A uniform rod is suspended horizontally from | vedclass

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

Question

(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, $\sigma = \frac{l_{1}}{l_{1} - l_{2}}$.

Similar Questions

$A$ block of steel of size $5 \text{ cm} \times 5 \text{ cm} \times 5 \text{ cm}$ is weighed in water. If the relative density of steel is $7$,its apparent weight is:

An empty glass jar is submerged in a tank of water with the open mouth of the jar downwards, so that the air inside the jar is trapped and cannot escape. As the jar is pushed down slowly, the magnitude of the net buoyant force on the system of the volume of gas trapped in the jar and the jar:

The spring balance $A$ reads $2 \, kg$ with a block $m$ suspended from it. $A$ balance $B$ reads $5 \, kg$ when a beaker filled with liquid is put on the pan of the balance. The two balances are now arranged such that the hanging mass is inside the liquid as shown in the figure. In this situation:

At what depth does a body remain in equilibrium in a liquid? Explain.

Vedclass Test Series

Exam Paper Generator

Online Exam Module