Deriving the dimensional formula of physical quantities and understanding the principle of dimensional homogeneity are essential concepts in dimensional analysis. These concepts help express physical quantities in terms of fundamental units like mass, length, and time, and ensure that physical equations are dimensionally consistent. The principle of dimensional homogeneity states that all terms in a valid physical equation must have the same dimensions, making it a powerful tool to check the correctness of equations and derive relationships between different physical quantities.
To Derive Relation Between Various Different Physical Quantities – Principle of Dimensional Homogeneity
In deriving the relation between the various physical quantities, again the principle of homogeneity of dimensional equation is used. To derive a physical relation, we first of all explore the possible physical quantities upon which the given physical quantity may depend. By assuming the powers of the dependent physical quantities, the relation between the given physical quantity and the quantities on which it depends, is written. By making use of dimensional formulae of the physical quantities involved, the relation is expressed in terms of the fundamental units of mass, length and time. When the powers of M, L and T are equated on both sides of the dimensional equation, we get three equations from which the values of the three unknown powers can be calculated. Setting the values of the powers, the required relation is obtained.
Numerical Problems Based on Principle of Dimensional Homogeneity
Problem. Using the method of dimensions, derive an expression for the centripetal force F acting on a particle of mass m moving with velocity v in a circle of radius r.
Solution. Suppose that the force F acting on the particle varies as ath power of mass m, bth power of velocity v and cth power of radius r of the circle. Then,
$$
F = k m^a v^b r^c \quad \dots(i)
$$
$$
\text{[L.H.S.]} = [F] = [M L T^{-2}]
$$
$$
\text{[R.H.S.]} = [k m^a v^b r^c] = [M]^a [L T^{-1}]^b [L]^c = [M^a L^{b+c} T^{-b}]
$$
Comparing the dimensions of M, L and T on the two sides, we get
$$
a = 1; \quad b + c = 1 \quad \text{and} \quad b = 2
$$
It follows that
$$
c = 1 – b = 1 – 2 = -1
$$
Substituting for a, b and c in the equation (i), we have
$$
F = k \frac{m v^2}{r}
$$
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 $c$.
Sol. Here, $T \propto p^{a} d^{b} E^{c}$ …(i)
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
$$[T] = [M L^{-1} T^{-2}]^{a} [M L^{-3}]^{b} [M L^{2} T^{-2}]^{c}$$
$$[M^{0} L^{0} T^{1}] = [M^{a+b+c} L^{-a-3b+2c} T^{-2a-2c}]$$
Comparing the dimensions of M, L and T on the two sides, we have
$$a + b + c = 0$$
$$-a – 3b + 2c = 0$$
$$-2a – 2c = 1$$
On solving the above three equations in $a, b$ and $c$ we obtain
$$a = -\frac{5}{6}, \quad b = \frac{1}{2}, \quad c = \frac{1}{3}$$
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.] $= [t] = [T] = [M^0 L^0 T]$
[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
$$a + b = 0 \quad \text{and} \quad -2b = 1 \quad \text{or} \quad b = -\frac{1}{2}$$
It follows that
$$a = -b = \frac{1}{2}$$
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
$$-2b = -1 \quad \text{or} \quad b = \frac{1}{2}$$
Substituting for $b$ in the equation (iv), we get
$$a + \frac{1}{2} = 0 \quad \text{or} \quad a = -\frac{1}{2}$$
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 ($v$) of a vibrating string depends upon the load (F) applied, length of the string (l) and mass per unit length (m), prove that
$$v = \dfrac{1}{2l} \sqrt{\frac{F}{m}} $$
Solution. Let
$$v \propto F^a \quad \dots(i)$$
$$v\propto m^c \quad \dots(ii)$$
$$v\propto l^b \quad \dots(iii)$$
Combining the equations (i), (ii) and (iii), we get
$$v = k F^a l^b m^c, \quad \dots(iv)$$
where $k$ is the dimensionless constant of proportionality.
In the equation (iv), substituting the dimensions of $v$, $F$, $l$ 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
$$-2a = -1 \quad \text{or} \quad a = \frac{1}{2},$$
$$a + c = 0 \quad \text{or} \quad c = -a = -\frac{1}{2}$$
and $a + b – c = 0 \quad \text{or} \quad b = c – a = -\frac{1}{2} – \frac{1}{2} = -1$
In the equation (iv), putting the values of $a$, $b$ and $c$, we obtain
$$v = k F^{1/2} l^{-1} m^{-1/2}$$
$$v = \frac{k}{l} \sqrt{\frac{F}{m}}$$
The constant of proportionality $k$ is found to be equal to $1/2$. Therefore, the above relation becomes
$$v = \frac{1}{2l} \sqrt{\frac{F}{m}}$$
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
$$S \propto u^x t^y$$
or
$$S = k_1 u^x t^y,$$
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. Derive the relation for the distance travelled by a body in the $n$th second of its motion.
Solution. Let us assume that the distance $S$ travelled by a body in the $n$th second depends on its initial velocity $u$, acceleration $a$ and the time $n$ (the $n$th second). That is,
$$S \propto u^{x} a^{y} n^{z}$$
$$S = k u^{x} a^{y} n^{z} \quad …(i)$$
where $k$ is a dimensionless constant of proportionality.
Writing the dimensions of both sides, we have
$$[\mathrm{L}] = [\mathrm{L}\mathrm{T}^{-1}]^{x} [\mathrm{L}\mathrm{T}^{-2}]^{y} [\mathrm{T}]^{z}$$
$$[\mathrm{L}] = [\mathrm{L}^{x+y} \mathrm{T}^{-x-2y+z}]$$
From the principle of homogeneity of dimensions, we have
$$x + y = 1 \quad \text{and} \quad -x – 2y + z = 0$$
We have three unknowns but only two equations. So we cannot find $x$, $y$ and $z$ uniquely. However, for the $n$th second, $n$ is a number. So we can take $z = 1$ (since the $n$th second is a particular interval of time). Then the above equations become
$$x + y = 1 \quad \text{and} \quad -x – 2y + 1 = 0 \quad \text{or} \quad x + 2y = 1$$
Solving, we get $x = 1, y = 0$.
Substituting in equation (i), we have
$$S = k u^{1} a^{0} n^{1} = k u n$$
The constant $k$ comes out to be 1. Therefore, the distance travelled in the $n$th second is
$$S_{nth} = u + \frac{a}{2}(2n – 1)$$
Problem. 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.
Solution. Suppose the relationship is given by:
$$v = k \lambda^a \rho^b g^c$$
Where $k$ is a non-dimensional constant.
Substituting the dimensional formulae for all quantities:
Velocity ($v$): $[M^0 L T^{-1}]$
Wavelength ($\lambda$): $[L]$
Density ($\rho$): $[M L^{-3}]$
Acceleration due to gravity ($g$): $[L T^{-2}]$
The equation becomes:
$$[M^0 L T^{-1}] = [L]^a [M L^{-3}]^b [L T^{-2}]^c$$
$$[M^0 L T^{-1}] = [M^b L^{a – 3b + c} T^{-2c}]$$
By equating the powers of $M, L,$ and $T$ from both sides:
For $M$: $b = 0$
For $T$: $-2c = -1 \implies c = \frac{1}{2}$
For $L$: $a – 3b + c = 1$
Substitute $b = 0$ and $c = \frac{1}{2}$ into the $L$ equation:
$$a – 3(0) + \frac{1}{2} = 1$$
$$a = 1 – \frac{1}{2} = \frac{1}{2}$$
Substituting the values of $a, b,$ and $c$ back into the original equation:
$$v = k \lambda^{1/2} \rho^0 g^{1/2}$$
$$v = k \sqrt{\lambda g}$$
The required relation is $v = k \lambda^{1/2} g^{1/2}$. Since $b = 0$, the velocity of water waves does not depend on the density of water.
Problem. The speed of light $c$, gravitational constant $G$ and Planck’s constant $h$ are taken as the fundamental units in a system. Find the dimensions of length and time in this new system of units.
Solution. Dimensions of the given quantities:
Speed of light ($c$): $[LT^{-1}]$ ….(i)
Gravitational constant ($G$): $[M^{-1}L^{3}T^{-2}]$ ….(ii)
Planck’s constant ($h$): $[ML^{2}T^{-1}]$ ….(iii)
1. Finding the Dimension of Length ($L$)
To eliminate Mass ($M$), we multiply equation (ii) and (iii):
$$Gh = [M^{-1}L^{3}T^{-2}] \times [ML^{2}T^{-1}] = [L^{5}T^{-3}]$$
From equation (i), we have $c^{3} = [L^{3}T^{-3}]$.
Dividing the result of $Gh$ by $c^{3}$:
$$\frac{Gh}{c^{3}} = \frac{[L^{5}T^{-3}]}{[L^{3}T^{-3}]} = L^{2}$$
Taking the square root to find $L$:
$$L = \sqrt{\frac{Gh}{c^{3}}} = G^{1/2}h^{1/2}c^{-3/2}$$
2. Finding the Dimension of Time ($T$)
From equation (i), we know that $c = \frac{\text{Length}}{\text{Time}}$, therefore:
$$T = \frac{\text{Length}}{\text{Speed (c)}} = \frac{G^{1/2}h^{1/2}c^{-3/2}}{c}$$
Using the laws of exponents:
$$T = G^{1/2}h^{1/2}c^{-3/2-1} = G^{1/2}h^{1/2}c^{-5/2}$$
Problem. Pressure $P$ varies as $P = \frac{\alpha}{\beta} \exp \left( -\frac{\alpha Z}{k\theta} \right)$, where $Z$ denotes the distance, $k$ is Boltzmann’s constant, $\theta$ is absolute temperature, and $\alpha, \beta$ are constants. Find the dimensions of $\beta$.
Solution. 1. Dimensional Analysis of the Exponent:
The term in the exponent must be dimensionless. Therefore:
$$\left[ \frac{\alpha Z}{k \theta} \right] = [M^0 L^0 T^0]$$
$$\alpha = \frac{[k \theta]}{[Z]}$$
2. Finding Dimensions of $\alpha$:
From the exponential term, we can rearrange to find $\alpha$ in terms of energy (since $k\theta$ has dimensions of energy $[ML^2T^{-2}]$):
$$\alpha = \frac{[ML^2T^{-2}]}{[L]} = [MLT^{-2}]$$
3. Finding Dimensions of $\beta$:
The exponential function $\exp(x)$ is a non-dimensional constant. Therefore, the dimensions of $P$ must be equal to the dimensions of the ratio $\dfrac{\alpha}{\beta}$:
$$[P] = \left[ \frac{\alpha}{\beta} \right]$$
$$\beta = \frac{[\alpha]}{[P]}$$
4. Substituting Dimensional Formulae:
Dimensions of $\alpha$: $[MLT^{-2}]$
Dimensions of Pressure ($P$): $[ML^{-1}T^{-2}]$
$$\beta = \frac{[MLT^{-2}]}{[ML^{-1}T^{-2}]}$$
$$\beta = [M^{1-1} L^{1-(-1)} T^{-2-(-2)}]$$
$$\beta = [M^0 L^2 T^0]$$
Problem. A force $F$ is given by $F = at + bt^2$, where $t$ is time. What are the dimensions of $a$ and $b$?
Solution. Dimensions of Force ($F$): $[MLT^{-2}]$
1. Finding Dimensions of $a$
According to the principle of dimensional homogeneity, the dimensions of the term $at$ must equal the dimensions of $F$:
$$[at] = [F]$$
$$[a][T] = [MLT^{-2}]$$
To find $[a]$, we divide by $[T]$:
$$[a] = \frac{[MLT^{-2}]}{[T]}$$
$$[a] = [MLT^{-3}]$$
2. Finding Dimensions of $b$
Similarly, the dimensions of the term $bt^2$ must also equal the dimensions of $F$:
$$[bt^2] = [F]$$
$$[b][T^2] = [MLT^{-2}]$$
To find $[b]$, we divide by $[T^2]$:
$$[b] = \frac{[MLT^{-2}]}{[T^2]}$$
$$[b] = [MLT^{-4}]$$
Problem. The position of a particle at time $t$ is given by the relation:
$$x(t) = \frac{v_0}{\alpha} (1 – e^{-\alpha t})$$
where $v_0$ is a constant and $\alpha > 0$. Find the dimensions of $v_0$ and $\alpha$.
Solution.
1. Finding Dimensions of $\alpha$:
The exponent $(-\alpha t)$ must be dimensionless:
$$[\alpha t] = [M^0 L^0 T^0]$$
$$[\alpha] [T] = [1]$$
$$[\alpha] = [T^{-1}]$$
2. Finding Dimensions of $v_0$:
The term $(1 – e^{-\alpha t})$ is a result of subtracting one non-dimensional constant from another, so the entire bracketed term is dimensionless.
Therefore, the dimensions of position $x$ must be equal to the dimensions of the leading coefficient $\dfrac{v_0}{\alpha}$:
$$[x] = \left[ \frac{v_0}{\alpha} \right]$$
$$[L] = \frac{[v_0]}{[T^{-1}]}$$
To find $[v_0]$, we multiply both sides by $[T^{-1}]$:
$$[v_0] = [L] [T^{-1}]$$
$$[v_0] = [LT^{-1}]$$
Problem. Find the dimensions of the physical quantity $X$ in the equation:
$$\text{Force} = \frac{X}{\text{Density}}$$
Solution. According to the uses of dimensional equations, we can rearrange the formula to find the dimensions of an unknown quantity.
Rearrange the equation for $X$:$$X = \text{Force} \times \text{Density}$$
Substitute the Dimensional Formulae:
Force: $[MLT^{-2}]$
Density: $[ML^{-3}]$$$[X] = [MLT^{-2}] \times [ML^{-3}]$$
The dimensions of $X$ are $[M^2 L^{-2} T^{-2}]$.
Problem. The number of particles $n$ crossing a unit area perpendicular to the $x$-axis in unit time is given by the formula:
$$n = -D \frac{n_2 – n_1}{x_2 – x_1}$$
where $n_1$ and $n_2$ are the number of particles per unit volume for the values of $x$ meant to $x_2$ and $x_1$. Find the dimensions of $D$ (the diffusion constant).
Solution. 1. Dimensional Formula of $n$ (Particle Flux):
By definition, $n$ is the number of particles passing through a unit area in unit time.
$$[n] = \frac{[\text{Number of particles}]}{[\text{Area}] \times [\text{Time}]}$$
Since “Number of particles” is a non-dimensional constant, its dimension is $[1]$.
$$[n] = \frac{1}{[L^2][T]} = [L^{-2} T^{-1}]$$
2. Dimensional Formula of Concentration ($n_1, n_2$):
These represent the number of particles per unit volume.
$$[n_1] = [n_2] = \frac{[\text{Number of particles}]}{[\text{Volume}]}$$
$$[n_1] = \frac{1}{[L^3]} = [L^{-3}]$$
Therefore, the dimension of the difference $[n_2 – n_1]$ is also $[L^{-3}]$.
3. Dimensional Formula of Distance ($x_1, x_2$):
$$[x_2 – x_1] = [L]$$
4. Finding Dimensions of $D$:
Using the principle of dimensional homogeneity, we rearrange the given formula for $D$:
$$[D] = \frac{[n] \times [x_2 – x_1]}{[n_2 – n_1]}$$
Substitute the dimensional formulae:
$$[D] = \frac{[L^{-2} T^{-1}] \times [L]}{[L^{-3}]}$$
$$[D] = \frac{[L^{-1} T^{-1}]}{[L^{-3}]}$$
Using the laws of exponents:
$$[D] = [L^{-1 – (-3)} T^{-1}]$$
$$[D] = [L^2 T^{-1}]$$
The dimensions of the diffusion constant $D$ are $[L^2 T^{-1}]$.
Problem. If $E, m, l$, and $G$ denote energy, mass, angular momentum, and gravitational constant respectively, find the dimensions of the quantity:
$$\frac{E l^2}{m^5 G^2}$$
Solution.
1. Dimensional Formulae of the Quantities:
Energy ($E$): $[ML^2 T^{-2}]$
Mass ($m$): $[M]$
Angular Momentum ($l$): $[ML^2 T^{-1}]$
Gravitational Constant ($G$): $[M^{-1} L^3 T^{-2}]$
2. Substituting Dimensions into the Expression:
$$\text{Dimensions} = \frac{[E] \times [l]^2}{[m]^5 \times [G]^2}$$
$$\text{Dimensions} = \frac{[ML^2 T^{-2}] \times [ML^2 T^{-1}]^2}{[M]^5 \times [M^{-1} L^3 T^{-2}]^2}$$
3. Simplifying the Numerator and Denominator:
Numerator:$$[ML^2 T^{-2}] \times [M^2 L^4 T^{-2}] = [M^3 L^6 T^{-4}]$$
Denominator:$$[M^5] \times [M^{-2} L^6 T^{-4}] = [M^3 L^6 T^{-4}]$$
4. Final Calculation:
$$\frac{[M^3 L^6 T^{-4}]}{[M^3 L^6 T^{-4}]} = [M^{3-3} L^{6-6} T^{-4-(-4)}] = [M^0 L^0 T^0]$$
The dimensions of the quantity $\dfrac{E l^2}{m^5 G^2}$ are $[M^0 L^0 T^0]$.
This means the resulting quantity is a non-dimensional constant, as it has no units or dimensions in mass, length, or time.
Problem. The equation of a wave is given by $y = A \sin \omega \left( \frac{x}{v} – k \right)$, where $\omega$ is the angular velocity, $x$ is distance, and $v$ is the linear velocity. Find the dimension of $k$.
Solution.
1. Dimensional Analysis of the Argument:
In the expression $\sin \omega \left( \frac{x}{v} – k \right)$, the term inside the parentheses must be such that when multiplied by $\omega$, the entire argument is dimensionless. However, a simpler approach is to look at the terms being subtracted inside the parentheses. According to the principle of homogeneity, only quantities with the same dimensions can be added or subtracted.
$$\left[ \frac{x}{v} \right] = [k]$$
2. Substituting Dimensional Formulae:
Distance ($x$): $[L]$
Linear Velocity ($v$): $[LT^{-1}]$
$$[k] = \frac{[L]}{[LT^{-1}]}$$
3. Final Calculation:
Using the laws of exponents:
$$[k] = [L^{1-1} T^{-(-1)}]$$
$$[k] = [T^1]$$
Problem. The potential energy $U$ of a particle varies with distance $x$ from a fixed origin as:
$$U = \frac{A \sqrt{x}}{x^2 + B}$$
where $A$ and $B$ are dimensional constants. Find the dimensional formula for the product $AB$.
Solution.
Finding Dimensions of $B$:
According to the principle of dimensional homogeneity, only physical quantities with the same dimensions can be added or subtracted. In the denominator, $x^2$ is added to $B$.
$$[B] = [x^2]$$
Since $x$ represents distance ($[L]$):
$$[B] = [L^2]$$
Finding Dimensions of $A$:
The dimensions of potential energy $U$ are the same as work or energy: $[ML^2 T^{-2}]$.
$$[U] = \frac{[A] [\sqrt{x}]}{[x^2 + B]}$$
Substituting the known dimensions:
$$[ML^2 T^{-2}] = \frac{[A] [L^{1/2}]}{[L^2]}$$
Rearranging to solve for $[A]$:
$$[A] = [ML^2 T^{-2}] \times \frac{[L^2]}{[L^{1/2}]}$$
$$[A] = [ML^2 T^{-2}] \times [L^{3/2}]$$
$$[A] = [M L^{7/2} T^{-2}]$$
Finding Dimensions of the product $AB$:
Now, we multiply the dimensions of $A$ and $B$:
$$[AB] = [A] \times [B]$$
$$[AB] = [M L^{7/2} T^{-2}] \times [L^2]$$
Using the laws of exponents ($7/2 + 2 = 11/2$):
$$[AB] = [M L^{11/2} T^{-2}]$$
Problem. In the following integration formula:
$$\int \frac{dx}{\sqrt{2ax – x^2}} = a^n \sin^{-1} \left( \frac{x}{a} – 1 \right)$$
Find the value of $n$ using dimensional analysis.
Solution : According to the principle of homogeneity, the argument of a trigonometric function (like $\sin^{-1}$) must be a non-dimensional constant.
In the term $(\frac{x}{a} – 1)$, both $x$ and $a$ must have the same dimensions for their ratio to be dimensionless.
Since $x$ represents a position or length, its dimension is $[L]$.
Therefore, the dimension of $a$ is $[L]$.
Dimensions of the Left Hand Side (LHS)
The symbol $\int$ is a summation and does not change the dimensions. We focus on the differential $dx$ and the denominator:
- $dx$: This is a small change in length, so $[dx] = [L]$.
- $\sqrt{2ax – x^2}$:
- Dimension of $2ax$ is $[L][L] = [L^2]$.
- Dimension of $x^2$ is $[L^2]$.
- The square root of $[L^2]$ is $[L]$.
LHS Dimensions:
$$\text{LHS} = \frac{[L]}{[L]} = [L^0 M^0 T^0]$$
The LHS is dimensionless.
Dimensions of the Right Hand Side (RHS)
The term $\sin^{-1} \dots$ is an angle and is a non-dimensional variable.
The dimension of the RHS is determined solely by the leading constant $a^n$.
Since $a$ has the dimension $[L]$, the dimension of the RHS is $[L]^n$.
RHS Dimensions:
$$\text{RHS} = [L]^n$$
Equating Dimensions
For the equation to be dimensionally correct, the dimensions of the LHS must equal the dimensions of the RHS:
$$\text{LHS} = \text{RHS}$$
$$[L]^0 = [L]^n$$
Comparing the powers of $L$:
$$n = 0$$
Problem. In a new system of units, Velocity ($v$), Acceleration ($A$), and Force ($F$) are chosen as the fundamental physical quantities instead of the standard Mass ($M$), Length ($L$), and Time ($T$). Find the dimensional formula for Angular Momentum in terms of these new fundamental units $v, A,$ and $F$.
Solution
1. Identify Standard Dimensions
First, we express the standard dimensions of all quantities involved:
Velocity ($v$): $[LT^{-1}]$
Acceleration ($A$): $[LT^{-2}]$
Force ($F$): $[MLT^{-2}]$
Angular Momentum ($L_{ang}$): $[ML^2T^{-1}]$ (derived from $r \times p$ or $mvr$)
2. Set up the Dimensional Equation
Assume that Angular Momentum ($L_{ang}$) is proportional to the product of powers of the new fundamental quantities:
$$[L_{ang}] = F^a v^b A^c$$
Now, substitute the standard $M, L, T$ dimensions into this expression:
$$[M^1 L^2 T^{-1}] = [MLT^{-2}]^a [LT^{-1}]^b [LT^{-2}]^c$$
$$[M^1 L^2 T^{-1}] = [M^a L^{a+b+c} T^{-2a-b-2c}]$$
3. Equate the Powers (Principle of Homogeneity)
According to the principle of dimensional homogeneity, the exponents of $M, L,$ and $T$ must be equal on both sides:
- For $M$: $a = 1$
- For $L$: $a + b + c = 2$
- For $T$: $-2a – b – 2c = -1$
4. Solve for the Unknowns ($a, b, c$)
Substitute the value of $a = 1$ into the other equations:
From the $L$ equation: $1 + b + c = 2 \implies \mathbf{b + c = 1}$
From the $T$ equation: $-2(1) – b – 2c = -1 \implies -b – 2c = 1 \implies \mathbf{b + 2c = -1}$
Now, subtract the first simplified equation ($b + c = 1$) from the second ($b + 2c = -1$):
$$(b + 2c) – (b + c) = -1 – 1$$
$$c = -2$$
Substitute $c = -2$ back into the equation $b + c = 1$:
$$b + (-2) = 1 \implies b = 3$$
Substituting the values of $a = 1$, $b = 3$, and $c = -2$ into the original assumed equation:
$$[L_{ang}] = F^1 v^3 A^{-2}$$
Dimensions of Angular Momentum: $[F v^3 A^{-2}]$
Problem. If the velocity of light ($c$), gravitational constant ($G$), and Planck’s constant ($h$) are chosen as the fundamental units in a new system of measurement, find the dimensional formula for mass ($M$) in this new system.
Solution :
1. Standard Dimensional Formulae
First, we identify the dimensions of the given quantities in the standard $M, L, T$ system:
Mass ($M$): $[M^1 L^0 T^0]$
Velocity of light ($c$): $[LT^{-1}]$
Gravitational constant ($G$): $[M^{-1} L^3 T^{-2}]$
Planck’s constant ($h$): $[ML^2 T^{-1}]$
2. Establish the Relationship
Assume that Mass ($M$) is proportional to the product of powers of $c, G,$ and $h$:
$$M \propto c^x G^y h^z \implies M = k c^x G^y h^z$$
where $k$ is a non-dimensional constant.
3. Substitute Dimensions
Substituting the $M, L, T$ dimensions into the equation:
$$[M^1 L^0 T^0] = [LT^{-1}]^x [M^{-1} L^3 T^{-2}]^y [ML^2 T^{-1}]^z$$
$$[M^1 L^0 T^0] = [M^{-y+z} L^{x+3y+2z} T^{-x-2y-z}]$$
4. Equate the Powers (Principle of Homogeneity)
By equating the exponents of $M, L,$ and $T$ from both sides:
For $M$: $-y + z = 1$
For $L$: $x + 3y + 2z = 0$
For $T$: $-x – 2y – z = 0$
5. Solving the Equations
From (1): $z = 1 + y$
Add equations (2) and (3) to eliminate $x$:$$(x + 3y + 2z) + (-x – 2y – z) = 0 + 0$$$$y + z = 0$$
Substitute $z = 1 + y$ into the result ($y + z = 0$):$$y + (1 + y) = 0 \implies 2y = -1 \implies \mathbf{y = -1/2}$$
Now, find $z$:$$z = 1 + (-1/2) \implies \mathbf{z = 1/2}$$
Finally, find $x$ using equation (3):$$-x – 2(-1/2) – (1/2) = 0$$$$-x + 1 – 1/2 = 0 \implies -x + 1/2 = 0 \implies \mathbf{x = 1/2}$$
Substituting the values of $x = 1/2, y = -1/2,$ and $z = 1/2$ back into the original equation:
$$M = c^{1/2} G^{-1/2} h^{1/2}$$
The dimensions of mass in this new system of units are:
$[c^{1/2} G^{-1/2} h^{1/2}]$
Problem. The time period ($T$) of vibration of a liquid drop depends on the surface tension ($S$), radius ($r$) of the drop, and the density ($\rho$) of the liquid. Derive the expression for the time period $T$ using dimensional analysis.
Solution :
1. Standard Dimensional Formulae
Identify the dimensions of the variables involved in the standard $M, L, T$ system:
Time Period ($T$): $[M^0 L^0 T^1]$
Surface Tension ($S$): $\frac{\text{Force}}{\text{Length}} = \frac{[MLT^{-2}]}{[L]} = [MT^{-2}]$
Radius ($r$): $[L]$
Density ($\rho$): $\frac{\text{Mass}}{\text{Volume}} = [ML^{-3}]$
2. Establish the Relationship
Assume the time period $T$ is proportional to the product of powers of $S, r,$ and $\rho$:
$$T \propto S^x r^y \rho^z \implies T = k S^x r^y \rho^z$$
where $k$ is a non-dimensional constant.
3. Substitute Dimensions
Substituting the $M, L, T$ dimensions into the equation:
$$[M^0 L^0 T^1] = [MT^{-2}]^x [L]^y [ML^{-3}]^z$$
$$[M^0 L^0 T^1] = [M^{x+z} L^{y-3z} T^{-2x}]$$
4. Equate the Powers (Principle of Homogeneity)
Equate the exponents of $M, L,$ and $T$ from both sides:
For $T$: $-2x = 1 \implies \mathbf{x = -1/2}$
For $M$: $x + z = 0 \implies z = -x \implies \mathbf{z = 1/2}$
For $L$: $y – 3z = 0 \implies y = 3z \implies y = 3(1/2) \implies \mathbf{y = 3/2}$
Substitute the values of $x = -1/2, y = 3/2,$ and $z = 1/2$ back into the original equation:
$$T = k S^{-1/2} r^{3/2} \rho^{1/2}$$
Rewriting the expression in radical form:
$$T = k \sqrt{\frac{\rho r^3}{S}}$$
The time period of vibration is given by the expression:
$$T = k \sqrt{\frac{\rho r^3}{S}}$$
Problem. If $P$ represents radiation pressure, $C$ represents the speed of light, and $Q$ represents radiation energy striking a unit area per second, find the non-zero integers $x, y,$ and $z$ such that the expression $P^x C^y Q^z$ is dimensionless.
Solution :
1. Identify Dimensional Formulae
First, we determine the dimensions of each physical quantity in the standard $M, L, T$ system:
Radiation Pressure ($P$): Pressure is Force per unit Area.$$[P] = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$$
Speed of Light ($C$): Speed is Distance per unit Time.$$[C] = [LT^{-1}]$$
Radiation Energy per unit area per second ($Q$): This is the intensity of radiation.$$[Q] = \frac{[\text{Energy}]}{[\text{Area}] \times [\text{Time}]} = \frac{[ML^2T^{-2}]}{[L^2][T]} = [MT^{-3}]$$
2. Set up the Dimensional Equation
For the expression $P^x C^y Q^z$ to be dimensionless, its dimensions must equal $[M^0 L^0 T^0]$:
$$[P]^x [C]^y [Q]^z = [M^0 L^0 T^0]$$
$$[ML^{-1}T^{-2}]^x [LT^{-1}]^y [MT^{-3}]^z = [M^0 L^0 T^0]$$
Combine the terms by summing the exponents of $M, L,$ and $T$:
$$[M^{x+z} L^{-x+y} T^{-2x-y-3z}] = [M^0 L^0 T^0]$$
3. Equate the Powers (Principle of Homogeneity)
According to the principle of dimensional homogeneity, the exponents on both sides must match:
For $M$: $x + z = 0 \implies z = -x$
For $L$: $-x + y = 0 \implies y = x$
For $T$: $-2x – y – 3z = 0$
4. Solve for $x, y, z$
Substitute the relations $z = -x$ and $y = x$ into the $T$ equation to verify consistency:
$$-2x – (x) – 3(-x) = -3x + 3x = 0$$
The equations are consistent for any value of $x$. Since the problem asks for non-zero integers, let’s choose the simplest set by setting $x = 1$:
If $x = 1$, then $y = 1$ and $z = -1$.
This means the quantity $\frac{PC}{Q}$ is dimensionless.
Frequently Asked Questions (FAQs) Based on Principle of Homogeneity of Dimensions
What is the fundamental principle used to derive physical relations?
The derivation process relies on the Principle of Homogeneity of Dimensions. This principle states that a physical equation is correct only if the dimensions of all terms on both sides of the equation are the same. For a derived relation, this means the combined dimensions of the independent variables must match the dimensions of the dependent variable.
What are the basic steps in deriving a new physical formula?
Identify Variables: Determine which physical quantities the target quantity depends on.
Set up the Equation: Write the target quantity as proportional to the product of the other quantities, each raised to an unknown power (e.g., $x, y, z$).
Substitute Dimensions: Replace every physical quantity with its standard dimensional formula ($M, L, T$).
Solve for Powers: Equate the exponents of $M, L,$ and $T$ on both sides to create a system of algebraic equations.
Finalize the Relation: Substitute the calculated values of the powers back into the original proportional equation.
Why can we usually only solve for three unknown powers?
In mechanics, we primarily deal with three fundamental dimensions: Mass ($M$), Length ($L$), and Time ($T$). Equating the exponents of these three base units provides a maximum of three independent linear equations. If a quantity depends on more than three variables, the system of equations becomes “underdetermined,” and unique solutions for the powers cannot be found using dimensional analysis alone.
Can dimensional analysis determine the numerical constant in a formula?
No. Dimensional analysis can only establish the functional relationship (the variables and their exponents). It cannot determine the value of dimensionless constants (like $2\pi, \frac{1}{2},$ or $e$). These constants must be determined through experimental measurement or derived through more advanced theoretical methods.
What are the limitations of this method?
While powerful, this method has specific constraints:
It fails if a quantity depends on more than three independent variables.
It cannot derive relations involving trigonometric, logarithmic, or exponential functions, as these arguments must be dimensionless.
It does not distinguish between physical quantities that have the same dimensions (for example, work and torque both have the dimensions $[ML^2T^{-2}]$).
It cannot be used to derive equations where terms are added or subtracted (e.g., $v = u + at$).
How do you handle cases where a variable has no dimensions?
If one of the variables is a non-dimensional constant (like an angle or a ratio), it is ignored during the dimensional balancing process because its dimensions are $[M^0L^0T^0]$. However, you must remember that it might still be a part of the actual physical relationship.
Summary
To strengthen your understanding of Class 11 physics dimensional analysis, you should also explore topics like Derivation of Dimensional Formula of Physical Quantities, which helps in expressing physical quantities in terms of fundamental units. It is equally important to study the Principle of Dimensional Homogeneity, which is used to check the correctness of physical equations and derive relations between different quantities. For better clarity and exam preparation, you can also practice related numerical problems and JEE Main Previous Year Questions (PYQs) available on this website.