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Question

While measuring the acceleration due to gravity by a simple pendulum, a student makes a positive error of $$1\%$$ in the length of the pendulum and a negative error of $$3\%$$ in the value of time period. His percentage error in the measurement of g by the relation $$g = 4 \pi^2(l /T^2)$$ will be


A
2%
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B
4%
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C
7%
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D
10%
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Solution

The correct option is D $$7\%$$
Error always gets added so,
$$g=4{ \pi  }^{ 2 }\left( \dfrac { l }{ { T }^{ 2 } }  \right) $$
$$\dfrac { \Delta g }{ g } =\dfrac { \Delta 1 }{ L } +\dfrac { 2\Delta T }{ T } $$
$$\dfrac { \Delta g }{ g } \times 100=1+2\left( 3 \right) $$
$$\dfrac { \Delta g }{ g } \times 100=7$$
Percentage error $$=7$$%

Physics

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