Question

The area of a cross-section of steel wire is $\text{0.1 c}\text{m}^{2}$and Young’s modulus of steel is $2 \times 10^{11}\text{ N }\text{m}^{2}.$ The force required to stretch by 0.1% of its length is

• A. 1000 N
• B. 2000 N
• C. 4000 N
• D. 5000 N

Answer 1

Answer: (B)

2000 N

Answer 2

: Here,

$A = 0.1cm^{2} = 0.1 \times 10^{- 4}m^{2}$

$Y = 2 \times 10^{11}Nm^{- 2}$

$\frac{\Delta L}{L} = 0.1\% = \frac{0.1}{100} = 0.1 \times 10^{- 2}$

As $Y = \frac{F/A}{\Delta L/L}$

$\therefore F = Y\frac{\Delta L}{L}A$

$= 2 \times 10^{11}Nm^{- 2} \times 0.1 \times 10^{- 2} \times 0.1 \times 10^{- 4}m^{2}$

$= 2 \times 10^{3}N = 2000N$