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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 

a
half
b
double
c
four times
d
one-fourth

detailed solution

Correct option is C

We have,  S=wπr2=w′π(2r)2∴ Load, w′=4w

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