Electricity: Resistance & Resistivity

Numerical Problems

These numerical problems are based on the chapter Electricity class 10 science of NCERT book and CBSE syllabus.

Example 1: What will be the resistivity of a metal wire of 2 m length and 0.6 mm in diameter, if the resistance of the wire is 50 Ω.

Solution: Given, Resistance ( R ) = 50 Ω, Length ( l ) = 2 m

Diameter = 0.6 mm

Hence, radius `= 0.3 mm = 3 xx 10^-4 m`

Resistivity (ρ) = ?

Now, area of cross section of wire `= π r^2`

Or, `A = 3.14 xx (3 xx 10^-4)^2`

Or, `A = 28.26 xx 10^-8 m^2`

`= 2.826 xx 10^-7 m^2`

We know that

`ρ=(RA)/(l)`

Or, `ρ=(50Ωxx2.826xx10^-7m^2)/(2m)`

Or, `ρ=25xx2.826xx10^-7Ωm`

`=70.65xx10^-7Ωm`

Or, `ρ=7.065xx10^-6Ωm`

Example 2: The resistance of an electric wire of an alloy is 10 Ω. If the thickness of wire is 0.001 meter, and length is 1 m, find its resistivity.

Solution: Given, Resistance ( R ) = 10 Ω, Length ( l ) = 1 m

Diameter = 0.001 m

Therefore, radius = 0.0005 m

Resistivity (ρ) =?

Now, area of cross section of wire `= π r^2`

Or, `A = 3.14 xx (0.005)^2 m^2`

Or, A = 0.00007850 m²

We know that

`ρ=(RA)/(l)`

Or, `ρ=(10Ωxx0.0000785m^2)/(1m)`

`=10Ωxx0.0000785 m`

`=0.000785 Ωm`

`=7.85xx10^-4Ωm`

Example 3: The resistivity of a metal wire is 10 x 10⁻⁸ Ω m at 20°C. Find the resistance of the same wire of 2 meter length and 0.3 mm thickness.

Solution: Given, Resistivity (ρ) = 10 x 10⁻⁸ Ω m, Length ( l ) = 2 m, Diameter = 0.3 mm

Resistance (R) =?

Now, Radius of wire `= text(Diameter)/(2) = (0.3 mm) / (2) = 0.15 mm = 1.5 xx 10^-5 m`

Now, area of cross section of wire `= π r^2`

Or, `A = 3.14 xx (1.5 xx 10^-5)^2`

Or, `A = 70.65 xx 10^-10m^2`

We know that

`R=ρ(l)/(A)`

Or, `R=(10xx10^-8Ω\xx2m)/(7.65xx10^-9m^2)`

`=(10xx2Ω)/(7.65xx10)`

`=(2Ω)/(7.65)=0.26Ω`

Example 4: The area of cross section of wire becomes half when its length is stretched to double. How the resistance of wire is affected in new condition?

Solution: Let the area of cross section of wire = A

Let length of wire before stretching = L

Let Resistance of wire = R

After stretching of wire, let

Area of cross section = A / 2

Length = 2L

Resistance = R1

Thus, ratio of resistance before stretching to resistance after stretching can be given as follows:

Or, `R:R_1=(ρL)/(A): (ρ2L)/(A/2)`

Or, `R:R_1=(ρL)/(A): (4ρ L)/(A)`

Or, `R:R_1=1:4`

This means R = 1 and R1 = 4

Thus, resistance increases four times after stretching of wire.