Units & Measurements — HOTS Questions

A physical quantity P is related to four measurable quantities a, b, c and d as follows:

$\text P = \dfrac{\text a^3 \text b^2}{\sqrt {\text c}\text d}$

The percentage errors of measurement in a, b, c and d are 1% ,3%, 4% and 2% respectively. Find the percentage error in the quantity P.

If the value of P calculated from the above relation comes out to be 3.763, to what value should you round off the result?

Answer

Given,

• $\text P = \dfrac{\text a^3 \text b^2}{\sqrt {\text c}\text d}$
• Percentage error in a = 1%
• Percentage error in b = 3%
• Percentage error in c = 4%
• Percentage error in d = 2%

When a quantity is expressed as a product or quotient of powers of measured quantities, the maximum percentage error in it is obtained by adding the percentage errors of all the quantities, each multiplied by the magnitude of its power. The sign of the power is not taken into account, since the errors always add up in the worst case.

Here the powers of a, b, c and d are 3, 2, $\dfrac{1}{2}$ and 1 respectively.

Therefore, the maximum percentage error in P is

$\dfrac{\Delta \text P}{\text P} \times 100 = 3\left(\dfrac{\Delta \text a}{\text a} \times 100\right) + 2\left(\dfrac{\Delta \text b}{\text b} \times 100\right) + \dfrac{1}{2}\left(\dfrac{\Delta \text c}{\text c} \times 100\right) + 1\left(\dfrac{\Delta \text d}{\text d} \times 100\right)$

Substituting the values,

$\dfrac{\Delta \text P}{\text P} \times 100 = 3(1) + 2(3) + \dfrac{1}{2}(4) + 1(2) \\[1em] = 3 + 6 + 2 + 2 \\[1em] = 13\$

The percentage error in P is 13%, which has two significant figures. The calculated result cannot be more accurate than the error itself, so the value of P must also be rounded off to two significant figures. Rounding off 3.763 to two significant figures gives 3.8.

Hence, the percentage error in P is 13% and the result should be rounded off to P = 3.8.

Precise measurements of physical quantities are need of science. For example, to ascertain the speed of an aircraft, one must have an accurate method to find its positions at closely separated instants of time. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modern science where precise measurements of length, time, mass, etc., are needed. Also, wherever you can, give a quantitative idea of the precision needed.

Answer

In modern science, precise measurements of length, time and mass are needed in many fields. Some examples, along with the order of precision required, are given below.

Precise measurement of length :

(a) In the study of crystals by X-ray diffraction, the spacing between atoms in a crystal has to be measured. This requires a precision of the order of 10^-10 m (1 Å).

(b) In electron microscopy, the size and structure of very small objects are measured to a precision of the order of 10^-10 m to 10^-11 m.

(c) In astronomy, the distances of stars and galaxies are measured, requiring the measurement of extremely large lengths of the order of 10¹⁶ m or more.

Precise measurement of time :

(a) In the working of atomic clocks (caesium clocks), time is measured to a precision of the order of 10^-13 s.

(b) The life-time of unstable elementary particles is measured to a precision of the order of 10^-24 s.

(c) In the measurement of the frequency of light emitted by atoms, time intervals of the order of 10^-15 s are involved.

Precise measurement of mass :

(a) The masses of atoms and molecules are measured using a mass spectrograph, to a precision of the order of 10^-27 kg (the atomic mass unit is about 1.66 × 10^-27 kg).

(b) The masses of elementary particles such as electrons and protons are measured to a precision of the order of 10^-30 kg to 10^-27 kg.

Thus, precise measurements of length, time and mass are essential in modern science, and the precision required ranges from extremely small to extremely large values depending on the physical quantity being measured.

Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the followings (where an estimate is difficult to obtain, try to get an upper bound on the quantity):

(a) the total mass of rain-bearing clouds over India during the Monsoon,

(b) the mass of an elephant.

Answer

(a) Total mass of rain-bearing clouds over India during the Monsoon.

The mass of the rain-bearing clouds can be estimated from the total rainfall over India during the Monsoon.

• Area of India, A ≈ 3.3 × 10¹² m²
• Average height (depth) of rainfall during the Monsoon, h ≈ 1 m (about 100 cm)
• Density of water, ρ = 10³ kg m^-3

The volume of the rain water is

$\text V = \text A \times \text h = 3.3 \times 10^{12} \times 1 = 3.3 \times 10^{12} \text { m}^3$

Since the whole of this water comes from the clouds, the mass of the clouds is

$\text m = \text V \times \rho \\[1em] = 3.3 \times 10^{12} \times 10^3 \\[1em] = 3.3 \times 10^{15} \text { kg}$

Hence, the total mass of the rain-bearing clouds over India during the Monsoon is of the order of 10¹⁵ kg.

(b) Mass of an elephant.

The mass of an elephant can be estimated by the principle of floatation, using a boat.

First, take a boat of known base area A floating in a river, and mark the level of water on its side. Then take the elephant into the boat and again mark the new level of water on its side.

• Let A = base area of the boat
• Let h = extra depth to which the boat sinks after the elephant is loaded (the distance between the two marks)
• Density of water, ρ = 10³ kg m^-3

By the principle of floatation, the weight of the elephant is equal to the weight of the extra water displaced. Hence the mass of the elephant is equal to the mass of the extra volume of water displaced :

$\text m = \text{volume of water displaced} \times \rho \\[1em] = (\text A \times \text h) \times \rho$

By measuring the base area A of the boat and the extra depth h to which it sinks, the mass of the elephant can be estimated.

Two atomic (caesium) clocks allowed to run for 100 years differ from each other only by about 0.02 s. Find the degree of accuracy of the standard cesium clock in measuring a time-interval of 1 s.

Answer

Given,

• Time-interval for which the clocks are run = 100 years
• Difference between the two clocks in this interval = 0.02 s

In 100 years there are 25 leap years, so the total number of days is

$(100 \times 365) + 25 = 36525 \text { days}$

Expressing this time-interval in seconds,

$\text t = 36525 \times 24 \times 60 \times 60 \\[1em] = 3.16 \times 10^{9}\ \text s$

The two clocks differ by 0.02 s in a time-interval of 3.16 × 10⁹ s. Hence the error in measuring a time-interval of 1 s is

$\dfrac{\Delta \text t}{\text t} = \dfrac{0.02\ \text s}{3.16 \times 10^{9}\ \text s} \\[1em] = 6.3 \times 10^{-12}$

The degree of accuracy is the reciprocal of this fractional error,

$\dfrac{1}{6.3 \times 10^{-12}} = 1.6 \times 10^{11} \approx 10^{11}$

Hence, the caesium clock measures a time-interval of 1 s with an accuracy of about 1 part in 10¹¹.

The diameter of a sodium atom is about 2.5 Å. Estimate the average mass density of a sodium atom, knowing that the atomic mass of sodium is 23 and Avogadro's number is 6.023 x 10²³. Compare this density with the density of sodium in crystalline phase which is 97O kg m^-3. Are they of the same order of magnitude? If so why?

Answer

Given,

• Diameter of a sodium atom, 2r = 2.5 Å = 2.5 × 10^-10 m, so radius r = 1.25 × 10^-10 m
• Atomic mass of sodium = 23, so 1 mole of sodium has a mass of 23 g = 23 × 10^-3 kg
• Avogadro's number, N_A = 6.023 × 10²³
• Density of sodium in the crystalline phase = 970 kg m^-3

Treating the sodium atom as a sphere of radius r, its volume is

$\text V = \dfrac{4}{3}\pi \text r^3 \\[1em] = \dfrac{4}{3} \times 3.14 \times (1.25 \times 10^{-10}\ \text m)^3 \\[1em] = 8.18 \times 10^{-30}\ \text m^3$

One mole of sodium contains N_A atoms and has a mass of 23 × 10^-3 kg. Hence the mass of one sodium atom is

$\text m = \dfrac{23 \times 10^{-3}\ \text{kg}}{6.023 \times 10^{23}} \\[1em] = 3.82 \times 10^{-26}\ \text{kg}$

Therefore, the average mass density of a sodium atom is

$\rho_\text a = \dfrac{\text m}{\text V} \\[1em] = \dfrac{3.82 \times 10^{-26}\ \text{kg}}{8.18 \times 10^{-30}\ \text m^3} \\[1em] = 4.67 \times 10^{3}\ \text{kg m}^{-3}$

Comparing this with the density of sodium in the crystalline phase,

$\rho_\text{crystal} = 970\ \text{kg m}^{-3} = 0.97 \times 10^{3}\ \text{kg m}^{-3}$

Both densities are of the same order of magnitude, that is, 10³ kg m^-3.

Hence, the average mass density of a sodium atom is 4.67 × 10³ kg m^-3, which is of the same order of magnitude as the density of sodium in the crystalline phase.

The two densities are of the same order of magnitude because in a sodium crystal the atoms are closely packed, so that the interatomic separation is of the same order as the size of a sodium atom, and very little empty space is left between the atoms.

Nuclear sizes obey roughly the following empirical relation:

$\text r = \text r_\text o \text A^{1/3},$

where r is the radius of a nucleus, A is the mass number of the nucleus and r_o is a constant equal to about 1.2 fermi (1 fermi or 1 F = 10^-15 m).

Using above relation, show that nuclear mass density is nearly same for different nuclei. Given: 1 atomic mass unit = 1.66 x 10^-27 kg.

Answer

Given,

• r = r_oA^1/3, where r_o = 1.2 fermi = 1.2 × 10^-15 m
• 1 atomic mass unit = 1.66 × 10^-27 kg

A nucleus of mass number A has a mass of nearly A atomic mass units, so

$\text m = \text A \times (1.66 \times 10^{-27}\ \text{kg})$

Treating the nucleus as a sphere of radius r, its volume is

$\text V = \dfrac{4}{3}\pi \text r^3 = \dfrac{4}{3}\pi (\text r_\text o \text A^{1/3})^3 \\[1em] = \dfrac{4}{3}\pi \text r_\text o^3 \text A$

Substituting the value of r_o,

$\text V = \dfrac{4}{3} \times 3.14 \times (1.2 \times 10^{-15}\ \text m)^3 \times \text A \\[1em] = (7.23 \times 10^{-45}\ \text m^3)\ \text A$

Therefore, the nuclear mass density is

$\rho = \dfrac{\text m}{\text V} = \dfrac{\text A \times (1.66 \times 10^{-27}\ \text{kg})}{(7.23 \times 10^{-45}\ \text m^3)\ \text A} \\[1em] = 2.29 \times 10^{17}\ \text{kg m}^{-3}$

The mass number A cancels out from the numerator and the denominator, so the nuclear mass density does not depend on A.

Hence, the nuclear mass density is nearly the same (about 2.29 × 10¹⁷ kg m^-3) for all nuclei.

In a submarine equipped with a SONAR the time delay between generation of a ultrasonic probe wave and the reception of its echo after reflection from an enemy submarine is found to be 77.0 s. What is the distance of the submarine? (Speed of sound in water is 1450 m s^-1.)

Answer

Given,

• Time-interval between the transmission of the ultrasonic wave and the reception of its echo, t = 77.0 s
• Speed of sound in water, v = 1450 m s^-1

Let x be the distance of the enemy submarine. In the time-interval t, the ultrasonic wave travels from the transmitter to the enemy submarine and back to the detector, that is, a total distance of 2x. Hence

$\text v = \dfrac{2\text x}{\text t}$

Therefore,

$\text x = \dfrac{\text v \times \text t}{2} \\[1em] = \dfrac{1450 \times 77.0}{2} \\[1em] = 55825\ \text m \\[1em] = 55.8 \times 10^{3}\ \text m = 55.8\ \text{km}$

Hence, the distance of the enemy submarine is 55.8 km.

The farthest objects in our universe are quasars. What is the distance of a quasar from which light takes 3.0 billion years to reach us? Express the result in km and in ly.

Answer

Given,

• Time taken by light to reach us from the quasar, t = 3.0 billion years = 3.0 × 10⁹ years
• Speed of light in vacuum, c = 3 × 10⁸ m s^-1

Expressing the time-interval in seconds,

$\text t = 3.0 \times 10^{9} \times 365 \times 24 \times 60 \times 60 \\[1em] = 9.46 \times 10^{16}\ \text s$

The distance of the quasar is

$\text s = \text c \times \text t \\[1em] = (3 \times 10^{8}\ \text{m s}^{-1}) \times (9.46 \times 10^{16}\ \text s) \\[1em] = 2.838 \times 10^{25}\ \text m \\[1em] = 2.838 \times 10^{22}\ \text{km}$

To express this distance in light years, we use 1 ly = 9.46 × 10¹⁵ m,

$\text s = \dfrac{2.838 \times 10^{25}\ \text m}{9.46 \times 10^{15}\ \text m} \\[1em] = 3.0 \times 10^{9}\ \text{ly}$

Hence, the distance of the quasar is 2.838 × 10²² km or 3.0 × 10⁹ ly.

A man walking in rain with speed v must slant his umbrella forward making an angle θ with the vertical. A student derives the following relation between θ and v:

$\text {tan θ} = \text v,$

and checks that the relation has a correct limit: as v ⟶ 0, θ ⟶ 0 as expected. (It has been assumed that there is no wind and that the rain falls vertically for a stationary man.) Can be the above relation correct? If not, guess the correct relation.

Answer

According to the principle of homogeneity of dimensions, the dimensions of the quantities on both sides of a physical relation must be the same.

In the given relation,

$\text{tan}\ \theta = \text v$

the left hand side, tan θ, is a trigonometric ratio and is therefore dimensionless,

$[\text{tan}\ \theta] = [\text M^0\text L^0\text T^0]$

whereas the right hand side is the speed of the man, whose dimensional formula is

$[\text v] = [\text{L}\text{T}^{-1}]$

Since the dimensions of the two sides are not the same, the relation is dimensionally incorrect.

Hence, the given relation tan θ = v is not correct.

To make the relation dimensionally correct, the right hand side must also be dimensionless. This is achieved by dividing the speed v of the man by another speed, namely the speed v' with which the rain falls vertically. Hence a correct form of the relation is

$\text{tan}\ \theta = \dfrac{\text v}{\text v'}$

where v' is the speed of the falling rain.

A book with many printing errors contains four different formulae for the displacement y of a particle under going a certain periodic motion:

(a) $\text y = \text {a sin} \dfrac{2\text {π t}}{\text T}$

(b) $\text y = \text {a sin vt}$

(c) $\text y = \dfrac {\text a}{\text T} \text {sin} (\text t/\text a)$

(d) $\text y = \dfrac {\text a}{\sqrt 2} \left(\text {sin} \dfrac{2\text {π t}}{\text T} + \text {cos} \dfrac{2\text {π t}}{\text T}\right)$

(where, $\text a$ = maximum displacement of the particle, $\text v$ = speed of the particle, $\text T$ = time period of motion).

Rule out the wrong formulae on dimensional ground.

Answer

The quantity y is a displacement, so the dimensional formula of the left hand side in each case is

$[\text y] = [\text L]$

By the principle of homogeneity of dimensions, the right hand side must also have the dimension [L]. In addition, the argument of a trigonometric function is an angle and must always be dimensionless.

The dimensional formulae of the given quantities are

• [a] = [L] (maximum displacement)
• [v] = [LT^-1] (speed)
• [T] = [T] (time period)

(a) $\text y = \text a\ \text{sin} \dfrac{2\pi \text t}{\text T}$

The argument is $\dfrac{2\pi \text t}{\text T}$, and

$\left[\dfrac{\text t}{\text T}\right] = \dfrac{[\text T]}{[\text T]} = [\text M^0\text L^0\text T^0]$

so the angle is dimensionless. The dimension of the right hand side is [L], which is the same as that of the left hand side.

Hence, this formula is correct.

(b) $\text y = \text a\ \text{sin vt}$

The argument is vt, and

$[\text{vt}] = [\text{LT}^{-1}][\text T] = [\text L]$

so the angle has the dimension of length and is not dimensionless.

Hence, this formula is wrong.

(c) $\text y = \dfrac{\text a}{\text T}\ \text{sin}\left(\dfrac{\text t}{\text a}\right)$

The argument is $\dfrac{\text t}{\text a}$, and

$\left[\dfrac{\text t}{\text a}\right] = \dfrac{[\text T]}{[\text L]} = [\text L^{-1}\text T]$

so the angle is not dimensionless. Also, the dimension of the right hand side is

$\left[\dfrac{\text a}{\text T}\right] = \dfrac{[\text L]}{[\text T]} = [\text{LT}^{-1}]$

which is not the same as [L].

Hence, this formula is wrong.

(d) $\text y = \dfrac{\text a}{\sqrt 2}\left(\text{sin}\dfrac{2\pi \text t}{\text T} + \text{cos}\dfrac{2\pi \text t}{\text T}\right)$

The argument $\dfrac{2\pi \text t}{\text T}$ is dimensionless, and $\sqrt 2$ is a pure number, so the dimension of the right hand side is

$\left[\dfrac{\text a}{\sqrt 2}\right] = [\text L]$

which is the same as that of the left hand side.

Hence, this formula is correct.

Therefore, the formulae (b) and (c) are ruled out on dimensional grounds, while the formulae (a) and (d) are dimensionally correct.