JEE MAIN - Physics (2022 - 28th July Evening Shift - No. 27)
In an experiment to find acceleration due to gravity (g) using simple pendulum, time period of $$0.5 \mathrm{~s}$$ is measured from time of 100 oscillation with a watch of $$1 \mathrm{~s}$$ resolution. If measured value of length is $$10 \mathrm{~cm}$$ known to $$1 \mathrm{~mm}$$ accuracy, The accuracy in the determination of $$\mathrm{g}$$ is found to be $$x \%$$. The value of $$x$$ is ___________.
Answer
5
Explanation
$$T = 2\pi \sqrt {{l \over g}} $$
$${{dg} \over g} \times 100 = {{2dT} \over T} \times 100 + {{dl} \over l} \times 100$$
$$ = 2 \times {1 \over {50}} \times 100 + {1 \over {100}} \times 100 = 5\% $$
Comments (0)
