Question

A cube of side 10 cm is suspended from one end of a fine string of length 27 cm, and a mass of 200 grams is connected to the other end of the string. When the cube is half immersed in water, the system remains in balance. Find the density of the cube.

• A. 800 kg/m\(^3\)
• B. 500 kg/m\(^3\)
• C. 700 kg/m\(^3\)
• D. 600 kg/m\(^3\)

Hint:

In problems involving buoyancy, use the principle of flotation, where the buoyant force is equal to the weight of the displaced fluid. In equilibrium, the weight of the object is balanced by the buoyant force.

Answer 1

The Correct Option is C

In this problem, the cube is in equilibrium when half of it is submerged in water.
The forces acting on the cube are: 1.
The buoyant force, which is equal to the weight of the displaced water. 2.
The weight of the cube, which is balanced by the mass hanging from the other side.
Step 1: Buoyant Force
The buoyant force \( F_B \) on the cube when it is half immersed is given by: \[ F_B = \rho_{\text{water}} \times g \times V_{\text{displaced}} \] where:
\( \rho_{\text{water}} = 1000 \, \text{kg/m}^3 \) (density of water),
\( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity),
\( V_{\text{displaced}} \) is the volume of the cube submerged in water, which is half of the cube's volume.
The volume of the cube is: \[ V_{\text{cube}} = \text{side}^3 = 10^3 = 1000 \, \text{cm}^3 = 1 \times 10^{-3} \, \text{m}^3 \] Thus, the volume of the submerged part of the cube is: \[ V_{\text{displaced}} = \frac{1}{2} V_{\text{cube}} = \frac{1}{2} \times 1 \times 10^{-3} = 5 \times 10^{-4} \, \text{m}^3 \] The buoyant force is then: \[ F_B = 1000 \times 9.8 \times 5 \times 10^{-4} = 4.9 \, \text{N} \]
Step 2: Weight of the Cube
The weight of the cube is equal to the mass of the cube times the acceleration due to gravity.
The mass of the cube is: \[ m_{\text{cube}} = \rho_{\text{cube}} \times V_{\text{cube}} \] where \( \rho_{\text{cube}} \) is the density of the cube, and \( V_{\text{cube}} = 1 \times 10^{-3} \, \text{m}^3 \) is the volume of the cube. The weight of the cube is: \[ W_{\text{cube}} = m_{\text{cube}} \times g = \rho_{\text{cube}} \times 1 \times 10^{-3} \times 9.8 \] Step 3: Force Balance
For the system to remain in equilibrium, the buoyant force \( F_B \) must balance the weight of the cube and the hanging mass.
The mass hanging from the string is 200 grams, which is \( 0.2 \, \text{kg} \), and its weight is: \[ W_{\text{hanging}} = 0.2 \times 9.8 = 1.96 \, \text{N} \] Thus, the total weight supported by the buoyant force is: \[ W_{\text{total}} = W_{\text{cube}} + W_{\text{hanging}} = 4.9 + 1.96 = 6.86 \, \text{N} \] Equating the buoyant force and the total weight: \[ F_B = W_{\text{cube}} + W_{\text{hanging}} = 6.86 \, \text{N} \] \[ 6.86 = \rho_{\text{cube}} \times 1 \times 10^{-3} \times 9.8 \] Solving for \( \rho_{\text{cube}} \): \[ \rho_{\text{cube}} = \frac{6.86}{1 \times 10^{-3} \times 9.8} = \frac{6.86}{0.0098} = 700 \, \text{kg/m}^3 \] Thus, the density of the cube is \( 700 \, \text{kg/m}^3 \).