Question:
In an experiment, on the measurement of g using a simple pendulum, the time period was measured with an accuracy of 0.2% while the length was meausred with an accuracy of 0.5%. The percentage accuracy in the value of g thus obtained is
In an experiment, on the measurement of g using a simple pendulum, the time period was measured with an accuracy of 0.2% while the length was meausred with an accuracy of 0.5%. The percentage accuracy in the value of g thus obtained is
Updated On: Jun 7, 2022
- 0.0025
- 0.001
- 0.007
- 0.009
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The Correct Option is D
Solution and Explanation
Given $ \frac{\Delta T}{T} = 0.2\% , \frac{\Delta L}{L} = 0.5\%$
Since $T = 2 \pi \sqrt{\frac{L}{g}} $
$\Rightarrow \:\:\:\:\: g = \frac{4 \pi^2 L}{T^2}$
$ \therefore \:\:\:\: \frac{\Delta g}{g} = \frac{\Delta L}{L} + \frac{2 \Delta T}{T}$
$ = 0.5\% + 2(0.2 \%)$
$ = 0.9\%$
Since $T = 2 \pi \sqrt{\frac{L}{g}} $
$\Rightarrow \:\:\:\:\: g = \frac{4 \pi^2 L}{T^2}$
$ \therefore \:\:\:\: \frac{\Delta g}{g} = \frac{\Delta L}{L} + \frac{2 \Delta T}{T}$
$ = 0.5\% + 2(0.2 \%)$
$ = 0.9\%$
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