Q#27 (Past Exam Paper – June 2017 Paper 22 Q1)
(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]
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.
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