An aluminium wire and steel wire of the same length and cross section are joined end to end.The composite wire is hung from a rigid support and a load is suspended from the free end. The young's modulus of steel is $20 / 7$ times the aluminium. The ratio of increase of length of steel and aluminium is

20 / 7

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400 / 49

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7 / 20

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49 / 400

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Solution

The correct option is C $7 / 20$
$Y = \frac{F l}{A △ l}$

$Y α \frac{1}{△ l}$
Now take the ratio,
$\frac{Y_{1}}{Y_{2}} = \frac{△ l_{2}}{△ l_{1}}$

$∴ \frac{△ l_{2}}{△ l_{1}} = \frac{Y_{1}}{Y_{2}} = \frac{20}{7}$

$o r \frac{△ l_{1}}{△ l_{2}} = \frac{7}{20}$

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An aluminium wire and steel wire of the same length and cross section are joined end to end.The composite wire is hung from a rigid support and a load is suspended from the free end. The young's modulus of steel is 20/7 times the aluminium. The ratio of increase of length of steel and aluminium is

The correct option is C 7/20Y=FlA△lYα1△l Now take the ratio,Y1Y2=△l2△l1∴△l2△l1=Y1Y2=207or△l1△l2=720

The correct option is C $7 / 20$
$Y = \frac{F l}{A △ l}$

$Y α \frac{1}{△ l}$
Now take the ratio,
$\frac{Y_{1}}{Y_{2}} = \frac{△ l_{2}}{△ l_{1}}$

$∴ \frac{△ l_{2}}{△ l_{1}} = \frac{Y_{1}}{Y_{2}} = \frac{20}{7}$

$o r \frac{△ l_{1}}{△ l_{2}} = \frac{7}{20}$