A metallic wire of uniform area of cross section has a resistance R, resistivity \( \rho \) and power rating P at V volts. The wire is uniformly stretched to reduce the radius to half the original radius. The values of resistance, resistivity and power rating at V volts are now denoted by R', \( \rho' \) and P' respectively. The corresponding values are correctly related as ________________.
Fill in the blank with the correct answer from the options given below.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \rho' = \rho,\ R' = 16R,\ P' = (1/16)\,P \)
Resistivity is a property of the material (at a given temperature), so stretching does not change it: \( \rho' = \rho \).
When the wire is stretched, its volume stays constant. Halving the radius makes the area one quarter, so the length becomes four times.
\( r' = r/2 \;\Rightarrow\; A' = A/4 \)
\( A'L' = AL \;\Rightarrow\; L' = 4L \)
\( \displaystyle R' = \frac{\rho L'}{A'} = \frac{\rho\,(4L)}{A/4} = 16\,\frac{\rho L}{A} = 16R \)
At the same voltage \( V \), the power is \( V^{2}/R \), so it falls in the same ratio as the resistance rises.
\( \displaystyle P' = \frac{V^{2}}{R'} = \frac{V^{2}}{16R} = \frac{1}{16}\,P \)