Units and Measurement Questions and Answers

Answer: A

The dimension of X is \(M^{0}L^{2}T^{-1}\).

Rearrange for X; the number \(6\pi\) is dimensionless. Force is \(MLT^{-2}\), and the coefficient of viscosity (from \(F = \eta A\frac{dv}{dx}\)) is \(ML^{-1}T^{-1}\).

\[ \begin{aligned} [X] &= \frac{[F]}{[\eta]} = \frac{MLT^{-2}}{ML^{-1}T^{-1}} \\ &= M^{0}L^{2}T^{-1} \end{aligned} \]

So X has the dimensions of area per second (in Stokes' law \(F = 6\pi\eta rv\), X stands for the product rv). \(ML^{2}T^{-2}\) is energy, \(ML^{2}T^{-1}\) is angular momentum, and the other two have the wrong power of T or L.

Answer: A

One torr is 1 mm of Hg.

The torr is defined as \(\frac{1}{760}\) of a standard atmosphere. Since one atmosphere supports 760 mm of mercury, one torr is the pressure of a 1 mm mercury column, about 133.3 Pa (the two differ only in the seventh significant figure).

76 cm (760 mm) of Hg is one whole atmosphere, not one torr. 1 cm, 76 mm and 100 mm of Hg are 10, 76 and 100 torr.

Answer: E

\(\frac{mB}{kT}\) is dimensionless: \(M^{0}L^{0}T^{0}\).

The energy of a magnetic dipole in a field is \(U = -mB\), so mB has the dimensions of energy. kT is also an energy (the thermal energy scale). Their ratio is a pure number, as it must be in expressions like the Langevin function for paramagnetism.

Every other option carries non-zero powers of M, L or T, which a ratio of two energies cannot have.

Answer: A

The SI unit of surface tension is \(Nm^{-1}\).

Surface tension is force per unit length acting along the surface, so its unit is newton per metre. Equivalently it is surface energy per unit area, J m⁻², which is the same thing.

N m⁻² is the unit of pressure, N m is the unit of work or torque, and N m² and N alone have no such meaning here.

Answer: C

Energy divided by Planck's constant has the dimensions of frequency.

From \(E = h\nu\), \(\frac{E}{h} = \nu\). Dimensionally, h is energy times time (J s), so

\[ \begin{aligned} \frac{[E]}{[h]} &= \frac{[\text{ML}^{2}\text{T}^{-2}]}{[\text{ML}^{2}\text{T}^{-1}]} \\ &= [\text{T}^{-1}] \end{aligned} \]

That is the dimension of frequency. Time is the reciprocal. Planck's constant itself has the same dimensions as angular momentum, which is why angular momentum is a natural trap, but the ratio E/h is not angular momentum. Velocity and linear momentum include length and mass, which cancel here.

Answer: C

The percentage error is 0.2%.

The uncertainty of a reading is taken as the least count of the instrument, here 1 mm = 0.1 cm:

\[ \begin{aligned} \frac{\Delta l}{l} \times 100 \\ &= \frac{0.1\ \text{cm}}{50\ \text{cm}} \times 100 \\ &= 0.2\% \end{aligned} \]

The units must match before dividing. 2% comes from using 1 mm against 50 cm without converting; 0.1% halves the least count; 0.5% and 5% have no route from these numbers.

Answer: B

The sum is 200 m.

Convert first: 200 mm = 0.200 m, so the arithmetic sum is 200.2 m. In addition, the result keeps no more decimal places than the least precise quantity. 200 m has no decimal places, so the sum is rounded to a whole number of metres:

\[ \begin{aligned} 200\ \text{m} + 0.200\ \text{m} &= 200.2\ \text{m} \\ &\approx 200\ \text{m} \end{aligned} \]

200.2 m is the unrounded sum, which claims a precision the first measurement does not have. 200.200 m claims even more. 202 m and 202.2 m treat 200 mm as 2 m.

Answer: D

The dimensions are \(M^{0}L^{-1}T^{1}\).

For an electromagnetic wave \(E = cB\), so \(\frac{B}{E} = \frac{1}{c}\), the reciprocal of a speed. Speed has dimensions \([LT^{-1}]\), so

\[ \left[\frac{B}{E}\right] = [L^{-1}T] \]

Equivalently, \(F = qE = qvB\) shows \(\frac{E}{B}\) is a velocity. \(M^{0}L^{1}T^{-1}\) is the dimension of \(\frac{E}{B}\), the inverted ratio.

Numerically, \(\frac{B}{E} = \frac{1}{c} \approx 3.3 \times 10^{-9}\) s m^-1 for a wave in vacuum, which carries the same dimensions.

Answer: D

Planck's constant has the dimensions of angular momentum.

From \(E = h\nu\), \(h = \frac{E}{\nu}\), so its unit is J s and its dimensions are \([ML^{2}T^{-2}] \times [T] = [ML^{2}T^{-1}]\). Angular momentum \(L = mvr\) has dimensions \([M][LT^{-1}][L] = [ML^{2}T^{-1}]\), the same.

Energy is \([ML^{2}T^{-2}]\), power \([ML^{2}T^{-3}]\), angular frequency \([T^{-1}]\) and linear momentum \([MLT^{-1}]\); none matches. Bohr's rule \(mvr = \frac{nh}{2\pi}\) is a reminder that \(h\) and angular momentum share units.

Answer: A

The number has 8 significant figures.

The power of ten does not affect the count; only the digits of 50000.040 matter. The rules:

• All non-zero digits are significant: 5 and 4.
• Zeros between non-zero digits are significant: the four zeros of 50000 and the first zero after the decimal point.
• Trailing zeros after a decimal point are significant: the final 0.

That gives 5, 0, 0, 0, 0, 0, 4, 0, eight figures. Dropping the final zero gives 7, and treating the zeros of 50000 as placeholders gives far fewer, but here they sit between the 5 and the 4 and so count.