Answer :
Answer:
Explanation:
Given
Length of cylinder [tex]L=6 cm[/tex]
diameter of cylinder [tex]d=1 cm[/tex]
Area of cross-section [tex]A=\frac{\pi\times d^2}{4}[/tex]
[tex]A=\frac{\pi \times 10^{-4}}{4}=7.855\times 10^{-5} m^2[/tex]
[tex]Emf =2.5 V[/tex]
current [tex]I=5 mA[/tex]
[tex]Resistance=\frac{V}{I}=\frac{2.5}{5\times 10^{-3}}[/tex]
[tex]R=500 \Omega [/tex]
R is also given by
[tex]R=\rho \frac{L}{A}[/tex]
where [tex]\rho =resistivity [/tex]
[tex]\rho =\frac{RA}{L}=\frac{500\times 7.85\times 10^{-3}}{6}[/tex]
[tex]\rho =0.654 \Omega -m[/tex]
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
[tex]\rho =\frac{R\times A}{l}[/tex]
[tex]\rho =\frac{500\times 7.85\times 10^{-5}}{0.06}[/tex]
ρ = 0.654 ohm - metre