Units and Measurement Questions and Answers

Answer: B

The pair is \(\text{kg m}^{2}\text{/s}^{2}\) and Ws; both measure energy (work).

Work = force × distance, so its unit is N m = (kg m/s²) × m = kg m²/s², which is the joule. Energy is also power × time, so a watt-second is a joule too.

The other pairs mix quantities: kg m/s² is the newton, a unit of force, while N m and W s are units of energy; and J/s is the watt, a unit of power, not energy.

Answer: A

The scalar quantity is mass.

A scalar has magnitude only, while a vector has both magnitude and direction. Mass, the amount of matter in a body, is a scalar measured in kilograms.

Weight is the force of gravity on a body and acts towards the centre of the earth, so it is a vector. Momentum, mass times velocity, has the direction of the velocity. Gravity, as a force or as the acceleration due to gravity, also has direction.

Answer: C

The set with only units of distance is cubit, metre, light year.

The metre is the SI unit of length. The cubit is an old unit of length based on the forearm, from the elbow to the tip of the middle finger. The light year is the distance light travels in one year, about \(9.46 \times 10^{15}\) m, used for astronomical distances despite the word 'year' in its name.

The other sets each contain an intruder: the hertz is the unit of frequency, and the year is a unit of time.

Answer: B

The result is \(4.6 \times 10^{-5}\).

Write both numbers with the same power of ten and add:

\[ \begin{aligned} &0.27 \times 10^{-5} + 4.3 \times 10^{-5} \\ &= 4.57 \times 10^{-5} \end{aligned} \]

In addition, the result keeps only as many decimal places as the least precise term. In units of \(10^{-5}\), 4.3 is known to one decimal place, so the sum is rounded to one decimal place: \(4.6 \times 10^{-5}\).

\(4.57 \times 10^{-5}\) is the unrounded sum, \(4.5 \times 10^{-5}\) truncates instead of rounding, \(7.0 \times 10^{-5}\) adds the mantissas without matching the powers, and \(4.57 \times 10^{-6}\) has the wrong power.

Answer: D

The product \(\lambda\tau\) is dimensionless: \(L^{\circ}M^{\circ}T^{\circ}\) (all powers zero).

The decay constant has dimensions of inverse time, \([T^{-1}]\), and the mean life is a time, \([T]\). In fact \(\tau = 1/\lambda\), so

\[ [\lambda\tau] = [T^{-1}][T] = [L^{0}M^{0}T^{0}] \]

The product is the pure number 1. Every other option contains a non-zero power of L, M or T, which a dimensionless quantity cannot have.

Answer: D

\(|\hat{A} - \hat{B}|\) equals 1.

For unit vectors at angle \(\theta\),

\[ \begin{aligned} |\hat{A} - \hat{B}|^{2} &= 1 + 1 - 2\cos\theta \\ &= 2 - 2\cos 60^{\circ} = 1 \end{aligned} \]

so the magnitude is 1. Geometrically, the two unit vectors and their difference form an equilateral triangle, since the angle between them is 60°.

\(\frac{1}{2}\) is \(\cos 60^{\circ}\) itself, and the other fractions come from forgetting the square root or the factor 2.

Answer: D

The sum, rounded correctly, is 31.1.

In addition and subtraction the result keeps as many decimal places as the least precise number. Here 12.1100 has four decimal places, 18.0 has one and 1.012 has three, so the answer may carry only one decimal place.

\[ \begin{aligned} 12.1100 + 18.0 + 1.012 &= 31.1220 \\ &\approx 31.1 \end{aligned} \]

31.1220 and 31.122 report precision that 18.0 does not have, and 31.12 keeps one decimal place too many. 31 drops a decimal place that 18.0 does justify. The rule about counting significant figures applies to multiplication and division, not to sums.

Answer: A

One barn is \(10^{-28}\ \text{m}^{2}\).

The barn is a unit of area used for nuclear cross-sections, the effective target area a nucleus presents to an incoming particle. It is defined as \(10^{-24}\ \text{cm}^{2}\). Converting to square metres, with \(1\ \text{cm}^{2} = 10^{-4}\ \text{m}^{2}\):

\[ 10^{-24} \times 10^{-4} = 10^{-28}\ \text{m}^{2} \]

The size makes sense: a nucleus has a radius of order \(10^{-14}\) m, so its cross-sectional area is of order \(10^{-28}\ \text{m}^{2}\). \(10^{-15}\) is the scale of a nuclear radius in metres (a femtometre), not an area, and the other powers do not match the definition.

Answer: D

The dimensional formula is \(ML^{2}T^{-2}\).

Moment of inertia is mass times distance squared, \([ML^{2}]\), and angular velocity is per unit time, \([T^{-1}]\):

\[ [I\omega^{2}] = [ML^{2}][T^{-2}] = [ML^{2}T^{-2}] \]

This is the dimension of energy, as expected, since rotational kinetic energy is \(\frac{1}{2}I\omega^{2}\). \(ML^{2}T^{-1}\) is angular momentum \(I\omega\), \(MLT^{-2}\) is force, \(MLT^{-1}\) is linear momentum, and \(ML^{0}T^{-2}\) drops the length dimension.

Answer: B

The ratio is of the order of \(10^{3}\).

The Sun is about \(1.5 \times 10^{11}\) m from the Earth and the Moon about \(3.84 \times 10^{8}\) m:

\[ \frac{1.5 \times 10^{11}}{3.84 \times 10^{8}} \approx 390 \]

Since \(\log_{10}390 \approx 2.6\), the nearest power of ten is \(10^{3}\). The larger powers would put the Sun millions of times farther than the Moon, and \(10^{-4}\) inverts the ratio.