Question

The length of two wires are in the ratio $3:4$.Ratio of the diameters is $1:2$; young's modulus of the wires are in the ratio $3:2$; If they are subjected to same tensile force, the ratio of the elongation produced is

• A. 1 :1
• B. 1 :2
• C. 2 : 3
• D. 2 : 1

Answer 1

The correct option is D 2 : 1

$\frac{L_1}{L_2}=\frac{3}{4},\frac{D_1}{D_2}=\frac{1}{2},\frac{A_1}{A_2}=\frac{1}{4}$

$\frac{y_1}{y_2}=\frac{3}{2}$ same force

$\frac{\Delta L_1}{\Delta L_2}=\frac{F.L_1}{A_1{y}_1}\times \frac{A_2{y}_2}{F{L}_2}$$=\frac{3}{4}\times \frac{4}{1}\times \frac{02}{3}$

$\frac{\Delta L_1}{\Delta L_2}=\frac{2}{1}$