Dimensional Analysis, Uses and Limitations Questions in English

(B) The dimensional formula for pressure is $[M^1 L^{-1} T^{-2}]$.
Given: $M_1 = 50 \ g = 0.05 \ kg$,$L_1 = 10 \ cm = 0.1 \ m$,$T_1 = 5 \ s$.
In $SI$ units: $M_2 = 1 \ kg$,$L_2 = 1 \ m$,$T_2 = 1 \ s$.
The conversion formula is $n_2 = n_1 [M_1/M_2]^1 [L_1/L_2]^{-1} [T_1/T_2]^{-2}$.
Substituting the values: $n_2 = 1 \times [0.05/1]^1 \times [0.1/1]^{-1} \times [5/1]^{-2}$.
$n_2 = 0.05 \times (1/0.1) \times (1/25)$.
$n_2 = 0.05 \times 10 \times 0.04 = 0.5 / 25 = 1/50 \ \text{Pascal}$.

(D) In the argument of the trigonometric functions,the dimensions must be dimensionless.
For $\sin \left(\frac{2 \pi ct}{\lambda}\right)$,the term $\frac{ct}{\lambda}$ must be dimensionless,meaning $[ct] = [\lambda]$. Thus,the unit of $ct$ is the same as $\lambda$.
For $\cos \left(\frac{2 \pi x}{\lambda}\right)$,the term $\frac{x}{\lambda}$ must be dimensionless,meaning $[x] = [\lambda]$. Thus,the unit of $x$ is the same as $\lambda$.
Now,let us check the dimensions of the terms in option $C$:
$\left[\frac{2 \pi c}{\lambda}\right] = \frac{[L T^{-1}]}{[L]} = [T^{-1}]$
$\left[\frac{2 \pi x}{\lambda t}\right] = \frac{[L]}{[L][T]} = [T^{-1}]$
Since both have dimensions of $[T^{-1}]$,option $C$ is correct.
Now,let us check option $D$:
$\left[\frac{c}{\lambda}\right] = [T^{-1}]$
$\left[\frac{x}{\lambda}\right] = [M^0 L^0 T^0] = 1$ (dimensionless).
Since $[T^{-1}] \neq 1$,the statement in option $D$ is wrong.

(D) Dimensional analysis is a powerful tool for checking the consistency of equations,but it has limitations.
$1$. It cannot derive expressions involving exponential,logarithmic,or trigonometric functions because these functions are dimensionless.
$2$. It cannot derive equations containing additive or subtractive constants.
$3$. Option $A$ involves an exponential function.
$4$. Option $B$ involves a trigonometric function.
$5$. Option $C$ involves a sum of two terms.
$6$. Option $D$,$L = mvr$,is a product of physical quantities and can be verified or derived using dimensional analysis as it relates to the definition of angular momentum.

(A) According to the principle of homogeneity of dimensions,the dimensions of each term in an equation must be the same.
Given the force equation: $F = a \sqrt{x} - bt$.
The dimensions of force $F$ are $[M^1 L^1 T^{-2}]$.
For the first term: $[F] = [a \sqrt{x}]$
$[M^1 L^1 T^{-2}] = [a] [L^{1/2}]$
$[a] = [M^1 L^{1/2} T^{-2}]$
Substituting the $SI$ units for mass $(kg)$,length $(m)$,and time $(s)$: Unit of $a = kg\ m^{1/2}\ s^{-2}$.
For the second term: $[F] = [bt]$
$[M^1 L^1 T^{-2}] = [b] [T^1]$
$[b] = [M^1 L^1 T^{-3}]$
Substituting the $SI$ units: Unit of $b = kg\ m\ s^{-3}$.
Thus,the units of $a$ and $b$ are $kg\ m^{1/2}\ s^{-2}$ and $kg\ m\ s^{-3}$ respectively.

(B) Dimensional analysis is a powerful tool for checking the consistency of equations,but it has significant limitations.
$1$. It cannot derive formulas containing additive or subtractive terms (e.g.,$s = ut + \frac{1}{2}at^2$) because dimensional analysis only deals with products and quotients of physical quantities.
$2$. It cannot determine dimensionless constants (like $1, 2, \pi, e$,etc.) because these constants have no dimensions and thus do not appear in the dimensional formula.
Both statements are true. The reason explains why dimensional analysis is limited,but the inability to derive additive relations is a specific consequence of the method's structure,not directly caused by the inability to find constants. Therefore,the reason is a true statement but not the direct explanation for the assertion.

(A) The principle of homogeneity of dimensions states that a physical equation is dimensionally correct only if the dimensions of all the terms on both sides of the equation are the same.
This principle is the fundamental basis for checking the dimensional consistency of any physical equation.
If the dimensions of the terms are not identical,the equation cannot be physically correct,as we cannot add or subtract physical quantities with different dimensions.
Therefore,the assertion is true,and the reason correctly explains why we use the principle of homogeneity to verify equations.

(A) The dimensional formula for power is $[P] = [M L^2 T^{-3}]$.
Let the new units be $M' = a \ kg$,$L' = b \ m$,and $T' = c \ s$.
Therefore,$1 \ kg = \frac{1}{a} M'$,$1 \ m = \frac{1}{b} L'$,and $1 \ s = \frac{1}{c} T'$.
Given power $P = 6 \ W = 6 \ kg \cdot m^2 \cdot s^{-3}$.
Substituting the new units into the expression for power:
$P = 6 \times (\frac{1}{a} M') \times (\frac{1}{b} L')^2 \times (\frac{1}{c} T')^{-3}$
$P = 6 \times \frac{1}{a} \times \frac{1}{b^2} \times c^3 \times (M' L'^2 T'^{-3})$
$P = \frac{6 c^3}{a b^2}$ new units.
Thus,the magnitude of $6 \ W$ in the new system is $\frac{6 c^3}{a b^2}$.

(B) Let energy $E$ be expressed as $E = k P^x A^y T^z$,where $k$ is a dimensionless constant.
Dimensional formula of energy $[E] = [M^1 L^2 T^{-2}]$.
Dimensional formula of momentum $[P] = [M^1 L^1 T^{-1}]$.
Dimensional formula of area $[A] = [L^2]$.
Dimensional formula of time $[T] = [T^1]$.
Substituting these into the equation:
$[M^1 L^2 T^{-2}] = [M^1 L^1 T^{-1}]^x [L^2]^y [T^1]^z$
$[M^1 L^2 T^{-2}] = [M^x L^{x+2y} T^{-x+z}]$
Comparing the powers of $M$,$L$,and $T$ on both sides:
For $M$: $x = 1$
For $L$: $x + 2y = 2$
For $T$: $-x + z = -2$
Substituting $x = 1$ into the $L$ equation: $1 + 2y = 2 \implies 2y = 1 \implies y = 1/2$.
Substituting $x = 1$ into the $T$ equation: $-1 + z = -2 \implies z = -1$.
Thus,the dimensional formula for energy is $[P^1 A^{1/2} T^{-1}]$.

(A) Given the relation $P = \frac{a - t^2}{bx}$.
According to the principle of homogeneity of dimensions,the dimensions of $a$ must be equal to the dimensions of $t^2$.
Therefore,$[a] = [T^2]$.
Now,the dimensions of the entire expression $\frac{a - t^2}{bx}$ must be equal to the dimensions of pressure $P$.
$[P] = [ML^{-1} T^{-2}]$.
So,$[P] = \frac{[a]}{[b][x]} \implies [ML^{-1} T^{-2}] = \frac{[T^2]}{[b][L]}$.
Rearranging for $[b]$,we get $[b] = \frac{[T^2]}{[ML^{-1} T^{-2}][L]} = \frac{[T^2]}{[MT^{-2}]} = [M^{-1} T^4]$.
Finally,the dimensions of $\frac{a}{b}$ are $\frac{[T^2]}{[M^{-1} T^4]} = [M^1 L^0 T^{-2}]$.

(A) The dimensions of the given constants are:
$G = [M^{-1} L^{3} T^{-2}]$
$h = [M L^{2} T^{-1}]$
$c = [L T^{-1}]$
Let the expression be $T = G^{a} h^{b} c^{d}$.
Substituting the dimensions:
$[T^{1}] = [M^{-1} L^{3} T^{-2}]^{a} [M L^{2} T^{-1}]^{b} [L T^{-1}]^{d}$
$[M^{0} L^{0} T^{1}] = [M^{-a+b} L^{3a+2b+d} T^{-2a-b-d}]$
Comparing the powers on both sides:
For $M$: $-a + b = 0 \implies a = b$
For $L$: $3a + 2b + d = 0 \implies 3a + 2a + d = 0 \implies d = -5a$
For $T$: $-2a - b - d = 1 \implies -2a - a - (-5a) = 1 \implies 2a = 1 \implies a = 1/2$
Thus,$a = 1/2, b = 1/2, d = -5/2$.
The expression is $G^{1/2} h^{1/2} c^{-5/2} = \left[\frac{G h}{c^{5}}\right]^{\frac{1}{2}}$.

(A) The dimensions of the given physical quantities are:
$[E] = [M L^{2} T^{-2}]$
$[M] = [M]$
$[L] = [M L^{2} T^{-1}]$
$[G] = [M^{-1} L^{3} T^{-2}]$
Now,substitute these into the expression $\left(\frac{E L^{2}}{G^{2} M^{5}}\right)$:
Dimensions $= \frac{[M L^{2} T^{-2}] \cdot [M L^{2} T^{-1}]^{2}}{[M^{-1} L^{3} T^{-2}]^{2} \cdot [M]^{5}}$
$= \frac{[M L^{2} T^{-2}] \cdot [M^{2} L^{4} T^{-2}]}{[M^{-2} L^{6} T^{-4}] \cdot [M^{5}]}$
$= \frac{[M^{3} L^{6} T^{-4}]}{[M^{3} L^{6} T^{-4}]}$
$= [M^{0} L^{0} T^{0}]$
Since the resulting dimension is dimensionless,it corresponds to the dimensions of an angle (which is a dimensionless quantity).

(B) The refractive index $\mu$ is a dimensionless quantity,so its dimension is $[M^0 L^0 T^0]$.
According to the principle of homogeneity of dimensions,the dimensions of each term in an equation must be the same.
Therefore,the dimensions of $\frac{B}{\lambda^2}$ must be equal to the dimensions of $\mu$ (which is dimensionless).
$\left[ \frac{B}{\lambda^2} \right] = [M^0 L^0 T^0]$
$[B] = [\lambda^2] \times [M^0 L^0 T^0]$
Since $\lambda$ is wavelength,its dimension is $[L]$.
$[B] = [L^2]$
The dimension $[L^2]$ corresponds to the dimension of area.

(D) The argument of a trigonometric function must be dimensionless.
Given $F = A \sin(Ct) + B \cos(Dx)$.
For the term $\sin(Ct)$,the argument $Ct$ must be dimensionless.
$[Ct] = [M^0 L^0 T^0] \implies [C][T] = [T^0] \implies [C] = [T^{-1}]$.
For the term $\cos(Dx)$,the argument $Dx$ must be dimensionless.
$[Dx] = [M^0 L^0 T^0] \implies [D][L] = [L^0] \implies [D] = [L^{-1}]$.
Now,the dimensions of $\frac{C}{D}$ are:
$\left[ \frac{C}{D} \right] = \frac{[T^{-1}]}{[L^{-1}]} = [L T^{-1}]$.
These are the dimensions of velocity.
Therefore,the correct option is $D$.

(A) Given the equation $P = \frac{c - t^{2}}{DS}$.
According to the principle of homogeneity of dimensions,the dimensions of $c$ must be equal to the dimensions of $t^{2}$.
Therefore,$[c] = [t^{2}] = [T^{2}]$.
Now,the dimension of pressure $P$ is $[P] = [M L^{-1} T^{-2}]$.
Given $S$ is distance,so $[S] = [L]$.
Substituting these into the equation $P = \frac{c}{DS} - \frac{t^{2}}{DS}$,we consider the term $\frac{c}{DS}$:
$[P] = \frac{[c]}{[D][S]} \Rightarrow [M L^{-1} T^{-2}] = \frac{[T^{2}]}{[D][L]}$.
Solving for $[D]$:
$[D] = \frac{[T^{2}]}{[M L^{-1} T^{-2}][L]} = \frac{[T^{2}]}{[M T^{-2}]} = [M^{-1} T^{4}]$.
We need to find the dimensions of $\left(\frac{D}{c}\right)$:
$\left[\frac{D}{c}\right] = \frac{[M^{-1} T^{4}]}{[T^{2}]} = [M^{-1} T^{2}] = [L^{0} M^{-1} T^{2}]$.

(A) According to the principle of homogeneity of dimensions,only quantities with the same dimensions can be added or subtracted.
In the expression $A = B + \frac{C}{D+E}$,since $B$ is added to the term $\frac{C}{D+E}$,the dimensions of $A$ must be equal to the dimensions of $B$.
Given $[B] = [L^{1} M^{0} T^{-1}]$,therefore $[A] = [L^{1} M^{0} T^{-1}]$.
Also,in the denominator $(D+E)$,$D$ and $E$ must have the same dimensions,so $[D] = [E]$.
The dimension of the entire term $\frac{C}{D+E}$ must be equal to the dimension of $B$.
$[B] = \frac{[C]}{[D+E]} \implies [D+E] = \frac{[C]}{[B]}$.
Substituting the given dimensions: $[D+E] = \frac{[L^{1} M^{0} T^{0}]}{[L^{1} M^{0} T^{-1}]} = [T^{1}]$.
Since $[D] = [E]$,we have $[D] = [T^{1}]$ and $[E] = [T^{1}]$.

(A) The dimension of force is $[F] = [M L T^{-2}]$.
The dimension of density is $[d] = [M L^{-3} T^0]$.
From the given relation,$F = \frac{y}{\sqrt{d}}$,we can write $y = F \sqrt{d}$.
Substituting the dimensions of $F$ and $d$:
$[y] = [M L T^{-2}] \times [M L^{-3}]^{1/2}$
$[y] = [M L T^{-2}] \times [M^{1/2} L^{-3/2}]$
$[y] = [M^{1 + 1/2} L^{1 - 3/2} T^{-2}]$
$[y] = [M^{3/2} L^{-1/2} T^{-2}]$.

(A) The speed of light in vacuum is given by the relation $c = \frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$.
Squaring both sides,we get $c^{2} = \frac{1}{\mu_{0} \varepsilon_{0}}$.
Therefore,$\mu_{0} \varepsilon_{0} = \frac{1}{c^{2}}$.
The dimensional formula for the speed of light $c$ is $[M^{0} L^{1} T^{-1}]$.
Substituting this into the expression,we get $\mu_{0} \varepsilon_{0} = \frac{1}{(M^{0} L^{1} T^{-1})^{2}}$.
$\mu_{0} \varepsilon_{0} = \frac{1}{M^{0} L^{2} T^{-2}} = M^{0} L^{-2} T^{2}$.

(B) The speed of light in vacuum is given by the relation $c = \frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$.
Squaring both sides,we get $c^{2} = \frac{1}{\mu_{0} \varepsilon_{0}}$.
Since $c$ represents velocity,its dimensional formula is $[M^{0} L^{1} T^{-1}]$.
Therefore,the dimensional formula of $\frac{1}{\mu_{0} \varepsilon_{0}}$ is the square of the dimensional formula of velocity:
$\frac{1}{\mu_{0} \varepsilon_{0}} = [M^{0} L^{1} T^{-1}]^{2} = M^{0} L^{2} T^{-2}$.
Thus,the correct option is $B$.

(B) The dimensions of Planck's constant $(h)$ are $[M L^{2} T^{-1}]$.
The dimensions of moment of inertia $(I)$ are $[M L^{2}]$.
The ratio of the dimensions of Planck's constant to the moment of inertia is given by $\frac{[h]}{[I]} = \frac{[M L^{2} T^{-1}]}{[M L^{2}]} = [T^{-1}]$.
The dimension of frequency is $[T^{-1}]$.
The dimension of velocity is $[L T^{-1}]$.
The dimension of angular momentum is $[M L^{2} T^{-1}]$.
The dimension of time is $[T]$.
Therefore,the ratio of the dimensions of Planck's constant and the moment of inertia has the dimensions of frequency.

(C) Dimensional analysis is a powerful tool for checking the consistency of physical equations,but it has limitations. It cannot determine dimensionless constants (like $6 \pi$) or the presence of transcendental functions (like trigonometric,exponential,or logarithmic functions) because their arguments must be dimensionless.
$1$. $x=A \cos (\omega t)$: While dimensionally consistent,the presence of the trigonometric function $\cos$ cannot be derived using dimensional analysis.
$2$. $N=N_0 e^{-\lambda t}$: While dimensionally consistent,the exponential form cannot be derived using dimensional analysis.
$3$. $F=6 \pi \eta r \nu$ (Stokes' Law): The dimensions of the right-hand side are $[M L^{-1} T^{-1}] \cdot [L] \cdot [L T^{-1}] = [M L T^{-2}]$,which is the dimension of force. Dimensional analysis can relate $F$ to $\eta$,$r$,and $\nu$ as $F \propto \eta^a r^b \nu^c$. By comparing dimensions,we can find $a=1, b=1, c=1$,leading to $F \propto \eta r \nu$. Thus,the functional dependence can be deduced.
Therefore,option $C$ is the correct choice.

(B) According to the principle of homogeneity of dimensions,only physical quantities with the same dimensions can be added or subtracted.
$(A)$ $PQ/R$: Multiplication and division of physical quantities with different dimensions are allowed.
$(B)$ $(P-Q)/R$: Since $P$ and $Q$ have different dimensions,the subtraction $(P-Q)$ is physically meaningless.
$(C)$ $(PR-Q^2)/R$: Here,$PR$ and $Q^2$ must have the same dimensions for the subtraction to be valid. While $P, Q, R$ have different dimensions,specific combinations might be invalid,but $(P-Q)$ is fundamentally invalid regardless of the denominator.
$(D)$ $PQ-R$: Since $PQ$ and $R$ have different dimensions,the subtraction $(PQ-R)$ is physically meaningless.
However,in the context of standard multiple-choice questions,$(P-Q)$ is the most direct violation of the principle of homogeneity.

(B) The given equation is $\text{Force} = \frac{\alpha}{\text{density} + \beta^3}$.
According to the principle of homogeneity of dimensions,quantities added together must have the same dimensions.
Therefore,the dimension of $\beta^3$ must be equal to the dimension of density.
Dimension of density = $[M L^{-3} T^0]$.
So,$[\beta^3] = [M L^{-3} T^0]$.
Taking the cube root,$[\beta] = [M^{1/3} L^{-1} T^0]$.
Now,the dimension of force is $[M L T^{-2}]$.
From the equation,$\alpha = \text{Force} \times (\text{density} + \beta^3)$.
Since $(\text{density} + \beta^3)$ has the same dimension as density,$[\alpha] = [M L T^{-2}] \times [M L^{-3} T^0] = [M^2 L^{-2} T^{-2}]$.
Thus,the dimensions are $[M^2 L^{-2} T^{-2}]$ and $[M^{1/3} L^{-1} T^0]$.

(A) The dimensions of the given physical quantities are:
$E = [M L^2 T^{-2}]$
$m = [M]$
$L = [M L^2 T^{-1}]$
$G = [M^{-1} L^3 T^{-2}]$
Substituting these into the expression $\frac{E L^2}{m^5 G^2}$:
$\left[\frac{E L^2}{m^5 G^2}\right] = \frac{[M L^2 T^{-2}] [M L^2 T^{-1}]^2}{[M]^5 [M^{-1} L^3 T^{-2}]^2}$
$= \frac{[M L^2 T^{-2}] [M^2 L^4 T^{-2}]}{[M^5] [M^{-2} L^6 T^{-4}]}$
$= \frac{[M^3 L^6 T^{-4}]}{[M^3 L^6 T^{-4}]}$
$= [M^0 L^0 T^0]$
Since the dimensions are $[M^0 L^0 T^0]$,the quantity is dimensionless,which corresponds to the dimensions of an $\text{Angle}$.

(A) Given the equation: $F = a \sqrt{x} + b t^2$.
According to the principle of homogeneity of dimensions,the dimensions of each term on both sides of the equation must be the same.
Therefore,$[F] = [a \sqrt{x}]$ and $[F] = [b t^2]$.
From $[F] = [a \sqrt{x}]$,we get $[a] = \frac{[F]}{[\sqrt{x}]} = \frac{[MLT^{-2}]}{[L^{1/2}]} = [ML^{1/2}T^{-2}]$.
From $[F] = [b t^2]$,we get $[b] = \frac{[F]}{[t^2]} = \frac{[MLT^{-2}]}{[T^2]} = [MLT^{-4}]$.
Now,we find the dimensional formula for $\frac{a}{b}$:
$\left[\frac{a}{b}\right] = \frac{[ML^{1/2}T^{-2}]}{[MLT^{-4}]} = [M^{1-1} L^{1/2-1} T^{-2-(-4)}] = [M^0 L^{-1/2} T^2]$.
Thus,the correct option is $A$.

(C) Let the time period $T$ be proportional to $R^a M^b G^c$,so $T = K R^a M^b G^c$.
Writing the dimensional formulas for each quantity:
$[T] = [T]^1$
$[R] = [L]^1$
$[M] = [M]^1$
$[G] = [M^{-1} L^3 T^{-2}]$
Substituting these into the equation: $[T]^1 = [L]^a [M]^b [M^{-1} L^3 T^{-2}]^c = [M]^{b-c} [L]^{a+3c} [T]^{-2c}$.
Comparing the powers of $M$,$L$,and $T$ on both sides:
For $T$: $-2c = 1 \Rightarrow c = -1/2$.
For $M$: $b - c = 0 \Rightarrow b = c = -1/2$.
For $L$: $a + 3c = 0 \Rightarrow a = -3c = -3(-1/2) = 3/2$.
Substituting these values back into the expression: $T = K R^{3/2} M^{-1/2} G^{-1/2} = K \sqrt{\frac{R^3}{GM}}$.

(B) According to the principle of homogeneity of dimensions,only physical quantities of the same dimension can be added or subtracted.
In the given equation $(P+\frac{a}{V^2})(V-b)=RT$,the term $\frac{a}{V^2}$ is added to pressure $P$.
Therefore,the dimensions of $\frac{a}{V^2}$ must be equal to the dimensions of pressure $P$.
$[P] = [\frac{a}{V^2}] \implies [a] = [P][V^2]$.
The dimension of pressure $P$ is $[ML^{-1} T^{-2}]$ and the dimension of volume $V$ is $[L^3]$.
Substituting these,we get $[a] = [ML^{-1} T^{-2}] \times [L^3]^2 = [ML^{-1} T^{-2}] \times [L^6] = [ML^5 T^{-2}]$.

(D) The dimensional formula for speed of light $c$ is $[L T^{-1}]$.
Given $c = 3 \times 10^8 \,m/s$ ... $(1)$
The dimensional formula for acceleration due to gravity $g$ is $[L T^{-2}]$.
Given $g = 10 \,m/s^2$ ... $(2)$
To find the unit of time $[T]$, we divide the dimensions of speed by the dimensions of acceleration:
$\frac{[L T^{-1}]}{[L T^{-2}]} = [T]$
Substituting the given values:
$[T] = \frac{3 \times 10^8}{10} = 3 \times 10^7 \,s$
Therefore, the unit of time in this system is $3 \times 10^7 \,s$.

(C) The given formula is $X = \frac{\pi}{3}(a^2 - b^2) h d$.
Since $a, b,$ and $h$ are lengths,their dimensional formula is $[L]$.
Thus,$(a^2 - b^2)$ has dimensions $[L^2]$.
The term $h$ has dimensions $[L]$.
The term $d$ is density,which is defined as $\frac{\text{mass}}{\text{volume}}$,so its dimensions are $[M L^{-3}]$.
Substituting these dimensions into the formula:
$[X] = [L^2] \cdot [L] \cdot [M L^{-3}] = [L^3] \cdot [M L^{-3}] = [M]$.
Since the resulting dimension is $[M]$,the physical quantity is mass.

(C) The energy of the system is given by the equation $E(t) = \alpha t - \beta t^3$.
According to the principle of homogeneity of dimensions,the dimensions of each term in an equation must be the same as the dimensions of the physical quantity on the other side of the equation.
Since $E$ represents energy,its dimensional formula is $[ML^2 T^{-2}]$.
For the first term,the dimension of $\alpha t$ must be equal to the dimension of energy:
$[\alpha][T] = [ML^2 T^{-2}]$
$[\alpha] = [ML^2 T^{-2}] / [T] = [ML^2 T^{-3}]$
For the second term,the dimension of $\beta t^3$ must be equal to the dimension of energy:
$[\beta][T^3] = [ML^2 T^{-2}]$
$[\beta] = [ML^2 T^{-2}] / [T^3] = [ML^2 T^{-5}]$
Thus,the dimensions of $\alpha$ and $\beta$ are $[ML^2 T^{-3}]$ and $[ML^2 T^{-5}]$ respectively.

(A) The dimensional formula for Energy is $[M^1 L^2 T^{-2}]$.
The dimensional formula for speed is $[M^0 L^1 T^{-1}]$.
Multiplying these two,we get:
$[\text{Energy} \times \text{speed}] = [M^1 L^2 T^{-2}] \times [M^0 L^1 T^{-1}]$
$= [M^{1+0} L^{2+1} T^{-2-1}]$
$= [M^1 L^3 T^{-3}]$
Comparing this with $[M^a L^b T^c]$,we get $a = 1, b = 3, c = -3$.

(C) According to the principle of homogeneity,a physical equation is dimensionally correct only if the dimensions of all terms on both sides of the equation are identical. The argument of an exponential function must be dimensionless. In all given options,the term $\frac{V}{V_0}$ is dimensionless,so the exponential terms are valid.
Now,we check the dimensions of the coefficients:
For $(A)$: $[I] = [I_0]$,which is correct.
For $(B)$: $[I] = [I_0]$,which is correct.
For $(C)$: $[I] = [I_0 V_0]$. Since $[V_0]$ is potential,$[I] \neq [I_0 V_0]$. Thus,$(C)$ is dimensionally incorrect.
For $(D)$: $[I] = [I_0] \times [\frac{V}{V_0}]$. Since $[\frac{V}{V_0}]$ is dimensionless,$[I] = [I_0]$,which is correct.
Therefore,option $(C)$ is the dimensionally incorrect expression.

(A) The speed of ripples $v$ depends on surface tension $\sigma$,density $\rho$,and wavelength $\lambda$.
We can write the relation as $v = k \sigma^a \rho^b \lambda^c$,where $k$ is a dimensionless constant.
The dimensional formula for speed $v$ is $[M^0 L T^{-1}]$.
The dimensional formula for surface tension $\sigma$ is $[M L^0 T^{-2}]$.
The dimensional formula for density $\rho$ is $[M L^{-3} T^0]$.
The dimensional formula for wavelength $\lambda$ is $[M^0 L T^0]$.
Substituting these into the equation:
$[M^0 L T^{-1}] = [M L^0 T^{-2}]^a [M L^{-3} T^0]^b [L]^c$
$[M^0 L T^{-1}] = [M]^{a+b} [L]^{-3b+c} [T]^{-2a}$
Equating the powers of $M, L,$ and $T$ on both sides:
$a + b = 0$
$-3b + c = 1$
$-2a = -1$
From $-2a = -1$,we get $a = 1/2$.
Substituting $a = 1/2$ into $a + b = 0$,we get $b = -1/2$.
Substituting $b = -1/2$ into $-3b + c = 1$,we get $-3(-1/2) + c = 1 \Rightarrow 1.5 + c = 1 \Rightarrow c = -0.5 = -1/2$.
Thus,$v \propto \sigma^{1/2} \rho^{-1/2} \lambda^{-1/2}$.
Therefore,$v \propto \sqrt{\frac{\sigma}{\rho \lambda}}$.
Squaring both sides,$v^2 \propto \frac{\sigma}{\rho \lambda}$.

(C) We can express the permeability of vacuum as $\mu_0 \propto e^a m^b c^c h^d$ or $\mu_0 = k e^a m^b c^c h^d$ ...$(i)$,where $k$ is a dimensionless constant.
Dimensions of the quantities are:
$\mu_0 = [M L T^{-2} A^{-2}]$
$e = [A T]$
$m = [M]$
$c = [L T^{-1}]$
$h = [M L^2 T^{-1}]$
Substituting these into Eq. $(i)$:
$[M L T^{-2} A^{-2}] = [A T]^a [M]^b [L T^{-1}]^c [M L^2 T^{-1}]^d$
$[M L T^{-2} A^{-2}] = [M]^{b+d} [L]^{c+2d} [T]^{a-c-d} [A]^a$
Comparing powers on both sides:
$a = -2$
$b + d = 1$
$c + 2d = 1$
$a - c - d = -2$
From $a = -2$,substituting into $a - c - d = -2$ gives $-2 - c - d = -2$,so $c + d = 0$,which means $c = -d$.
Substituting $c = -d$ into $c + 2d = 1$ gives $-d + 2d = 1$,so $d = 1$.
Then $c = -1$ and $b = 1 - d = 1 - 1 = 0$.
Thus,$\mu_0 = k e^{-2} m^0 c^{-1} h^1 = k \frac{h}{c e^2}$.
Therefore,$\mu_0$ is proportional to $\frac{h}{c e^2}$.

(D) The dimensional formula of the coefficient of thermal conductivity $[k]$ is $[M^1 L^1 T^{-3} K^{-1}]$.
The dimensional formula of the universal gravitational constant $[G]$ is $[M^{-1} L^3 T^{-2}]$.
Taking the ratio,$\frac{[k]}{[G]} = \frac{[M^1 L^1 T^{-3} K^{-1}]}{[M^{-1} L^3 T^{-2}]} = [M^{1-(-1)} L^{1-3} T^{-3-(-2)} K^{-1}] = [M^2 L^{-2} T^{-1} K^{-1}]$.
Comparing this with the given dimensional formula $[M^{2a} L^{4b} T^{2c} K^d]$:
$2a = 2 \implies a = 1$
$4b = -2 \implies b = -\frac{1}{2}$
$2c = -1 \implies c = -\frac{1}{2}$
$d = -1$
Now,calculating the required expression: $\frac{a+b}{c+b} - d = \frac{1 + (-1/2)}{-1/2 + (-1/2)} - (-1) = \frac{1/2}{-1} + 1 = -\frac{1}{2} + 1 = \frac{1}{2}$.

(D) The dimensional formula for energy per unit volume is $[M L^{-1} T^{-2}]$.
The dimensional formula for angular momentum is $[M L^2 T^{-1}]$.
Since the dimensions of energy per unit volume and angular momentum are different,they cannot be added or subtracted.
Therefore,Assertion $(A)$ is false.
The principle of homogeneity states that only physical quantities having the same dimensions can be added or subtracted.
Therefore,Reason $(R)$ is true.

(D) The dimensional formula for electric permittivity $\epsilon_0$ is $[M^{-1} L^{-3} T^4 A^2]$.
The dimensional formula for length $L$ is $[L]$.
The dimensional formula for potential difference $\Delta V$ is $[M L^2 T^{-3} A^{-1}]$.
The dimensional formula for time interval $\Delta t$ is $[T]$.
Substituting these into the expression $P = \epsilon_0 L \frac{\Delta V}{\Delta t}$:
$[P] = [M^{-1} L^{-3} T^4 A^2] \cdot [L] \cdot \frac{[M L^2 T^{-3} A^{-1}]}{[T]}$
$[P] = [M^{-1} L^{-3} T^4 A^2] \cdot [L] \cdot [M L^2 T^{-3} A^{-1} T^{-1}]$
$[P] = [M^{-1+1} L^{-3+1+2} T^{4-3-1} A^{2-1}]$
$[P] = [M^0 L^0 T^0 A^1] = [A]$
Since the dimensional formula of $P$ is $[A]$,which represents electric current,the correct option is $D$.

(B) Given the formula: $r = \sqrt{\frac{64 IA}{\pi Bv}}$.
Squaring both sides,we get: $r^2 = \frac{64 IA}{\pi Bv}$.
Rearranging for '$A$': $A = \frac{r^2 \pi Bv}{64 I}$.
Now,substitute the dimensions of each physical quantity:
$[r] = [L]$,$[B] = [M T^{-2} I^{-1}]$,$[v] = [L T^{-1}]$,$[I] = [I]$.
Substituting these into the expression for '$A$':
$[A] = \frac{[L]^2 [M T^{-2} I^{-1}] [L T^{-1}]}{[I]} = [M L^3 T^{-3} I^{-2}]$.
Comparing this with the dimensions of resistivity $(\rho)$:
Resistance $R = \rho \frac{L}{Area} \implies \rho = R \frac{Area}{L}$.
$[R] = [M L^2 T^{-3} I^{-2}]$,$[Area] = [L^2]$,$[L] = [L]$.
$[\rho] = [M L^2 T^{-3} I^{-2}] \cdot \frac{[L^2]}{[L]} = [M L^3 T^{-3} I^{-2}]$.
Since the dimensions of '$A$' match the dimensions of resistivity,'$A$' represents resistivity.

(C) Given: $A = [LT^{-1}]$ (velocity),$B = [LT^{-2}]$ (acceleration),$C = [ML^2T^{-2}A^{-2}]$ (inductance),$D = [M^{-1}L^{-2}T^4A^2]$ (capacitance).
We need to find the dimensions of $A^{-1} BCD$.
Substitute the dimensions:
$A^{-1} = [L^{-1}T]$
$B = [LT^{-2}]$
$C = [ML^2T^{-2}A^{-2}]$
$D = [M^{-1}L^{-2}T^4A^2]$
Now,calculate the product:
$A^{-1} BCD = [L^{-1}T] \cdot [LT^{-2}] \cdot [ML^2T^{-2}A^{-2}] \cdot [M^{-1}L^{-2}T^4A^2]$
Group the terms:
$= [L^{-1} \cdot L \cdot L^2 \cdot L^{-2}] \cdot [T \cdot T^{-2} \cdot T^{-2} \cdot T^4] \cdot [M \cdot M^{-1}] \cdot [A^{-2} \cdot A^2]$
$= [L^0] \cdot [T^1] \cdot [M^0] \cdot [A^0]$
$= [T]$
Thus,the expression has the dimensions of Time.

(B) Let the mass $M$ be expressed as $M = C^a h^b G^c$.
The dimensions are:
$C = [LT^{-1}]$
$h = [ML^2T^{-1}]$
$G = [M^{-1}L^3T^{-2}]$
Substituting these into the equation:
$[M^1L^0T^0] = [LT^{-1}]^a [ML^2T^{-1}]^b [M^{-1}L^3T^{-2}]^c$
$[M^1L^0T^0] = M^{b-c} L^{a+2b+3c} T^{-a-b-2c}$
Comparing the powers of $M, L, T$ on both sides:
$b - c = 1$ $(i)$
$a + 2b + 3c = 0$ (ii)
$-a - b - 2c = 0$ (iii)
Adding (ii) and (iii): $b + c = 0$,so $b = -c$.
Substituting $b = -c$ into $(i)$: $-c - c = 1 \implies -2c = 1 \implies c = -1/2$.
Then $b = 1/2$.
Substituting $b = 1/2$ and $c = -1/2$ into (iii): $-a - 1/2 - 2(-1/2) = 0 \implies -a - 1/2 + 1 = 0 \implies a = 1/2$.
Thus,$M = C^{1/2} h^{1/2} G^{-1/2}$.

(B) The dimensional formula for force is $[M L T^{-2}]$.
Given,$N_1 = 10$,$M_1 = 1 \,kg$,$L_1 = 1 \,m$,$T_1 = 1 \,s$.
In the new system,$M_2 = A \,kg$,$L_2 = B \,m$,$T_2 = C \,s$.
Using the conversion formula $N_2 = N_1 \left( \frac{M_1}{M_2} \right)^1 \left( \frac{L_1}{L_2} \right)^1 \left( \frac{T_1}{T_2} \right)^{-2}$.
Substituting the values:
$N_2 = 10 \left( \frac{1}{A} \right)^1 \left( \frac{1}{B} \right)^1 \left( \frac{1}{C} \right)^{-2}$.
$N_2 = 10 \cdot A^{-1} \cdot B^{-1} \cdot C^2$.
Thus,the value of $10 \,N$ in the new system is $10 A^{-1} B^{-1} C^2$.

(B) The given expression is $10 \ g \ cm \ s^{-1} = x \ N \ s$.
First,convert the $CGS$ unit $g \ cm \ s^{-1}$ to $SI$ units $(kg \ m \ s^{-1})$:
$1 \ g = 10^{-3} \ kg$
$1 \ cm = 10^{-2} \ m$
So,$1 \ g \ cm \ s^{-1} = 10^{-3} \ kg \times 10^{-2} \ m \times s^{-1} = 10^{-5} \ kg \ m \ s^{-1}$.
Thus,$10 \ g \ cm \ s^{-1} = 10 \times 10^{-5} \ kg \ m \ s^{-1} = 10^{-4} \ kg \ m \ s^{-1}$.
We know that $1 \ N = 1 \ kg \ m \ s^{-2}$,so $1 \ N \ s = 1 \ kg \ m \ s^{-1}$.
Equating the two,we get $x \ N \ s = 10^{-4} \ kg \ m \ s^{-1}$.
Therefore,$x = 10^{-4}$.

(C) The given physical quantity is $X = \frac{2 k^3 l^2}{m \sqrt{n}}$.
To find the relative error in $X$,we use the formula for propagation of errors:
$\frac{\Delta X}{X} = 3 \left( \frac{\Delta k}{k} \right) + 2 \left( \frac{\Delta l}{l} \right) + 1 \left( \frac{\Delta m}{m} \right) + \frac{1}{2} \left( \frac{\Delta n}{n} \right)$.
Given percentage errors are $\frac{\Delta k}{k} \times 100 = 1 \%$,$\frac{\Delta l}{l} \times 100 = 2 \%$,$\frac{\Delta m}{m} \times 100 = 3 \%$,and $\frac{\Delta n}{n} \times 100 = 4 \%$.
Substituting these values into the percentage error formula:
$\frac{\Delta X}{X} \times 100 = 3(1 \%) + 2(2 \%) + 1(3 \%) + \frac{1}{2}(4 \%)$.
$\frac{\Delta X}{X} \times 100 = 3 \% + 4 \% + 3 \% + 2 \% = 12 \%$.
Thus,the percentage uncertainty in $X$ is $12 \%$.

(D) Given the relation: $z = a b^2 c^{-2}$.
According to the theory of propagation of errors,the relative error in $z$ is given by:
$\frac{\Delta z}{z} = \frac{\Delta a}{a} + 2 \frac{\Delta b}{b} + 2 \frac{\Delta c}{c}$.
To find the percentage error,multiply the entire equation by $100$:
$\frac{\Delta z}{z} \times 100 = \left( \frac{\Delta a}{a} \times 100 \right) + 2 \left( \frac{\Delta b}{b} \times 100 \right) + 2 \left( \frac{\Delta c}{c} \times 100 \right)$.
Substituting the given percentage errors:
$\frac{\Delta z}{z} \times 100 = 2.1 \% + 2(1.3 \%) + 2(2.2 \%)$.
$\frac{\Delta z}{z} \times 100 = 2.1 \% + 2.6 \% + 4.4 \% = 9.1 \%$.
Therefore,the percentage error in $z$ is $9.1 \%$.

(C) The given expression is $Q V = k P T L^\alpha$.
The dimensions of the physical quantities are:
$[Q] = M L^{-1} T^{-1}$
$[V] = L^3$
$[P] = M L^{-1} T^{-2}$
$[T] = T$
$[L] = L$
$k$ is a dimensionless constant,so $[k] = 1$.
Equating the dimensions on both sides:
$[Q][V] = [k][P][T][L]^\alpha$
$(M L^{-1} T^{-1})(L^3) = (1)(M L^{-1} T^{-2})(T)(L^\alpha)$
$M L^2 T^{-1} = M L^{-1+\alpha} T^{-1}$
Comparing the powers of $L$ on both sides:
$2 = -1 + \alpha$
$\alpha = 3$.

(D) For an equation to be dimensionally correct,the exponent of an exponential function must be dimensionless.
In the expression $E=E_0 e^{-t / t_0}$,the term $-t / t_0$ is the ratio of two times,which is dimensionless.
Since $E$ and $E_0$ both represent energy,they have the same dimensional formula $[M L^2 T^{-2}]$.
Therefore,the equation $E=E_0 e^{-t / t_0}$ is dimensionally consistent because the exponential factor $e^{-t / t_0}$ is a dimensionless constant.
Other options are dimensionally incorrect because their exponents are not dimensionless or the dimensions on both sides do not match.

(A) The van der Waals gas equation is $(P+\frac{a}{V^2})(V-b)=n R T$.
According to the principle of homogeneity,dimensions of terms added or subtracted must be the same.
$1$. Dimension of $\frac{a}{V^2}$ must be equal to the dimension of $P$ (pressure).
$[P] = [ML^{-1} T^{-2}]$ and $[V] = [L^3]$.
So,$[a] = [P] \times [V^2] = [ML^{-1} T^{-2}] \times [L^3]^2 = [ML^5 T^{-2}]$.
$2$. Dimension of $b$ must be equal to the dimension of $V$ (volume).
$[b] = [V] = [L^3]$.
$3$. Now,the ratio $\frac{b}{a}$ has dimensions:
$\frac{[b]}{[a]} = \frac{[L^3]}{[ML^5 T^{-2}]} = [M^{-1} L^{-2} T^2]$.

(C) According to the principle of homogeneity of dimensions,only physical quantities with the same dimensions can be added or subtracted.
Given that $A, B$ and $C$ have different dimensional formulae,the expressions $(A-C)$ and $(AC-B)$ are physically meaningless because they involve the addition or subtraction of quantities with different dimensions.
However,the question asks for a combination that can never represent a proper physical quantity. In option $(C)$,the term $(A-C)$ is dimensionally invalid. In option $(D)$,the term $(AC-B)$ is dimensionally invalid.
Looking at standard physics problems of this type,the expression $\frac{A-C}{B}$ is often cited as invalid because the numerator $(A-C)$ is undefined. Similarly,$(AC-B)$ is undefined. Given the options,$(C)$ is the most standard representation of an invalid dimensional operation where subtraction is performed on quantities of different dimensions.

(A) The given equation is $m \frac{d^2 x}{dt^2} + b \frac{dx}{dt} + kx = 0$.
According to the principle of homogeneity of dimensions,each term in the equation must have the same dimensions.
Thus,the dimensions of $m \frac{d^2 x}{dt^2}$,$b \frac{dx}{dt}$,and $kx$ are equal.
Equating the dimensions of $b \frac{dx}{dt}$ and $kx$:
$[b] [v] = [k] [x] \implies [b] = [k] [x] / [v] = [k] [x] / ([x] / [t]) = [k] [t]$.
Therefore,$[b] = [k] [T]$.
Now,consider the expression $\frac{b}{\sqrt{km}}$.
Substituting $[b] = [k] [T]$ into the expression:
$\frac{[b]}{\sqrt{[k][m]}} = \frac{[k][T]}{\sqrt{[k][m]}}$.
Since $k = F/x = ma/x$,the dimensions of $k$ are $[M T^{-2}]$.
Also,the angular frequency of a damped oscillator is related to $\omega = \sqrt{k/m}$,so $\sqrt{k/m}$ has dimensions of $[T^{-1}]$.
Alternatively,using $[b] = [k] [T]$:
$\frac{b}{\sqrt{km}} = \frac{k T}{\sqrt{km}} = \sqrt{\frac{k}{m}} T = [T^{-1}] [T] = [M^0 L^0 T^0]$.