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

Let's analyze the changes in resistance, resistivity, and power rating when the radius of the wire is halved.

1. Resistivity (ρ):

Resistivity is an intrinsic property of the material and depends only on the material and temperature. Stretching the wire does not change the material or temperature. Therefore:

ρ' = ρ

2. Resistance (R):

Resistance is given by:

R = ρL/A

Where L is the length and A is the cross-sectional area (A = πr²).

When the radius is halved (r' = r/2), the new area (A') is:

A' = π(r/2)² = πr²/4 = A/4

Since the volume of the wire remains constant during stretching, we have:

V = AL = A'L'

AL = (A/4)L'

L' = 4L

The new resistance (R') is:

R' = ρL'/A'

R' = ρ(4L)/(A/4)

R' = 16(ρL/A)

R' = 16R

3. Power Rating (P):

Power is given by:

P = V²/R

Since the voltage (V) is constant:

P' = V²/R'

P' = V²/(16R)

P' = P/16

Therefore:

ρ' = ρ

R' = 16R

P' = P/16

The correct answer is:

Option 3: ρ' = ρ, R' = 16R, P' = P/16