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Question

A simple pendulum is being used to determine the value of gravitational acceleration $$g$$ at a certain place. The length of the pendulum is $$25.0cm$$ and a stopwatch with $$1s$$ resolution measures the time taken for $$40$$ oscillation to be $$50s$$. The accuracy in $$g$$ is:

A
$$3.40$$%
B
$$5.40$$%
C
$$2.40$$%
D
$$4.40$$%
Solution
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Correct option is D. $$4.40$$%
For simple pendulum

$$T = 2\pi \sqrt{\dfrac{l}{g}}$$

so, $$\dfrac{\Delta T}{T} = \dfrac{1}{2} \left(\dfrac{\Delta l}{l} + \dfrac{\Delta g}{g}\right)$$

Givn $$\Delta T = 1 \ sec$$ $$T = 50 sec$$

$$\Delta l = 0.1 cm$$

$$l = 25 cm$$

$$\dfrac{\Delta g}{g} = \dfrac{2\Delta T}{T} + \dfrac{\Delta l}{l}$$

$$\dfrac{\Delta g}{g} = 2\times \dfrac{1}{50} + \dfrac{0.1}{25} = \dfrac{1.1}{25}$$

$$\dfrac{\Delta g}{g} = \dfrac{1.1}{25}$$

$$\% \dfrac{\Delta g}{g} = \dfrac{1.1}{25} \times 100 = 4.4 \%$$

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