Dimensional Analysis, Uses and Limitations Questions in English

(D) According to the principle of homogeneity of dimensions,every term in an equation must have the same dimensions.
Since $S$ is measured in metres $(m)$,the term $ct^2$ must also have the dimensions of length $(L)$.
Therefore,the dimensions of $c$ are given by $[c] = [S] / [t^2] = L / T^2 = L T^{-2}$.
Since $S$ is in metres $(m)$ and $t$ is in seconds $(s)$,the unit of $c$ is $m/s^2$ or $m s^{-2}$.

(C) According to the principle of homogeneity of dimensions,the dimensions of each term in a physical equation must be the same.
Given the equation $x = at + bt^2$,where $x$ is distance and $t$ is time.
Therefore,the dimensions of $bt^2$ must be equal to the dimensions of $x$.
$[bt^2] = [x]$
$[b] = [x] / [t^2]$
Since $x$ is in kilometres $(km)$ and $t$ is in seconds $(s)$,the unit of $b$ is $km/s^2$.

(B) According to the principle of homogeneity of dimensions,only physical quantities with the same dimensions can be added or subtracted.
In the term $\left( P + \frac{a}{V^2} \right)$,$P$ (pressure) is added to $\frac{a}{V^2}$.
Therefore,the dimensions of $P$ must be equal to the dimensions of $\frac{a}{V^2}$.
$[P] = \left[ \frac{a}{V^2} \right]$
$[a] = [P] \times [V^2]$
Since pressure $P$ is force per unit area,its unit is $Dyne/cm^2$ (in $CGS$ system).
Volume $V$ has units of $cm^3$,so $V^2$ has units of $cm^6$.
Unit of $a = (Dyne/cm^2) \times (cm^6) = Dyne \times cm^4$.

(B) The dimensional formula for Pressure is $P = \frac{F}{A} = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$.
Energy per unit volume is $\frac{E}{V} = \frac{[ML^2T^{-2}]}{[L^3]} = [ML^{-1}T^{-2}]$.
Since the dimensions of Pressure and Energy per unit volume are identical,option $(b)$ is correct.

(A) According to the principle of dimensional homogeneity,only physical quantities with the same dimensions can be added or subtracted.
In the given equation $\left( {P + \frac{a}{{{V^2}}}} \right)\,(V - b) = RT$,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 $P$.
$[\frac{a}{{{V^2}}}] = [P]$
$[a] = [P] \times [V^2]$
We know that the dimensions of pressure $P$ are $[M{L^{ - 1}}{T^{ - 2}}]$ and the dimensions of volume $V$ are $[L^3]$.
So,$[a] = [M{L^{ - 1}}{T^{ - 2}}] \times [L^3]^2$
$[a] = [M{L^{ - 1}}{T^{ - 2}}] \times [L^6]$
$[a] = [M{L^5}{T^{ - 2}}]$

(D) The dimensional formula for frequency $f$ is $[M^0 L^0 T^{-1}]$.
The dimensional formula for mass $m$ is $[M^1 L^0 T^0]$.
The dimensional formula for spring constant $K$ is $[M^1 L^0 T^{-2}]$.
Given the relation $f = C m^x K^y$,substituting the dimensions on both sides:
$[M^0 L^0 T^{-1}] = [M^1]^x [M^1 T^{-2}]^y$
$[M^0 L^0 T^{-1}] = [M^{x+y} T^{-2y}]$
Comparing the powers of $M$ and $T$ on both sides:
For $M$: $x + y = 0$
For $T$: $-2y = -1 \implies y = \frac{1}{2}$
Substituting $y = \frac{1}{2}$ into $x + y = 0$,we get $x + \frac{1}{2} = 0 \implies x = -\frac{1}{2}$.
Thus,$x = -\frac{1}{2}$ and $y = \frac{1}{2}$.

(C) The dimension of velocity $v$ is $[M^0 L^1 T^{-1}]$.
The dimension of wavelength $\lambda$ is $[M^0 L^1 T^0]$.
The dimension of acceleration due to gravity $g$ is $[M^0 L^1 T^{-2}]$.
The dimension of density $\rho$ is $[M^1 L^{-3} T^0]$.
Let the relation be $v = k \lambda^a g^b \rho^c$,where $k$ is a dimensionless constant.
Equating dimensions on both sides:
$[L T^{-1}] = [L]^a [L T^{-2}]^b [M L^{-3}]^c$
$[M^0 L^1 T^{-1}] = [M^c L^{a+b-3c} T^{-2b}]$
Comparing the powers of $M, L,$ and $T$:
For $M$: $c = 0$
For $T$: $-2b = -1 \implies b = 0.5$
For $L$: $a + b - 3c = 1 \implies a + 0.5 - 0 = 1 \implies a = 0.5$
Substituting the values,$v \propto \lambda^{0.5} g^{0.5} \rho^0$.
Therefore,$v \propto \sqrt{g \lambda}$,which implies $v^2 \propto g \lambda$.

(B) According to the principle of homogeneity of dimensions,the dimensions of all terms added or subtracted in an equation must be the same.
In the given equation $Y = A \sin \omega \left( \frac{x}{v} - k \right)$,the term $k$ is subtracted from $\frac{x}{v}$.
Therefore,the dimension of $k$ must be equal to the dimension of $\frac{x}{v}$.
Dimension of $x$ (distance) is $[L]$.
Dimension of $v$ (velocity) is $[LT^{-1}]$.
Thus,the dimension of $\frac{x}{v} = \frac{[L]}{[LT^{-1}]} = [T]$.
Hence,the dimension of $k$ is $[T]$.

(A) The dimensional formula for the given quantities are:
$T = [T^1]$
$P = [M^1 L^{-1} T^{-2}]$
$D = [M^1 L^{-3} T^0]$
$S = [M^1 L^0 T^{-2}]$
Substituting these into the equation $T = P^a D^b S^c$:
$[M^0 L^0 T^1] = [M^1 L^{-1} T^{-2}]^a [M^1 L^{-3}]^b [M^1 T^{-2}]^c$
$[M^0 L^0 T^1] = M^{a+b+c} L^{-a-3b} T^{-2a-2c}$
Comparing the powers of $M, L,$ and $T$ on both sides:
$a + b + c = 0$ $(1)$
$-a - 3b = 0 \implies a = -3b$ $(2)$
$-2a - 2c = 1 \implies a + c = -1/2$ $(3)$
From $(2)$,substitute $a = -3b$ into $(1)$:
$-3b + b + c = 0 \implies c = 2b$
Substitute $a = -3b$ and $c = 2b$ into $(3)$:
$-3b + 2b = -1/2 \implies -b = -1/2 \implies b = 1/2$
Then $a = -3(1/2) = -3/2$ and $c = 2(1/2) = 1$.
Thus,the values are $a = -3/2, b = 1/2, c = 1$.

(B) Given the relation: $v \propto g^p h^q$.
The dimensions of the quantities are:
Velocity $(v)$: $[L T^{-1}]$
Acceleration due to gravity $(g)$: $[L T^{-2}]$
Height $(h)$: $[L]$
Substituting these dimensions into the given relation:
$[L T^{-1}] = [L T^{-2}]^p [L]^q$
$[L T^{-1}] = L^p T^{-2p} L^q$
$[L T^{-1}] = L^{p+q} T^{-2p}$
Comparing the powers of $L$ and $T$ on both sides:
For $T$: $-2p = -1 \Rightarrow p = \frac{1}{2}$
For $L$: $p + q = 1 \Rightarrow \frac{1}{2} + q = 1 \Rightarrow q = \frac{1}{2}$
Thus,the values are $p = \frac{1}{2}$ and $q = \frac{1}{2}$.

(A) Let the terminal velocity $v_T$ be expressed as $v_T = k \cdot m^a \eta^b r^c g^d$,where $k$ is a dimensionless constant.
The dimensional formulas are:
$[v_T] = [L T^{-1}]$
$[m] = [M]$
$[\eta] = [M L^{-1} T^{-1}]$
$[r] = [L]$
$[g] = [L T^{-2}]$
Substituting these into the equation:
$[L T^{-1}] = [M]^a [M L^{-1} T^{-1}]^b [L]^c [L T^{-2}]^d$
$[M^0 L^1 T^{-1}] = [M^{a+b} L^{-b+c+d} T^{-b-2d}]$
Comparing the powers of $M, L,$ and $T$ on both sides:
For $M$: $a + b = 0 \Rightarrow a = -b$
For $T$: $-b - 2d = -1 \Rightarrow b + 2d = 1$
For $L$: $-b + c + d = 1$
Using Stokes' Law,the terminal velocity is $v_T = \frac{2r^2(\rho - \sigma)g}{9\eta}$. Since the mass $m$ of the ball is proportional to $r^3$ (i.e.,$m \propto r^3$),we substitute $r \propto m^{1/3}$ into the relation. Dimensional analysis shows that $v_T \propto \frac{mg}{\eta r}$ is the only relation consistent with the dimensions of velocity.

(B) The speed of light in free space 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,the expression for the product is: ${\mu _0}{\varepsilon _0} = \frac{1}{c^2}$.
Since the dimension of velocity $c$ is $[L{T^{ - 1}}]$,the dimensions of ${\mu _0}{\varepsilon _0}$ are:
$[{\mu _0}{\varepsilon _0}] = \frac{1}{[L{T^{ - 1}}]^2} = \frac{1}{L^2 T^{-2}} = {L^{ - 2}}{T^2}$.

(C) Given the equation: $\text{Force} = \frac{X}{\text{Density}}$
Rearranging for $X$,we get: $X = \text{Force} \times \text{Density}$
The dimensional formula for Force is $[F] = [M^1 L^1 T^{-2}]$
The dimensional formula for Density is $[\rho] = [M^1 L^{-3}]$
Therefore,the dimensions of $X$ are: $[X] = [M^1 L^1 T^{-2}] \times [M^1 L^{-3}]$
$[X] = [M^{1+1} L^{1-3} T^{-2}]$
$[X] = [M^2 L^{-2} T^{-2}]$

(D) The definition of acceleration $(A)$ in terms of length $(L)$ and time $(T)$ is given by the formula: $A = L T^{-2}$.
To find the dimensions of length $(L)$ in terms of the given fundamental quantities,we rearrange the formula:
$L = A T^2$.
Since $A$ and $T$ are chosen as fundamental quantities in the Martian system,the dimension of length is represented as $A T^2$.
Therefore,the correct option is $(d)$.

(C) The foundations of dimensional analysis were established by the French mathematician Joseph Fourier in his work 'Théorie analytique de la chaleur' (The Analytical Theory of Heat) published in $1822$. He introduced the concept of dimensions to represent the physical nature of quantities,which remains a fundamental tool in physics today.

(B) Let the angular momentum $L$ be expressed as $L = k v^x A^y F^z$,where $k$ is a dimensionless constant.
Substituting the dimensions of each quantity:
$[M L^2 T^{-1}] = [L T^{-1}]^x [L T^{-2}]^y [M L T^{-2}]^z$
$[M L^2 T^{-1}] = M^z L^{x+y+z} T^{-x-2y-2z}$
Comparing the exponents of $M, L$,and $T$ on both sides:
For $M$: $z = 1$
For $L$: $x + y + z = 2$
For $T$: $-x - 2y - 2z = -1$
Substituting $z = 1$ into the other equations:
$x + y = 1$ (Equation $i$)
$-x - 2y = 1$ (Equation $ii$)
Adding $(i)$ and $(ii)$: $-y = 2 \Rightarrow y = -2$
Substituting $y = -2$ into $x + y = 1$: $x - 2 = 1 \Rightarrow x = 3$
Thus,the dimensional formula is $[L] = [F v^3 A^{-2}]$.

(A) Given the formula for the viscous force (Stokes' Law): $F = 6\pi \eta av$.
To find the dimensions of $\eta$,we rearrange the formula: $\eta = \frac{F}{6\pi av}$.
Since $6\pi$ is a dimensionless constant,the dimensional formula is: $[\eta] = \frac{[F]}{[a][v]}$.
Substituting the dimensions of force $[F] = M L T^{-2}$,radius $[a] = L$,and velocity $[v] = L T^{-1}$:
$[\eta] = \frac{M L T^{-2}}{L \cdot L T^{-1}} = \frac{M L T^{-2}}{L^2 T^{-1}}$.
Simplifying the expression: $[\eta] = M L^{1-2} T^{-2-(-1)} = M L^{-1} T^{-1}$.

(A) According to the principle of dimensional homogeneity,physical quantities can only be added or subtracted if they have the same dimensions.
However,physical quantities with different dimensions can be multiplied or divided to form new physical quantities.
Therefore,the operation $A/B$ is physically meaningful,while $A + B$ and $A - B$ are not.

(A) The dimensions of the given physical quantities are:
$v = [LT^{-1}]$
$r = [L]$
$g = [LT^{-2}]$
Now,check the dimensions of the expression $\frac{v^2}{rg}$:
Dimension of $v^2 = [LT^{-1}]^2 = [L^2 T^{-2}]$
Dimension of $rg = [L] \cdot [LT^{-2}] = [L^2 T^{-2}]$
Therefore,the dimension of $\frac{v^2}{rg} = \frac{[L^2 T^{-2}]}{[L^2 T^{-2}]} = [M^0 L^0 T^0]$.
This expression is dimensionless. This quantity represents $\tan \theta$ in the context of the angle of banking for a vehicle on a curved road.

(B) The dimensional formula for energy is $[M{L^2}{T^{ - 2}}]$.
Solar constant is defined as the energy received per unit area per unit time.
Dimensional formula of Solar constant = $\frac{[M{L^2}{T^{ - 2}}]}{[L^2][T]} = [M^1 T^{-3}]$.
Therefore,the correct option is $B$.

(B) According to the principle of dimensional homogeneity,the dimensions of each term in an equation must be the same.
Since $F = at + bt^2$,the dimensions of $F$ must be equal to the dimensions of $at$ and $bt^2$.
The dimension of force $F$ is $[MLT^{-2}]$.
For term $at$: $[a][t] = [MLT^{-2}] \implies [a] = [MLT^{-2}] / [T] = [MLT^{-3}]$.
For term $bt^2$: $[b][t^2] = [MLT^{-2}] \implies [b] = [MLT^{-2}] / [T^2] = [MLT^{-4}]$.
Thus,the dimensions of $a$ and $b$ are $[MLT^{-3}]$ and $[MLT^{-4}]$ respectively.

(B) Let the dimension of the gravitational constant be $[G] = [c^x g^y p^z]$.
The dimensions of the given quantities are:
$[G] = [M^{-1} L^3 T^{-2}]$
$[c] = [L T^{-1}]$
$[g] = [L T^{-2}]$
$[p] = [M L^{-1} T^{-2}]$
Substituting these into the equation:
$[M^{-1} L^3 T^{-2}] = [L T^{-1}]^x [L T^{-2}]^y [M L^{-1} T^{-2}]^z$
$[M^{-1} L^3 T^{-2}] = [M^z L^{x+y-z} T^{-x-2y-2z}]$
Comparing the powers of $M, L,$ and $T$ on both sides:
For $M$: $z = -1$
For $L$: $x + y - z = 3 \implies x + y - (-1) = 3 \implies x + y = 2$
For $T$: $-x - 2y - 2z = -2 \implies -x - 2y - 2(-1) = -2 \implies -x - 2y = -4 \implies x + 2y = 4$
Solving the equations:
$(x + 2y) - (x + y) = 4 - 2 \implies y = 2$
$x + 2 = 2 \implies x = 0$
Thus,$[G] = [c^0 g^2 p^{-1}]$.

(A) Let the time period $T$ be proportional to $S^x r^y \rho^z$,so $T = k S^x r^y \rho^z$.
Substituting the dimensions:
$[T] = [M^0 L^0 T^1]$
$[S] = [M L^0 T^{-2}]$
$[r] = [L]$
$[\rho] = [M L^{-3} T^0]$
Equating dimensions on both sides:
$[M^0 L^0 T^1] = [M L^0 T^{-2}]^x [L]^y [M L^{-3} T^0]^z$
$[M^0 L^0 T^1] = [M^{x+z} L^{y-3z} T^{-2x}]$
Comparing powers:
For $M$: $x + z = 0 \Rightarrow z = -x$
For $T$: $-2x = 1 \Rightarrow x = -1/2$
Therefore,$z = 1/2$
For $L$: $y - 3z = 0 \Rightarrow y = 3z = 3(1/2) = 3/2$
Substituting these values back into the expression:
$T = k S^{-1/2} r^{3/2} \rho^{1/2} = k \sqrt{\frac{\rho r^3}{S}}$

(A) Let the force $F$ be expressed as $F = P^{\alpha} V^{\beta} T^{\gamma}$.
The dimensional formula for force is $[M^1 L^1 T^{-2}]$.
The dimensional formula for pressure $P$ is $[M^1 L^{-1} T^{-2}]$.
The dimensional formula for velocity $V$ is $[L^1 T^{-1}]$.
The dimensional formula for time $T$ is $[T^1]$.
Substituting these into the equation:
$[M^1 L^1 T^{-2}] = [M^1 L^{-1} T^{-2}]^{\alpha} [L^1 T^{-1}]^{\beta} [T^1]^{\gamma}$
$[M^1 L^1 T^{-2}] = [M^{\alpha} L^{-\alpha + \beta} T^{-2\alpha - \beta + \gamma}]$
Comparing the powers of $M$,$L$,and $T$ on both sides:
For $M$: $\alpha = 1$
For $L$: $-\alpha + \beta = 1 \Rightarrow -1 + \beta = 1 \Rightarrow \beta = 2$
For $T$: $-2\alpha - \beta + \gamma = -2 \Rightarrow -2(1) - 2 + \gamma = -2 \Rightarrow -4 + \gamma = -2 \Rightarrow \gamma = 2$
Thus,the dimensional formula for force is $P^1 V^2 T^2$ or $P V^2 T^2$.

(B) Let the dimension of mass be $[M] = [E]^x [v]^y [F]^z$.
Substituting the dimensions of each quantity:
$[M] = [M L^2 T^{-2}]^x [L T^{-1}]^y [M L T^{-2}]^z$
$[M] = M^{x+z} L^{2x+y+z} T^{-2x-y-2z}$
Comparing the powers of $M, L,$ and $T$ on both sides:
For $M$: $x + z = 1$
For $L$: $2x + y + z = 0$
For $T$: $-2x - y - 2z = 0$
From the third equation: $y = -2x - 2z = -2(x + z)$.
Since $x + z = 1$,we get $y = -2(1) = -2$.
Now,substitute $y = -2$ into the second equation: $2x - 2 + z = 0 \implies 2x + z = 2$.
We have the system:
$x + z = 1$
$2x + z = 2$
Subtracting the first from the second: $x = 1$.
Then $z = 1 - x = 1 - 1 = 0$.
Thus,$[M] = [E]^1 [v]^{-2} [F]^0 = [E v^{-2}]$.

(D) Given the equation: $x = Ay + B \tan(Cz)$.
According to the principle of dimensional homogeneity,the dimensions of each term in an equation must be the same.
$1$. For the term $B \tan(Cz)$,the argument of the trigonometric function $(Cz)$ must be dimensionless. Therefore,$[Cz] = [M^0 L^0 T^0]$,which implies $[C] = [z^{-1}]$. Thus,$C$ and $z^{-1}$ have the same dimensions.
$2$. The dimensions of $x$ must be equal to the dimensions of $Ay$ and $B$. Therefore,$[x] = [Ay] = [B]$. This means $x$ and $B$ have the same dimensions.
$3$. From $[x] = [Ay]$,we get $[y] = [x/A]$. Also,from $[x] = [B]$,we get $[A] = [B/x]$. Combining these,$[y] = [x/A] = [B/A]$. Thus,$y$ and $B/A$ have the same dimensions.
$4$. Comparing $x$ and $A$: Since $[x] = [Ay]$,the dimension of $A$ is $[x/y]$. Therefore,$x$ and $A$ do not have the same dimensions.
Hence,the correct option is $(d)$.

(C) Let mass $M$ be expressed as $M \propto c^x G^y h^z$.
Substituting the dimensions:
$[M] = M^1 L^0 T^0$
$[c] = L T^{-1}$
$[G] = M^{-1} L^3 T^{-2}$
$[h] = M L^2 T^{-1}$
Thus,$M^1 L^0 T^0 = (L T^{-1})^x (M^{-1} L^3 T^{-2})^y (M L^2 T^{-1})^z$
$M^1 L^0 T^0 = M^{-y+z} L^{x+3y+2z} T^{-x-2y-z}$
Comparing the powers of $M, L, T$ on both sides:
$1$) $-y + z = 1$
$2$) $x + 3y + 2z = 0$
$3$) $-x - 2y - z = 0$
Adding $(2)$ and $(3)$: $y + z = 0 \implies y = -z$.
Substituting $y = -z$ into $(1)$: $-(-z) + z = 1 \implies 2z = 1 \implies z = 1/2$.
Then $y = -1/2$.
Substituting $y$ and $z$ into $(3)$: $-x - 2(-1/2) - (1/2) = 0 \implies -x + 1 - 1/2 = 0 \implies x = 1/2$.
Therefore,the dimensions of mass are $c^{1/2} G^{-1/2} h^{1/2}$.

(A) Let the frequency $n$ be expressed as $n = k{\rho ^x}{a^y}{T^z}$.
Dimensional formulas are: $[n] = [T^{-1}]$,$[\rho] = [ML^{-3}]$,$[a] = [L]$,and $[T] = [MT^{-2}]$.
Substituting these into the equation: $[M^0L^0T^{-1}] = [ML^{-3}]^x [L]^y [MT^{-2}]^z$.
$[M^0L^0T^{-1}] = [M^{x+z} L^{-3x+y} T^{-2z}]$.
Comparing the powers on both sides:
For $M$: $x + z = 0 \implies x = -z$.
For $T$: $-2z = -1 \implies z = 1/2$. Thus,$x = -1/2$.
For $L$: $-3x + y = 0 \implies y = 3x = 3(-1/2) = -3/2$.
Wait,re-evaluating: If $n \propto \rho^x a^y T^z$,then $x = -1/2, y = -3/2, z = 1/2$.
Thus,$n = k \rho^{-1/2} a^{-3/2} T^{1/2} = k \frac{\sqrt{T}}{\sqrt{\rho} a^{3/2}}$.
Given the options provided,the correct dimensional relationship is $n = k \frac{\sqrt{T}}{\rho^{1/2} a^{3/2}}$. Since the question asks for the form matching option $(A)$,we identify the correct dimensional dependency.

(A) Let the mass $M$ be expressed as $M = k F^a L^b T^c$,where $k$ is a dimensionless constant.
Substituting the dimensions of each quantity:
$[M] = [M^1 L^0 T^0]$
$[F] = [M^1 L^1 T^{-2}]$
$[L] = [L]$
$[T] = [T]$
So,$[M^1 L^0 T^0] = [M^1 L^1 T^{-2}]^a [L]^b [T]^c$
$[M^1 L^0 T^0] = [M^a L^{a+b} T^{-2a+c}]$
Comparing the powers of $M, L,$ and $T$ on both sides:
For $M$: $a = 1$
For $L$: $a + b = 0 \implies 1 + b = 0 \implies b = -1$
For $T$: $-2a + c = 0 \implies -2(1) + c = 0 \implies c = 2$
Substituting these values back,we get the dimensional formula for mass as $M = F^1 L^{-1} T^2$ or $F L^{-1} T^2$.

(C) In the given wave equation $y = a \cos (\omega t - kx)$,the argument of the trigonometric function must be dimensionless.
Therefore,the dimensions of $kx$ must be equal to the dimensions of a dimensionless quantity (i.e.,$[M^0 L^0 T^0]$).
$[kx] = [M^0 L^0 T^0]$
$[k] [x] = [M^0 L^0 T^0]$
Since $[x]$ represents length,its dimension is $[L]$.
$[k] [L] = [M^0 L^0 T^0]$
$[k] = \frac{[M^0 L^0 T^0]}{[L]} = [M^0 L^{-1} T^0]$
Thus,the correct option is $C$.

(B) Given the equation $x = K{a^m}{t^n}$.
Applying the principle of homogeneity of dimensions,the dimensions on both sides must be equal.
The dimension of position $x$ is $[L]$.
The dimension of acceleration $a$ is $[LT^{-2}]$.
The dimension of time $t$ is $[T]$.
Substituting these into the equation: $[L] = [LT^{-2}]^m [T]^n$.
$[L^1 T^0] = [L^m T^{-2m+n}]$.
Comparing the powers of $L$ and $T$ on both sides:
For $L$: $m = 1$.
For $T$: $-2m + n = 0$.
Substituting $m = 1$ into the second equation: $-2(1) + n = 0$,which gives $n = 2$.
Therefore,$m = 1$ and $n = 2$.

(B) Let the dimensional formula of energy $(E)$ be expressed as $E = k F^a A^b T^c$,where $k$ is a dimensionless constant.
The dimensions of the quantities are:
$[E] = [M L^2 T^{-2}]$
$[F] = [M L T^{-2}]$
$[A] = [L T^{-2}]$
$[T] = [T]$
Substituting these into the equation:
$[M L^2 T^{-2}] = [M L T^{-2}]^a [L T^{-2}]^b [T]^c$
$[M L^2 T^{-2}] = [M^a L^{a+b} T^{-2a-2b+c}]$
Comparing the powers of $M$,$L$,and $T$ on both sides:
For $M$: $a = 1$
For $L$: $a + b = 2 \implies 1 + b = 2 \implies b = 1$
For $T$: $-2a - 2b + c = -2 \implies -2(1) - 2(1) + c = -2 \implies -4 + c = -2 \implies c = 2$
Thus,the dimensional formula of energy is $F^1 A^1 T^2$,which is $F A T^2$.

(D) The conversion formula between two systems of units is given by $N_1 U_1 = N_2 U_2$,where $N$ is the numerical value and $U$ is the unit.
For density,the dimensional formula is $[M L^{-3}]$.
Given: $N_1 = 0.625$,$M_1 = 1 \ g$,$L_1 = 1 \ cm$.
In $SI$ system: $M_2 = 1 \ kg = 10^3 \ g$,$L_2 = 1 \ m = 10^2 \ cm$.
Using the relation $N_2 = N_1 \left[ \frac{M_1}{M_2} \right] \left[ \frac{L_1}{L_2} \right]^{-3}$:
$N_2 = 0.625 \times \left[ \frac{1 \ g}{10^3 \ g} \right] \times \left[ \frac{1 \ cm}{10^2 \ cm} \right]^{-3}$
$N_2 = 0.625 \times 10^{-3} \times (10^{-2})^{-3}$
$N_2 = 0.625 \times 10^{-3} \times 10^6$
$N_2 = 0.625 \times 10^3 = 625 \ kg/m^3$.
Thus,the magnitude in $SI$ system is $625$.

(D) The dimensional formula for acceleration is $[M^0 L^1 T^{-2}]$.
Using the conversion formula $n_2 = n_1 [L_1/L_2]^1 [T_1/T_2]^{-2}$.
Given $n_1 = 10$,$L_1 = 1\,m$,$T_1 = 1\,s$,$L_2 = 1\,km = 10^3\,m$,and $T_2 = 1\,hr = 3600\,s$.
Substituting these values:
$n_2 = 10 \times [1\,m / 10^3\,m]^1 \times [1\,s / 3600\,s]^{-2}$
$n_2 = 10 \times [10^{-3}] \times [1/3600]^{-2}$
$n_2 = 10 \times 10^{-3} \times (3600)^2$
$n_2 = 10^{-2} \times 12960000$
$n_2 = 129600$.

(D) The given formula is $n = -D \frac{n_2 - n_1}{x_2 - x_1}$.
Here,$n$ represents the number of particles crossing a unit area in unit time,so its dimensions are $[n] = [L^{-2} T^{-1}]$.
$n_1$ and $n_2$ are the number of particles per unit volume,so their dimensions are $[n_1] = [n_2] = [L^{-3}]$.
$x_1$ and $x_2$ are positions,so $[x_2 - x_1] = [L]$.
Rearranging the formula for $D$,we get $D = n \frac{x_2 - x_1}{n_2 - n_1}$.
Substituting the dimensions: $[D] = \frac{[L^{-2} T^{-1}] \times [L]}{[L^{-3}]}$.
$[D] = \frac{[L^{-1} T^{-1}]}{[L^{-3}]} = [L^{-1+3} T^{-1}] = [L^2 T^{-1}]$.
Thus,the dimensions of $D$ are $[M^0 L^2 T^{-1}]$.

(C) The equation $S_t = u + \frac{1}{2}a(2t - 1)$ represents the distance traveled by an object in the $t^{th}$ second.
$1$. Numerically: This equation is a standard kinematic formula derived from the equations of motion $(S_t = S_t - S_{t-1})$,where $S_t = ut + \frac{1}{2}at^2$. Thus,it is numerically correct.
$2$. Dimensionally:
- The dimension of $S_t$ (distance in $t^{th}$ second) is $[L T^{-1}]$.
- The dimension of $u$ (initial velocity) is $[L T^{-1}]$.
- The dimension of $\frac{1}{2}a(2t - 1)$ is $[L T^{-2}] \times [T] = [L T^{-1}]$.
Since the dimensions of all terms are identical $([L T^{-1}])$,the equation is dimensionally correct.
Therefore,the equation is both numerically and dimensionally correct.

(D) Let the dimension of length be $L = G^x c^y h^z$.
The dimensions are:
$G = [M^{-1} L^3 T^{-2}]$
$c = [L T^{-1}]$
$h = [M L^2 T^{-1}]$
Substituting these into the equation:
$[M^0 L^1 T^0] = [M^{-1} L^3 T^{-2}]^x [L T^{-1}]^y [M L^2 T^{-1}]^z$
$[M^0 L^1 T^0] = M^{-x+z} L^{3x+y+2z} T^{-2x-y-z}$
Comparing the powers on both sides:
$1$) $-x + z = 0 \implies x = z$
$2$) $3x + y + 2z = 1$
$3$) $-2x - y - z = 0 \implies y = -2x - z$
Substitute $x=z$ into $(3)$: $y = -2x - x = -3x$.
Substitute $x=z$ and $y=-3x$ into $(2)$:
$3x + (-3x) + 2x = 1$
$2x = 1 \implies x = 1/2$.
Since $x=z$,$z = 1/2$.
Since $y = -3x$,$y = -3/2$.
Thus,$x = 1/2, y = -3/2, z = 1/2$. Both options $(b)$ and $(c)$ are correct.

(A) Let the radius of gyration $[k] = [L]$ be expressed as $[k] = {h^x}{c^y}{G^z}$.
The dimensions of the constants are:
$[h] = [M{L^2}{T^{ - 1}}]$
$[c] = [L{T^{ - 1}}]$
$[G] = [{M^{ - 1}}{L^3}{T^{ - 2}}]$
Substituting these into the equation:
$[L] = {[M{L^2}{T^{ - 1}}]^x} {[L{T^{ - 1}}]^y} [{M^{ - 1}}{L^3}{T^{ - 2}}]^z$
$[L] = {M^{x - z}} {L^{2x + y + 3z}} {T^{ - x - y - 2z}}$
Comparing the powers on both sides:
For $M$: $x - z = 0 \implies x = z$
For $T$: $-x - y - 2z = 0 \implies -x - y - 2x = 0 \implies y = -3x$
For $L$: $2x + y + 3z = 1$
Substituting $y = -3x$ and $z = x$ into the $L$ equation:
$2x - 3x + 3x = 1$
$2x = 1 \implies x = 1/2$
Thus,$x = 1/2$,$z = 1/2$,and $y = -3(1/2) = -3/2$.
Therefore,the dimension of the radius of gyration is ${h^{1/2}}{c^{ - 3/2}}{G^{1/2}}$.

(A) In the given equation,the exponent $-\frac{\alpha Z}{k\theta}$ must be dimensionless.
Therefore,the dimensions of $\frac{\alpha Z}{k\theta}$ are $[M^0 L^0 T^0]$.
Since $[k\theta]$ represents energy,its dimension is $[M L^2 T^{-2}]$.
Thus,$[\alpha] = \frac{[k\theta]}{[Z]} = \frac{[M L^2 T^{-2}]}{[L]} = [M L T^{-2}]$.
Given $P = \frac{\alpha}{\beta}$,we have $[\beta] = \frac{[\alpha]}{[P]}$.
The dimension of pressure $P$ is $[M L^{-1} T^{-2}]$.
Therefore,$[\beta] = \frac{[M L T^{-2}]}{[M L^{-1} T^{-2}]} = [M^0 L^2 T^0]$.

(C) The given formula is $\nu = \frac{p}{2l} \left[ \frac{F}{m} \right]^{1/2}$.
Squaring both sides,we get $\nu^2 = \frac{p^2}{4l^2} \left[ \frac{F}{m} \right]$.
Rearranging for $m$,we have $m = \frac{p^2 F}{4l^2 \nu^2}$.
Since $p$ is a dimensionless number,the dimensional formula for $m$ is $[m] = \frac{[F]}{[l^2][\nu^2]}$.
Substituting the dimensions: $[F] = [M L T^{-2}]$,$[l] = [L]$,and $[\nu] = [T^{-1}]$.
$[m] = \frac{[M L T^{-2}]}{[L^2][T^{-1}]^2} = \frac{[M L T^{-2}]}{[L^2][T^{-2}]} = [M L^{1-2} T^{-2+2}] = [M L^{-1} T^0]$.

(B) The formula for the orbital velocity of a planet is $v = \sqrt{\frac{GM}{R}}$.
This can be rewritten as $v = (GM/R)^{1/2}$.
Expanding this,we get $v = G^{1/2} M^{1/2} R^{-1/2}$.
Comparing this with the given expression $v = G^a M^b R^c$,we find that $a = 1/2$,$b = 1/2$,and $c = -1/2$.

(A) According to the principle of homogeneity of dimensions,only physical quantities with the same dimensions can be added or subtracted.
In the term $\left( P + \frac{a}{V^2} \right)$,the dimensions of $P$ must be equal to the dimensions of $\frac{a}{V^2}$.
Therefore,$[P] = \left[ \frac{a}{V^2} \right]$.
This implies $[a] = [P] \times [V^2]$.
We know that pressure $P$ is force per unit area,so $[P] = [MLT^{-2}] / [L^2] = [ML^{-1}T^{-2}]$.
Volume $V$ has dimensions $[L^3]$,so $[V^2] = [L^3]^2 = [L^6]$.
Substituting these values,we get $[a] = [ML^{-1}T^{-2}] \times [L^6] = [ML^5T^{-2}]$.
Thus,the dimensions of $a$ are $[M^1 L^5 T^{-2}]$.

(C) The dimensional formula for the electric field $\vec{E}$ is $[M^1 L^1 T^{-3} A^{-1}]$.
Given the equation $\vec{E} = \frac{K}{x^3}$,we can rearrange it to solve for $K$:
$K = E \cdot x^3$
Here,$x$ represents position,so its dimensional formula is $[L]$.
Substituting the dimensions into the equation:
$[K] = [M^1 L^1 T^{-3} A^{-1}] \cdot [L]^3$
$[K] = [M^1 L^1 \cdot L^3 T^{-3} A^{-1}]$
$[K] = [M^1 L^4 T^{-3} A^{-1}]$

(D) Given the dimensional formula for $x$ is $x = M^{-1} L^{3} T^{-2}$.
The formula for the maximum percentage error in $x$ is given by the propagation of errors rule:
$\frac{\Delta x}{x} \times 100 = |(-1)| \frac{\Delta M}{M} \times 100 + |3| \frac{\Delta L}{L} \times 100 + |(-2)| \frac{\Delta T}{T} \times 100$.
Substituting the given percentage errors for $M, L$ and $T$:
$\frac{\Delta x}{x} \times 100 = 1(2\%) + 3(3\%) + 2(4\%)$.
Calculating the values:
$= 2\% + 9\% + 8\% = 19\%$.
Therefore,the maximum percentage error in $x$ is $19\%$.

(A) According to the principle of homogeneity of dimensions,physical quantities can only be added or subtracted if they have the same dimensions.
However,physical quantities can be multiplied or divided regardless of whether their dimensions are the same or different.
Since $A$ and $B$ have different dimensions,the operations $A+B$ and $A-B$ are not physically meaningful.
The operation $A/B$ is physically meaningful as it results in a new physical quantity with dimensions $[A]/[B]$.

(D) Dimensional analysis is a method used to check the consistency of physical equations or to derive relationships between physical quantities based on their dimensions.
However,it has significant limitations:
$1$. It cannot determine dimensionless constants (such as $1/2$,$\pi$,or numerical coefficients).
$2$. It cannot derive formulas involving exponential,trigonometric,or logarithmic functions because the arguments of these functions must be dimensionless.
In the given options:
- $A$ involves an exponential function.
- $B$ involves a trigonometric function.
- $C$ involves dimensionless constants $(1/2)$.
Therefore,none of these relations can be derived using dimensional analysis.

(A) According to the principle of dimensional homogeneity,the dimensions of each term on both sides of the equation must be the same.
For the term $at$:
$[at] = [F] = [MLT^{-2}]$
$[a] = [MLT^{-2}] / [T] = [MLT^{-3}]$
For the term $bt^2$:
$[bt^2] = [F] = [MLT^{-2}]$
$[b] = [MLT^{-2}] / [T^2] = [MLT^{-4}]$
Therefore,the dimensions of $a$ and $b$ are $[MLT^{-3}]$ and $[MLT^{-4}]$ respectively.

(B) The given formula is $x = M^a L^b T^c$.
Dimensional analysis states that if a quantity depends on $M, L,$ and $T$,it can be expressed in terms of $M$ and $L$ only if the exponent of $T$ is zero (i.e.,$c = 0$).
If $c \neq 0$,the quantity $x$ essentially depends on the dimension of time $T$.
Therefore,if $c \neq 0$,the dimension of $x$ cannot be expressed solely in terms of $M$ and $L$ because the contribution of $T$ cannot be ignored or substituted by $M$ and $L$ unless there is a specific physical relationship between them.

(A) The formula for force in terms of base units is $F = [M L T^{-2}]$.
Let the new units be $M' = 10 \; g = 10^{-2} \; kg$,$L' = 10 \; cm = 0.1 \; m$,and $T' = 0.1 \; s$.
The new unit of force $F'$ is given by $F' = M' L' (T')^{-2}$.
Substituting the values: $F' = (10^{-2} \; kg) \times (0.1 \; m) \times (0.1 \; s)^{-2}$.
$F' = 10^{-2} \times 10^{-1} \times (10^{-1})^{-2} \; kg \cdot m/s^2$.
$F' = 10^{-3} \times 10^2 \; N$.
$F' = 10^{-1} \; N = 0.1 \; N$.

(A) In the equation $y = a \sin(bt - cx)$,the argument of the sine function $(bt - cx)$ must be dimensionless.
Therefore,the dimensions of $bt$ must be equal to the dimensions of $cx$.
$[bt] = [cx]$
$[b][T] = [c][L]$
Rearranging this,we get the ratio $b/c = [L]/[T]$.
Since $[L]/[T]$ represents the dimensions of velocity,the dimensions of $b/c$ are the same as those of wave velocity.