The Young's modulus of a steel wire of length `6m` and cross–sectional area `3mm^2` , is `2 xx 10^11 N/m^2 `. the wire is suspended from its support on a given planet. A block of mass 4 kg is attached to the free end of the wire. The acceleration due to gravity on the planet is `1/4` of its value on the earth. The elongation of wire is (Take g on the earth = `10 m/s^2 `) :

Options

A. `0.1 cm`
B. `1mm`
C. `1 cm`
D. `0.1 mm`

Answer

To solve the problem, we need to find the elongation of a steel wire when a block of mass is attached to it. We will use the formula for elongation based on Young's modulus.

### Given Data:
- Length of the wire, \( L = 6 \, \text{m} \)
- Cross-sectional area, \( A = 3 \, \text{mm}^2 = 3 \times 10^{-6} \, \text{m}^2 \)
- Young's modulus, \( Y = 2 \times 10^{11} \, \text{N/m}^2 \)
- Mass of the block, \( m = 4 \, \text{kg} \)
- Acceleration due to gravity on the planet, \( g' = \frac{1}{4} g \) where \( g = 10 \, \text{m/s}^2 \)

### Step 1: Calculate the effective acceleration due to gravity on the planet.
\[ g' = \frac{1}{4} \times 10 \, \text{m/s}^2 = 2.5 \, \text{m/s}^2 \]

### Step 2: Calculate the force exerted by the block on the wire.
The force \( F \) due to the weight of the block is given by:
\[ F = m \cdot g' = 4 \, \text{kg} \times 2.5 \, \text{m/s}^2 = 10 \, \text{N} \]

### Step 3: Calculate the stress in the wire.
Stress \( \sigma \) is defined as force per unit area:
\[ \sigma = \frac{F}{A} = \frac{10 \, \text{N}}{3 \times 10^{-6} \, \text{m}^2} = \frac{10}{3 \times 10^{-6}} = \frac{10 \times 10^6}{3} \, \text{N/m}^2 \approx 3.33 \times 10^6 \, \text{N/m}^2 \]

### Step 4: Calculate the elongation of the wire using Young's modulus.
The formula for elongation \( \Delta L \) is given by:
\[ \Delta L = \frac{\sigma L}{Y} \]
Substituting the values we have:
\[ \Delta L = \frac{(3.33 \times 10^6) \times 6}{2 \times 10^{11}} \]
\[ \Delta L = \frac{19.98 \times 10^6}{2 \times 10^{11}} = \frac{19.98}{2} \times 10^{-5} = 9.99 \times 10^{-6} \, \text{m} \]
\[ \Delta L \approx 1.0 \times 10^{-5} \, \text{m} = 0.00001 \, \text{m} = 0.01 \, \text{cm} \]

### Final Answer:
The elongation of the wire is approximately \( 1.0 \times 10^{-5} \, \text{m} \) or \( 0.01 \, \text{cm} \).

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