Solution
Case 1: Equilibrium in Water
Initially, the rod is suspended from the center (length $l/2$) and is horizontal while submerged in water. This implies the net torque about the center is zero.
Let $V_A, V_B$ be the volumes and $\rho_A, \rho_B$ be the densities of the balls. $\rho_w$ is the density of water.
Effective weight in water = Weight – Buoyant Force = $V\rho g – V\rho_w g = V(\rho – \rho_w)g$.
Torque balance about the center:
$$W_{eff, A} \times \frac{l}{2} = W_{eff, B} \times \frac{l}{2}$$ $$V_A(\rho_A – \rho_w) = V_B(\rho_B – \rho_w)$$ $$\frac{V_A}{V_B} = \frac{\rho_B – \rho_w}{\rho_A – \rho_w} \quad \text{…(i)}$$Case 2: Equilibrium in Air
When the system is in air, the buoyant force is negligible (or zero). To keep the rod horizontal, ball B is shifted to a new position. Let the distance of ball B from the center pivot be $x$. Ball A remains at distance $l/2$.
Torque balance in air:
$$m_A g \times \frac{l}{2} = m_B g \times x$$ $$\rho_A V_A \frac{l}{2} = \rho_B V_B x$$ $$x = \frac{l}{2} \left( \frac{\rho_A V_A}{\rho_B V_B} \right)$$Substituting $\frac{V_A}{V_B}$ from equation (i):
$$x = \frac{l}{2} \frac{\rho_A}{\rho_B} \left( \frac{\rho_B – \rho_w}{\rho_A – \rho_w} \right)$$Finding the Shift
The question asks for the distance ball B should be shifted. Originally it was at $l/2$. The shift $S$ is:
$$S = \frac{l}{2} – x = \frac{l}{2} \left[ 1 – \frac{\rho_A(\rho_B – \rho_w)}{\rho_B(\rho_A – \rho_w)} \right]$$Simplifying the term in brackets:
$$1 – \frac{\rho_A\rho_B – \rho_A\rho_w}{\rho_B\rho_A – \rho_B\rho_w} = \frac{(\rho_B\rho_A – \rho_B\rho_w) – (\rho_A\rho_B – \rho_A\rho_w)}{\rho_B(\rho_A – \rho_w)}$$ $$= \frac{\rho_A\rho_w – \rho_B\rho_w}{\rho_B(\rho_A – \rho_w)} = \frac{\rho_w(\rho_A – \rho_B)}{\rho_B(\rho_A – \rho_w)}$$So the shift is:
$$S = \frac{l}{2} \left[ \frac{\rho_w(\rho_A – \rho_B)}{\rho_B(\rho_A – \rho_w)} \right]$$Numerical Substitution
Given: $l = 40 \text{ cm} \implies l/2 = 20 \text{ cm}$.
$\rho_A = 7140 \text{ kg/m}^3$, $\rho_B = 1740 \text{ kg/m}^3$, $\rho_w = 1000 \text{ kg/m}^3$.