Question

Two wires are made of the same material and have the same volume. The first wire has cross-sectional area $ A $ and the second wire has cross-sectional area $ 3A $. If the length of the first wire is increased by $ \Delta l $ on applying a force $ F $, how much force is needed to stretch the second wire by the same amount?

• A. \( 4F \)
• B. \( 6F \)
• C. \( 9F \)
• D. \( F \)

Hint:

To stretch a wire by the same amount, the force required is inversely proportional to the cross-sectional area of the wire. A wire with a larger cross-sectional area requires more force to achieve the same extension.

Answer 1

The Correct Option is C

The extension of a wire is given by the formula: \[ \Delta l = \frac{F L}{A Y} \] where \( F \) is the force applied, \( L \) is the original length of the wire, \( A \) is the cross-sectional area, and \( Y \) is the Young's modulus of the material. 
Since both wires are made of the same material, \( Y \) is the same for both wires. 
The first wire has cross-sectional area \( A \), and the second wire has cross-sectional area \( 3A \). 
The length of both wires is the same. 
Let's compare the forces required to stretch the two wires by the same amount: \[ \Delta l_1 = \frac{F L}{A Y}, \quad \Delta l_2 = \frac{F_2 L}{3A Y} \] Since the length increase \( \Delta l_1 = \Delta l_2 \), we have: \[ \frac{F L}{A Y} = \frac{F_2 L}{3A Y} \] Solving for \( F_2 \): \[ F_2 = 3F \] 
Thus, to stretch the second wire by the same amount, a force of \( 9F \) is required. 
Thus, the correct answer is \( 9F \).