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Pendulum of length 10 cm (1 mm accuracy) with period 0.5 s timed on a 1 s watch: percentage error in g

The same question, asked 2 times: 2021 Paper 9*, 28 Jul 2022. Every wording below, oldest first.

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  1. Question 1

    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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  2. Question 2

    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