A weight of 200 ,kg is suspended by vertical wire of length 600.5, cm . area of cross-section of wir

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8.Mechanical Properties of Solids

easy

A weight of $200 \,kg$ is suspended by vertical wire of length $600.5\, cm$. The area of cross-section of wire is $1\,m{m^2}$. When the load is removed, the wire contracts by $0.5 \,cm$. The Young's modulus of the material of wire will be

A

$2.35 \times {10^{12}}\,N/{m^2}$

B

$1.35 \times {10^{10}}\,N/{m^2}$

C

$13.5 \times {10^{11}}\,N/{m^2}$

D

$23.5 \times {10^9}\,N/{m^2}$

Solution

(a) $F = 2000N,\;L = 6m,\;l = 0.5\;cm,A = {10^{ – 6}}{m^2}$

$Y = \frac{{FL}}{{Al}} = \frac{{2000 \times 6}}{{{{10}^{ – 6}} \times 0.5 \times {{10}^{ – 2}}}} = 2.35 \times {10^{12}}\;N/{m^2}$

Standard 11

Physics

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