The breaking force for a wire of diameter D of a material is F. The breaking force for a wire of the same material of radius D is?

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2 F

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\frac{F}{4}

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4 F

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Solution

The correct option is D $4 F$
Breaking stress $= \frac{B r e a k i n g f o r c e}{A r e a} =$ constant
$\frac{F}{(\frac{π D^{2}}{4})} = \frac{F^{'}}{π D^{2}}$
$F^{'} = 4 F$ .

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