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A cable is replaced by another one of the same length and material but of twice the diameter. The maximum load that the new wire can support without exceeding the elastic limit, as compared to the load that the original wire could support, is

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Half

Double

Four times

One-fourth

answer is C.

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Detailed Solution

$F \propto A (∵ Y = \frac{F L}{A e})$

$\Rightarrow$ $\frac{F_{1}}{F_{2}} = {(\frac{D_{1}^{}}{D_{2}^{}})}^{2} = {(\frac{D_{1}^{}}{2 D_{1}^{}})}^{2} = \frac{1}{4}$

$\Rightarrow$ $F_{2} = 4 F_{1}$

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