Question

A wire of length L and radius r fixed at one end and a force F applied to the other end produces and extension $l$ . The extension produced in another wire of the same material of length 2Land radius 2r by a force 2 F is:

• A. $l$
• B. $2l$
• C. $\frac{l}{2}$
• D. $4l$

Answer 1

Answer: (A)

$l$

$Y=\frac{FL}{Al}$
Y - Young's Modulus
F - Applied Force
A - Cross Section Area
l - Elongation
For 1st Wire
$Y=\frac{FL}{\left(\pi \right)\left(r^2\right)l}$ ---- (M)
For 2nd Wire
z - elongation in 2nd wire
$Y=\frac{2F2L}{\left(\pi \right)\left(2^2\right)\left(r^2\right)z}$ ------ (N)
Dividing M by M
$\Rightarrow 1=\frac{z}{l}$
Therefore $z=l$