Dimensional Analysis, Uses and Limitations Questions in English

(C) The dimensional formula for force is $[F] = [M^1 L^1 T^{-2}]$.
Given units are:
$[F] = 100 \, N$
$[L] = 10 \, m$
$[T] = 100 \, s$
Substituting these into the dimensional formula:
$100 = M \times (10) \times (100)^{-2}$
$100 = M \times 10 \times \frac{1}{10000}$
$100 = M \times \frac{1}{1000}$
$M = 100 \times 1000 = 10^{5} \, kg$.

(B) The dimensional formula for energy is $[M L^2 T^{-2}]$.
Let the two systems be $1$ $(SI)$ and $2$ (new system).
In system $1$: $M_1 = 1 \ kg$,$L_1 = 1 \ m$,$T_1 = 1 \ s$,and $n_1 = 5$.
In system $2$: $M_2 = \alpha \ kg$,$L_2 = \beta \ m$,$T_2 = \gamma \ s$,and $n_2 = ?$.
Using the principle of homogeneity,$n_1 u_1 = n_2 u_2$,so $n_2 = n_1 (M_1/M_2)^1 (L_1/L_2)^2 (T_1/T_2)^{-2}$.
Substituting the values: $n_2 = 5 \times (1/\alpha)^1 \times (1/\beta)^2 \times (1/\gamma)^{-2}$.
$n_2 = 5 \times (1/\alpha) \times (1/\beta^2) \times (\gamma^2)$.
$n_2 = 5 \alpha^{-1} \beta^{-2} \gamma^2$.

(A) The volume of a liquid flowing out per second of a pipe is given by $V = \frac{\pi p r^4}{8 \eta l}$.
Dimensional formula of $LHS$:
$[V] = \frac{[Volume]}{[Time]} = \frac{[L^3]}{[T]} = [L^3 T^{-1}]$.
Dimensional formula of $RHS$:
$[p] = [M L^{-1} T^{-2}]$
$[r] = [L]$
$[\eta] = [M L^{-1} T^{-1}]$
$[l] = [L]$
Substituting these into the $RHS$ expression:
$[RHS] = \frac{[M L^{-1} T^{-2}] \cdot [L^4]}{[M L^{-1} T^{-1}] \cdot [L]} = \frac{[M L^3 T^{-2}]}{[M T^{-1}]} = [L^3 T^{-1}]$.
Since $[LHS] = [RHS]$,the equation is dimensionally correct.

(N/A) Given expression is $P = E l^2 m^{-5} G^{-2}$.
Dimensions of the physical quantities are:
$E$ (Energy) = $[M^1 L^2 T^{-2}]$
$l$ (Angular momentum) = $[M^1 L^2 T^{-1}]$
$m$ (Mass) = $[M^1]$
$G$ (Gravitational constant) = $[M^{-1} L^3 T^{-2}]$
Substituting these dimensions into the expression for $P$:
$[P] = [M^1 L^2 T^{-2}] \times [M^1 L^2 T^{-1}]^2 \times [M^1]^{-5} \times [M^{-1} L^3 T^{-2}]^{-2}$
Expanding the powers:
$[P] = [M^1 L^2 T^{-2}] \times [M^2 L^4 T^{-2}] \times [M^{-5}] \times [M^2 L^{-6} T^4]$
Combining the exponents for each base:
For $M$: $1 + 2 - 5 + 2 = 0$
For $L$: $2 + 4 + 0 - 6 = 0$
For $T$: $-2 - 2 + 0 + 4 = 0$
Thus,$[P] = [M^0 L^0 T^0]$.
Therefore,$P$ is a dimensionless quantity.

(A) We know that the dimensions are:
Dimensions of $(h) = [M^1 L^2 T^{-1}]$
Dimensions of $(c) = [L^1 T^{-1}]$
Dimensions of $(G) = [M^{-1} L^3 T^{-2}]$
To express mass $(M)$,length $(L)$,and time $(T)$ in terms of $c, h, G$:
Let $M = k c^a h^b G^d$. Substituting dimensions:
$[M^1 L^0 T^0] = [L T^{-1}]^a [M L^2 T^{-1}]^b [M^{-1} L^3 T^{-2}]^d$
Equating powers:
$M: b - d = 1$
$L: a + 2b + 3d = 0$
$T: -a - b - 2d = 0$
Solving these equations gives $b = 1/2, d = -1/2, a = 1/2$.
Thus,$M = k \sqrt{\frac{hc}{G}}$.
Similarly,for length $(L)$ and time $(T)$,we solve for exponents to get:
$L = k \sqrt{\frac{hG}{c^3}}$ and $T = k \sqrt{\frac{hG}{c^5}}$.

(A) According to Kepler's third law,$T^2 \propto r^3$,which implies $T \propto r^{3/2}$.
We assume $T$ is a function of $r$,$R$,and $g$ such that $T = k r^{3/2} R^a g^b$,where $k$ is a dimensionless constant.
Substituting the dimensions of each term: $[T] = [L]^{3/2} [L]^a [LT^{-2}]^b$.
$[M^0 L^0 T^1] = [L^{3/2 + a + b} T^{-2b}]$.
Comparing the powers of $T$: $1 = -2b \Rightarrow b = -1/2$.
Comparing the powers of $L$: $3/2 + a + b = 0 \Rightarrow 3/2 + a - 1/2 = 0 \Rightarrow a + 1 = 0 \Rightarrow a = -1$.
Substituting $a$ and $b$ back into the equation: $T = k r^{3/2} R^{-1} g^{-1/2}$.
Therefore,$T = \frac{k}{R} \sqrt{\frac{r^3}{g}}$.

(A) Let the length $l$ be expressed as $l = k c^x h^y G^z$,where $k$ is a dimensionless constant and $x, y, z$ are exponents.
The dimensions of the quantities are:
$[l] = [M^0 L^1 T^0]$
$[c] = [L T^{-1}]$
$[h] = [M L^2 T^{-1}]$
$[G] = [M^{-1} L^3 T^{-2}]$
Substituting these into the equation:
$[M^0 L^1 T^0] = [L T^{-1}]^x [M L^2 T^{-1}]^y [M^{-1} L^3 T^{-2}]^z$
$[M^0 L^1 T^0] = [M^{y-z} L^{x+2y+3z} T^{-x-y-2z}]$
Equating the powers of $M, L,$ and $T$ on both sides:
$1$) $y - z = 0 \implies y = z$
$2$) $x + 2y + 3z = 1$
$3$) $-x - y - 2z = 0 \implies x = -y - 2z$
Substituting $y = z$ into $(3)$: $x = -z - 2z = -3z$
Substituting $x = -3z$ and $y = z$ into $(2)$:
$-3z + 2z + 3z = 1 \implies 2z = 1 \implies z = 1/2$
Thus,$y = 1/2$ and $x = -3/2$.
Therefore,$l \propto c^{-3/2} h^{1/2} G^{1/2} = \sqrt{\frac{hG}{c^3}}$.

(B) Let the time $T$ be expressed as $T = k c^x h^y G^z$,where $k$ is a dimensionless constant.
The dimensional formulas are:
$[T] = [M^0 L^0 T^1]$
$[c] = [L T^{-1}]$
$[h] = [M L^2 T^{-1}]$
$[G] = [M^{-1} L^3 T^{-2}]$
Substituting these into the equation:
$[M^0 L^0 T^1] = [L T^{-1}]^x [M L^2 T^{-1}]^y [M^{-1} L^3 T^{-2}]^z$
$[M^0 L^0 T^1] = [M^{y-z} L^{x+2y+3z} T^{-x-y-2z}]$
Comparing the powers of $M$,$L$,and $T$ on both sides:
$1) y - z = 0 \implies y = z$
$2) x + 2y + 3z = 0$
$3) -x - y - 2z = 1$
Substitute $y = z$ into equations $(2)$ and $(3)$:
$x + 5z = 0 \implies x = -5z$
$-(-5z) - z - 2z = 1 \implies 5z - 3z = 1 \implies 2z = 1 \implies z = 1/2$
Thus,$y = 1/2$ and $x = -5/2$.
Therefore,the expression for time is $T = k \sqrt{\frac{h G}{c^5}}$.

(A) The Lorentz force on a charged particle is given by $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$.
For a particle moving in a magnetic field $\vec{B}$ with velocity $\vec{v}$ perpendicular to the field,the magnetic force provides the centripetal force:
$qvB = \frac{mv^2}{R}$
Rearranging this,we get the cyclotron frequency $\omega = \frac{v}{R} = \frac{qB}{m}$.
The dimensions of $\omega$ are:
$[\omega] = \frac{[q][B]}{[m]} = \frac{[I][T][M][I]^{-1}[T]^{-2}}{[M]} = [T]^{-1}$.
Thus,the quantity $\frac{qB}{m}$ has the dimension $T^{-1}$.
Additionally,the ratio of the electric field to the magnetic field $\frac{E}{B}$ has the dimension of velocity $[L][T]^{-1}$,and the quantity $\frac{eE}{mv}$ or $\frac{eB}{m}$ can be used to construct various dimensionless ratios depending on the physical context.

(N/A) We know that work done or energy is defined as force multiplied by displacement,so $1 \text{ Joule} = 1 \text{ Newton} \times 1 \text{ meter}$.
Substituting this into the expression:
$\frac{\text{Joule}}{\text{meter}^2} = \frac{\text{Newton} \times \text{meter}}{\text{meter}^2}$.
Canceling one meter from the numerator and denominator,we get:
$\frac{\text{Joule}}{\text{meter}^2} = \frac{\text{Newton}}{\text{meter}}$.

(B) Let the dimensional formula for energy be $[E] = [P]^x [A]^y [T]^z$.
Substituting the dimensions of each quantity:
$[M L^2 T^{-2}] = [M L T^{-1}]^x [L^2]^y [T]^z$.
Equating the powers of $M, L,$ and $T$ on both sides:
$M^1 L^2 T^{-2} = M^x L^{x+2y} T^{-x+z}$.
Comparing the exponents:
For $M$: $x = 1$.
For $L$: $x + 2y = 2$. Substituting $x=1$,we get $1 + 2y = 2$,which implies $2y = 1$,so $y = 1/2$.
For $T$: $-x + z = -2$. Substituting $x=1$,we get $-1 + z = -2$,which implies $z = -1$.
Thus,the dimensional formula for energy is $[P^1 A^{1/2} T^{-1}]$ or $[P A^{1/2} T^{-1}]$.

(A) Let the dimension of Young's modulus $Y$ be expressed as $Y = F^{x} A^{y} V^{z}$.
The dimensional formula for Young's modulus is $[M^{1} L^{-1} T^{-2}]$.
The dimensional formula for Force $F$ is $[M^{1} L^{1} T^{-2}]$.
The dimensional formula for Area $A$ is $[L^{2}]$.
The dimensional formula for Speed $V$ is $[L^{1} T^{-1}]$.
Substituting these into the equation:
$[M^{1} L^{-1} T^{-2}] = [M^{1} L^{1} T^{-2}]^{x} [L^{2}]^{y} [L^{1} T^{-1}]^{z}$
$[M^{1} L^{-1} T^{-2}] = [M]^{x} [L]^{x + 2y + z} [T]^{-2x - z}$
Comparing the powers of $M$,$L$,and $T$ on both sides:
For $M$: $x = 1$
For $T$: $-2x - z = -2 \Rightarrow -2(1) - z = -2 \Rightarrow z = 0$
For $L$: $x + 2y + z = -1 \Rightarrow 1 + 2y + 0 = -1 \Rightarrow 2y = -2 \Rightarrow y = -1$
Thus,the dimension of Young's modulus is $F^{1} A^{-1} V^{0}$.

(C) Given the expression: $x = \frac{I F v^{2}}{W L^{4}}$
Dimensions of the quantities are:
$I = [M L^{2}]$
$F = [M L T^{-2}]$
$v = [L T^{-1}]$
$W = [M L^{2} T^{-2}]$
$L = [L]$
Substituting these into the expression for $x$:
$[x] = \frac{[M L^{2}] [M L T^{-2}] [L T^{-1}]^{2}}{[M L^{2} T^{-2}] [L]^{4}}$
$[x] = \frac{[M^{2} L^{3} T^{-2}] [L^{2} T^{-2}]}{[M L^{6} T^{-2}]}$
$[x] = \frac{[M^{2} L^{5} T^{-4}]}{[M L^{6} T^{-2}]}$
$[x] = [M L^{-1} T^{-2}]$
Now,checking the dimensions of Energy density:
Energy density = $\frac{\text{Energy}}{\text{Volume}} = \frac{[M L^{2} T^{-2}]}{[L^{3}]} = [M L^{-1} T^{-2}]$
Since the dimensions of $x$ match the dimensions of Energy density,the correct option is $C$.

(B) The given relation is $z = \frac{a^2 b^{2/3}}{c^{1/2} d^3}$.
The relative error in $z$ is given by the formula: $\frac{\Delta z}{z} = 2 \frac{\Delta a}{a} + \frac{2}{3} \frac{\Delta b}{b} + \frac{1}{2} \frac{\Delta c}{c} + 3 \frac{\Delta d}{d}$.
Given percentage errors are: $\frac{\Delta a}{a} \times 100 = 2\%$,$\frac{\Delta b}{b} \times 100 = 1.5\%$,$\frac{\Delta c}{c} \times 100 = 4\%$,and $\frac{\Delta d}{d} \times 100 = 2.5\%$.
Substituting these values into the error formula:
$\frac{\Delta z}{z} \times 100 = 2(2\%) + \frac{2}{3}(1.5\%) + \frac{1}{2}(4\%) + 3(2.5\%)$.
$\frac{\Delta z}{z} \times 100 = 4\% + 1\% + 2\% + 7.5\% = 14.5\%$.
Thus,the percentage error in $z$ is $14.5\%$.

(A) Given the formula $X = 3 Y Z^{2}$.
We know the dimensions of capacitance $C$ in $MKSQ$ system are $[X] = [C] = [M^{-1} L^{-2} T^{2} Q^{2}]$.
We know the dimensions of magnetic induction $B$ in $MKSQ$ system are $[Z] = [B] = [M T^{-1} Q^{-1}]$.
Rearranging the formula for $Y$,we get $Y = \frac{X}{3 Z^{2}}$.
Ignoring the dimensionless constant $3$,the dimensional formula is $[Y] = \frac{[X]}{[Z^{2}]}$.
Substituting the dimensions: $[Y] = \frac{[M^{-1} L^{-2} T^{2} Q^{2}]}{[M T^{-1} Q^{-1}]^{2}}$.
$[Y] = \frac{[M^{-1} L^{-2} T^{2} Q^{2}]}{[M^{2} T^{-2} Q^{-2}]}$.
$[Y] = [M^{-1-2} L^{-2} T^{2-(-2)} Q^{2-(-2)}] = [M^{-3} L^{-2} T^{4} Q^{4}]$.

(B) The exponent of the exponential function must be dimensionless,so $\frac{x^{2}}{\alpha kT}$ is dimensionless.
$[\alpha] = \frac{[x^{2}]}{[kT]} = \frac{L^{2}}{M L^{2} T^{-2}} = M^{-1} T^{2}$.
Since the exponential term $e^{-\frac{x^{2}}{\alpha kT}}$ is dimensionless,the dimension of work $W$ is given by $[W] = [\alpha][\beta]^{2}$.
$[W] = M L^{2} T^{-2}$.
Substituting the dimensions: $M L^{2} T^{-2} = (M^{-1} T^{2}) [\beta]^{2}$.
$[\beta]^{2} = \frac{M L^{2} T^{-2}}{M^{-1} T^{2}} = M^{2} L^{2} T^{-4}$.
Taking the square root: $[\beta] = M L T^{-2}$.

(B) The term $kT$ represents energy,so its dimension is $[ML^{2}T^{-2}]$.
The exponent $\frac{-\beta x^{2}}{kT}$ must be dimensionless.
Therefore,$[\beta][x^{2}] = [kT] \implies [\beta][L^{2}] = [ML^{2}T^{-2}]$.
This gives $[\beta] = [MT^{-2}]$.
The dimension of work $W$ is $[ML^{2}T^{-2}]$.
Given $W = \alpha^{2} \beta e^{\frac{-\beta x^{2}}{kT}}$,and since the exponential term is dimensionless,we have $[W] = [\alpha^{2}][\beta]$.
$[ML^{2}T^{-2}] = [\alpha^{2}][MT^{-2}]$.
$[\alpha^{2}] = \frac{[ML^{2}T^{-2}]}{[MT^{-2}]} = [L^{2}]$.
Therefore,$[\alpha] = [L] = [M^{0}LT^{0}]$.

(D) The dimensions of the given quantities are:
$E = [ML^2 T^{-2}]$
$L = [ML^2 T^{-1}]$
$m = [M]$
$G = [M^{-1} L^3 T^{-2}]$
Given the formula $P = E L^2 m^{-5} G^{-2}$,we substitute the dimensions:
$[P] = [ML^2 T^{-2}] \cdot [ML^2 T^{-1}]^2 \cdot [M]^{-5} \cdot [M^{-1} L^3 T^{-2}]^{-2}$
$[P] = [ML^2 T^{-2}] \cdot [M^2 L^4 T^{-2}] \cdot [M^{-5}] \cdot [M^2 L^{-6} T^4]$
$[P] = [M^{1+2-5+2} L^{2+4-6} T^{-2-2+4}]$
$[P] = [M^0 L^0 T^0]$
Thus,the dimensions of $P$ are dimensionless.

(A) The dimensional formula for density is $[M L^{-3}]$.
Let the dimension of density be expressed as $[F^{a} L^{b} T^{c}]$.
Since force $F = [M L T^{-2}]$,we can write $[M L T^{-2}]^{a} [L]^{b} [T]^{c} = [M L^{-3}]$.
Equating the powers of $M, L,$ and $T$:
For $M$: $a = 1$.
For $L$: $a + b = -3$.
For $T$: $-2a + c = 0$.
Substituting $a = 1$ into the equations:
$1 + b = -3 \implies b = -4$.
$-2(1) + c = 0 \implies c = 2$.
Thus,the dimension of density is $[F^{1} L^{-4} T^{2}]$.

(A) To check dimensional correctness,we analyze the dimensions of both sides for each equation:
$A$. Poiseuille's Law: $v = \frac{\pi p a^4}{8 \eta L}$. Here,$v$ represents volume flow rate $(L^3 T^{-1})$. The $RHS$ dimensions are $[M L^{-1} T^{-2}][L^4] / ([M L^{-1} T^{-1}][L]) = [M L^3 T^{-2}] / [M T^{-1}] = [L^3 T^{-1}]$. This is dimensionally correct.
$B$. Capillary rise: $h = \frac{2 s \cos \theta}{\rho r g}$. Dimensions: $[L] = [M T^{-2}] / ([M L^{-3}][L][L T^{-2}]) = [M T^{-2}] / [M T^{-2}] = [L]$. This is dimensionally correct.
$C$. Displacement current density: $J = \varepsilon \frac{\partial E}{\partial t}$. Dimensions of $J$ are $[I L^{-2}]$. Dimensions of $\varepsilon \frac{\partial E}{\partial t}$ are $[M^{-1} L^{-3} T^4 I^2] \cdot ([M L T^{-3} I^{-1}] / [T]) = [I L^{-2}]$. This is dimensionally correct.
$D$. Work done: $W = \Gamma \theta$. Dimensions of $W$ are $[M L^2 T^{-2}]$. Dimensions of $\Gamma$ (torque) are $[M L^2 T^{-2}]$ and $\theta$ (angle) is dimensionless. Thus,$W = \Gamma$ is dimensionally correct,but the equation $W = \Gamma \theta$ implies work equals torque times angle,which is dimensionally correct as $[M L^2 T^{-2}] = [M L^2 T^{-2}] \cdot [1]$. However,in the context of the provided options,all are dimensionally correct. Re-evaluating $A$: $v$ is volume,not volume flow rate. Thus,$A$ is dimensionally incorrect.

(A) Let the dimension of mass be represented as $[M] = [F]^a [T]^b [V]^c$.
Substituting the dimensions of the base quantities:
$[M^1 L^0 T^0] = [M^1 L^1 T^{-2}]^a [T^1]^b [L^1 T^{-1}]^c$
$[M^1 L^0 T^0] = [M^a L^{a+c} T^{-2a+b-c}]$
Comparing the powers of $M, L,$ and $T$ on both sides:
For $M$: $a = 1$
For $L$: $a + c = 0 \implies 1 + c = 0 \implies c = -1$
For $T$: $-2a + b - c = 0 \implies -2(1) + b - (-1) = 0 \implies -2 + b + 1 = 0 \implies b = 1$
Therefore,the dimensions of mass are $[F^1 T^1 V^{-1}]$,which is written as $[F T V^{-1}]$.

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

(D) The argument of the logarithmic function must be dimensionless,so $\frac{\mu k R}{J \beta^{2}}$ must be dimensionless.
Given $S = \frac{dQ}{T}$,the dimensions of entropy $S$ are $[M L^{2} T^{-2} K^{-1}]$.
The gas constant $R$ has dimensions $[M L^{2} T^{-2} K^{-1} mol^{-1}]$ and Boltzmann constant $k$ has dimensions $[M L^{2} T^{-2} K^{-1}]$.
Since $\frac{\mu k R}{J \beta^{2}}$ is dimensionless,$[\beta^{2}] = \frac{[\mu][k][R]}{[J]}$.
Since $J$ (mechanical equivalent of heat) is a conversion factor between energy units,it is dimensionless. Thus,$[\beta^{2}] = [mol] \cdot [M L^{2} T^{-2} K^{-1}] \cdot [M L^{2} T^{-2} K^{-1} mol^{-1}] = [M^{2} L^{4} T^{-4} K^{-2}]$.
Therefore,$[\beta] = [M L^{2} T^{-2} K^{-1}]$,which is the same as the dimensions of $S, k,$ and $\mu R$.
From $S = \alpha^{2} \beta$,since $S$ and $\beta$ have the same dimensions,$\alpha^{2}$ must be dimensionless,meaning $\alpha$ is dimensionless.
Checking the options:
$(A)$ $S, \beta, k, \mu R$ have same dimensions: Correct.
$(B)$ $\alpha$ and $J$ have same dimensions: Incorrect,as $\alpha$ is dimensionless and $J$ is dimensionless,but the statement implies they share a non-trivial dimension or are being compared incorrectly. Actually,$\alpha$ is dimensionless and $J$ is dimensionless,so they have the same dimensions. Wait,let's re-evaluate: $\alpha$ is dimensionless,$J$ is dimensionless. So $(B)$ is correct.
$(C)$ $S$ and $\alpha$ have different dimensions: Correct,as $S$ has dimensions and $\alpha$ is dimensionless.
$(D)$ $\alpha$ and $k$ have the same dimensions: Incorrect,as $\alpha$ is dimensionless and $k$ has dimensions $[M L^{2} T^{-2} K^{-1}]$. Thus,$(D)$ is the incorrect option.

(A) Let the dimension of mass be $m = k \cdot t^a \cdot u^b \cdot I^c$,where $k$ is a dimensionless constant.
Substituting the dimensions of each quantity:
$[M^1 L^0 T^0] = [T]^a [L T^{-1}]^b [M L^2 T^{-1}]^c$
$[M^1 L^0 T^0] = [M^c] [L^{b+2c}] [T^{a-b-c}]$
Comparing the powers of $M$,$L$,and $T$ on both sides:
For $M$: $c = 1$
For $L$: $b + 2c = 0 \implies b + 2(1) = 0 \implies b = -2$
For $T$: $a - b - c = 0 \implies a - (-2) - 1 = 0 \implies a + 1 = 0 \implies a = -1$
Therefore,the dimension of mass is $[t^{-1} u^{-2} I^1]$.

(A) The argument of trigonometric functions like $\cos$ and $\sin$ must be dimensionless.
For $\cos(Bx)$,the term $Bx$ must be dimensionless,so $[B] = [x]^{-1} = [L^{-1}]$.
For $\sin(Dt)$,the term $Dt$ must be dimensionless,so $[D] = [t]^{-1} = [T^{-1}]$.
Since $F = A \cos(Bx) + C \sin(Dt)$,the dimensions of $A$ must be equal to the dimensions of force $F$,so $[A] = [MLT^{-2}]$.
Now,we calculate the dimensions of $\frac{AD}{B}$:
$[\frac{AD}{B}] = \frac{[A][D]}{[B]} = \frac{[MLT^{-2}][T^{-1}]}{[L^{-1}]}$.
$[\frac{AD}{B}] = [MLT^{-3}][L] = [ML^{2}T^{-3}]$.

(C) The argument of the logarithmic function must be dimensionless. Therefore,$\frac{kt}{\beta x} = 1$,which implies $\beta = \frac{kt}{x}$.
Since $kt$ has the dimensions of energy $([ML^{2}T^{-2}])$ and $x$ is distance $([L])$,the dimensions of $\beta$ are $[\beta] = \frac{[ML^{2}T^{-2}]}{[L]} = [MLT^{-2}]$.
Given that $P$ is a dimensionless quantity,the expression $P = \frac{\alpha}{\beta} \times (\text{dimensionless term})$ implies that $[P] = \frac{[\alpha]}{[\beta]}$.
Since $[P] = [M^{0}L^{0}T^{0}]$,we have $[\alpha] = [\beta] = [MLT^{-2}]$.

(B) Given: $v_{2} = \frac{n}{m^{2}} v_{1}$ and $a_{2} = \frac{a_{1}}{mn}$.
Since $v = \frac{L}{T}$ and $a = \frac{L}{T^{2}}$,we have:
$\frac{L_{2}}{T_{2}} = \frac{n}{m^{2}} \frac{L_{1}}{T_{1}}$ --- $(1)$
$\frac{L_{2}}{T_{2}^{2}} = \frac{1}{mn} \frac{L_{1}}{T_{1}^{2}}$ --- $(2)$
Dividing $(1)$ by $(2)$:
$\frac{T_{1}^{2}}{T_{2}} = \frac{n}{m^{2}} \times mn \times T_{1} = \frac{n^{2}}{m} T_{1}$
$\frac{T_{2}}{T_{1}} = \frac{m}{n^{2}}$ (This implies $T_{1} = \frac{n^{2}}{m} T_{2}$).
Substituting $T_{2} = \frac{m}{n^{2}} T_{1}$ into $(1)$:
$\frac{L_{2}}{(\frac{m}{n^{2}}) T_{1}} = \frac{n}{m^{2}} \frac{L_{1}}{T_{1}}$
$L_{2} = \frac{n}{m^{2}} \times \frac{m}{n^{2}} L_{1} = \frac{1}{mn} L_{1}$
$L_{1} = mn L_{2}$.
Checking the options,the correct relation for time is $T_{1} = \frac{n^{2}}{m} T_{2}$.

(C) According to the principle of dimensional homogeneity,only quantities with the same dimensions can be added or subtracted.
$1$. From the term $\left[ P + \frac{a}{V^2} \right]$,the dimensions of $P$ must be equal to the dimensions of $\frac{a}{V^2}$.
$[P] = \left[ \frac{a}{V^2} \right] \implies [a] = [P][V^2]$.
$2$. From the term $[V - b]$,the dimensions of $b$ must be equal to the dimensions of $V$.
$[b] = [V]$.
$3$. Now,find the dimensions of the ratio $\frac{a}{b}$:
$\left[ \frac{a}{b} \right] = \frac{[P][V^2]}{[V]} = [P][V]$.
Therefore,the ratio $\frac{a}{b}$ is dimensionally equal to $PV$.

(A) The dimensional formula for the coefficient of viscosity $\eta$ is $[M^1 L^{-1} T^{-1}]$.
Let the dimensional formula be expressed as $\eta = P^x A^y T^z$.
We know the dimensions of the fundamental quantities are:
$[P] = [M^1 L^1 T^{-1}]$
$[A] = [L^2]$
$[T] = [T^1]$
Substituting these into the equation:
$[M^1 L^{-1} T^{-1}] = [M^1 L^1 T^{-1}]^x [L^2]^y [T^1]^z$
$[M^1 L^{-1} T^{-1}] = M^x L^{x+2y} T^{-x+z}$
Comparing the powers of $M, L,$ and $T$ on both sides:
For $M: x = 1$
For $L: x + 2y = -1 \implies 1 + 2y = -1 \implies 2y = -2 \implies y = -1$
For $T: -x + z = -1 \implies -1 + z = -1 \implies z = 0$
Thus,the dimensional formula is $[P^1 A^{-1} T^0]$.

(D) The argument of the $\sin$ function must be dimensionless,so $[\frac{\alpha x}{kt}] = [M^{0}L^{0}T^{0}]$.
Given $x = [L]$,$k = [ML^{2}T^{-2}\theta^{-1}]$,and $t = [\theta]$.
Thus,$[\alpha] = [\frac{kt}{x}] = \frac{[ML^{2}T^{-2}\theta^{-1}][\theta]}{[L]} = [MLT^{-2}]$.
Energy density $u$ is energy per unit volume,so $[u] = [ML^{-1}T^{-2}]$.
From the expression $u = \frac{\alpha}{\beta}$,we have $[\beta] = \frac{[\alpha]}{[u]}$.
$[\beta] = \frac{[MLT^{-2}]}{[ML^{-1}T^{-2}]} = [L^{2}]$.
Therefore,the dimensions of $\beta$ are $[M^{0}L^{2}T^{0}]$.

(D) Efficiency $\eta$ is a dimensionless quantity. Therefore,$[\eta] = [M^0 L^0 T^0]$.
In the expression $\eta = \frac{\alpha \beta}{\sin \theta} \log_{e} \frac{\beta x}{k T}$,the argument of the logarithm must be dimensionless. Thus,$[\frac{\beta x}{k T}] = [M^0 L^0 T^0]$.
Since $[k T] = \text{Energy} = [M L^2 T^{-2}]$ and $[x] = [L]$,we have $[\beta] = \frac{[M L^2 T^{-2}]}{[L]} = [M L T^{-2}]$,which is the dimension of force.
Now,since $\eta$ is dimensionless and $\sin \theta$ is dimensionless,the term $\frac{\alpha \beta}{\sin \theta}$ must be dimensionless. Thus,$[\alpha \beta] = [M^0 L^0 T^0]$.
Since $[\beta] = [M L T^{-2}]$,then $[\alpha] = [M^{-1} L^{-1} T^2]$.
Checking the options:
$A$: $[\beta] = [M L T^{-2}]$ (Force),which is correct.
$B$: $[\alpha^{-1} x] = [M L T^{-2} \cdot L] = [M L^2 T^{-2}]$ (Energy),which is correct.
$C$: $[\eta^{-1} \sin \theta] = [1] = [\alpha \beta]$,which is correct.
$D$: $[\alpha] = [M^{-1} L^{-1} T^2]$ and $[\beta] = [M L T^{-2}]$. These are not the same. Thus,this is the incorrect option.

(C) The dimension of time period $T$ is $[T^1]$.
The given formula is $T = k \sqrt{\frac{\rho r^3}{S}}$.
Dimensions of density $\rho = [M L^{-3}]$.
Dimensions of radius $r = [L]$.
Dimensions of surface tension $S = [M T^{-2}]$.
Substituting these into the formula: $[RHS] = [M L^{-3} L^3]^{1/2} / [M T^{-2}]^{1/2} = [M^{1/2}] / [M^{1/2} T^{-1}] = [T^1]$.
Wait,let us re-evaluate the expression provided in the assertion: $T = k \sqrt{\rho r^3 / S}$.
$[RHS] = \sqrt{\frac{[M L^{-3}] [L^3]}{[M T^{-2}]}} = \sqrt{\frac{[M]}{[M T^{-2}]}} = \sqrt{[T^2]} = [T^1]$.
Since the dimensions of $LHS$ and $RHS$ match,the assertion $(A)$ is actually true. However,based on the provided options and the logic that the assertion claims the formula is dimensionally correct,and the reason claims it is not,we conclude $(A)$ is true and $(R)$ is false.

(A) According to the principle of homogeneity of dimensions,an equation is dimensionally correct only if the dimensions of each term on both sides are the same.
For option $A$: $u^2 = 2a(gt - 1)$. The term $gt$ has dimensions $[LT^{-2}][T] = [LT^{-1}]$,which is the dimension of velocity. However,the number $1$ is dimensionless. We cannot subtract a dimensionless quantity from a quantity with dimensions $[LT^{-1}]$. Thus,this equation is dimensionally incorrect.
For option $B$: $s - ut = \frac{1}{2}at^2$. Dimensions of $s$ are $[L]$,$ut$ are $[LT^{-1}][T] = [L]$,and $\frac{1}{2}at^2$ are $[LT^{-2}][T^2] = [L]$. All terms have the same dimensions.
For option $C$: $u = v - at$. Dimensions of $u$ are $[LT^{-1}]$,$v$ are $[LT^{-1}]$,and $at$ are $[LT^{-2}][T] = [LT^{-1}]$. All terms have the same dimensions.
For option $D$: $v^2 - u^2 = 2as$. Dimensions of $v^2$ are $[L^2T^{-2}]$,$u^2$ are $[L^2T^{-2}]$,and $2as$ are $[LT^{-2}][L] = [L^2T^{-2}]$. All terms have the same dimensions.
Therefore,option $A$ is dimensionally incorrect.

(A) The dimensional formula for energy is $[M^1 L^2 T^{-2}]$.
Let the $SI$ system be system $1$ and the new system be system $2$.
In $SI$ system: $M_1 = 1 \,kg$,$L_1 = 1 \,m$,$T_1 = 1 \,s$.
In the new system: $M_2 = 10 \,kg$,$L_2 = 10 \,m$,$T_2 = 1 \,minute = 60 \,s$.
The conversion formula is $n_2 = n_1 [M_1/M_2]^a [L_1/L_2]^b [T_1/T_2]^c$.
Here $n_1 = 1$,$a = 1$,$b = 2$,$c = -2$.
Substituting the values: $n_2 = 1 \times [1 \,kg / 10 \,kg]^1 \times [1 \,m / 10 \,m]^2 \times [1 \,s / 60 \,s]^{-2}$.
$n_2 = [1/10] \times [1/10]^2 \times [1/60]^{-2}$.
$n_2 = [1/10] \times [1/100] \times [60]^2$.
$n_2 = [1/1000] \times 3600 = 3600 / 1000 = 3.6$.
$3.6$ can be written as $36 \times 10^{-1}$.

(A) The dimensional formula for force is $[F] = [M L T^{-2}]$.
Given units are: Force $F' = 1 \,kN = 10^3 \,N$,Length $L' = 1 \,km = 10^3 \,m$,and Time $T' = 100 \,s$.
We know that $F = M L T^{-2}$,so $M = F L^{-1} T^2$.
Substituting the given units into the formula for mass:
$M' = F' \cdot (L')^{-1} \cdot (T')^2$
$M' = (10^3 \,N) \cdot (10^3 \,m)^{-1} \cdot (100 \,s)^2$
$M' = 10^3 \cdot 10^{-3} \cdot (10^2)^2 \,kg$
$M' = 1 \cdot 10^4 \,kg = 10000 \,kg$.
Therefore,the unit of mass is $10000 \,kg$.

(A) The potential energy $U$ is given by $U = \frac{A \sqrt{x}}{x^2 + B}$.
According to the principle of homogeneity of dimensions,the dimensions of terms added or subtracted must be the same. Therefore,the dimensions of $B$ must be equal to the dimensions of $x^2$.
Since $x$ is distance,$[x] = [L]$,so $[B] = [L^2]$.
The dimensions of potential energy $U$ are $[ML^2 T^{-2}]$.
From the equation,$A = \frac{U(x^2 + B)}{\sqrt{x}}$.
Substituting the dimensions:
$[A] = \frac{[ML^2 T^{-2}] [L^2]}{[L^{1/2}]} = [ML^2 T^{-2} L^2 L^{-1/2}] = [ML^{7/2} T^{-2}]$.
Now,we need the dimensions of $AB$:
$[AB] = [A][B] = [ML^{7/2} T^{-2}] [L^2] = [ML^{11/2} T^{-2}]$.
Thus,the correct option is $A$.

(D) Dimensional analysis is a powerful tool for checking the consistency of equations,but it has significant limitations.
$1$. It cannot derive equations involving trigonometric,exponential,or logarithmic functions because these functions are dimensionless.
$2$. It cannot determine dimensionless constants (like $1/2$ or $\pi$) in a formula.
$3$. It cannot derive equations containing more than three physical quantities if they are independent.
In the given options:
- $v=u+at$ is a linear relation that can be checked for dimensional consistency.
- $k=\frac{1}{2}mv^2$ is a relation where the constant $1/2$ cannot be determined,but the dimensions can be checked.
- $y=A\sin(\omega t+kx)$ involves a trigonometric function. Dimensional analysis cannot be used to derive or verify the structure of such transcendental equations because the argument of the sine function must be dimensionless,and the function itself does not have a simple power-law form.
Therefore,the relation $y=A\sin(\omega t+kx)$ cannot be deduced using dimensional analysis.

(D) The dimensional formula for time period $T$ is $[M^0 L^0 T^1]$.
The dimensional formula for pressure $p$ is $[M^1 L^{-1} T^{-2}]$.
The dimensional formula for density $D$ is $[M^1 L^{-3} T^0]$.
The dimensional formula for surface tension $S$ is $[M^1 L^0 T^{-2}]$.
Given the relation $T = p^a D^b S^c$,we equate the dimensions on both sides:
$[M^0 L^0 T^1] = [M^1 L^{-1} T^{-2}]^a [M^1 L^{-3} T^0]^b [M^1 L^0 T^{-2}]^c$
$[M^0 L^0 T^1] = [M^{a+b+c} L^{-a-3b} T^{-2a-2c}]$
Comparing the powers of $M, L,$ and $T$ on both sides:
$1$) $a + b + c = 0$
$2$) $-a - 3b = 0 \implies a = -3b$
$3$) $-2a - 2c = 1$
Substitute $a = -3b$ into equation $(1)$:
$-3b + b + c = 0 \implies c = 2b$
Substitute $a = -3b$ and $c = 2b$ into equation $(3)$:
$-2(-3b) - 2(2b) = 1$
$6b - 4b = 1 \implies 2b = 1 \implies b = \frac{1}{2}$
Now,find $a$ and $c$:
$a = -3b = -3(\frac{1}{2}) = -\frac{3}{2}$
$c = 2b = 2(\frac{1}{2}) = 1$
Thus,the values are $a = -\frac{3}{2}, b = \frac{1}{2}, c = 1$.

(A) Given,drag force $F_d$ depends on air density $\sigma$,velocity of ball $v$,and cross-sectional area $A$.
$F_d \propto \sigma^a v^b A^c$
Using dimensional analysis: $[MLT^{-2}] = [ML^{-3}]^a [LT^{-1}]^b [L^2]^c$.
Equating exponents: $a=1$,$-3a+b+2c=1$,and $-b=-2 \Rightarrow b=2$.
Substituting $a=1, b=2$ into the second equation: $-3(1)+2+2c=1 \Rightarrow 2c=2 \Rightarrow c=1$.
Thus,$F_d = k \sigma v^2 A$.
At terminal velocity $v_T$,$mg = F_d = k \sigma v_T^2 A$.
Since $A = \pi R^2$ and $m = \frac{4}{3} \pi R^3 \rho_{ball}$,we have $R^2 \propto m^{2/3}$.
Substituting $A \propto m^{2/3}$ into the force equation: $mg \propto v_T^2 m^{2/3} \Rightarrow v_T^2 \propto m^{1/3} \Rightarrow v_T \propto m^{1/6}$.
Therefore,the ratio of terminal velocities is $\frac{v_1}{v_2} = \left(\frac{m_1}{m_2}\right)^{1/6} = \left(\frac{250}{125}\right)^{1/6} = 2^{1/6}$.

(D) Given the formula for power: $P = G c^{-5} m^{x} R^{y} \omega^{z} \quad \dots (i)$
The dimensions of the physical quantities are:
Power $P = [M L^{2} T^{-3}]$
Gravitational constant $G = [M^{-1} L^{3} T^{-2}]$
Speed of light $c = [L T^{-1}]$
Mass $m = [M]$
Radius $R = [L]$
Angular velocity $\omega = [T^{-1}]$
Substituting these dimensions into equation $(i)$:
$[M L^{2} T^{-3}] = [M^{-1} L^{3} T^{-2}] [L T^{-1}]^{-5} [M]^{x} [L]^{y} [T^{-1}]^{z}$
$[M L^{2} T^{-3}] = [M^{-1} L^{3} T^{-2}] [L^{-5} T^{5}] [M^{x}] [L^{y}] [T^{-z}]$
$[M L^{2} T^{-3}] = [M^{x-1}] [L^{3-5+y}] [T^{-2+5-z}]$
$[M L^{2} T^{-3}] = [M^{x-1}] [L^{y-2}] [T^{3-z}]$
Comparing the powers of $M, L,$ and $T$ on both sides:
For $M$: $x - 1 = 1 \Rightarrow x = 2$
For $L$: $y - 2 = 2 \Rightarrow y = 4$
For $T$: $3 - z = -3 \Rightarrow z = 6$
Thus,the values are $x=2, y=4, z=6$.

(B) Dimensional analysis is the most efficient way to solve this problem. The radius $R$ depends on energy $E$,density $\rho$,and time $t$.
Let $R = k E^a \rho^b t^c$,where $k$ is a dimensionless constant.
The dimensions are: $[R] = L$,$[E] = ML^2T^{-2}$,$[\rho] = ML^{-3}$,$[t] = T$.
Substituting these into the equation:
$L = (ML^2T^{-2})^a (ML^{-3})^b (T)^c$
$L^1 = M^{a+b} L^{2a-3b} T^{-2a+c}$
Equating the powers of $M, L,$ and $T$:
$a + b = 0 \Rightarrow b = -a$
$2a - 3b = 1 \Rightarrow 2a - 3(-a) = 1 \Rightarrow 5a = 1 \Rightarrow a = 1/5$
Thus,$b = -1/5$.
$-2a + c = 0 \Rightarrow c = 2a = 2(1/5) = 2/5$.
Therefore,$R \propto E^{1/5} \rho^{-1/5} t^{2/5}$,which implies $R \propto t^{2/5}$.

(C) Let the resistivity $\rho$ be expressed as $\rho = k \cdot h^a \cdot m_e^b \cdot c^c \cdot e^d \cdot \varepsilon_0^f$,where $k$ is a dimensionless constant.
The dimensions of the physical quantities are:
$\rho = [M L^3 T^{-3} A^{-2}]$
$h = [M L^2 T^{-1}]$
$m_e = [M]$
$c = [L T^{-1}]$
$e = [A T]$
$\varepsilon_0 = [M^{-1} L^{-3} T^4 A^2]$
Substituting these into the equation:
$[M L^3 T^{-3} A^{-2}] = [M L^2 T^{-1}]^a [M]^b [L T^{-1}]^c [A T]^d [M^{-1} L^{-3} T^4 A^2]^f$
Equating the powers of $M, L, T, A$ on both sides:
$M: a + b - f = 1$
$L: 2a + c - 3f = 3$
$T: -a - c + d + 4f = -3$
$A: d + 2f = -2$
Solving this system of equations,we find $a=2, b=-1, c=-1, d=-2, f=0$.
Thus,$\rho = k \frac{h^2}{m_e c e^2}$.

(A) Given,power radiated $P$ is $P = \mu_{0}^{x} m^{y} \omega^{z} c^{u}$.
Substituting the dimensions of the physical quantities:
$[P] = [ML^{2}T^{-3}]$
$[\mu_{0}] = [MLT^{-2}A^{-2}]$
$[m] = [L^{2}A]$
$[\omega] = [T^{-1}]$
$[c] = [LT^{-1}]$
Equating dimensions: $[ML^{2}T^{-3}] = [MLT^{-2}A^{-2}]^{x} [L^{2}A]^{y} [T^{-1}]^{z} [LT^{-1}]^{u}$.
Comparing powers of $M, L, T, A$:
$M: x = 1$
$A: -2x + y = 0 \implies y = 2x = 2$
$L: x + 2y + u = 2 \implies 1 + 2(2) + u = 2 \implies u = -3$
$T: -2x - z - u = -3 \implies -2(1) - z - (-3) = -3 \implies -2 - z + 3 = -3 \implies z = 4$.
Thus,$x=1, y=2, z=4, u=-3$.

(B) Given,$A = G^\alpha M^\beta c^\gamma$.
The dimensions of the quantities are:
$[A] = [L]^2$
$[G] = [M^{-1} L^3 T^{-2}]$
$[M] = [M]$
$[c] = [L T^{-1}]$
Substituting these into the equation:
$[L]^2 = [M^{-1} L^3 T^{-2}]^\alpha [M]^\beta [L T^{-1}]^\gamma$
$[L]^2 = M^{-\alpha + \beta} L^{3\alpha + \gamma} T^{-2\alpha - \gamma}$
Comparing the powers of $M, L,$ and $T$ on both sides:
For $M: -\alpha + \beta = 0 \implies \beta = \alpha \quad (i)$
For $L: 3\alpha + \gamma = 2 \quad (ii)$
For $T: -2\alpha - \gamma = 0 \implies \gamma = -2\alpha \quad (iii)$
Substituting $(iii)$ into $(ii)$:
$3\alpha - 2\alpha = 2 \implies \alpha = 2$
Since $\beta = \alpha$,$\beta = 2$.
Since $\gamma = -2\alpha$,$\gamma = -2(2) = -4$.
Thus,$\alpha = 2, \beta = 2$ and $\gamma = -4$.

(C) The Stefan-Boltzmann constant $\sigma$ has dimensions $[M T^{-3} K^{-4}]$.
The dimensions of the constants are:
$h = [M L^2 T^{-1}]$
$k_B = [M L^2 T^{-2} K^{-1}]$
$c = [L T^{-1}]$
Given $\sigma = h^\alpha k_B^\beta c^\gamma$,we equate the dimensions:
$[M T^{-3} K^{-4}] = [M L^2 T^{-1}]^\alpha [M L^2 T^{-2} K^{-1}]^\beta [L T^{-1}]^\gamma$
$[M T^{-3} K^{-4}] = [M^{\alpha+\beta} L^{2\alpha+2\beta+\gamma} T^{-\alpha-2\beta-\gamma} K^{-\beta}]$
Comparing the exponents of $M, L, T,$ and $K$:
$1$) $M: \alpha + \beta = 0$ (Wait,$\sigma$ is energy per unit area per unit time per unit temperature to the fourth power,so $[M T^{-3} K^{-4}]$ is correct. The mass dimension is $1$,so $\alpha + \beta = 1$)
$2$) $K: -\beta = -4 \implies \beta = 4$
$3$) $M: \alpha + \beta = 1 \implies \alpha + 4 = 1 \implies \alpha = -3$
$4$) $T: -\alpha - 2\beta - \gamma = -3 \implies -(-3) - 2(4) - \gamma = -3 \implies 3 - 8 - \gamma = -3 \implies -5 - \gamma = -3 \implies \gamma = -2$
Thus,$\alpha = -3, \beta = 4, \gamma = -2$.

(C) The $EOTVOS$ number $E_0$ is dimensionless. We check the dimensions of the given options to identify the correct expression.
For option $(c)$,the expression is $\frac{\rho g L^2}{\sigma}$.
The dimensions are:
$[\rho] = [M L^{-3}]$
$[g] = [L T^{-2}]$
$[L^2] = [L^2]$
$[\sigma] = [M T^{-2}]$
Substituting these into the expression:
$\left[ \frac{\rho g L^2}{\sigma} \right] = \frac{[M L^{-3}] \cdot [L T^{-2}] \cdot [L^2]}{[M T^{-2}]}$
$= \frac{[M L^0 T^{-2}]}{[M T^{-2}]}$
$= [M^0 L^0 T^0]$
Since the dimensions cancel out,the expression is dimensionless.

(A) The frequency of vibration $v$ of a liquid drop is governed by surface tension $\sigma$,density $\rho$,and radius $R$. The given ratio is $\frac{v}{\sqrt{\sigma / \rho R^3}}$.
Dimensions of frequency $v = [T^{-1}]$.
Dimensions of surface tension $\sigma = [MT^{-2}]$.
Dimensions of density $\rho = [ML^{-3}]$.
Dimensions of radius $R = [L]$.
Let the ratio be $X = \frac{v}{\sqrt{\sigma / \rho R^3}}$.
Dimensions of $\sqrt{\frac{\sigma}{\rho R^3}} = \sqrt{\frac{MT^{-2}}{ML^{-3} \cdot L^3}} = \sqrt{\frac{MT^{-2}}{M}} = [T^{-1}]$.
Thus,the ratio $\frac{v}{\sqrt{\sigma / \rho R^3}}$ is dimensionless.
We check the dimensions of option $(A)$:
$\frac{k \rho g R^2}{\sigma} = \frac{[ML^{-3}] \cdot [LT^{-2}] \cdot [L^2]}{[MT^{-2}]} = \frac{[ML^0T^{-2}]}{[MT^{-2}]} = [M^0L^0T^0]$.
Since this expression is dimensionless,it is consistent with the dimensional analysis of the given ratio.

(D) The potential energy of the particle is given by $V(x) = -\alpha x + \beta \sin(x / \gamma)$.
For the equation to be dimensionally consistent,the dimensions of each term must be the same as the dimensions of potential energy $[V] = [ML^2T^{-2}]$.
$1$. For the term $\alpha x$: $[\alpha][x] = [V] \implies [\alpha][L] = [ML^2T^{-2}] \implies [\alpha] = [MLT^{-2}]$.
$2$. For the term $\beta \sin(x / \gamma)$: The argument of the sine function must be dimensionless,so $[x / \gamma] = [M^0L^0T^0] \implies [L] / [\gamma] = 1 \implies [\gamma] = [L]$.
$3$. Also,$[\beta] = [V] = [ML^2T^{-2}]$.
Now,check the dimensions of the given options:
For option $(d)$: $\frac{[\alpha][\gamma]}{[\beta]} = \frac{[MLT^{-2}][L]}{[ML^2T^{-2}]} = \frac{[ML^2T^{-2}]}{[ML^2T^{-2}]} = [M^0L^0T^0]$.
Since this combination is dimensionless,the correct option is $(d)$.

(B) The dimension of resistance $R$ is given by $[R] = [ML^2T^{-3}A^{-2}]$.
Planck's constant $h$ has dimensions $[h] = [ML^2T^{-1}]$.
Electric charge $e$ has dimensions $[e] = [AT]$.
Let the dimension of $R_H$ be proportional to $h^a e^b$.
$[ML^2T^{-3}A^{-2}] = [ML^2T^{-1}]^a [AT]^b = [M^a L^{2a} T^{-a+b} A^b]$.
Comparing the powers of $M, L, T,$ and $A$ on both sides:
For $M$: $a = 1$.
For $A$: $b = -2$.
Substituting these values,we get the dimension of $R_H$ as $[h^1 e^{-2}] = [h/e^2]$.
Thus,the dimension of $R_H$ is the same as $\frac{h}{e^2}$.

(A) The given equation is $\frac{V}{t} = k \left( \frac{p}{l} \right)^a \eta^b r^c$.
The dimensions of the quantities are:
$[V/t] = [L^3 T^{-1}]$
$[p/l] = [M L^{-1} T^{-2} / L] = [M L^{-2} T^{-2}]$
$[\eta] = [M L^{-1} T^{-1}]$
$[r] = [L]$
Substituting these into the equation:
$[L^3 T^{-1}] = [M L^{-2} T^{-2}]^a [M L^{-1} T^{-1}]^b [L]^c$
$[L^3 T^{-1}] = M^{a+b} L^{-2a-b+c} T^{-2a-b}$
Equating the powers of $M, L,$ and $T$ on both sides:
For $M$: $a + b = 0 \Rightarrow b = -a$
For $T$: $-2a - b = -1$
Substituting $b = -a$ into the $T$ equation: $-2a - (-a) = -1 \Rightarrow -a = -1 \Rightarrow a = 1$.
Since $b = -a$,we get $b = -1$.
For $L$: $-2a - b + c = 3$
Substituting $a = 1$ and $b = -1$: $-2(1) - (-1) + c = 3 \Rightarrow -2 + 1 + c = 3 \Rightarrow -1 + c = 3 \Rightarrow c = 4$.
Thus,the values are $a=1, b=-1, c=4$.