JEE Main 2025 · Physics · Units and Measurements · easy

In an experiment to find g with a simple pendulum, the length is measured as L = 1.000 m using a metre scale of least count 1 mm, and the time for 20…

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In an experiment to find

g

with a simple pendulum, the length is measured as

L = 1.000\ \mathrm{m}

using a metre scale of least count

1\ \mathrm{mm}

, and the time for

20

oscillations is

40.0\ \mathrm{s}

using a stopwatch of least count

0.1\ \mathrm{s}

. The maximum percentage error in the value of

g

Show answer and solution

Answer: (C) $0.6\%$

From

T = 2\pi\sqrt{L/g}

we get

g = 4\pi^2 L/T^2

, so

\dfrac{\Delta g}{g} = \dfrac{\Delta L}{L} + 2\dfrac{\Delta T}{T}

. Here

\Delta L/L = 1\ \mathrm{mm}/1000\ \mathrm{mm} = 0.1\%

. The error in the total time of

20

oscillations is

0.1\ \mathrm{s}

and the same fractional error applies to

T

, so

\Delta T/T = 0.1/40 = 0.25\%

. Because

T

appears squared, its contribution is doubled:

0.5\%

. Total

= 0.1 + 0.5 = 0.6\%

. Forgetting the factor

2

gives the trap value

0.35\%

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