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 with