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 \(=10\, m / s ^2\)) :
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 \(=10\, m / s ^2\)) :
Show 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.
\(0.1\,cm\)
\(1 \,mn\)
\(1 \,cm\)
\(0.1 \,mm\)
The Correct Option is A
Solution and Explanation
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).
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