A force of {10^3} newton stretches length of a hanging wire by 1 millimetre. force required to stret

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8.Mechanical Properties of Solids

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A force of ${10^3}$ newton stretches the length of a hanging wire by $1$ millimetre. The force required to stretch a wire of same material and length but having four times the diameter by $1$ millimetre is

A

$4 \times {10^3}N$

B

$16 \times {10^3}N$

C

$\frac{1}{4} \times {10^3}N$

D

$\frac{1}{{16}} \times {10^3}N$

Solution

(b) $F = Y \times A \times \frac{l}{L}$ $⇒$ $F \propto {r^2}$ $(Y,l$ and $L$ are constant$)$

If diameter is made four times then force required will be $16$ times. i.e. $16 \times 10^3 N$

Standard 11

Physics

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