Question

Under the same load, wire A having length $50 m$ and cross section $25 \times 10^{-5} m ^2$ stretches uniformly by the same amount as another wire $B$ of length $60 m$ and a cross section of $30 \times 10^{-5}$ $m ^2$ stretches The ratio of the Young's modulus of wire $A$ to that of wire $B$ will be :

• A. $1: 4$
• B. $1: 1$
• C. $3=10$
• D. $1: 2$

Hint:

Young's modulus is inversely proportional to the product of the length and cross-sectional area for equal strain.

Answer 1

The Correct Option is B

To determine the ratio of the Young's modulus of wire A to that of wire B, we'll use the formula for Young's modulus:

\(Y = \frac{F \cdot L}{A \cdot \Delta L}\)

where:

• \(F\) is the force applied,
• \(L\) is the original length of the wire,
• \(A\) is the cross-sectional area, and
• \(\Delta L\) is the extension in the wire.

Given that under the same load, wire A extends uniformly by the same amount as wire B, we have \(\Delta L_A = \Delta L_B\) and also \(F_A = F_B\).

For wire A, Young's modulus \((Y_A)\) can be expressed as:

\(Y_A = \frac{F \cdot L_A}{A_A \cdot \Delta L}\)

For wire B, Young's modulus \((Y_B)\) can be expressed as:

\(Y_B = \frac{F \cdot L_B}{A_B \cdot \Delta L}\)

According to the problem, the extensions are equal, thus the equations become:

\(Y_A = \frac{F \cdot 50}{25 \times 10^{-5} \cdot \Delta L}\)

\(Y_B = \frac{F \cdot 60}{30 \times 10^{-5} \cdot \Delta L}\)

The ratio of Young's modulii becomes:

\(\frac{Y_A}{Y_B} = \frac{\left(\frac{F \cdot 50}{25 \times 10^{-5} \cdot \Delta L}\right)}{\left(\frac{F \cdot 60}{30 \times 10^{-5} \cdot \Delta L}\right)}\)

After simplification, we get:

\(\frac{Y_A}{Y_B} = \frac{50 \times 30 \times 10^{-5}}{25 \times 60 \times 10^{-5}}\)

This simplifies further to:

\(\frac{Y_A}{Y_B} = \frac{1500}{1500} = 1:1\)

Hence, the correct ratio of the Young's modulus of wire A to wire B is 1:1.