Question

# The period of oscillation of a simple pendulum is $$\displaystyle T = 2 \pi \sqrt {\dfrac {L} {g}}$$. Measured of $$L$$ is $$20.0\ cm$$ known to $$1\ mm$$ accuracy and time for $$100$$ oscillations of the pendulum is found to be $$90\ s$$ using a wrist watch of $$1\ s$$ resolution. The accuracy in the determination of $$g$$?

A
2.5%
B
2.7%
C
2%
D
3%

Solution

## The correct option is A $$2.7\%$$Given, $$\displaystyle T = 2 \pi \sqrt {\frac {L} {g}}$$Squaring on both sides and rearranging, $$g=4\pi^2\dfrac{L}{T^2}$$Now, $$\displaystyle \frac{\Delta g}{g}=\frac{\Delta L}{L}+2\frac{\Delta T}{T}$$$$\displaystyle \frac{\Delta g}{g}=\frac{10^{-3}}{0.2}+\frac{2\times 1}{90}$$$$\displaystyle \frac{\Delta g}{g}\times 100=\frac{1}{2}+\frac{20}{9}$$$$=2.7$$%Physics

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