A wire of length L is drawn such that its diameter is reduced to half of its original diameter. If the initial resistance of the wire were 10Ω, its new resistance would bea)40Ωb)80Ωc)120Ωd)160ΩCorrect answer is option 'D'. Can you explain this answer?

Question

A wire of length L is drawn such that its diameter is reduced to half of its original diameter. If the initial resistance of the wire were 10&Omega;, its new resistance would bea)40&Omega;b)80&Omega;c)120&Omega;d)160&Omega;Correct answer is option 'D'. Can you explain this answer?

Answer

We have kept the volume of the wire constant i.e.

Answer

Answer

The resistance of a wire is given by the formula:

R = (ρ * L) / A

Where R is the resistance, ρ is the resistivity of the material, L is the length of the wire, and A is the cross-sectional area of the wire.

Let's assume the initial diameter of the wire is D. Therefore, the initial radius is D/2.

The initial cross-sectional area of the wire is given by:

A_initial = π * (D/2)^2 = π * (D^2) / 4

The final diameter of the wire is reduced to half of its original diameter, so the final diameter is D/2.

The final cross-sectional area of the wire is given by:

A_final = π * (D/4)^2 = π * (D^2) / 16

The initial resistance of the wire is given as 10, so we can rearrange the formula to solve for ρ:

ρ = (R * A) / L

Substituting the values into the formula, we have:

ρ_initial = (10 * (π * (D^2) / 4)) / L

ρ_final = (10 * (π * (D^2) / 16)) / L

Since the resistivity ρ of the material remains constant, we can equate the two expressions:

(10 * (π * (D^2) / 4)) / L = (10 * (π * (D^2) / 16)) / L

Simplifying the equation, we have:

(π * (D^2) / 4) = (π * (D^2) / 16)

Cancelling out common terms, we have:

4 = 16

This equation is not valid, which means there is a contradiction in our assumption. Therefore, the resistance of the wire cannot remain the same if its diameter is reduced to half of its original diameter.

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