Two identical wires are suspended from the same rigid support but one is of copper and the other is of iron. Young’s modulus of iron is thrice that of copper. The weights to be added on copper and iron wires so that the ends are on the same level must be in the ratio of

1 : 3

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2 : 1

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3 : 1

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4 : 1

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Solution

The correct option is A 1 : 3
$Y \propto F$
Therefore $\frac{F_{c u}}{F_{F A}} = \frac{Y_{c u}}{Y_{F A}} = \frac{1}{3}$

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Two identical wires are suspended from the same rigid support but one is of copper and the other is of iron. Young’s modulus of iron is thrice that of copper. The weights to be added on copper and iron wires so that the ends are on the same level must be in the ratio of

The correct option is A 1 : 3Y ∝ F Therefore FcuFFA=YcuYFA=13

The correct option is A 1 : 3
$Y \propto F$
Therefore $\frac{F_{c u}}{F_{F A}} = \frac{Y_{c u}}{Y_{F A}} = \frac{1}{3}$