Question

A potential difference 'V' is applied across a conductor of length 'L' and diameter 'd'. How is
the resistance and resistivity of the conductor affected, when (i) 'V' is halved (ii) d' is doubled ?
Justify your answer in both the cases​

Answer 1

Answer:

Explanation:

R = p L / A, and R = V / I

A= Area of Circular cross-section = (\pi) r^2 = (\pi) (d/2)^2 = (\pi) d^2 / 4

R = 4 p L / (\pi) d^2

where R = Resistance, p (rho) = resistivity of material of conductor, L = length of conductor, A = area of cross-section of conductor, d=diameter of cross-section of conductor, V = potential difference and I = current.

(i) 'V' is halved.

When V is halved, Resistance is also halved if current I remains constant. By using Ohm's law, R = V / I, and R is proportional to V.

Now since resistivity, p = R A / L,

As R is halved, so resistivity p will also be halved as p is proportional to R.

(ii) 'd' is doubled.

R = 4 p L / (\pi) d^2

Since, R is proportional to 1 / d^2, so when diameter 'd' is doubled then R will become 1/4 R.

Since, R is proportional to resistivity p, so p will also become 1/4 p.