Question:
Three students S1, S2 and S3 perform an experiment for determining the acceleration due to gravity (g) using a simple pendulum. They use different lengths of pendulum and record time for different number of oscillations. The observations are as shown in the table.
Student No. Length of Pendulum (cm) No. of oscillations (n) Total time for n oscillations Time period (s) 1 64.0 8 128.0 16.0 2 64.0 4 64.0 16.0 3 20.0 4 36.0 9.0
(Least count of length = 0.1 cm and Least count for time = 0.1 s)
If \(E_1\), \(E_2\) and \(E_3\) are the percentage errors in 'g' for students 1, 2 and 3 respectively, then the minimum percentage error is obtained by student no. ________.
Three students S1, S2 and S3 perform an experiment for determining the acceleration due to gravity (g) using a simple pendulum. They use different lengths of pendulum and record time for different number of oscillations. The observations are as shown in the table.
(Least count of length = 0.1 cm and Least count for time = 0.1 s)
If \(E_1\), \(E_2\) and \(E_3\) are the percentage errors in 'g' for students 1, 2 and 3 respectively, then the minimum percentage error is obtained by student no. ________.
| Student No. | Length of Pendulum (cm) | No. of oscillations (n) | Total time for n oscillations | Time period (s) |
| 1 | 64.0 | 8 | 128.0 | 16.0 |
| 2 | 64.0 | 4 | 64.0 | 16.0 |
| 3 | 20.0 | 4 | 36.0 | 9.0 |
(Least count of length = 0.1 cm and Least count for time = 0.1 s)
If \(E_1\), \(E_2\) and \(E_3\) are the percentage errors in 'g' for students 1, 2 and 3 respectively, then the minimum percentage error is obtained by student no. ________.
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\(g = 4\pi^2 L/T^2\), so \(\dfrac{\Delta g}{g} = \dfrac{\Delta L}{L} + 2\dfrac{\Delta t}{t}\) with \(\Delta L = 0.1\) cm, \(\Delta t = 0.1\) s on the total time.
Updated On: Nov 4, 2026
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Correct Answer: 1
Solution and Explanation
From \(T = 2\pi\sqrt{L/g}\): \(g = \dfrac{4\pi^2 L}{T^2}\), and \(T = t/n\) so the relative error in T equals the relative error in the total time t.
\(\dfrac{\Delta g}{g} = \dfrac{\Delta L}{L} + 2\dfrac{\Delta t}{t}\), with \(\Delta L = 0.1\) cm and \(\Delta t = 0.1\) s.
Student 1: \(\dfrac{0.1}{64} + 2\cdot\dfrac{0.1}{128} = 0.0015625 + 0.0015625 = 0.003125\), so \(E_1 = 0.3125\%\).
Student 2: \(\dfrac{0.1}{64} + 2\cdot\dfrac{0.1}{64} = 0.0015625 + 0.003125 = 0.0046875\), so \(E_2 = 0.469\%\).
Student 3: \(\dfrac{0.1}{20} + 2\cdot\dfrac{0.1}{36} = 0.005 + 0.00556 = 0.01056\), so \(E_3 = 1.056\%\).
The minimum percentage error is \(E_1\): student number 1.
Answer: 1
\(\dfrac{\Delta g}{g} = \dfrac{\Delta L}{L} + 2\dfrac{\Delta t}{t}\), with \(\Delta L = 0.1\) cm and \(\Delta t = 0.1\) s.
Student 1: \(\dfrac{0.1}{64} + 2\cdot\dfrac{0.1}{128} = 0.0015625 + 0.0015625 = 0.003125\), so \(E_1 = 0.3125\%\).
Student 2: \(\dfrac{0.1}{64} + 2\cdot\dfrac{0.1}{64} = 0.0015625 + 0.003125 = 0.0046875\), so \(E_2 = 0.469\%\).
Student 3: \(\dfrac{0.1}{20} + 2\cdot\dfrac{0.1}{36} = 0.005 + 0.00556 = 0.01056\), so \(E_3 = 1.056\%\).
The minimum percentage error is \(E_1\): student number 1.
Answer: 1
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