Question

# When the load on a wire is increased from $$3 \mathrm { kg }-wt$$ to $$8\mathrm { kg } -wt$$, the elongation increases from $$0.61 \mathrm { mm }$$to $$1.02 \mathrm { mm }$$ . The required work done during the extension of the wire, is

A
16×103J
B
8×102J
C
20×102J
D
31×103J

Solution

## The correct option is D $$31 \times 10 ^ { - 3 } \mathrm { J }$$\begin{aligned} &\Delta l_{1}=0.61 \mathrm{~mm} \quad \Delta \mathrm{l}_{2}=1.02 \mathrm{~mm}\\ &\text { (work done) }=\bigcup_{1}-\bigcup_{2} \ldots . .(i)\\ \end{aligned} \begin{aligned} U &=\frac{1}{2} \times \text { stress } x \text { strain } x \text { volume } \\ &=\frac{1}{2} \times \frac{F}{A-1} \times \frac{\Delta L}{-L} \times A \cdot K \\ U &=\frac{1}{2} \times F \times \Delta L \end{aligned} \begin{aligned} U_{1} &=\frac{1}{2} \times 3 \times 10 \times 0.61 \times 10^{-3} \\ &=9.15 \times 10^{-3} \mathrm{~J} \\ &=\text { work done dussing } \\ & \text { case } 1 \end{aligned} \begin{aligned} U_{2} &=\frac{1}{2} \times 8 \times 10 \times 1.02 \times 10^{-3} \\ U_{2} &=40.8 \times 10^{-3} \mathrm{~J} \\=& \text { work done during } \\ & \text { case II } \end{aligned} \begin{aligned} (\text { work done) }& U_{2}-U_{1} \text { during case II to case } 1 &=(40.8-9.15) \times 10^{-3} \mathrm{~J} \\ &=31.65 \times 10^{-3} \mathrm{~J} \end{aligned}Physics

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