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Question

When the load applied to a suspended wire is increased from 3 kg-wt to 5 kg-wt; the elongation increases from 0.6 mm to 1 mm. How much work is done during the extention of the wire.


Solution

Initial load, $$P_1=3kg.wt =30N$$
$$\Rightarrow$$ Final load, $$P_2=5kg.wt=50N$$
$$\Rightarrow$$ Initial deformation, $$S_1=0.6mm$$
$$\Rightarrow$$ Final deformation, $$S_2=1mm$$
$$\Rightarrow$$ Initial W.D. , $$W_1=\dfrac{1}{2}.P_1S_1$$

                                 $$=\dfrac{1}{2}\times 30\times 0.6$$

                                 $$=9N.mm$$
Hence, the answer is $$9N.mm.$$


Physics

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