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

To find the elongation of the wire, we need to understand the relationship between force, elongation, and Young's modulus. The formula for elongation \(\Delta L\) of a wire under an applied force is given by:

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

where:

• \(F\) is the force applied,
• \(L\) is the original length of the wire,
• \(A\) is the cross-sectional area, and
• \(Y\) is Young's modulus.

Let's calculate each parameter:

1. Original length of the wire, \(L = 6 \, \text{m}\)
2. Cross-sectional area, \(A = 3 \, \text{mm}^2 = 3 \times 10^{-6} \, \text{m}^2\)
3. Young's modulus, \(Y = 2 \times 10^{11} \, \text{N/m}^2\)

The force \(F\) applied to the wire is the gravitational force due to the block, which is given by:

\[F = m \cdot g'\]

where \(m = 4 \, \text{kg}\) is the mass of the block and \(g'\) is the acceleration due to gravity on the planet. Since the gravity on the planet is \(\frac{1}{4}\) of that on the Earth, we have:

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

Therefore, the force becomes:

\[F = 4 \, \text{kg} \times 2.5 \, \text{m/s}^2 = 10 \, \text{N}\]

Substitute the values into the formula for elongation:

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

Calculate the above expression:

\[\Delta L = \frac{60}{6 \times 10^5} = \frac{1}{10^4} \, \text{m} = 0.0001 \, \text{m}\]

Converting meters to centimeters:

\(0.0001 \, \text{m} = 0.1 \, \text{cm}\)

Thus, the elongation of the wire is \(0.1 \, \text{cm}\), which matches the correct option.