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JEE PHYSICS MCQ: OSCILLATIONS

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Application +4 | -1

The percentage error in the measurement of

{g}

is:

(Given that

{g}=\frac{4 \pi^2 {L}}{{T}^2}, {~L}=(10 \pm 0.1) \mathrm{cm}, \, {T}=(100 \pm 1) \mathrm{s})

View answer and explanation

Correct option: D

3 \%

Option D (

3 \%

) is correct because the percentage error in g is calculated by summing the percentage errors in L and T, considering the power of T.

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