First slide

Elastic behaviour of solids

Question

The area of cross section of a steel wire $(Y = 2 .0 \times 10^{11} N / m^{2})$ is 0.1 cm². The force required to double its length will be

Moderate

A

$2 \times 10^{12} N$
B

$2 \times 10^{11} N$
C

$2 \times 10^{10} N$
D

$2 \times 10^{6} N$

Solution

When the length of wire is doubled, then INCREASE IN LENGTH OF THE WIRE=LENGTH OF THE WIRE l = L and strain = 1 $∴ Y strain = \frac{F}{A}$
$∴ Force = Y \times A = 2 \times 10^{11} \times 0 .1 \times 10^{- 4} = 2 \times 10^{6} N$

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