Question:
In a simple pendulum experiment for determination of acceleration due to gravity (g), time taken for $20$ oscillations is measured by using a watch of $1$ second least count. The mean value of time taken comes out to be $30\, s$. The length of pendulum is measured by using a meter scale of least count $1\, mm$ and the value obtained is $55.0\, cm$. The percentage error in the determination of $g$ is close to :
In a simple pendulum experiment for determination of acceleration due to gravity (g), time taken for $20$ oscillations is measured by using a watch of $1$ second least count. The mean value of time taken comes out to be $30\, s$. The length of pendulum is measured by using a meter scale of least count $1\, mm$ and the value obtained is $55.0\, cm$. The percentage error in the determination of $g$ is close to :
Updated On: Jul 3, 2024
- 0.70%
- 0.20%
- 3.50%
- 6.80%
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The Correct Option is D
Solution and Explanation
$T = \frac{30 \sec}{20} \Delta T = \frac{1}{20} \sec $
$ L = 55 cm \Delta L = 1 mm = 0.1 cm $
$ g = \frac{4\pi^{2}L}{T^{2}} $
percentage error in g is
$ \frac{\Delta g}{g} \times100\% = \left(\frac{\Delta L}{L} + \frac{2\Delta T}{T}\right) 100\% $
$ = \left(\frac{0.1 }{55} + \frac{2\left(\frac{1}{20}\right)}{\frac{30}{20}}\right) 100\% \simeq 6.8 \%$
$ L = 55 cm \Delta L = 1 mm = 0.1 cm $
$ g = \frac{4\pi^{2}L}{T^{2}} $
percentage error in g is
$ \frac{\Delta g}{g} \times100\% = \left(\frac{\Delta L}{L} + \frac{2\Delta T}{T}\right) 100\% $
$ = \left(\frac{0.1 }{55} + \frac{2\left(\frac{1}{20}\right)}{\frac{30}{20}}\right) 100\% \simeq 6.8 \%$
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