{0} \Delta V .\) A proton is released from rest in uniform electric field of magnitude \(4.0 \times 10^{8} \mathrm{Vm}^{-1}\) directed along the positive X-axis. The proton undergoes a displacement of 0.25 m in the direction of E.
Mass of a proton = 1.66 x 10^-27 kg and charge of proton = 1.6 x10^-19 C

(i) The change in electric potential of the proton between the points A and B is

\(\text { (a) }-1 \times 10^{8} \mathrm{~V}\)	\(\text { (b) } 1 \times 10^{8} \mathrm{~V}\)
\(\text { (c) } 6.4 \times 10^{-19} \mathrm{~V}\)	\(\text { (d) }-6.4 \times 10^{-19} \mathrm{~V}\)

(ii) The change in electric potential energy of the proton for displacement from A to B is

\(\text { (a) } 1.6 \times 10^{11} \mathrm{~J}\)	\(\text { (b) } 0.5 \times 10^{23} \mathrm{~J}\)
\(\text { (c) }-1.6 \times 10^{-11} \mathrm{~J}\)	\(\text { (d) } 3.2 \times 10^{22} \mathrm{~J}\)

(iii) The mutual electrostatic potential energy between two protons which are at a distance of 9 x 10^-15m, in \({ }_{92} \mathrm{U}^{235}\) nucleus is

\(\text { (a) } 1.56 \times 10^{-14} \mathrm{~J}\)	\(\text { (b) } 5.5 \times 10^{-14} \mathrm{~J}\)
\(\text { (c) } 2.56 \times 10^{-14} \mathrm{~J}\)	\(\text { (d) } 4.56 \times 10^{-14} \mathrm{~J}\)

(iv) If a system consists of two charges 4 mC and -3mC with no external field placed at (-5 em, 0, 0) and (5 em, 0, 0) respectively. The amount of work required to separate the two charges infinitely away from each other is

(a) -1.1 J	(b) 2 J
(c) 2.5 J	(d) 3 J

(v) As the proton moves from P to Q, then

(a) the potential energy of proton decreases	(b) the potential energy of proton increases
(c) the proton loses kinetic energy	(d) total energy of the proton increases

13.

Gauss's law and Coulomb's law, although expressed in different forms, are equivalent ways of describing the relation between charge and electric field in static conditions. Gauss's law is \(\varepsilon_{0} \phi=q_{\text {encl }}\), when
q_encl is the net charge inside an imaginary closed surface called Gaussian surface. \(\phi=\oint \vec{E} \cdot d \vec{A}\) gives the electric flux through the Gaussian surface. The two equations hold only when the net charge is in vacuum or air.

(I) If there is only one type of charge in the universe, then \((\vec{E} \rightarrow \text { Electric field, } d \vec{s} \rightarrow \text { Area vector })\)

(a) \(\oint \vec{E} \cdot d \vec{s} \neq 0\) on any surface

(b) \(\oint \vec{E} \cdot d \vec{s}\) could not be defined

(c) \(\oint \vec{E} \cdot d \vec{s}=\infty\) if charge is inside

(d) \(\oint \vec{E} \cdot d \vec{s}=0\) if charge is outside, \(\oint \vec{E} \cdot d \vec{s}=\frac{q}{\varepsilon_{0}}\) if charge is inside

(ii) What is the nature of Gaussian surface involved in Gauss law of electrostatic?

(a) Magnetic

(b) Scalar

(c) Vector

(d) Electrical

(iii) A charge 10 \(\mu \)C is placed at the centre of a hemisphere of radius R = 10 cm as shown. The electric flux through the hemisphere (in MKS units) is

(a) 20 x 10⁵

(b) 10 x 10⁵

(c) 6 x 10⁵

(d) 2 x 10⁵

(iv) The electric flux through a closed surface area S enclosing charge Q is \(\phi\). If the surface area is doubled, then the flux is

\(\text { (a) } 2 \phi\)

\(\text { (b) } \phi / 2\)

\(\text { (c) } \phi / 4\)

\(\text { (d) } \phi\)

(v) A Gaussian surface encloses a dipole. 'The electric flux through this surface is

\(\text { (a) } \frac{q}{\varepsilon_{0}}\)

\(\text { (b) } \frac{2 q}{\varepsilon_{0}}\)

\(\text { (c) } \frac{q}{2 \varepsilon_{0}}\)

(d) zero

Assertion and reason

10 x 1 = 10

14.

Assertion : The energy of a charged particle moving in a uniform magnetic field remains constant.
Reason : Work done by the magnetic field on the charge is zero.
Codes:
A) Both A and R are true and R is the correct explanation of A
B) Both A and R are true but R is NOT the correct explanation of A
C) A is true but R is false
D) A is false and R is true

15.

Assertion : When cells are connected in parallel to the external load, the effective e.m.f increases.
Reason : Because effective internal resistance of cells decreases.
Codes:
(A) If both Assertion & Reason are true & the Reason is a correct explanation of the Assertion.
(B) If both Assertion and Reason are true but Reason is not a correct explanation of the Assertion.
(C) If Assertion is true but the Reason is false
(D) If Assertion & Reason both are false.

16.

ASSERTION: Positive charge always moves from a higher potential point to a lower potential point.

REASON: Electric potential is a vector quantity

Codes:

A) If both assertion and reason are true and reason is the correct explanation of assertion.

B) If both assertion and reason are true and reason is not the correct explanation of assertion.

C) If assertion is true but reason is false.

D) If both assertion and reason are false.
E) If assertion is false but reason is true.

17.

Assertion: Two equipotential surfaces can be orthogonal.
Reason: Electric field lines are normal to the equipotential surface.
Codes:
(a) If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
(b) If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
(c) If the Assertion is correct but Reason is incorrect.
(d) If both the Assertion and Reason are incorrect.

18.

Assertion : In a simple battery circuit, the point of the lowest potential is positive terminal of the battery.
Reason : The current flows towards the point of the higher potential, as it does in such a circuit from the negative to the positive terminal.
Codes:
(a) If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
(b) If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
(c) If the Assertion is correct but Reason is incorrect.
(d) If both the Assertion and Reason are incorrect.

19.

Assertion (A) : When radius of a current carrying loop is doubled, its magnetic moment becomes four times.
Reason (R) : The magnetic moment of a current carrying loop is directly proportional to the area of the loop.
(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) Assertion is true but Reason is false.
(d) Assertion is false but Reason is true.

20.

Assertion(A) : A positive point charge initially at rest in a uniform electric field starts moving along electric lines of force. (Neglect all other forces except electric forces).
Reason (R) : A point charge released from rest in an electric field always moves along the line of
force.
(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) Assertion is true but Reason is false.
(d) Assertion is false but Reason is true.

21.

Assertion (A) : When two long parallel wires, hanging freely are connected in parallel to a battery, they come closer to each other.
Reason (R) : Wires carrying current in opposite direction repel each other.
Codes:
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is NOT the correct explanation of A
(c) A is true but R is false
(d) A is false and R is also false

22.

Assertion (A) : Circuits containing capacitors should be handled cautiously even when there is no current.
Reason (R) : The capacitors are very delicate and so quickly breakdown.
Codes:
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is NOT the correct explanation of A
(c) A is true but R is false
(d) A is false and R is also false

23.

Assertion (A) : A point charge is brought in an electric field. The field at a nearby point is increase, whatever be the nature of the charge.
Reason (R) : The electric field is independent of the nature of charge.
Codes:
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is NOT the correct explanation of A
(c) A is true but R is false
(d) A is false and R is also false

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Answers Key:

Multiple Choice Question

10 x 1 = 10

(a)
0

(b)
on the surface of the sphere A.

(a)
\(\frac{Q}{4 \pi \varepsilon_{0} r}\) for points outside and \(\frac{Q}{4 \pi \varepsilon_{0} R}\) for points inside the shell.

(d)
zero

(b)
zero

(a)
4\(\Omega\)

10.

Case Study Questions

3 x 4 = 12

11.

(i) (c): According to Joule's law of heating, Heat produced in a conductor, H= I²Rt
where, I = Current flowing through the conductor
R = Resistance of the conductor
t = Time for which current flows through the conductor.
\(\therefore \quad H \propto I^{2}\)
(ii) (a): If the coil is cut into half, its resistance is also halved.
As \(H=\frac{V^{2}}{R} t \quad \therefore \quad H^{\prime}=2\)
(iii) (b): \(P=\frac{V^{2}}{R} \text { or } R=\frac{V^{2}}{P}\)
The bulbs are joined in series. Current in both the bulbs will same
\(\therefore\) The heat produced in them is given by H = I²Rt
or \(H \propto R \Rightarrow H \propto \frac{1}{P}\)
Therefore the bulb with low wattage or high resistance will glow brighter or we can say the 25 W bulb will glow brighter than the 100 W bulb.
(iv) (d): \(R=100 \Omega ; I=1 \mathrm{~A} ; t=5 \mathrm{~min} .=5 \times 60=300 \mathrm{~s}\)
change in internal energy = heat generated in coil
\(=I^{2} R t=\left((1)^{2} \times 100 \times 300\right) \mathrm{J}\)
= 30000 J = 30 kJ
(v) (d): Here, P = 100 W, t = 1 min = 60 s
Heat developed in time t
H = P x t = (100 W)( 60 s) = 6000 J

12.

(i) (a) : As \(\Delta V=-E \Delta \psi=-\left(4.0 \times 10^{8} \mathrm{~V} / \mathrm{m}\right)(0.25 \mathrm{~m})\)= -10⁸V
(ii) (c) : As \(\Delta U=q_{0} \Delta V=\left(1.6 \times 10^{-19}\right) \times\left(-1.0 \times 10^{8} \mathrm{~V}\right)\)= \(-1.6 \times 10^{-11} \mathrm{~V}\)
(iii) (c) : Here, \(q_{1}=q_{2}=1.6 \times 10^{-19} \mathrm{C}, r=9 \times 10^{-15} \mathrm{~m}\)
\(U=\frac{9 \times 10^{9} \times 1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{9 \times 10^{-15}}=2.56 \times 10^{-14} \mathrm{~J}\)
(iv) (a): Here, \(q_{1}=4 \mu \mathrm{C}, q_{2}=-3 \mu \mathrm{C}\)
r = 10 cm = 0.1 m
Electrostatic potential energy,
\(U=\frac{1}{4 \pi \varepsilon_{o}} \frac{q_{1} q_{2}}{r}=9 \times 10^{9} \times \frac{4 \times 10^{-6} \times(-3) \times 10^{-6}}{0.1}=-1.1 \mathrm{~J}\)
(v) (a) : As proton moves in the direction of the electric field, then its potential energy decreases.

13.

(I) (d): If there is only one type of charge in the universe then it will produce electric field somehow. Hence Gauss's law is valid.
(ii) (c)
(iii) (c) : According to Gauss's theorem,
Electric flux through the sphere = \(\frac{q}{\varepsilon_{0}}\)
\(\therefore\) Electric flux through the hemisphere = \(\frac{1}{2} \frac{q}{\varepsilon_{0}}\)
= \(\frac{10 \times 10^{-6}}{2 \times 8.854 \times 10^{-12}}=0.56 \times 10^{6} \mathrm{~N} \mathrm{~m}^{2} \mathrm{C}^{-1}\)
\(\approx 0.6 \times 10^{6} \mathrm{Nm}^{2} \mathrm{C}^{-1}=6 \times 10^{5} \mathrm{~N} \mathrm{~m}^{2} \mathrm{C}^{-1}\)
(iv) (d): As flux is the total number of lines passing through the surface, for a given charge, it is always the charge enclosed \(Q / \varepsilon_{0}\).If area is doubled, the flux remains the same.
(v) (d): As net charge on a dipole is \((-q+q)=0\)
Thus, when a gaussian surface encloses a dipole, as per Gauss's theorem, electric flux through the surface,
\(\oint \vec{E} \cdot d \vec{S}=\frac{q}{\varepsilon_{0}}=0\)

Assertion and reason

10 x 1 = 10

14.

A) Both A and R are true and R is the correct explanation of A

15.

(D) If Assertion & Reason both are false.

16.

C) If assertion is true but reason is false.

17.

(d) If both the Assertion and Reason are incorrect.
Explanation:
Two equipotential surfaces never intersect each other so they cannot be orthogonal.

18.

(d) If both the Assertion and Reason are incorrect.

19.

(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

20.

(c) If the field lines are curved, then the charged particle follows the straight line path along the direction of tangent drawn to electric field lines at its starting point.

21.

(b): The wires are parallel to each other but the direction of current in it is in same direction so they attract each other. If the current in the wires is in opposite direction then wires repel each other. When the currents are in opposite directions, the magnetic forces are reversed and the wires repels each other

22.

(c): A charged capacitor, after removing the battery, does not discharge itself. If this capacitor is touched by someone, he may feel shock due to large charge still present on the capacitor. Hence it should be handled cautiously otherwise this may cause asevere shock.

23.

(d): Electric field at the nearby point will be resultant of existing field and field due to the charge brought. It may increase or decrease if the charge is positive or negative depending on the position of the point with respect to the charge brought.