Question

A cube having a side of 10 cm with unknown mass and 200 gm mass were hung at two ends of an uniform rigid rod of 27 cm long. The rod along with masses was placed on a wedge keeping the distance between wedge point and 200 gm weight as 25 cm. Initially the masses were not at balance. A beaker is placed beneath the unknown mass and water is added slowly to it. At given point the masses were in balance and half volume of the unknown mass was inside the water. (Take the density of the unknown mass is more than that of the water, the mass did not absorb water and water density is 1 gm/cm$^3$.) The unknown mass is ______ kg.

Hint:

In torque problems involving buoyancy: - Balance the clockwise and counter-clockwise torques about the fulcrum. - Account for the buoyant force when part of the object is submerged. - Ensure consistent units (e.g., convert grams to kilograms if needed).

Answer 1

Correct Answer: 3

We are given the equation: 

\( 25 \times 0.2 \times g = 2 \times (m - \rho \times v) \times g \)

First, simplify the equation:

\( m - \rho \times v = 2.5 \, \text{kg} \)

Next, substitute the values for \( \rho \times v \):

\( \rho \times v = \frac{1 \times 10^{-3} \, \text{kg}}{\text{cm}^3} \times \frac{10^3 \, \text{cm}^3}{2} = \frac{1}{2} \, \text{kg} \)

Now, solve for \( m \):

\( m = 3 \, \text{kg} \)

Answer 2

Step 1: Determine the distances from the wedge (fulcrum).
- Total length of the rod = 27 cm.
- Distance from the wedge to the 200 gm mass = 25 cm (given).
- Therefore, distance from the wedge to the unknown mass = 27 cm - 25 cm = 2 cm.

Step 2: Calculate the volume of the cube and the buoyant force when half-submerged.

- Side of the cube = 10 cm.
- Volume of the cube, \( V = 10^3 = 1000 \, \text{cm}^3 \).
- Half volume submerged, \( V_{\text{sub}} = 500 \, \text{cm}^3 \).
- Buoyant force, \( F_b = \rho_{\text{water}} \times V_{\text{sub}} \times g = 1 \, \text{gm/cm}^3 \times 500 \, \text{cm}^3 \times g = 500 \, \text{gm} \times g \).

Step 3: Set up the torque equilibrium equation about the wedge.
Let \( M \) be the unknown mass in grams.
- Torque due to the 200 gm mass: \( 200 \, \text{gm} \times g \times 25 \, \text{cm} \).
- Torque due to the unknown mass: \( (M \times g - F_b) \times 2 \, \text{cm} = (M \times g - 500 \, \text{gm} \times g) \times 2 \, \text{cm} \).
For equilibrium, the torques must balance: \[ 200 \times g \times 25 = (M \times g - 500 \times g) \times 2 \] Cancel \( g \) from both sides: \[ 200 \times 25 = (M - 500) \times 2 \] Simplify: \[ 5000 = 2M - 1000 \quad \Rightarrow \quad 2M = 6000 \quad \Rightarrow \quad M = 3000 \, \text{gm} = 3 \, \text{kg}. \]