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Percentage error in g from a simple pendulum's length and timed oscillations

This type of question, asked 4 times: 2021 Paper 2, 2021 Paper 9, 28 Jul 2022, 31 Jan 2024. All 4 questions below, oldest first.

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  1. Question 1

    2021 Paper 2

    13
    The period of oscillation of a simple pendulum is T=2πLgT= 2\pi\sqrt{\frac{L}{g}}. Measured value of ' LL ' is 1.0 m1.0\text{ }m from meter scale having a minimum division of 1 mm1\text{ }mm and time of one complete oscillation is 1.95 s1.95\text{ }s measured from stopwatch of 0.01 s0.01\text{ }s resolution. The percentage error in the determination of ' gg ' will be:

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  2. Question 2

    2021 Paper 9

    7
    The time period of a simple pendulum is given by T=2πlgT = 2\pi\sqrt{\frac{\mathcal{l}}{g}}. The measured value of the length of pendulum is 10 cm10\text{ }cm known to a 1 mm1\text{ }mm accuracy. The time for 200 oscillations of the pendulum is found to be 100 second using a clock of 1 s1\text{ }s resolution. The percentage accuracy in the determination of ' gg ' using this pendulum is ' xx '. The value of ' xx ' to the nearest integer is:-

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  3. Question 3

    28 Jul 2022

    120
    In an experiment to find acceleration due to gravity (g) using simple pendulum, time period of 0.5 s0.5\text{ }s is measured from time of 100 oscillation with a watch of 1 s1\text{ }s resolution. If measured value of length is 10 cm10\text{ }cm known to 1 mm1\text{ }mm accuracy. The accuracy in the determination of gg is found to be x%x\%. The value of xx is

  4. Question 4

    31 Jan 2024

    93
    The measured value of the length of a simple pendulum is 20 cm20\text{ }cm with 2 mm2\text{ }mm accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is N%N\%. The value of NN is:

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