The period of oscillation of a simple pendulum is T=2 l g . Measu… - Vidyadip

MCQ

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 ........ $\%$

✓
$3$
B
$1 $
C
$5$
D
$2 $

✓

Answer

Correct option: A.

$3$

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\% $

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