Elastic behaviour of solids

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Question

The work done in stretching a wire by 0.1 mm is 4 J. Find the work done (in J) in stretching another wire of same material, but with double the radius and half the length by 0.1 mm in joules.

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Solution

Work done $W = \frac{1}{2} Fx$ (F is force, x is elongation)
Here elongation is same and $W \propto F$
But elongation is given by $= \frac{Fl}{AY}$
As x, Y are constants $F \propto \frac{A}{l}$
If F₁ and F₂ are the forces in the first and second cases
$\begin{array}{l} \frac{F_{1}}{F_{2}} = \frac{A_{1}}{A_{2}} \times \frac{l_{2}}{l_{1}} = {(\frac{r_{1}}{2 r_{1}})}^{2} \times (\frac{l_{1} / 2}{l_{1}}) = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8} \\ Then \frac{W_{1}}{W_{2}} = \frac{F_{1}}{F_{2}} = \frac{1}{8} \\ W_{2} = 8 W_{1} = 8 \times 4 = 32 J \end{array}$

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Elastic behaviour of solids

Remember concepts with our Masterclasses.

The work done in stretching a wire by 0.1 mm is 4 J. Find the work done (in J) in stretching another wire of same material, but with double the radius and half the length by 0.1 mm in joules.

Work done W=12Fx (F is force, x is elongation)Here elongation is same and W∝FBut elongation is given by =FlAYAs x, Y are constants F∝AlIf F1 and F2 are the forces in the first and second cases F1F2=A1A2×l2l1=r12r12×l1/2l1=14×12=18Then W1W2=F1F2=18W2=8W1=8×4=32 J

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