A conductor of radius r, length l and resistivity p has resistance R. It is melted down and formed into a new conductor, also cylindrical,

with one half the length of the original conductor. The resistance of the new conductor is

(A)16/R

(B)4/R

(C) 4R

(D) 16R

The option is not there.

Explanation:

Given:

Before melting:

Radius, r

Length, l

Resistance, R

Resistivity, p

After melting:

Length, l2 = 1/2 × l

Using the resistivity relation,

p × l = R × A

Where A = area

Since only the length of the conductor changed, volume also will be affected.

Volume, V = area × l

Area, A2 = V/(1/2l)

= 2 × V/l

(R × A)/l = (R2 × A2)/(l/2)

(R × V/l)/l = (R2 × 2 × V/l)/(l/2)

(R × V)/l^2 = 4 × (R2 × V)/l^2

R = 4 × R2

R2 = R/4

Where

R2 = resistance of the new conductor

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