m_1 एवं m_2 द्रव्यर्मान की धातु की दो छोटी गे | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(d)

Each ball is in equilibrium under three forces.

$(i)$ Electrostatic repulsion force $F_e$, equal on both balls along line joining centre to

Vedclass · Accepted Answer

$1.7$

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

Similar Questions