A simple pendulum is being used to determine the value of gravitational acceleration at a certain place. The length of the pendulum is $25.0 c m$ measured with accuracy of $0.1 c m$ and a stopwatch with $1 s$ resolution measures the time taken for $40$ oscillations to be $50 s$ . The accuracy in $g$

Question

Toppr Admin · Accepted Answer

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-160- -160-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 -

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: