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Percentage error in g measured with a simple pendulum (or in X from a square-root formula)

This type of question, asked 5 times: 22 Apr 2025 S1, 11 Apr 2026 S1, 13 Apr 2026 S1, 12 May 2026 S1, 14 May 2026 S2. All 5 questions below, oldest first.

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  1. Question 1

    22 Apr 2025 S1

    36
    The period of oscillating simple pendulum is T=2πl gT = 2\pi\sqrt{\frac{l}{\text{ }g}} where length ' ll ' is 100 cm with error 1 mm. Period is 2 second. The time of 100 oscillations is measured by a stopwatch of least count 0.1 s. The percentage error in gravitational acceleration ' gg ' is

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  2. Question 2

    11 Apr 2026 S1

    101
    The period of oscillation of a simple pendulum is given by T=2πlgT = 2\pi\sqrt{\dfrac{l}{g}} where ll is about 8080 cm and is known to have 0.10.1 cm accuracy. The period is about 1.51.5 s. The time of 5050 oscillations is measured by a stop watch of least count 0.10.1 s. The percentage error in gg is nearly

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  3. Question 3

    13 Apr 2026 S1

    101
    The period of oscillation of a simple pendulum is T=2π(L/g)1/2T = 2\pi(L/g)^{1/2}. Measured value of 'L' is 10 cm known to 1 mm accuracy and time for 100 oscillations is 50 second, using a watch of 1 second resolution. The percentage error in the measurement of 'g' is

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  4. Question 4

    12 May 2026 S1

    102
    A particular physical quantity is given by X=2πabX = 2\pi\sqrt{\dfrac{a}{b}} and the measured value of 'a' is 4040 unit with 0.010.01 unit accuracy. The value of 'b' is 9810\dfrac{98}{10} unit with 11 unit accuracy. The % error in XX is nearly

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  5. Question 5

    14 May 2026 S2

    117
    The period of oscillation of a simple pendulum is T=2πlgT = 2\pi\sqrt{\dfrac{l}{g}} where ll is about 100100 cm and has 11 mm accuracy. The period is about 22 s. The time of 100100 oscillations is measured by a stopwatch of least count 0.10.1 s. The percentage error in 'gg' is (gravitation acceleration)

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