The radius of a wire has been reduced to one-third. If its volume remains the same, then the increase in its length is :

Question

Accepted Answer

To determine why Option 4 is the correct answer, we need to analyze the relationship between the radius, length, and volume of the wire.

1. **Understanding the Problem:**
   - The volume of a cylindrical wire is given by the formula $ V = \pi r^2 L $, where $ r $ is the radius and $ L $ is the length.
   - We are told that the radius of the wire is reduced to one-third of its original size, i.e., $ r' = \frac{r}{3} $.
   - The volume remains constant, so the new volume $ V' = \pi (r')^2 L' = \pi \left(\frac{r}{3}\right)^2 L' $.

2. **Equating the Volumes:**
   - Since the volume remains the same, we equate the original and new volumes:
     $$
     \pi r^2 L = \pi \left(\frac{r}{3}\right)^2 L'
     $$
   - Simplifying the equation:
     $$
     r^2 L = \left(\frac{r^2}{9}\right) L'
     $$
   - Solving for $ L' $:
     $$
     L' = 9L
     $$

3. **Calculating the Increase in Length:**
   - The new length $ L' $ is 9 times the original length $ L $.
   - Therefore, the increase in length is $ L' - L = 9L - L = 8L $.
   - This means the length increases by a factor of 9, which corresponds to Option 4.

4. **Explanation of Incorrect Options:**
   - **Option 1 (2 times):** This option suggests the length doubles, which would imply a volume change if the radius is reduced to one-third. This does not satisfy the condition of constant volume.
   - **Option 2 (3 times):** A threefold increase in length would not compensate for the reduction in radius to maintain the same volume.
   - **Option 3 (6 times):** A sixfold increase is insufficient to maintain the same volume when the radius is reduced to one-third.

In conclusion, the correct answer is Option 4, as the length must increase by a factor of 9 to maintain the same volume when the radius is reduced to one-third. This ensures that the volume remains constant, satisfying the conditions of the problem.