A wire of length $$L$$ and radius $$r$$ is clamped at one end. If its other end is pulled by a force $$F$$, its length increases by $$l$$. If the radius of the wire and the applied force both are reduced to half of their original values keeping original length constant, the increase in length will become:

We need to find how the elongation changes when both the force and radius are halved.

From Young's modulus we have $$Y = \frac{F/A}{\Delta l/L} = \frac{FL}{\pi r^2 \Delta l}$$ which gives $$\Delta l = \frac{FL}{\pi r^2 Y}.$$

When the force is halved ($$F' = F/2$$) and the radius is halved ($$r' = r/2$$), while $$L$$ and $$Y$$ remain the same, the new elongation becomes $$\Delta l' = \frac{F'L}{\pi (r')^2 Y} = \frac{(F/2)L}{\pi (r/2)^2 Y} = \frac{(F/2)L}{\pi r^2 Y / 4} = \frac{4FL}{2\pi r^2 Y} = \frac{2FL}{\pi r^2 Y} = 2\Delta l.$$

The new elongation is twice the original; halving the radius (which quarters the cross-sectional area) has a larger effect than halving the force.

The correct answer is Option (4): 2 times.