Question

A homogenous aluminum ball of volume V is suspended on a weightless thread from one end of a homogeneous rod of mass M. Rod is placed on the edge of a tumbler so that half of the ball is submerged in water when system is in equilibrium. The densities of aluminium and water are r_A and r_W

respectively then –

• A. $\frac{y}{x}$ = $\frac{2Mg + 2\lbrack\rho_{A}gV–\rho_{W}g\frac{V}{2}\rbrack}{Mg}$
• B. $\frac{y}{x}$ = 1 + $\frac{2\left\lbrack \rho_{A}–\frac{\rho_{W}}{2} \right\rbrack V}{M}$
• C. $\frac{y}{x}$ = 1
• D. $\frac{y}{x}$ = 2

Answer 1

Answer: (B)

$\frac{y}{x}$ = 1 + $\frac{2\left\lbrack \rho_{A}–\frac{\rho_{W}}{2} \right\rbrack V}{M}$

Answer 2

Taking torque,

$\left( V\rho_{A}g - \frac{V}{2}\rho_{\omega}g \right)$x = Mg × $\left\lbrack y - \left( \frac{x + y}{2} \right) \right\rbrack$