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.

A thermometer can be read to an accuracy of ± 0.5 °C. This thermometer is used to measure a temperature rise from 40 °C to 100 °C.

Question 26

A thermometer can be read to an accuracy of ± 0.5 °C. This thermometer is used to measure a temperature rise from 40 °C to 100 °C.

What is the percentage uncertainty in the measurement of the temperature rise?

A 0.5 %                       B 0.8 %                       C 1.3 %                       D 1.7 %

Reference: Past Exam Paper – November 2003 Paper 1 Q4 & June 2014 Paper 12 Q5

Solution:

Answer: D.

Initial temperature: T₁ = 40 °C

Final temperature: T₂ = 100 °C

Temperature rise: T = T₂ – T₁ = 100 – 40 = 60 °C

The measurement of both the initial and final temperatures have an uncertainty of ± 0.5°C.

ΔT = ΔT₂ + ΔT₁

ΔT = 0.5 + 0.5 = 1.0 °C

Percentage uncertainty in the measurement of the temperature rise:

%uncertainty in T = [ΔT / T] × 100%

%uncertainty in T = (1/60) × 100% = 1.7%

A radioactive substance contains a number of identical nuclei that emit β-particles. Which property of these nuclei remains unaltered by the emission?

Question 21

A radioactive substance contains a number of identical nuclei that emit β-particles.

Which property of these nuclei remains unaltered by the emission?

A charge

B neutron number

C nucleon number

D proton number

Reference: Past Exam Paper – June 2015 Paper 12 Q39

Solution:

Answer: C.

Beta particle is ⁰_-1β.

The nucleon number of β-particles is zero. So, the nucleon number of the nucleus formed is the same as the parent nucleus.

Since the β-particle has a negative charge, the daughter nucleus will need to have a charge greater than the parent daughter so that the charge is conserved. Similarly, it will have one more proton than the parent nucleus.

The neutron number will also change as the proton number has changed.

Nucleon number = Neutron number + Proton number