Question

To estimate 'g' (from g = $$4\, \pi^2\, \displaystyle \frac{L}{T^2}$$), error in measurement of L is $$\pm 2\, \%$$ and error in measurement of T is $$\pm\, 3\, \%$$. The error in estimated 'g' will be

A
±8%
B
±6%
C
±3%
D
±5%

Solution

The correct option is A $$\pm \, 8\, \%$$Acceleration due to gravity   $$g = \dfrac{4\pi^2 \ L}{T^2}$$Percentage error in $$g$$ is   $$\dfrac{\Delta g}{g}\times 100 = \dfrac{\Delta L}{L}\times 100 +\times \dfrac{\Delta T}{T}\times 100$$$$\implies \ \dfrac{\Delta g}{g}\times 100 = 2+2\times 3 = \pm 8$$%Maths

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