Question

Two spheres having same radius and mass are suspended by two strings of equal length from the same point in such a way that their surfaces touch each other. On depositing charge $2\times {10}^{-6}C$ on them they repel each other in such a way that in equilibrium the angle between their strings becomes ${60}^o$. If the distance from the point of suspension to the centre of the sphere is $10cm$. Find the mass of each sphere. $(K=9\times {10}^{9}SIandg=10m{s}^{-1})$

• A. $0.6235kg$
• B. $0.3117kg$
• C. $0.1559kg$
• D. $1.2468kg$

Answer 1

The correct option is A $0.6235kg$

Given,
$q_1=q_2=2\times {10}^{-6}C$
$I=10cm=r=10\times {10}^{-2}m$
$m=$?, $g=10m/s^2$
$F=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{q_1{q}_2}{r^2}$
$F=9\times {10}^9\times \frac{{(2\times {10}^{-6})}^2}{{(10\times {10}^{-2})}^2}$
$\tan \theta =\frac{F}{mg}$
$\tan {30}^o=\frac{9\times {10}^9\times 4\times {10}^{-12}}{100\times {10}^{-4}\times m\times 10}$
$\frac{1}{\sqrt{3}}=\frac{9\times {10}^9\times 4\times {10}^{-12}}{1000\times {10}^{-4}\times m}$
$m=\sqrt{3}\times 9\times {10}^9\times 4\times {10}^{-12}\times 10$
$=1.732\times 9\times 4\times {10}^{-2}$
$=36\times 1.7320\times {10}^{-2}$
$=62.35\times {10}^{-2}g$
$m=0.6235kg$