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$\frac{1}{{16}} \times {10^3}N$
Medium
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B$16 \times {10^3}N$
b (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. $16 \times 10^3 N$
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