Dimensional Analysis, Uses and Limitations Questions in English

(A) Given the formula $x = \frac{a^{1/2} b^2}{c^3}$.
The relative error in $x$ is given by the expression:
$\frac{\Delta x}{x} = \frac{1}{2} \frac{\Delta a}{a} + 2 \frac{\Delta b}{b} + 3 \frac{\Delta c}{c}$.
Substituting the given percentage errors:
$\frac{\Delta x}{x} \times 100 = \frac{1}{2} (\pm 1\%) + 2 (\pm 3\%) + 3 (\pm 2\%)$.
Calculating the maximum possible percentage error:
$= \pm 0.5\% \pm 6\% \pm 6\% = \pm 12.5\%$.

(B) The standard integral formula is $\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1} \frac{x}{a}$.
According to the principle of homogeneity of dimensions,terms that are added or subtracted must have the same dimensions.
In the denominator,the terms $a^2$ and $x^n$ are subtracted,so they must have the same dimensions.
Since both $x$ and $a$ represent distance,their dimension is $[L]$.
Therefore,$[a^2] = [L^2]$ and $[x^n] = [L^n]$.
Equating the dimensions: $[L^2] = [L^n]$.
This implies $n = 2$.

(B) The dimensions of the given quantities are:
$[v] = [L T^{-1}]$
$[r] = [L]$
$[g] = [L T^{-2}]$
Now,check the dimensions of each option:
Option $A$: $[v^2 r / g] = \frac{[L^2 T^{-2}] [L]}{[L T^{-2}]} = [L^2]$
Option $B$: $[v^2 / (rg)] = \frac{[L^2 T^{-2}]}{[L] [L T^{-2}]} = \frac{[L^2 T^{-2}]}{[L^2 T^{-2}]} = [M^0 L^0 T^0] = 1$ (Dimensionless)
Option $C$: $[v^2 / (g/r)] = [v^2 r / g] = [L^2]$
Option $D$: $[v^2 r g] = [L^2 T^{-2}] [L] [L T^{-2}] = [L^4 T^{-4}]$
Therefore,the expression $\frac{v^2}{rg}$ is dimensionless.

(B) To determine the Planck length $l_p$,we use dimensional analysis with the constants $G$ (gravitational constant),$h$ (Planck's constant),and $c$ (speed of light).
Dimensions are:
$[G] = M^{-1}L^3T^{-2}$
$[h] = ML^2T^{-1}$
$[c] = LT^{-1}$
Let $l_p \propto G^a h^b c^d$.
Substituting dimensions: $L^1 = (M^{-1}L^3T^{-2})^a (ML^2T^{-1})^b (LT^{-1})^d$.
Equating powers of $M, L, T$:
$M: -a + b = 0 \implies a = b$
$T: -2a - b - d = 0 \implies -3a = d$
$L: 3a + 2b + d = 1 \implies 3a + 2a - 3a = 1 \implies 2a = 1 \implies a = 1/2$.
Thus,$a = 1/2, b = 1/2, d = -3/2$.
Therefore,$l_p = \sqrt{\frac{Gh}{c^3}}$.

(D) In the equation $AD = C \ln(BD)$,the argument of the logarithmic function must be dimensionless. Therefore,the dimension of $BD$ must be $1$ (dimensionless),i.e.,$[BD] = [M^0 L^0 T^0]$.
This implies $[B] = [D]^{-1}$.
Substituting this into the equation,we have $[AD] = [C] \times [1]$,so $[A][D] = [C]$.
For a physical quantity to be meaningful in addition or subtraction,the dimensions of the terms must be identical.
Checking option $D$: $\frac{A - C}{D}$. Here,$A$ and $C$ are being subtracted. Since $[A] = [C][D]^{-1}$ and $[C] = [A][D]$,the dimensions of $A$ and $C$ are different. Thus,$A - C$ is not a meaningful operation.

(D) The dimension of capacitance $C$ is $[M^{-1} L^{-2} T^4 I^2]$.
We analyze the dimensions of the given constants:
Charge $e = [I T]$
Bohr radius $a_0 = [L]$
Planck's constant $h = [M L^2 T^{-1}]$
Speed of light $c = [L T^{-1}]$
Now,check the dimensions of option $C$ $(u = \frac{e^2 c}{h a_0})$:
Dimensions of $\frac{e^2 c}{h a_0} = \frac{[I T]^2 [L T^{-1}]}{[M L^2 T^{-1}] [L]} = \frac{[I^2 T^2 L T^{-1}]}{[M L^3 T^{-1}]} = [M^{-1} L^{-2} T^2 I^2]$.
Wait,let us re-evaluate the dimensions of capacitance. Capacitance $C = \frac{Q}{V} = \frac{Q}{W/Q} = \frac{Q^2}{W} = \frac{[I^2 T^2]}{[M L^2 T^{-2}]} = [M^{-1} L^{-2} T^4 I^2]$.
Checking option $D$ $(u = \frac{e^2 a_0}{hc})$:
Dimensions of $\frac{e^2 a_0}{hc} = \frac{[I^2 T^2] [L]}{[M L^2 T^{-1}] [L T^{-1}]} = \frac{[I^2 T^2 L]}{[M L^3 T^{-2}]} = [M^{-1} L^{-2} T^4 I^2]$.
Since the dimensions of $u = \frac{e^2 a_0}{hc}$ match the dimensions of capacitance,option $D$ is correct.

(C) Let $\mu_0$ be related to $e, m, c,$ and $h$ as follows: $\mu_0 = k e^a m^b c^c h^d$.
The dimensional formula for $\mu_0$ is $[M L T^{-2} A^{-2}]$.
The dimensional formulas for the given quantities are: $e = [A T]$,$m = [M]$,$c = [L T^{-1}]$,and $h = [M L^2 T^{-1}]$.
Substituting these into the equation:
$[M L T^{-2} A^{-2}] = [A T]^a [M]^b [L T^{-1}]^c [M L^2 T^{-1}]^d$
$[M L T^{-2} A^{-2}] = [M^{b+d} L^{c+2d} T^{a-c-d} A^a]$
Comparing the powers of $M, L, T,$ and $A$ on both sides:
$A: a = -2$
$M: b + d = 1$
$L: c + 2d = 1$
$T: a - c - d = -2$
Substituting $a = -2$ into the $T$ equation: $-2 - c - d = -2 \implies c + d = 0 \implies c = -d$.
Substituting $c = -d$ into the $L$ equation: $-d + 2d = 1 \implies d = 1$.
Since $d = 1$,then $c = -1$.
Since $b + d = 1$,then $b + 1 = 1 \implies b = 0$.
Thus,$\mu_0 \propto e^{-2} m^0 c^{-1} h^1 = \frac{h}{c e^2}$.

(D) Let $\frac{Q}{A} = \eta^a \left( \frac{S\Delta \theta}{h} \right)^b \left( \frac{1}{\rho g} \right)^c$.
Dimensions of $\frac{Q}{A}$ (Heat flux) are $[M T^{-3}]$.
Dimensions of $\eta$ are $[M L^{-1} T^{-1}]$.
Dimensions of $\frac{S\Delta \theta}{h}$ are $[L^2 T^{-2} K^{-1} \cdot K \cdot L^{-1}] = [L T^{-2}]$.
Dimensions of $\frac{1}{\rho g}$ are $[(M L^{-3})^{-1} (L T^{-2})^{-1}] = [M^{-1} L^3 \cdot L^{-1} T^2] = [M^{-1} L^2 T^2]$.
Equating dimensions: $[M T^{-3}] = [M L^{-1} T^{-1}]^a [L T^{-2}]^b [M^{-1} L^2 T^2]^c$.
$[M T^{-3}] = [M^{a-c} L^{-a+b+2c} T^{-a-2b+2c}]$.
Comparing powers:
$a - c = 1$
$-a + b + 2c = 0$
$-a - 2b + 2c = -3$
Solving these equations:
From the first,$a = 1 + c$.
Substituting into the third: $-(1+c) - 2b + 2c = -3 \Rightarrow -1 + c - 2b = -3 \Rightarrow c - 2b = -2$.
Substituting into the second: $-(1+c) + b + 2c = 0 \Rightarrow c + b = 1$.
Adding $c - 2b = -2$ and $2(c + b) = 2$ gives $3c = 0$,so $c = 0$.
Then $b = 1$ and $a = 1$.
Therefore,$\frac{Q}{A} = \eta \frac{S\Delta \theta}{h}$.

(B) physical quantity has the same value in different systems of units if it is dimensionless. We check the dimensions of the given expression $\frac{e^2}{2\pi \varepsilon _0 G m_e^2}$.
The dimensions of the constants are:
$e = [M^0 L^0 T^1 A^1]$
$\varepsilon _0 = [M^{-1} L^{-3} T^4 A^2]$
$G = [M^{-1} L^3 T^{-2}]$
$m_e = [M^1 L^0 T^0]$
Substituting these into the expression:
$\frac{[T^2 A^2]}{[M^{-1} L^{-3} T^4 A^2] [M^{-1} L^3 T^{-2}] [M^2]} = \frac{[T^2 A^2]}{[M^{-1-1+2} L^{-3+3} T^{4-2} A^2]} = \frac{[T^2 A^2]}{[M^0 L^0 T^2 A^2]} = 1$
Since the expression is dimensionless,its value remains constant across all systems of units.

(A) The classical electron radius $r_e$ is defined by equating the electrostatic potential energy of an electron to its rest mass energy.
$mc^2 = \frac{1}{4\pi \varepsilon_0} \frac{e^2}{r_e}$
Rearranging for $r_e$:
$r_e = \frac{e^2}{4\pi \varepsilon_0 mc^2}$
Substituting the values of the constants ($e \approx 1.6 \times 10^{-19} \ C$,$\varepsilon_0 \approx 8.85 \times 10^{-12} \ F/m$,$m \approx 9.1 \times 10^{-31} \ kg$,$c \approx 3 \times 10^8 \ m/s$),we get $r_e \approx 2.8 \times 10^{-15} \ m$,which is the order of magnitude for the size of an atom/nucleus related constant.

(C) The dimensions of the physical quantities are: Time period $t = [T]$,Density $d = [ML^{-3}]$,Radius $r = [L]$,Surface tension $s = [MT^{-2}]$.
Given formula: $t = r^b s^{c/2} d^{a/4}$.
Substituting dimensions: $[T] = [L]^b [MT^{-2}]^{c/2} [ML^{-3}]^{a/4}$.
Equating powers of $M$: $0 = c/2 + a/4 \implies a = -2c$.
Equating powers of $T$: $1 = -c \implies c = -1$.
Substituting $c = -1$ into $a = -2c$,we get $a = 2$.
Equating powers of $L$: $0 = b - 3(a/4) = b - 3(2/4) = b - 3/2$.
Therefore,$b = 3/2$.

(A) Surface tension is defined as force per unit length: $S = \frac{F}{\ell}$.
We know that energy $K = F \cdot \ell$,so $F = \frac{K}{\ell}$.
Substituting this into the surface tension formula: $S = \frac{K}{\ell^2}$.
Since velocity $V = \frac{\ell}{T}$,we have $\ell = V \cdot T$.
Substituting $\ell$ in the expression for $S$: $S = \frac{K}{(V \cdot T)^2} = \frac{K}{V^2 T^2}$.
Therefore,the dimensional formula for surface tension in terms of $K, V, T$ is $[K V^{-2} T^{-2}]$.

(A) The formula for conversion between two systems of units is $n_2 = n_1 \left( \frac{M_1}{M_2} \right)^a \left( \frac{L_1}{L_2} \right)^b \left( \frac{T_1}{T_2} \right)^c$.
Here,the dimension of density is $[M^1 L^{-3} T^0]$,so $a=1, b=-3, c=0$.
Given: $n_1 = 128$,$M_1 = 1 \ kg = 1000 \ g$,$L_1 = 1 \ m = 100 \ cm$.
New units: $M_2 = 50 \ g$,$L_2 = 25 \ cm$.
Substituting the values:
$n_2 = 128 \times \left( \frac{1000 \ g}{50 \ g} \right)^1 \times \left( \frac{100 \ cm}{25 \ cm} \right)^{-3}$
$n_2 = 128 \times (20) \times (4)^{-3}$
$n_2 = 128 \times 20 \times \frac{1}{64}$
$n_2 = 2 \times 20 = 40$.

(D) Young's modulus $(Y)$ is defined as the ratio of stress to strain. Since strain is dimensionless,the dimension of $Y$ is the same as that of stress (Force/Area).
$[Y] = [F] / [L^2]$.
We are given fundamental units $V, A, F$.
We know $F = M \cdot A$,so $M = F \cdot A^{-1}$.
Also,$V = L \cdot T^{-1}$ and $A = L \cdot T^{-2}$.
Dividing $A$ by $V$: $A/V = (L \cdot T^{-2}) / (L \cdot T^{-1}) = T^{-1}$,so $T = V \cdot A^{-1}$.
Now,$L = V \cdot T = V \cdot (V \cdot A^{-1}) = V^2 \cdot A^{-1}$.
Therefore,$L^2 = (V^2 \cdot A^{-1})^2 = V^4 \cdot A^{-2}$.
Substituting these into the expression for $Y$:
$[Y] = [F] / [L^2] = F / (V^4 \cdot A^{-2}) = F \cdot V^{-4} \cdot A^2$.

(A) Let the linear momentum $P$ be expressed as $P = k S^a I^b h^c$,where $k$ is a dimensionless constant.
Dimensional formulas are:
$P = [MLT^{-1}]$
$S = [MT^{-2}]$
$I = [ML^2]$
$h = [ML^2T^{-1}]$
Substituting these into the equation:
$[MLT^{-1}] = [MT^{-2}]^a [ML^2]^b [ML^2T^{-1}]^c$
$[MLT^{-1}] = M^{a+b+c} L^{2b+2c} T^{-2a-c}$
Comparing the powers of $M, L,$ and $T$ on both sides:
$a + b + c = 1$ $(1)$
$2b + 2c = 1$ $(2)$
$-2a - c = -1$ $(3)$
From $(2)$,$b + c = 1/2$. Substituting this into $(1)$:
$a + 1/2 = 1 \implies a = 1/2$
From $(3)$,$c = 1 - 2a = 1 - 2(1/2) = 0$.
Substituting $c = 0$ into $(2)$,$2b = 1 \implies b = 1/2$.
Thus,the dimensional formula for linear momentum is $S^{1/2} I^{1/2} h^0$.

(B) Given the formula $X = 5YZ^2$,we can express $Y$ as $Y = \frac{X}{5Z^2}$.
The dimension of capacitance $X$ is $[M^{-1} L^{-2} T^4 A^2]$.
The dimension of magnetic field $Z$ is $[M^1 L^0 T^{-2} A^{-1}]$.
Substituting these into the expression for $Y$:
$Y = \frac{[M^{-1} L^{-2} T^4 A^2]}{[M^1 L^0 T^{-2} A^{-1}]^2}$
$Y = \frac{[M^{-1} L^{-2} T^4 A^2]}{[M^2 L^0 T^{-4} A^{-2}]}$
$Y = [M^{-1-2} L^{-2-0} T^{4-(-4)} A^{2-(-2)}]$
$Y = [M^{-3} L^{-2} T^8 A^4]$.

(B) The principle of homogeneity of dimensions states that only quantities with the same dimensions can be added or subtracted.
Since $B$ is added to $x^2$,the dimensions of $B$ must be equal to the dimensions of $x^2$.
$[B] = [x^2] = L^2$.
The potential energy $U$ has the dimensions of work,which is $[U] = M L^2 T^{-2}$.
The given equation is $U = \frac{A \sqrt{x}}{x^2 + B}$.
Considering the dimensions of the denominator,$[x^2 + B] = L^2$.
Thus,$[U] = \frac{[A] [x]^{1/2}}{L^2}$.
Substituting the dimensions: $M L^2 T^{-2} = \frac{[A] L^{1/2}}{L^2}$.
$[A] = (M L^2 T^{-2}) \times (L^{3/2}) = M L^{7/2} T^{-2}$.
Now,calculate the dimensions of $A/B$:
$[A/B] = \frac{M L^{7/2} T^{-2}}{L^2} = M L^{7/2 - 2} T^{-2} = M L^{3/2} T^{-2}$.

(B) According to the principle of homogeneity of dimensions,physical quantities can only be added or subtracted if they have the same dimensions.
$1$. In option $A$,$A$ and $\frac{A^3}{B}$ can have the same dimensions if the dimensions of $B$ are equal to the dimensions of $A^2$. Thus,this expression could represent a physical quantity.
$2$. In option $B$,the argument of an exponential function must be dimensionless. Since $A$ and $B$ have different dimensions,the ratio $\frac{A}{B}$ is not dimensionless. Therefore,$\exp \left( -\frac{A}{B} \right)$ is physically meaningless.
$3$. In option $C$,$AB^2$ represents a product of physical quantities,which is always a valid physical quantity.
$4$. In option $D$,$\frac{A}{B^4}$ represents a quotient of physical quantities,which is also a valid physical quantity.
Thus,the expression that cannot denote a physical quantity is $\exp \left( -\frac{A}{B} \right)$.

(D) Let the dimensional formula for energy $(E)$ be expressed as $E = k P^a A^b T^c$,where $k$ is a dimensionless constant.
Substituting the dimensions of each quantity:
$[M L^2 T^{-2}] = [M L T^{-1}]^a [L^2]^b [T]^c$
$[M L^2 T^{-2}] = [M^a L^{a+2b} T^{-a+c}]$
Comparing the powers of $M$,$L$,and $T$ on both sides:
For $M$: $a = 1$
For $L$: $a + 2b = 2 \implies 1 + 2b = 2 \implies 2b = 1 \implies b = 1/2$
For $T$: $-a + c = -2 \implies -1 + c = -2 \implies c = -1$
Substituting these values back into the expression,we get $E = [P^1 A^{1/2} T^{-1}]$.
Thus,the correct option is $D$.

(B) Given the equation $v = \sqrt{ab} + bt + \frac{c}{d + t}$.
$1$. For the term $bt$: Since $v$ is velocity $([LT^{-1}])$ and $t$ is time $([T])$,the dimension of $bt$ must be $[LT^{-1}]$. Thus,$[b][T] = [LT^{-1}] \Rightarrow [b] = [LT^{-2}]$,which represents acceleration.
$2$. For the term $\sqrt{ab}$: The dimension of $\sqrt{ab}$ must be velocity $([LT^{-1}])$. Squaring both sides,$[ab] = [L^2T^{-2}]$. Since $[b] = [LT^{-2}]$,we have $[a][LT^{-2}] = [L^2T^{-2}] \Rightarrow [a] = [L]$,which represents distance.
$3$. For the term $d+t$: Since we can only add quantities of the same dimension,$d$ must have the dimension of time $([T])$.
$4$. For the term $\frac{c}{d+t}$: The dimension of this term must be velocity $([LT^{-1}])$. Since $[d+t] = [T]$,we have $\frac{[c]}{[T]} = [LT^{-1}] \Rightarrow [c] = [L]$,which represents distance.
Therefore,the order is $a$ (distance),$b$ (acceleration),$c$ (distance),$d$ (time).

(A) According to the principle of homogeneity of dimensions,quantities added together must have the same dimensions.
Since $\beta$ is added to $\sqrt{d}$,the dimensions of $\beta$ must be equal to the dimensions of $\sqrt{d}$.
Density $d = [M L^{-3}]$,so $\sqrt{d} = [M^{1/2} L^{-3/2}]$.
Thus,$[\beta] = [M^{1/2} L^{-3/2}]$.
The denominator $(\beta + \sqrt{d})$ will also have the same dimensions: $[\beta + \sqrt{d}] = [M^{1/2} L^{-3/2}]$.
The given equation is $F = \frac{\alpha}{\beta + \sqrt{d}}$.
Rearranging for $\alpha$,we get $\alpha = F \times (\beta + \sqrt{d})$.
Force $F = [M L T^{-2}]$.
Substituting the dimensions: $[\alpha] = [M L T^{-2}] \times [M^{1/2} L^{-3/2}]$.
$[\alpha] = [M^{1 + 1/2} L^{1 - 3/2} T^{-2}] = [M^{3/2} L^{-1/2} T^{-2}]$.

(A) According to the principle of homogeneity of dimensions,the dimensions of each term added or subtracted in an equation must be the same.
Given the equation $F = a \sin(ct) + b \cos(dx)$,the dimensions of the terms $a \sin(ct)$ and $b \cos(dx)$ must be equal to the dimensions of force $F$.
Since the arguments of trigonometric functions ($sin$ and $cos$) are dimensionless,the dimensions of $a$ and $b$ must be equal to the dimensions of force $F$.
Dimension of force $F = [M^1L^1T^{-2}]$.
Therefore,$[a] = [M^1L^1T^{-2}]$ and $[b] = [M^1L^1T^{-2}]$.
Now,the dimension of $a/b$ is $[a]/[b] = [M^1L^1T^{-2}] / [M^1L^1T^{-2}] = [M^0L^0T^0]$.

(B) The argument of a trigonometric function must be dimensionless.
Since $qt$ is the argument of $\tan$,we have $[qt] = [M^{0}L^{0}T^{0}]$.
Given that $t$ is time,$[t] = [T]$.
Therefore,$[q][T] = [1] \Rightarrow [q] = [T^{-1}]$.
In the equation $y = pq \tan(qt)$,the term $\tan(qt)$ is dimensionless.
Thus,the dimensions of $y$ must be equal to the dimensions of the product $pq$.
$[y] = [pq]$.
Since $y$ represents position,$[y] = [L]$.
$[p][T^{-1}] = [L]$.
$[p] = [L][T] = [L^{1}T^{1}]$.

(A) The time period $T$ is given by $T \propto P^a d^b E^c$.
Writing the dimensional formulas for each quantity:
$[T] = [M^0 L^0 T^1]$
$[P] = [M L^{-1} T^{-2}]$
$[d] = [M L^{-3} T^0]$
$[E] = [M L^2 T^{-2}]$
Substituting these into the equation:
$[M^0 L^0 T^1] = [M L^{-1} T^{-2}]^a [M L^{-3}]^b [M L^2 T^{-2}]^c$
$[M^0 L^0 T^1] = [M^{a+b+c} L^{-a-3b+2c} T^{-2a-2c}]$
Comparing the exponents of $M, L,$ and $T$ on both sides:
$1) a + b + c = 0$
$2) -a - 3b + 2c = 0$
$3) -2a - 2c = 1$
From equation $(3)$,$a + c = -1/2$,so $c = -1/2 - a$.
Substitute $c$ into equation $(1)$: $a + b + (-1/2 - a) = 0 \implies b = 1/2$.
Substitute $b = 1/2$ and $c = -1/2 - a$ into equation $(2)$:
$-a - 3(1/2) + 2(-1/2 - a) = 0$
$-a - 3/2 - 1 - 2a = 0$
$-3a = 5/2 \implies a = -5/6$.
Finally,$c = -1/2 - (-5/6) = -3/6 + 5/6 = 2/6 = 1/3$.
Thus,$a = -5/6, b = 1/2, c = 1/3$.

(D) The given equation is $F = A \cos(Bx) + C \sin(Dt).$
According to the principle of homogeneity of dimensions,the argument of any trigonometric function must be dimensionless.
Therefore,the dimensions of $Bx$ must be $[M^0 L^0 T^0]$.
$[B] \cdot [x] = [M^0 L^0 T^0]$
$[B] \cdot [L] = [M^0 L^0 T^0]$
$[B] = [L^{-1}]$
Similarly,the dimensions of $Dt$ must be $[M^0 L^0 T^0]$.
$[D] \cdot [t] = [M^0 L^0 T^0]$
$[D] \cdot [T] = [M^0 L^0 T^0]$
$[D] = [T^{-1}]$
Now,the dimensional formula of $D/B$ is:
$[D/B] = \frac{[T^{-1}]}{[L^{-1}]} = [L^1 T^{-1}]$

(B) Given the relation: $P = \frac{a - t^2}{bx}$.
According to the principle of homogeneity of dimensions,quantities added or subtracted must have the same dimensions. Therefore,the dimension of $a$ must be equal to the dimension of $t^2$.
$[a] = [T^2]$.
Now,consider the dimension of the entire expression:
$[P] = \frac{[a - t^2]}{[b][x]} = \frac{[T^2]}{[b][L]}$.
We know that pressure $P$ has dimensions $[M L^{-1} T^{-2}]$.
Substituting this into the equation:
$[M L^{-1} T^{-2}] = \frac{[T^2]}{[b][L]}$.
Solving for $[b]$:
$[b] = \frac{[T^2]}{[M L^{-1} T^{-2}][L]} = \frac{[T^2]}{[M T^{-2}]} = [M^{-1} T^4]$.
Now,find the dimensions of $a/b$:
$[a/b] = \frac{[T^2]}{[M^{-1} T^4]} = [M^1 L^0 T^{-2}]$.
Thus,the correct option is $B$.

(A) Let time $T \propto c^{x} G^{y} h^{z}$.
$\Rightarrow T = k c^{x} G^{y} h^{z}$.
Taking dimensions on both sides: $[M^{0} L^{0} T^{1}] = [L T^{-1}]^{x} [M^{-1} L^{3} T^{-2}]^{y} [M L^{2} T^{-1}]^{z}$.
$[M^{0} L^{0} T^{1}] = [M^{-y+z} L^{x+3y+2z} T^{-x-2y-z}]$.
Equating powers of $M, L, T$ on both sides:
$-y + z = 0 \implies z = y \quad \dots(1)$
$x + 3y + 2z = 0 \quad \dots(2)$
$-x - 2y - z = 1 \quad \dots(3)$
Adding $(2)$ and $(3)$: $(x + 3y + 2z) + (-x - 2y - z) = 0 + 1 \implies y + z = 1$.
Since $z = y$,we have $2y = 1 \implies y = 1/2$.
Thus,$z = 1/2$.
Substituting into $(2)$: $x + 3(1/2) + 2(1/2) = 0 \implies x + 3/2 + 1 = 0 \implies x = -5/2$.
Therefore,$[T] = [G^{1/2} h^{1/2} c^{-5/2}]$.

(B) Given the expression for $X$ is $X = \frac{A^2 B^{1/2}}{C^{1/3} D^3}$.
The relative error in $X$ is given by the formula: $\frac{\Delta X}{X} = 2 \frac{\Delta A}{A} + \frac{1}{2} \frac{\Delta B}{B} + \frac{1}{3} \frac{\Delta C}{C} + 3 \frac{\Delta D}{D}$.
To find the maximum percentage error,we multiply by $100$:
$\frac{\Delta X}{X} \times 100 = 2 \left( \frac{\Delta A}{A} \times 100 \right) + \frac{1}{2} \left( \frac{\Delta B}{B} \times 100 \right) + \frac{1}{3} \left( \frac{\Delta C}{C} \times 100 \right) + 3 \left( \frac{\Delta D}{D} \times 100 \right)$.
Substituting the given percentage errors $(1\%, 2\%, 3\%, 4\%)$:
$\frac{\Delta X}{X} \times 100 = 2(1\%) + \frac{1}{2}(2\%) + \frac{1}{3}(3\%) + 3(4\%)$.
Calculating the values:
$= 2\% + 1\% + 1\% + 12\% = 16\%$.
Thus,the maximum percentage error in $X$ is $16\%$.

(D) The dimension of stopping potential $V_{0}$ is equivalent to the dimension of potential difference,which is $\frac{\text{Work}}{\text{Charge}} = \frac{ML^{2}T^{-2}}{AT} = ML^{2}T^{-3}A^{-1}$.
Let $V_{0} = h^{x} c^{y} G^{z} A^{w}$.
Substituting the dimensions: $[ML^{2}T^{-3}A^{-1}] = [ML^{2}T^{-1}]^{x} [LT^{-1}]^{y} [M^{-1}L^{3}T^{-2}]^{z} [A]^{w}$.
Equating the powers of $A$: $w = -1$.
Equating the powers of $M$: $x - z = 1$.
Equating the powers of $L$: $2x + y + 3z = 2$.
Equating the powers of $T$: $-x - y - 2z = -3$.
Solving these equations: From $x - z = 1$,we get $x = 1 + z$.
Adding the equations for $L$ and $T$: $(2x + y + 3z) + (-x - y - 2z) = 2 + (-3) \Rightarrow x + z = -1$.
Now we have $x - z = 1$ and $x + z = -1$. Adding them gives $2x = 0 \Rightarrow x = 0$. Then $z = -1$.
Substituting $x=0$ and $z=-1$ into $2x + y + 3z = 2$: $0 + y - 3 = 2 \Rightarrow y = 5$.
Thus,$V_{0} = h^{0} c^{5} G^{-1} A^{-1}$.

(C) The dimensions of the constants are:
$[h] = M^1 L^2 T^{-1}$
$[c] = L^1 T^{-1}$
$[G] = M^{-1} L^3 T^{-2}$
Substituting these into the expression for $f$:
$[f] = \sqrt{\frac{(M^1 L^2 T^{-1}) \times (L^1 T^{-1})^5}{M^{-1} L^3 T^{-2}}}$
$[f] = \sqrt{\frac{M^1 L^2 T^{-1} \times L^5 T^{-5}}{M^{-1} L^3 T^{-2}}}$
$[f] = \sqrt{\frac{M^1 L^7 T^{-6}}{M^{-1} L^3 T^{-2}}}$
$[f] = \sqrt{M^{1-(-1)} L^{7-3} T^{-6-(-2)}}$
$[f] = \sqrt{M^2 L^4 T^{-4}}$
$[f] = M^1 L^2 T^{-2}$
The dimension $M^1 L^2 T^{-2}$ corresponds to the dimension of Energy.

(N/A) The dimensions of the $LHS$ are:
$[M][LT^{-1}]^{2} = [M][L^{2}T^{-2}] = [ML^{2}T^{-2}]$
The dimensions of the $RHS$ are:
$[M][LT^{-2}][L] = [M][L^{2}T^{-2}] = [ML^{2}T^{-2}]$
Since the dimensions of the $LHS$ and $RHS$ are the same,the equation is dimensionally correct.

(A, C, E) The dimension of kinetic energy $K$ is $[M L^{2} T^{-2}]$.
According to the principle of homogeneity of dimensions,every term in an equation must have the same dimensions.
For $(a)$,the dimension is $[M^{2} (L T^{-1})^{3}] = [M^{2} L^{3} T^{-3}]$,which is incorrect.
For $(b)$,the dimension is $[M (L T^{-1})^{2}] = [M L^{2} T^{-2}]$,which is correct.
For $(c)$,the dimension is $[M (L T^{-2})] = [M L T^{-2}]$,which is incorrect.
For $(d)$,the dimension is $[M (L T^{-1})^{2}] = [M L^{2} T^{-2}]$,which is correct.
For $(e)$,the terms $m v^{2}$ and $m a$ have different dimensions ($[M L^{2} T^{-2}]$ and $[M L T^{-2}]$ respectively),so they cannot be added. This is incorrect.
Therefore,formulae $(a)$,$(c)$,and $(e)$ are ruled out based on dimensional analysis.

(N/A) The dependence of time period $T$ on the quantities $l, g$,and $m$ as a product may be written as:
$T = k l^{x} g^{y} m^{z}$
where $k$ is a dimensionless constant and $x, y$,and $z$ are the exponents.
By considering dimensions on both sides,we have:
$[M^0 L^0 T^1] = [L]^x [L T^{-2}]^y [M]^z$
$[M^0 L^0 T^1] = M^z L^{x+y} T^{-2y}$
On equating the dimensions on both sides,we have:
$z = 0$
$x + y = 0$
$-2y = 1$
Solving these,we get $y = -1/2$,$x = 1/2$,and $z = 0$.
Substituting these values,we get $T = k l^{1/2} g^{-1/2} m^0$.
Thus,$T = k \sqrt{\frac{l}{g}}$.
Note that the value of the constant $k$ cannot be obtained by the method of dimensions. Experimentally,$k = 2\pi$,so $T = 2\pi \sqrt{\frac{l}{g}}$.

(A) Given that, $1 \; \text{calorie} = 4.2 \; (1 \; kg) (1 \; m^2) (1 \; s^{-2})$.
Let the new units be $M' = \alpha \; kg$, $L' = \beta \; m$, and $T' = \gamma \; s$.
Then, $1 \; kg = \frac{1}{\alpha} \; M' = \alpha^{-1} \; M'$.
$1 \; m = \frac{1}{\beta} \; L' = \beta^{-1} \; L'$, so $1 \; m^2 = \beta^{-2} \; (L')^2$.
$1 \; s = \frac{1}{\gamma} \; T' = \gamma^{-1} \; T'$, so $1 \; s^{-2} = (\gamma^{-1})^{-2} \; (T')^{-2} = \gamma^2 \; (T')^{-2}$.
Substituting these into the expression for a calorie:
$1 \; \text{calorie} = 4.2 \times (\alpha^{-1} \; M') \times (\beta^{-2} \; (L')^2) \times (\gamma^2 \; (T')^{-2})$.
Therefore, the magnitude of a calorie in the new system is $4.2 \; \alpha^{-1} \beta^{-2} \gamma^2$.

(B, C) Correct: $y = a \sin \left(\frac{2 \pi t}{T}\right)$
Dimension of $y = [L]$. Dimension of $a = [L]$. The argument of $\sin$ is $\frac{2 \pi t}{T}$,which is dimensionless. Thus,the formula is dimensionally correct.
$(b)$ Incorrect: $y = a \sin v t$
Dimension of $v t = [LT^{-1}] \times [T] = [L]$. The argument of the trigonometric function must be dimensionless,but here it has dimensions of length. Thus,it is dimensionally incorrect.
$(c)$ Incorrect: $y = \left(\frac{a}{T}\right) \sin \left(\frac{t}{a}\right)$
Dimension of $\frac{a}{T} = [LT^{-1}] \neq [L]$. Also,the argument $\frac{t}{a} = [TL^{-1}]$ is not dimensionless. Thus,it is dimensionally incorrect.
$(d)$ Correct: $y = (a \sqrt{2}) \left(\sin \frac{2 \pi t}{T} + \cos \frac{2 \pi t}{T}\right)$
Dimension of $y = [L]$. Dimension of $a = [L]$. The argument $\frac{2 \pi t}{T}$ is dimensionless. Thus,the formula is dimensionally correct.

(B) Given the relation,$m = \frac{m_{0}}{(1 - v^{2})^{1/2}}$.
Dimension of $m = [M^{1} L^{0} T^{0}]$.
Dimension of $m_{0} = [M^{1} L^{0} T^{0}]$.
Dimension of $v = [M^{0} L^{1} T^{-1}]$.
Dimension of $v^{2} = [M^{0} L^{2} T^{-2}]$.
Dimension of $c = [M^{0} L^{1} T^{-1}]$.
The given formula will be dimensionally correct only when the dimension of the $L$.$H$.$S$. is the same as that of the $R$.$H$.$S$.
This is only possible when the factor $(1 - v^{2})^{1/2}$ is dimensionless,i.e.,$(1 - v^{2})$ must be dimensionless.
Since $1$ is dimensionless,$v^{2}$ must also be dimensionless. This is only possible if $v^{2}$ is divided by $c^{2}$,as the ratio $v^{2}/c^{2}$ is dimensionless.
Hence,the correct relation is $m = \frac{m_{0}}{(1 - v^{2}/c^{2})^{1/2}}$.

(N/A) One relation consisting of fundamental constants that gives the age of the Universe is:
$t = \left(\frac{e^2}{4 \pi \varepsilon_0}\right)^2 \times \frac{1}{m_p m_e^2 c^3 G}$
Where:
$t =$ Age of Universe
$e = 1.6 \times 10^{-19} \; C$
$\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \; N m^2 / C^2$
$m_p = 1.67 \times 10^{-27} \; kg$
$m_e = 9.1 \times 10^{-31} \; kg$
$c = 3 \times 10^8 \; m/s$
$G = 6.67 \times 10^{-11} \; N m^2 kg^{-2}$
Substituting these values:
$t = \frac{(1.6 \times 10^{-19})^4 \times (9 \times 10^9)^2}{(1.67 \times 10^{-27}) \times (9.1 \times 10^{-31})^2 \times (3 \times 10^8)^3 \times (6.67 \times 10^{-11})}$
Calculating this yields a value of approximately $6 \times 10^{17} \; s$, which is roughly $19$ billion years.
If the coincidence with the age of the universe is significant, it would imply that the fundamental constants might not be strictly constant over time, but could vary as the universe evolves.

(N/A) The given ratio is $\frac{k e^{2}}{G m_{e} m_{p}}$.
Where,$G$ is the gravitational constant with unit $N \, m^{2} \, kg^{-2}$.
$m_{e}$ and $m_{p}$ are the masses of an electron and a proton,respectively,with unit $kg$.
$e$ is the electric charge with unit $C$.
$k = \frac{1}{4 \pi \varepsilon_{0}}$ is the Coulomb constant with unit $N \, m^{2} \, C^{-2}$.
Substituting the units:
$\frac{[N \, m^{2} \, C^{-2}] [C^{2}]}{[N \, m^{2} \, kg^{-2}] [kg] [kg]} = \frac{N \, m^{2}}{N \, m^{2}} = M^{0} L^{0} T^{0}$.
Thus,the ratio is dimensionless.
Using the values:
$k = 9 \times 10^{9} \, N \, m^{2} \, C^{-2}$
$e = 1.6 \times 10^{-19} \, C$
$G = 6.67 \times 10^{-11} \, N \, m^{2} \, kg^{-2}$
$m_{e} = 9.11 \times 10^{-31} \, kg$
$m_{p} = 1.67 \times 10^{-27} \, kg$
Calculating the value:
$\frac{9 \times 10^{9} \times (1.6 \times 10^{-19})^{2}}{6.67 \times 10^{-11} \times 9.11 \times 10^{-31} \times 1.67 \times 10^{-27}} \approx 2.3 \times 10^{39}$.
This ratio signifies the strength of the electrostatic force relative to the gravitational force between an electron and a proton.

(N/A) Dimensional analysis is a method of obtaining the solution to certain problems in physics by using the dimensional formulas of physical quantities.
Uses of Dimensional Analysis:
$(a)$ To convert the value of a physical quantity from one system of units to another.
$(b)$ To check the dimensional consistency (correctness) of a given physical equation.
$(c)$ To derive the relationship between different physical quantities in a physical equation.

(N/A) The unit of work in the $MKS$ system is Joule $(J)$ and in the $CGS$ system is erg. The relation between Joule and erg is derived as follows:
Dimensional formula for work is $W = [M^1 L^2 T^{-2}]$.
Let $n_1$ and $n_2$ be the numerical values and $u_1$ and $u_2$ be the units in the two systems.
$n_1 u_1 = n_2 u_2$
$n_2 = n_1 [M_1/M_2]^1 [L_1/L_2]^2 [T_1/T_2]^{-2}$
In $MKS$ system: $M_1 = 1 \text{ kg} = 10^3 \text{ g}$,$L_1 = 1 \text{ m} = 10^2 \text{ cm}$,$T_1 = 1 \text{ s}$.
In $CGS$ system: $M_2 = 1 \text{ g}$,$L_2 = 1 \text{ cm}$,$T_2 = 1 \text{ s}$.
$n_2 = 1 \times [10^3 \text{ g} / 1 \text{ g}]^1 \times [10^2 \text{ cm} / 1 \text{ cm}]^2 \times [1 \text{ s} / 1 \text{ s}]^{-2}$
$n_2 = 10^3 \times (10^2)^2 \times 1 = 10^3 \times 10^4 = 10^7$.
Therefore,$1 \text{ Joule} = 10^7 \text{ erg}$.

(A) The principle of homogeneity of dimensions states that physical quantities having the same dimensions can only be added or subtracted from one another.
This principle is used to check the dimensional consistency of a given equation. For an equation to be dimensionally consistent,every term on both sides of the equation must have the same dimensions.
Note: Dimensional consistency does not guarantee the physical correctness of an equation,as it cannot account for dimensionless constants or functions.
Example: Check the consistency of $x = x_{0} + v_{0}t + \frac{1}{2}at^{2}$.
Here,$x$ is the final position,$x_{0}$ is the initial position,$v_{0}$ is the initial velocity,$a$ is the acceleration,and $t$ is time.
$1$. Dimension of $LHS$ $(x)$: $[L^1] = [M^0 L^1 T^0]$.
$2$. Dimensions of $RHS$ terms:
- Dimension of $x_{0}$: $[L^1] = [M^0 L^1 T^0]$.
- Dimension of $v_{0}t$: $[L^1 T^{-1}] \times [T^1] = [L^1] = [M^0 L^1 T^0]$.
- Dimension of $\frac{1}{2}at^{2}$: Since $\frac{1}{2}$ is a dimensionless constant,the dimension is $[L^1 T^{-2}] \times [T^2] = [L^1] = [M^0 L^1 T^0]$.
Since all terms on both sides have the same dimensions,the equation is dimensionally consistent.

(N/A) Suppose,heat energy $H \propto I^{a} R^{b} t^{c}$.
$\therefore H = k I^{a} R^{b} t^{c} \dots (i)$ (where $k$ is a dimensionless constant).
Now,writing the dimensional formula on both sides:
$[H] = M^{1} L^{2} T^{-2}$
$[I] = A^{1}$
$[R] = M^{1} L^{2} T^{-3} A^{-2}$
$[t] = T^{1}$
Substituting dimensions in equation $(i)$:
$M^{1} L^{2} T^{-2} = (A^{1})^{a} (M^{1} L^{2} T^{-3} A^{-2})^{b} (T^{1})^{c}$
$M^{1} L^{2} T^{-2} = M^{b} L^{2b} T^{-3b+c} A^{a-2b}$
Equating the indices of $M, L, T,$ and $A$ on both sides:
For $M$: $b = 1$
For $L$: $2b = 2 \implies b = 1$
For $A$: $a - 2b = 0 \implies a - 2(1) = 0 \implies a = 2$
For $T$: $-3b + c = -2 \implies -3(1) + c = -2 \implies c = 1$
Thus,$a = 2, b = 1, c = 1$.
Substituting these values in equation $(i)$,we get:
$H = k I^{2} R t$
Experimentally,$k = 1$,therefore $H = I^{2} R t$.

(N/A) Dimensional analysis is a method used to check the consistency of physical equations,derive relationships between physical quantities,and convert units from one system to another based on the dimensions of the quantities involved.
Limitations of dimensional analysis:
$(1)$ In dimensional equations containing $M, L,$ and $T$,we can obtain at most three equations by equating the indices of $M, L,$ and $T$. Hence,this method is of no avail in deducing the exact form of a physical relation that depends on more than three independent physical quantities.
$(2)$ Information about dimensionless constants (such as $\pi, e,$ or numerical coefficients) cannot be obtained using this method.
$(3)$ Equations containing exponential,logarithmic,or trigonometric functions lie outside the scope of this method.
$(4)$ This method is not useful if the constant of proportionality is not dimensionless.

(N/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.
In other words,we can only add,subtract,or equate physical quantities that have the same dimensions.
For example,in the equation $v = u + at$,the dimensions of $v$,$u$,and $at$ must all be equal to $[LT^{-1}]$.

(A) According to the principle of homogeneity of dimensions,the dimensions of each term on both sides of an equation must be the same.
Given the equation $x = a + bt + ct^2$,where $x$ is in meters $(m)$ and $t$ is in seconds $(s)$.
$1$. For the term $a$: Since $x$ and $a$ are added,they must have the same unit. Therefore,the unit of $a$ is $m$.
$2$. For the term $bt$: The unit of $bt$ must be the same as the unit of $x$. So,$[bt] = m$. Thus,the unit of $b = m/s$.
$3$. For the term $ct^2$: The unit of $ct^2$ must be the same as the unit of $x$. So,$[ct^2] = m$. Thus,the unit of $c = m/s^2$.
Therefore,the units are $a: m, b: m/s, c: m/s^2$.

(A) The given unit is $W/m^2$ (Watt per square meter).
Power $(P)$ has the unit Watt $(W)$,and its dimensional formula is $[M^1 L^2 T^{-3}]$.
Area $(A)$ has the unit $m^2$,and its dimensional formula is $[L^2]$.
The physical quantity is Intensity or Power density,defined as $I = P/A$.
Dimensional formula = $[M^1 L^2 T^{-3}] / [L^2] = [M^1 L^0 T^{-3}]$.
Therefore,the correct dimensional formula is $[M^1 L^0 T^{-3}]$.

(A) According to the principle of homogeneity of dimensions,the dimensions of each term in an equation must be the same.
Given the equation $y = x^2 r + M^1 L^1 T^{-2}$,the dimensions of $x^2 r$ must be equal to the dimensions of $M^1 L^1 T^{-2}$.
Let $[x^2]$ be the dimensional formula of $x^2$ and $[r] = L^1$ (as $r$ is displacement).
So,$[x^2] \cdot [r] = M^1 L^1 T^{-2}$.
$[x^2] \cdot L^1 = M^1 L^1 T^{-2}$.
$[x^2] = \frac{M^1 L^1 T^{-2}}{L^1}$.
$[x^2] = M^1 L^0 T^{-2}$.

(A) The dimensional formula for pressure $P$ is $[P] = [M^{1} L^{-1} T^{-2}]$.
The dimensional formula for force $F$ is $[F] = [M^{1} L^{1} T^{-2}]$.
Given the equation $P = FK$,we can write the dimensional equation as $[P] = [F][K]$.
Substituting the dimensions: $[M^{1} L^{-1} T^{-2}] = [M^{1} L^{1} T^{-2}] [K]$.
Therefore,$[K] = \frac{[M^{1} L^{-1} T^{-2}]}{[M^{1} L^{1} T^{-2}]}$.
$[K] = M^{1-1} L^{-1-1} T^{-2-(-2)} = M^{0} L^{-2} T^{0}$.