First slide

Elastic Modulie

Question

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

Moderate

A

1000 N
B

2000N
C

4000 N
D

5000 N

Solution

Here,
A = 0.1 ${cm}^{2} = 0.1 \times 10^{- 4} m^{2}$
Y = 2 $\times 1011 {Nm}^{- 2}$
$\frac{∆ L}{L} = 0.1 % = \frac{0.1}{100} = 0.1 \times 10^{- 2}$
As Y = $\frac{\frac{F}{A}}{\frac{∆ L}{L}} F = Y \frac{∆ L}{L} A$
= 2 $\times 10^{11} {Nm}^{- 2} \times 0.1 \times 10^{- 7} \times 0.1 \times 10^{- 4} m^{2}$
= 2 $\times 10^{3} N = 2000 N$

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