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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
$2.0$
$3.0$
$0.58$
$1.7$
Solution
(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,
Now, using Lami's theorem for ball $1$ and ball $2$ , we have
$\frac{F_e}{\sin 150^{\circ}}=\frac{m_1 g}{\sin 135^{\circ}} \text { (For ball 1) }$
$\text { and } \frac{F_e}{\sin 120^{\circ}} =\frac{m_2 g}{\sin 135^{\circ}}(\text { For ball 2) }$
Dividing these equations, we get
$\frac{\sin 120^{\circ}}{\sin 150^{\circ}}=\frac{m_1}{m_2}$
$\Rightarrow \quad \frac{m_1}{m_2} =\frac{\sin \left(180^{\circ}-60^{\circ}\right)}{\sin \left(180^{\circ}-30^{\circ}\right)}$
$=\frac{\sin 60^{\circ}}{\sin 30^{\circ}}=\frac{\sqrt{3} / 2}{1 / 2}=\sqrt{3}$
So, $\quad \frac{m_1}{m_2} \approx 1.73$
