A wire of cross-sectional area 1.5 mm2 and length 2.5 m has a resistance of 0.030 Ω. Calculate the resistivity of the material of the wire in nΩ m.

Question 27

(a) State two SI base units other than kilogram, metre and second. [1]

(b) Determine the SI base units of resistivity. [3]

(c) (i) A wire of cross-sectional area 1.5 mm² and length 2.5 m has a resistance of 0.030 Ω.

Calculate the resistivity of the material of the wire in nΩ m. [3]

(ii)

1. State what is meant by precision.

2. Explain why the precision in the value of the resistivity is improved by using a

micrometer screw gauge rather than a metre rule to measure the diameter of the

wire.

 [2]

[Total: 9]

Reference: Past Exam Paper – June 2017 Paper 22 Q1

Solution:

(a) kelvin, mole, ampere, candela

any two

(b)

{Resistance of wire: R = ρL / A

Resistivity ρ = RA / L

Now, to determine the unit of resistance R.

Power: P = I²R

R = P / I²

Power = energy / time = force × distance / time = mas / t

Units of power: kg ms^-2 m s^-1 = kg m² s^-3 

Unit of current = A

Units of R: kg m² s^-3 A^-2

Resistivity ρ = RA / L

Unit of resistivity: kg m² s^-3 A^-2 m² m^-1 = kg m³ s^-3 A^-2

}

(c) (i)

ρ = (RA / L)                                        

{The area should be in m². To convert from mm² to m², multiply by 10^-6 (that is, divide by 1 000 000).}

ρ = (0.03 × 1.5×10^-6) / 2.5 (= 1.8 × 10^-8)

ρ = 18 nΩ m                                       

(ii)

1. Precision is determined by the range in the measurements

2. A metre rule measures to ± 1 mm while a micrometer measures up to ± 0.01 mm. So, there is less percentage uncertainty.