In case of measurement of $' g'$ , if error in measurement of length of pendulum is $2 %$ , the percentage error in time period is $1 %$ . The maximum error in measurement of $g$ is

1 %

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2 %

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4 %

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Solution

The correct option is C $4 %$
$T = 2 π \sqrt{\frac{l}{g}}$
$g = \frac{4 π^{2} l}{T^{2}}$

Total % error $= 1 \times 2$ % $+ 2 \times 1$ % $= 4$ %

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In case of measurement of 'g', if error in measurement of length of pendulum is 2%, the percentage error in time period is 1%. The maximum error in measurement of g is

The correct option is C 4%T=2π√lg g=4π2lT2 Total % error =1×2% +2×1%=4%

In case of measurement of $' g'$ , if error in measurement of length of pendulum is $2 %$ , the percentage error in time period is $1 %$ . The maximum error in measurement of $g$ is

The correct option is C $4 %$
$T = 2 π \sqrt{\frac{l}{g}}$
$g = \frac{4 π^{2} l}{T^{2}}$

Total % error $= 1 \times 2$ % $+ 2 \times 1$ % $= 4$ %