Question

A simple pendulum with a bob of mass $40g$ and charge $+2\mu C$ makes $20$ oscillation in $44s$. A vertical electric field magnitude $4.2\times 10^{4}N{C}^{-1}$ pointing downward is applied. The time taken by the pendulum to make $15$ oscillation in the electric field is (acceleration due to gravity $=10m{s}^{-2}$ )

• A. 30 s
• B. 60 s
• C. 90 s
• D. 15 s

Answer 1

Answer: (A)

Mass, $m=40g=0.04kg,q=2\times 10^{-6}C$
Now, $ma=qE$
$\Rightarrow a=\frac{q}{m}E$
$=\frac{2\times 10^{-6}}{0.04}\times 4.2\times 10^{4}m/s^2$
$=2.1m/s^2$(downward)
So, effective acceleration on bob,
$a_e=a+g=12.1m/s^2$
In the absence of electric field,
$T=2\pi \sqrt{\frac{l}{g}}=2\pi \sqrt{\frac{l}{10}}$
In the presence of electric field,
$T^{\prime }=2\pi \sqrt{\frac{l}{a_e}}=2\pi \sqrt{\frac{l}{12.1}}$
$\frac{T}{T^{\prime }}=\sqrt{\frac{12.1}{10}}=\frac{11}{10}$
$\Rightarrow T^{\prime }=\frac{10}{11}T$
Given, $T=\frac{44}{20}$
$\Rightarrow T^{\prime }=\frac{10}{11}\times \frac{44}{20}=2s$
So, time taken in 15 oscillations
$=2\times 15=30s$