The wire of a Young's modules appartus is elongated by 2 mm when a brick is suspended from .it When the brick is immersed in water the wire contracts by 0.6mm Calculate the density of the brick given that the density of water is `1000 kg m^(-3)`

Options

A. `3333 kgm^(-3)`
B. `4210 kg m^(-3)`
C. `5000 kgm^(-3)`
D. `2000 kgm^(-3)`

Explanation

To solve the problem step by step, we will analyze the elongation of the wire when the brick is suspended in air and when it is immersed in water.

### Step 1: Understand the forces acting on the brick
When the brick is suspended in air, the force acting on the wire is the weight of the brick, which can be expressed as:
\[ F_1 = mg \]
where \( m \) is the mass of the brick and \( g \) is the acceleration due to gravity.

### Step 2: Analyze the situation when the brick is immersed in water
When the brick is immersed in water, the apparent weight of the brick is reduced due to the buoyant force. The buoyant force can be expressed as:
\[ F_b = \rho_l V g \]
where \( \rho_l \) is the density of water, \( V \) is the volume of the brick, and \( g \) is the acceleration due to gravity.

The effective force acting on the wire when the brick is in water can be expressed as:
\[ F_2 = mg - F_b = mg - \rho_l V g \]

### Step 3: Relate the elongations to the forces
The elongation of the wire is directly proportional to the force applied. Let \( \Delta L_1 \) be the elongation when the brick is in air and \( \Delta L_2 \) be the elongation when the brick is in water. We have:
\[ \frac{\Delta L_1}{\Delta L_2} = \frac{F_1}{F_2} \]

Given:
- \( \Delta L_1 = 2 \, \text{mm} \)
- The wire contracts by \( 0.6 \, \text{mm} \) when the brick is immersed in water, so the new elongation \( \Delta L_2 = 2 \, \text{mm} - 0.6 \, \text{mm} = 1.4 \, \text{mm} \).

### Step 4: Substitute the forces into the ratio
Substituting the expressions for \( F_1 \) and \( F_2 \):
\[ \frac{2}{1.4} = \frac{mg}{mg - \rho_l V g} \]

### Step 5: Cancel out \( g \) and rearrange
Cancelling \( g \) from both sides:
\[ \frac{2}{1.4} = \frac{m}{m - \rho_l V} \]

### Step 6: Express mass in terms of density and volume
The mass of the brick can be expressed as:
\[ m = \rho_s V \]
where \( \rho_s \) is the density of the brick.

Substituting this into the equation gives:
\[ \frac{2}{1.4} = \frac{\rho_s V}{\rho_s V - \rho_l V} \]

### Step 7: Simplify the equation
Cancelling \( V \) (assuming \( V \neq 0 \)):
\[ \frac{2}{1.4} = \frac{\rho_s}{\rho_s - \rho_l} \]

### Step 8: Cross-multiply and solve for \( \rho_s \)
Cross-multiplying gives:
\[ 2(\rho_s - \rho_l) = 1.4 \rho_s \]
\[ 2\rho_s - 2\rho_l = 1.4\rho_s \]
\[ 0.6\rho_s = 2\rho_l \]
\[ \rho_s = \frac{2\rho_l}{0.6} \]

### Step 9: Substitute the value of \( \rho_l \)
Given \( \rho_l = 1000 \, \text{kg/m}^3 \):
\[ \rho_s = \frac{2 \times 1000}{0.6} = \frac{2000}{0.6} \approx 3333.33 \, \text{kg/m}^3 \]

### Final Answer
The density of the brick is approximately:
\[ \rho_s \approx 3333 \, \text{kg/m}^3 \]

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