JEE Main 2023 · Physics · Units and Measurements · easy

In a simple pendulum experiment the length is read as 100.0 cm with a scale of least count 0.1 cm, and 50 complete oscillations are timed as 100.0 s…

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In a simple pendulum experiment the length is read as

100.0\ \mathrm{cm}

with a scale of least count

0.1\ \mathrm{cm}

, and

50

complete oscillations are timed as

100.0\ \mathrm{s}

with a stopwatch of least count

0.1\ \mathrm{s}

. The maximum percentage error in the computed value of

g

Show answer and solution

Answer: (B) $0.3\%$

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

, so

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

. The error is in the total time of

50

oscillations:

\Delta t/t=0.1/100=0.1\%

, and the period has the same fractional error because the count

50

is exact.

\Delta L/L=0.1/100=0.1\%

. Hence

\Delta g/g=0.1+2(0.1)=0.3\%

. Those who divide the time error by

50

again get

0.12\%

and those who forget the square get

0.2\%

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