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

The given equation is:

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

Simplifying the equation yields:

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

Substituting the values for \( \rho \times v \):

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

Solving for \( m \):

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