A copper wire of length 4.0m and area of cross-section 1.2,c{m^2} is stretched with a force of 4.8 ?

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8.Mechanical Properties of Solids

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A copper wire of length $4.0m$ and area of cross-section $1.2\,c{m^2}$ is stretched with a force of $4.8 \times {10^3}$ $N.$ If Young’s modulus for copper is $1.2 \times {10^{11}}\,N/{m^2},$ the increase in the length of the wire will be

A

$1.33 \,mm$

B

$1.33 \,cm$

C

$2.66\, mm$

D

$2.66\, cm$

Solution

(a) $l = \frac{{FL}}{{AY}} = \frac{{4.8 \times {{10}^3} \times 4}}{{1.2 \times {{10}^{ – 4}} \times 1.2 \times {{10}^{11}}}} = 1.33\;mm$

Standard 11

Physics

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