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%
B
4%
C
7%
D
10%

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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