Solveeit Logo
Login

Question

Physics Question on Stress and Strain

The force required to stretch a steel wire of 1 scm1\,s cm cross-section to 1.11.1 times, its length would be (Y=2×1011Nm−2Y=2 \times 10^ {11} Nm ^{-2}).

A

2×107N2 \times 10^{7} N

B

2×10−7N2 \times 10^{-7} N

C

2×10−6N2 \times 10^{-6} N

D

4.2×107N4.2 \times 10^{7} N

Answer

2×107N2 \times 10^{7} N

Explanation

Solution

Young's modulus Y=(F/A)/(ΔL/L)Y =( F / A ) /(\Delta L / L ) As the length of the wire is doubled, the change in length is equal to its original length. If Lis the original length, then ΔL=L\Delta L = L. ThereforeY =F/A= F / A F=Y×A=2×1011(N/m2)×(10−2m)2 =2×107N\begin{aligned} F = Y \times A =2 \times 1011( N / m2) \times (10-2 m )2 \\\ =2 \times 10^{7} N \end{aligned}