In simple pendulum experiment determination |EMFJEEMain2019

JEE MAIN 2019

08-04-2019 S2

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:

Select the correct option:

6.8%

0.2%

3.5%

0.7%

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$\begin{aligned} & \mathrm{T}=2 \pi \sqrt{\frac{1}{\mathrm{~g}}} \Rightarrow \mathrm{~g}=4 \pi^2 \frac{\mathrm{l}}{\mathrm{T}^2} \\ & \frac{\Delta \mathrm{~g}}{\mathrm{~g}}=\frac{\Delta \mathrm{l}}{\mathrm{l}}+\frac{2 \Delta \mathrm{~T}}{\mathrm{~T}}=\left(\frac{0.1}{55} \times \frac{2 \times 1}{30}\right) \times 100 \\ & \approx 6.8 \%\end{aligned}$

Question Tags

JEE Main

Physics

Medium

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