Question

Two small metal balls of different masses $m_1$ and $m_2$ are connected by strings of equal length to a fixed point. When the balls are given equal charges, the angles that the two strings make with the vertical are $30^{\circ }$ and $60^{\circ }$, respectively. The ratio $m_1/m_2$ is close to

• A. $1.7$
• B. $3.0$
• C. $0.58$
• D. $2.0$

Answer 1

Answer: (D)

Each ball is in equilibrium under three forces.
(i) Electrostatic repulsion force $F_e$, equal on both balls along line joining centre to centre of balls.
(ii) Weight $M_{1}g$ and $M_{2}g$, different on each ball vertically downwards through centre of ball.
(iii) Tensions $T_1$ and $T_2$, different on each ball along the string.
As, angle between strings is given $30^{\circ }+60^{\circ }=90^{\circ }$ and strings are of equal length, strings forms an isosceles triangle as shown below,

Free body diagrams of balls are

Now, using Lami’s theorem for ball $1$ and ball $2,$ we have
$\frac{F_e}{\sin 150^{\circ }}=\frac{m_{1}g}{\sin 135^{\circ }}($ For ball 1)
and $\frac{F_e}{\sin 120^{\circ }}=\frac{m_{2}g}{\sin 135^{\circ }}$ (For ball 2)
Dividing these equations, we get
$\frac{\sin 120^{\circ }}{\sin 150^{\circ }}=\frac{m_1}{m_2}$
$\Rightarrow \frac{m_1}{m_2}=\frac{\sin \left(180^{\circ }-60^{\circ }\right)}{\sin \left(180^{\circ }-30^{\circ }\right)}$
$=\frac{\sin 60^{\circ }}{\sin 30^{\circ }}=\frac{\sqrt{3}/2}{1/2}=\sqrt{3}$
So, $\frac{m_1}{m_2}\approx 2.0$