Question

The Young's modulus of a steel wire of length \(6 m\) and cross-sectional area \(3 \,mm ^2\), is \(2 \times 10^{11}\) \(N / m ^2\). 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 \(\frac{1}{4}\) of its value on the earth The elongation of wire is (Take \(g\) on the earth \(=10\, m / s ^2\)) :

• A. \(0.1\,cm\)
• B. \(1 \,mn\)
• C. \(1 \,cm\)
• D. \(0.1 \,mm\)

Hint:

Remember the formula for elongation and ensure all units are consistent (SI units are preferred). Pay attention to details like the effective gravitational acceleration on a different planet.

Answer 1

The Correct Option is A

Calculate the Effective Acceleration Due to Gravity:

The acceleration due to gravity on the planet is \(\frac{1}{4}\) of the Earth’s gravity (\(g = 10 \, \text{m/s}^2\)):

\[ g_\text{planet} = \frac{1}{4} g = \frac{1}{4} \times 10 = 2.5 \, \text{m/s}^2. \]

Calculate the Tension in the Wire:

The tension (\(F\)) in the wire is equal to the weight of the block:

\[ F = m g_\text{planet} = 4 \times 2.5 = 10 \, \text{N}. \]

Convert the Cross-sectional Area to \(m^2\):

Given area \(A = 3 \, \text{mm}^2\). Convert this to \(m^2\):

\[ A = 3 \times (10^{-3})^2 = 3 \times 10^{-6} \, \text{m}^2. \]

Use the Formula for Elongation:

The elongation (\(\Delta L\)) of a wire under tension is given by:

\[ \Delta L = \frac{F L}{A Y}, \]

where \(F\) is the tension, \(L\) is the original length, \(A\) is the cross-sectional area, and \(Y\) is Young’s modulus.

Substitute the Values and Calculate Elongation:

Substituting the given values, we get:

\[ \Delta L = \frac{10 \times 6}{3 \times 10^{-6} \times 2 \times 10^{11}} \]

\[ \Delta L = \frac{60}{6 \times 10^5} = 10^{-5} \, \text{m} = 0.1 \times 10^{-3} \, \text{m} = 0.1 \, \text{mm}. \]

Conclusion: The elongation of the wire is \(0.1 \, \text{mm}\) (Option 4).