Respuesta :
Answer:
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.)
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
Answer:
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
