Question

An inclined plane of length $5.60m$ making an angle of ${45}^o$ 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. None of these

Answer 1

The correct option is A $1s$

The electric field$E=100Vm-1$. For the downward motion of the particle of mass

1kg.$\mathrm{mg}\sin {45}^{\circ }-qE\cos {45}^{\circ }-\mu (\mathrm{mg}\cos {45}^{\circ }+qE\sin {45}^{\circ })=ma$

$\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}{\sqrt{0}0}\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$

$\Rightarrow \frac{10}{\sqrt{2}}-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}-\frac{1}{10\sqrt{2}}=a$

$\Rightarrow a=\frac{79}{10\sqrt{2}}=\frac{7.9}{\sqrt{2}}$

$\therefore a=5\cdot 6m/s^2$

$\therefore$ time$\left(t\right)=\sqrt{\frac{2s}{a}}$

$=\sqrt{\frac{2\times 5.6}{5.6}}$

$=\sqrt{2}sec$

$=1.416ecAn$