CUET · MATHS · PYQ PAPER 2024
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 \(R ^{\prime}, \square\) ' and \(P ^{\prime}\) respectively. The corresponding values are correctly related as ___________ . Fill in the blank with the correct answer from the options given below
- A \(\rho^{\prime}=2 \rho, R^{\prime}=2 R, P^{\prime}=2 P\)
- B \(\rho^{\prime}=1 / 2 \rho, R^{\prime}=1 / 2 R, P^{\prime}=1 / 2 P\)
- C \(\rho^{\prime}=\rho, R^{\prime}=16 R, P^{\prime}=1 / 16 P\)
- D \(\rho^{\prime}=\rho, R^{\prime}=1 / 16 R, P^{\prime}=16 P\)
Answer & Solution
Correct Answer
(C) \(\rho^{\prime}=\rho, R^{\prime}=16 R, P^{\prime}=1 / 16 P\)
Step-by-step Solution
Detailed explanation
\(\rho' = \rho\) \(\pi r^2 L = \pi (r/2)^2 L' \Rightarrow L' = 4L\) \(R' = \rho \frac{L'}{\pi (r/2)^2} = \rho \frac{4L}{\pi r^2/4} = 16 \left(\rho \frac{L}{\pi r^2}\right) = 16R\) \(P' = \frac{V^2}{R'} = \frac{V^2}{16R} = \frac{1}{16} P\)…
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