The period of oscillation of a simple pendulum is T=2pi sqrt {frac{l}{g}} . Measured value of L is 20.0; cm known to 1; mm accuracy

Q: The period of oscillation of a simple pendulum is $T=2\pi \sqrt {\frac{l}{g}} $. Measured value of $L$ is $20.0\; cm$ known to $1\; mm$ accuracy and time for $100$ oscillations of the pendulum is found to be $90\ s$ using a wrist watch of $1\; s$ resolution. The accuracy in the determination of $g$ is ........ $\%$

a AS, $g = 4\,{\pi ^2}\frac{l}{{{T^2}}}$ So, $\frac{{\Delta g}}{g} \times 100 = \frac{{\Delta l}}{L} \times 100 + 2\frac{{\Delta T}}{T} \times 100$ $ = \frac{{0.1}}{{20}} \times 100 + 2 \times \frac{1}{{90}} \times 100 = 2.72 \simeq 3\% $

SECTION - A [PHYSICS MCQ]

GSEB - English Medium

The period of oscillation of a simple pendulum is $T=2\pi \sqrt {\frac{l}{g}} $. Measured value of $L$ is $20.0\; cm$ known to $1\; mm$ accuracy and time for $100$ oscillations of the pendulum is found to be $90\ s$ using a wrist watch of $1\; s$ resolution. The accuracy in the determination of $g$ is ........ $\%$

A$3$
B$1 $
C$5$
D$2 $

JEE MAIN 2015, Easy

STD 11 Science
JEE
STD 11 - 1. units,dimensions and measurement
Physics

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