Question

A metallic wire of uniform area of cross-section has a resistance \(R,\) resistivity \(ρ,\) 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 \(R'\), \(ρ'\), and \(P' \)respectively. The corresponding values are correctly related as __________.
Fill in the blank with the correct answer from the options given below

• A. \(ρ' = 2ρ, R' = 2R, P' = 2P\)
• B. \(ρ' = \frac{1}{2} ρ, R' = \frac{1}{2} R, P' = \frac{1}{2} P\)
• C. \(ρ' = ρ, R' = 16R, P' = \frac{1}{16} P\)
• D. \(ρ' = ρ, R' = \frac{1}{16} R, P' = 16P\)

Answer 1

The Correct Option is C

The resistivity \(ρ\) of a material is an intrinsic property and does not change when the wire is deformed. Therefore, \(ρ' = ρ\).

The resistance \(R\) of a wire is given by \(R = \frac{ρL}{A}\), where \(L\) is the length and \(A\) is the cross-sectional area. When the radius of the wire is halved, the new area \(A'\) becomes \(\frac{\pi(r/2)^2}{πr^2} = \frac{1}{4}\). Since the volume remains constant \(AL = A'L'\), and using \(A' = \frac{1}{4}A\), we find \(L' = 4L\).

The new resistance \(R'\) is given by: \(R' = \frac{ρL'}{A'} = \frac{ρ(4L)}{\frac{1}{4}A} = 16R\).

Hence, \(R' = 16R\).

The power \(P\) is given by \(P = \frac{V^2}{R}\). After stretching, \(P'\) is given by \(P' = \frac{V^2}{R'} = \frac{V^2}{16R} = \frac{1}{16}P\). Thus, \(P' = \frac{1}{16}P\).

Value	Original	Stretched
Resistivity ρ	ρ	ρ
Resistance R	R	16R
Power P	P	\(\frac{1}{16}P\)

Hence, the correct relationship is: \(ρ' = ρ, R' = 16R, P' = \frac{1}{16} P\).