Question

A bob of a simple pendulum of mass $40$gm with a positive charge $4\times {10}^{-6}$ is oscillating with a time period $T_1$ .An electric field of intensity $3.6\times {10}^{4}N/C$ is applied vertically upwards.Now the time period is $T_2$ the value of $\frac{T_2}{T_1}$ is $(g=10m/s^2)$ :

• A. $0.16$
• B. $0.64$
• C. $1.25$
• D. $0.8$

Answer 1

Answer: (C)

$1.25$

We know that time period of pendulum is given by

$T_1=2\pi \sqrt{\frac{l}{g}}$

when E will be applied, effective g will becomes $g-\frac{Eq}{m}$

$\Rightarrow T_2=2\pi \sqrt{\left(\frac{l}{g-\frac{Eq}{m}}\right)}$

$\frac{T_2}{T_1}=\sqrt{\frac{g}{g-\frac{Eq}{m}}}$

$\frac{T_2}{T_1}=\sqrt{\frac{10}{10-\frac{3.6\times {10}^4\times 4\times {10}^{-6}}{40\times {10}^{-3}}}}$

$\frac{T_2}{T_1}=\sqrt{\frac{10}{6.4}}$

$\frac{T_2}{T_1}=\frac{10}{8}$

$\frac{T_2}{T_1}=1.25$