Question

A particular conductor has 3.0 × 10^27 mobile electrons per cubic meter. The material is in the shape of a cylinder of length 6.0 cm and diameter 1.0 cm. When the bar is connected to a battery with an emf of 2.5 V, a current of 5.0 mA begins to flow.
What is the resistivity of the bar?

Answer 1

Answer:

Explanation:

Given

Length of cylinder $L=6 cm$

diameter of cylinder $d=1 cm$

Area of cross-section $A=\frac{\pi\times d^2}{4}$

$A=\frac{\pi \times 10^{-4}}{4}=7.855\times 10^{-5} m^2$

$Emf =2.5 V$

current $I=5 mA$

$Resistance=\frac{V}{I}=\frac{2.5}{5\times 10^{-3}}$

$R=500 \Omega$

R is also given by

$R=\rho \frac{L}{A}$

where $\rho =resistivity$

$\rho =\frac{RA}{L}=\frac{500\times 7.85\times 10^{-3}}{6}$

$\rho =0.654 \Omega -m$        

Answer 2

Answer:

0.654 ohm - metre

Explanation:

length, l = 6 cm

diameter, d = 1 cm

radius, r = 0.5 cm

Voltage, V = 2.5 V

current, i = 5 mA

According to Ohm's law

V = i x R

2.5 = 0.005 x R

R = 500 ohm

Let ρ be the resistivity of material of conductor.

Let A be the area of crossection of the conductor.

A = πr²

A = 3.14 x 0.005 x 0.005 = 7.85 x 10^-5 m^2

$\rho =\frac{R\times A}{l}$

$\rho =\frac{500\times 7.85\times 10^{-5}}{0.06}$

ρ = 0.654 ohm - metre