Question

Two wires of cross sectional area \(1 \text{ cm}^2\) and \(2 \text{ cm}^2\) and lengths 20 cm and 30 cm are connected to the same load. If their extensions are same, find the ratio of their Young's modulus :

• A. \(\frac{4}{2}\)
• B. \(\frac{4}{3}\)
• C. \(\frac{1}{2}\)
• D. \(\frac{3}{2}\)

Hint:

When variables like load and extension are "same", isolate the remaining variables in the formula. Here, \(Y \cdot \frac{A}{L} = \text{constant}\).

Answer 1

The Correct Option is B

To solve this problem, we need to find the ratio of Young's modulus of two wires which are connected to the same load and have the same extension. The Young's modulus \(Y\) is defined by the formula:

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

 

where:

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

Given that the extensions \(\Delta L\) are the same for both wires, we can equate the two forms of Young's modulus for each wire:

\(Y_1 = \frac{F \cdot L_1}{A_1 \cdot \Delta L}\) and \(Y_2 = \frac{F \cdot L_2}{A_2 \cdot \Delta L}\)

 

Since the extensions and the force \(F\) are the same, we get:

\(\frac{Y_1}{Y_2} = \frac{L_1 \cdot A_2}{L_2 \cdot A_1}\)

 

Substituting the given values:

• \(L_1 = 20 \text{ cm}\)
• \(L_2 = 30 \text{ cm}\)
• \(A_1 = 1 \text{ cm}^2\)
• \(A_2 = 2 \text{ cm}^2\)

Plug in these values:

\(\frac{Y_1}{Y_2} = \frac{20 \cdot 2}{30 \cdot 1} = \frac{40}{30} = \frac{4}{3}\)

 

Therefore, the ratio of their Young's modulus \(Y_1 : Y_2\) is \(\frac{4}{3}\).