A uniform rod is suspended horizontally from | vedclass

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(C) For the rod to be in horizontal equilibrium, the net torque about the mid-point must be zero.Initially, the torque balance is: $w \cdot l = w_{1}

Vedclass · Accepted Answer

$\frac{l_{1}}{l_{1} - l_{2}}$

Vedclass · Answer

(C) For the rod to be in horizontal equilibrium, the net torque about the mid-point must be zero.Initially, the torque balance is: $w \cdot l = w_{1} \cdot l_{1}$.When the metal piece of weight $w$ is immersed in water, it experiences an upward buoyant force $F_{B}$. The effective weight becomes $w' = w - F_{B}$.The buoyant force is $F_{B} = V \rho_{w} g$, where $V$ is the volume of the metal and $\rho_{w}$ is the density of water. Since $w = V \rho_{metal} g$, we have $F_{B} = w \cdot \frac{\rho_{w}}{\rho_{metal}} = \frac{w}{\sigma}$, where $\sigma$ is the specific gravity of the metal.Thus, the new effective weight is $w' = w(1 - \frac{1}{\sigma})$.For the new equilibrium, the torque balance is: $w' \cdot l = w_{1} \cdot l_{2}$.Substituting $w'$, we get: $w(1 - \frac{1}{\sigma}) l = w_{1} l_{2}$.From the initial condition, $w = \frac{w_{1} l_{1}}{l}$.Substituting $w$ into the new equilibrium equation: $\frac{w_{1} l_{1}}{l} (1 - \frac{1}{\sigma}) l = w_{1} l_{2}$.$l_{1} (1 - \frac{1}{\sigma}) = l_{2}$.$1 - \frac{1}{\sigma} = \frac{l_{2}}{l_{1}}$.$\frac{1}{\sigma} = 1 - \frac{l_{2}}{l_{1}} = \frac{l_{1} - l_{2}}{l_{1}}$.Therefore, $\sigma = \frac{l_{1}}{l_{1} - l_{2}}$.

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