find the ratio of resistances of copper rods X and Y of length 30 cm and 10cm respectively and having radii 2cm and 1 centimetre respectively
Answers
Answer:
Radius of first wire = 2 cm
So,
Area of cross section of first wire =
\pi( {r}^{2} ) = \pi( {2}^{2} ) = 4\pi \: cm ^{2} π(r
2
)=π(2
2
)=4πcm
2
Radius of second wire = 1 cm
So,
Area of cross section of second wire=
\pi( {r}^{2} ) = \pi( {1}^{2} ) = \pi \: cm ^{2} π(r
2
)=π(1
2
)=πcm
2
General formula of resistance R =
\rho \: \frac{length}{area} ρ
area
length
Resistivity of both wires is same because both wires are made of same copper material.
So, Resistance of first wire =
r _1 = \rho \: (\frac{30}{4\pi} ) \: ohmr
1
=ρ(
4π
30
)ohm
and Resistance of second wire =
r _2 = \rho \: (\frac{10}{\pi} ) \: ohmr
2
=ρ(
π
10
)ohm
So, the ratio of both resistance =
\frac{r_1 }{r _2} = \frac{\rho \: (\frac{30}{4\pi} ) }{ (\rho \frac{10}{\pi}) } = \frac{30}{40} = \frac{3}{4}
r
2
r
1
=
(ρ
π
10
)
ρ(
4π
30
)
=
40
30
=
4
3