CBSE Class 10 Physics Electricity MCQs Set 19

Multiple Choice Questions (MCQs) for Class 10 Physics: Chapter 11 Electricity

Explore reliable objective questions for Chapter 11 Electricity tailored for Class 10 learners. Utilizing these Physics multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

Practice Chapter 11 Electricity MCQs for Class 10 Physics

Navigate directly to the 50 objective questions for Chapter 11 Electricity using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question 1: The electric potential at a point P is -10 V and that at a point Q is 20 V. What is the work done (in J) by the electric force in moving a proton slowly from Q to P? (e = 1.6 × 10-19 C)
(a) 48 × 10-19
(b) 16 × 10-19
(c) 24 × 10-18
(d) 12 × 10-19
Show Answer & Explanation

Answer: (a) 48 × 10-19

Explanation:
1. Work done by the electric force W = q(VQ - VP) = 1.6 × 10-19 × (20 + 10).
2. W = 48 × 10-19 J (positive because a proton moves to a lower potential).

Teacher's Note:
1. Positive charges move spontaneously from high to low potential.

Question 2: A point charge q is placed at the centre of a square PQRS. WP, WQ, WR and WS are the work done in bringing a test charge from infinity to the vertices P, Q, R and S respectively. Then
(a) WP = WQ > WS = WR
(b) WQ > WP > WR > WS
(c) WP = WQ = WR = WS
(d) WQ < WP < WR < WS
Show Answer & Explanation

Answer: (c) WP = WQ = WR = WS

Explanation:
1. All four vertices are at the same distance from the central charge, so the potential V = q/(4πε0r) is the same at each.
2. Work done = (test charge) × V, so WP = WQ = WR = WS.

Teacher's Note:
1. Equal distance from a point charge means equal potential.

Question 3: A particle of mass 20 g and charge -20 μC is at rest in a uniform electric field of 100 N C-1. What is its velocity (in m s-1) after 3 s?
(a) 3 × 104
(b) 300
(c) 600
(d) 0.3
Show Answer & Explanation

Answer: (d) 0.3

Explanation:
1. F = qE = 20 × 10-6 × 100 = 2 × 10-3 N.
2. a = F/m = 2 × 10-3/0.02 = 0.1 m s-2.
3. v = at = 0.1 × 3 = 0.3 m s-1.

Teacher's Note:
1. Convert grams to kilograms and μC to C.

Question 4: Two charged particles, each of mass m and charge q, hang from a common point by massless strings of equal length l. In equilibrium they are 2x apart. Which relation is correct?
(a) \( \frac{x^3}{\sqrt{l^2 - x^2}} = \frac{q^2}{4\pi\varepsilon_0 mg} \)
(b) \( \frac{x^3}{\sqrt{l^2 - x^2}} = \frac{q^2}{16\pi\varepsilon_0 mg} \)
(c) \( \frac{x}{l - x} = \frac{q\,mg}{4\pi\varepsilon_0} \)
(d) \( \frac{x^2}{l^2} = \frac{mg}{4\pi\varepsilon_0} \)
Show Answer & Explanation

Answer: (b) \( \frac{x^3}{\sqrt{l^2 - x^2}} = \frac{q^2}{16\pi\varepsilon_0 mg} \)

Explanation:
1. Each string makes an angle θ with the vertical, with tanθ = x/\( \sqrt{l^2 - x^2} \).
2. Balance of forces: tanθ = F/mg with F = q2/(4πε0(2x)2) = q2/(16πε0x2).
3. So x/\( \sqrt{l^2 - x^2} \) = q2/(16πε0x2mg), i.e. x3/\( \sqrt{l^2 - x^2} \) = q2/(16πε0mg).

Teacher's Note:
1. Resolve the tension into horizontal and vertical parts.
2. Separation is 2x, which gives the factor 16.

Question 5: Three long parallel wires A, B and C are equally spaced 10 cm apart, with B in the middle, and each carries 5 A in the same direction. What is the net force per unit length on wire B due to A and C?
(a) 8 × 10-5 N m-1
(b) 2 × 10-4 N m-1
(c) 6 × 10-3 N m-1
(d) zero
Show Answer & Explanation

Answer: (d) zero

Explanation:
1. Parallel currents attract, so A pulls B towards A and C pulls B towards C.
2. The two forces per unit length are equal (μ0i2/2πr = 2.5 × 10-5 N m-1 each) and opposite, so the net force is zero.

Teacher's Note:
1. Equal and opposite attractions cancel for the middle wire.

Question 6: What is the magnetic field at the centre of a long hollow cylinder carrying a current i?
(a) zero
(b) \( \frac{\mu_0 i}{2r} \)
(c) \( \frac{\mu_0 i}{2\pi r} \)
(d) \( \frac{\mu_0 i}{r} \)
Show Answer & Explanation

Answer: (a) zero

Explanation:
1. By Ampere's law, a closed loop inside the hollow region encloses no current.
2. So the magnetic field inside the hollow cylinder is zero.

Teacher's Note:
1. The field is zero inside a long hollow current-carrying cylinder.

Question 7: An electron and a proton move in the same direction on two circular paths with a common centre, of radii a and b respectively (b > a), each with frequency f. If the magnitude of each charge is q, what is the resultant magnetic field at the centre?
(a) \( \frac{\mu_0 qf}{2}\left(\frac{b-a}{ab}\right) \)
(b) \( \frac{\mu_0 f}{2q}\,ab \)
(c) \( \frac{\mu_0 qf}{2}\left(\frac{ab}{a+b}\right) \)
(d) \( \frac{\mu_0 qf}{2}\left(\frac{ab}{a-b}\right) \)
Show Answer & Explanation

Answer: (a) \( \frac{\mu_0 qf}{2}\left(\frac{b-a}{ab}\right) \)

Explanation:
1. Each moving charge is a current i = qf, giving a field μ0qf/(2r) at the centre.
2. Because the charges are opposite and move in the same direction, the currents are opposite and the fields subtract.
3. B = (μ0qf/2)(1/a - 1/b) = (μ0qf/2)(b - a)/(ab).

Teacher's Note:
1. Opposite charges moving the same way give opposite currents.

Question 8: The ratio of the electrostatic to the gravitational force between two electrons is about 4 × 1042. What is the approximate ratio between two protons? (mass of a proton = 2000 times that of an electron)
(a) 4 × 1042
(b) 1036
(c) 1
(d) 1019
Show Answer & Explanation

Answer: (b) 1036

Explanation:
1. The electrostatic force is the same (same magnitude of charge) while the gravitational force is (2000)2 = 4 × 106 times larger.
2. The ratio = 4 × 1042/(4 × 106) = 1036.

Teacher's Note:
1. Fe depends on charge and Fg on mass.

Question 9: Which of the following is/are units of electric field?
(a) N m-1
(b) N C-1
(c) V m-1
(d) both (b) and (c)
Show Answer & Explanation

Answer: (d) both (b) and (c)

Explanation:
1. E = F/q has the unit N C-1.
2. Also E = V/d has the unit V m-1, and the two units are equivalent.

Teacher's Note:
1. 1 N C-1 = 1 V m-1.

Question 10: A charge of 1 C gives rise to 1/ε0 lines of force. What is the electric field strength (in N C-1) at a distance r m from a point charge q C?
(a) \( \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2} \)
(b) \( \frac{1}{\varepsilon_0}\frac{q}{r} \)
(c) \( \frac{3}{4\pi\varepsilon_0}\frac{q}{r^3} \)
(d) \( \frac{qr}{4\pi\varepsilon_0} \)
Show Answer & Explanation

Answer: (a) \( \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2} \)

Explanation:
1. The flux q/ε0 is spread over a sphere of area 4πr2.
2. E = (q/ε0)/(4πr2) = q/(4πε0r2).

Teacher's Note:
1. Coulomb's law field of a point charge falls as 1/r2.

Question 11: Two coils A and B of resistances 60 Ω and 100 Ω are connected in parallel to a battery. If the heat developed in coil B is 420 J in 20 s, how much heat (in J) is produced in coil A in 60 s?
(a) 4200
(b) 2100
(c) 1640
(d) 1540
Show Answer & Explanation

Answer: (b) 2100

Explanation:
1. For coil B: V2t/R = 420 gives V2 × 20/100 = 420, so V2 = 2100.
2. For coil A: H = V2t/R = 2100 × 60/60 = 2100 J.

Teacher's Note:
1. Parallel coils have the same potential difference.

Question 12: Two long parallel wires A and B, a distance r apart, carry currents i1 and i2 (i1 > i2) in the same direction. Arrange the steps to find the resultant magnetic field at a point P between the wires, at distance r1 from A.
(A) Use the right-hand thumb rule to find the directions of B1 and B2 at P.
(B) Find B1 due to wire A using B1 = μ0i1/2πr1.
(C) If they have the same direction the resultant is B1 + B2; if opposite, B1 - B2.
(D) Find B2 due to wire B using B2 = μ0i2/[2π(r - r1)].
(a) A C B D
(b) B D A C
(c) B A D C
(d) A B C D
Show Answer & Explanation

Answer: (b) B D A C

Explanation:
1. Calculate the two magnitudes B1 (B) and B2 (D).
2. Find their directions (A) and then combine them (C).
3. The order is B, D, A, C.

Teacher's Note:
1. Between wires carrying currents in the same direction the two fields are opposite, so they subtract.

Question 13: Three identical charges +q are at the vertices of an equilateral triangle ABC of side a. Arrange the steps to find the net force on the charge at C.
(A) Determine the resultant force.
(B) Let the three charges +q be placed at the vertices of triangle ABC of side a.
(C) The force on C due to the charge at A is F1 = q2/(4πε0a2), directed away from A along AC.
(D) The force on C due to the charge at B is F2 = q2/(4πε0a2), directed away from B along BC.
(a) B C D A
(b) B D A C
(c) B A D C
(d) B C A D
Show Answer & Explanation

Answer: (a) B C D A

Explanation:
1. Set up the arrangement (B), find the force F1 due to A (C) and the force F2 due to B (D).
2. Then combine them vectorially (A): F = 2F1cos30° = \( \sqrt{3} \)q2/(4πε0a2).

Teacher's Note:
1. Forces between like charges are repulsive.

Question 14: An electric field E = 10 N C-1 is applied between two metal plates 20 cm apart. What is the potential difference (in V) between the plates?
(a) 200
(b) 5
(c) 2
(d) 0.2
Show Answer & Explanation

Answer: (c) 2

Explanation:
1. V = Ed = 10 × 0.2 = 2 V.

Teacher's Note:
1. Convert cm to m.

Question 15: An electron revolves in a circular path of radius r m with frequency f Hz. If the magnitude of its charge is q C, what is the current flowing?
(a) qf
(b) q/f
(c) \( \frac{qf}{2\pi r} \)
(d) \( \frac{2\pi r q}{f} \)
Show Answer & Explanation

Answer: (a) qf

Explanation:
1. The charge q passes a given point f times every second.
2. Current I = qf.

Teacher's Note:
1. Current = charge × number of revolutions per second.

Question 16: A charged particle of charge q moves along a circle of radius x m with frequency n Hz. What is the magnetic field at the centre of the circle?
(a) \( \frac{\mu_0\,2qn}{x} \)
(b) \( \frac{\mu_0\,qn}{2x} \)
(c) \( \frac{\mu_0\,qn}{2\pi x} \)
(d) \( \frac{\mu_0\,qn}{2\pi x^2} \)
Show Answer & Explanation

Answer: (b) \( \frac{\mu_0\,qn}{2x} \)

Explanation:
1. The moving charge is equivalent to a current i = qn.
2. B = μ0i/(2x) = μ0qn/(2x).

Teacher's Note:
1. A circulating charge behaves like a current loop.

Question 17: A straight long wire carries 10 A. What is the magnetic field at a distance of 5 cm from the wire?
(a) 4 × 10-5 T
(b) 5 × 10-7 T
(c) 15 × 10-3 T
(d) 20 × 10-5 T
Show Answer & Explanation

Answer: (a) 4 × 10-5 T

Explanation:
1. B = μ0i/(2πr) = 2 × 10-7 × 10/0.05 = 4 × 10-5 T.

Teacher's Note:
1. Use B = (μ0/2π)(i/r).

Question 18: Three identical wires, each of length l, are joined to form an equilateral triangle ABC with A at the lower left, B at the top and C at the lower right. A current i enters at A and leaves at C. What are the directions of the magnetic fields due to AB, BC and AC at the centre of the triangle? (⊗ is into the plane and ⊙ is out of the plane)
(a) ⊗, ⊙, ⊗
(b) ⊙, ⊙, ⊗
(c) ⊙, ⊗, ⊙
(d) ⊗, ⊗, ⊙
Show Answer & Explanation

Answer: (d) ⊗, ⊗, ⊙

Explanation:
1. The current takes two paths: A to B to C (clockwise as seen from the centre) and A to C along the bottom side (anticlockwise).
2. By the right-hand rule, a clockwise current gives a field into the plane, so AB and BC give ⊗.
3. The direct path AC is anticlockwise, so it gives ⊙.

Teacher's Note:
1. Clockwise current: field into the page at the centre.

Question 19: The area of a circular conducting loop in a uniform magnetic field of 0.2 T (perpendicular to it) is increased from 0.1 cm2 to 0.2 cm2 in 0.1 s. If the resistance of the loop is 5 Ω, what is the induced current?
(a) 4 × 10-4 A
(b) 4 × 10-6 A
(c) 3 × 10-3 A
(d) 6 × 10-3 A
Show Answer & Explanation

Answer: (b) 4 × 10-6 A

Explanation:
1. emf = B(ΔA)/Δt = 0.2 × (0.1 × 10-4)/0.1 = 2 × 10-5 V.
2. I = emf/R = 2 × 10-5/5 = 4 × 10-6 A.

Teacher's Note:
1. 1 cm2 = 10-4 m2.

Question 20: A potential difference V is applied across a wire of length l and specific resistance ρ, and n electrons pass any cross-section in time t. If e is the charge of an electron, what is the area of cross-section of the wire?
(a) \( \frac{Vl\rho}{net} \)
(b) \( \frac{Vtl}{ne\rho} \)
(c) \( \frac{\rho l n e}{Vt} \)
(d) \( \frac{\rho t n e}{Vl} \)
Show Answer & Explanation

Answer: (c) \( \frac{\rho l n e}{Vt} \)

Explanation:
1. Current I = ne/t.
2. V = IR = I ρ l/A, so A = ρ l I/V = ρ l n e/(V t).

Teacher's Note:
1. Combine I = q/t with R = ρl/A.

Question 21: A cell of emf 6 V is connected across a resistor of 11.5 Ω and the current is 0.5 A. What is the heat developed in the cell (in its internal resistance) in 20 s?
(a) 4.2 J
(b) 0.5 J
(c) 2.5 J
(d) 6.5 J
Show Answer & Explanation

Answer: (c) 2.5 J

Explanation:
1. E = I(R + r) gives 6 = 0.5(11.5 + r), so r = 0.5 Ω.
2. Heat in the cell = I2rt = 0.25 × 0.5 × 20 = 2.5 J.

Teacher's Note:
1. The cell's internal resistance also dissipates heat.

Question 22: The magnetic fields at two points P1 and P2, at distances r1 and r2 from a long straight current-carrying conductor, are 25 T and 5 T respectively. Arrange the steps to find r1 : r2.
(A) Let the fields be B1 and B2 at distances r1 and r2.
(B) Divide the expressions to get B1/B2 = r2/r1, since i is the same.
(C) Write B1 and B2 using B = μ0i/2πr.
(D) Substitute the values of B1 and B2 to get r1 : r2.
(a) A C B D
(b) A D C B
(c) C B A D
(d) C B D A
Show Answer & Explanation

Answer: (a) A C B D

Explanation:
1. Define B1 and B2 (A) and write their expressions (C).
2. Divide them (B) and substitute the values (D): r1/r2 = B2/B1 = 5/25 = 1/5.

Teacher's Note:
1. B is inversely proportional to the distance from a long wire.

Question 23: Arrange the steps to find a point between two like charges +q1 and +q2, d apart, where the net electric field is zero.
(A) The field at P due to q1 is E1 = q1/(4πε0x2).
(B) Let P be the point of zero field, at distance x from q1 and (d - x) from q2.
(C) The field at P due to q2 is E2 = q2/(4πε0(d - x)2).
(D) Let the two like charges +q1 and +q2 be d apart.
(E) Equate E1 and E2 (equal and opposite at P) and find x.
(a) A D C B E
(b) A C D B E
(c) D B A C E
(d) D B E A C
Show Answer & Explanation

Answer: (c) D B A C E

Explanation:
1. Set up the charges (D) and the point P (B).
2. Write E1 (A) and E2 (C), then equate them (E) to find x.

Teacher's Note:
1. Between two like charges the fields are opposite in direction.

Free study material for Science

Multiple Choice Questions (MCQs) for Class 10 Physics Chapter 11 Electricity

About Chapter 11 Electricity MCQs for Class 10 Physics

Test your conceptual understanding of Chapter 11 Electricity with these targeted multiple-choice questions. Designed in alignment with the latest CBSE curriculum for Class 10 Physics, these problem sets build accuracy and prepare students for objective exams.

How to Verify Your MCQ Answers

Built using the official NCERT book for Class 10, these Physics objective sets provide reliable academic guidance. Pair your practice with our recommended NCERT solutions to master optimal problem-solving approaches.

Enhance Speed with Online MCQ Tests

Wrap up your chapter revision by testing your knowledge against standard question formats. Everything on our platform is provided free of charge.

FAQs

Where can I access latest CBSE Class 10 Physics Electricity MCQs Set 19?

You can get most exhaustive CBSE Class 10 Physics Electricity MCQs Set 19 for free on StudiesToday.com. These MCQs for Class 10 Physics are updated for the 2026-27 academic session as per CBSE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Physics Class 10 material?

Yes, our CBSE Class 10 Physics Electricity MCQs Set 19 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the CBSE paper is now competency-based.

How do practicing Physics MCQs help in scoring full marks in Class 10 exams?

By solving our CBSE Class 10 Physics Electricity MCQs Set 19, Class 10 students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Physics.

Do you provide answers and explanations for CBSE Class 10 Physics Electricity MCQs Set 19?

Yes, Physics MCQs for Class 10 have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused CBSE exams.

Can I practice these Physics Class 10 MCQs online?

Yes, you can also access online interactive tests for CBSE Class 10 Physics Electricity MCQs Set 19 on StudiesToday.com as they provide instant answers and score to help you track your progress in Physics.