Question

An inclined plane of length $5.60m$ making an angle of ${45}^{\circ }$ with the horizontal is placed in an uniform electric field $E=100V{m}^{-1}$. A particle of mass $1kg$ and charge ${10}^{-2}C$ is allowed to slide down from rest position from maximum height of slope. If the coefficient of friction is $0.1$, the time taken by the particle to reach the bottom is ____

• A. $1s$
• B. $1.41s$
• C. $2s$
• D. $Noneofthese$

Answer 1

The correct option is B $1.41s$

The situation can be digrameted as below.
The electric field $E=100V{m}^{-1}$
For the downward motion of the particle of mass $1kg$.
$mg\sin {45}^{\circ }-qE\cos {45}^{\circ }-\mu (mg\cos {45}^{\circ }+qE\sin {45}^{\circ })=ma$
$\Rightarrow 1\times 10\times \frac{1}{\sqrt{2}}-\frac{1}{100}\times 100\times \frac{1}{\sqrt{2}}-\frac{1}{10}$
$(1\times 10\times \frac{1}{\sqrt{2}}+\frac{1}{100}\times 100\times \frac{1}{\sqrt{2}})=1\times a$
$\Rightarrow \frac{10}{\sqrt{2}}-\frac{1}{\sqrt{2}}-\frac{1}{10}(\frac{10}{\sqrt{2}}+\frac{1}{\sqrt{2}})=a$
$\Rightarrow \frac{10}{\sqrt{2}}-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}-\frac{1}{10\sqrt{2}}=a$
$\Rightarrow a=\frac{79}{10\sqrt{2}}=\frac{7.9}{\sqrt{2}}$
$\Rightarrow a=5.6m/s^2$
Now, time taken by the particle to cover $5.6m$ distance along the incline is
$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2\times 5.6}{5.6}}=\sqrt{2}=1.41s$.