COMEDK 2023 Shift 1 Paper

To find the expression for the charge on the metal ball, we need to consider the forces acting on the ball when it is just suspended in the liquid due to the electric field. The forces involved are:
The weight of the ball (gravitational force), F‌gravity ‌=mg, where m= mass of the ball =ρV=ρ(‌

4

3

πr3), and g is the acceleration due to gravity.
The buoyant force due to the liquid, which can be given by Archimedes' principle,
F‌buoyant ‌= Volume of displaced liquid ×g× density of liquid =Vσg=(‌

4

3

πr3)σg.
The electric force in the upward direction which keeps the ball suspended, F‌electric ‌=qE, where E is the electric field strength.
For the ball to be just suspended, the net force acting on it must be zero. This implies:
F‌electric ‌=F‌gravity ‌−F‌buoyant ‌
Substituting the expressions we have for each:
qE=ρ(‌

4

3

πr3)g−σ(‌

4

3

πr3)g
Isolating q, we find:
q=‌

(ρ( 4 3 πr3)g−σ( 4 3 πr3)g)

E

Simplifying the expression:
q=‌

( 4 3 πr3)(ρ−σ)g

E

So, the correct expression for the charge on the metal ball, as per the given options, is option B
q=‌

[4πr3(ρ−σ)g]

3E