Question

In an experiment of simple pendulum, the errors in the measurement of length of the pendulum $(L)$ and time period $(T)$ are $3 \%$ and $2 \%$ respectively. The maximum percentage error in the value of $\frac{L}{T^{2}}$ is

• A. $5 \%$
• B. $7 \%$
• C. $8 \%$
• D. $1 \%$

Answer 1

Answer: (B)

The time period of simple pendulum is $T=2 \pi \sqrt{\frac{L}{g}}$
Squaring both sides, we get
$T^{2}=4 \pi^{2} \frac{L}{g}$ or $g=4 \pi^{2} \frac{L}{T^{2}} \ldots$ (i)
The maximum percentage error in $g$ is
$\frac{\Delta g}{g} \times 100=\frac{\Delta L}{L} \times 100+2\left(\frac{\Delta T}{T}\right) \times 100$
$=3 \%+2 \times 2 \%=7 \%$
From (i), we get $\frac{L}{T^{2}}=\frac{g}{4 \pi^{2}}$
The maximum percentage error in
$\frac{L}{T^{2}}$ is $\frac{\Delta\left(\frac{L}{T^{2}}\right)}{\frac{L}{T^{2}}} \times 100=\frac{\Delta g}{g} \times 100=7 \%$