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8.Mechanical Properties of Solids
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A uniform wire of length $L$ and radius $r$ is twisted by an angle $\alpha$. If modulus of rigidity of the wire is $\eta$, then the elastic potential energy stored in wire, is .........
A
$\frac{\pi \eta r^4 \alpha}{2 L^2}$
B
$\frac{\pi \eta r^4 \alpha}{4 L^2}$
C
$\frac{\pi \eta r^4 \alpha^2}{4 L}$
D
$\frac{\pi \eta r^4 \alpha^2}{2 L}$
Solution
(c)
$U=$ work done
We know
Work done $=\frac{\pi S r^4 \phi^2}{4 L}$ $\left\{\begin{array}{l}\text { Where, } \\ \phi=\text { Angle of twist }=\alpha \\ S=\text { Modulus of rigidity }=4\end{array}\right.$
Substituting values
$U=\frac{\pi \eta r^4 \alpha^2}{4 L}$
Standard 11
Physics
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