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, $\frac{m_{1}}{m_{2}} \approx 2.0$