- Home
- Standard 11
- Physics
1.Units, Dimensions and Measurement
medium
The period of oscillation of a simple pendulum is $T =2 \pi \sqrt{\frac{ L }{ g }} .$ Measured value of $ L $ is $1.0\, m$ from meter scale having a minimum division of $1 \,mm$ and time of one complete oscillation is $1.95\, s$ measured from stopwatch of $0.01 \,s$ resolution. The percentage error in the determination of $g$ will be ..... $\%.$
A
$1.13$
B
$1.03$
C
$1.33$
D
$1.30$
(JEE MAIN-2021)
Solution
$T =2 \pi \sqrt{\frac{\ell}{ g }}$
$g =\frac{4 \pi^{2} \ell}{ T ^{2}}$
$\frac{\Delta g }{ g }=\frac{\Delta \ell}{\ell}+\frac{2 \Delta T }{ T }$
$\frac{\Delta g }{ g }=\frac{1 \times 10^{-3}}{1}+2 \times \frac{0.01}{1.95}$
$\frac{\Delta g }{ g }=0.0113$ or $1.13\, \%$
Standard 11
Physics
