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Byju's Answer

Standard XII

Physics

Young's Modulus of Elasticity

Two wires A a...

Question

Two wires A and B are of same materials. Their lengths are in the ratio 1 : 2 and diameters are in the ratio 2 : 1 when stretched by force $F_{A}$ and $F_{B}$ respectively they get equal increase in their lengths. Then the ratio $\frac{F_{A}}{F_{B}}$ should be

1 : 2

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1 : 1

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2 : 1

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8 : 1

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Solution

The correct option is D 8 : 1
$Y = \frac{F L}{π r^{2} l}$ $∴ F = Y π r^{2} \frac{l}{L}$
$\frac{F_{A}}{F_{B}} = \frac{Y_{A}}{Y_{B}} {(\frac{r_{A}}{r_{B}})}^{2} (\frac{l_{A}}{l_{B}}) (\frac{L_{B}}{L_{A}}) = 1 \times {(\frac{2}{1})}^{2} \times (1) \times (\frac{2}{1}) = 8$

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Two wires A and B are of same materials. Their lengths are in the ratio 1 : 2 and diameters are in the ratio 2 : 1 when stretched by force FA and FB respectively they get equal increase in their lengths. Then the ratio FAFB should be

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Two wires A and B are of same materials. Their lengths are in the ratio 1 : 2 and diameters are in the ratio 2 : 1 when stretched by force $F_{A}$ and $F_{B}$ respectively they get equal increase in their lengths. Then the ratio $\frac{F_{A}}{F_{B}}$ should be

The correct option is D 8 : 1
$Y = \frac{F L}{π r^{2} l}$ $∴ F = Y π r^{2} \frac{l}{L}$
$\frac{F_{A}}{F_{B}} = \frac{Y_{A}}{Y_{B}} {(\frac{r_{A}}{r_{B}})}^{2} (\frac{l_{A}}{l_{B}}) (\frac{L_{B}}{L_{A}}) = 1 \times {(\frac{2}{1})}^{2} \times (1) \times (\frac{2}{1}) = 8$