The radius of an iron rod decreased to one-fourth. If its volume remains constant , the length will becomes :

Options

A. 2 times
B. 8 times
C. 4 times
D. 16 times

Answer

To solve the problem step by step, we will use the formula for the volume of a cylinder and the relationship between the dimensions of the cylinder before and after the change in radius.

### Step-by-Step Solution:

1. **Understand the Volume of a Cylinder**:
The volume \( V \) of a cylinder is given by the formula:
\[ V = \pi r^2 h \]
where \( r \) is the radius and \( h \) is the height (or length) of the cylinder.

2. **Initial Dimensions**:
Let the initial radius of the iron rod be \( r \) and the initial height (length) be \( h_1 \). Thus, the initial volume \( V_1 \) is:
\[ V_1 = \pi r^2 h_1 \]

3. **New Dimensions**:
The radius of the iron rod decreases to one-fourth of the original radius. Therefore, the new radius \( r_2 \) is:
\[ r_2 = \frac{r}{4} \]
Let the new height (length) be \( h_2 \). The new volume \( V_2 \) is:
\[ V_2 = \pi r_2^2 h_2 = \pi \left(\frac{r}{4}\right)^2 h_2 = \pi \frac{r^2}{16} h_2 \]

4. **Setting Volumes Equal**:
Since the volume remains constant, we can set the initial volume equal to the new volume:
\[ V_1 = V_2 \]
Substituting the expressions for \( V_1 \) and \( V_2 \):
\[ \pi r^2 h_1 = \pi \frac{r^2}{16} h_2 \]

5. **Canceling Common Terms**:
We can cancel \( \pi \) and \( r^2 \) from both sides (assuming \( r \neq 0 \)):
\[ h_1 = \frac{1}{16} h_2 \]

6. **Solving for \( h_2 \)**:
Rearranging the equation gives us:
\[ h_2 = 16 h_1 \]
This means the new height (length) of the iron rod will be 16 times the original height.

### Final Answer:
The length of the iron rod will become \( 16 \) times the original length.

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