Question

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.

• A. 3 times
• B. 3/2 times
• C. 4 times
• D. 2 times

Answer 1

Answer: (D)

2 times

Answer 2

The increase in length $l$ of a wire when subjected to a force $F$ is given by:

$l = \frac{FL}{AY}$

where:
- $L$ is the original length of the wire,
- $A = \pi r^2$ is the cross-sectional area of the wire,
- $Y$ is Young’s modulus of the material of the wire.

If both the force $F$ and the radius $r$ are reduced to half, let’s see how $l$ changes.

Step 1: New Force:
$F' = \frac{F}{2}$

Step 2: New Radius and Area:
Since the radius is reduced to half:
$r' = \frac{r}{2}$
The new cross-sectional area $A'$ is:
$A' = \pi (r')^2 = \pi \left(\frac{r}{2}\right)^2 = \frac{\pi r^2}{4} = \frac{A}{4}$

Step 3: New Increase in Length $l'$:
Substituting the new values of $F'$ and $A'$:
$l' = \frac{F'L}{A'Y} = \frac{\frac{F}{2} \cdot L}{\frac{A}{4} \cdot Y} = \frac{FL}{AY} \cdot 2 = 2l$

Thus, the increase in length will become 2 times the original increase in length.

The Correct Answer is: 2 Times