A cylindrical wire of length L and radius r has resistance R. The resistance of another wire of the same material but of twice its length and one-fourth its radius is:

A cylindrical wire of length L and radius r has resistance R. The resistance of another wire of the same material but of twice its length and one-fourth its radius is: Correct Answer 32R

Given:

The resistance of a wire in Ω

ρ is the resistivity of the material in Ω-m

L is length in meters

A is the cross-sectional area in m²

r is radius of wire in m

Formula:

Resistance = ρL/A, Area = πr², Volume = πr²h

Concept:

If you stretch it to double the length, the volume stays constant.

Cylinder Volume = πr²h

So if h is doubled, for V to stay constant, the area gets cut in half.

From the equation above, doubling the length will multiply the R by 2

Also from that equation, cutting the area in half will multiply the R by 2

So net result is the resistance is 4x the original value.

Calculation:

Resistance of original wire = ρL/A

⇒ ρL / πr²

Resistance of new wire = (ρ2L)/ π(r²/16)

⇒ (ρ2L) × 16/ π(r²)

Now, compare both the resistance

So it becomes 32 times