A simple pendulum with a bob of mass $40 \mathrm{~g}$ and charge $+2 \mu \mathrm{C}$ makes 20 oscillation in…
A simple pendulum with a bob of mass $40 \mathrm{~g}$ and charge $+2 \mu \mathrm{C}$ makes 20 oscillation in $44 \mathrm{~s}$. A vertical electric field magnitude $4.2 \times 10^4 \mathrm{NC}^{-1}$ pointing downward is applied. The time taken by the pendulum to make 15 oscillation in the electric field is (acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
30 s
60 s
90 s
15 s
Solution
Mass, $m=40 \mathrm{~g}=0.04 \mathrm{~kg}, q=2 \times 10^{-6} \mathrm{C}$
Now, $\quad m a=q E$
$
\begin{aligned}
\Rightarrow \quad a & =\frac{q}{m} E \\
& =\frac{2 \times 10^{-6}}{0.04} \times 4.2 \times 10^4 \mathrm{~m} / \mathrm{s}^2 \\
& =2.1 \mathrm{~m} / \mathrm{s}^2(\text { downward })
\end{aligned}
$
So, effective acceleration on bob,
$
a_e=a+g=12.1 \mathrm{~m} / \mathrm{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,
$
\begin{aligned}
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
\end{aligned}
$
Given, $\quad T=\frac{44}{20} \Rightarrow T^{\prime}=\frac{10}{11} \times \frac{44}{20}=2 \mathrm{~s}$
So, time taken in 15 oscillations
$
=2 \times 15=30 \mathrm{~s}
$