Question

Two wires of same material and radius have their lengths in ratio 1:2. If these wires are stretched by the same force, the strain produced in the two wires will be in the ratio

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

Answer 1

The correct option is C. 1:2.

We know that, the strain developed in a material is given by,

$l=\frac{FL}{AY}$

Where, $l$ = Change in the length

$L$ = Original length

$F$ = Force applied

$A$ = Area of cross section

$Y$ = Young's modulus

Therefore, for constant $F$ and $Y$ we have,

$l\propto \frac{L}{A}=\frac{L}{r^2}$

$\frac{l_1}{l_2}=\frac{L_1}{L_2}\times \frac{r_2^2}{r_1^2}=\frac{L_1}{L_2}$ (Since, $r_1=r_2$).

$or\frac{l_1}{l_2}=\frac{1}{2}$