Units & Measurements

(a) The volume of a cube of side 1 cm is equal to 10^-6 m³.

(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to 1.51 x 10⁴ mm².

(c) A vehicle moving with a speed of 18 kmh^-1 covers 5 m in 1 s.

(d) The relative density of lead is 11.3. Its density is 11.3 g cm^-3 or 11.3 x 10³ kg m^-3.

Reason —

(a) Given,

• Side of the cube, a = 1 cm

Since 1 cm = 10^-2 m, the volume of the cube is

$$\begin{aligned}\text V = \text a^3 = (1\ \text{cm})^3 \\[1em] = (10^{-2}\ \text m)^3 \\[1em] = 10^{-6}\ \text m^3\end{aligned}$$

(b) Given,

• Radius of the cylinder, r = 2.0 cm = 20 mm
• Height of the cylinder, h = 10.0 cm = 100 mm

The total surface area of a solid cylinder of radius r and height h is

$\text A = 2\pi \text{rh} + 2\pi \text r^2 = 2\pi \text r(\text h + \text r)$

Substituting the values,

$$\begin{aligned}\text A = 2 \times 3.14 \times 20 \times (100 + 20) \\[1em] = 125.6 \times 120 \\[1em] = 15072\ \text{mm}^2 \\[1em] = 1.51 \times 10^4\ \text{mm}^2\end{aligned}$$

(c) Given,

• Speed of the vehicle, v = 18 km h^-1

Since 1 km = 1000 m and 1 h = 3600 s, converting km h^-1 into m s^-1,

$$\begin{aligned}\text v = 18 \times \frac{1\ \text{km}}{1\ \text h} \\[1em] = 18 \times \frac{1000\ \text m}{3600\ \text s} \\[1em] = 18 \times \frac{5\ \text m}{18\ \text s} \\[1em] = 5\ \text{m s}^{-1}\end{aligned}$$

Hence, the vehicle covers 5 m in 1 s.

(d) Given,

• Relative density of lead = 11.3

The relative density of a substance is the ratio of its density to the density of water, so

$\text{Relative density of lead} = \frac{\text{Density of lead}}{\text{Density of water}}$

Taking the density of water as 1 g cm^-3,

$\text{Density of lead} = 11.3 \times 1\ \text{g cm}^{-3} = 11.3\ \text{g cm}^{-3}$

Since 1 g = 10^-3 kg and 1 cm = 10^-2 m,

$$\begin{aligned}\text{Density of lead} = 11.3 \times \frac{10^{-3}\ \text{kg}}{(10^{-2}\ \text m)^3} \\[1em] = 11.3 \times 10^{-3} \times 10^{6}\ \text{kg m}^{-3} \\[1em] = 11.3 \times 10^{3}\ \text{kg m}^{-3}\end{aligned}$$

(a) 1 kg m² s^-2 = 10⁷ g cm² s^-2.

(b) 1 m = 1.06 x 10^-16 light year (ly).

(c) 3 m s^-2 = 3.888 x 10⁴ km h^-2.

(d) G = 6.67 x 10^-11 N m² kg^-2 = 6.67 x 10^-8 cm³ s^-2 g^-1.

Reason —

(a) Since 1 kg = 10³ g and 1 m = 10² cm,

$$\begin{aligned}1\ \text{kg m}^2\text{ s}^{-2} = (10^3\ \text g)(10^2\ \text{cm})^2\ \text s^{-2} \\[1em] = 10^3 \times 10^4\ \text{g cm}^2\text{ s}^{-2} \\[1em] = 10^{7}\ \text{g cm}^2\text{ s}^{-2}\end{aligned}$$

(b) One light year is the distance travelled by light in vacuum in one year, and

$1\ \text{ly} = 9.46 \times 10^{15}\ \text m$

Therefore,

$$\begin{aligned}1\ \text m = \frac{1}{9.46 \times 10^{15}}\ \text{ly} \\[1em] = 1.06 \times 10^{-16}\ \text{ly}\end{aligned}$$

(c) Since 1 m = 10^-3 km and 1 s = $\frac{1}{3600}$ h,

$$\begin{aligned}3\ \text{m s}^{-2} = 3 \times (10^{-3}\ \text{km}) \times \left(\frac{1}{3600}\ \text h\right)^{-2} \\[1em] = 3 \times 10^{-3} \times (3600)^2\ \text{km h}^{-2} \\[1em] = 3 \times 10^{-3} \times 1.296 \times 10^{7}\ \text{km h}^{-2} \\[1em] = 3.888 \times 10^{4}\ \text{km h}^{-2}\end{aligned}$$

(d) Writing the newton in terms of the fundamental units, 1 N = 1 kg m s^-2. Therefore,

$$\begin{aligned}\text G = 6.67 \times 10^{-11}\ \text{N m}^2\text{ kg}^{-2} \\[1em] = 6.67 \times 10^{-11}\ (\text{kg m s}^{-2})\ \text m^2\ \text{kg}^{-2} \\[1em] = 6.67 \times 10^{-11}\ \text{m}^3\text{ s}^{-2}\text{ kg}^{-1}\end{aligned}$$

Since 1 m = 10² cm and 1 kg = 10³ g,

$$\begin{aligned}\text G = 6.67 \times 10^{-11} (10^2\ \text{cm})^3\ \text s^{-2}\ (10^3\ \text g)^{-1} \\[1em] = 6.67 \times 10^{-11} \times 10^{6} \times 10^{-3}\ \text{cm}^3\text{ s}^{-2}\text{ g}^{-1} \\[1em] = 6.67 \times 10^{-8}\ \text{cm}^3\text{ s}^{-2}\text{ g}^{-1}\end{aligned}$$

Given,

• Magnitude of a calorie in SI units, n₁ = 4.2 (since 1 calorie = 4.2 J)
• First system of units (SI) : unit of mass M₁ = 1 kg, unit of length L₁ = 1 m, unit of time T₁ = 1 s
• New system of units : unit of mass M₂ = α kg, unit of length L₂ = β m, unit of time T₂ = γ s

The dimensional formula of heat (energy) is

$[\text{energy}] = [\text{M}^1\text{L}^2\text{T}^{-2}]$

so the dimensions in mass, length and time are a = 1, b = 2 and c = -2.

The magnitude of a physical quantity remains the same in every system of units, that is,

$\text n_1\text u_1 = \text n_2\text u_2$

Hence the numerical value in the new system is

$\text n_2 = \text n_1 \left[\frac{\text M_1}{\text M_2}\right]^\text a \left[\frac{\text L_1}{\text L_2}\right]^\text b \left[\frac{\text T_1}{\text T_2}\right]^\text c$

Substituting the values,

$$\begin{aligned}\text n_2 = 4.2 \left[\frac{1\ \text{kg}}{\alpha\ \text{kg}}\right]^1 \left[\frac{1\ \text m}{\beta\ \text m}\right]^2 \left[\frac{1\ \text s}{\gamma\ \text s}\right]^{-2} \\[1em] = 4.2 \left[\frac{1}{\alpha}\right] \left[\frac{1}{\beta}\right]^2 \left[\frac{1}{\gamma}\right]^{-2} \\[1em] = 4.2\ \alpha^{-1}\ \beta^{-2}\ \gamma^{2}\end{aligned}$$

Hence, the magnitude of a calorie in terms of the new units is 4.2 α^-1 β^-2 γ².

A dimensional quantity has a definite unit, and its measured value depends on the unit chosen. Calling such a quantity large or small has no meaning by itself, because the same quantity may appear large when compared with one standard and small when compared with another. A quantity can be called large or small only when it is compared with a standard (or reference) quantity of the same kind. Hence the statements are reframed by including a suitable standard of comparison.

(a) An atom is very small compared to the objects we see around us in daily life. So the statement should be reframed as : Atoms are very small objects compared to a pinhead (or any object of everyday size).

(b) The speed of a jet plane is great compared to the speed of an ordinary vehicle such as a car or a train. So the statement should be reframed as : A jet plane moves with a speed greater than that of a superfast train (or an ordinary car).

(c) The mass of Jupiter is very large compared to the mass of other planets. So the statement should be reframed as : The mass of Jupiter is very large compared to the mass of the Earth (or any other planet).

(d) The number of air molecules in this room is large compared to the number of molecules present in a small region, such as the air contained in a test tube. So the statement should be reframed as : The air inside this room has a large number of molecules compared to the number contained in a test tube.

(e) This statement is already correct, since the mass of a proton is compared with the mass of an electron, which serves as the standard of comparison. Both are of the same kind (mass), so no reframing is necessary.

A proton is much more massive than an electron.

(f) This statement is also correct, since the speed of sound is compared with the speed of light, which serves as the standard of comparison. Both are of the same kind (speed), so no reframing is necessary.

The speed of sound is much smaller than the speed of light.

Given,

• Time taken by light to travel from the sun to the earth, t = 8 min 20 s = (8 × 60) + 20 = 500 s
• Speed of light in vacuum, c = 3 × 10⁸ m s^-1

The new unit of length is so chosen that the speed of light in vacuum is unity, that is, light travels 1 new unit of length in 1 s. Hence

$1\ \text{new unit of length} = 3 \times 10^8\ \text m$

The distance between the sun and the earth in SI units is

$$\begin{aligned}\text s = \text c \times \text t \\[1em] = (3 \times 10^8\ \text{m s}^{-1}) \times (500\ \text s) \\[1em] = 1.5 \times 10^{11}\ \text m\end{aligned}$$

The magnitude of a physical quantity remains the same in every system of units, that is, n₁u₁ = n₂u₂. Here

• n₁ = 1.5 × 10¹¹, u₁ = 1 m
• u₂ = 3 × 10⁸ m

Therefore,

$$\begin{aligned}\text n_2 = \frac{\text n_1 \text u_1}{\text u_2} = \frac{1.5 \times 10^{11}\ \text m}{3 \times 10^{8}\ \text m} \\[1em] = 0.5 \times 10^{3} \\[1em] = 500\end{aligned}$$

Hence, the distance between the sun and the earth is 500 new units of length.

The precision of a measuring instrument is determined by its least count. The smaller the least count, the greater is the precision. Hence the most precise device is the one having the smallest least count.

(a) For the vernier callipers,

$$\begin{aligned}\text{Least count} = \frac{\text{Value of one main scale division}}{\text{Number of divisions on the sliding scale}} \\[1em] = \frac{1\ \text{mm}}{20} \\[1em] = 5 \times 10^{-2}\ \text{mm} = 5 \times 10^{-5}\ \text m\end{aligned}$$

(b) For the screw gauge,

$$\begin{aligned}\text{Least count} = \frac{\text{Pitch}}{\text{Number of divisions on the circular scale}} \\[1em] = \frac{1\ \text{mm}}{100} \\[1em] = 10^{-2}\ \text{mm} = 10^{-5}\ \text m\end{aligned}$$

(c) For the optical instrument, the least count is of the order of the wavelength of light, which for visible light is about 5000 Å,

$\text{Least count} = 5000 \times 10^{-10}\ \text m = 5 \times 10^{-7}\ \text m$

Comparing the three least counts, the optical instrument has the smallest least count.

Hence, the optical instrument is the most precise device for measuring length.

Given,

• Magnification of the microscope, m = 100
• Observed (apparent) width of the hair in the field of view = 3.5 mm

The magnification of a microscope is the ratio of the apparent width to the actual width, that is,

$\text{Magnification} = \frac{\text{Apparent width of the hair}}{\text{Actual width of the hair}}$

Therefore,

$$\begin{aligned}\text{Actual width of the hair} = \frac{\text{Apparent width of the hair}}{\text{Magnification}} \\[1em] = \frac{3.5\ \text{mm}}{100} \\[1em] = 0.035\ \text{mm}\end{aligned}$$

Hence, the estimated thickness of the hair is 0.035 mm.

(a) Wind the thread closely and uniformly on a cylindrical body such as a pencil, so that the successive turns touch one another without any gap or overlap. Let the number of turns be n. Measure with the metre scale the length L of the portion of the pencil covered by these n turns. Then the diameter of the thread is

$\text d = \frac{\text L}{\text n}$

The estimate can be improved by increasing the number of turns n.

(b) The least count of a screw gauge is

$\text{Least count} = \frac{\text{Pitch}}{\text{Number of divisions on the circular scale}}$

so increasing the number of divisions on the circular scale reduces the least count and thus increases the precision. However, this cannot be done indefinitely, because on increasing the number of divisions the divisions come so close together that they can no longer be resolved by the eye, and the reading itself becomes uncertain.

Hence, the accuracy of the screw gauge cannot be increased arbitrarily by increasing the number of divisions on the circular scale.

(c) The readings of the diameter differ from one another because of random errors, which occur irregularly and are equally likely to be positive or negative. Random errors are minimised by taking a large number of observations of the same quantity and then taking their arithmetic mean, since the positive and negative errors tend to cancel one another.

Hence, the mean of 100 measurements is a more reliable estimate of the diameter than the mean of only 5 measurements.

Given,

• Area of the house on the slide (area of the object), A_o = 1.75 cm²
• Area of the house on the screen (area of the image), A_i = 1.55 m² = 1.55 × 10⁴ cm²

The areal magnification is the ratio of the area of the image to the area of the object,

$$\begin{aligned}\text{Areal magnification} = \frac{\text A_\text i}{\text A_\text o} = \frac{1.55 \times 10^4\ \text{cm}^2}{1.75\ \text{cm}^2} \\[1em] = 8857.14\end{aligned}$$

Since area is proportional to the square of the length, the areal magnification is the square of the linear magnification. Therefore,

$$\begin{aligned}\text{Linear magnification} = \sqrt{\text{Areal magnification}} \\[1em] = \sqrt{8857.14} \\[1em] = 94.1\end{aligned}$$

Hence, the linear magnification of the projector-screen arrangement is 94.1.

Note: The value 8857.14 is the areal magnification, not the linear magnification. The linear magnification is its square root, 94.1.

(i) One.

Reason — All the initial zeros on the right of the decimal point but on the left of the first non-zero digit are not significant, since they only fix the position of the decimal point. Hence in 0.007 only the digit 7 is significant.

(ii) Three.

Reason — All the non-zero digits are significant. In 2.64 × 10²⁴ the significant figures are 2, 6 and 4; the power of ten only fixes the position of the decimal point and is not counted.

(iii) Four.

Reason — All the zeros to the right of the decimal point which come after a non-zero digit are significant. Hence in 0.2370 the significant figures are 2, 3, 7 and 0.

(iv) Four.

Reason — All the non-zero digits are significant, and the trailing zero to the right of the decimal point is also significant. Hence in 6.320 the significant figures are 6, 3, 2 and 0.

(v) Four.

Reason — All the zeros lying between two non-zero digits are significant. Hence in 6.032 the significant figures are 6, 0, 3 and 2.

(vi) Four.

Reason — The initial zeros on the right of the decimal point but on the left of the first non-zero digit are not significant, whereas the zero lying between the digits 6 and 3 is significant. Hence in 0.0006032 the significant figures are 6, 0, 3 and 2.

Given,

• Length, l = 4.234 m (4 significant figures)
• Breadth, b = 1.005 m (4 significant figures)
• Thickness, t = 2.01 cm = 0.0201 m (3 significant figures)

Area of the sheet.

The total surface area of the rectangular sheet is

$\text A = 2(\text{lb} + \text{bt} + \text{tl})$

Substituting the values,

$$\begin{aligned}\text A = 2[(4.234 \times 1.005) + (1.005 \times 0.0201) + (0.0201 \times 4.234)] \\[1em] = 2[4.25517 + 0.0202005 + 0.0851034] \\[1em] = 2 \times 4.3604739 \\[1em] = 8.7209478 \text { m}^2\end{aligned}$$

In multiplication, the result is rounded off to the least number of significant figures in the given data, which is 3. Hence,

$\text A = 8.72 \text { m}^2$

Volume of the sheet.

The volume of the rectangular sheet is

$\text V = \text l \times \text b \times \text t$

Substituting the values,

$$\begin{aligned}\text V = 4.234 \times 1.005 \times 0.0201 \\[1em] = 0.0855289... \\[1em] = 0.0855 \text { m}^3\end{aligned}$$

Here also the result is rounded off to 3 significant figures, the least number of significant figures in the given data.

Hence, the area of the sheet is 8.72 m² and the volume of the sheet is 0.0855 m³.

Given,

• Mass of the box, m = 2.3 kg
• Mass of the first gold piece, m₁ = 20.15 g = 0.02015 kg
• Mass of the second gold piece, m₂ = 20.17 g = 0.02017 kg

(a) The total mass of the box is

$$\begin{aligned}\text M = \text m + \text m_1 + \text m_2 \\[1em] = 2.3 + 0.02015 + 0.02017 \\[1em] = 2.34032 \text { kg}\end{aligned}$$

In addition, the result is rounded off to the same number of decimal places as the quantity having the least number of decimal places. Here the mass of the box, 2.3 kg, has only one decimal place. Hence

$\text M = 2.3 \text { kg}$

Hence, the total mass of the box is 2.3 kg.

(b) The difference in the masses of the two gold pieces is

$$\begin{aligned}\text m_2 - \text m_1 = 20.17 - 20.15 \\[1em] = 0.02 \text { g}\end{aligned}$$

Both the masses are given up to two decimal places, so the difference is also written up to two decimal places.

Hence, the difference in the mass of the pieces is 0.02 g.

The given relation is

$\text m = \frac{\text m_\text o}{(1 - \text v^2)^{1/2}}$

According to the principle of homogeneity of dimensions, only quantities of the same dimensions can be added to or subtracted from one another. In the bracket, the number 1 is dimensionless, so the quantity subtracted from it, that is v², must also be dimensionless.

But the speed v is not dimensionless; its dimensional formula is

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

Hence v² by itself cannot be subtracted from 1. To make this term dimensionless, v² must be divided by a quantity having the same dimensions as v². The speed of light c has the same dimensions as v,

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

so the correct dimensionless term is $\frac{\text v^2}{\text c^2}$, since

$\left[\frac{\text v^2}{\text c^2}\right] = \frac{[\text{L}^2\text{T}^{-2}]}{[\text{L}^2\text{T}^{-2}]} = [\text{M}^0\text{L}^0\text{T}^0]$

which is dimensionless.

Therefore, the constant c must be placed with v so that v² is replaced by $\frac{\text v^2}{\text c^2}$, and the correct relation is

$\text m = \frac{\text m_\text o}{\left(1 - \frac{\text v^2}{\text c^2}\right)^{1/2}}$

Hence, the missing constant c should be placed as $\frac{\text v^2}{\text c^2}$ in place of v².

Given,

• Size (radius) of a hydrogen atom, r = 0.5 Å = 0.5 × 10^-10 m
• Avogadro's number, N_A = 6.023 × 10²³ mol^-1

Assuming the hydrogen atom to be a sphere of radius r, the volume of one atom is

$\text V_{\text{atom}} = \frac{4}{3}\pi \text r^3$

Substituting the values,

$$\begin{aligned}\text V_{\text{atom}} = \frac{4}{3} \times 3.14 \times (0.5 \times 10^{-10}\ \text m)^3 \\[1em] = \frac{4}{3} \times 3.14 \times 1.25 \times 10^{-31} \\[1em] = 5.24 \times 10^{-31}\ \text m^3\end{aligned}$$

One mole of hydrogen contains N_A atoms, so the total atomic volume of one mole of hydrogen atoms is

$$\begin{aligned}\text V_{\text{total}} = \text N_\text A \times \text V_{\text{atom}} \\[1em] = (6.023 \times 10^{23}) \times (5.24 \times 10^{-31}) \\[1em] = 3.16 \times 10^{-7}\ \text m^3\end{aligned}$$

Hence, the total atomic volume of a mole of hydrogen atoms is 3.16 × 10^-7 m³.

Given,

• Molar volume of hydrogen at STP, V_molar = 22.4 L = 22.4 × 10^-3 m³
• Size (diameter) of a hydrogen molecule = 1 Å, so radius r = 0.5 Å = 0.5 × 10^-10 m
• Avogadro's number, N_A = 6.023 × 10²³ mol^-1

Treating the hydrogen molecule as a sphere of radius r, the atomic volume of one mole of hydrogen is

$\text V_{\text{atomic}} = \frac{4}{3}\pi \text r^3 \times \text N_\text A$

Substituting the values,

$$\begin{aligned}\text V_{\text{atomic}} = \frac{4}{3} \times 3.14 \times (0.5 \times 10^{-10}\ \text m)^3 \times (6.023 \times 10^{23}) \\[1em] = 3.152 \times 10^{-7}\ \text m^3\end{aligned}$$

Therefore, the required ratio is

$$\begin{aligned}\frac{\text V_{\text{molar}}}{\text V_{\text{atomic}}} = \frac{22.4 \times 10^{-3}\ \text m^3}{3.152 \times 10^{-7}\ \text m^3} \\[1em] = 7.1 \times 10^{4}\end{aligned}$$

Hence, the ratio of the molar volume to the atomic volume of a mole of hydrogen is 7.1 × 10⁴.

This ratio is so large because in a gas the intermolecular separation is very much greater than the size of a molecule. The molecules of a gas occupy only a very small part of the total volume of the gas, the rest being empty space.

This observation is explained by the change in the line of sight, that is, the line joining the object to the eye of the observer.

For a nearby object such as a tree or an electric pole, the distance from the eye is small. As the train moves, the line of sight of the tree turns through a large angle in a short time, so its direction changes rapidly. Hence the tree appears to move rapidly in a direction opposite to the motion of the train.

For a distant object such as a hill top, the moon or a star, the distance from the eye is extremely large. As the train moves, the line of sight of such an object turns through an extremely small angle, so its direction remains practically unchanged. Hence the distant object appears to be stationary, that is, it appears to move along with the observer.

Hence, the nearby objects appear to move rapidly in the backward direction because the direction of the line of sight changes rapidly, while the distant objects appear stationary because the direction of the line of sight remains practically unchanged.

Given,

• Mass of the sun, M = 2.0 × 10³⁰ kg
• Radius of the sun, R = 7.0 × 10⁸ m

Treating the sun as a sphere of radius R, its volume is

$\text V = \frac{4}{3}\pi \text R^3$

The mass density of the sun is

$\rho = \frac{\text{Mass of the sun}}{\text{Volume of the sun}} = \frac{\text M}{\frac{4}{3}\pi \text R^3}$

Substituting the values,

$$\begin{aligned}\rho = \frac{2.0 \times 10^{30}\ \text{kg}}{\frac{4}{3} \times 3.14 \times (7.0 \times 10^8\ \text m)^3} \\[1em] = \frac{2.0 \times 10^{30}}{1.437 \times 10^{27}} \\[1em] = 1.39 \times 10^{3}\ \text{kg m}^{-3}\end{aligned}$$

This value lies in the range of the densities of solids and liquids (about 10³ kg m^-3), and is very much greater than the densities of ordinary gases (about 1 kg m^-3). Although the sun is a hot plasma, its enormous gravitational attraction compresses the matter to such a high density.

Hence, the mass density of the sun is 1.39 × 10³ kg m^-3, which lies in the range of the densities of solids and liquids.

Given,

• $\text P = \frac{\text a^3 \text b^2}{\sqrt {\text c}\text d}$
• 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, $\frac{1}{2}$ and 1 respectively.

Therefore, the maximum percentage error in P is

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

Substituting the values,

$$\begin{aligned}\frac{\Delta \text P}{\text P} \times 100 = 3(1) + 2(3) + \frac{1}{2}(4) + 1(2) \\[1em] = 3 + 6 + 2 + 2 \\[1em]\end{aligned}$$

= 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.

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.

(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

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

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 :

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

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.

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,

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

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

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

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

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

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

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

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

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

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

Therefore, the average mass density of a sodium atom is

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

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.

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

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

Substituting the value of r_o,

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

Therefore, the nuclear mass density is

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

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.

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

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

Therefore,

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

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

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,

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

The distance of the quasar is

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

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

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

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

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

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

where v' is the speed of the falling rain.

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

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

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

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

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

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

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

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

which is not the same as [L].

Hence, this formula is wrong.

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

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

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

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.

Millisecond

Reason — The light year, the astronomical unit (AU) and the parsec are all units of length used for measuring large astronomical distances. One light year is the distance travelled by light in vacuum in one year, one astronomical unit is the average distance of the sun from the earth, and one parsec is the distance at which the average radius of the earth's orbit around the sun subtends an angle of 1". The millisecond, on the other hand, is a sub-multiple of the second and is therefore a unit of time, not of distance.

Ampere

Reason — Electric current is one of the seven base quantities of the SI system, and its base unit is the ampere (A). The other options are derived units of different quantities : the volt is the unit of potential difference, the coulomb is the unit of electric charge, and the ohm is the unit of resistance.

candela

Reason — Luminous intensity is one of the seven base quantities in the SI system, and its SI base unit is the candela (cd). The other options are different photometric units : the lumen is the unit of luminous flux, the lux is the unit of illuminance, and the watt is the unit of power.

pascal

Reason — Pressure is defined as the thrust acting per unit area, so its SI unit is the newton per square metre, which is given the special name pascal (Pa). Thus 1 Pa = 1 N m^-2. The bar, the atmosphere and the torr are also units of pressure, but they are practical units and not SI units.

kelvin

Reason — Thermodynamic temperature is one of the seven base quantities of the SI system, and its base unit is the kelvin (K). The kelvin is defined in terms of the Boltzmann constant, which is a fundamental constant of nature, so the unit is invariant and easily reproducible. The celsius, fahrenheit and rankine are other temperature scales and are not SI base units.

Steradian

Reason — The SI system has two supplementary units : the radian (rad) for plane angle and the steradian (sr) for solid angle. Hence the solid angle is measured in steradian. The degree and the revolution are practical units of plane angle and not of solid angle.

Becquerel

Reason — The activity of a radioactive substance is the number of disintegrations taking place per unit time, and its SI unit is the becquerel (Bq), where 1 Bq = 1 disintegration per second. The gray is the SI unit of absorbed dose, the sievert is the SI unit of equivalent dose, and the curie is an older practical unit of activity.

henry

Reason — The SI unit of inductance is the henry (H). The other options are the SI units of different electrical quantities : the farad is the unit of capacitance, the weber is the unit of magnetic flux, and the ohm is the unit of resistance.

[ML²T^-3]

Reason — Power is defined as the work done per unit time,

$\text{Power} = \frac{\text{Work}}{\text{Time}}$

The dimensional formula of work is [ML²T^-2] and that of time is [T]. Therefore,

$$\begin{aligned}[\text{Power}] = \frac{[\text{ML}^2\text{T}^{-2}]}{[\text T]} \\[1em] = [\text{ML}^2\text{T}^{-3}]\end{aligned}$$

a dimensionless quantity indicating the type of flow.

Reason — The Reynolds number is the ratio of the inertial force to the viscous force acting on a flowing fluid. Since it is a ratio of two quantities of the same kind, it is a pure number having neither units nor dimensions. Its value indicates the nature of the flow — a small value corresponds to streamline (laminar) flow and a large value corresponds to turbulent flow.

newton

Reason — Friction, air resistance, tension and thrust are all forces, and the SI unit of force is the newton (N), where 1 N = 1 kg m s^-2. Since tension is a force, it is also measured in newton. The other options are the SI units of different quantities : the joule is the unit of work (or energy), the watt is the unit of power, and the henry is the unit of inductance.

Energy

Reason — The SI unit of energy is the joule (J), where 1 J = 1 kg m² s^-2. Since the SI is a rational system, the same unit joule is used for all forms of energy — mechanical, heat and electrical. The other options have different SI units : force is measured in newton, power in watt, and momentum in kg m s^-1.

Newton

Reason — The seven base units of the SI system are the metre, the kilogram, the second, the ampere, the kelvin, the mole and the candela. The mole, the second and the kilogram are therefore base units. The newton is not a base unit; it is a derived unit obtained from the base units as 1 N = 1 kg m s^-2.

magnetic field strength

Reason — The tesla (T) is the SI unit of magnetic field (magnetic flux density), and 1 T = 1 Wb m^-2. The other options are measured in different units : magnetic flux is measured in weber (Wb), magnetic moment in A m² (or J T^-1), and magnetic permeability in H m^-1 (or N A^-2).

in a physically meaningful equation, all terms must have the same dimensions.

Reason — According to the principle of homogeneity of dimensions, every term on both sides of a physically meaningful equation must have the same dimensions. This is because only quantities of the same kind can be added to, subtracted from or equated with one another. The principle is used to check the dimensional correctness of a physical relation.

Strain

Reason — Strain is defined as the ratio of the change in dimension to the original dimension. Since it is the ratio of two quantities of the same kind, it is a pure number and therefore has neither units nor dimensions. The remaining quantities have definite dimensional formulae : gravitational potential [M⁰L²T^-2], velocity [LT^-1] and work [ML²T^-2].

both determining the units of any physical quantities and deriving relationships between them.

Reason — Once the dimensional formula of a physical quantity is known, the units of that quantity in any system can be written down directly from it. The dimensional method is also used to derive the relation between physical quantities, by writing the required quantity as a product of powers of the quantities on which it depends and then equating the dimensions on both sides. However, the dimensional method cannot verify the numerical coefficients appearing in an equation, since pure numbers are dimensionless.

Dimensions are independent of the system of units.

Reason — The dimensions of a physical quantity are the powers to which the base quantities must be raised in order to represent that quantity. They depend only on the nature of the quantity and not on the units in which it is measured. For example, the dimensional formula of force is [MLT^-2] whether the force is expressed in newton, dyne or pound-force. Hence the dimensions remain the same in every system of units.

determine whether a physical equation is dimensionally correct.

Reason — Dimensional analysis is used to check the dimensional consistency of a physical relation, that is, to verify that every term of the relation has the same dimensions. For example, in the equation of motion

$\text s = \text{ut} + \frac{1}{2}\text{at}^2$

the dimensions of each term are

$[\text s] = [\text L],\quad [\text{ut}] = [\text{LT}^{-1}][\text T] = [\text L],\quad [\text{at}^2] = [\text{LT}^{-2}][\text T^2] = [\text L]$

Since all the terms have the dimension of length, the equation is dimensionally correct.

However, dimensional analysis cannot determine the dimensionless constant $\frac{1}{2}$ appearing in the second term, nor can it tell whether a quantity is a scalar or a vector, nor give the numerical value of a physical quantity. Dimensional correctness is therefore only a necessary, and not a sufficient, condition for a relation to be physically correct.

It can be used to check the plausibility of derived equations.

Reason — Dimensional analysis is used to test whether a derived equation is dimensionally consistent, and hence whether it is plausible. It cannot give the exact magnitude of a physical quantity, since dimensionless constants do not appear in a dimensional equation. It is not restricted to mechanical quantities, as electrical, thermal and other quantities also have dimensional formulae. Further, dimensions are independent of the system of units, so no particular unit system is needed to apply the method.

It has neither units nor dimensions.

Reason — A dimensionless quantity is one whose dimensional formula is [M⁰L⁰T⁰]. Such a quantity is a pure number and has neither units nor dimensions, so its numerical value is the same in every system of units. Strain, relative density, refractive index, the Reynolds number and mathematical constants such as π are examples of dimensionless quantities.

pressure

Reason — The modulus of elasticity is defined as the ratio of stress to strain. Strain is dimensionless, so the modulus of elasticity has the same dimensions as stress, which is force per unit area. Pressure is also defined as force per unit area. Hence

$[\text{Modulus of elasticity}] = [\text{Stress}] = \frac{[\text{MLT}^{-2}]}{[\text L^2]} = [\text{ML}^{-1}\text{T}^{-2}]$

which is the same as the dimensional formula of pressure.

4.00

Reason — In a measurement, the digits measured accurately together with the first doubtful digit are called the significant figures. The measurement having the maximum number of significant figures is the most accurate.

• 4.0 has 2 significant figures, so it is measured up to an accuracy of 0.1
• 4.00 has 3 significant figures, so it is measured up to an accuracy of 0.01

Since 4.00 has more significant figures, it is the more accurate measurement.

Power - Watt

Reason — The SI unit of power is the watt (W), where 1 W = 1 J s^-1. Hence this pair is correctly matched. The remaining pairs are wrongly matched : the SI unit of force is the newton (not the joule), the SI unit of work is the joule (not the newton), and the SI unit of energy is the joule (the pascal is the unit of pressure).

$\text s = \text {ut} + \frac{1}{2}\text {at}^2$

Reason — By the principle of homogeneity of dimensions, an equation is dimensionally homogeneous when every term of the equation has the same dimensions. Checking each option :

Option 3 : $\text s = \text{ut} + \frac{1}{2}\text{at}^2$

$$\begin{aligned}[\text s] = [\text L] \\[1em] [\text{ut}] = [\text{LT}^{-1}][\text T] = [\text L] \\[1em] [\text{at}^2] = [\text{LT}^{-2}][\text T^2] = [\text L]\end{aligned}$$

All the terms have the dimension [L], so the equation is dimensionally homogeneous.

Option 2 : v = u + at²

$[\text v] = [\text{LT}^{-1}],\quad [\text u] = [\text{LT}^{-1}],\quad [\text{at}^2] = [\text{LT}^{-2}][\text T^2] = [\text L]$

The term at² has the dimension [L] while the other terms have [LT^-1], so the equation is not dimensionally homogeneous.

Option 4 : F = ma + bt

$[\text F] = [\text{MLT}^{-2}],\quad [\text{ma}] = [\text M][\text{LT}^{-2}] = [\text{MLT}^{-2}]$

The dimensions of the constant b are not specified, so the term bt cannot be assumed to have the dimensions of force. Hence the equation is not dimensionally homogeneous.

Option 1 : $\text K = \frac{1}{2}\text{mv}^2$

$[\text K] = [\text{ML}^2\text{T}^{-2}],\quad \left[\frac{1}{2}\text{mv}^2\right] = [\text M][\text{LT}^{-1}]^2 = [\text{ML}^2\text{T}^{-2}]$

Both the terms have the dimension [ML²T^-2], so this equation is also dimensionally homogeneous.

Note: Both option 1 and option 3 are dimensionally homogeneous, which means that the dimensions on both sides of each equation are the same. Therefore, both options are correct, although the question is intended to have only one correct answer. The question or its options may contain a printing error.

can be added or subtracted directly.

Reason — By the principle of homogeneity of dimensions, only quantities having the same dimensions can be added to or subtracted from one another. However, quantities having the same dimensions need not have the same units or the same physical meaning. For example, work and torque both have the dimensional formula [ML²T^-2], but work is a scalar quantity measured in joule while torque is a vector quantity measured in N m.

Determining the numerical value of dimensionless constants.

Reason — The three chief applications of dimensional analysis are : to check the dimensional correctness of a physical relation, to derive the relation between physical quantities, and to convert the value of a physical quantity from one system of units to another. Dimensional analysis cannot determine the numerical value of a dimensionless constant such as π or $\frac{1}{2}$, because pure numbers have the dimensional formula [M⁰L⁰T⁰] and therefore do not appear in a dimensional equation.

Momentum and impulse

Reason — Momentum is the product of mass and velocity, while impulse is the product of force and time. Their dimensional formulae are

$$\begin{aligned}[\text{Momentum}] = [\text M][\text{LT}^{-1}] = [\text{MLT}^{-1}] \\[1em] [\text{Impulse}] = [\text{MLT}^{-2}][\text T] = [\text{MLT}^{-1}]\end{aligned}$$

and their SI units are kg m s^-1 and N s, which are the same unit since 1 N s = 1 kg m s^-1. Hence this pair has both the same units and the same dimensions.

The other pairs do not satisfy both conditions :

• Work and power — [ML²T^-2] and [ML²T^-3], with units joule and watt. Different dimensions and different units.
• Torque and energy — both [ML²T^-2], but the units are N m and joule respectively, and torque is a vector while energy is a scalar. Same dimensions but different units.
• Pressure and energy — [ML^-1T^-2] and [ML²T^-2], with units pascal and joule. Different dimensions and different units.

5%

Reason —

Given,

• Percentage error in the measurement of mass = 2%
• Percentage error in the measurement of volume = 3%

The density of the object is

$\rho = \frac{\text m}{\text V}$

When a quantity is expressed as a product or quotient 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. Here the powers of m and V are each 1. Therefore,

$$\begin{aligned}\frac{\Delta \rho}{\rho} \times 100 = \frac{\Delta \text m}{\text m} \times 100 + \frac{\Delta \text V}{\text V} \times 100 \\[1em] = 2 + 3 \\[1em]\end{aligned}$$

= 5%

4

Reason — The rules for determining significant figures give :

• The non-zero digit 3 is significant.
• The zero lying to the right of a non-zero digit but on the left of the decimal point is significant, so the zero in 30 is significant.
• All the zeros to the right of the decimal point after a non-zero digit are significant, so both the zeros after the decimal point are significant.

Hence in 30.00 the significant figures are 3, 0, 0 and 0, that is, 4 significant figures.

1

Reason — To find the order of magnitude, the quantity is written in the form N × 10^x, where N is a number between 1 and 10 and x is a positive or negative integer. If N is equal to or smaller than $\sqrt{10}$ = 3.16, the order of magnitude is 10^x; if N is greater than 3.16, the order of magnitude is 10^x+1.

Here,

$11 = 1.1 \times 10^{1}$

Since N = 1.1 is smaller than 3.16, the order of magnitude is 10¹, that is, 1.

0.005 mm

Reason —

Given,

• Pitch of the screw = 0.5 mm
• Number of divisions on the circular scale = 100

The least count of a screw gauge is given by

$\text{Least count} = \frac{\text{Pitch}}{\text{Number of divisions on the circular scale}}$

Substituting the values,

$$\begin{aligned}\text{Least count} = \frac{0.5\ \text{mm}}{100} \\[1em] = 0.005\ \text{mm}\end{aligned}$$

4

Reason — The rules for determining significant figures give :

• All the initial zeros on the right of the decimal point but on the left of the first non-zero digit are not significant, since they only fix the position of the decimal point. Hence the zeros in 0.00 are not significant.
• All the zeros to the right of the decimal point after a non-zero digit are significant. Hence the two trailing zeros are significant.

Hence in 0.004500 the significant figures are 4, 5, 0 and 0, that is, 4 significant figures.

0.1 mm

Reason —

Given,

• 10 vernier scale divisions coincide with 9 main scale divisions
• Value of one main scale division = 1 mm

The least count of a vernier callipers is the difference between the value of one main scale division and one vernier scale division,

$\text{Least count} = 1\ \text{M.S.D.} - 1\ \text{V.S.D.}$

Since 10 V.S.D. = 9 M.S.D., one vernier scale division is

$1\ \text{V.S.D.} = \frac{9}{10}\ \text{M.S.D.} = 0.9\ \text{mm}$

Therefore,

$$\begin{aligned}\text{Least count} = 1\ \text{mm} - 0.9\ \text{mm} \\[1em] = 0.1\ \text{mm}\end{aligned}$$

3.25 cm

Reason —

Given,

• Height difference (sagitta), h = 0.5 cm
• Distance between the legs, l = 3 cm

The radius of curvature measured by a spherometer is given by

$\text R = \frac{\text l^2}{6\text h} + \frac{\text h}{2}$

Substituting the values,

$$\begin{aligned}\text R = \frac{(3)^2}{6 \times 0.5} + \frac{0.5}{2} \\[1em] = \frac{9}{3} + 0.25 \\[1em] = 3 + 0.25 \\[1em] = 3.25 \text { cm}\end{aligned}$$

2.53 mm

Reason —

Given,

• Main scale reading = 2.5 mm
• Circular scale reading = 30 divisions
• Least count = 0.01 mm

The total reading of a screw gauge is given by

$\text{Total reading} = \text{Main scale reading} + (\text{Circular scale reading} \times \text{Least count})$

Substituting the values,

$$\begin{aligned}\text{Total reading} = 2.5 + (30 \times 0.01) \\[1em] = 2.5 + 0.30 \\[1em] = 2.53\ \text{mm}\end{aligned}$$

the expression showing how and which of the base quantities represent the dimensions of a physical quantity.

Reason — The dimensional formula of a physical quantity is the expression which shows which of the base quantities, and with what powers, are contained in that quantity. It is written in terms of the symbols of the base quantities, such as [M] for mass, [L] for length and [T] for time. For example, the dimensional formula of force is [MLT^-2], which shows that force contains mass to the power 1, length to the power 1 and time to the power −2.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The second is the SI base unit of time.

Reason (R) is also correct: The second is defined in terms of the frequency of the radiation emitted in the transition between two hyperfine energy levels of the ground state of the caesium-133 atom. One second is equal to 9,192,631,770 periods of this radiation. This definition fixes the exact standard by which the second is measured, so it explains what the SI unit of time is. Hence the Reason correctly explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Electric current is one of the seven base quantities of the SI system, and the ampere (A) is its base unit.

Reason (R) is also correct: The ampere is defined as that constant current which, when maintained in two straight parallel conductors of infinite length and negligible cross-section placed 1 m apart in vacuum, produces a force of 2 × 10^-7 N per metre of length between them. This gives the standard by which the ampere is fixed, so the Reason correctly explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Note: The SI definition of the ampere was revised on 20 May 2019. Under the revised definition, the ampere is fixed by taking the numerical value of the elementary charge to be e = 1.602176634 × 10^-19 C. If the revised definition is followed, the Reason is false and the answer becomes option 3. The answer given above follows the definition used in the prescribed textbook.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Thermodynamic temperature is one of the seven base quantities of the SI system, and the kelvin (K) is its base unit.

Reason (R) is also correct: The kelvin is defined as $\frac{1}{273.16}$ of the thermodynamic temperature of the triple point of water. This fixes the standard by which the kelvin is measured, so the Reason correctly explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Note: The SI definition of the kelvin was revised on 20 May 2019, and it is now defined by fixing the numerical value of the Boltzmann constant. The answer given above follows the definition used in the prescribed textbook.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Luminous intensity is one of the seven base quantities of the SI system, and the candela (cd) is its base unit.

Reason (R) is also correct: The candela is defined as the luminous intensity, in a given direction, of a source emitting monochromatic radiation of a specified frequency and of a specified radiant intensity in that direction. This gives the standard by which the candela is fixed, so the Reason correctly explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: Mass is one of the seven base quantities of the SI system, and the kilogram (kg) is its base unit.

Reason (R) is false: The kilogram was formerly defined as the mass of the international prototype, a platinum-iridium cylinder kept at Sèvres in France. Under the revised SI, the kilogram is no longer defined in this way; it is defined by fixing the numerical value of the Planck constant as h = 6.62607015 × 10^-34 J s. The base units of the SI are now defined in terms of fundamental constants of nature so that they are invariant and easily reproducible, and are not tied to any physical object.

Therefore, assertion is true but reason is false.

Note: Some editions of the prescribed text still give the older definition of the kilogram in terms of the platinum-iridium prototype. If that definition is followed, the Reason is true and the answer becomes option 1.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: In 0.00450, the significant figures are 4, 5 and the trailing zero. The trailing zero lies to the right of the decimal point after a non-zero digit, so it is significant since it arises due to measurement. Hence there are three significant figures.

Reason (R) is also correct: All the initial zeros on the right of the decimal point but on the left of the first non-zero digit are not significant, since they only fix the position of the decimal point. Leaving out these leading zeros in 0.00450 leaves exactly three significant digits, so the Reason correctly explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The least count of a measuring instrument is the smallest quantity that the instrument is capable of measuring accurately.

Reason (R) is also correct: For a vernier callipers, the least count is given by

$\text{Least count} = 1\ \text{M.S.D.} - 1\ \text{V.S.D.}$

that is, the difference between the values of one main scale division and one vernier scale division. This difference fixes the smallest measurable value, so the Reason correctly explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation

Assertion (A) is correct: An instrument of smaller least count gives a smaller absolute error, and since the relative error depends on the absolute error, the relative error is also reduced.

Reason (R) is also correct: The relative error is defined as

$\text{Relative error} = \frac{\text{Absolute error}}{\text{True value}}$

However, the Reason only states the definition of relative error. It does not by itself explain why the use of a more precise instrument reduces the error; that follows from the fact that a smaller least count gives a smaller absolute error. Hence the Reason is not the correct explanation of the Assertion.

Therefore, both assertion and reason are true but reason is not the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Random errors arise from small changes in the conditions of the experiment and from incorrect judgement of the observer in taking readings. Such causes are unknown and uncontrollable, so the exact cause of a random error cannot be traced.

Reason (R) is also correct: Since random errors occur irregularly and are equally likely to be positive or negative, they are minimised by taking a large number of readings of the same quantity and then taking their arithmetic mean, the positive and negative errors tending to cancel one another. This follows directly from the unpredictable nature of these errors, so the Reason correctly explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: If z = x + y, where the limiting errors in x and y are ±Δx and ±Δy, then

$\text z \pm \Delta \text z = (\text x \pm \Delta \text x) + (\text y \pm \Delta \text y)$

so that the maximum possible error in z is

$\Delta \text z = \Delta \text x + \Delta \text y$

Reason (R) is also correct: In addition (and in subtraction), the absolute errors combine directly, that is, the limiting error in the final result is the sum of the absolute errors in the quantities involved. This linear propagation is exactly what leads to the result stated in the Assertion, so the Reason correctly explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: The absolute error in a measurement is the difference between the measured value of the quantity and its true value.

Reason (R) is false: The absolute error is taken as the magnitude of this difference, the sign being ignored. Hence the absolute error is always positive or zero, and can never be negative.

Therefore, assertion is true but reason is false.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The metre was originally defined as a fraction of the length of the earth's meridian. It is now defined as the length of the path travelled by light in vacuum in $\frac{1}{299,792,458}$ part of a second.

Reason (R) is also correct: The speed of light in vacuum is a fundamental constant of nature and has the same value at all places and at all times. A unit defined in terms of such a constant is invariant and easily reproducible, which is why the metre is now defined in this way. Hence the Reason correctly explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The precision of an instrument is determined by its least count — the smaller the least count, the greater is the precision. An instrument of smaller least count records a greater number of significant figures, so the precision is reflected in the number of significant figures recorded.

Reason (R) is also correct: A greater number of significant figures means that the measurement has been made with finer divisions, that is, with a smaller least count. This is precisely why more significant figures indicate higher precision, so the Reason correctly explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: When a quantity is measured a large number of times, the arithmetic mean of all the readings is taken as the most accurate value of the quantity, that is, as the true value.

Reason (R) is also correct: Random errors are equally likely to be positive or negative, so on taking the arithmetic mean the positive and negative errors tend to cancel one another and the mean comes very close to the correct value. This is exactly why the mean is the best estimate, so the Reason correctly explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: With the main scale alone, a length can be read only up to one main scale division. The vernier scale reduces the least count of the instrument, so a length can be read to a fraction of a main scale division, giving a more precise measurement.

Reason (R) is also correct: The divisions of the vernier scale are slightly smaller than those of the main scale, so that

$\text{Least count} = 1\ \text{M.S.D.} - 1\ \text{V.S.D.}$

which is a fraction of one main scale division. This finer resolution is the reason for the greater precision, so the Reason correctly explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The standard deviation measures how much the individual readings deviate from the arithmetic mean, and therefore indicates the spread or dispersion of the set of measurements.

Reason (R) is also correct: A small standard deviation means that the readings are closely clustered about the mean, that is, the spread is small; a large standard deviation means the readings are widely scattered. This is exactly what makes the standard deviation a measure of spread, so the Reason correctly explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Surface tension is the force acting per unit length and surface energy is the energy per unit area. Writing their dimensional formulae,

$$\begin{aligned}[\text{Surface tension}] = \frac{[\text{MLT}^{-2}]}{[\text L]} = [\text{ML}^0\text T^{-2}] \\[1em] [\text{Surface energy}] = \frac{[\text{ML}^2\text T^{-2}]}{[\text L^2]} = [\text{ML}^0\text T^{-2}]\end{aligned}$$

Hence both have the same dimensional formula [ML⁰T^-2].

Reason (R) is also correct: The SI unit of surface tension is N m^-1 and that of surface energy is J m^-2. Since 1 J = 1 N m,

$\text{J m}^{-2} = \text{N m} \times \text m^{-2} = \text{N m}^{-1}$

so both have the same SI unit. Two quantities having the same unit must have the same dimensions, so the Reason correctly explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The dimensional formula of force is [MLT^-2].

Reason (R) is also correct: From Newton's second law, F = m × a. Writing the dimensional formulae,

$[\text F] = [\text M][\text{LT}^{-2}] = [\text{MLT}^{-2}]$

The dimensional formula of force is obtained directly from this definition, so the Reason correctly explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The SI unit of power is the watt (W), and 1 W = 1 J s^-1.

Reason (R) is also correct: Power is defined as the work done per unit time,

$\text{Power} = \frac{\text{Work}}{\text{Time}}$

Since the SI is a coherent system, the unit of power follows directly from this definition as the joule per second, without any numerical factor. Hence the Reason correctly explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Systematic errors occur in one direction and follow a definite rule, for example the calibration error or the zero error of an instrument. Since the rule governing them can be identified, they can be removed by applying proper corrections, in particular by re-calibrating the instrument.

Reason (R) is also correct: Calibration consists in comparing the readings of the instrument with a standard reference and adjusting the instrument so that it gives correct readings. This removes the fixed bias of the instrument, which is exactly the source of the systematic error. Hence the Reason correctly explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: The argument of a trigonometric function is an angle and must therefore be dimensionless. Checking the two terms,

$$\begin{aligned}[\omega \text t] = [\text T^{-1}][\text T] = [\text M^0\text L^0\text T^0] \\[1em] [\text{kx}] = [\text L^{-1}][\text L] = [\text M^0\text L^0\text T^0]\end{aligned}$$

Both terms are dimensionless, and hence (ωt − kx) is dimensionless.

Reason (R) is false: The dimensional formula of the angular frequency ω is [M⁰L⁰T^-1], and not [M⁰L⁰T] as stated.

Therefore, assertion is true but reason is false.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: Dimensional constants are quantities which have a fixed value and also possess dimensions, for example the universal gravitational constant G and Planck's constant h.

Reason (R) is false: Dimensional constants have dimensions — this is why they are called dimensional constants. For example, the dimensional formula of G is [M^-1L³T^-2] and that of h is [ML²T^-1]. It is the dimensionless constants, such as π and e, which have no dimensions.

Therefore, assertion is true but reason is false.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Writing the dimensional formula of each term,

$$\begin{aligned}[\text x] = [\text L],\quad [\text x_\text o] = [\text L] \\[1em] [\text u_\text o \text t] = [\text{LT}^{-1}][\text T] = [\text L] \\[1em] \left[\frac{1}{2}\text{at}^2\right] = [\text{LT}^{-2}][\text T^2] = [\text L]\end{aligned}$$

Since every term has the dimension [L], the equation is dimensionally homogeneous and hence dimensionally correct.

Reason (R) is also correct: By the principle of homogeneity of dimensions, every term of a physically meaningful equation must have the same dimensions, so dimensional analysis is used to check the dimensional consistency of an equation. It is by applying this very method that the equation in the Assertion is found to be correct, so the Reason correctly explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are false.

Explanation

Assertion (A) is false: Repeated measurements minimise random errors, not systematic errors. Systematic errors occur in one direction and remain the same in every measurement, so taking the arithmetic mean of a large number of readings does not remove them. They are removed by applying proper corrections, such as correcting the zero error or re-calibrating the instrument.

Reason (R): Systematic errors do arise from definite and consistent causes such as a defective instrument, a zero error, an imperfect experimental technique or the personal bias of the observer, and they follow a definite rule. This statement, as it stands, is correct.

Since the assertion is false while the reason states the nature of systematic errors correctly, the correct combination is "assertion false, reason true", which is not listed among the four options. Of the given choices, option 4 is marked as the answer, on the understanding that the assertion is false.

Note: The given options do not include the case “Assertion is false, but Reason is true.” The Assertion is false because repeated measurements help to reduce random errors, not systematic errors. The Reason is true because it correctly describes systematic errors. Therefore, the required correct option is missing from the question.

If both assertion and reason are false.

Explanation

Assertion (A) is false: In the product of 2.5 × 10² and 4.56, the number 2.5 × 10² has 2 significant figures and 4.56 has 3 significant figures. In multiplication, the result carries the least number of significant figures among the given numbers, that is, 2. Hence the product has 2 significant figures, not three, so the assertion is false.

Reason (R): When two numbers are multiplied, the number of significant figures in the result is equal to that of the number having the fewest significant figures. This rule, as stated, is correct.

Since the assertion is false while the reason states the rule correctly, the correct combination is "assertion false, reason true", which is not listed among the four options. Of the given choices, option 4 is marked as the answer, on the understanding that the assertion is false.

Note: The given options do not include the case “Assertion is false, but Reason is true.” The Assertion is false because the product has 2 significant figures, not 3. The Reason is true because it correctly states the rule for determining significant figures. Therefore, the required correct option is missing from the question.

If both assertion and reason are false.

Explanation

Assertion (A) is false: By the principle of homogeneity of dimensions, only quantities having the same dimensions can be added to one another. Force and pressure are quantities of different kinds, so force cannot be added to pressure.

Reason (R) is also false: The dimensional formulae of the two quantities are

$$\begin{aligned}[\text{Force}] = [\text{MLT}^{-2}] \\[1em] [\text{Pressure}] = \frac{[\text{MLT}^{-2}]}{[\text L^2]} = [\text{ML}^{-1}\text T^{-2}]\end{aligned}$$

which are not the same.

Therefore, both assertion and reason are false.

Length, mass and time are chosen as base quantities in mechanics because : They are independent of one another, that is, none of them can be expressed in terms of the other two.

No, all constants are not dimensionless. Only pure numbers such as π and e are dimensionless constants. Physical constants such as the gravitational constant G, whose dimensional formula is [M^-1L³T^-2], and Planck's constant h, whose dimensional formula is [ML²T^-1], have definite dimensions.

The S.I. is called a coherent system of units because all the derived units in this system can be obtained from the base units by simple multiplication or division, without introducing any numerical factor (other than unity).

For example, the S.I. unit of force (newton) is obtained from the base units as 1 N = 1 kg m s^-2, and the S.I. unit of work (joule) is obtained as 1 J = 1 kg m² s^-2. In each case the derived unit follows directly from the base units with no extra numerical factor.

Hence, the S.I. is a coherent system of units.

No, the magnitude of a physical quantity does not change with a change in the system of units. Only the numerical value and the size of the unit change, in such a way that their product remains the same, that is,

$\text n_1\text u_1 = \text n_2\text u_2$

If the size of the unit is small, the numerical value is large, and vice versa.

No, the light year and the parsec are units of distance and not of time. One light year is the distance travelled by light in vacuum in one year, and one parsec is the distance at which the average radius of the earth's orbit around the sun subtends an angle of 1" (second of arc).

The parsec is larger. Since 1 ly = 9.46 × 10¹⁵ m and 1 parsec = 3.08 × 10¹⁶ m,

$\frac{1\ \text{parsec}}{1\ \text{ly}} = \frac{3.08 \times 10^{16}\ \text m}{9.46 \times 10^{15}\ \text m} = 3.26$

Hence, 1 parsec = 3.26 light years.

Since 1 kg = 10³ g and 1 g = 10³ mg,

$$\begin{aligned}1\ \text{kg} = 10^3 \times 10^3\ \text{mg} \\[1em] = 10^{6}\ \text{mg}\end{aligned}$$

Hence, the kilogram is 10⁶ times larger than the milligram.

The second has been defined in terms of the period of vibration of the caesium-133 atom because a good unit should be precisely defined, easily reproducible and should not change with time, place or physical conditions such as temperature and pressure. The vibrations of the caesium-133 atom satisfy all these conditions, since the fundamental properties of caesium atoms remain constant everywhere. Hence the second so defined is invariant and easily reproducible.

No. In one second there are 10⁶ microseconds, whereas in one year there are

$365 \times 24 \times 60 \times 60 = 3.15 \times 10^{7}\ \text s$

Since 3.15 × 10⁷ is greater than 10⁶, the number of seconds in a year is greater than the number of microseconds in a second.

Since 1 millisecond = 10^-3 s and 1 microsecond = 10^-6 s,

$$\begin{aligned}\frac{1\ \text{millisecond}}{1\ \text{microsecond}} = \frac{10^{-3}\ \text s}{10^{-6}\ \text s} \\[1em] = 10^{3}\end{aligned}$$

Hence, a millisecond is 10³ times greater than a microsecond.

No, the inertial mass and the gravitational mass of an object are not different. They represent different physical concepts — the inertial mass is a measure of the inertia of the body, while the gravitational mass determines the gravitational force on it — but experimentally they are found to be equal in magnitude.

The atomic clock is the most accurate clock. The latest hydrogen maser clock is accurate to 1 s in 3 × 10⁷ years.

The number of significant figures remains unchanged, that is, it is the same as in the original measurement. This is because pure numbers are not obtained by measurement and have unlimited accuracy, so they are not counted while deciding the number of significant figures in the result.

The accuracy of a result obtained by calculation can never be greater than the accuracy of the original measurements. Therefore the number of significant figures in the final result should not be more than the number of significant figures in the least accurate quantity given, and the non-significant figures are dropped by rounding off.

The measurement 2.0 cm has two significant figures and is accurate up to the first place of decimal only, whereas the measurement 2.000 cm has four significant figures and is accurate up to the third place of decimal. Since the measurement having the maximum number of significant figures is the most accurate, 2.000 cm is the more accurate measurement of the two.

Energy and torque (moment of force). Both have the dimensional formula [ML²T^-2], which is the same as that of work.

No, a mechanical quantity has the same dimensions in all systems of units. The dimensions depend only on the nature of the quantity and not on the units chosen; only the units in which the quantity is expressed may change.

Yes. The plane angle and the solid angle have the units radian (rad) and steradian (sr) respectively, yet both are dimensionless quantities, since each is the ratio of two quantities of the same kind.

No. If a quantity has dimensions, it must also have a unit. The dimensions show which base quantities are contained in the quantity, and every base quantity has a unit, so the quantity is necessarily measured in a unit derived from those base units.

No, the magnitude of a dimensionless quantity does not depend upon the system of units used. A dimensionless quantity is a pure number having no units, so its numerical value remains the same in every system of units.

The torr is a practical unit of pressure. One torr is the pressure exerted by a mercury column of height 1 mm, that is, 1 torr = 1 mm of mercury.

The order of the mass of our universe is 10⁵⁵ kg.

Spring balance.

The order of the age of the earth is 10¹⁷ s.

Random errors occur irregularly and are equally likely to be positive or negative. They can therefore be minimised by taking a large number of readings of the same quantity and then taking their arithmetic mean, since the positive and negative errors tend to cancel one another.

Angular displacement (angle) is the ratio of the arc length to the radius, that is, the ratio of two lengths. Hence it is dimensionless and its dimensional formula is [M⁰L⁰T⁰].

Avogadro's number is a dimensionless constant. Other examples of dimensionless constants are the pure numbers π and e.

$$\begin{aligned}0.000003\ \text{kg} = \frac{3}{10^{6}}\ \text{kg} \\[1em] = 3 \times 10^{-6}\ \text{kg}\end{aligned}$$

(a) Radius of the earth,

(b) Height of Mount Everest,

(c) Distance travelled by sound in air in 3 s,

(d) Length of a playground,

(e) Thickness of a cardboard,

(f) Mean free path of an air molecule,

(g) Size of an atomic nucleus.

(i) Two — the initial zeros are not significant, while the digits 5 and 0 are significant.

(ii) One — the initial zeros are not significant, so only the digit 9 is significant.

(iii) Two — the initial zero is not significant, so the digits 3 and 9 are significant.

(iv) Three — the digits 6, 3 and 7 are significant; the power of ten is not counted.

(v) Four — the initial zeros are not significant, while the digits 9, 2, 0 and 3 are significant.

(vi) Three — the digits 1, 9 and 9 are significant.

(vii) Two — the digits 9 and 1 are significant.

(viii) Three — the digits 3, 0 and 8 are significant, the zero lying between two non-zero digits.

(ix) Three — the digits 2, 9 and 9 are significant.

(x) Three — the digits 1, 6 and 0 are significant, the trailing zero after the decimal point being significant.

(i) Velocity → [M⁰LT^-1]

Force → [MLT^-2]

Momentum → [MLT^-1]

Energy → [ML²T^-2]

Surface tension → [ML⁰T^-2]

(ii) Impulse → [MLT^-1]

Moment of force → [ML²T^-2]

Angle → [M⁰L⁰T⁰]

(iii) Pressure → [ML^-1T^-2]

Kinetic energy → [ML²T^-2]

Potential energy → [ML²T^-2]

Frequency → [M⁰L⁰T^-1]

Strain → [M⁰L⁰T⁰]

(iv) Work → [ML²T^-2]

Acceleration → [M⁰LT^-2]

Gravitational constant (G) → [M^-1L³T^-2]

(v) Stress → [ML^-1T^-2]

Power → [ML²T^-3]

(vi) Moment of inertia → [ML²T⁰]

(vii) Young's modulus of elasticity or Modulus of rigidity → [ML^-1T^-2]

(viii) Angular momentum → [ML²T^-1]

(ix) Velocity gradient → [M⁰L⁰T^-1]

(x) Force constant → [ML⁰T^-2]

(xi) Coefficient of viscosity → [ML^-1T^-1]

(xii) Gravitational potential → [M⁰L²T^-2]

(xiii) Gravitational potential energy → [ML²T^-2]

(xiv) Latent heat → [M⁰L²T^-2]

(xv) Specific heat → [M⁰L²T^-2Θ^-1]

(xvi) Coefficient of thermal conductivity → [MLT^-3Θ^-1]

(xvii) Boltzmann's constant → [ML²T^-2Θ^-1]

(xviii) Gas constant → [ML²T^-2Θ^-1 mol^-1]

(xix) Planck's constant → [ML²T^-1]

(i) 1.01 — the digit to be dropped is 8, which is more than 5, so the preceding digit 0 is increased by 1.

(ii) 37.0 — the digit to be dropped is 9, which is more than 5, so the preceding digit 9 is increased by 1, giving 37.0.

(iii) 0.00514 — the digit to be dropped is 5, and the preceding digit 3 is odd, so it is increased by 1.

(iv) 6.22 — the digit to be dropped is 5, and the preceding digit 2 is even, so it is retained unchanged.

(v) 0.0383 — the digit to be dropped is 8, which is more than 5, so the preceding digit 2 is increased by 1.

(vi) 3.15 × 10⁷ — the digit to be dropped is 3, which is less than 5, so the preceding digit 5 is retained unchanged.

Given,

• X × velocity = power

Writing the dimensional formulae of the known quantities,

$[\text{Power}] = [\text{ML}^2\text{T}^{-3}],\quad [\text{Velocity}] = [\text{LT}^{-1}]$

Therefore,

$$\begin{aligned}[\text X] = \frac{[\text{Power}]}{[\text{Velocity}]} \\[1em] = \frac{[\text{ML}^2\text{T}^{-3}]}{[\text{LT}^{-1}]} \\[1em] = [\text{MLT}^{-2}]\end{aligned}$$

This is the dimensional formula of force.

Hence, the unknown quantity X is force.

The orders of magnitude are :

• Size of our galaxy ⟶ 10²⁰ m
• Height of an average man ⟶ 10⁰ m
• Mean free path of an air molecule ⟶ 10^-6 m

3.000 cm is the most accurate measurement. The measurement having the maximum number of significant figures is the most accurate, and 3.000 cm has four significant figures whereas 3.0 cm has two and 3.00 cm has three.

2000 g expresses the most accurate measurement. The zeros to the right of a non-zero digit are significant when they arise due to measurement, so 2000 g has four significant figures, whereas 2.00 kg has three and 2.0 × 10³ g has only two. Since the measurement having the maximum number of significant figures is the most accurate, 2000 g is the most accurate of the three.

Given,

• f = x²

When a quantity is raised to a power, its fractional error is multiplied by that power. For f = x², taking the fractional error on both sides,

$\frac{\Delta \text f}{\text f} = 2 \times \frac{\Delta \text x}{\text x}$

Hence, the fractional error in f is 2 times the fractional error in x.

The digit 0 may or may not be significant, depending on its position in the number.

The zero is significant when it lies between two non-zero digits (as in 60.5 g), and also when it lies to the right of the decimal point after a non-zero digit, since it then arises due to actual measurement (as in 6.50 cm).

The zero is not significant when it only fixes the position of the decimal point, that is, when it lies to the right of the decimal point but to the left of the first non-zero digit (as in 0.065 m).

The quantity occurring with the highest power, say n, in the formula should be measured with the maximum accuracy. This is because the fractional error in such a quantity is multiplied by n in the final result, so any error in its measurement is magnified n times.

The measurement occurring with the highest power in the formula is responsible for the maximum error, since its fractional error is multiplied by that power. If all the quantities occur with equal powers, then the measurement smallest in magnitude is responsible for the maximum error, because for the same absolute error the fractional error is greatest for the smallest measurement.

The dimensional formula of Planck's constant is [ML²T^-1]. Comparing with the given quantities,

• Torque → [ML²T^-2]
• Work → [ML²T^-2]
• Angular momentum → [ML²T^-1]
• Coefficient of viscosity → [ML^-1T^-1]

Hence, angular momentum has the same dimensions as Planck's constant.

Yes, variables can be dimensionless. For example, angle, strain and specific gravity (relative density) are dimensionless variables, since each of them is the ratio of two quantities of the same kind.

Work is defined as the product of force and displacement,

$\text{Work} = \text{Force} \times \text{displacement}$

The force is the product of mass and acceleration,

$\text{Force} = \text{mass} \times \text{acceleration}$

Writing the units of each quantity in terms of the fundamental units,

• Unit of mass = kg
• Unit of acceleration = m s^-2
• Unit of displacement = m

Therefore, the SI unit of work is

$$\begin{aligned}\text{Unit of work} = (\text{kg} \times \text{m s}^{-2}) \times \text{m} \\[1em] = \text{kg m}^2 \text{ s}^{-2}\end{aligned}$$

This unit is called the joule (J).

Hence, the SI unit of work in terms of the fundamental units is kg m² s^-2.

Given,

• Volume occupied by 1 mole of air at NTP = 22.4 L = 22.4 × 10^-3 m³
• Number of molecules in 1 mole (Avogadro's number), N_A = 6.023 × 10²³

At NTP, 22.4 × 10^-3 m³ of air contains 6.023 × 10²³ molecules. Hence the number of molecules contained in 1 m³ of air is

$\frac{6.023 \times 10^{23}}{22.4 \times 10^{-3}}$

If the length, breadth and height of the room are measured, its volume V is obtained, and the number of air molecules in the room at NTP is

$\text N = \frac{6.023 \times 10^{23}}{22.4 \times 10^{-3}} \times \text V$

Hence, the number of air molecules in a room of volume V m³ at NTP is $\frac{6.023 \times 10^{23}}{22.4 \times 10^{-3}} \times \text V$.

By the principle of homogeneity of dimensions, the dimensions of both sides of a correct formula must be the same, and only quantities of the same dimensions can be added to one another.

The dimensional formula of kinetic energy is

$[\text E] = [\text{ML}^2\text{T}^{-2}]$

Writing the dimensions of the right hand side of each formula,

(i) $[\text m^2\text v^3] = [\text M^2][\text{LT}^{-1}]^3 = [\text M^2\text L^3\text T^{-3}]$ — different from [ML²T^-2], so this formula is wrong.

(ii) $\left[\frac{1}{2}\text{mv}^2\right] = [\text M][\text{LT}^{-1}]^2 = [\text{ML}^2\text{T}^{-2}]$ — same as [ML²T^-2], so this formula is dimensionally correct.

(iii) $[\text{ma}] = [\text M][\text{LT}^{-2}] = [\text{MLT}^{-2}]$ — different from [ML²T^-2], so this formula is wrong.

(iv) $\left[\frac{3}{16}\text{mv}^2\right] = [\text M][\text{LT}^{-1}]^2 = [\text{ML}^2\text{T}^{-2}]$ — same as [ML²T^-2], so this formula is dimensionally correct.

(v) Here the term $\frac{1}{2}\text{mv}^2$ has the dimension [ML²T^-2] while the term ma has the dimension [MLT^-2]. Two quantities of different dimensions cannot be added, so this formula is wrong.

Hence, the formulae (i), (iii) and (v) are ruled out on dimensional grounds.

Dimensional analysis cannot decide between (ii) and (iv), since both are dimensionally correct; it cannot determine the value of the dimensionless constant. From the definition of kinetic energy, the correct formula is (ii), that is, E = $\frac{1}{2}$mv².

Given,

• s = ct³

By the principle of homogeneity of dimensions, the dimensions of both sides must be the same,

$[\text s] = [\text c][\text t^3]$

The dimensional formula of displacement is [L] and that of time is [T]. Therefore,

$$\begin{aligned}[\text L] = [\text c][\text T^3] \\[1em] \Rightarrow [\text c] = \frac{[\text L]}{[\text T^3]} \\[1em] = [\text{LT}^{-3}]\end{aligned}$$

Hence, the dimensional formula for c is [LT^-3].

Given,

• A + B = C
• [A] = [C] = [ML^-1T^-2]

By the principle of homogeneity of dimensions, only quantities having the same dimensions can be added to one another. Since B is added to A, it must have the same dimensions as A.

Hence, the dimensions of B are [M L^-1 T^-2].

Given,

• $\text v = \text{at} + \frac{\text b}{\text t}$

By the principle of homogeneity of dimensions, each term on the right hand side must have the same dimensions as the velocity v,

$[\text v] = [\text{at}] = \left[\frac{\text b}{\text t}\right] = [\text{LT}^{-1}]$

For a :

$$\begin{aligned}[\text a][\text T] = [\text{LT}^{-1}] \\[1em] \Rightarrow [\text a] = [\text{LT}^{-1}][\text T^{-1}] \\[1em] = [\text{LT}^{-2}]\end{aligned}$$

For b :

$$\begin{aligned}\frac{[\text b]}{[\text T]} = [\text{LT}^{-1}] \\[1em] \Rightarrow [\text b] = [\text{LT}^{-1}][\text T] \\[1em] = [\text L]\end{aligned}$$

Hence, the dimensional formula for a is [LT^-2] and for b is [L].

Given,

• $\text v = \text a + \text{bt} + \frac{\text c}{\text d + \text t}$

By the principle of homogeneity of dimensions, each term on the right hand side must have the same dimensions as the velocity v, and only quantities of the same dimensions can be added to one another.

For a : Since a is added to the other terms which have the dimensions of velocity,

$[\text a] = [\text v] = [\text{LT}^{-1}]$

For b :

$$\begin{aligned}[\text b][\text T] = [\text{LT}^{-1}] \\[1em] \Rightarrow [\text b] = [\text{LT}^{-2}]\end{aligned}$$

For d : Since d is added to the time t,

$[\text d] = [\text t] = [\text T]$

For c :

$$\begin{aligned}\frac{[\text c]}{[\text d + \text t]} = [\text{LT}^{-1}] \\[1em] \Rightarrow [\text c] = [\text{LT}^{-1}][\text T] \\[1em] = [\text L]\end{aligned}$$

Hence, [a] = [LT^-1], [b] = [LT^-2], [c] = [L] and [d] = [T].

Given,

• PV = RT

By the principle of homogeneity of dimensions,

$[\text P][\text V] = [\text R][\text T]$

The dimensional formulae of pressure, volume and thermodynamic temperature are [ML^-1T^-2], [L³] and [θ] respectively. Therefore,

$$\begin{aligned}[\text{ML}^{-1}\text{T}^{-2}][\text L^3] = [\text R][\theta] \\[1em] \Rightarrow [\text{ML}^2\text{T}^{-2}] = [\text R][\theta] \\[1em] \Rightarrow [\text R] = [\text{ML}^2\text{T}^{-2}\theta^{-1}]\end{aligned}$$

Hence, the dimensional formula for R is [M L² T^-2 θ^-1].

Given,

• v = At² + Bt + C, where v is in m s^-1 and t is in s

By the principle of homogeneity, only quantities of the same kind can be added, so each term on the right hand side must have the same unit as the velocity v, that is, m s^-1.

For A :

$$\begin{aligned}\text{Unit of A} \times \text s^2 = \text{m s}^{-1} \\[1em] \Rightarrow \text{Unit of A} = \text{m s}^{-1} \times \text s^{-2} = \text{m s}^{-3}\end{aligned}$$

For B :

$$\begin{aligned}\text{Unit of B} \times \text s = \text{m s}^{-1} \\[1em] \Rightarrow \text{Unit of B} = \text{m s}^{-1} \times \text s^{-1} = \text{m s}^{-2}\end{aligned}$$

For C : Since C is added directly to the other terms,

$\text{Unit of C} = \text{m s}^{-1}$

Hence, the unit of A is m s^-3, the unit of B is m s^-2 and the unit of C is m s^-1.

Given,

• $\frac{1}{2}\text{mv}^2 = \text h\nu - \text W$

By the principle of homogeneity, only quantities of the same kind can be added to or subtracted from one another, and both sides of the equation must represent the same physical quantity.

The left hand side, $\frac{1}{2}$mv², is the kinetic energy of the emitted electron, and hν is the energy of the incident photon. Hence W, which is subtracted from hν, must also be an energy — it is the work function of the metal.

Hence, the unit of W is the joule (J).

Let the speed of sound v depend upon the modulus of volume elasticity E and the density d as

$$\begin{aligned}\text v \propto \text E^\text a\ \text d^\text b \\[1em] \Rightarrow \text v = \text k\ \text E^\text a\ \text d^\text b \quad \text{...............(i)}\end{aligned}$$

where k is a dimensionless constant.

Writing the dimensions of both sides,

$$\begin{aligned}[\text M^0\text{LT}^{-1}] = [\text{ML}^{-1}\text{T}^{-2}]^\text a\ [\text{ML}^{-3}\text T^{0}]^\text b \\[1em] = [\text M^{\text a + \text b}\ \text L^{-\text a - 3\text b}\ \text T^{-2\text a}]\end{aligned}$$

Equating the powers of M, L and T on both sides,

• a + b = 0
• −a − 3b = 1
• −2a = −1

From the third equation, a = $\frac{1}{2}$, and from the first equation, b = −a = −$\frac{1}{2}$.

Substituting these values in equation (i),

$$\begin{aligned}\text v = \text k\ \text E^{1/2}\ \text d^{-1/2} \\[1em] = \text k\sqrt{\frac{\text E}{\text d}}\end{aligned}$$

Experimentally, the value of k is found to be 1.

Hence, the speed of sound in a medium is $\text v = \sqrt{\frac{\text E}{\text d}}$.

Let the velocity v of the transverse wave depend upon the tension F and the mass per unit length m as

$$\begin{aligned}\text v \propto \text F^\text a\ \text m^\text b \\[1em] \Rightarrow \text v = \text k\ \text F^\text a\ \text m^\text b \quad \text{...............(i)}\end{aligned}$$

where k is a dimensionless constant.

Writing the dimensions of both sides,

$$\begin{aligned}[\text M^0\text{LT}^{-1}] = [\text{MLT}^{-2}]^\text a\ [\text{ML}^{-1}\text T^{0}]^\text b \\[1em] = [\text M^{\text a + \text b}\ \text L^{\text a - \text b}\ \text T^{-2\text a}]\end{aligned}$$

Equating the powers of M, L and T on both sides,

• a + b = 0
• a − b = 1
• −2a = −1

From the third equation, a = $\frac{1}{2}$, and from the first equation, b = −a = −$\frac{1}{2}$.

Substituting these values in equation (i),

$$\begin{aligned}\text v = \text k\ \text F^{1/2}\ \text m^{-1/2} \\[1em] = \text k\sqrt{\frac{\text F}{\text m}}\end{aligned}$$

Experimentally, the value of k is found to be 1.

Hence, the velocity of a transverse wave in a stretched wire is $\text v = \sqrt{\frac{\text F}{\text m}}$.

Let the time period T of the drop depend upon the density d, the radius r and the surface tension S as

$$\begin{aligned}\text T \propto \text d^\text a\ \text r^\text b\ \text S^\text c \\[1em] \Rightarrow \text T = \text k\ \text d^\text a\ \text r^\text b\ \text S^\text c \quad \text{...............(i)}\end{aligned}$$

where k is a dimensionless constant.

Writing the dimensions of both sides,

$$\begin{aligned}[\text M^0\text L^0\text T] = [\text{ML}^{-3}\text T^{0}]^\text a\ [\text M^0\text{LT}^{0}]^\text b\ [\text{ML}^{0}\text T^{-2}]^\text c \\[1em] = [\text M^{\text a + \text c}\ \text L^{-3\text a + \text b}\ \text T^{-2\text c}]\end{aligned}$$

Equating the powers of M, L and T on both sides,

• a + c = 0
• −3a + b = 0
• −2c = 1

From the third equation, c = −$\frac{1}{2}$.

From the first equation, a = −c = $\frac{1}{2}$.

From the second equation, b = 3a = $\frac{3}{2}$.

Substituting these values in equation (i),

$$\begin{aligned}\text T = \text k\ \text d^{1/2}\ \text r^{3/2}\ \text S^{-1/2} \\[1em] = \text k\sqrt{\frac{\text{dr}^3}{\text S}}\end{aligned}$$

Experimentally, the value of k is found to be 1.

Hence, the time period of the drop is $\text T = \sqrt{\frac{\text{dr}^3}{\text S}}$.

Let the force F depend upon the radius r, the terminal velocity v and the coefficient of viscosity η as

$$\begin{aligned}\text F \propto \text r^\text a\ \text v^\text b\ \eta^\text c \\[1em] \Rightarrow \text F = \text k\ \text r^\text a\ \text v^\text b\ \eta^\text c \quad \text{...............(i)}\end{aligned}$$

where k is a dimensionless constant.

Writing the dimensions of both sides,

$$\begin{aligned}[\text{MLT}^{-2}] = [\text M^0\text{LT}^{0}]^\text a\ [\text M^0\text{LT}^{-1}]^\text b\ [\text{ML}^{-1}\text T^{-1}]^\text c \\[1em] = [\text M^{\text c}\ \text L^{\text a + \text b - \text c}\ \text T^{-\text b - \text c}]\end{aligned}$$

Equating the powers of M, L and T on both sides,

• c = 1
• a + b − c = 1
• −b − c = −2, that is, b + c = 2

From the third equation, b = 2 − c = 2 − 1 = 1.

From the second equation, a = 1 − b + c = 1 − 1 + 1 = 1.

Substituting these values in equation (i),

$$\begin{aligned}\text F = \text k\ \text r\ \text v\ \eta \\[1em] = \text k\ \eta\ \text{rv}\end{aligned}$$

Experimentally, the value of k is found to be 6π.

Hence, the force acting on the ball is F = 6πηrv, which is Stokes' law.

Given,

• T ∝ P^a d^b E^c, that is, T = k P^a d^b E^c, where k is a dimensionless constant

Writing the dimensions of both sides,

$$\begin{aligned}[\text M^0\text L^0\text T] = [\text{ML}^{-1}\text T^{-2}]^\text a\ [\text{ML}^{-3}\text T^{0}]^\text b\ [\text{ML}^{2}\text T^{-2}]^\text c \\[1em] = [\text M^{\text a + \text b + \text c}\ \text L^{-\text a - 3\text b + 2\text c}\ \text T^{-2\text a - 2\text c}]\end{aligned}$$

Equating the powers of M, L and T on both sides,

• a + b + c = 0   ...............(i)
• −a − 3b + 2c = 0   ...............(ii)
• −2a − 2c = 1   ...............(iii)

From equation (iii),

$\text a + \text c = -\frac{1}{2}$

Substituting this in equation (i),

$\text b = -(\text a + \text c) = \frac{1}{2}$

Substituting b = $\frac{1}{2}$ in equation (ii),

$$\begin{aligned}-\text a - \frac{3}{2} + 2\text c = 0 \\[1em] \Rightarrow -\text a + 2\text c = \frac{3}{2}\end{aligned}$$

Also, from equation (iii), a = $-\frac{1}{2} - \text c$. Substituting this,

$$\begin{aligned}\frac{1}{2} + \text c + 2\text c = \frac{3}{2} \\[1em] \Rightarrow 3\text c = 1 \\[1em] \Rightarrow \text c = \frac{1}{3}\end{aligned}$$

Therefore,

$\text a = -\frac{1}{2} - \frac{1}{3} = -\frac{5}{6}$

Hence, a = $-\frac{5}{6}$, b = $\frac{1}{2}$ and c = $\frac{1}{3}$.

Given,

$\text Z = \frac{\text A^4 \text B^{1/3}}{\text {CD}^{3/2}}$

When a quantity is expressed as a product or quotient of powers of measured quantities, the maximum fractional error in it is obtained by adding the fractional 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 4, $\frac{1}{3}$, 1 and $\frac{3}{2}$ respectively.

Therefore, the maximum fractional error in Z is

$\frac{\Delta \text Z}{\text Z} = 4\frac{\Delta \text A}{\text A} + \frac{1}{3}\frac{\Delta \text B}{\text B} + \frac{\Delta \text C}{\text C} + \frac{3}{2}\frac{\Delta \text D}{\text D}$

Hence, the maximum fractional error in Z is $\frac{\Delta \text Z}{\text Z} = 4\frac{\Delta \text A}{\text A} + \frac{1}{3}\frac{\Delta \text B}{\text B} + \frac{\Delta \text C}{\text C} + \frac{3}{2}\frac{\Delta \text D}{\text D}$.

A dimensional quantity has a definite unit, and its measured value depends on the unit chosen. Calling such a quantity large or small has no meaning by itself, because the same quantity may appear large when compared with one standard and small when compared with another. A quantity can be called large or small only when it is compared with a standard (or reference) quantity of the same kind.

For example : The statement "the mass of the earth is very large" is meaningless as it stands. The mass of the earth is very large compared with the mass of a ship, but it is very small compared with the mass of the sun.

The statement becomes meaningful when a standard of comparison is specified, for example : The mass of the earth is larger than the mass of the moon.

The different measurements give slightly different readings because of random errors, which arise from causes such as :

(i) the rod may not be perfectly uniform in diameter at all places, and

(ii) the rod may be held between the stud and the screw with a different pressure in each measurement.

Random errors occur irregularly and are equally likely to be positive or negative, so they are minimised by taking a large number of readings of the same quantity and then taking their arithmetic mean, since the positive and negative errors tend to cancel one another.

Hence, the mean of ten measurements is closer to the actual value of the diameter than the mean of five measurements.

Any phenomenon which repeats itself regularly can be used as a standard of time. Some such repetitive phenomena are :

• the oscillations of a simple pendulum
• the oscillations of a loaded spring
• the electrically sustained vibrations of a quartz crystal
• the characteristic frequency of the radiation emitted by certain atoms
• the rotation of the earth about its own axis

The axial rotation of the earth was used as the standard of time for centuries.

The most suitable standard is the atomic clock. The caesium (atomic) clock and the hydrogen maser clock are the most precise clocks available. Two hydrogen maser clocks could run for 30,000,000 years before their readings would differ by 1 s. The second is therefore now defined in terms of the vibrations of the caesium-133 atom, since this standard is precisely defined, easily reproducible and does not change with time, place or physical conditions.

Given,

• Clock A : average time 11 : 59 AM, range of variation 162 s
• Clock B : average time 10 AM, range of variation 31 s

The average reading of clock A is closer to the standard time, so clock A is the more accurate of the two. However, its readings vary over a range of 162 s, whereas the readings of clock B vary over a range of only 31 s. Hence clock B is the more precise of the two.

For measuring time-intervals precisely, what matters is the consistency of the readings and not the difference from the standard time, because the constant difference of clock B is a systematic (zero) error which follows a definite rule and can be corrected by applying a proper correction. The larger variation of clock A is a random error, which cannot be removed in this way.

Hence, clock B is to be preferred for an experiment requiring precision time-interval measurements.

The accuracy of a measurement is judged from its fractional (relative) error,

$\text{Fractional error} = \frac{\text{Absolute error}}{\text{Measured value}}$

For the measurement 2.5 m : the measurement is made up to one place of decimal, so the absolute error is 0.1 m,

$\frac{\Delta \text l}{\text l} = \frac{0.1}{2.5} = 0.04$

For the measurement 2.54 m : the measurement is made up to two places of decimal, so the absolute error is 0.01 m,

$\frac{\Delta \text l}{\text l} = \frac{0.01}{2.54} = 0.004$

Hence, the measurement 2.54 m is more accurate, since it has the smaller fractional error.

The accuracy of a measurement is judged from its fractional (relative) error,

$\text{Fractional error} = \frac{\text{Absolute error}}{\text{Measured value}}$

5.00 cm — the measurement is made up to two places of decimal, so the absolute error is 0.01 cm,

$\frac{0.01}{5.00} = 0.002$

0.005 mm — the measurement is made up to three places of decimal, so the absolute error is 0.001 mm,

$\frac{0.001}{0.005} = 0.2$

50.00 cm — the measurement is made up to two places of decimal, so the absolute error is 0.01 cm,

$\frac{0.01}{50.00} = 0.0002$

Hence, the measurement 50.00 cm is the most accurate, since it has the smallest fractional error.

(i) When F, L and T are taken as the fundamental quantities.

From Newton's second law, force is the product of mass and acceleration,

$$\begin{aligned}\text F = \text m \times \text a \\[1em] \Rightarrow \text m = \frac{\text F}{\text a}\end{aligned}$$

The dimensional formula of acceleration is [LT^-2]. Therefore,

$$\begin{aligned}[\text M] = \frac{[\text F]}{[\text{LT}^{-2}]} = [\text F][\text L^{-1}\text T^{2}] \\[1em] = [\text F\ \text L^{-1}\ \text T^{2}] \quad \text{...............(i)}\end{aligned}$$

Hence, the dimensions of mass are [F L^-1 T²].

(ii) When E, L and T are taken as the fundamental quantities.

Energy is the work done, which is the product of force and distance,

$$\begin{aligned}\text E = \text F \times \text L \\[1em] \Rightarrow \text F = \frac{\text E}{\text L}\end{aligned}$$

so that

$[\text F] = [\text E][\text L^{-1}] = [\text E\ \text L^{-1}]$

Substituting this in equation (i),

$$\begin{aligned}[\text M] = [\text E\ \text L^{-1}][\text L^{-1}\text T^{2}] \\[1em] = [\text E\ \text L^{-2}\ \text T^{2}]\end{aligned}$$

Hence, the dimensions of mass are [E L^-2 T²].

Given,

• [L] = [G^x c^y h^z]

The dimensional formulae of the three constants are

• Universal gravitational constant, [G] = [M^-1L³T^-2]
• Speed of light in vacuum, [c] = [LT^-1]
• Planck's constant, [h] = [ML²T^-1]

Therefore,

$$\begin{aligned}[\text G^\text x\ \text c^\text y\ \text h^\text z] = [\text M^{-1}\text L^{3}\text T^{-2}]^\text x\ [\text{LT}^{-1}]^\text y\ [\text{ML}^{2}\text T^{-1}]^\text z \\[1em] = [\text M^{-\text x + \text z}\ \text L^{3\text x + \text y + 2\text z}\ \text T^{-2\text x - \text y - \text z}]\end{aligned}$$

Since this represents the dimensions of length,

$[\text M^0\text{LT}^0] = [\text M^{-\text x + \text z}\ \text L^{3\text x + \text y + 2\text z}\ \text T^{-2\text x - \text y - \text z}]$

Equating the powers of M, L and T on both sides,

• −x + z = 0, so z = x
• 3x + y + 2z = 1
• −2x − y − z = 0

Substituting x = z in the third equation,

$$\begin{aligned}-2\text z - \text y - \text z = 0 \\[1em] \Rightarrow \text y = -3\text z\end{aligned}$$

Substituting x = z and y = −3z in the second equation,

$$\begin{aligned}3\text z - 3\text z + 2\text z = 1 \\[1em] \Rightarrow 2\text z = 1 \\[1em] \Rightarrow \text z = \frac{1}{2}\end{aligned}$$

Therefore,

$\text x = \text z = \frac{1}{2}, \quad \text y = -3 \times \frac{1}{2} = -\frac{3}{2}$

Hence, x = $\frac{1}{2}$, y = $-\frac{3}{2}$ and z = $\frac{1}{2}$.

Given,

• Velocity (v), force (F) and time (T) are taken as the fundamental quantities.

Energy is the work done, which is the product of force and distance,

$\text{Energy} = \text{Force} \times \text{distance}$

The distance is the product of velocity and time,

$\text{distance} = \text{velocity} \times \text{time}, \quad \text{that is,}\quad [\text L] = [\text{vT}]$

Therefore,

$$\begin{aligned}[\text E] = [\text F][\text v][\text T] \\[1em] = [\text F\ \text v\ \text T]\end{aligned}$$

Hence, the dimensional formula for energy is [F v T].

The given relation is

$\text T = 2\pi\sqrt{\frac{\text l}{\text g}}$

By the principle of homogeneity of dimensions, the dimensions of both sides of the relation must be the same.

Left hand side. The dimensional formula of the time period is

$[\text{L.H.S.}] = [\text T]$

Right hand side. The factor 2π is a pure number and is dimensionless. The dimensional formulae of the length l and the acceleration due to gravity g are [L] and [LT^-2] respectively. Therefore,

$$\begin{aligned}[\text{R.H.S.}] = \sqrt{\frac{[\text L]}{[\text{LT}^{-2}]}} = \sqrt{\frac{1}{[\text T^{-2}]}} \\[1em] = \sqrt{[\text T^{2}]} = [\text T]\end{aligned}$$

Since the dimensions of the two sides are the same,

$[\text{L.H.S.}] = [\text{R.H.S.}] = [\text T]$

Hence, the equation $\text T = 2\pi\sqrt{\frac{\text l}{\text g}}$ is dimensionally correct.

Given,

• Velocity (v), force (F) and time (T) are taken as the fundamental quantities.

From Newton's second law, force is the product of mass and acceleration,

$\text F = \text m \times \text a$

and acceleration is the rate of change of velocity,

$\text a = \frac{\text v}{\text T}$

Therefore,

$$\begin{aligned}\text F = \text m \times \frac{\text v}{\text T} \\[1em] \Rightarrow \text m = \frac{\text F \times \text T}{\text v}\end{aligned}$$

Writing the dimensions in terms of the new fundamental quantities,

$$\begin{aligned}[\text m] = [\text F][\text T][\text v]^{-1} \\[1em] = [\text F\ \text v^{-1}\ \text T]\end{aligned}$$

Hence, the fundamental formula for mass is [F v^-1 T].

(i) The closeness of a measurement to the true value of a physical quantity is known as its accuracy. An instrument which gives repeated readings close to the true value of the quantity is called an accurate instrument.

(ii) The accuracy of a measurement depends mainly on the systematic errors present in the instrument, such as the calibration error and the zero error. Since these errors follow a definite rule, they can be identified and removed. Hence the accuracy can be increased by re-calibrating the instrument and by applying proper corrections to the observed readings.

(iii) The precision of a measurement is determined by the least count of the measuring instrument — the smaller the least count, the greater is the precision. Hence a more precise measurement can be obtained by using an instrument of smaller least count.

(i) In the measurement of a quantity, the digits which are measured accurately together with the first doubtful digit are called the significant figures. Since any measuring instrument can measure accurately only up to a certain limit, the last digit of an observation is always doubtful.

(ii) Given,

• Length of the lamina, l = 5.245 m (4 significant figures)
• Breadth of the lamina, b = 1.24 m (3 significant figures)

The area of the rectangular lamina is

$\text A = \text l \times \text b$

Substituting the values,

$$\begin{aligned}\text A = 5.245 \times 1.24 \\[1em] = 6.5038\ \text m^2\end{aligned}$$

In multiplication, the result is rounded off to the least number of significant figures in the given data, which is 3. Hence

$\text A = 6.50\ \text m^2$

Hence, the area of the lamina is 6.50 m².

(iii) Given,

• Mass of the block, m = 4.50 kg (3 significant figures)
• Volume of the block, V = 1.204 m³ (4 significant figures)

The density of the material of the block is

$\text{Density} = \frac{\text{Mass}}{\text{Volume}}$

Substituting the values,

$$\begin{aligned}\text{Density} = \frac{4.50\ \text{kg}}{1.204\ \text m^3} \\[1em] = 3.7375\ \text{kg m}^{-3}\end{aligned}$$

In division, the result is rounded off to the least number of significant figures in the given data, which is 3. Hence

$\text{Density} = 3.74\ \text{kg m}^{-3}$

Hence, the density of the material of the block is 3.74 kg m^-3.

(i) angular momentum

Reason — Planck's constant is defined by the relation E = hν, where E is the energy and ν is the frequency. Therefore

$\text h = \frac{\text E}{\nu}$

Writing the dimensional formulae,

$[\text h] = \frac{[\text{ML}^2\text{T}^{-2}]}{[\text T^{-1}]} = [\text{ML}^2\text{T}^{-1}]$

The angular momentum is the product of the moment of inertia and the angular velocity, L = Iω, so

$[\text L] = [\text{ML}^2][\text T^{-1}] = [\text{ML}^2\text{T}^{-1}]$

Both have the same dimensional formula [ML²T^-1]. The other options have different dimensions : energy [ML²T^-2], momentum [MLT^-1] and power [ML²T^-3].

(ii) [F V^-1 T]

Reason —

Given,

• Force (F), velocity (v) and time (T) are taken as the fundamental quantities.

From Newton's second law, force is the product of mass and acceleration,

$\text F = \text m \times \text a$

and acceleration is the rate of change of velocity,

$\text a = \frac{\text v}{\text T}$

Therefore,

$$\begin{aligned}\text F = \text m \times \frac{\text v}{\text T} \\[1em] \Rightarrow \text m = \frac{\text F \times \text T}{\text v}\end{aligned}$$

Writing the dimensions in terms of the new fundamental quantities,

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

(i) V

Reason — In the given equation, the constant b is subtracted from the volume V in the bracket (V − b). By the principle of homogeneity of dimensions, only quantities having the same dimensions can be subtracted from one another. Hence

$[\text b] = [\text V] = [\text M^0\text L^3\text T^0]$

(ii) PV²

Reason — In the given equation, the term $\frac{\text a}{\text V^2}$ is added to the pressure P in the bracket. By the principle of homogeneity of dimensions, only quantities having the same dimensions can be added. Hence

$$\begin{aligned}\left[\frac{\text a}{\text V^2}\right] = [\text P] \\[1em] \Rightarrow [\text a] = [\text P][\text V^2]\end{aligned}$$

Therefore the dimensional formula of a is the same as that of PV².

(iii) energy

Reason — Expanding the given equation,

$$\begin{aligned}\left(\text P + \frac{\text a}{\text V^2}\right)(\text V - \text b) = \text{RT} \\[1em] \Rightarrow \text{PV} - \text{Pb} + \frac{\text a}{\text V} - \frac{\text{ab}}{\text V^2} = \text{RT}\quad \text{...............(1)}\end{aligned}$$

By the principle of homogeneity of dimensions, every term on both sides of the equation must have the same dimensions. Hence

$$\begin{aligned}[\text{RT}] = [\text{PV}] = [\text{ML}^{-1}\text{T}^{-2}][\text L^3] \\[1em] = [\text{ML}^2\text{T}^{-2}]\end{aligned}$$

which is the dimensional formula of energy.

(iv) [M⁰ L⁶ T⁰]

Reason — From equation (1) obtained in part (iii), the term $\frac{\text{ab}}{\text V^2}$ appears along with RT, so by the principle of homogeneity of dimensions

$\left[\frac{\text{ab}}{\text V^2}\right] = [\text{RT}]$

Therefore,

$$\begin{aligned}\left[\frac{\text{ab}}{\text{RT}}\right] = [\text V^2] = [\text L^3]^2 \\[1em] = [\text M^0\text L^6\text T^0]\end{aligned}$$

Fundamental units : The units of the fundamental (base) quantities are called fundamental units. The fundamental quantities are those which are independent of one another, that is, none of them can be expressed in terms of the others. In the S.I. system there are seven fundamental units — the metre (m), the kilogram (kg), the second (s), the ampere (A), the kelvin (K), the mole (mol) and the candela (cd).

Derived units : The units of the derived quantities are called derived units. A derived quantity is obtained from a combination of the base quantities through multiplication or division, and hence its unit is obtained from the fundamental units in the same way. Since the S.I. is a coherent system, the derived units are obtained from the fundamental units without introducing any numerical factor.

Examples of derived units :

• Unit of velocity = m s^-1
• Unit of force = newton (N), where 1 N = 1 kg m s^-2
• Unit of work (or energy) = joule (J), where 1 J = 1 kg m² s^-2

Error in measurement : The difference between the measured value of a quantity and its true value is called the error in measurement. However carefully a measurement is made, some error is always present, so no measurement can be perfectly accurate.

The errors in measurement are mainly of two kinds : systematic errors and random errors.

Systematic errors :

The errors which tend to occur in one direction, either positive or negative, are called systematic errors. Such errors follow a definite rule, so they can be identified and corrected.

The main sources of systematic errors are :

(i) Instrumental errors — These arise due to imperfect design or zero error of the measuring instrument, for example a zero error in a screw gauge or a wrongly marked scale.

(ii) Errors due to imperfect technique (or procedure) — These arise due to the limitations of the experimental method, for example neglecting the effect of temperature, pressure or buoyancy of air.

(iii) Personal errors — These arise due to the carelessness or individual bias of the observer, for example an error due to improper setting of the eye while taking a reading (parallax error).

Elimination : Systematic errors can be minimised or removed by improving the design of the instrument, by correcting for the zero error, by using a better experimental technique, and by removing personal bias in taking the observations.

Random errors :

The errors which occur irregularly and are random in magnitude and direction are called random errors. They arise due to unknown and uncontrollable causes, such as small changes in the conditions of the experiment, so they cannot be traced to a definite source.

Because these errors occur equally in the positive and negative directions, they can be reduced by taking a large number of observations of the same quantity and then taking their arithmetic mean. The arithmetic mean is taken as the true value, since the positive and negative errors tend to cancel one another.

Elimination : Random errors can be minimised by repeating the measurement a large number of times and taking the arithmetic mean of all the readings as the most accurate value of the quantity.

Significant figures : In the measurement of a quantity, the digits which are measured accurately together with the first doubtful digit are called the significant figures. Since any measuring instrument can measure accurately only up to a certain limit, the last digit of every observation is always doubtful. The measurement having the maximum number of significant figures is the most accurate.

For example, if the length of the side of a cube read by a vernier callipers is 2.58 cm, the last digit 8 is doubtful, and the measurement has three significant figures.

Rule for addition and subtraction :

In adding or subtracting measured quantities, the result should be rounded off to the same number of decimal places as are contained in the least accurate quantity, that is, the quantity having the least number of decimal places.

Example (addition) : Adding the lengths 27.8 cm, 7.324 cm and 0.66 cm,

$27.8 + 7.324 + 0.66 = 35.784\ \text{cm}$

The first length, 27.8 cm, has only one digit after the decimal point, so the sum is rounded off to one decimal place, giving 35.8 cm.

Example (subtraction) : Subtracting 9.7 cm from 12.192 cm,

$12.192 - 9.7 = 2.492\ \text{cm}$

The second length is known only up to one place of decimal, so the difference is rounded off to one decimal place, giving 2.5 cm.

Rule for multiplication and division :

In multiplying or dividing measured quantities, the result should be rounded off to the same number of significant figures as are contained in the least accurate quantity, that is, the quantity having the least number of significant figures.

Example (multiplication) : If a man runs 200.8 m in 20.6 s, his average speed is

$\text v = \frac{200.8\ \text m}{20.6\ \text s} = 9.74757...\ \text{m s}^{-1}$

The measurement of time has the least number of significant figures, namely three, so the result is rounded off to three significant figures, giving 9.75 m s^-1.

Example (multiplication) : If the diameter of a wire is 2.40 cm, its circumference is

$\pi \text d = 3.142 \times 2.40 = 7.5408\ \text{cm}$

The diameter has only three significant figures, so the result is rounded off to three significant figures, giving 7.54 cm.

Note that pure numbers, which are not obtained by measurement, have unlimited accuracy and are not counted while deciding the number of significant figures in the result.

When the final result depends on a number of measured quantities, the error in each measurement affects the result. The way in which the individual errors combine depends on the mathematical operation involved in the formula.

Let a physical quantity Z be given by

$\text Z = \frac{\text A^\text m \text B^\text n}{\text C^\text p}$

where A, B and C are the measured quantities and m, n and p are their powers.

Taking the logarithm of both sides,

$\text{log Z} = \text m\ \text{log A} + \text n\ \text{log B} - \text p\ \text{log C}$

and differentiating partially, the maximum fractional error in Z is obtained by adding the fractional errors of all the quantities, each multiplied by the magnitude of its power. The sign of the power is not taken into account, since in the worst case all the errors add up. Hence

$\frac{\Delta \text Z}{\text Z} = \text m\frac{\Delta \text A}{\text A} + \text n\frac{\Delta \text B}{\text B} + \text p\frac{\Delta \text C}{\text C}$

The maximum percentage error is obtained by multiplying the maximum fractional error by 100,

$\frac{\Delta \text Z}{\text Z} \times 100 = \text m\left(\frac{\Delta \text A}{\text A} \times 100\right) + \text n\left(\frac{\Delta \text B}{\text B} \times 100\right) + \text p\left(\frac{\Delta \text C}{\text C} \times 100\right)$

Example : If Z = A²B, then

$\frac{\Delta \text Z}{\text Z} = 2\frac{\Delta \text A}{\text A} + \frac{\Delta \text B}{\text B}$

and the percentage error in Z is

$\frac{\Delta \text Z}{\text Z} \times 100 = 2\left(\frac{\Delta \text A}{\text A} \times 100\right) + \left(\frac{\Delta \text B}{\text B} \times 100\right)$

Thus, the percentage error in the result is obtained by adding the percentage errors of all the measured quantities, each multiplied by the magnitude of its power in the formula.

Given,

$\text g = 4\pi^2\frac{\text l}{\text T^2}$

The factor 4π² is a pure number, so it contributes no error to the result. The measured quantities are the length l of the pendulum and its time period T, occurring with the powers 1 and 2 respectively.

The maximum fractional error is obtained by adding the fractional errors of all the measured quantities, each multiplied by the magnitude of its power. The sign of the power is not taken into account, since in the worst case the errors add up. Therefore,

$\frac{\Delta \text g}{\text g} = \frac{\Delta \text l}{\text l} + 2\frac{\Delta \text T}{\text T}$

Hence, the maximum fractional error in the value of g is $\frac{\Delta \text g}{\text g} = \frac{\Delta \text l}{\text l} + 2\frac{\Delta \text T}{\text T}$.

Dimensions of a physical quantity : The dimensions of a physical quantity are the powers to which the base quantities must be raised in order to represent that quantity. The expression which shows which of the base quantities, and with what powers, are contained in a given quantity is called its dimensional formula.

For example, velocity is displacement per unit time, so its dimensional formula is

$[\text v] = \frac{[\text L]}{[\text T]} = [\text M^0\text{LT}^{-1}]$

which shows that velocity contains length to the power 1 and time to the power −1, and does not contain mass.

The dimensions of a quantity depend only on its nature and are independent of the system of units in which it is measured.

Principle of homogeneity of dimensions : According to this principle, the dimensions of all the terms on both sides of a physically meaningful equation must be the same. This is because only quantities of the same kind can be added to, subtracted from or equated with one another.

Example : Consider the equation of motion

$\text s = \text{ut} + \frac{1}{2}\text{at}^2$

The dimensions of each term are

$$\begin{aligned}[\text s] = [\text L] \\[1em] [\text{ut}] = [\text{LT}^{-1}][\text T] = [\text L] \\[1em] \left[\frac{1}{2}\text{at}^2\right] = [\text{LT}^{-2}][\text T^2] = [\text L]\end{aligned}$$

since the factor $\frac{1}{2}$ is a pure number and is dimensionless. As every term has the dimension [L], the equation is dimensionally homogeneous and hence dimensionally correct.

Uses of dimensional equations :

(i) To check the correctness of a physical relation. By the principle of homogeneity of dimensions, every term of a correct relation must have the same dimensions. If the dimensions of the two sides do not agree, the relation is certainly wrong.

Example : In the relation v = u + at, each term has the dimension [LT^-1], so the relation is dimensionally correct.

(ii) To derive the relation between physical quantities. If a quantity depends on certain other quantities, the relation between them can be obtained by writing the quantity as a product of powers of those quantities and equating the dimensions on both sides.

Example : The time period T of a simple pendulum is found to depend on its length l and the acceleration due to gravity g, and dimensional analysis gives $\text T \propto \sqrt{\frac{\text l}{\text g}}$.

(iii) To convert the value of a physical quantity from one system of units to another. Using n₁u₁ = n₂u₂ together with the dimensional formula of the quantity, its numerical value in the new system can be found.

Example : The value of 1 joule in the C.G.S. system is found to be 10⁷ erg.

(iv) To find the dimensions of unknown constants appearing in a relation. If the relation is known, the dimensions of a constant occurring in it can be obtained by the principle of homogeneity.

Example : From the relation E = hν, the dimensional formula of Planck's constant is found to be [ML²T^-1].

Limitations of dimensional analysis :

(i) It cannot determine the value of a dimensionless constant appearing in a relation, such as 2, π or $\frac{1}{2}$.

(ii) It cannot be used for relations involving trigonometric, exponential or logarithmic functions, since their arguments are dimensionless.

(iii) It cannot distinguish between two quantities having the same dimensions, for example work and torque, both of which have the dimensional formula [ML²T^-2].

(iv) It cannot be used to derive a relation which contains more than one term added to or subtracted from another, such as s = ut + $\frac{1}{2}$at².

(v) It cannot be used when the quantity depends on more than three other quantities, since only three equations are obtained from the dimensions of mass, length and time.

(vi) Dimensional correctness is only a necessary condition and not a sufficient condition for a relation to be physically correct.

Dimensional balance of an equation : An equation is said to be dimensionally balanced (or dimensionally homogeneous) when the dimensions of all the terms on the left hand side and the right hand side of the equation are the same. This follows from the principle of homogeneity of dimensions, according to which only quantities of the same kind can be added to, subtracted from or equated with one another.

Checking the correctness of an equation : To check the correctness of a physical relation, the dimensional formula of every term on both sides is written down. If the dimensions of the two sides are the same, the equation is dimensionally correct; if they are different, the equation is certainly wrong.

Example : Consider the relation for the time period of a simple pendulum,

$\text T = 2\pi\sqrt{\frac{\text l}{\text g}}$

Left hand side. The dimensional formula of the time period is

$[\text{L.H.S.}] = [\text T] \quad \text{...............(i)}$

Right hand side. The factor 2π is a pure number and is dimensionless. The dimensional formulae of the length l and the acceleration due to gravity g are [L] and [LT^-2] respectively. Therefore,

$$\begin{aligned}[\text{R.H.S.}] = \sqrt{\frac{[\text L]}{[\text{LT}^{-2}]}} = \sqrt{[\text T^{2}]} \\[1em] = [\text T] \quad \text{...............(ii)}\end{aligned}$$

From (i) and (ii),

$[\text{L.H.S.}] = [\text{R.H.S.}] = [\text T]$

Since the dimensions of both sides are the same, the equation is dimensionally balanced.

Hence, the relation $\text T = 2\pi\sqrt{\frac{\text l}{\text g}}$ is dimensionally correct.

It should be noted that dimensional analysis cannot verify the numerical factor 2π appearing in the relation, since pure numbers are dimensionless. Hence dimensional correctness is only a necessary, and not a sufficient, condition for the relation to be physically correct.

Given,

• Mean wavelength of yellow light = 5893 Å

Since 1 Å = 10^-10 m,

$$\begin{aligned}5893\ \text{Å} = 5893 \times 10^{-10}\ \text m \\[1em] = 5.893 \times 10^{-7}\ \text m\end{aligned}$$

Also, 1 nm = 10^-9 m = 10 × 10^-10 m = 10 Å, so 1 Å = 0.1 nm. Therefore,

$$\begin{aligned}5893\ \text{Å} = 5893 \times 0.1\ \text{nm} \\[1em] = 589.3\ \text{nm}\end{aligned}$$

Hence, 5893 Å = 5.893 × 10^-7 m = 589.3 nm.

Given,

• 1 teragram (Tg) = 10¹² g
• 1 metric ton = 10³ kg = 10⁶ g

Therefore,

$$\begin{aligned}\frac{1\ \text{teragram}}{1\ \text{metric ton}} = \frac{10^{12}\ \text g}{10^{6}\ \text g} \\[1em] = 10^{6}\end{aligned}$$

Hence, 1 teragram = 10⁶ metric tons.

Given,

• Mass of one proton = 1.67 × 10^-27 kg
• Total mass = 1 g = 10^-3 kg

The number of protons is

$$\begin{aligned}\text N = \frac{\text{Total mass}}{\text{Mass of one proton}} \\[1em] = \frac{10^{-3}\ \text{kg}}{1.67 \times 10^{-27}\ \text{kg}} \\[1em] = 0.599 \times 10^{24} \\[1em] = 5.99 \times 10^{23}\end{aligned}$$

To find the order of magnitude, the number is written in the form N × 10^x. Here 5.99 × 10²³, and since 5.99 is greater than $\sqrt{10}$ = 3.16, the order of magnitude is 10²³⁺¹ = 10²⁴.

Hence, the number of protons is 5.99 × 10²³ and its order of magnitude is 10²⁴.

Given,

• Mass of one hydrogen atom = 1.0 u = 1.66 × 10^-27 kg
• Total mass of hydrogen = 1.0 kg

The number of hydrogen atoms is

$$\begin{aligned}\text N = \frac{\text{Total mass}}{\text{Mass of one atom}} \\[1em] = \frac{1.0\ \text{kg}}{1.66 \times 10^{-27}\ \text{kg}} \\[1em] = 0.60 \times 10^{27} \\[1em] = 6.0 \times 10^{26}\end{aligned}$$

Hence, the number of hydrogen atoms required is 6.0 × 10²⁶.

Given,

• Density of water at 4 °C = 1.00 g cm^-3

Since 1 g = 10^-3 kg and 1 cm³ = 10^-6 m³,

$$\begin{aligned}1.00\ \text{g cm}^{-3} = \frac{1.00 \times 10^{-3}\ \text{kg}}{10^{-6}\ \text m^3} \\[1em] = 1.00 \times 10^{3}\ \text{kg m}^{-3}\end{aligned}$$

Hence, the density of water in SI units is 1.00 × 10³ kg m^-3.

Given,

• Acceleration of the body = 10 m s^-2

Since 1 m = 10^-3 km and 1 s = $\frac{1}{3600}$ h,

$$\begin{aligned}10\ \text{m s}^{-2} = \frac{10 \times 10^{-3}\ \text{km}}{\left(\frac{1}{3600}\ \text h\right)^2} \\[1em] = 10^{-2} \times (3600)^2\ \text{km h}^{-2} \\[1em] = 10^{-2} \times 1.296 \times 10^{7}\ \text{km h}^{-2} \\[1em] = 1.296 \times 10^{5}\ \text{km h}^{-2}\end{aligned}$$

Hence, the acceleration of the body is 1.296 × 10⁵ km h^-2.

Given,

• G = 6.67 × 10^-11 N m² kg^-2

Writing the newton in terms of the fundamental units, 1 N = 1 kg m s^-2. Therefore,

$$\begin{aligned}\text G = 6.67 \times 10^{-11}\ (\text{kg m s}^{-2})\ \text m^2\ \text{kg}^{-2} \\[1em] = 6.67 \times 10^{-11}\ \text{kg}^{-1}\ \text m^3\ \text s^{-2}\end{aligned}$$

Since 1 kg = 10³ g and 1 m³ = 10⁶ cm³,

$$\begin{aligned}\text G = 6.67 \times 10^{-11} \times (10^{3}\ \text g)^{-1} \times (10^{6}\ \text{cm}^3)\ \text s^{-2} \\[1em] = 6.67 \times 10^{-11} \times 10^{-3} \times 10^{6}\ \text g^{-1}\ \text{cm}^3\ \text s^{-2} \\[1em] = 6.67 \times 10^{-8}\ \text g^{-1}\ \text{cm}^3\ \text s^{-2}\end{aligned}$$

Hence, the value of G is 6.67 × 10^-8 g^-1 cm³ s^-2.

Given,

• Stefan's constant, σ = 5.67 × 10^-5 erg s^-1 cm^-2 K^-4
• 1 J = 10⁷ erg, so 1 erg = 10^-7 J
• 1 cm = 10^-2 m, so 1 cm^-2 = (10^-2 m)^-2 = 10⁴ m^-2

To convert the value into SI units, we replace the erg by joule and the cm^-2 by m^-2, while the units s^-1 and K^-4 remain unchanged.

$\sigma = 5.67 \times 10^{-5}\ \frac{\text{erg}}{\text{s cm}^2\text{ K}^4}$

Substituting 1 erg = 10^-7 J and 1 cm^-2 = 10⁴ m^-2,

$$\begin{aligned}\sigma = 5.67 \times 10^{-5} \times (10^{-7}\ \text{J}) \times (10^{4}\ \text{m}^{-2})\ \text{s}^{-1}\text{K}^{-4} \\[1em] = 5.67 \times 10^{-5} \times 10^{-7} \times 10^{4}\ \text{J s}^{-1}\text{m}^{-2}\text{K}^{-4} \\[1em] = 5.67 \times 10^{-5 - 7 + 4}\ \text{J s}^{-1}\text{m}^{-2}\text{K}^{-4} \\[1em] = 5.67 \times 10^{-8}\ \text{J s}^{-1}\text{m}^{-2}\text{K}^{-4}\end{aligned}$$

Since 1 J s^-1 = 1 W, this may also be written as

$\sigma = 5.67 \times 10^{-8}\ \text{W m}^{-2}\text{K}^{-4}$

Hence, the value of Stefan's constant in SI units is 5.67 × 10^-8 W m^-2 K^-4.

Given,

• Speed of light, c = 3.0 × 10⁸ m s^-1
• 1 AU = 1.50 × 10⁸ km = 1.50 × 10¹¹ m

Since 1 AU = 1.50 × 10¹¹ m,

$1\ \text m = \frac{1}{1.50 \times 10^{11}}\ \text{AU}$

and since 1 min = 60 s,

$1\ \text s^{-1} = 60\ \text{min}^{-1}$

Therefore,

$$\begin{aligned}\text c = 3.0 \times 10^{8}\ \text{m s}^{-1} \\[1em] = 3.0 \times 10^{8} \times \frac{1}{1.50 \times 10^{11}}\ \text{AU} \times 60\ \text{min}^{-1} \\[1em] = \frac{3.0 \times 60}{1.50 \times 10^{3}}\ \text{AU min}^{-1} \\[1em] = \frac{180}{1500}\ \text{AU min}^{-1} \\[1em] = 0.12\ \text{AU min}^{-1}\end{aligned}$$

Hence, the speed of light is 0.12 AU min^-1.

Given,

• Length of the first rod, l₁ = (35.2 ± 0.1) cm
• Length of the second rod, l₂ = (16.8 ± 0.2) cm

When two quantities are added, the limiting error in the final result is the sum of the absolute errors in the quantities involved. Hence

$$\begin{aligned}\text l = \text l_1 + \text l_2 = 35.2 + 16.8 = 52.0\ \text{cm} \\[1em] \Delta \text l = \Delta \text l_1 + \Delta \text l_2 = 0.1 + 0.2 = 0.3\ \text{cm}\end{aligned}$$

Hence, the sum of the lengths of the two rods is (52.0 ± 0.3) cm.

Given,

• Initial temperature = (30.6 ± 0.2) °C
• Final temperature = (68.3 ± 0.1) °C

When two quantities are subtracted, the limiting error in the final result is again the sum of the absolute errors in the quantities involved. Hence

$$\begin{aligned}\text{Rise in temperature} = 68.3 - 30.6 = 37.7\ ^{\circ}\text C \\[1em] \text{Error} = 0.1 + 0.2 = 0.3\ ^{\circ}\text C\end{aligned}$$

Hence, the rise in temperature is (37.7 ± 0.3) °C.

Given,

• Radius of the sphere, r = 5.3 cm
• Absolute error in the radius, Δr = 0.1 cm

The volume of a sphere is

$\text V = \frac{4}{3}\pi \text r^3$

The factor $\frac{4}{3}\pi$ is a pure number and contributes no error. Since the radius occurs with the power 3, the maximum percentage error in the volume is

$\frac{\Delta \text V}{\text V} \times 100 = 3 \times \frac{\Delta \text r}{\text r} \times 100$

Substituting the values,

$$\begin{aligned}\frac{\Delta \text V}{\text V} \times 100 = 3 \times \frac{0.1}{5.3} \times 100 \\[1em] = 3 \times 1.887 \\[1em]\end{aligned}$$

= 5.66 ≈ 5.7%

Hence, the percentage error in the volume of the sphere is 5.7%.

Given,

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

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 2, 3, 1 and $\frac{1}{2}$ respectively.

Therefore, the maximum percentage error in S is

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

Substituting the values,

$$\begin{aligned}\frac{\Delta \text S}{\text S} \times 100 = 2(4) + 3(2) + 1(3) + \frac{1}{2}(1) \\[1em] = 8 + 6 + 3 + 0.5 \\[1em]\end{aligned}$$

= 17.5%

Hence, the maximum percentage error in the value of S is 17.5%.

Given,

• Percentage error in the radius, $\frac{\Delta \text r}{\text r} \times 100$ = 1.5%

The area of a circle is

$\text A = \pi \text r^2$

The constant π contributes no error, and the radius occurs with the power 2. Hence the maximum percentage error in the area is

$$\begin{aligned}\frac{\Delta \text A}{\text A} \times 100 = 2 \times \frac{\Delta \text r}{\text r} \times 100 \\[1em] = 2 \times 1.5 \\[1em]\end{aligned}$$

= 3.0%

Hence, the maximum percentage error in the measurement of the area of the circle is 3.0%.

Given,

• Length, l = 24.0 cm
• Width, b = 18.6 cm
• Height, h = 6.0 cm

Each measurement is made up to one place of decimal, so the absolute error in each is

$\Delta \text l = \Delta \text b = \Delta \text h = 0.1\ \text{cm}$

The volume of the box is

$\text V = \text{lbh}$

Since each of the three quantities occurs with the power 1, the maximum percentage error in the volume is

$\frac{\Delta \text V}{\text V} \times 100 = \frac{\Delta \text l}{\text l} \times 100 + \frac{\Delta \text b}{\text b} \times 100 + \frac{\Delta \text h}{\text h} \times 100$

Substituting the values,

$$\begin{aligned}\frac{\Delta \text V}{\text V} \times 100 = \frac{0.1}{24.0} \times 100 + \frac{0.1}{18.6} \times 100 + \frac{0.1}{6.0} \times 100 \\[1em] = 0.417 + 0.538 + 1.67 \\[1em]\end{aligned}$$

= 2.625 ≈ 2.6%

Hence, the maximum probable error in measuring the volume of the box is 2.6%.

Given,

• Percentage change in the length, $\frac{\Delta \text l}{\text l} \times 100$ = 4%
• The acceleration due to gravity g remains unchanged, so $\frac{\Delta \text g}{\text g} \times 100$ = 0

The time period of a simple pendulum is

$\text T = 2\pi\sqrt{\frac{\text l}{\text g}}$

The factor 2π is a pure number, and the length occurs with the power $\frac{1}{2}$. Hence the percentage change in the time period is

$\frac{\Delta \text T}{\text T} \times 100 = \frac{1}{2}\left(\frac{\Delta \text l}{\text l} \times 100\right) + \frac{1}{2}\left(\frac{\Delta \text g}{\text g} \times 100\right)$

Substituting the values,

$$\begin{aligned}\frac{\Delta \text T}{\text T} \times 100 = \frac{1}{2}(4) + 0 \\[1em]\end{aligned}$$

= 2%

Hence, the percentage change in the time period of the simple pendulum is 2%.

Given,

• $\text S = \frac{\text a^3\text b^2}{\text c}$
• Percentage error in a = 1%
• Percentage error in b = 2%
• Percentage error in c = 3%

The maximum percentage error is obtained by adding the percentage errors of all the quantities, each multiplied by the magnitude of its power. Here the powers of a, b and c are 3, 2 and 1 respectively. Therefore,

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

Substituting the values,

$$\begin{aligned}\frac{\Delta \text S}{\text S} \times 100 = 3(1) + 2(2) + 1(3) \\[1em] = 3 + 4 + 3 \\[1em]\end{aligned}$$

= 10%

Hence, the maximum possible percentage error in the value of S is 10%.

Given,

• $\text S = \frac{\text a\text b^2}{\text c^3}$
• Percentage error in a = 1%
• Percentage error in b = 2%
• Percentage error in c = 3%

The maximum percentage error is obtained by adding the percentage errors of all the quantities, each multiplied by the magnitude of its power. Here the powers of a, b and c are 1, 2 and 3 respectively. Therefore,

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

Substituting the values,

$$\begin{aligned}\frac{\Delta \text S}{\text S} \times 100 = 1(1) + 2(2) + 3(3) \\[1em] = 1 + 4 + 9 \\[1em]\end{aligned}$$

= 14%

Hence, the maximum possible percentage error in the value of S is 14%.

Given,

• E = hν

By the principle of homogeneity of dimensions,

$[\text E] = [\text h][\nu]$

The dimensional formula of energy is [ML²T^-2] and that of frequency is [T^-1]. Therefore,

$$\begin{aligned}[\text{ML}^2\text T^{-2}] = [\text h][\text T^{-1}] \\[1em] \Rightarrow [\text h] = [\text{ML}^2\text T^{-2}][\text T] \\[1em] = [\text{ML}^2\text T^{-1}]\end{aligned}$$

The unit of h is obtained in the same way,

$\text{Unit of h} = \frac{\text{Unit of E}}{\text{Unit of }\nu} = \frac{\text J}{\text s^{-1}} = \text{J s}$

Hence, the dimensional formula of Planck's constant h is [M L² T^-1] and its SI unit is J s.

Given,

• $\text V = \frac{\pi \text p \text r^4}{8\eta \text l}$, where V is the rate of flow of the liquid

By the principle of homogeneity of dimensions, the dimensions of both sides must be the same. The numbers 8 and π are pure numbers and are dimensionless.

The rate of flow is the volume flowing per unit time, so

$[\text V] = \frac{[\text L^3]}{[\text T]} = [\text L^3\text T^{-1}]$

Writing the dimensions of the right hand side,

$[\text L^3\text T^{-1}] = \frac{[\text{ML}^{-1}\text T^{-2}][\text L^4]}{[\eta][\text L]}$

Therefore,

$$\begin{aligned}[\eta] = \frac{[\text{ML}^{-1}\text T^{-2}][\text L^4]}{[\text L^3\text T^{-1}][\text L]} \\[1em] = \frac{[\text{ML}^{3}\text T^{-2}]}{[\text L^4\text T^{-1}]} \\[1em] = [\text{ML}^{-1}\text T^{-1}]\end{aligned}$$

Hence, the dimensional formula of the coefficient of viscosity η is [M L^-1 T^-1].

Given,

• Velocity of the body = 36 km h^-1

Since 1 km = 10³ m and 1 h = 3600 s,

$$\begin{aligned}36\ \text{km h}^{-1} = 36 \times \frac{10^3\ \text m}{3600\ \text s} \\[1em] = 36 \times \frac{5}{18}\ \text{m s}^{-1} \\[1em] = 10\ \text{m s}^{-1}\end{aligned}$$

Hence, the velocity of the body is 10 m s^-1.

Given,

• Velocity of sound in air = 332 m s^-1

Since 1 m = 10^-3 km and 1 s = $\frac{1}{3600}$ h, so that 1 s^-1 = 3600 h^-1,

$$\begin{aligned}332\ \text{m s}^{-1} = 332 \times (10^{-3}\ \text{km}) \times (3600\ \text h^{-1}) \\[1em] = 332 \times 3.6\ \text{km h}^{-1} \\[1em] = 1195.2\ \text{km h}^{-1} \\[1em] \approx 1195\ \text{km h}^{-1}\end{aligned}$$

Hence, the velocity of sound in air is 1195 km h^-1.

Given,

• Acceleration of the body = 5 km h^-2

Since 1 km = 10³ m = 10⁵ cm and 1 h = 3600 s,

$$\begin{aligned}5\ \text{km h}^{-2} = 5 \times \frac{10^{5}\ \text{cm}}{(3600\ \text s)^2} \\[1em] = 5 \times \frac{10^{5}}{1.296 \times 10^{7}}\ \text{cm s}^{-2} \\[1em] = \frac{5}{129.6}\ \text{cm s}^{-2} \\[1em] \approx 0.0386\ \text{cm s}^{-2}\end{aligned}$$

Hence, the acceleration of the body is 0.0386 cm s^-2 in the C.G.S. system.

Given,

• 1 J = 1 N m (SI unit of work)
• 1 erg = 1 dyne cm (C.G.S. unit of work)

Since 1 N = 10⁵ dyne and 1 m = 10² cm,

$$\begin{aligned}1\ \text J = 1\ \text{N m} \\[1em] = (10^{5}\ \text{dyne}) \times (10^{2}\ \text{cm}) \\[1em] = 10^{7}\ \text{dyne cm} \\[1em] = 10^{7}\ \text{erg}\end{aligned}$$

Hence, 1 J = 10⁷ erg.

Given,

• 1 atmospheric pressure = 10⁵ N m^-2

Since 1 N = 10⁵ dyne and 1 m² = 10⁴ cm², so that 1 m^-2 = 10^-4 cm^-2,

$$\begin{aligned}1\ \text{atm} = 10^{5} \times (10^{5}\ \text{dyne}) \times (10^{-4}\ \text{cm}^{-2}) \\[1em] = 10^{5 + 5 - 4}\ \text{dyne cm}^{-2} \\[1em] = 10^{6}\ \text{dyne cm}^{-2}\end{aligned}$$

Hence, 1 atmospheric pressure = 10⁶ dyne cm^-2 in the C.G.S. system.

Let the energy E depend upon the mass m, the frequency n and the amplitude a as

$$\begin{aligned}\text E \propto \text m^\text x\ \text n^\text y\ \text a^\text z \\[1em] \Rightarrow \text E = \text k\ \text m^\text x\ \text n^\text y\ \text a^\text z \quad \text{...............(i)}\end{aligned}$$

where k is a dimensionless constant.

Writing the dimensions of both sides,

$$\begin{aligned}[\text{ML}^2\text T^{-2}] = [\text M]^\text x\ [\text T^{-1}]^\text y\ [\text L]^\text z \\[1em] = [\text M^\text x\ \text L^\text z\ \text T^{-\text y}]\end{aligned}$$

Equating the powers of M, L and T on both sides,

• x = 1
• z = 2
• −y = −2, that is, y = 2

Substituting these values in equation (i),

$\text E = \text k\ \text m\ \text n^2\ \text a^2$

Hence, E ∝ m n² a².

Let the velocity v depend upon the length l, the tension F and the mass per unit length m as

$$\begin{aligned}\text v \propto \text l^\text a\ \text F^\text b\ \text m^\text c \\[1em] \Rightarrow \text v = \text k\ \text l^\text a\ \text F^\text b\ \text m^\text c \quad \text{...............(i)}\end{aligned}$$

where k is a dimensionless constant.

Writing the dimensions of both sides,

$$\begin{aligned}[\text M^0\text{LT}^{-1}] = [\text L]^\text a\ [\text{MLT}^{-2}]^\text b\ [\text{ML}^{-1}\text T^{0}]^\text c \\[1em] = [\text M^{\text b + \text c}\ \text L^{\text a + \text b - \text c}\ \text T^{-2\text b}]\end{aligned}$$

Equating the powers of M, L and T on both sides,

• b + c = 0
• a + b − c = 1
• −2b = −1

From the third equation, b = $\frac{1}{2}$, and from the first equation, c = −b = −$\frac{1}{2}$.

Substituting these in the second equation,

$$\begin{aligned}\text a + \frac{1}{2} - \left(-\frac{1}{2}\right) = 1 \\[1em] \Rightarrow \text a + 1 = 1 \\[1em] \Rightarrow \text a = 0\end{aligned}$$

Substituting these values in equation (i),

$$\begin{aligned}\text v = \text k\ \text l^{0}\ \text F^{1/2}\ \text m^{-1/2} \\[1em] = \text k\sqrt{\frac{\text F}{\text m}}\end{aligned}$$

Hence, the velocity of the transverse wave is $\text v = \text k\sqrt{\frac{\text F}{\text m}}$, and it does not depend on the length of the string.

Given,

• Time period of one heart beat = 0.8 s
• Total time = 60 years

The number of beats per second is

$\frac{1}{0.8} = 1.25\ \text{beats per second}$

Expressing the total time in seconds,

$$\begin{aligned}60\ \text{years} = 60 \times 365 \times 24 \times 60 \times 60\ \text s \\[1em] = 1.89 \times 10^{9}\ \text s\end{aligned}$$

Therefore, the total number of beats is

$$\begin{aligned}\text N = 1.25 \times 1.89 \times 10^{9} \\[1em] = 2.36 \times 10^{9}\end{aligned}$$

Hence, the heart beats about 2.36 × 10⁹ times in the life of 60 years of a man.

Given,

• $\mu = \text A + \frac{\text B}{\lambda^2}$

The refractive index μ is the ratio of two speeds and is therefore dimensionless,

$[\mu] = [\text M^0\text L^0\text T^0]$

By the principle of homogeneity of dimensions, each term on the right hand side must have the same dimensions as μ.

For A :

$[\text A] = [\mu] = [\text M^0\text L^0\text T^0]$

so A is dimensionless and has no unit.

For B :

$$\begin{aligned}\frac{[\text B]}{[\lambda^2]} = [\text M^0\text L^0\text T^0] \\[1em] \Rightarrow [\text B] = [\text L^2] = [\text M^0\text L^2\text T^0]\end{aligned}$$

so the SI unit of B is m².

Hence, A is dimensionless and has no unit, while the dimensional formula of B is [M⁰ L² T⁰] and its SI unit is m².

Given,

• $\text v = \text a + \text{bt} + \frac{\text c}{\text d + \text t}$

By the principle of homogeneity of dimensions, each term on the right hand side must have the same dimensions as the velocity v, and only quantities of the same dimensions can be added to one another.

For a : Since a is added to the other terms which have the dimensions of velocity,

$[\text a] = [\text v] = [\text{LT}^{-1}]$

For b :

$$\begin{aligned}[\text b][\text T] = [\text{LT}^{-1}] \\[1em] \Rightarrow [\text b] = [\text{LT}^{-2}]\end{aligned}$$

For d : Since d is added to the time t,

$[\text d] = [\text t] = [\text T]$

For c :

$$\begin{aligned}\frac{[\text c]}{[\text T]} = [\text{LT}^{-1}] \\[1em] \Rightarrow [\text c] = [\text L]\end{aligned}$$

Hence, the dimensions of a, b, c and d are [L T^-1], [L T^-2], [L] and [T] respectively.

Given,

• v = a + bt

By the principle of homogeneity of dimensions, each term on the right hand side must have the same dimensions as the velocity v.

For a :

$[\text a] = [\text v] = [\text{LT}^{-1}]$

For b :

$$\begin{aligned}[\text b][\text T] = [\text{LT}^{-1}] \\[1em] \Rightarrow [\text b] = [\text{LT}^{-2}]\end{aligned}$$

Therefore,

$\left[\frac{\text a}{\text b}\right] = \frac{[\text{LT}^{-1}]}{[\text{LT}^{-2}]} = [\text T]$

Hence, the dimensional formula of $\frac{\text a}{\text b}$ is [T].

Given,

• $\text E = \frac{\text b - \text x^2}{\text{at}}$, where E is energy, x is distance and t is time

By the principle of homogeneity of dimensions, only quantities of the same dimensions can be subtracted from one another. Since b is subtracted by x²,

$[\text b] = [\text x^2] = [\text L^2]$

Also, the dimensions of both sides of the relation must be the same,

$[\text E] = \frac{[\text b - \text x^2]}{[\text a][\text t]}$

Substituting the dimensional formulae,

$[\text{ML}^2\text T^{-2}] = \frac{[\text L^2]}{[\text a][\text T]}$

Therefore,

$$\begin{aligned}[\text a] = \frac{[\text L^2]}{[\text{ML}^2\text T^{-2}][\text T]} \\[1em] = \frac{[\text L^2]}{[\text{ML}^2\text T^{-1}]} \\[1em] = [\text M^{-1}\text T]\end{aligned}$$

Hence,

$$\begin{aligned}[\text a \times \text b] = [\text M^{-1}\text T][\text L^2] \\[1em] = [\text M^{-1}\text L^2\text T]\end{aligned}$$

Hence, the dimensional formula of the constant a × b is [M^-1 L² T].

Given,

• Force = 36 units in the system having metre, kilogram and minute as the fundamental units

The dimensional formula of force is [MLT^-2], so in this system the unit of force is kg m min^-2.

Since 1 kg = 10³ g, 1 m = 10² cm and 1 min = 60 s,

$$\begin{aligned}\text F = 36\ \text{kg m min}^{-2} \\[1em] = 36 \times (10^{3}\ \text g)(10^{2}\ \text{cm})(60\ \text s)^{-2} \\[1em] = 36 \times \frac{10^{5}}{3600}\ \text{g cm s}^{-2} \\[1em] = \frac{36 \times 10^{5}}{3600}\ \text{g cm s}^{-2} \\[1em] = 10^{3}\ \text{g cm s}^{-2} \\[1em] = 10^{3}\ \text{dyne}\end{aligned}$$

Hence, the value of the force in the C.G.S. system is 10³ dyne.

Let the centripetal force F depend upon the mass m, the radius r and the speed v as

$$\begin{aligned}\text F \propto \text m^\text a\ \text r^\text b\ \text v^\text c \\[1em] \Rightarrow \text F = \text k\ \text m^\text a\ \text r^\text b\ \text v^\text c \quad \text{...............(i)}\end{aligned}$$

where k is a dimensionless constant.

Writing the dimensions of both sides,

$$\begin{aligned}[\text{MLT}^{-2}] = [\text M]^\text a\ [\text L]^\text b\ [\text{LT}^{-1}]^\text c \\[1em] = [\text M^\text a\ \text L^{\text b + \text c}\ \text T^{-\text c}]\end{aligned}$$

Equating the powers of M, L and T on both sides,

• a = 1
• b + c = 1
• −c = −2, that is, c = 2

From the second equation, b = 1 − c = 1 − 2 = −1.

Substituting these values in equation (i),

$$\begin{aligned}\text F = \text k\ \text m\ \text r^{-1}\ \text v^{2} \\[1em] = \text k\frac{\text{mv}^2}{\text r}\end{aligned}$$

Experimentally, the value of k is found to be 1.

Hence, the centripetal force is $\text F = \frac{\text{mv}^2}{\text r}$.

Let the frequency n depend upon the length l, the density ρ and the Young's modulus Y as

$$\begin{aligned}\text n \propto \text l^\text a\ \rho^\text b\ \text Y^\text c \\[1em] \Rightarrow \text n = \text k\ \text l^\text a\ \rho^\text b\ \text Y^\text c \quad \text{...............(i)}\end{aligned}$$

where k is a dimensionless constant.

Writing the dimensions of both sides,

$$\begin{aligned}[\text M^0\text L^0\text T^{-1}] = [\text L]^\text a\ [\text{ML}^{-3}]^\text b\ [\text{ML}^{-1}\text T^{-2}]^\text c \\[1em] = [\text M^{\text b + \text c}\ \text L^{\text a - 3\text b - \text c}\ \text T^{-2\text c}]\end{aligned}$$

Equating the powers of M, L and T on both sides,

• b + c = 0
• a − 3b − c = 0
• −2c = −1

From the third equation, c = $\frac{1}{2}$, and from the first equation, b = −c = −$\frac{1}{2}$.

Substituting these in the second equation,

$$\begin{aligned}\text a - 3\left(-\frac{1}{2}\right) - \frac{1}{2} = 0 \\[1em] \Rightarrow \text a + \frac{3}{2} - \frac{1}{2} = 0 \\[1em] \Rightarrow \text a = -1\end{aligned}$$

Substituting these values in equation (i),

$$\begin{aligned}\text n = \text k\ \text l^{-1}\ \rho^{-1/2}\ \text Y^{1/2} \\[1em] = \frac{\text k}{\text l}\sqrt{\frac{\text Y}{\rho}}\end{aligned}$$

Hence, the frequency of the tuning fork is $\text n = \frac{\text k}{\text l}\sqrt{\frac{\text Y}{\rho}}$.

Let the frequency n depend upon the radius r, the density ρ and the surface tension S as

$$\begin{aligned}\text n \propto \text r^\text a\ \rho^\text b\ \text S^\text c \\[1em] \Rightarrow \text n = \text k\ \text r^\text a\ \rho^\text b\ \text S^\text c \quad \text{...............(i)}\end{aligned}$$

where k is a dimensionless constant.

Writing the dimensions of both sides,

$$\begin{aligned}[\text M^0\text L^0\text T^{-1}] = [\text L]^\text a\ [\text{ML}^{-3}]^\text b\ [\text{ML}^{0}\text T^{-2}]^\text c \\[1em] = [\text M^{\text b + \text c}\ \text L^{\text a - 3\text b}\ \text T^{-2\text c}]\end{aligned}$$

Equating the powers of M, L and T on both sides,

• b + c = 0
• a − 3b = 0
• −2c = −1

From the third equation, c = $\frac{1}{2}$, and from the first equation, b = −c = −$\frac{1}{2}$.

From the second equation,

$\text a = 3\text b = 3\left(-\frac{1}{2}\right) = -\frac{3}{2}$

Substituting these values in equation (i),

$$\begin{aligned}\text n = \text k\ \text r^{-3/2}\ \rho^{-1/2}\ \text S^{1/2} \\[1em] = \text k\sqrt{\frac{\text S}{\rho \text r^3}}\end{aligned}$$

Hence, the frequency of the oscillating liquid drop is $\text n = \text k\sqrt{\frac{\text S}{\rho \text r^3}}$.

Let the time period T depend upon the height h, the density ρ and the acceleration due to gravity g as

$$\begin{aligned}\text T \propto \text h^\text a\ \rho^\text b\ \text g^\text c \\[1em] \Rightarrow \text T = \text k\ \text h^\text a\ \rho^\text b\ \text g^\text c \quad \text{...............(i)}\end{aligned}$$

where k is a dimensionless constant.

Writing the dimensions of both sides,

$$\begin{aligned}[\text M^0\text L^0\text T] = [\text L]^\text a\ [\text{ML}^{-3}]^\text b\ [\text{LT}^{-2}]^\text c \\[1em] = [\text M^{\text b}\ \text L^{\text a - 3\text b + \text c}\ \text T^{-2\text c}]\end{aligned}$$

Equating the powers of M, L and T on both sides,

• b = 0
• a − 3b + c = 0
• −2c = 1

From the third equation, c = −$\frac{1}{2}$. Since b = 0, the second equation gives a = −c = $\frac{1}{2}$.

Substituting these values in equation (i),

$$\begin{aligned}\text T = \text k\ \text h^{1/2}\ \rho^{0}\ \text g^{-1/2} \\[1em] = \text k\sqrt{\frac{\text h}{\text g}}\end{aligned}$$

Hence, the time period of the oscillations is $\text T = \text k\sqrt{\frac{\text h}{\text g}}$, and it does not depend on the density of the liquid.

Given,

• Mass of the earth = 5.98 × 10²⁴ kg
• Average mass of one atom = 40 u

Since 1 u = 1.66 × 10^-27 kg,

$$\begin{aligned}\text{Mass of one atom} = 40 \times 1.66 \times 10^{-27} \\[1em] = 6.64 \times 10^{-26}\ \text{kg}\end{aligned}$$

Therefore, the number of atoms in the earth is

$$\begin{aligned}\text N = \frac{\text{Mass of the earth}}{\text{Mass of one atom}} \\[1em] = \frac{5.98 \times 10^{24}}{6.64 \times 10^{-26}} \\[1em] = 0.90 \times 10^{50} \\[1em] = 9.0 \times 10^{49}\end{aligned}$$

Hence, there are about 9.0 × 10⁴⁹ atoms in the earth.

Given,

• New unit of speed = c = 3 × 10⁸ m s^-1
• New unit of time = 1 year = 3.154 × 10⁷ s

Since length is the product of speed and time,

$$\begin{aligned}\text{New unit of length} = \text{speed} \times \text{time} \\[1em] = (3 \times 10^{8}) \times (3.154 \times 10^{7}) \\[1em] = 9.46 \times 10^{15}\ \text m\end{aligned}$$

This is the distance travelled by light in vacuum in one year.

Hence, the new unit of length is 9.46 × 10¹⁵ m, which is called the light year.

Given,

• Number of protons and neutrons in the universe ≈ 10⁸²
• Number of protons and neutrons in the sun ≈ 10⁵⁷

Assuming all the stars to be of the same mass as the sun, the number of stars is

$$\begin{aligned}\text N = \frac{\text{Number of nucleons in the universe}}{\text{Number of nucleons in the sun}} \\[1em] = \frac{10^{82}}{10^{57}} \\[1em] = 10^{25}\end{aligned}$$

Hence, the order of magnitude of the number of stars in the universe is 10²⁵.

Given,

• 185 (3 significant figures)
• 1.52 (3 significant figures)

Carrying out the multiplication,

$185 \times 1.52 = 281.2$

In multiplication, the result is rounded off to the least number of significant figures in the given data, which is 3 here. The digit to be dropped is 2, which is less than 5, so the preceding digit is retained unchanged.

Hence, the value of the product 185 × 1.52 is 281.

Given,

• Side of the square = 1.6 m (2 significant figures)

The area of the square is

$$\begin{aligned}\text A = (\text{side})^2 = 1.6 \times 1.6 \\[1em] = 2.56\ \text m^2\end{aligned}$$

In multiplication, the result is rounded off to the least number of significant figures in the given data, which is 2 here. The digit to be dropped is 6, which is more than 5, so the preceding digit 5 is increased by 1.

Hence, the area of the square is 2.6 m².

Given,

• Length of the path-strip, l = 10.53 m (4 significant figures)
• Width of the path-strip, b = 0.97 m (2 significant figures)

The area of the path-strip is

$$\begin{aligned}\text A = \text l \times \text b = 10.53 \times 0.97 \\[1em] = 10.2141\ \text m^2\end{aligned}$$

In multiplication, the result is rounded off to the least number of significant figures in the given data, which is 2 here.

Hence, the area of the path-strip is 10 m².

Given,

• Length of the block, l = 12.1 cm (3 significant figures)
• Breadth of the block, b = 6.3 cm (2 significant figures)
• Height of the block, h = 8.4 mm = 0.84 cm (2 significant figures)

The volume of the block is

$$\begin{aligned}\text V = \text l \times \text b \times \text h = 12.1 \times 6.3 \times 0.84 \\[1em] = 64.0332\ \text{cm}^3\end{aligned}$$

In multiplication, the result is rounded off to the least number of significant figures in the given data, which is 2 here.

Hence, the volume of the block is 64 cm³.

Given,

• Length of the block, l = 20 m = 2000 cm (2 significant figures)
• Width of the block, b = 25 cm (2 significant figures)
• Thickness of the block, t = 13.53 cm (4 significant figures)

The volume of the block is

$$\begin{aligned}\text V = \text l \times \text b \times \text t = 2000 \times 25 \times 13.53 \\[1em] = 676500\ \text{cm}^3\end{aligned}$$

In multiplication, the result is rounded off to the least number of significant figures in the given data, which is 2 here. Hence

$\text V = 6.8 \times 10^{5}\ \text{cm}^3$

Since 1 cm³ = 10^-6 m³,

$$\begin{aligned}\text V = 6.8 \times 10^{5} \times 10^{-6}\ \text m^3 \\[1em] = 0.68\ \text m^3\end{aligned}$$

Hence, the volume of the block is 6.8 × 10⁵ cm³ or 0.68 m³.

Given,

• Diameter of the circle, d = 5.2 cm (2 significant figures)

The circumference of the circle is

$$\begin{aligned}\text C = \pi \text d = 3.14 \times 5.2 \\[1em] = 16.328\ \text{cm}\end{aligned}$$

The constant π is a pure number and has unlimited accuracy, so it is not counted while deciding the number of significant figures. The result is therefore rounded off to 2 significant figures, the number of significant figures in the diameter.

Hence, the circumference of the circle is 16 cm.

Given,

• Diameter of the sphere, d = 4.24 cm (3 significant figures), so radius r = 2.12 cm
• π = 3.142

The surface area of the sphere is

$$\begin{aligned}\text A = 4\pi \text r^2 \\[1em] = 4 \times 3.142 \times (2.12)^2 \\[1em] = 4 \times 3.142 \times 4.4944 \\[1em] = 56.48\ \text{cm}^2\end{aligned}$$

The number 4 and the constant π are pure numbers and have unlimited accuracy, so the result is rounded off to 3 significant figures, the number of significant figures in the diameter.

Hence, the surface area of the sphere is 56.5 cm².

Given,

• Length of the cylinder, l = 1.0 × 10^-1 m (2 significant figures)
• Diameter of the cylinder, d = 2.40 × 10^-3 m (3 significant figures)
• Radius of the cylinder, r = 1.20 × 10^-3 m (3 significant figures)

Cross-sectional area of the cylinder.

$$\begin{aligned}\text A = \pi \text r^2 \\[1em] = 3.14 \times (1.20 \times 10^{-3})^2 \\[1em] = 3.14 \times 1.44 \times 10^{-6} \\[1em] = 4.5216 \times 10^{-6}\ \text m^2\end{aligned}$$

The constant π has unlimited accuracy, so the result is rounded off to 3 significant figures, the number of significant figures in the radius. Hence

$\text A = 4.52 \times 10^{-6}\ \text m^2$

Volume of the cylinder.

$$\begin{aligned}\text V = \text A \times \text l \\[1em] = (4.52 \times 10^{-6}) \times (1.0 \times 10^{-1}) \\[1em] = 4.52 \times 10^{-7}\ \text m^3\end{aligned}$$

Here the length has only 2 significant figures, which is the least in the given data, so the result is rounded off to 2 significant figures.

Hence, the cross-sectional area of the cylinder is 4.52 × 10^-6 m² and its volume is 4.5 × 10^-7 m³.

Given,

• Surface area of the leaf, A = 3.41 m² (3 significant figures)
• Width of the leaf, b = 1.034 m (4 significant figures)

The length of the leaf is

$$\begin{aligned}\text l = \frac{\text A}{\text b} = \frac{3.41}{1.034} \\[1em] = 3.2979...\ \text m\end{aligned}$$

In division, the result is rounded off to the least number of significant figures in the given data, which is 3 here.

Hence, the length of the leaf is 3.30 m.

Given,

• 2.91 (3 significant figures)
• 0.3842 (4 significant figures)
• 0.080 (2 significant figures)

Carrying out the calculation,

$$\begin{aligned}\frac{2.91 \times 0.3842}{0.080} = \frac{1.118}{0.080} \\[1em] = 13.975\end{aligned}$$

In multiplication and division, the result is rounded off to the least number of significant figures in the given data, which is 2 here.

Hence, $\frac{2.91 \times 0.3842}{0.080} = 14$.

Number of seconds in a day.

$$\begin{aligned}24 \times 60 \times 60 = 86400\ \text s \\[1em] = 0.864 \times 10^{5}\ \text s\end{aligned}$$

Writing the number in the form N × 10^x gives 8.64 × 10⁴. Since N = 8.64 is greater than $\sqrt{10}$ = 3.16, the order of magnitude is 10⁴⁺¹ = 10⁵.

Number of seconds in a year.

$365 \times 24 \times 60 \times 60 = 3.15 \times 10^{7}\ \text s$

Here N = 3.15, which is smaller than 3.16, so the order of magnitude is 10⁷.

Hence, the order of magnitude of the number of seconds in a day is 10⁵ and that of the number of seconds in a year is 10⁷.

Given,

• Mass of the block, m = 39.3 g, so Δm = 0.1 g
• Length of the block, l = 5.12 cm, so Δl = 0.01 cm
• Breadth of the block, b = 2.56 cm, so Δb = 0.01 cm
• Thickness of the block, t = 0.37 cm, so Δt = 0.01 cm

The density of the material of the block is

$\rho = \frac{\text m}{\text{lbt}}$

Each of the four quantities occurs with the power 1, so the maximum permissible percentage error in the density is

$\frac{\Delta \rho}{\rho} \times 100 = \frac{\Delta \text m}{\text m} \times 100 + \frac{\Delta \text l}{\text l} \times 100 + \frac{\Delta \text b}{\text b} \times 100 + \frac{\Delta \text t}{\text t} \times 100$

Substituting the values,

$$\begin{aligned}\frac{\Delta \rho}{\rho} \times 100 = \frac{0.1}{39.3} \times 100 + \frac{0.01}{5.12} \times 100 + \frac{0.01}{2.56} \times 100 + \frac{0.01}{0.37} \times 100 \\[1em] = 0.254 + 0.195 + 0.391 + 2.70 \\[1em]\end{aligned}$$

= 3.54%

Hence, the maximum permissible percentage error in the determination of the density is 3.54%.

Given,

• Length of the pendulum, l = 90.6 × 10^-2 m = 0.906 m (3 significant figures)
• Time period of oscillation, T = 1.91 s (3 significant figures)

The time period of a simple pendulum is

$$\begin{aligned}\text T = 2\pi\sqrt{\frac{\text l}{\text g}} \\[1em] \Rightarrow \text T^2 = 4\pi^2\frac{\text l}{\text g} \\[1em] \Rightarrow \text g = 4\pi^2\frac{\text l}{\text T^2}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text g = \frac{4 \times (3.1416)^2 \times 0.906}{(1.91)^2} \\[1em] = \frac{35.7676}{3.6481} \\[1em] = 9.8044\ \text{m s}^{-2}\end{aligned}$$

The constant 4π² has unlimited accuracy, and both the measured quantities have 3 significant figures, so the result is rounded off to 3 significant figures.

Hence, the value of the acceleration due to gravity is 9.80 m s^-2.

Given,

• Mass of the brass ball, m = 29.150 × 10^-3 kg (5 significant figures)
• Diameter of the brass ball, d = 1.92 × 10^-2 m (3 significant figures)
• Radius of the brass ball, r = 9.6 × 10^-3 m

Treating the ball as a sphere, its volume is

$$\begin{aligned}\text V = \frac{4}{3}\pi \text r^3 \\[1em] = \frac{4}{3} \times 3.14 \times (9.6 \times 10^{-3})^3 \\[1em] = 4.18667 \times 8.84736 \times 10^{-7} \\[1em] = 3.70419 \times 10^{-6}\ \text m^3\end{aligned}$$

Therefore, the density of brass is

$$\begin{aligned}\rho = \frac{\text m}{\text V} \\[1em] = \frac{29.150 \times 10^{-3}}{3.70419 \times 10^{-6}} \\[1em] = 7.8695 \times 10^{3}\ \text{kg m}^{-3}\end{aligned}$$

In division, the result is rounded off to the least number of significant figures in the given data. The diameter has 3 significant figures, which is the least, so the result is rounded off to 3 significant figures.

Hence, the density of brass is 7.87 × 10³ kg m^-3.

By the principle of homogeneity of dimensions, an equation is dimensionally correct when the dimensions of both sides are the same. Pure numbers such as $\frac{1}{2}$, 2, 3, 4 and π, and dimensionless quantities such as cos θ, contribute no dimensions.

(i) K = $\frac{1}{2}$mv²

$$\begin{aligned}[\text{L.H.S.}] = [\text K] = [\text{ML}^2\text T^{-2}] \\[1em] [\text{R.H.S.}] = [\text M][\text{LT}^{-1}]^2 = [\text{ML}^2\text T^{-2}]\end{aligned}$$

Since [L.H.S.] = [R.H.S.], the equation is dimensionally correct.

(ii) Fs = $\frac{1}{2}$mv² − $\frac{1}{2}$mu²

$$\begin{aligned}[\text{L.H.S.}] = [\text F][\text s] = [\text{MLT}^{-2}][\text L] = [\text{ML}^2\text T^{-2}] \\[1em] [\text{R.H.S.}] = [\text M][\text{LT}^{-1}]^2 = [\text{ML}^2\text T^{-2}]\end{aligned}$$

Both terms on the right hand side have the same dimensions, so they can be subtracted. Since [L.H.S.] = [R.H.S.], the equation is dimensionally correct.

(iii) $\rho = \frac{3\text g}{4\pi \text R_\text e \text G}$

$$\begin{aligned}[\text{L.H.S.}] = [\rho] = [\text{ML}^{-3}] \\[1em] [\text{R.H.S.}] = \frac{[\text{LT}^{-2}]}{[\text L][\text M^{-1}\text L^3\text T^{-2}]} = \frac{[\text{LT}^{-2}]}{[\text M^{-1}\text L^4\text T^{-2}]} \\[1em] = [\text{LT}^{-2}][\text{ML}^{-4}\text T^{2}] = [\text{ML}^{-3}]\end{aligned}$$

Since [L.H.S.] = [R.H.S.], the equation is dimensionally correct.

(iv) $\text v_\text e = \sqrt{\frac{2\text{GM}_\text e}{\text R_\text e}}$

$$\begin{aligned}[\text{L.H.S.}] = [\text v_\text e] = [\text{LT}^{-1}] \\[1em] [\text{R.H.S.}] = \sqrt{\frac{[\text M^{-1}\text L^3\text T^{-2}][\text M]}{[\text L]}} = \sqrt{[\text L^2\text T^{-2}]} = [\text{LT}^{-1}]\end{aligned}$$

Since [L.H.S.] = [R.H.S.], the equation is dimensionally correct.

(v) $\text h = \frac{2\text S\ \text{cos}\ \theta}{\text r\rho \text g}$

$$\begin{aligned}[\text{L.H.S.}] = [\text h] = [\text L] \\[1em] [\text{R.H.S.}] = \frac{[\text{MT}^{-2}]}{[\text L][\text{ML}^{-3}][\text{LT}^{-2}]} = \frac{[\text{MT}^{-2}]}{[\text{ML}^{-1}\text T^{-2}]} = [\text L]\end{aligned}$$

Since [L.H.S.] = [R.H.S.], the equation is dimensionally correct.

(vi) $\text v = \sqrt{\frac{\text B}{\rho}}$

$$\begin{aligned}[\text{L.H.S.}] = [\text v] = [\text{LT}^{-1}] \\[1em] [\text{R.H.S.}] = \sqrt{\frac{[\text{ML}^{-1}\text T^{-2}]}{[\text{ML}^{-3}]}} = \sqrt{[\text L^2\text T^{-2}]} = [\text{LT}^{-1}]\end{aligned}$$

Since [L.H.S.] = [R.H.S.], the equation is dimensionally correct.

(vii) $\lambda = \frac{\text h}{\text{mv}}$

$$\begin{aligned}[\text{L.H.S.}] = [\lambda] = [\text L] \\[1em] [\text{R.H.S.}] = \frac{[\text{ML}^2\text T^{-1}]}{[\text M][\text{LT}^{-1}]} = \frac{[\text{ML}^2\text T^{-1}]}{[\text{MLT}^{-1}]} = [\text L]\end{aligned}$$

Since [L.H.S.] = [R.H.S.], the equation is dimensionally correct.

Given,

• Measured values of the refractive index : 1.54, 1.45, 1.53, 1.56, 1.44 and 1.54
• Number of observations = 6

(i) Mean refractive index. The arithmetic mean of a large number of readings is taken as the true value of the quantity,

$$\begin{aligned}\overline{\text n} = \frac{\text{Sum of all the readings}}{\text{Number of readings}} \\[1em] = \frac{1.54 + 1.45 + 1.53 + 1.56 + 1.44 + 1.54}{6} \\[1em] = \frac{9.06}{6} = 1.51\end{aligned}$$

Hence, the mean refractive index of the glass is 1.51.

(ii) Mean absolute error. The absolute error in each reading is the magnitude of its difference from the mean,

$\Delta \text n_\text i = |\text n_\text i - \overline{\text n}|$

• Δn₁ = |1.54 − 1.51| = 0.03
• Δn₂ = |1.45 − 1.51| = 0.06
• Δn₃ = |1.53 − 1.51| = 0.02
• Δn₄ = |1.56 − 1.51| = 0.05
• Δn₅ = |1.44 − 1.51| = 0.07
• Δn₆ = |1.54 − 1.51| = 0.03

The mean absolute error is

$$\begin{aligned}\overline{\Delta \text n} = \frac{0.03 + 0.06 + 0.02 + 0.05 + 0.07 + 0.03}{6} \\[1em] = \frac{0.26}{6} = 0.043 \approx 0.04\end{aligned}$$

Hence, the mean absolute error is 0.04.

(iii) Fractional error.

$$\begin{aligned}\frac{\overline{\Delta \text n}}{\overline{\text n}} = \frac{0.04}{1.51} \\[1em] = 0.026 \approx 0.03\end{aligned}$$

Hence, the fractional error is 0.03.

(iv) Percentage error.

$\frac{\overline{\Delta \text n}}{\overline{\text n}}$ × 100 = 0.03 × 100 = 3%

Hence, the percentage error is 3%.

(v) Refractive index with error limits.

Hence, the refractive index of the glass is n = 1.51 ± 0.04.

Given,

• Least count of the screw gauge = 0.001 cm, so the absolute error Δd = 0.001 cm
• Diameter of the wire, d = 0.225 cm

The percentage error in the measurement is

$$\begin{aligned}\frac{\Delta \text d}{\text d} \times 100 = \frac{0.001}{0.225} \times 100 \\[1em] = 0.444 \\[1em]\end{aligned}$$

≈ 0.4%

Hence, the percentage error in the measurement is 0.4%.

Given,

• Mass of the diamond = 4.37 g = 0.00437 kg
• Mass of the box = 1.5 kg

The total mass is

$$\begin{aligned}\text M = 1.5 + 0.00437 \\[1em] = 1.50437\ \text{kg}\end{aligned}$$

In addition, the result is rounded off to the same number of decimal places as are contained in the least accurate quantity. Here the mass of the box, 1.5 kg, has only one place of decimal.

Hence, the total weight is 1.5 kg.

Given,

• Order of the size of the universe ≈ 10²⁶ m
• Order of the size of the earth ≈ 10⁷ m

If the universe shrinks to the size of the earth, the shrinking factor is

$\frac{10^{7}}{10^{26}} = 10^{-19}$

Every length shrinks in the same ratio, so the new size of the earth is

$10^{7} \times 10^{-19} = 10^{-12}\ \text m$

Hence, the earth would then be of the order of 10^-12 m in size.

In addition, the numbers must first be expressed in the same power of ten, and the sum is then rounded off to the same number of decimal places as are contained in the least accurate quantity.

(i) $3.8 \times 10^{-8} + 4.2 \times 10^{-6}$

$$\begin{aligned}= 0.038 \times 10^{-6} + 4.2 \times 10^{-6} \\[1em] = 4.238 \times 10^{-6}\end{aligned}$$

Since 4.2 has only one digit after the decimal point, the sum is 4.2 × 10^-6.

(ii) $3.8 \times 10^{-7} + 4.2 \times 10^{-6}$

$$\begin{aligned}= 0.38 \times 10^{-6} + 4.2 \times 10^{-6} \\[1em] = 4.58 \times 10^{-6}\end{aligned}$$

Since 4.2 has only one digit after the decimal point, the sum is 4.6 × 10^-6.

(iii) $4.22 \times 10^{5} + 3.11 \times 10^{7} + 6.003 \times 10^{6}$

$$\begin{aligned}= 0.422 \times 10^{6} + 31.1 \times 10^{6} + 6.003 \times 10^{6} \\[1em] = 37.525 \times 10^{6} \\[1em] = 3.7525 \times 10^{7}\end{aligned}$$

Since 31.1 × 10⁶ has only one digit after the decimal point, the sum is rounded off to 37.5 × 10⁶. Hence the sum is 3.75 × 10⁷.

(iv) $1.294\ \text{cm} + 35.1\ \text{cm}$

$= 36.394\ \text{cm}$

Since 35.1 cm has only one digit after the decimal point, the sum is 36.4 cm.

(v) $348.2\ \text{km} + 105\ \text{km} + 143.8\ \text{km}$

$= 597.0\ \text{km}$

Since 105 km has no digit after the decimal point, the sum is 597 km.

(vi) $1.5\ \text{kg} + 264\ \text g + 52\ \text{mg}$

$$\begin{aligned}= 1.5\ \text{kg} + 0.264\ \text{kg} + 0.000052\ \text{kg} \\[1em] = 1.764052\ \text{kg}\end{aligned}$$

Since 1.5 kg has only one digit after the decimal point, the sum is 1.8 kg.

In subtraction, the numbers must first be expressed in the same power of ten, and the difference is then rounded off to the same number of decimal places as are contained in the least accurate quantity.

(i) $5.0 \times 10^{-4} - 2.5 \times 10^{-6}$

$$\begin{aligned}= 5.0 \times 10^{-4} - 0.025 \times 10^{-4} \\[1em] = 4.975 \times 10^{-4}\end{aligned}$$

Since 5.0 has only one digit after the decimal point, the difference is 5.0 × 10^-4.

(ii) $18.4\ \text{cm} - 18.132\ \text{cm}$

$= 0.268\ \text{cm}$

Since 18.4 cm has only one digit after the decimal point, the difference is 0.3 cm.

(iii) $104.4\ \text g - 2.34\ \text g$

$= 102.06\ \text g$

Since 104.4 g has only one digit after the decimal point, the difference is 102.1 g.

(iv) $4.0 \times 10^{2}\ \text{kg} - 47.72\ \text{kg}$

$$\begin{aligned}= 4.0 \times 10^{2}\ \text{kg} - 0.4772 \times 10^{2}\ \text{kg} \\[1em] = 3.5228 \times 10^{2}\ \text{kg}\end{aligned}$$

Since 4.0 × 10² kg has only one digit after the decimal point, the difference is 3.5 × 10² kg.

(v) $172.4\ \text{kg} - 98.767\ \text g$

$$\begin{aligned}= 172.4\ \text{kg} - 0.098767\ \text{kg} \\[1em] = 172.301233\ \text{kg}\end{aligned}$$

Since 172.4 kg has only one digit after the decimal point, the difference is 172.3 kg.

Given, the new fundamental units are

• Unit of mass = 10 kg
• Unit of length = 1 dm = 0.1 m
• Unit of time = 1 min = 60 s

The dimensional formula of power is [ML²T^-3], so the new unit of power is

$$\begin{aligned}1\ \text{new unit} = (10\ \text{kg})(0.1\ \text m)^2(60\ \text s)^{-3} \\[1em] = \frac{10 \times 0.01}{2.16 \times 10^{5}}\ \text{kg m}^2\text s^{-3} \\[1em] = \frac{0.1}{2.16 \times 10^{5}}\ \text W \\[1em] = 4.63 \times 10^{-7}\ \text W\end{aligned}$$

Therefore,

$1\ \text W = \frac{1}{4.63 \times 10^{-7}} = 2.16 \times 10^{6}\ \text{new units}$

Hence,

$$\begin{aligned}1\ \text{MW} = 10^{6}\ \text W \\[1em] = 10^{6} \times 2.16 \times 10^{6}\ \text{new units} \\[1em] = 2.16 \times 10^{12}\ \text{new units}\end{aligned}$$

Hence, 1 MW = 2.16 × 10¹² new units of power.

Given,

• $\text h = \frac{1}{2}\text{gt}$

By the principle of homogeneity of dimensions, the dimensions of both sides must be the same. The factor $\frac{1}{2}$ is a pure number and is dimensionless.

$$\begin{aligned}[\text{L.H.S.}] = [\text h] = [\text L] \\[1em] [\text{R.H.S.}] = [\text g][\text t] = [\text{LT}^{-2}][\text T] = [\text{LT}^{-1}]\end{aligned}$$

Since [L.H.S.] ≠ [R.H.S.], the given equation is dimensionally incorrect.

To make the right hand side have the dimension [L], it must be multiplied by a further factor of [T]. This is achieved by taking the square of the time, so the correct equation is

$\text h = \frac{1}{2}\text{gt}^2$

for which

$[\text{R.H.S.}] = [\text{LT}^{-2}][\text T^2] = [\text L] = [\text{L.H.S.}]$

Hence, the correct equation is h = $\frac{1}{2}$gt².

Given,

• $\text t = \frac{2\pi \text l}{\text g}$

By the principle of homogeneity of dimensions, the dimensions of both sides must be the same. The factor 2π is a pure number and is dimensionless.

$$\begin{aligned}[\text{L.H.S.}] = [\text t] = [\text T] \\[1em] [\text{R.H.S.}] = \frac{[\text L]}{[\text{LT}^{-2}]} = [\text T^{2}]\end{aligned}$$

Since [L.H.S.] ≠ [R.H.S.], the given equation is dimensionally incorrect.

To make the right hand side have the dimension [T], the square root of $\frac{\text l}{\text g}$ must be taken. Hence the correct form is

$\text T = 2\pi\sqrt{\frac{\text l}{\text g}}$

for which

$[\text{R.H.S.}] = \sqrt{[\text T^2]} = [\text T] = [\text{L.H.S.}]$

Hence, the correct form of the equation is $\text T = 2\pi\sqrt{\frac{\text l}{\text g}}$.

Given,

• Percentage error in the mass, $\frac{\Delta \text m}{\text m} \times 100$ = 3%
• Percentage error in the length, $\frac{\Delta \text l}{\text l} \times 100$ = 2%

The density of the cube is

$\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{\text m}{\text l^3}$

Here the mass occurs with the power 1 and the length with the power 3, so the maximum percentage error in the density is

$\frac{\Delta \rho}{\rho} \times 100 = \frac{\Delta \text m}{\text m} \times 100 + 3\left(\frac{\Delta \text l}{\text l} \times 100\right)$

Substituting the values,

$$\begin{aligned}\frac{\Delta \rho}{\rho} \times 100 = 3 + 3(2) \\[1em] = 3 + 6 \\[1em]\end{aligned}$$

= 9%

Hence, the maximum error in the density is 9%.

3%

Reason —

Given,

• Measured value of length $\text L$ = 20.0 cm
• Error in length $\Delta \text L$ = 1 mm = 0.1 cm
• Time for 100 oscillations = 90 s
• Error in time $\Delta \text t$ = 1 s (resolution of wrist watch)

So, time period for one oscillation,

$\text T = \frac{90}{100} = 0.9 \text { s}$

And error in time period,

$\Delta \text T = \frac{\Delta \text t}{100} = \frac{1}{100} = 0.01 \text { s}$

As the formula for time period is,

$\text T = 2\pi\sqrt{\frac{\text L}{\text g}}$

Squaring both sides and solving for g,

$\text g = \frac{4\pi^2 \text L}{\text T^2}$

Then

Maximum percentage error in g is given by,

$\frac{\Delta \text g}{\text g}$ × 100% = $\frac{\Delta \text L}{\text L}$ × 100% + 2 × $\frac{\Delta \text T}{\text T}$ × 100%

= $\frac{0.1}{20.0}$ × 100% + 2 × $\frac{0.01}{0.9}$ × 100%

= 0.5% + 2 × 1.11%

= 0.5% + 2.22%

≈ 2.72% ≈ 3%

2.87 and 2.83

Reason —

Given,

• 1 Main Scale Division (MSD) = $\frac{1}{10}$ cm = 0.1 cm

For Calliper C₁:

• 10 VSD = 9 MSD

$\text {1 VSD} = \frac{9}{10} \text { MSD} = \frac{9}{10} \times 0.1 = 0.09 \text { cm}$

Least count of C₁,

$\text {LC}_1 = \text {1 MSD} - \text {1 VSD} = 0.1 - 0.09 = 0.01 \text { cm}$

From the figure,

• Main Scale Reading (MSR) = 2.8 cm
• Coinciding vernier division (VD) = 7

$$\begin{aligned}\text {Reading of C}_1 = \text {MSR} + \text {VD} \times \text {LC}_1 \\[1em] = 2.8 + 7 \times 0.01 \\[1em] = 2.8 + 0.07 \\[1em] = 2.87 \text { cm}\end{aligned}$$

For Calliper C₂:

• 10 VSD = 11 MSD (Backward Vernier)

$\text {1 VSD} = \frac{11}{10} \text { MSD} = \frac{11}{10} \times 0.1 = 0.11 \text { cm}$

Least count of C₂,

$\text {LC}_2 = \text {1 VSD} - \text {1 MSD} = 0.11 - 0.10 = 0.01 \text { cm}$

Since C₂ is a backward vernier (VSD > MSD), the reading is given by,

$\text {Reading} = \text {MSR} - \text {VD} \times \text {LC}_2$

From the figure,

• Main Scale Reading (MSR) = 2.9 cm
• Coinciding vernier division (VD) = 7

$$\begin{aligned}\text {Reading of C}_2 = \text {MSR} - \text {VD} \times \text {LC}_2 \\[1em] = 2.9 - 7 \times 0.01 \\[1em] = 2.9 - 0.07 \\[1em] = 2.83 \text { cm}\end{aligned}$$

92 ± 2 s

Reason — Given,

• Measured values: 90 s, 91 s, 95 s, 92 s
• Minimum division of measuring clock = 1 s

Mean time is given by,

$$\begin{aligned}\text {T}_{\text{mean}} = \frac{90 + 91 + 95 + 92}{4} \\[1em] = \frac{368}{4} \\[1em] = 92 \text { s}\end{aligned}$$

Absolute errors in each measurement,

$$\begin{aligned}|\Delta \text T_1| = |90 - 92| = 2 \text { s} \\[1em] |\Delta \text T_2| = |91 - 92| = 1 \text { s} \\[1em] |\Delta \text T_3| = |95 - 92| = 3 \text { s} \\[1em] |\Delta \text T_4| = |92 - 92| = 0 \text { s}\end{aligned}$$

Mean absolute error is given by,

$$\begin{aligned}\Delta \text T_{\text{mean}} = \frac{|\Delta \text T_1| + |\Delta \text T_2| + |\Delta \text T_3| + |\Delta \text T_4|}{4} \\[1em] = \frac{2 + 1 + 3 + 0}{4} \\[1em] = \frac{6}{4} \\[1em] = 1.5 \text { s}\end{aligned}$$

Since the minimum division of the measuring clock is 1 s, the error must be rounded to the nearest integer,

$\Delta \text T_{\text{mean}} \approx 2 \text { s}$

0.80 mm

Reason —

Given,

• Pitch of screw gauge = 0.5 mm
• Number of circular scale divisions = 50
• At zero: 45th division coincides and zero of main scale is barely visible
• Main Scale Reading (MSR) = 0.5 mm
• Coinciding Circular Division = 25

Least count of screw gauge,

$\text {LC} = \frac{\text {Pitch}}{\text {Number of divisions}} = \frac{0.5}{50} = 0.01 \text { mm}$

Since the zero of the main scale is barely visible (not fully covered), it means the thimble has not yet reached the zero mark. This indicates a negative zero error,

$$\begin{aligned}\text {Zero Error} = -(50 - 45) \times \text {LC} \\[1em] = -5 \times 0.01 \\[1em] = -0.05 \text { mm}\end{aligned}$$

Observed reading of the sheet,

$$\begin{aligned}\text {Observed Reading} = \text {MSR} + \text {Circular Division} \times \text {LC} \\[1em] = 0.5 + 25 \times 0.01 \\[1em] = 0.5 + 0.25 \\[1em] = 0.75 \text { mm}\end{aligned}$$

Corrected (Actual) thickness is given by,

$$\begin{aligned}\text {Actual Reading} = \text {Observed Reading} - \text {Zero Error} \\[1em] = 0.75 - (-0.05) \\[1em] = 0.75 + 0.05 \\[1em] = 0.80 \text { mm}\end{aligned}$$

$\text R$

Reason —

Given,

• Total mass of the sphere = constant
• Instantaneous density ρ is uniform throughout the volume
• $\frac {1}{\rho}\frac{\text {dρ}}{\text {dt}}$ = constant

Mass of the expanding sphere is given by,

$\text M = \rho \times \frac{4}{3}\text {πR}^3$

Since mass M remains constant, differentiating both sides with respect to time,

$$\begin{aligned}\frac{\text {dM}}{\text {dt}} = 0 \\[1em] \frac{\text d}{\text {dt}}\left(\rho \times \frac{4}{3}\text {πR}^3\right) = 0 \\[1em] \frac{4}{3}\text {πR}^3 \times \frac{\text {dρ}}{\text {dt}} + ρ \times \frac{\text d}{\text {dt}}\left(\frac{4}{3}\text {πR}^3\right)=0\\[1em] \frac{4}{3}\text {πR}^3\frac{\text {dρ}}{\text {dt}} + \frac{4}{3}πρ \times 3\text R^2 \frac{\text {dR}}{\text {dt}} \\[1em]\end{aligned}$$

Dividing throughout by $\frac{4}{3}\text {πρR}^3$,

$\frac{1}{\rho}\frac{\text {dρ}}{\text {dt}} + \frac{3}{\text R}\frac{\text {dR}}{\text {dt}} = 0$

Since velocity of any point on the surface $\text v = \frac{\text {dR}}{\text {dt}}$, we get,

$\frac{1}{\rho}\frac{\text {dρ}}{\text {dt}} = -\frac{3\text v}{\text R}$

Since $\frac{1}{\rho}\frac{\text {dρ}}{\text {dt}}$ = constant, we have,

$$\begin{aligned}\frac{\text v}{\text R} = \text {constant} \\[1em] \text v \propto \text R\end{aligned}$$

1%

Reason —

Given,

• Error in time measurement $\Delta \text T$ = 0.01 s
• Depth of well $\text L$ = 20 m
• Acceleration due to gravity $\text g$ = 10 m s^-2
• Velocity of sound, $\text v$ = 300 m s^-1

The total time T consists of two parts — time $\text t_1$ for stone to fall and time $\text t_2$ for sound to travel back,

$$\begin{aligned}\text L = 0 \times \text t_1 + \frac{1}{2}\text {gt}_1^2 \implies \text t_1 = \sqrt{\frac{2\text L}{\text g}} \\[1em] \text t_2 = \frac{\text {Depth of well}}{\text {Velocity of sound}} \\[1em] \text t_2 = \frac{\text L}{\text v}\end{aligned}$$

So total time is,

$\text T = \text t_1 + \text t_2 = \sqrt{\frac{2\text L}{\text g}} + \frac{\text L}{\text v}$

Differentiating T with respect to L to find the relation between δT and δL,

$$\begin{aligned}\frac{\text {dT}}{\text {dL}} = \sqrt{\frac{2}{\text g}}\times \frac{1}{2\sqrt{\text L}}+ \frac{1}{\text v}\\[1em] \frac{\text {dT}}{\text {dL}} = \frac{1}{\sqrt{2\text {Lg}}} + \frac{1}{\text v}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\frac{\text {dT}}{\text {dL}} = \frac{1}{\sqrt{2 \times 20 \times 10}} + \frac{1}{300} \\[1em] = \frac{1}{\sqrt{400}} + \frac{1}{300} \\[1em] = \frac{1}{20} + \frac{1}{300} \\[1em] = \frac{15}{300} + \frac{1}{300} \\[1em] = \frac{16}{300}\end{aligned}$$

Then

Error in depth δL is given by,

$\Delta \text L = \frac{\Delta \text T}{\left(\frac{\text {dT}}{\text {dL}}\right)} = \frac{0.01}{\frac{16}{300}} = \frac{0.01 \times 300}{16} = \frac{3}{16} \approx 0.1875 \text { m}$

Fractional error in measurement,

$\frac{\Delta \text L}{\text L} = \frac{0.1875}{20} \approx 0.009375 \approx 1%$