Units and Measurement Questions and Answers

Answer: D

The quantity with dimensions is the gravitational constant.

From \(F = \frac{G m_{1} m_{2}}{r^{2}}\), \(G = \frac{F r^{2}}{m_{1} m_{2}}\), which has dimensions \([\text{M}^{-1}\text{L}^{3}\text{T}^{-2}]\) and unit N m\(^{2}\) kg\(^{-2}\).

The other four are ratios of like quantities and are dimensionless: strain is change in length over length, Poisson's ratio is lateral strain over longitudinal strain, an angle in radians is arc over radius, and relative density is density over the density of water.

Answer: D

The one that is not a unit of pressure is the newton.

The newton is the SI unit of force (1 N = 1 kg m s^-2). Pressure is force per unit area, measured in pascal (N m^-2).

The other four all measure pressure: the standard atmosphere (1 atm = 101325 Pa), the bar (10⁵ Pa), the torr and the millimetre of mercury (each about 133.3 Pa, and equal to each other to within a tiny fraction).

Answer: C

The dimensionally incorrect expression is \(\frac{1}{2}mv^{2}\).

Power has dimensions \([ML^{2}T^{-3}]\). \(\frac{1}{2}mv^{2}\) has \([M][LT^{-1}]^{2} = [ML^{2}T^{-2}]\), the dimensions of kinetic energy, not power.

The others all reduce to \([ML^{2}T^{-3}]\):

• mav: \([M][LT^{-2}][LT^{-1}]\) (force times velocity)
• \(\frac{mv^{2}}{t}\): energy per unit time
• \(ma^{2}t\): \([M][L^{2}T^{-4}][T]\)
• \(\frac{mas}{t}\): work (force times displacement) per unit time

Numerical factors like \(\frac{1}{2}\) have no dimensions, so they do not decide the answer.

Answer: A

The result is \(8.4 \times 10^{-3}\).

In addition, the answer keeps only as many decimal places as the least precise term. Write both terms as decimals:

\[ \begin{aligned} 14.0 \times 10^{-5} &= 0.000140 \\ 8.3 \times 10^{-3} &= 0.0083 \\ \text{sum} &= 0.008440 \end{aligned} \]

0.0083 is known only to the fourth decimal place, so the sum is rounded to four decimal places: 0.0084, that is \(8.4 \times 10^{-3}\).

\(8.44 \times 10^{-3}\) and \(8.440 \times 10^{-3}\) are the unrounded sum, which claims more precision than the data have. \(8.314 \times 10^{-3}\) and \(2.43 \times 10^{-3}\) do not come from adding the two terms.

Answer: D

The mismatched pair is Energy density - \(J\,m^{-2}\).

Energy density is energy per unit volume, so its SI unit is \(J\,m^{-3}\), not \(J\,m^{-2}\).

The others are right. Compressibility is the reciprocal of bulk modulus, so \((N\,m^{-2})^{-1} = N^{-1}m^{2}\). Latent heat is energy per unit mass, \(J\,kg^{-1}\). Intensity is power per unit area, \(W\,m^{-2}\). Impulse is force times time, \(N\,s\).

Answer: C

The dimensions of A are \(M^{0}L^{0}T^{2}\).

Inductive reactance and capacitive reactance are both measured in ohms, so B and C have the same dimensions and \(\frac{B}{C}\) is dimensionless. Angular frequency D has dimensions \(T^{-1}\). Hence

\[ \begin{aligned} [A] &= \frac{[B]}{[C][D]^{2}} = \frac{1}{T^{-2}} \\ &= M^{0}L^{0}T^{2} \end{aligned} \]

\(M^{0}L^{0}T^{-2}\) multiplies by \(D^{2}\) instead of dividing. The options with nonzero powers of M or L forget that the two reactances cancel completely.

Answer: E

The dimensions of \(\frac{b}{a}\) are \(T^{-1/2}\).

Each term on the right must have the dimension of length, like x.

\[ \begin{aligned} [a]\,T &= L \Rightarrow [a] = LT^{-1} \\ [b]\,T^{3/2} &= L \Rightarrow [b] = LT^{-3/2} \\ \left[\tfrac{b}{a}\right] &= \frac{LT^{-3/2}}{LT^{-1}} = T^{-1/2} \end{aligned} \]

Dividing the other way, a by b, gives \(T^{1/2}\), the usual slip. The integer powers \(T^{-2}\), \(T^{-1}\) and \(T\) would need an integer exponent on t in the second term.

Answer: D

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

By Newton's second law the rate of change of momentum is force. Momentum is mass times velocity, \([MLT^{-1}]\), and dividing by time gives \([MLT^{-2}]\), the dimensions of force.

\(MLT^{-1}\) is momentum itself, not its rate of change. \(ML^{2}T^{-2}\) is energy, \(ML^{-1}T^{-1}\) is the coefficient of viscosity, and \(M^{0}LT^{-1}\) is velocity.

Answer: B

The key gives restoring force, which matches the units this formula gives σ.

Square the formula and solve for σ; \(k\) has no dimensions:

\[ \begin{aligned} \sigma &= \frac{\rho r^{4}}{T^{2}} \\ [\sigma] &= \frac{ML^{-3} \cdot L^{4}}{T^{2}} = MLT^{-2} \end{aligned} \]

\(MLT^{-2}\) is the dimension of force, so σ has the units of a force (newton), which is the 'restoring force' option. Surface tension would be \(MT^{-2}\) (force per length), viscosity \(ML^{-1}T^{-1}\) and vapour pressure \(ML^{-1}T^{-2}\), so none of those fits; 'surface' is an area, \(L^{2}\).

Answer: A

\(X\) is \(2.6 \times 10^{4}\).

Convert metres to kilometres and seconds to hours: 1 m = \(10^{-3}\) km and 1 s = \(\frac{1}{3600}\) h, so \(1 \text{ s}^{-2} = 3600^{2} \text{ h}^{-2}\).

\[ \begin{aligned} 2.0 \text{ m s}^{-2} &= 2.0 \times 10^{-3} \times 3600^{2} \\ &= 2.0 \times 12960 \\ &= 25920 \text{ km h}^{-2} \end{aligned} \]

The given value 2.0 has two significant figures, so the answer is rounded to two: \(2.6 \times 10^{4}\).

\(2.592 \times 10^{4}\) is the unrounded value, which claims more precision than the data, and the \(10^{5}\) options misplace the decimal point.