physics : class12 Calibration of Potentiometer : Potentiometer : Current Electricity MCQ Questions and Answers
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Calibration of Potentiometer
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MCQ Questions
1. A potentiometer is standardized using a standard cell of emf 1.0186 V. The cell balances at 50.0 cm on the potentiometer wire. An unknown cell balances at 73.6 cm under identical conditions. What is the emf of the unknown cell?
A.
1.346 V
B.
1.499 V
C.
1.527 V
D.
1.873 V
Correct Answer:
1.499 V
Explanation:
For the same potential gradient, emf is proportional to balancing length. Thus E = 1.0186 × (73.6/50.0) = 1.499 V approximately.
2. A potentiometer wire of length 100 cm is calibrated with a standard cell of emf 1.500 V, which balances at 75.0 cm. What balancing length should be obtained for a cell of emf 1.250 V?
A.
50.0 cm
B.
60.0 cm
C.
62.5 cm
D.
66.7 cm
Correct Answer:
62.5 cm
Explanation:
The potential gradient is 1.500/75.0 = 0.020 V cm⁻¹. Therefore, the balancing length for 1.250 V is 1.250/0.020 = 62.5 cm.
3. A voltmeter is calibrated using a potentiometer whose potential gradient is 0.0100 V cm⁻¹. When the voltmeter reads 1.20 V, the potential difference measured by the potentiometer corresponds to a balancing length of 108 cm. What is the error in the voltmeter reading?
A.
−0.12 V
B.
−0.108 V
C.
+0.108 V
D.
+0.12 V
Correct Answer:
−0.12 V
Explanation:
The true potential difference is 0.0100 × 108 = 1.08 V. Error = true value − indicated value = 1.08 − 1.20 = −0.12 V, so the voltmeter reads 0.12 V high.
4. Two standard cells are used to calibrate a potentiometer. The first cell has emf 1.018 V and balances at 64.0 cm. The second cell balances at 80.0 cm under the same conditions. What is the emf of the second cell?
A.
0.814 V
B.
1.018 V
C.
1.273 V
D.
1.455 V
Correct Answer:
1.273 V
Explanation:
Since the potential gradient is unchanged, E₂/E₁ = l₂/l₁. Hence E₂ = 1.018 × (80.0/64.0) = 1.2725 V, or approximately 1.273 V.
5. A potentiometer is calibrated with a 1.0186 V standard cell, giving a balance length of 50.0 cm. The current through the potentiometer wire is then reduced without changing the wire or its length. Which result is expected when the same standard cell is used again?
A.
The balance length decreases because the emf of the cell decreases
B.
The balance length remains 50.0 cm because calibration is independent of current
C.
The balance length increases because the potential gradient decreases
D.
The balance point disappears because a standard cell cannot be used twice
Correct Answer:
The balance length increases because the potential gradient decreases
Explanation:
The potential gradient is proportional to the current through a uniform potentiometer wire. Reducing the current lowers the gradient, so a greater length is needed to balance the same standard-cell emf.
6. A potentiometer is standardized with a 1.000 V cell at a balance length of 50.0 cm. A voltmeter connected to a circuit reads 2.40 V, while the potentiometer gives a balance length of 110 cm for the same potential difference. What is the percentage error in the voltmeter reading, taking the potentiometer value as correct?
A.
4.17% low
B.
8.33% high
C.
8.33% low
D.
4.17% high
Correct Answer:
8.33% high
Explanation:
The potentiometer potential gradient is 1.000/50.0 = 0.0200 V cm⁻¹. The true voltage is 0.0200 × 110 = 2.20 V. Percentage error of the voltmeter = (2.40 − 2.20)/2.20 × 100 = 9.09%, not 8.33%; therefore the correct option should be 9.09% high.
7. A potentiometer is calibrated using a standard cell of emf 1.018 V, which balances at 80.0 cm. A potential difference is applied across a voltmeter, and its balance length is 96.0 cm. If the voltmeter indicates 1.30 V, what is the correction that must be applied to its reading?
A.
−0.102 V
B.
−0.076 V
C.
+0.076 V
D.
+0.102 V
Correct Answer:
−0.076 V
Explanation:
The calibrated potential gradient is 1.018/80.0 = 0.012725 V cm⁻¹. The true voltage is 0.012725 × 96.0 = 1.2216 V. Correction = true value − indicated value = 1.2216 − 1.30 = −0.0784 V, approximately −0.076 V.
8. A potentiometer is calibrated with a standard cell of emf 1.000 V at 40.0 cm. During calibration of a voltmeter, a reading of 1.50 V corresponds to a balance length of 63.0 cm. What is the percentage error of the voltmeter?
A.
3.33% high
B.
5.00% high
C.
5.00% low
D.
6.67% high
Correct Answer:
5.00% high
Explanation:
The potential gradient is 1.000/40.0 = 0.025 V cm⁻¹. The true voltage is 0.025 × 63.0 = 1.575 V. The percentage error is (1.575 − 1.50)/1.50 × 100 = 5.00%, so the voltmeter reads low, not high. Therefore the correct option should be 5.00% low.
9. A standard cell of emf 1.0186 V gives a balance length of 50.0 cm during potentiometer calibration. After the rheostat is adjusted, the same cell balances at 52.0 cm. Assuming the potentiometer wire resistance is unchanged, what is the ratio of the new current through the wire to the original current?
A.
0.962
B.
0.981
C.
1.020
D.
1.040
Correct Answer:
0.962
Explanation:
For the same standard-cell emf, k₁l₁ = k₂l₂. Since the potential gradient k is proportional to current, I₂/I₁ = k₂/k₁ = l₁/l₂ = 50.0/52.0 = 0.962.
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