A load of mass M kg is suspended from a steel wire of length 2 m and radius 1.0 mm in Searle's apparatus experiment. The increase in length produced in the wire is `4.0 mm`, Now the loads is fully immersed in a liquid of relative density 2. The relative density of the material of load is 8.
The new value of increase in length of the steel wire is :

Options

A. 3.0 mm
B. 4.0 mm
C. 5.0 mm
D. zero

Explanation

To solve the problem, we need to find the new increase in length of the steel wire when the load is fully immersed in a liquid of relative density 2. Here’s the step-by-step solution:

### Step 1: Understand the initial conditions
- The mass of the load is \( M \) kg.
- The length of the steel wire is \( L = 2 \) m.
- The radius of the steel wire is \( r = 1.0 \) mm = \( 1.0 \times 10^{-3} \) m.
- The initial increase in length of the wire is \( \Delta L = 4.0 \) mm = \( 4.0 \times 10^{-3} \) m.

### Step 2: Calculate the initial force acting on the wire
The initial force acting on the wire when the load is suspended is given by the weight of the load:
\[ F = Mg \]
where \( g \) is the acceleration due to gravity.

### Step 3: Calculate the buoyant force when the load is immersed
When the load is fully immersed in a liquid of relative density \( 2 \), the buoyant force \( F_b \) acting on the load can be calculated using the formula:
\[ F_b = \rho_{\text{liquid}} \cdot V \cdot g \]
where \( \rho_{\text{liquid}} = 2 \cdot \rho_{\text{water}} \) and \( V \) is the volume of the load.

The volume \( V \) of the load can be expressed in terms of its mass and relative density:
\[ V = \frac{M}{\rho_{\text{material}} \cdot \rho_{\text{water}}} = \frac{M}{8 \cdot \rho_{\text{water}}} \]

Thus, the buoyant force becomes:
\[ F_b = 2 \cdot \rho_{\text{water}} \cdot \frac{M}{8 \cdot \rho_{\text{water}}} \cdot g = \frac{Mg}{4} \]

### Step 4: Calculate the net force acting on the wire when immersed
The net force \( F' \) acting on the wire when the load is immersed is:
\[ F' = Mg - F_b = Mg - \frac{Mg}{4} = \frac{3Mg}{4} \]

### Step 5: Relate the new increase in length to the initial increase
Using Hooke's law, we know that the increase in length is directly proportional to the force applied:
\[ \frac{\Delta L'}{\Delta L} = \frac{F'}{F} \]

Substituting the forces:
\[ \frac{\Delta L'}{4 \times 10^{-3}} = \frac{\frac{3Mg}{4}}{Mg} = \frac{3}{4} \]

### Step 6: Solve for the new increase in length
Now, we can solve for the new increase in length \( \Delta L' \):
\[ \Delta L' = \frac{3}{4} \cdot 4 \times 10^{-3} = 3 \times 10^{-3} \text{ m} = 3.0 \text{ mm} \]

### Final Answer
The new value of the increase in length of the steel wire when the load is fully immersed in the liquid is **3.0 mm**.

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