A conductor of resistance R and conductance G is stretched to twice its original length keeping its volume constant. Calculate the new resistance and the conductance of the conductor.

Answer

To solve the problem, we need to find the new resistance and conductance of a conductor that has been stretched to twice its original length while keeping its volume constant.

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- Let the original length of the conductor be \( L \).
- Let the original cross-sectional area be \( A \).
- The original resistance is \( R \) and the conductance is \( G \).
- We know that \( G = \frac{1}{R} \).

2. **Volume Conservation**:
- The volume of the conductor is given by \( V = A \times L \).
- After stretching, the new length becomes \( 2L \).
- Let the new cross-sectional area be \( A' \).
- The volume after stretching is \( V' = A' \times 2L \).
- Since the volume is constant, we have:
\[ A \times L = A' \times 2L \]
- Canceling \( L \) from both sides (assuming \( L \neq 0 \)):
\[ A = 2A' \]
- Rearranging gives:
\[ A' = \frac{A}{2} \]

3. **Calculate the New Resistance**:
- The resistance \( R \) of a conductor is given by:
\[ R = \rho \frac{L}{A} \]
- For the new resistance \( R' \):
\[ R' = \rho \frac{2L}{A'} \]
- Substituting \( A' = \frac{A}{2} \):
\[ R' = \rho \frac{2L}{\frac{A}{2}} = \rho \frac{2L \times 2}{A} = \rho \frac{4L}{A} \]
- Recognizing that \( \rho \frac{L}{A} = R \):
\[ R' = 4R \]

4. **Calculate the New Conductance**:
- The new conductance \( G' \) is given by:
\[ G' = \frac{1}{R'} \]
- Substituting \( R' = 4R \):
\[ G' = \frac{1}{4R} \]
- Since \( G = \frac{1}{R} \), we can express \( G' \) in terms of \( G \):
\[ G' = \frac{1}{4} G \]

### Final Results:
- The new resistance \( R' = 4R \).
- The new conductance \( G' = \frac{1}{4} G \).