What will be the change in resistance of a constantan wire when its radius is made half and length reduced to one-fourth of its original length ?

Answer

To solve the problem of finding the change in resistance of a constantan wire when its radius is halved and its length is reduced to one-fourth, we can follow these steps:

### Step 1: Understand the formula for resistance
The resistance \( R \) of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
where:
- \( R \) is the resistance,
- \( \rho \) is the resistivity of the material,
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.

### Step 2: Determine the cross-sectional area
The cross-sectional area \( A \) of a wire with radius \( r \) is given by:
\[ A = \pi r^2 \]

### Step 3: Calculate the new dimensions
Given that the radius is made half, we have:
\[ r' = \frac{r}{2} \]
The new cross-sectional area \( A' \) will be:
\[ A' = \pi (r')^2 = \pi \left(\frac{r}{2}\right)^2 = \pi \frac{r^2}{4} = \frac{A}{4} \]

Also, the length is reduced to one-fourth:
\[ L' = \frac{L}{4} \]

### Step 4: Substitute into the resistance formula
Now we can find the new resistance \( R' \):
\[ R' = \frac{\rho L'}{A'} = \frac{\rho \left(\frac{L}{4}\right)}{\left(\frac{A}{4}\right)} = \frac{\rho L}{4} \cdot \frac{4}{A} = \frac{\rho L}{A} = R \]

### Step 5: Conclusion
Thus, the new resistance \( R' \) is equal to the original resistance \( R \):
\[ R' = R \]
This means there is no change in resistance.

### Final Answer
The change in resistance of the constantan wire is zero; \( R' = R \).

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