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8.Mechanical Properties of Solids
medium
The work per unit volume to stretch the length by $1\%$ of a wire with cross sectional area of $1\,m{m^2}$ will be. $[Y = 9 \times {10^{11}}\,N/{m^2}]$
A
$9 \times {10^{11}}\,J$
B
$4.5 \times {10^7}\,J$
C
$9 \times {10^7}J$
D
$4.5 \times {10^{11}}\,J$
Solution
(b) $U = \frac{1}{2} \times Y \times {({\rm{Strain}})^2} = \frac{1}{2} \times 9 \times {10^{11}} \times {\left( {\frac{1}{{100}}} \right)^2}$
$ = 4.5 \times {10^7}\;J$
Standard 11
Physics
