Question

A particle of mass 1 Kg and carrying 0.01 C is at rest on an inclined plane of angle ${30}^{\circ }$ with horizontal when an electric field of $\frac{490}{\sqrt{3}}N{C}^{-1}$ applied parallel to horizontal, the coefficient of friction is

• A. $0.5$
• B. $\frac{1}{\sqrt{3}}$
• C. $\frac{\sqrt{3}}{2}$
• D. $\frac{\sqrt{3}}{7}$

Answer 1

The correct option is D $\frac{\sqrt{3}}{7}$

$Eqcos\theta +f=mgsin\theta$
$f=\mu N=mgsin\theta -Eqcos\theta$
$\mu [mgcos\theta +Eqsin\theta ]=mgsin\theta -Eqcos\theta$
$\mu =\frac{mgsin\theta -Eqcos\theta }{mgcos\theta +Eqsin\theta }$
$=\frac{1\times 9.8\times \frac{1}{2}-\frac{490}{\sqrt{3}}\times 0.01\times \frac{\sqrt{3}}{2}}{1\times 9.8\times \frac{\sqrt{3}}{2}+\frac{490}{\sqrt{3}}\times 0.01\times \frac{1}{2}}$
$=\frac{(9.8-4.9)\frac{1}{2}}{(9.8+\frac{4.9}{3})\frac{\sqrt{3}}{2}}=\frac{4.9(2-1)}{4.9(2+\frac{1}{3})\sqrt{3}}$
$\mu =\frac{2}{7\sqrt{3}}=\frac{\sqrt{3}}{7}$