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The below diagram shows different copper conductors of diameters d and 2d and lengths, which are multiples of l. The increasing order of their resistances will be:

The below diagram shows different copper conductors of diameters d and 2d and lengths, which are multiples of l. The increasing order of their resistances will be. Physics Competency Focused Practice Questions Class 10 Solutions.
  1. A < B < C < D
  2. A < C < D < B
  3. A < D < C < B
  4. D < A < C < B

Current Electricity

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Answer

A < C < D < B

Reason

R=ρlAR=ρlπr2R=ρ4lπd2\text{R} = \rho \frac{l}{A} \\[1em] \text{R} = \rho\dfrac{l}{\pi r^2} \\[1em] \text{R}=\rho\frac{4l}{\pi d^2} \\[1em]

Here, ρ\rho is specific resistance , l is length of conductor, d is diameter of conductor.

Conductor A:

RA=ρ×4lπ(2d)2RA=ρlπd2RA = \dfrac{\rho\times 4l}{\pi (2d)^2} \\[1em] RA = \dfrac{\rho l}{\pi d^2}

Conductor B:

RB=ρ×4×3lπd2RB=12ρlπd2RB = \dfrac{\rho\times 4\times 3l}{\pi d^2} \\[1em] RB = \dfrac{12\rho l}{\pi d^2}

Conductor C:

RC=ρ×4×2lπ(2d2)RC=2ρlπd2RC = \dfrac{\rho\times 4\times 2l}{\pi (2d^2)} \\[1em] RC = \dfrac{2\rho l}{\pi d^2}

Conductor D:

RD=ρ×4×lπd2RD=4ρlπd2RD = \dfrac{\rho\times 4\times l}{\pi d^2} \\[1em] RD = \dfrac{4\rho l}{\pi d^2}

so, A < C < D < B

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