The ratio of diameters of two wires of same materials is n : 1.The length of each wire is 4 m. On applying the same load, the increase in length of thin wire will be $(n > 1)$

n² times

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n times

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2n times

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(2n + 1) times

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Solution

The correct option is A n² times
$Y = \frac{\frac{F}{A}}{\frac{△ l}{l}} = \frac{F l}{A △ l}$
or $Y = \frac{F l X 4}{π D^{2} X △ l} o r △ l \propto \frac{1}{D^{2}} o r \frac{△ L_{2}}{△ L_{1}} = \frac{D_{1}^{2}}{D_{1}^{2}} = \frac{n^{2}}{1}$

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The ratio of diameters of two wires of same materials is n : 1.The length of each wire is 4 m. On applying the same load, the increase in length of thin wire will be (n>1)

The correct option is A n2 timesY=FA△ ll=FlA △ l or Y=FlX 4π D2X△ lor△ l ∝1D2or△ L2△ L1=D21D21=n21

The ratio of diameters of two wires of same materials is n : 1.The length of each wire is 4 m. On applying the same load, the increase in length of thin wire will be $(n > 1)$

The correct option is A n² times
$Y = \frac{\frac{F}{A}}{\frac{△ l}{l}} = \frac{F l}{A △ l}$
or $Y = \frac{F l X 4}{π D^{2} X △ l} o r △ l \propto \frac{1}{D^{2}} o r \frac{△ L_{2}}{△ L_{1}} = \frac{D_{1}^{2}}{D_{1}^{2}} = \frac{n^{2}}{1}$