Dimensional Analysis, Units and Errors Numerical Problems and Solutions : Practice Questions and Answers

Dimensional analysis, units, measurements, and errors form the foundation of quantitative science, ensuring that physical quantities are expressed accurately and consistently. These concepts are essential for solving numerical problems in physics and engineering, as they help verify equations, convert units, and estimate uncertainties in measurements. Mastery of these topics not only improves problem-solving skills but also builds a deeper understanding of real-world scientific applications.

Solved Numerical Problems

Type A. On Checking the correctness of a relation

Problem. Check the dimensional consistency of the equation :

$$
F S = \frac{1}{2} m v^2 – \frac{1}{2} m u^2,
$$

where $S$ is distance moved, $u$ and $v$ are the initial and final velocities of a body of mass $m$ and $F$ is the force acting on it.

Solution. We have, $F S = \frac{1}{2} m v^2 – \frac{1}{2} m u^2$

Dimensional formula of $FS = [M L T^{-2}] \times [L] = [M L^2 T^{-2}]$

Dimensional formula of $\frac{1}{2} m v^2 = [M] [L T^{-1}]^2 = [M L^2 T^{-2}]$

Dimensional formula of $\frac{1}{2} m u^2 = [M] [L T^{-1}]^2 = [M L^2 T^{-2}]$

As the dimensional formula of all the terms in the given equation is same, the equation is a correct one.

---

Problem. In van der Waals’ equation :

$$
\left( P + \frac{a}{V^2} \right) (V – b) = R T
$$

what are the dimensions of constants ‘a’ and ‘b’ ?

Solution. Here, $\left( P + \dfrac{a}{V^2} \right) (V – b) = R T$

According to the principle of homogeneity of dimensions, $\dfrac{a}{V^2}$ and $b$ should have dimensions of $P$ (pressure) and $V$ (volume) respectively.

Thus, $\dfrac{a}{V^2} =$ pressure

or $a =$ pressure $\times V^2 = [M L^{-1} T^{-2}] [L^3] = [M L^5 T^{-2}]$

Also, $b =$ volume $= [L^3] = [M^0 L^3 T^0]$

---

Problem. Check the correctness of relation by dimensional analysis :

$$
\lambda = \frac{h}{m v}
$$

where the letters have their usual meanings.

Solution. Here, $\lambda = \dfrac{h}{m v}$

Dimensional formula of wavelength $\lambda = [L]$

Therefore, dimensional formula of L.H.S.

$\lambda = [L] = [M^0 L T^0]$

Dimensional formula of Planck’s constant $h$,

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

Dimensional formula of mass $m = [M]$

Dimensional formula of velocity $v = [L T^{-1}]$

Therefore, dimensional formula of R.H.S.

$\dfrac{h}{m v} = \dfrac{[M L^2 T^{-1}]}{[M] \times [L T^{-1}]} = [M^0 L T^0]$

As dimensional formula of L.H.S. is the same as that of R.H.S., the given relation is correct.

---

Problem. Check the correctness of the relation $\tau = I \alpha$, where $\tau$ is the torque acting on a body, $I$ is the moment of inertia and $\alpha$ is the angular acceleration.

Solution. We have, $\tau = I \alpha$

Now, $\tau =$ force $\times$ distance $= [M L T^{-2}] [L] = [M L^2 T^{-2}]$

Therefore, dimensional formula of L.H.S. $= [M L^2 T^{-2}]$

$I =$ mass $\times$ distance$^2 = [M] [L]^2 = [M L^2]$

$\alpha =$ angular acceleration $= \dfrac{\text{angle}}{(\text{time})^2} = [T^{-2}]$

Therefore, dimensional formula of R.H.S.

$I \alpha = [M L^2] [T^{-2}] = [M L^2 T^{-2}]$

Since the dimensional formulae of L.H.S. and R.H.S. are the same, the given relation is correct.

---

Problem. Test by using dimensional analysis, the accuracy of the relation :

$$
v_c = \dfrac{k \eta}{r \rho}
$$

Solution. Here, $v_c = \dfrac{k \eta}{r \rho}$

Dimensional formula of $v_c$ (critical velocity) $= [L T^{-1}]$

Therefore, dimensional formula of L.H.S.

$v_c = [L T^{-1}] = [M^0 L T^{-1}]$

Dimensional formula of $\eta$ (coefficient of viscosity) $= [M L^{-1} T^{-1}]$

Dimensional formula of $r$ (radius) $= [L]$

Dimensional formula of $\rho$ (density) $= [M L^{-3}]$

The constant $k$ is dimensionless.

Therefore, dimensional formula of R.H.S.

$\dfrac{k \eta}{r \rho} = \dfrac{[M L^{-1} T^{-1}]}{[L][M L^{-3}]} = [M^0 L T^{-1}]$

As the dimensional formula of L.H.S. is the same as that of R.H.S., the given relation is correct.

---

Type B. On Deriving a Physical Relation

Problem. Consider a simple pendulum. The period of oscillation of the simple pendulum depends on its length and acceleration due to gravity. Derive the expression for period of oscillation by method of dimensions.

Solution. Suppose that period of oscillation $t$ of the simple pendulum depends on $a$th power of its length $l$ and $b$th power of acceleration due to gravity. Then,

$$
t = k l^a g^b \quad \dots(i)
$$

[L.H.S.] $= [t] = [T] = [M^0 L^0 T^1]$

[R.H.S.] $= [k l^a g^b] = [L]^a [L T^{-2}]^b = [M^0 L^{a+b} T^{-2b}]$

Comparing the dimensions of $M$, $L$ and $T$ on the two sides, we get

$$
a + b = 0 \quad \text{and} \quad -2b = 1 \quad \text{or} \quad b = -1/2
$$

It follows that

$$
a = -b = 1/2
$$

Substituting for $a$ and $b$ in the equation (i), we have

$$
t = k \sqrt{\frac{l}{g}}
$$

---

Problem. The velocity ($v$) of sound through a medium may be assumed to depend on (i) the density ($\rho$) of the medium and (ii) modulus of elasticity ($E$). If the dimensions for elasticity (ratio of stress to strain) are $[M L^{-1} T^{-2}]$, deduce by the method of dimensions the formula for the velocity of sound.

Solution. Let

$$
v \propto \rho^a \quad \dots(i)
$$

$$
v\propto E^b \quad \dots(ii)
$$

Combining the equations (i) and (ii), we get

$$
v = k \rho^a E^b, \quad \dots(iii)
$$

where $k$ is a dimensionless constant of proportionality.

Writing the dimensional formula of $v$, $\rho$ and $E$ in the equation (iii), we get

$$
[L T^{-1}] = [M L^{-3}]^a [M L^{-1} T^{-2}]^b
$$

or

$$
[M^0 L T^{-1}] = [M^{a+b} L^{-3a-b} T^{-2b}]
$$

Comparing the dimensions of $M$, $L$ and $T$ on the two sides, we get

$$
a + b = 0 \quad \dots(iv)
$$

$$
-3a – b = 1 \quad \dots(v)
$$

and

$$
-2b = -1 \quad \text{or} \quad b = 1/2
$$

Substituting for $b$ in the equation (iv), we get

$$
a + 1/2 = 0 \quad \text{or} \quad a = -1/2
$$

In the equation (iii), putting the values of $a$ and $b$, we get

$$
v = k \rho^{-1/2} E^{1/2}
$$

or

$$
v = k \sqrt{\frac{E}{\rho}}
$$

The constant of proportionality $k$ is found to be 1. Therefore, the above relation becomes

$$
\therefore \quad v = \sqrt{\frac{E}{\rho}}
$$

---

Problem. Assuming that the frequency ($\nu$) of a vibrating string depends upon the load ($F$) applied, length of the string ($l$) and mass per unit length ($m$), prove that

$$
\nu = \frac{1}{2l} \sqrt{\frac{F}{m}}
$$

Sol. Let

$$
\nu\propto F^a \quad \dots(i)
$$

$$
\nu\propto m^c \quad \dots(ii)
$$

$$
\nu\propto l^b \quad \dots(iii)
$$

Combining the equations (i), (ii) and (iii), we get

$$
\nu = k F^a l^b m^c, \quad \dots(iv)
$$

where $k$ is the dimensionless constant of proportionality.

In the equation (iv), substituting the dimensions of $\nu$, $F$, $l$ and $m$, we get

$$
[T^{-1}] = [M L T^{-2}]^a [L]^b [M L^{-1}]^c
$$

or

$$
[M^0 L^0 T^{-1}] = [M^{a+c} L^{a+b-c} T^{-2a}]
$$

Comparing the dimensions of $M$, $L$ and $T$ on the two sides, we get

$$
-2a = -1 \quad \text{or} \quad a = 1/2,
$$

$$
a + c = 0 \quad \text{or} \quad c = -a = -1/2
$$

and $a + b – c = 0 \quad \text{or} \quad b = c – a = -1/2 – 1/2 = -1$

In the equation (iv), putting the values of $a$, $b$ and $c$, we obtain

$$
\nu = k F^{1/2} l^{-1} m^{-1/2}
$$

or

$$
\nu = \frac{k}{l} \sqrt{\frac{F}{m}}
$$

The constant of proportionality $k$ is found to be equal to $1/2$. Therefore, the above relation becomes

$$
\nu = \frac{1}{2l} \sqrt{\frac{F}{m}}
$$

---

Problem. By the method of dimensional analysis, derive the relation :

$$
S = u t + \frac{1}{2} a t^2,
$$

where the letters have their usual meanings.

Solution. The distance ($S$) covered by a body depends upon :
initial velocity ($u$);
acceleration ($a$) and
time ($t$), for which it travels.

By the method of dimensions, expression for the distance ($S$) covered by a body is found in the following two steps :

(a) When the body has initial velocity but no acceleration :

Let

$$
S \propto u^x t^y \quad \text{or} \quad S = k_1 u^x t^y, \quad \dots(i)
$$

where $k_1$ is a dimensionless constant of proportionality.

Writing the dimensions of $S$, $u$ and $t$ in the equation (i), we have

$$
[L] = [L T^{-1}]^x [T]^y
$$

or

$$
[L] = [L^x T^{-x + y}]
$$

From the principle of homogeneity of dimensions, we have

$$
x = 1;
$$

and

$$
-x + y = 0 \quad \text{or} \quad y = x = 1
$$

Substituting for $x$ and $y$ in the equation (i), we have

$$
S = k_1 u t
$$

(b) When the body has acceleration but no initial velocity :

Let

$$
S \propto a^x t^y
$$

or

$$
S = k_2 a^x t^y,
$$

where $k_2$ is a dimensionless constant of proportionality.

Writing the dimensions of $S$, $a$ and $t$ in the equation (ii), we have

$$
[L] = [L T^{-2}]^x [T]^y
$$

or

$$
[L] = [L^x T^{-2x + y}]
$$

From the principle of homogeneity of dimensions, we have

$$
x = 1;
$$

and $-2x + y = 0$ or $y = 2x – 2 \times 1 = 2$

In the equation (ii), substituting for $x$ and $y$, we have

$$
S = k_2 a t^2
$$

In the real situation, when the body possesses both initial velocity and acceleration, the expression for $S$ must contain both $k_1 u t$ and $k_2 a t^2$ i.e.

$$
S = k_1 u t + k_2 a t^2
$$

The constants $k_1$ and $k_2$ come out to be 1 and $1/2$ respectively. Therefore, the required relation is

$$
S = u t + \frac{1}{2} a t^2
$$

---

Problem. Match the physical quantities given in Column I with their dimensions expressed in terms of Mass ($M$), Length ($L$), Time ($T$), and Charge ($Q$) given in Column II. Write the final matched pairs in a tabular format.

Column I	Column II
(A) Angular momentum	(i) $M L^{2} T^{-2}$
(B) Latent heat	(ii) $M L^{3} T^{-1} Q^{-2}$
(C) Torque	(iii) $M L^{2} Q^{-2}$
(D) Capacitance	(iv) $M^{-1} L^{-2} T^{2} Q^{2}$
(E) Inductance	(v) $M L^{2} T^{-1}$
(F) Resistivity	(vi) $L^{2} T^{-2}$

Solution : The dimensions of physical quantities mention in the above table as below :

Angular Momentum ($L = mvr$) : $[M] \times [L T^{-1}] \times [L] = \mathbf{M L^2 T^{-1}}$

Torque ($\tau = F \times r$) : $[M L T^{-2}] \times [L] = \mathbf{M L^2 T^{-2}}$

Latent Heat ($L = \dfrac{Q}{m}$) : Energy divided by mass : $\dfrac{M L^2 T^{-2}}{M} = \mathbf{L^2 T^{-2}}$

Capacitance ($C = \dfrac{Q}{V}$) : Since Potential ($V$) is Work/Charge ($W/Q$) : $C = \dfrac{Q}{W/Q} = \dfrac{Q^2}{W} = \dfrac{Q^2}{M L^2 T^{-2}} = \mathbf{M^{-1} L^{-2} T^2 Q^2}$

Inductance ($L$) : Using Energy $U = \frac{1}{2}LI^2$, where Current $I = Q/T$:$L = \dfrac{2 \times \text{Energy}}{I^2} = \dfrac{M L^2 T^{-2}}{(Q/T)^2} = \dfrac{M L^2 T^{-2}}{Q^2 T^{-2}} = \mathbf{M L^2 Q^{-2}}$

Resistivity ($\rho$) : $\rho = \dfrac{R \times A}{l} = \dfrac{(V/I) \times A}{l} = \dfrac{(W/Q) \times A}{(Q/T) \times l}$$\dfrac{(M L^2 T^{-2} Q^{-1}) \times L^2}{Q T^{-1} \times L} = \mathbf{M L^3 T^{-1} Q^{-2}}$

Correct Matches

Physical Quantity	Dimensional Formula
Angular momentum	$M L^{2} T^{-1}$
Latent heat	$L^{2} T^{-2}$
Torque	$M L^{2} T^{-2}$
Capacitance	$M^{-1} L^{-2} T^{2} Q^{2}$
Inductance	$M L^{2} Q^{-2}$
Resistivity	$M L^{3} T^{-1} Q^{-2}$

---

Problem. A gas bubble from an explosion under water, oscillates with a period $T$ proportional to $p^a d^b E^c$, where $p$ is the static pressure, $d$ is the density of water and $E$ is the total energy of the explosion. Find the values of $a$, $b$ and $c$.

Solution. Here, $T \propto p^a d^b E^c$

Now, $[T] = [\text{time period}] = [T]$;

$[p] = [\text{pressure}] = [M L^{-1} T^{-2}]$;

$[d] = [\text{density}] = [M L^{-3}]$;

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

Equating the two sides of the equation (i) dimensionally, we have

$$
[T] = [M L^{-1} T^{-2}]^a [M L^{-3}]^b [M L^2 T^{-2}]^c
$$

or $[M^0 L^0 T^1] = [M^{a+b+c} L^{-a-3b+2c} T^{-2a-2c}]$

Comparing the dimensions of $M$, $L$ and $T$ on the two sides, we have

$$
a + b + c = 0,
$$

$$
-a – 3b + 2c = 0 \quad \text{and}
$$

$$
-2a – 2c = 1
$$

On solving the above three equations in $a$, $b$ and $c$, we obtain

$$
a = -\frac{5}{6}; \quad b = \frac{1}{2} \quad \text{and} \quad c = \frac{1}{3}
$$

---

Problem. The density of a material in cgs system is 8 g cm$^{-3}$. In a system of units, in which unit of length is 5 cm and unit of mass is 20 g, what is the density of the material?

Solution. Let cgs system and the new system be called as the systems 1 and 2 respectively.

If $n_2$ is the numerical value of the density of the material in the new system, then

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

Now, dimensional formula of density $= [M L^{-3} T^0]$

Therefore, $a = 1$; $b = -3$ and $c = 0$

Hence,

$$
n_2 = n_1 \left[ \frac{M_1}{M_2} \right]^1 \left[ \frac{L_1}{L_2} \right]^{-3} \left[ \frac{T_1}{T_2} \right]^0
$$

$$
n_2 = n_1 \left[ \frac{M_1}{M_2} \right] \left[ \frac{L_1}{L_2} \right]^{-3}
$$

Here, $M_1 = 1$ g; $L_1 = 1$ cm; $n_1 = 8$

and $M_2 = 20$ g; $L_2 = 5$ cm

$$
\therefore n_2 = 8 \left[ \frac{1 \text{ g}}{20 \text{ g}} \right] \times \left[ \frac{1 \text{ cm}}{5 \text{ cm}} \right]^{-3} = \frac{8 \times (5)^3}{20} = 50
$$

Hence, density of the material in new system $= 50$ units

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Chapter Important Links

In this chapter on Dimensional Analysis, Units, Measurements, and Errors, we explore the principles that ensure accuracy and consistency in physical calculations. You will learn how to apply dimensional formulas, convert between unit systems, and calculate measurement errors with precision. The chapter also includes a wide range of numerical problems with step-by-step solutions and practice questions to strengthen your understanding and prepare you for JEE exams.

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Frequently Asked Very Short Answer Questions (VSAQs)

Q. What are the dimensions of angular displacement?

Ans. Since angular displacement $= \dfrac{\text{length of arc}}{\text{radius}}$; it is dimensionless.

---

Q. What are the dimensions of angular momentum?

Ans. $[M L^2 T^{-1}]$

---

Q. What are the dimensional formulae for the following:
(i) pressure, (ii) power, (iii) density and (iv) angle.

Ans.
(i) $[M L^{-1} T^{-2}]$,
(ii) $[M L^2 T^{-3}]$,
(iii) $[M L^{-3} T^0]$ and (iv) dimensionless.

---

Q. Which physical quantities are represented by the following?
(a) Product of moment of inertia and angular velocity.
(b) Product of moment of inertia and angular acceleration.

Ans. (a) Angular momentum (b) Torque

---

Q. Which physical quantities are represented by the following?
(a) Rate of change of angular momentum.
(b) Moment of linear momentum.

Ans. (a) Torque (b) Angular momentum

---

Q. Find dimensional formula of latent heat.

Ans. $[\text{Latent heat}] = \left[ \dfrac{\text{quantity of heat}}{\text{mass}} \right] = \left[ \dfrac{\text{energy}}{\text{mass}} \right]$
$= \dfrac{[M L^2 T^{-2}]}{[M]} = [M^0 L^2 T^{-2}]$

---

Q. Write three non-dimensional variables.

Ans. Strain, angular displacement and trigonometric ratios.

---

Q. Name two physical quantities which have same dimensions as that of work.

Ans. Torque and energy.

---

Q. Write three mathematical variables.

Ans. Angle, T-ratio, coordinates of a point.

---

Q. Are all constants dimensionless?

Ans. No. For example, $G$ (gravitational constant); $h$ (Planck’s constant); $k$ (Boltzmann constant); etc are constants, but possess dimensions.

---

Q. Can a quantity have units but still be dimensionless?

Ans. Yes. For example, unit of angle is radian, but it is a dimensionless quantity.

---

Q. Can a quantity have dimensions but still have no units?

Ans. No.

---

Q. Can a quantity have neither units nor dimensions?

Ans. Yes. For example, strain has neither units nor dimensions.

---

Q. State the principle of homogeneity of dimensions.

Ans. It states that the dimensions of the fundamental quantities (mass, length and time) are same in each and every term on either side of a physical relation.

---

Q. Is there any difference between dimensional formula and dimensional equation ?

Ans. Yes, the two are different from each other.

---

Q. What is the advantage of expressing physical quantities in terms of dimensional equations ?

Ans. The dimensional formula of a physical quantity indicates the fundamental units on which the physical quantity depends. It further tells the powers of the fundamental units, on which the given physical quantity depends.

---

Q. Will the measure of a physical quantity depend upon the system of units used ?

Ans. The measure of a physical quantity does not depend on the system of units used. It is because, measure of a physical quantity $= n u$
If the size of the unit ($u$) is small, then numerical value ($n$) of the physical quantity will be large and vice-versa, but the measure $n u$ will be constant.

---

Q. Write four pairs of physical quantities, which have the same dimensional formula.

Ans.

1. Work and energy
2. Pressure and stress
3. Velocity gradient and frequency
4. Angular momentum and Planck’s constant.

---

Q. Pressure may be defined as the linear momentum per unit volume. Is it correct ? Check by the method of dimensions.

Ans. Now, pressure $= $ force $/$ area $= M L T^{-2} / L^2 = [M] L^{-1} T^{-2}$
Linear momentum per unit volume
$\dfrac{\text{momentum}}{\text{volume}} = \dfrac{M L T^{-1}}{L^3} = [M L^{-2} T^{-1}]$
As the dimensions of pressure and linear momentum per unit volume are not same, it is incorrect.

---

Q. What is the unit of pressure in SI ?

Ans. N m$^{-2}$

---

Q. The rotational kinetic energy of a body is given by $E = \frac{1}{2} I \omega^2$, where $\omega$ is angular velocity of the body. Use this equation to get dimensional formula for $I$.

Ans. Here, $E = \frac{1}{2} I \omega^2$
or
$I = \dfrac{2E}{\omega^2}$
$\therefore \quad [I] = \left[ \dfrac{2 \times \text{energy}}{(\text{angular velocity})^2} \right] = \dfrac{[M L^2 T^{-2}]}{[T^{-1}]^2} = [M L^2 T^0]$

---

Q. What are the dimensions of gravitational constant ?

Ans. We know, $F = G M_1 M_2 / r^2$
$\therefore \quad [G] = \left[ \dfrac{F \times r^2}{M_1 \times M_2} \right] = \dfrac{[M L T^{-2}] \times [L]^2}{[M] \times [M]} = [M^{-1} L^3 T^{-2}]$

---

Q. If $x = a + b t + c t^2$, where $x$ is in metres and $t$ in seconds, find the units of $b$.

Ans. Here, $x = a + b t + c t^2$
Since $x$ stands for length, all the three factors on the R.H.S. of the given relation should possess the dimension of length.
$\therefore \quad [b t] = [L] $
or
$[b] = \dfrac{[L]}{[T]} = [L T^{-1}]$
i.e. $b$ represents velocity and hence its unit is m s$^{-1}$

---

Q. In the equation :
$y = A \sin (\omega t – k x)$,
$t$ and $x$ stand for time and distance respectively. Obtain the dimensional formula of $\omega$ and $k$.

Ans. Since angle is dimensionless, both $\omega t$ and $k x$ have no dimensions.
$$\therefore \quad \omega = \dfrac{\text{dimensionless quantity}}{t} = \dfrac{1}{[T]} = [M^0 L^0 T^{-1}]$$

and $$k = \dfrac{\text{dimensionless quantity}}{x} = \dfrac{1}{[L]} = [M^0 L^{-1} T^0]$$

---

Q. Give the dimensional formula of thermal conductivity.

Ans. The coefficient of thermal conductivity is given by
$k = \dfrac{Q d}{A (\theta_1 – \theta_2) t}$,

where $Q$, $d$, $A$, $(\theta_1 – \theta_2)$ and $t$ stand for energy, distance, area, temperature difference and time.

---

Q. Write the dimensions of $a / b$ in the relation, $F = a \sqrt{x} + b t^2$, where $F$ is force, $x$ is distance and $t$ is time.

Ans. Here, $F = a \sqrt{x} + b t^2$
Since $F$ stands for force, both the factors on the R.H.S. of the given relation should possess the dimension of force.
$\therefore [a \sqrt{x}] = [F]$

or $[a] = \dfrac{[F]}{[\sqrt{x}]} = \dfrac{M L T^{-2}}{L^{1/2}} = [M L^{1/2} T^{-2}]$
Also, $[b t^2] = [F]$

or $[b] = \dfrac{[F]}{[t^2]} = \dfrac{M L T^{-2}}{T^2} = [M L T^{-4}]$

Hence, $[a/b] = \dfrac{[M L^{1/2} T^{-2}]}{[M L T^{-4}]} = [M L^{-1/2} T^2]$

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Q. Choose the correct formula for the displacement ($y$) of a particle executing periodic motion :
(i) $y = r \sin \frac{2\pi}{T} t$
(ii) $y = r \sin v t$
(iii) $y = r \sin \frac{2\pi}{T} t$
(iv) $y = r \sin \frac{2\pi}{v} t$,
where $T$, $r$, $v$ and $t$ stand for time period, amplitude, velocity and time respectively.

Ans. The argument of a trigonometric function (usually called angle) is a dimensionless quantity. Only in the first expression, the angle i.e. $\dfrac{2\pi}{T} t$ is dimensionless. In the other equations, the factors $v$, $t$, $\dfrac{2\pi}{r}$, $\dfrac{2\pi}{v}$ have the dimensions of $[L]$, $[L^{-1} T]$ and $[L^{-1} T^2]$ respectively. Hence, the displacement of the particle is given only by
$y = r \sin \dfrac{2\pi}{T} t$

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Q. The number of particles given by
$$n = -D \left( \dfrac{n_2 – n_1}{x_2 – x_1} \right)$$
are crossing a unit area perpendicular to X-axis in unit time. $n_1$ and $n_2$ are the number of particles per unit volume for the values of $x$ meant to be $x_1$ and $x_2$. What is the dimensional formula of diffusion constant $D$?

Ans. Here, $n = -D \left( \dfrac{n_2 – n_1}{x_2 – x_1} \right)$

or $D = -n \left( \dfrac{x_2 – x_1}{n_2 – n_1} \right)$

Dimensional formula of $n$ $= \left[ \dfrac{\text{number}}{\text{area} \times \text{time}} \right] = \dfrac{1}{[L^2][T]} = [L^{-2} T^{-1}]$

Dimensional formula of $n_1$ or $n_2$

$= \left[ \dfrac{\text{number}}{\text{volume}} \right] = \dfrac{1}{[L^3]} = [L^{-3}]$
Dimensional formula of $x_1$ or $x_2 = [L]$

$\therefore [D] = \dfrac{[L^{-2} T^{-1}][L]}{[L^{-3}]} = [M^0 L^2 T^{-1}]$

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Q. Calculate the dimensions of (i) force ($F$) and (ii) impulse ($I$) in terms of velocity ($v$), density ($\rho$) and frequency ($v$) as the fundamental units.

Ans. We know that
$[v] = [L T^{-1}]$, $[\rho] = [M L^{-3}]$ and $[v] = [T^{-1}]$

(i) Now, $[F] = [M L T^{-2}] = [M L^{-3} \times L^3 \times L T^{-2}]$

$= [M L^{-3}] [L T^{-1}]^4 / [T^{-1}]^2$

or $F = \dfrac{\rho v^4}{v^2} = \rho v^4 v^{-2}$

(ii) Now, $[I] = [M L T^{-1}] = [M L^{-3} \times L^3 \times L T^{-1}]$

$= [M L^{-3}] [L T^{-1}]^4 / [T^{-1}]^3$

or $I = \dfrac{\rho v^4}{v^3} = \rho v^4 v^{-3}$

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Q. The SI unit of energy is $J = \text{kg m}^2 \text{s}^{-2}$; that of speed $v$ is $\text{m s}^{-1}$ and of acceleration $a$ is $\text{m s}^{-2}$. Which of the formulae for kinetic energy ($K$) given below can you rule out on the basis of dimensional arguments ($m$ stands for the mass of the body).
(a) $K = m^2 v^3$
(b) $K = m a$
(c) $K = \frac{1}{2} m v^2$
(d) $K = \frac{1}{2} m v^2 + m a$

Ans. Here, $[K] = \text{kg m}^2 \text{s}^{-2}$, $[v] = \text{m s}^{-1}$ and $[a] = \text{m s}^{-2}$

(a) $[m^2 v^3] = (\text{kg})^2 \times (\text{m s}^{-1})^3 = \text{kg}^2 \text{m}^{-3} \text{s}^{-3}$

Therefore, formula $K = m^2 v^3$ is not possible.

(b) $[m a] = \text{kg} \times \text{m s}^{-2} = \text{kg m s}^{-2}$

Therefore, formula $K = m a$ is not possible.

(c) $\left[ \frac{1}{2} m v^2 \right] = \text{kg} \times (\text{m s}^{-1})^2 = \text{kg m}^2 \text{s}^{-2}$

Therefore, formula $K = \frac{1}{2} m v^2$ is possible.

(d) Since the dimensions (or units) of two terms in the formula $K = \frac{1}{2} m v^2 + m a$ are not same, this formula is also not possible.

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Q. If the velocity of light $c$, gravitational constant $G$ and Planck’s constant $h$ be chosen as fundamental units, find the dimensions of (i) mass, (ii) length and (iii) time in new system.

Ans. We know that
$[c] = [L T^{-1}]$, $[G] = [M^{-1} L^3 T^{-2}]$ and $[h] = [M L^2 T^{-1}]$

(i) Now, $[M] = \left[ \dfrac{h c}{G} \right]^{1/2} = [h^{1/2} c^{1/2} G^{-1/2}]$

(ii) $[L] = \left[ \dfrac{h G}{c^3} \right]^{1/2} = [h^{1/2} c^{-3/2} G^{1/2}]$

(iii) $[T] = \left[ \dfrac{h G}{c^5} \right]^{1/2} = [h^{1/2} c^{-5/2} G^{1/2}]$

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Q. The knowledge of Reynold number (R) tells, whether the flow of a liquid through a pipe will be streamlined or not. By the method of dimensions, find the expression for Reynold number, if it depends upon the velocity (v) of flow of the liquid, density ($\rho$) of the liquid, coefficient of viscosity ($\eta$) of the liquid and diameter (D) of the pipe. Given that R varies directly as the diameter D of the pipe.

Ans. Let $R = k v^a \rho^b \eta^c D^d$

Since R varies directly as diameter D of the pipe, $d = 1$

$\therefore R = k v^a \rho^b \eta^c D$

[L.H.S.] $= [R] = [\text{number}] = [M^0 L^0 T^0]$

[R.H.S.] $= [k v^a \rho^b \eta^c D] = [L T^{-1}]^a [M L^{-3}]^b [M L^{-1} T^{-1}]^c [L]$

$= [M^{b+c} L^{a-3b-c+1} T^{-a-c}]$

Comparing the dimensions of $M$, $L$ and $T$ on the two sides, we get
$b + c = 0$, $a – 3b – c + 1 = 0$ and $a + c = 0$

On solving, we get
$a = 1$, $b = 1$ and $c = -1$

Substituting for $a$, $b$ and $c$ in the equation (i), we get

$R = k \dfrac{v \rho D}{\eta}$

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FAQs Frequently Asked Short Answer Questions

1. What is meant by dimensions of a physical quantity? What are the dimensional formulae and equations?
2. Write down the dimensional formula of the following physical quantities : (i) work (ii) angular velocity (iii) pressure (iv) Planck’s constant
3. Find dimensional formula for angle, angular velocity, angular acceleration, torque and surface tension.
4. Obtain dimensional formula for universal gravitational constant (G), Planck’s constant (h) and universal gas constant (R).
5. What do you understand by dimensions of a physical quantity ? Discuss the principle of homogeneity of dimensions, giving an example.
6. Find the dimensional formula of Planck’s constant and universal gravitational constant.
7. What are the uses of dimensional analysis ? Derive the formula used for conversion of one system of units into another.
8. Define dimensional formula. Give uses of dimensional analysis.
9. Write down the limitations of dimensional analysis.
10. Name the various advantages and limitations of dimensional analysis.

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FAQs Frequently Asked Long Answer Questions

1. Define dimensions and dimensional formula of a physical quantity. What are the various uses of dimensional analysis ? Explain in brief. What are its limitations ?
2. What is meant by dimensions of a physical quantity and the dimensional equation? State giving illustrations the uses of dimensional equations.
3. State and explain, giving illustrations, the uses and limitations of dimensional equations.

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FAQs Frequently Asked Numerical Problems And Solutions For Practice

Type A. On Dimensional formula

1. In the Poiseuille’s equation :
   $V = \frac{\pi}{8} \frac{p r^4}{\eta l}$
   determine the dimensions of $\eta$.
   [Ans. $[M L^{-1} T^{-1}]$]
2. Deduce the dimensional formula of the following physical quantities :
   (i) Velocity gradient (ii) Young’s modulus (iii) Gravitational constant (iv) Surface tension (v) Universal gas constant.
   [Ans. (i) $[M^0 L^0 T^{-1}]$; (ii) $[M L^{-1} T^{-2}]$; (iii) $[M^{-1} L^3 T^{-2}]$; (iv) $[M L^0 T^{-2}]$; (v) $[M L^2 T^{-2} K^{-1}]$]
3. Show that the angular momentum has the same physical units as those of Planck’s constant. Planck’s constant is given by the relation $E = h\nu$, where $E$ is energy and $\nu$ is frequency.
4. Write the dimensions of $a$ and $b$ in the relation :
   $E = \frac{b – x^2}{a t}$
   where $E$, $x$ and $t$ represent energy, distance and time respectively. [Ans. $[M^{-1} L^0 T]$ $[M^0 L^2 T^0]$]
5. Write the dimensions of $a/b$ in the relation :
   $F = a\sqrt{x} + b t^2$
   where $F$ is force, $x$ is distance and $t$ is time.
   [Ans. $[M^0 L^{-1/2} T^2]$]

Type B. On Checking the correctness of a given physical relation

6. Check the correctness of the relation :
   $F = \frac{m v^2}{r}$
   where the letters have their usual meanings.
   [Ans. Correct]
7. Verify dimensionally the relation :
   $t = 2\pi \sqrt{\frac{l}{g}}$
   for the time period of a simple pendulum. Here, $l$ is the length of the pendulum and $g$ is acceleration due to gravity.
   [Ans. Correct]
8. The frequency $\nu$ of a string of length $l$ vibrating under tension $F$ is given by
   $\nu = \frac{1}{2l} \sqrt{\frac{F}{m}}$
   where $m$ is mass per unit length. Check whether the relation is correct or not.
   [Ans. Correct]
9. Check the dimensional consistency in case of the following equations :
   (i) $S = u t + \frac{1}{2} a t^2$
   (ii) $v = u + a t$
   (iii) $v^2 – u^2 = 2 a S$
   [Ans. (i) correct ; (ii) correct ; (iii) correct]
10. Check the correctness of the relation :
    $\rho = \frac{3g}{4\pi r G}$
    where the letters have their usual meanings.
    [Ans. Correct]
11. Check the correctness of the relation :
    $v = \sqrt{\frac{2GM}{R}}$,
    where $v$ is velocity, $G$ is gravitational constant, $M$ is the mass and $R$ is the radius of the earth.
    [Ans. Correct]
12. Check the correctness of the relation :
    $v = \sqrt{\frac{E}{\rho}}$,
    where $v$ is speed of the sound in a medium of elasticity $E$ and density $\rho$.
    [Ans. Correct]
13. Check the correctness of the relation :
    $h = \frac{r \rho g}{2 S \cos \theta}$,
    where $h$ is height, $r$ is radius, $\rho$ is density, $S$ is surface tension and $g$ is acceleration due to gravity.
    [Ans. Incorrect]
14. Time period of an oscillating drop of radius $r$, density $\rho$ and surface tension $S$ is
    $t = k \sqrt{\frac{\rho r^3}{S}}$
    Check the correctness of the relation.
    [Ans. Correct]

Type C. On Deriving a physical relation

15. Show dimensionally that the time period of a simple pendulum is given by
    $t = 2 \pi \sqrt{\frac{l}{g}}$,
    where $l$ is length of the pendulum and $g$ is acceleration due to gravity.
16. A body of mass $m$ is moving in a circle of radius $r$ with angular velocity $\omega$. Find expression for centripetal force $F$ acting on it using method of dimensional analysis. [Ans. $F = m r \omega^2$]
17. The escape velocity ($v$) of a body depends upon : (i) the acceleration due to gravity ($g$) on the planet, (ii) the radius ($R$) of the planet. Establish dimensionally the relationship between them.
    [Ans. $v = k\sqrt{gR}$]
18. The orbital velocity $v$ of a satellite may depend on its mass $m$, distance $r$ from the centre of earth and acceleration due to gravity $g$. Obtain an expression for the orbital velocity. [Ans. $v = k\sqrt{gr}$]
19. The velocity of a body which has fallen freely under gravity varies as $g^p h^q$, where $g$ is acceleration due to gravity at a place and $h$ is the height which the body has fallen from. Find the values of $p$ and $q$.
    [Ans. 1/2, 1/2]
20. Derive by the method of dimensions, an expression for the energy ($E$) of a body executing S.H.M., assuming that this energy depends upon : the mass ($m$), the frequency ($\nu$) and the amplitude of vibration ($r$).
    [Ans. $E = k m \nu^2 r^2$]
21. The velocity ($v$) of water waves may depend upon their wavelength $\lambda$, the density of water $\rho$ and the acceleration due to gravity $g$. Find the relation between these quantities by the method of dimensions. [Ans. $v = k\sqrt{g\lambda}$]
22. Assuming that the mass $m$ of the largest stone that can be moved by a flowing river depends upon the velocity $v$, the density $\rho$ and acceleration due to gravity $g$, show that $m$ varies with sixth power of the velocity of flow.
23. The wavelength $\lambda$ associated with a moving particle depends upon its mass $m$, its velocity $v$ and Planck’s constant $h$. Show dimensionally that
    $\lambda \propto \frac{h}{m v}$
24. Obtain by the method of dimensional analysis, an expression for the surface tension of a liquid rising in a capillary tube. Assume that the surface tension depends upon (i) mass ($m$) of the liquid, (ii) pressure ($p$) of the liquid and (iii) radius ($r$) of the capillary tube. The constant $k = 1/2$.
    [Ans. $S = \frac{p r}{2}$]
25. The coefficient of viscosity ($\eta$) of a gas depends on the mass ($m$), the effective diameter ($d$) and the mean speed ($v$) of the gas molecules. Use dimensional analysis to find the relation between them.
    [Ans. $\eta = k \frac{m v}{d^2}$]
26. A small steel ball of radius $r$ is allowed to fall under gravity through a column of a viscous liquid of coefficient of viscosity $\eta$. After some time, the velocity of the ball attains a constant value $v_T$, called the terminal velocity. The terminal velocity depends upon (i) the weight of the ball $mg$, (ii) the coefficient of viscosity $\eta$ and (iii) the radius of the ball $r$. By the method of dimensions, determine the relation expressing terminal velocity.
    [Ans. $v_T = k \frac{mg}{\eta r}$]

Type D. On Conversion from one system to another

27. If the value of Planck’s constant in SI is $6.62 \times 10^{-34}$ J s, what will be its value in cgs system ?
    [Ans. $6.62 \times 10^{-27}$ erg]
28. If the value of universal gravitational constant in SI is $6.6 \times 10^{-11}$ N m$^2$ kg$^{-2}$, then find its value in cgs system ?
    [Ans. $6.6 \times 10^{-8}$ dyne cm$^2$ g$^{-2}$]
29. Convert 1 dyne into newton.
    [Ans. $10^{-5}$ N]
30. Find the number of erg in 10 joule by the method of dimensional analysis. [Ans. $10^8$ erg]
31. The value of acceleration due to gravity is 980 cm s$^{-2}$. What will be its value, if the unit of length is kilometre and that of time is minute ?
    [Ans. 35.28 km min$^{-2}$]

Solutions/Hints Numerical Problems For Practice

1.

Here,
$$ V = \frac{\pi}{8} \cdot \frac{p r^4}{\eta l} $$
In the given formula, $V$ represents volume of liquid flowing out per unit time.
$$ \therefore \eta l = \left[ \frac{\pi}{8} \cdot \frac{p r^4}{V l} \right] = \frac{[M L^{-1} T^{-2}]}{[L^3 T^{-1}][L]} = [M L^{-1} T^{-4}] $$

2.

(i) [Velocity gradient] = $\left[ \frac{dv}{dx} \right] = \left[ \frac{L T^{-1}}{L} \right] = [M^0 L^0 T^{-1}]$
(ii) [Young’s modulus] = $[Y] = \left[ \frac{FL}{dL} \right] = \frac{[M L T^{-2}][L]}{[L^2][L]} = [M L^{-1} T^{-2}]$

(iii) [Gravitational constant]
$$ [G] = \left[ \frac{F r^2}{m_1 m_2} \right] = \frac{[M L T^{-2}][L^2]}{[M][M]} = [M^{-1} L^3 T^{-2}] $$

(iv) [Surface tension] = $[S] = [F][L] = [M L^0 T^{-2}]$

(v) In the formula,
$$ R = \frac{P V}{T} $$
$T$ represents temperature on absolute scale. It is measured in kelvin (K).
$$ \therefore \text{[Universal gas constant]} = [R] = \left[ \frac{P V}{T} \right] $$
$$ [R] = \frac{[M L^{-1} T^{-2}][L^3]}{[K]} = [M L^2 T^{-2} K^{-1}] $$

3.

[Angular momentum] = $[L] = [m v r] = [M][L T^{-1}][L] = [M L^2 T^{-1}]$
[Planck’s constant] = $[h] = [E][T^{-1}] = [M L^2 T^{-1}]$
Since angular momentum ($L$) and Planck’s constant ($h$) have the same dimensional formula, they have same physical units.

4.

Here,
$$ E = \frac{b – x^2}{a t} $$
From the principle of homogeneity of dimensions,
$$ [b] = [x^2] = [L^2] = [M^0 L^2 T^0] $$
Now,
$$ [a] = \left[ \frac{b – x^2}{E t} \right] = \frac{[L^2]}{[M L^2 T^{-2}][T]} = [M^{-1} L^0 T] $$

5.

Here,
$$ F = a \sqrt{x} + b t^2 $$
From the principle of homogeneity of dimensions,
$$ [a \sqrt{x}] = [F] $$
or
$$ [a] = \left[ \frac{F}{\sqrt{x}} \right] = \frac{[M L T^{-2}]}{[L]^{1/2}} = [M L^{1/2} T^{-2}] $$
Also,
$$ [b t^2] = [F] $$
or
$$ [b] = \left[ \frac{F}{t^2} \right] = \frac{[M L T^{-2}]}{[T^2]} = [M L T^{-4}] $$
$$ \therefore \left[ \frac{a}{b} \right] = \frac{[M L^{1/2} T^{-2}]}{[M L T^{-4}]} = [M^0 L^{-1/2} T^2] $$

7.

Here,
$$ t = 2\pi \sqrt{\frac{l}{g}} $$
[L.H.S.] = $[t] = [T] = [M^0 L^0 T]$
[R.H.S.] = $\left[ 2\pi \sqrt{\frac{l}{g}} \right] = \left[ \frac{L}{[L T^{-2}]} \right]^{1/2} = [M^0 L^0 T]$
Hence, the given formula is correct.

9.

(i) Here,
$$ S = u t + \frac{1}{2} a t^2 $$
[L.H.S.] = $[S] = [L]$
Now, $[u t] = [L T^{-1}][T] = [L]$ and $\left[ \frac{1}{2} a t^2 \right] = [L T^{-2}][T^2] = [L]$
$\therefore$ [R.H.S.] = $[L]$
Hence, the given relation is correct.

(ii) Here,
$$ v = u + a t $$
[L.H.S.] = $[v] = [L T^{-1}]$
Now, $[u] = [L T^{-1}]$ and $[a t] = [L T^{-2}][T] = [L T^{-2}]$
$\therefore$ [R.H.S.] = $[L T^{-1}]$
Hence, the given relation is correct.

(iii) Here,
$$ v^2 – u^2 = 2aS $$
It can be obtained that the dimensional formula of both L.H.S. and R.H.S. is $[L^2]$.
Hence, the given relation is correct.

10.

Here,
$$ \rho = \frac{3g}{4\pi r G} $$
[L.H.S.] = $[\rho] = [M L^{-3}] = [M L^{-3} T^0]$
[R.H.S.] = $\left[ \frac{3g}{4\pi r G} \right] = \frac{[L T^{-2}]}{[L][M^{-1} L^3 T^{-2}]} = [M L^{-3} T^0]$
Hence, the given relation is correct.

13.

Here,
$$ h = \frac{r \rho g}{2 S \cos \theta} $$
[L.H.S.] = $[h] = [L] = [M^0 L T^0]$
[R.H.S.] = $\left[ \frac{r \rho g}{2 S \cos \theta} \right] = \frac{[L][M L^{-3}][L T^{-2}]}{[M L^0 T^{-2}]} = [M^0 L^{-1} T^0]$
Hence, the given relation is incorrect.

14.

Here,
$$ t = k \sqrt{\frac{\rho r^3}{S}} $$
[L.H.S.] = $[t] = [T] = [M^0 L^0 T]$
[R.H.S.] = $\left[ k \sqrt{\frac{\rho r^3}{S}} \right] = \left[ \frac{[M L^{-3}][L^3]}{[M L^0 T^{-2}]} \right]^{1/2} = [M^0 L^0 T]$
Hence, the given relation is correct.

16.

Suppose that the force $F$ acting on the body varies as $a^{\text{th}}$ power of mass $m$, $b^{\text{th}}$ power of radius $r$ and $c^{\text{th}}$ power of angular velocity $\omega$. Then,
$$ F = k m^a r^b \omega^c $$
Dimensional formula of $F$, $m$, $r$ and $\omega$ are $[M L T^{-2}]$, $[M]$, $[L]$ and $[T^{-1}]$ respectively. Now, by proceeding as in problem no. 12, it can be obtained that
$$ a = 1,\ b = 1\ \text{and}\ c = 2 $$
$$ \therefore F = k m r \omega^2 $$

17.

Let
$$ v = k g^a R^b $$
[L.H.S.] = $[v] = [M^0 L T^{-1}]$
[R.H.S.] = $[k g^a R^b] = [L T^{-2}]^a [L]^b = [M^0 L^{a+b} T^{-2a}]$
Comparing the dimensions of $M$, $L$ and $T$ on the two sides, we get
$$ a + b = 1 \quad \text{and} \quad 2a = 1 \quad \text{or} \quad a = 1/2 $$
It follows that
$$ b = 1 – a = 1 – 1/2 = 1/2 $$
Substituting for $a$ and $b$ in the equation (i), we get
$$ v = k \sqrt{g R} $$

18.

Let
$$ v = k m^a r^b g^c $$
[L.H.S.] = $[v] = [M^0 L T^{-1}]$
[R.H.S.] = $[k m^a r^b g^c] = [M]^a [L]^b [L T^{-2}]^c = [M^a L^b T^{-2c}]$
Comparing the dimensions of $M$, $L$ and $T$ on the two sides,
$$ a = 0;\quad b + c = 1 \quad \text{and} \quad 2c = 1 \quad \text{or} \quad c = 1/2 $$
It follows that
$$ b = 1 – c = 1 – 1/2 = 1/2 $$
Substituting for $a$, $b$ and $c$ in the equation (i), we get
$$ v = k \sqrt{g r} $$

19.

Let
$$ v = k g^p h^q $$
[L.H.S.] = $[v] = [M^0 L T^{-1}]$
[R.H.S.] = $[k g^p h^q] = [L T^{-2}]^p [L]^q = [M^0 L^{p+q} T^{-2p}]$
Comparing the dimensions of $M$, $L$ and $T$ on the two sides, we get
$$ p + q = 1 \quad \text{and} \quad 2p = 1 \quad \text{or} \quad p = 1/2 $$
It follows that
$$ q = 1 – p = 1 – 1/2 = 1/2 $$
$$ \therefore v = k \sqrt{g h} $$

20.

Let
$$ E = k m^a \nu^b r^c $$
[L.H.S.] = $[E] = [M L^2 T^{-2}]$
[R.H.S.] = $[k m^a \nu^b r^c] = [M]^a [T^{-1}]^b [L]^c = [M^a L^c T^{-b}]$
Comparing the dimensions of $M$, $L$ and $T$ on the two sides, we get
$$ a = 1;\ c = 2 \quad \text{and} \quad b = 2 $$
Substituting for $a$, $b$ and $c$ in the equation (i), we get
$$ E = k m \nu^2 r^2 $$

21.

Let
$$ v = k \lambda^a \rho^b g^c $$
[L.H.S.] = $[v] = [M^0 L T^{-1}]$
[R.H.S.] = $[k \lambda^a \rho^b g^c] = [L]^a [M L^{-3}]^b [L T^{-2}]^c = [M^b L^{a-3b+c} T^{-2c}]$
Comparing the dimensions of $M$, $L$ and $T$ on the two sides, we get
$$ b = 0,\ a – 3b + c = 1 \quad \text{and} \quad 2c = 1 \quad \text{or} \quad c = 1/2 $$
It follows that
$$ a = 1 + 3b – c = 1 + 3 \times 0 – 1/2 = 1/2 $$
Substituting for $a$, $b$ and $c$ in the equation (i), we get
$$ v = k \sqrt{g \lambda} $$

22.

Let
$$ m = k v^a \rho^b g^c $$
[L.H.S.] = $[m] = [M L^0 T^0]$
[R.H.S.] = $[k v^a \rho^b g^c] = [L T^{-1}]^a [M L^{-3}]^b [L T^{-2}]^c = [M^b L^{a-3b+c} T^{-a-2c}]$
Comparing the dimensions of $M$, $L$ and $T$ on the two sides, we get
$$ b = 1,\ a – 3b + c = 0 \quad \text{and} \quad -a – 2c = 0 $$
On solving, we get
$$ c = -3 \quad \text{and} \quad a = 6 $$
Substituting for $a$, $b$ and $c$ in the equation (i), we get
$$ m = k \frac{v^6 \rho}{g^3} $$
Hence, mass of the stone varies as sixth power of the velocity of flow of the river.

23.

Suppose that the wavelength associated with a moving particle is given by
$$ \lambda = k m^a v^b h^c $$
Knowing that the dimensional formulae of $m$, $v$ and $h$ are $[M]$, $[L T^{-1}]$ and $[M L^2 T^{-1}]$ respectively, it can be obtained that
$$ a = -1,\ b = -1 \quad \text{and} \quad c = 1 $$
$$ \therefore \lambda = k \frac{h}{m v} $$

24.

Let
$$ S = k m^a p^b r^c $$
[L.H.S.] = $[S] = [M L^0 T^{-2}]$
[R.H.S.] = $[k m^a p^b r^c] = [M]^a [M L^{-1} T^{-2}]^b [L]^c = [M^{a+b} L^{-b+c} T^{-2b}]$
Comparing the dimensions of $M$, $L$ and $T$ on the two sides, we get
$$ a + b = 1;\ -b + c = 0 \quad \text{and} \quad 2b = 2 \quad \text{or} \quad b = 1 $$
It follows that
$$ a = 1 – b = 1 – 1 = 0 \quad \text{and} \quad c = b = 1 $$
Substituting for $a$, $b$ and $c$ in the equation (i), we have
$$ S = k p r $$
Since $k = 1/2$, we have
$$ S = \frac{p r}{2} $$

25.

Let
$$ \eta = k m^a v^b d^c $$
[L.H.S.] = $[\eta] = [M L^{-1} T^{-1}]$
[R.H.S.] = $[k m^a v^b d^c] = [M]^a [L T^{-1}]^b [L]^c = [M^a L^{b+c} T^{-b}]$
Comparing the dimensions of $M$, $L$ and $T$ on the two sides, we get
$$ a = 1,\ b + c = -1 \quad \text{and} \quad -b = -1 \quad \text{or} \quad b = 1 $$
It follows that
$$ c = -1 – b = -1 – 1 = -2 $$
Substituting for $a$, $b$ and $c$ in the equation (i), we get
$$ \eta = k \frac{m v}{d^2} $$

26.

Let
$$ v_T = k (m g)^a \eta^b r^c $$
[L.H.S.] = $[v_T] = [M^0 L T^{-1}]$
[R.H.S.] = $[k (m g)^a \eta^b r^c] = [M L T^{-2}]^a [M L^{-1} T^{-1}]^b [L]^c = [M^{a+b} L^{a-b+c} T^{-2a-b}]$
Comparing the dimensions of $M$, $L$ and $T$ on the two sides, we get
$$ a + b = 0;\ a – b + c = 1 \quad \text{and} \quad -2a – b = -1 $$
On solving, we get
$$ a = 1,\ b = -1 \quad \text{and} \quad c = -1 $$
Substituting for $a$, $b$ and $c$ in the equation (i), we get
$$ v_T = k \frac{m g}{\eta r} $$

31.

Here, $g = 980$ cm s⁻²
In new system of units: $[L] = 1$ km and $[T] = 1$ min
Now, $1$ km $= 10^5$ cm or $1$ cm $= 10^{-5}$ km
and $1$ min $= 60$ s or $1$ s $= 1/60$ min
$$ \therefore g = 980 \times (10^{-5}\ \text{km}) \times \left( \frac{1}{60\ \text{min}} \right)^{-2} $$
$$ = 980 \times 10^{-5} \times 60 \times 60 = 35.28\ \text{km min}^{-2} $$

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