Download CBSE 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.
Chapter-wise Objective Questions: Chapter 11 Electricity
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
(a) 30 kWh
(b) 60 kWh
(c) 90 kWh
(d) 150 kWh
Show Answer & Explanation
Answer: (c) 90 kWh
Explanation:
1. Tube lights: 5 x 40 W x 5 h = 1000 Wh = 1 kWh per day.
2. Press: 0.5 kW x 4 h = 2 kWh per day.
3. Total per day 3 kWh, so in 30 days 90 kWh.
Teacher's Note:
1. Find the daily energy first.
2. Then multiply by the number of days.
(a) 6 V
(b) 2 V
(c) 1 V
(d) 10 V
Show Answer & Explanation
Answer: (a) 6 V
Explanation:
1. Current is 2 A, so \( V = E - Ir = 10 - 2 \times 2 = 6 \) V.
2. This equals the voltage across the external 3 ohm resistor: 2 x 3 = 6 V.
Teacher's Note:
1. Terminal voltage is lower than EMF while the battery supplies current.
(a) \( 5\ \Omega \)
(b) \( 10\ \Omega \)
(c) \( 20\ \Omega \)
(d) \( 40\ \Omega \)
Show Answer & Explanation
Answer: (a) \( 5\ \Omega \)
Explanation:
1. The diameter divides the ring into two equal semicircles, each of resistance 10 ohm.
2. They are in parallel: \( 10/2 = 5\ \Omega \).
Teacher's Note:
1. Both halves connect the same two points, so they are in parallel.
(a) 2.5 J
(b) 36 000 J
(c) 9000 J
(d) 600 J
Show Answer & Explanation
Answer: (b) 36 000 J
Explanation:
1. Power \( P = VI = 5 \times 0.5 = 2.5 \) W.
2. Time = 4 x 3600 = 14 400 s, so \( E = 2.5 \times 14400 = 36000 \) J.
Teacher's Note:
1. Convert milliamp to amp and hours to seconds.
2. The resistance is 5/0.5 = 10 ohm.
(a) \( 16\ \Omega \)
(b) \( 31.25\ \Omega \)
(c) \( 62.5\ \Omega \)
(d) \( 650\ \Omega \)
Show Answer & Explanation
Answer: (b) \( 31.25\ \Omega \)
Explanation:
1. \( 1/R = 1/100 + 1/50 + 1/500 = (5+10+1)/500 = 16/500 \)
2. \( R = 31.25\ \Omega \), and the total current is \( 220/31.25 = 7.04 \) A.
3. The iron must have the same resistance as the combination: 31.25 ohm.
Teacher's Note:
1. Equal current at the same voltage means equal resistance.
2. Parallel combination is smaller than the smallest branch.
Assertion (A): Resistivity changes when a conductor is stretched or contracted.
Reason (R): Resistivity is a material property.
(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, but R is true.
Show Answer & Explanation
Answer: (d) A is false, but R is true.
Explanation:
1. Resistivity depends only on the material and temperature, not on the shape or size, so stretching does not change it. The assertion is false.
2. The reason is true.
Teacher's Note:
1. Resistance changes on stretching; resistivity does not.
(a) 3 : 4
(b) 4 : 3
(c) 3 : 1
(d) 1 : 3
Show Answer & Explanation
Answer: (a) 3 : 4
Explanation:
1. \( R = \rho l/(\pi r^2) \), so \( R \propto l/r^2 \) for the same material.
2. A: 30/4 = 7.5; B: 10/1 = 10. Ratio = 7.5 : 10 = 3 : 4.
3. Wire A has the smaller resistance.
Teacher's Note:
1. Resistance depends on the square of the radius in the denominator.
2. Doubling the radius cuts the resistance to one quarter.
(a) 10 J
(b) 20 J
(c) 40 J
(d) 80 J
Show Answer & Explanation
Answer: (c) 40 J
Explanation:
1. Potential difference = 40 - 20 = 20 V.
2. \( W = q\,\Delta V = 2 \times 20 = 40 \) J.
Teacher's Note:
1. Only the potential difference matters, not the individual potentials.
Assertion (A): The resistance of a conductor is proportional to the square of its length.
Reason (R): \( R = \rho \dfrac{l}{A} \)
(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, but R is true.
Show Answer & Explanation
Answer: (d) A is false, but R is true.
Explanation:
1. From \( R = \rho l/A \), resistance is directly proportional to length, not to its square, so the assertion is false (for a fixed area).
2. The reason is the correct formula, so it is true.
Teacher's Note:
1. Stretching a wire changes the area too, which is why R changes as l squared in that special case.
2. Do not mix up fixed-area and fixed-volume cases.
(a) The 100 W bulb
(b) The 25 W bulb
(c) Neither
(d) Both
Show Answer & Explanation
Answer: (b) The 25 W bulb
Explanation:
1. \( R_{25} = 220^2/25 = 1936\ \Omega \) and \( R_{100} = 484\ \Omega \).
2. Series current \( = 440/2420 \approx 0.18 \) A.
3. Rated current of the 25 W bulb is 25/220 = 0.114 A, which is exceeded; the 100 W bulb can take 0.45 A.
4. So the 25 W bulb fuses.
Teacher's Note:
1. In series the same current flows in both.
2. The lower-power bulb has the smaller safe current.
(a) \( 4\ \Omega \)
(b) \( 10\ \Omega \)
(c) \( 20\ \Omega \)
(d) \( 24\ \Omega \)
Show Answer & Explanation
Answer: (a) \( 4\ \Omega \)
Explanation:
1. Total resistance \( = 6/0.25 = 24\ \Omega \).
2. Rheostat resistance \( = 24 - 20 = 4\ \Omega \).
Teacher's Note:
1. Resistances in series add up.
Assertion (A): Current flows from the positive to the negative terminal of the battery in the external circuit.
Reason (R): This is the conventional direction of current.
(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, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Conventional current is taken as the direction of flow of positive charge, from the positive terminal to the negative terminal through the external circuit.
2. The reason is the convention that explains the assertion.
Teacher's Note:
1. Electrons move opposite to conventional current.
2. Inside the cell the current goes from negative to positive.
(A) Ohm's law, (B) R_eq = R1 + R2 + ... + Rn, (C) V = E - Ir, (D) 1/R_eq = 1/R1 + 1/R2 + ... + 1/Rn
(P) series combination, (Q) terminal potential difference, (R) V = IR, (S) parallel combination
(a) A-R, B-P, C-Q, D-S
(b) A-P, B-R, C-Q, D-S
(c) A-R, B-S, C-Q, D-P
(d) A-Q, B-P, C-R, D-S
Show Answer & Explanation
Answer: (a) A-R, B-P, C-Q, D-S
Explanation:
1. Ohm's law is V = IR (A-R).
2. The sum of resistances is for a series combination (B-P).
3. V = E - Ir gives the terminal potential difference (C-Q).
4. The reciprocal sum is for a parallel combination (D-S).
Teacher's Note:
1. Learn each formula with its circuit.
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 4 : 1
Show Answer & Explanation
Answer: (a) 1 : 1
Explanation:
1. (i) Current \( I = 6/(1+2) = 2 \) A, so \( P = I^2 R = 4 \times 2 = 8 \) W.
2. (ii) The 2 \( \Omega \) resistor has the full 4 V: \( P = V^2/R = 16/2 = 8 \) W.
3. The two powers are equal, so the ratio is 1 : 1.
Teacher's Note:
1. In a series circuit use I squared R.
2. In a parallel circuit each branch gets the full voltage, so use V squared over R.
(a) The 100 W bulb
(b) The 60 W bulb
(c) Both have the same resistance
(d) It depends on the supply
Show Answer & Explanation
Answer: (b) The 60 W bulb
Explanation:
1. For the same voltage, \( R = V^2/P \), so resistance is inversely proportional to power.
2. The 60 W bulb has the smaller power, so it has the greater resistance.
Teacher's Note:
1. Higher wattage at the same rated voltage means lower resistance.
2. Compare bulbs with R = V squared over P.
(a) 6.6 kWh and 10 A
(b) 6.6 kWh and 5 A
(c) 2.2 kWh and 10 A
(d) 7.3 kWh and 20 A
Show Answer & Explanation
Answer: (a) 6.6 kWh and 10 A
Explanation:
1. Energy = 2.2 kW x 3 h = 6.6 kWh.
2. Current \( I = P/V = 2200/220 = 10 \) A.
Teacher's Note:
1. Use P = VI to get the current.
(a) \( 10\ \Omega \)
(b) \( 12\ \Omega \)
(c) \( 15\ \Omega \)
(d) \( 20\ \Omega \)
Show Answer & Explanation
Answer: (c) \( 15\ \Omega \)
Explanation:
1. Currents in the first two: 30/10 = 3 A and 30/12 = 2.5 A.
2. Current in the third = 7.5 - 3 - 2.5 = 2 A, so \( R = 30/2 = 15\ \Omega \).
Teacher's Note:
1. Voltage across each parallel branch is the same.
2. Branch currents add to the total.
(a) \( 2\ \Omega \) and \( 14\ \Omega \)
(b) \( 4\ \Omega \) and \( 12\ \Omega \)
(c) \( 6\ \Omega \) and \( 10\ \Omega \)
(d) \( 8\ \Omega \) and \( 8\ \Omega \)
Show Answer & Explanation
Answer: (b) \( 4\ \Omega \) and \( 12\ \Omega \)
Explanation:
1. Series: \( R_1 + R_2 = 16 \). Parallel: \( R_1R_2/(R_1+R_2) = 3 \), so \( R_1R_2 = 48 \).
2. The numbers with sum 16 and product 48 are 4 and 12.
Teacher's Note:
1. Form the two equations and find a pair of numbers.
2. Check: 4 + 12 = 16 and 4 x 12 / 16 = 3.
(a) R1 = 2 R2
(b) R2 = 2 R1
(c) R2 = 4 R1
(d) R1 = 4 R2
Show Answer & Explanation
Answer: (b) R2 = 2 R1
Explanation:
1. \( R = V^2/P \): the resistance is inversely proportional to power.
2. The 100 W bulb has half the power, so twice the resistance: R2 = 2 R1.
Teacher's Note:
1. Higher power rating means lower resistance for the same voltage.
(a) 1 A
(b) 1.5 A
(c) 2 A
(d) 3 A
Show Answer & Explanation
Answer: (c) 2 A
Explanation:
1. The 3 A current divides in inverse proportion to the resistances, 2 : 1.
2. The smaller resistor takes more: 2 A in 10 ohm and 1 A in 20 ohm.
3. Check: both have 20 V across them.
Teacher's Note:
1. Current divides inversely as resistance in a parallel combination.
Assertion (A): Kirchhoff's current law states that the net current at a junction is zero.
Reason (R): This law is based on the principle of conservation of charge.
(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, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. The algebraic sum of currents at a junction is zero, so the assertion is true.
2. Charge cannot pile up at a junction, which is conservation of charge, and this explains the law.
Teacher's Note:
1. The current law is about charge; the loop law is about energy.
(a) 0.01 R
(b) 0.1 R
(c) 10 R
(d) 100 R
Show Answer & Explanation
Answer: (a) 0.01 R
Explanation:
1. Each part has resistance R/10.
2. Ten equal resistors in parallel give \( (R/10)/10 = R/100 = 0.01R \).
Teacher's Note:
1. Divide by n once for cutting and once for the parallel join.
2. So the result is R divided by n squared.
(a) 5 V
(b) 10 V
(c) 14 V
(d) 9 V
Show Answer & Explanation
Answer: (c) 14 V
Explanation:
1. In charging mode \( V = E + Ir = 10 + 2 \times 2 = 14 \) V.
Teacher's Note:
1. During charging the terminal voltage is higher than the EMF.
(a) 0.5 A
(b) 2 A
(c) 10 A
(d) 200 A
Show Answer & Explanation
Answer: (b) 2 A
Explanation:
1. \( I = Q/t = 20/10 = 2 \) A.
Teacher's Note:
1. Current is the rate of flow of charge.
(a) \( 6.25 \times 10^{18} \)
(b) \( 1.6 \times 10^{19} \)
(c) \( 6.25 \times 10^{19} \)
(d) \( 1.6 \times 10^{-19} \)
Show Answer & Explanation
Answer: (a) \( 6.25 \times 10^{18} \)
Explanation:
1. \( n = q/e = 1/(1.6 \times 10^{-19}) \)
2. \( n = 6.25 \times 10^{18} \) electrons.
Teacher's Note:
1. Charge is quantised: q = ne.
2. Divide the total charge by the charge of one electron.
(a) \( 5\ \Omega \)
(b) \( 10\ \Omega \)
(c) \( 20\ \Omega \)
(d) \( 5/4\ \Omega \)
Show Answer & Explanation
Answer: (a) \( 5\ \Omega \)
Explanation:
1. Between opposite corners there are two paths, each with two wires in series: 10 \( \Omega \).
2. Two 10 \( \Omega \) paths in parallel give 5 \( \Omega \).
Teacher's Note:
1. Treat each side of the square as one resistor.
2. Two identical parallel paths halve the resistance.
(a) 0.05 V
(b) 5 V
(c) 20 V
(d) 500 V
Show Answer & Explanation
Answer: (c) 20 V
Explanation:
1. \( V = W/q = 100/5 = 20 \) V.
Teacher's Note:
1. Potential difference is work per unit charge.
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Chapter 11 Electricity Objective Questions & Solutions for Class 10 Physics
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FAQs
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