- Home
- Standard 11
- Physics
1.Units, Dimensions and Measurement
hard
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)
Solution
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\% $
Standard 11
Physics
