Question

One end of thick horizontal copper wire of length \(2L\) and radius \(2R\) is welded to the end of another thin horizontal copper wire of length \(L\) and radius \(R\). When they are stretched by applying the same force at the two ends, the ratio of the elongation in the thick wire to that in the thin wire is

• A. \(1:2\)
• B. \(4:1\)
• C. \(2:1\)
• D. \(1:1\)

Hint:

Elongation is directly proportional to length and inversely proportional to cross-sectional area.

Answer 1

The Correct Option is A

Step 1: Use formula for elongation of a wire.
\[ \Delta L = \frac{FL}{AY} \] where \(F\) is force, \(L\) is length, \(A\) is cross-sectional area, and \(Y\) is Young’s modulus.

Step 2: Write elongation expressions.
For thick wire: \[ \Delta L_1 = \frac{F(2L)}{\pi(2R)^2Y} = \frac{2FL}{4\pi R^2Y} \] For thin wire: \[ \Delta L_2 = \frac{FL}{\pi R^2Y} \]

Step 3: Find ratio.
\[ \frac{\Delta L_1}{\Delta L_2} = \frac{1}{2} \]