Question

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

Answer 1

The option is not there.

Step-by-step 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

Answer 2

b) R/4 (There seems to an error in mentioning the multiple choices of this question, please see below explanation of correct calculations for this question.)

Step-by-step explanation:

dimension of the conductor before melting is l, r

reistivity is p

R=(p*l)/(pie*r2)

after reforming length is reduced to L=l/4

volume in both the cases will be same

i.e. pie * r^2 * l =pie * R^2 * L

r^2 * l = R^2 * (1/2)l

due to this radius will become R=sqrt(2) * r

now new reistance is given by Rx=(p * L)/(pie * R^2)

i.e. Rx=(p * l/2)/(pie * r^2 * 2)

after simplification RX=((p * l)/(pie * r^2))/4

i.e. Rx=R/4