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

The Correct Option is D

To solve this problem, we need to use the concept of Young's Modulus, which gives a relationship between the stress and strain in a material. The formula for elongation of a wire when a force is applied is given by: 

\(l = \frac{FL}{A \cdot Y}\)

where:

• \(F\) is the applied force.
• \(L\) is the original length of the wire.
• \(A\) is the cross-sectional area of the wire.
• \(Y\) is the Young's Modulus of the material.

The cross-sectional area \(A\) for a wire with radius \(r\) is given by:

\(A = \pi r^2\)

Initially, the elongation of the wire is:

\(l = \frac{FL}{\pi r^2 Y}\)

Now, if both the radius and the force are reduced to half of their original values, the new radius becomes \(\frac{r}{2}\) and the new force becomes \(\frac{F}{2}\).

The new cross-sectional area \(A'\) is:

\(A' = \pi \left(\frac{r}{2}\right)^2 = \frac{\pi r^2}{4}\)

The new elongation \(l'\) is given by:

\(l' = \frac{\frac{F}{2} \cdot L}{\frac{\pi r^2}{4} \cdot Y} = \frac{FL \cdot 4}{2 \cdot \pi r^2 \cdot Y} = 2 \cdot \frac{FL}{\pi r^2 Y}\)

This shows that the new elongation is twice the original elongation. Therefore, the increase in length will become:

The correct answer is: 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