Question

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

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

Answer 1

Answer: (A)

Mass, $m=40\, g =0.04\, kg ,\, q=2 \times 10^{-6} C$
Now, $m a =q E$
$\Rightarrow a =\frac{q}{m} E$
$=\frac{2 \times 10^{-6}}{0.04} \times 4.2 \times 10^{4}\, m / s ^{2}$
$=2.1\, m / s ^{2}$(downward)
So, effective acceleration on bob,
$a_{e}=a +g=12.1\, m / s ^{2}$
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,
$T'=2 \pi \sqrt{\frac{l}{a_{e}}}=2 \pi \sqrt{\frac{l}{12.1}}$
$\frac{T}{T'}=\sqrt{\frac{12.1}{10}}=\frac{11}{10}$
$\Rightarrow T'=\frac{10}{11} T$
Given, $T=\frac{44}{20}$
$\Rightarrow T'=\frac{10}{11} \times \frac{44}{20}=2\, s$
So, time taken in 15 oscillations
$=2 \times 15=30\, s$