Question

If the work done in stretching a wire by 1 mm is 2 J, the work necessary for stretching another wire of same material but with double radius of cross section and half the length by 1 mm is

• A. 16 J
• B. 8 J
• C. 4 J
• D. \( \frac{1}{4} \) J

Hint:

When comparing the work done in stretching two wires of the same material but different dimensions, use the ratio of the squares of their radii and the inverse of their lengths to determine the work done.

Answer 1

The Correct Option is A

Step 1: The work done in stretching a wire is given by the formula: \[ W = \frac{1}{2} \frac{F \Delta L}{Y} \] where \( F \) is the force applied, \( \Delta L \) is the elongation, and \( Y \) is Young's Modulus of the material. The force \( F \) is related to the tension in the wire, which depends on the cross-sectional area of the wire and the applied stress. The elongation \( \Delta L \) depends on the wire's length and Young's modulus. The work done in stretching a wire is proportional to the ratio of the square of the radius of the wire to the length of the wire. So, the work done on a wire is given by: \[ W \propto \frac{r^2}{L} \Delta L \] where \( r \) is the radius of the wire, \( L \) is the length of the wire, and \( \Delta L \) is the elongation. 

Step 2: If the new wire has double the radius and half the length, we can compare the work done on the new wire with the initial wire. Let the initial work done be \( W_1 = 2 \, \text{J} \) for a wire with radius \( r \) and length \( L \). For the new wire, the radius is \( 2r \) and the length is \( L/2 \). The work done on the new wire \( W_2 \) is: \[ W_2 = W_1 \times \left( \frac{2r}{r} \right)^2 \times \frac{L}{L/2} \] \[ W_2 = 2 \times 4 \times 2 = 16 \, \text{J} \] Thus, the work required is 16 J.