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  • A block of mass M is suspended from a wire of mass m, cross-sectional area A and length L. If all the energy stored in the wire is converted into heat, the rise in the temperature of the

A block of mass M is suspended from a wire of mass m, cross-sectional area A and length L. If all the energy stored in the wire is converted into heat, the rise in the temperature of the wire is ( Y= Young's modulus and \mathrm{s=} specific heat capacity of the material of the wire).

Option: 1

\mathrm{\frac{(M g)^2 L}{2 Y A m s}}


Option: 2

\mathrm{\frac{M g L m s}{2 Y A}}


Option: 3

\mathrm{\frac{2 Y A L}{(M g)^2 m s}}


Option: 4

\mathrm{\text { none of these }}


Answers (1)

best_answer

\mathrm{ \text { Stress }=\frac{M g}{A}, \text { strain }=\frac{\Delta L}{L}, \text { volume }=A L }

Energy stored in wire is

\mathrm{U=} energy stored per unit volume \mathrm{\times} volume

\mathrm{ \begin{aligned} & =\left(\frac{1}{2} \text { stress } \times \text { strain }\right) \times \text { volume } \\ & =\frac{1}{2} \times\left(\frac{M g}{A} \times \frac{\Delta L}{L}\right) \times A L \\ & =\frac{1}{2} M g \Delta L \\ & =\frac{(M g)^2 L}{2 A Y} \quad\left(\because \Delta L=\frac{M g L}{A Y}\right) \end{aligned} }
If \mathrm{\Delta T} is the rise in temperature, then \mathrm{U=m s \Delta T}, i.e.

\mathrm{ \begin{aligned} \frac{(M g)^2 L}{2 A Y} & =m s \Delta T \\ \Rightarrow \quad \Delta T & =\frac{(M g)^2 L}{2 A Y m s}, \text { which is choice (a). } \end{aligned} }

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Ritika Kankaria

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