H1.01.Measurement Tutorial Solutions

1 Measurement Tutorial Solutions Tutorial Questions SIOs / Notes Q1 Ans: A 1 Students need to recall the following base quantities and their units: mass (kg), length (m), time (s), current (A), temperature (K), amount of substance (mol). Q2 Ans: B Since Pressure = Force / area Therefore, [Pressure] = [Force] / [Area] = [mass] [acceleration] / [Area] = (kg) (m s-2) / m2 = kg m-1 s-2 1 Students need to express derived units as products or quotients of the base units. Q3 Ans: B Converting all to the SI units and their scientific notations: Area in m2 A 1 x 10-2 x 1 x 10-1 1 x 10-3 B 1 x 103 x 1 x 10-3 1 C 1 x 106 x 1 x 10-9 1 x 10-3 D 1 x 10-12 x 1 x 10-6 1 x 10-18 1 Students need to use the prefixes and their symbols. Q4 Ans: C Recall the four rules: (i) Only terms with the same units can be added or subtracted. => Q and RS have same units (ii) Units on both sides of an equation must be the same. => P, Q and RS have the same units (iii) The exponent of a term has no units. E.g. In e x, x has no units. => not applicable here (iv) The logarithm of a quantity has no units . => not applicable here Hence, option A is wrong. There is no basis for us to know if option B is correct. While option D is a correct statement, it does not relate to a homogeneous equation which is referred to in the question. Hence the best answer is option C. 1 Students need to use SI base units to check homogeneity of physical equations.

2 Q5 (a) (b) (c) No. of moles of iron atoms, massn= molar mass = -2 -2 50.4 x 10 = 9.0 mol5.6 x 10 The mass of an individual iron atom can be obtained by dividing its molar mass by the Avogadro’s number. Therefore, mass of 1 atom = mass of 1 mole of iron atoms Avogadro's number = 3 23 −56 x 10 6.02 x 10 = 9.30  10-26 kg The number of iron atoms in the cube = n x NA = (9.0) x (6.02x1023) = 5.42 x 1024 Q6 Ans: C [C] = J K-1 [α] = [C]/[T] = J K-1/ K = J K-2 [β] = [C]/[T]3 = J K-1/ K3 = J K-4 1 Students need to express derived units as products or quotients of the base units. Q7 Assume: - Average heart rate of a person = 70 per minute → heart beat = 1.2 beat per sec - Average lifespan of a person = 80 years Estimated number of a human heart beats in a life time: No. of times = 1.2 x (60 x 60 x 24 x365 x 80) ≈ 3 x 109 1 1 Students to note that estimates can be kept as 1 s.f. Q8 a. Joe is 180 cm tall. reasonable b. I rode my bike to campus at a speed of 50 m s-1. Not reasonable (50 m s-1 = 180 km/hr, average bike speed is abt 15.5 km/hr) c. I can throw a ball a distance of 2 km. Not reasonable (if a ball is throw at a speed of 60 km/hr, it can Students need to make reasonable estimates of physical quantities included within the syllabus.

3 travel to around 15 m) d. Joan’s newborn baby has a mass of 33 kg. Not reasonable (a new born is about 3.3 kg) e. I can throw a ball at a speed of 50 km/hr. Reasonable f. The atmospheric pressure is 100 Pa. Not reasonable (atm pressure is 105 Pa) g. The power of a LED light bulb is 200 W. Not reasonable (abt 2W to 5W) Q9 Ans: C Absolute uncertainty = 1% x 3.924 = 0.03924 = 0.04 V (Limit to 1 sf) Voltmeter reading should have the same precision as the absolute uncertainty. In this case, 2 d.p. Voltmeter reading with its uncertainty = (3.92±0.04) V 1 Students need to know how to calculate absolute uncertainty from the percentage uncertainty and express the quantity and its uncertainty correctly. Q10 Ans: C Since P = I2R The percentage uncertainty of P = Δ𝑃 P × 100% = 2Δ𝐼 I × 100% + Δ𝑅 R × 100% = 2 ( 0.05 2.5 ) × 100% + 2% = 6 % 1 Students need to assess the uncertainty in a derived quantity by simple addition of actual, fractional or percentage uncertainties or by simple numerical substitution. Q11 Ans: B Δ𝜌 ρ = 2 Δ𝑑 d + Δ𝑉 V + Δ𝐿 𝐿 + Δ𝐼 I Thus, compare each of the term to find the least uncertainty: For diameter, 2 Δ𝑑 d = 2 ( 0.01 1.20) = 0.017 For current, Δ𝐼 I = 0.05 1.50 = 0.033 For length, Δ𝐿 𝐿 = 1 100 = 0.010 For potential difference, Δ𝑉 V = 0.1 5.0 = 0.020 Hence, measurement in current gives rise to the least uncertainty in the value for the resistivity. 1 Students need to assess the uncertainty in a derived quantity by simple addition of actual, fractional or percentage uncertainties or by simple numerical substitution. Q12 Equation provided: 1 𝑓 = 1 𝑢 + 1 𝑣 Student should know when to use the uncertainty formula and

4 Apply the Extreme Values Method. Plugging in the highest value of u and v into the equation gives the highest value of f: 1 53 + 1 205 = 1 𝑓 and thus 𝑓 = 42.1 Plugging in the lowest value of u and v into the equation gives the lowest value of f: 1 47 + 1 195 = 1 𝑓 and thus 𝑓 = 37.9 So, f ranges from 37.9 to 42.1. The absolute uncertainty is found by: 𝐻𝑖𝑔ℎ𝑒𝑠𝑡 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑓−𝑙𝑜𝑤𝑒𝑠𝑡 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑓 2 = 42.1−37.9 2 = 2.1 Thus the absolute uncertainty of f is ±2 mm (1 s.f.). * Remember to include units! 1 1 1 when to apply the Extreme Value Method. In this case, 1 𝑓 = 𝑢 + 𝑣 𝑢𝑣 𝑓 = 𝑢𝑣 𝑢 + 𝑣 The product/quotient rule cannot be applied when the same variables appear in both numerator and denominator. Use the Extreme Values Method in this case. However, note that when using the Extreme Values method, use the equation in the original form given in the question. Do NOT use the equation in the form 𝑓 = 𝑢𝑣 𝑢 + 𝑣 whereby you will get uncertainty of 5 mm instead of 2 mm. This is because it doesn’t make sense to substitute different values of u in the numerator and denominator since at any instant the two values of u should be the same. Likewise for values of v. E.g. to get maximum f, you cannot substitute maximum u in the numerator and minimum u in the denominator. There is an alternative method using the addition/subtraction rule and product/quotient

5 rule. But this method proves more tedious: Method 2 -- Find the absolute uncertainties in 1/u and 1/v separately first. ∆(1 𝑢) (1 𝑢) = ∆(1) 1 + ∆𝑢 𝑢 ∆ (1 𝑢) = [0 + ∆𝑢 𝑢 ] . (1 𝑢) = ∆𝑢 𝑢2 = 1.2𝑥10−3𝑚𝑚−1 Similarly, ∆(1 𝑣) = ∆𝑣 𝑣2 = 1.25𝑥10−4𝑚𝑚−1 ∆ (1 𝑓) = ∆ (1 𝑢) + ∆ (1 𝑣) = 1.325𝑥10−3𝑚𝑚−1 ∆ (1 𝑓) = ∆𝑓 𝑓2 ∆𝑓 = ∆ (1 𝑓) . 𝑓2 = 2 𝑚𝑚 (1 𝑠. 𝑓. ) Absolute uncertainty must be rounded off to 1 s.f. Q13 (a)(i) Using the average values of h & t. 2 22 2 sm72.9 740.0 66.22 t h2g gt2 1h −=== = 1 1 It suffices to use the average / mean values of h & t. There is no need to compute maximum and minimum values of g and use the values to calculate the average value of g. Students should recall the meaning of number of significant values and quote answers accordingly.

6 (a)(ii) 1. %38.0%100266 1%100 == h h (2 s.f) Feedback for Assignment: • Several students wrote ∆h or ∆h/h on the left hand side of the equation, which resulted in a mismatch between the left hand side and right hand side. E.g. %38.0%100266 1 ==h (wrong!!) 2 Students should recall and apply the correct equation for % uncertainty. Strictly speaking, final answer should be to 1 s.f. since ∆h is in 1 s.f. However, this particular question stipulates expressing the answer in 2 s.f. (a)(ii)2. %68.0%100740.0 005.0%100 == t t (2 s.f.) 2 b) 2 2 t 2hg gt2 1h = = ( ) ( ) ( ) 2 2 sm0.29.7Δgg (1s.f.)sm0.20.1699.72100 0.6820.38Δg t Δt2h Δh g Δg − − = ==+= += Feedback for Assignment: • An important step is to make the unknown (g) the subject in the equation before applying the product/quotient rule. • Many students did not include