Question

The two wires A and B of the same material have their lengths in the ratio 1 : 2 and their diameters in the ratio 2 : 1. If they are stretched with the same force, the ratio of the increase in the length of A to that of B will be:

• A. 1 : 2
• B. 4 : 1
• C. 1 : 8
• D. 1 : 4

Hint:

The extension in the wire is inversely proportional to the cross-sectional area and directly proportional to the length of the wire.

Answer 1

The Correct Option is B

To solve the problem, let's consider the following parameters for the two wires A and B, given they are made from the same material:

• Length of A, \( L_A \) and B, \( L_B \) have a ratio of 1:2.
• Diameters are in the ratio: 2:1 for A and B.

The formula for elongation (\(\Delta L\)) in a wire when a force \( F \) is applied is given by: \[\Delta L = \frac{FL}{AE}\] where \( F \) is the force, \( L \) is the original length, \( A \) is the cross-sectional area, and \( E \) is the Young's modulus (same for both wires since material is same). Now, using the diameter to find the area, recall the area of a circle is: \[A = \frac{\pi d^2}{4}\] For wire A:

• Area \( A_A = \frac{\pi (2d)^2}{4} = \pi d^2\)
• Length \( L_A = L \)

For wire B:

• Area \( A_B = \frac{\pi (d)^2}{4} = \frac{\pi d^2}{4}\)
• Length \( L_B = 2L \)

Let’s find the ratio of elongations \(\Delta L_A\) to \(\Delta L_B\): \[\Delta L_A = \frac{FL_A}{A_AE} = \frac{FL}{\pi d^2 E}\] \[\Delta L_B = \frac{FL_B}{A_BE} = \frac{F(2L)}{(\pi d^2/4)E} = \frac{8FL}{\pi d^2 E}\] Therefore, the ratio of changes in length is: \[\frac{\Delta L_A}{\Delta L_B} = \frac{\frac{FL}{\pi d^2 E}}{\frac{8FL}{\pi d^2 E}} = \frac{1}{8}\] Hence, the ratio of increase in length of A to B when the same force is applied is: 4:1.