Q. 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) $\frac{1}{16} \times 10^{3}N$
Explanation
(b) $F = Y \times A \times \frac{l}{L}$
$\Rightarrow F \propto r^{2}\lbrack Y,l$ and $L$ are constant]
If diameter is made four times then force required will be 16 times, $i.e.16 \times 10^{3}N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.