A wire of length l is drawn such that its diameter is reduced to half of its original diameler. If the initial resistance of the wire were `10Omega`. its resistance would become

Options

A. `40 Omega`
B. `60Omega`
C. `120Omega`
D. `160Omega`

Answer

To solve the problem step by step, let's analyze the situation given in the question:

### Step 1: Understand the relationship between volume, length, and diameter
When a wire is drawn (stretched), its volume remains constant. The volume \( V \) of a cylindrical wire is given by the formula:
\[ V = A \cdot l \]
where \( A \) is the cross-sectional area and \( l \) is the length of the wire.

### Step 2: Calculate the initial volume of the wire
The initial diameter of the wire is \( D \). The cross-sectional area \( A \) of the wire can be calculated as:
\[ A = \frac{\pi D^2}{4} \]
Thus, the initial volume \( V \) can be expressed as:
\[ V = A \cdot l = \frac{\pi D^2}{4} \cdot l \]

### Step 3: Determine the new dimensions after stretching
After the wire is drawn, its diameter is reduced to half, so the new diameter \( D' \) is:
\[ D' = \frac{D}{2} \]
The new cross-sectional area \( A' \) becomes:
\[ A' = \frac{\pi (D')^2}{4} = \frac{\pi \left(\frac{D}{2}\right)^2}{4} = \frac{\pi D^2}{16} \]

### Step 4: Set up the equation for constant volume
Since the volume remains constant, we can equate the initial and final volumes:
\[ \frac{\pi D^2}{4} \cdot l = A' \cdot l' \]
Substituting for \( A' \):
\[ \frac{\pi D^2}{4} \cdot l = \frac{\pi D^2}{16} \cdot l' \]
Cancelling \( \pi D^2 \) from both sides:
\[ \frac{l}{4} = \frac{l'}{16} \]
Cross-multiplying gives:
\[ 16l = 4l' \implies l' = 4l \]
This means the new length of the wire is \( 4l \).

### Step 5: Calculate the new resistance
The resistance \( R \) of a wire is given by:
\[ R = \frac{\rho l}{A} \]
where \( \rho \) is the resistivity of the material.

For the initial resistance \( R \):
\[ R = \frac{\rho l}{\frac{\pi D^2}{4}} = \frac{4\rho l}{\pi D^2} \]
Given that \( R = 10 \, \Omega \), we can write:
\[ 10 = \frac{4\rho l}{\pi D^2} \]

For the new resistance \( R' \):
\[ R' = \frac{\rho l'}{A'} = \frac{\rho (4l)}{\frac{\pi D^2}{16}} = \frac{64\rho l}{\pi D^2} \]

### Step 6: Relate the new resistance to the initial resistance
Now substituting for \( R \):
\[ R' = 64 \cdot \frac{4\rho l}{\pi D^2} = 64 \cdot \left(\frac{\pi D^2}{4}\right) = 64 \cdot 10 \]
Thus:
\[ R' = 160 \, \Omega \]

### Final Answer
The new resistance of the wire after it has been drawn is:
\[ \boxed{160 \, \Omega} \]