What force is required to stretch a steel wire $1 c m^{2}$ in cross section to double its length? (Youngs modulus for steel is $2 \times 10^{11} P a$ )

Q. The are of cross-section of steel wire is

0.1 {c m}^{2}

and Young's modulus of steel is

2 \times 10^{11} N m^{- 2}

. The force required to stretch by

0.1

% of tis length is then

Join BYJU'S Learning Program

The force required to stretch a steel wire of 1cm2 cross - section to 1.1 times its length would be (Y=2×1011Nm−2)

The correct option is A 2×106NL2=1.1L1 ∴ Strain = lL1=L2−L1L1=1.1L1−L1L1=0.1. F=YAlL=2×1011×1×10−4×0.1=2×106N.

The force required to stretch a steel wire of $1 c m^{2}$ cross - section to 1.1 times its length would be $(Y = 2 \times 10^{11} N m^{- 2})$

The correct option is A $2 \times 10^{6} N$
$L_{2} = 1.1 L_{1}$ $∴$ Strain = $\frac{l}{L_{1}} = \frac{L_{2} - L_{1}}{L_{1}} = \frac{1.1 L_{1} - L_{1}}{L_{1}} = 0.1.$
$F = Y A \frac{l}{L} = 2 \times 10^{11} \times 1 \times 10^{- 4} \times 0.1 = 2 \times 10^{6} N .$