Dimensional Analysis, Uses and Limitations Questions in English

(C) Let the relationship be $v = k g^a \lambda^b \rho^c$,where $k$ is a dimensionless constant.
Writing the dimensions of each quantity:
$[v] = [L T^{-1}]$
$[g] = [L T^{-2}]$
$[\lambda] = [L]$
$[\rho] = [M L^{-3}]$
Substituting these into the equation: $[L T^{-1}] = [L T^{-2}]^a [L]^b [M L^{-3}]^c$
$[L^1 T^{-1} M^0] = [M^c L^{a+b-3c} T^{-2a}]$
Comparing the powers of $M, L,$ and $T$ on both sides:
For $M$: $c = 0$
For $T$: $-2a = -1 \implies a = 1/2$
For $L$: $a + b - 3c = 1 \implies 1/2 + b - 0 = 1 \implies b = 1/2$
Thus,$v \propto g^{1/2} \lambda^{1/2} \rho^0 \implies v \propto \sqrt{g \lambda} \implies v^2 \propto g \lambda$.

(D) The dimensional formula of force is $[F] = [M^1 L^1 T^{-2}]$.
Using the principle of dimensional consistency,$n_2 = n_1 \left[ \frac{M_1}{M_2} \right]^1 \left[ \frac{L_1}{L_2} \right]^1 \left[ \frac{T_1}{T_2} \right]^{-2}$.
Here,$n_1 = 1$,$M_1 = 1 \; \text{kg} = 10^3 \; \text{g}$,$L_1 = 1 \; \text{m} = 10^2 \; \text{cm}$,and $T_1 = T_2 = 1 \; \text{s}$.
Substituting these values: $n_2 = 1 \times \left[ \frac{10^3 \; \text{g}}{1 \; \text{g}} \right]^1 \times \left[ \frac{10^2 \; \text{cm}}{1 \; \text{cm}} \right]^1 \times \left[ \frac{1 \; \text{s}}{1 \; \text{s}} \right]^{-2}$.
$n_2 = 1 \times 10^3 \times 10^2 \times 1 = 10^5$.
Therefore,$1 \; \text{newton} = 10^5 \; \text{dyne}$.

(A) According to the principle of homogeneity of dimensions,only physical quantities with the same dimensions can be added or subtracted.
In the term $(P + \frac{a}{V^2})$,the dimensions of $\frac{a}{V^2}$ must be equal to the dimensions of pressure $P$.
So,$[\frac{a}{V^2}] = [P]$.
$[a] = [P] \times [V^2]$.
The dimensions of pressure $P$ are $[ML^{-1}T^{-2}]$ and the dimensions of volume $V$ are $[L^3]$.
Therefore,$[a] = [ML^{-1}T^{-2}] \times [L^3]^2$.
$[a] = [ML^{-1}T^{-2}] \times [L^6]$.
$[a] = [ML^5T^{-2}]$.

(A) The dimensional formula for energy $E$ is $[M^1 L^2 T^{-2}]$.
The dimensional formula for $G$ is $[M^{-1} L^3 T^{-2}]$.
The dimensional formula for $h$ is $[M^1 L^2 T^{-1}]$.
The dimensional formula for $c$ is $[L^1 T^{-1}]$.
Given $E = G^p h^q c^r$,we have:
$[M^1 L^2 T^{-2}] = [M^{-1} L^3 T^{-2}]^p [M^1 L^2 T^{-1}]^q [L^1 T^{-1}]^r$
$[M^1 L^2 T^{-2}] = [M^{-p+q} L^{3p+2q+r} T^{-2p-q-r}]$
Comparing the powers of $M, L$,and $T$ on both sides:
For $M$: $-p + q = 1 \implies q = 1 + p$
For $T$: $-2p - q - r = -2 \implies 2p + q + r = 2$
For $L$: $3p + 2q + r = 2$
Subtracting the $T$ equation from the $L$ equation: $(3p + 2q + r) - (2p + q + r) = 2 - 2 \implies p + q = 0 \implies q = -p$
Substituting $q = -p$ into $q = 1 + p$: $-p = 1 + p \implies 2p = -1 \implies p = -\frac{1}{2}$.
Then $q = -(-\frac{1}{2}) = \frac{1}{2}$.
Using $2p + q + r = 2$: $2(-\frac{1}{2}) + \frac{1}{2} + r = 2 \implies -1 + \frac{1}{2} + r = 2 \implies r = 2 + 1 - \frac{1}{2} = \frac{5}{2}$.
Thus,$p = -\frac{1}{2}, q = \frac{1}{2}, r = \frac{5}{2}$.

(B) According to the principle of dimensional homogeneity,the argument of a trigonometric function must be dimensionless.
Therefore,the dimensions of $k_1x$ and $k_2t$ must be $M^0 L^0 T^0$.
For $k_1x$ to be dimensionless,the dimension of $k_1$ must be the inverse of the dimension of $x$ (distance).
$[k_1] = [x]^{-1} = L^{-1}$. Thus,the unit of $k_1$ is $meter^{-1}$.
For $k_2t$ to be dimensionless,the dimension of $k_2$ must be the inverse of the dimension of $t$ (time).
$[k_2] = [t]^{-1} = T^{-1}$. Thus,the unit of $k_2$ is $s^{-1}$.
Therefore,the units of $k_1$ and $k_2$ are $meter^{-1}$ and $s^{-1}$ respectively.

(C) Dimensional analysis is a powerful tool for checking the consistency of physical equations.
$1$. If an equation is dimensionally incorrect,it must be physically incorrect because the units on both sides do not match.
$2$. If an equation is dimensionally correct,it does not guarantee that the equation is physically correct,as it may lack dimensionless constants or factors.
$3$. Therefore,a dimensionally correct formula may be correct or incorrect,but a dimensionally incorrect formula is always incorrect.
$4$. Thus,the statement '$A$ dimensionally incorrect formula may be correct' is false.

(B) The given equation is $P = P_0 \exp(-\alpha t^2)$.
In any exponential function of the form $e^x$,the exponent $x$ must be dimensionless.
Therefore,the term $\alpha t^2$ must be dimensionless.
This implies that the dimensions of $\alpha$ multiplied by the dimensions of $t^2$ must equal $1$ (dimensionless).
$[\alpha] [t^2] = [M^0 L^0 T^0] = 1$.
Since $[t] = [T]$,we have $[\alpha] [T^2] = 1$.
Thus,$[\alpha] = [T^{-2}]$.

(A) According to the principle of homogeneity of dimensions,the dimensions of terms added together must be the same. Therefore,the dimension of $\beta$ must be equal to the dimension of $\sqrt{d}$.
$[\beta] = [d]^{1/2} = [M^1 L^{-3} T^0]^{1/2} = [M^{1/2} L^{-3/2} T^0]$.
Now,for the force equation: $F = \frac{\alpha}{\beta + \sqrt{d}}$.
The dimension of the denominator $(\beta + \sqrt{d})$ is the same as $[d]^{1/2} = [M^{1/2} L^{-3/2} T^0]$.
Thus,$[F] = \frac{[\alpha]}{[M^{1/2} L^{-3/2} T^0]}$.
$[\alpha] = [F] \times [M^{1/2} L^{-3/2} T^0] = [M^1 L^1 T^{-2}] \times [M^{1/2} L^{-3/2} T^0] = [M^{3/2} L^{-1/2} T^{-2}]$.
Comparing these results with the given options,option $A$ is correct.

(C) The dimensional formula for gravitational acceleration $g$ is $[M^0 L^1 T^{-2}]$.
In the $C.G.S.$ system,the unit of length is $cm$ and time is $s$. Given $g = 980 \; cm/s^2$.
To convert to the $M.K.S.$ system,we use the conversion factors:
$1 \; m = 10^2 \; cm \Rightarrow 1 \; cm = 10^{-2} \; m$
$1 \; s = 1 \; s$
Substituting these into the expression:
$g = 980 \times (1 \; cm) / (1 \; s)^2$
$g = 980 \times (10^{-2} \; m) / (1 \; s)^2$
$g = 980 \times 10^{-2} \; m/s^2$
$g = 9.8 \; m/s^2$.

(C) The formula for the relative error in $w$ is given by the propagation of errors rule:
$\frac{\Delta w}{w} = 4 \frac{\Delta a}{a} + 3 \frac{\Delta b}{b} + 2 \frac{\Delta c}{c} + \frac{1}{2} \frac{\Delta d}{d}$
Given percentage errors are:
$\frac{\Delta a}{a} \times 100 = 1\%$
$\frac{\Delta b}{b} \times 100 = 2\%$
$\frac{\Delta c}{c} \times 100 = 3\%$
$\frac{\Delta d}{d} \times 100 = 4\%$
Substituting these values into the error equation:
$\frac{\Delta w}{w} \times 100 = 4(1\%) + 3(2\%) + 2(3\%) + \frac{1}{2}(4\%)$
$= 4\% + 6\% + 6\% + 2\%$
$= 18\%$
Therefore,the percentage error in $w$ is $18\%$.

(A) According to the principle of homogeneity of dimensions,the dimensions of each term in the equation must be the same.
$1$. For the term $at$: $[at] = [v] = [LT^{-1}]$.
Therefore,$[a] = \frac{[LT^{-1}]}{[T]} = [LT^{-2}]$.
$2$. For the term $\frac{b}{t + c}$: Since $c$ is added to $t$,the dimension of $c$ must be the same as the dimension of $t$.
Therefore,$[c] = [T]$.
$3$. For the term $\frac{b}{t + c}$: The dimension of the whole term must be equal to the dimension of velocity $[v]$.
$[\frac{b}{t + c}] = [v] \implies \frac{[b]}{[T]} = [LT^{-1}]$.
Therefore,$[b] = [LT^{-1}] \times [T] = [L]$.
Thus,the dimensions are $[a] = [LT^{-2}]$,$[b] = [L]$,and $[c] = [T]$.

(C) According to the principle of homogeneity of dimensions,the dimensions of each term in a physical equation must be the same.
Since $x$ represents distance,its dimension is $[L]$.
The term $b/t^2$ must also have the dimension of distance $[L]$.
Therefore,$[b/t^2] = [L]$.
$[b] = [L] \times [t^2] = [L][T^2]$.
Since $x$ is in metres $(m)$ and $t$ is in seconds $(s)$,the unit of $b$ is $m-s^2$.

(B) We know that the resonant frequency of an $LC$ circuit is given by $f = \frac{1}{2\pi \sqrt{LC}}$.
Therefore,the dimensions of $LC$ are $[LC] = [T^{2}]$.
Also,the time constant of an $LR$ circuit is $\tau = \frac{L}{R}$,so the dimensions of $\frac{L}{R}$ are $[T]$.
This implies that the dimensions of $\frac{R}{L}$ are $[T^{-1}]$.
Now,we need to find the dimensions of $C^{2}LR$.
We can rewrite the expression as $C^{2}LR = (LC) \cdot (RC)$.
Since $RC$ is the time constant of an $RC$ circuit,its dimensions are $[T]$.
Thus,$[C^{2}LR] = [LC] \cdot [RC] = [T^{2}] \cdot [T] = [T^{3}]$.
Alternatively,$[C^{2}LR] = [LC] \cdot [LC] \cdot [R/L] = [T^{2}] \cdot [T^{2}] \cdot [T^{-1}] = [T^{3}]$.
Therefore,the dimensional formula is $[M^{0}L^{0}T^{3}I^{0}]$.

(B) According to the principle of homogeneity of dimensions,only quantities with the same dimensions can be added or subtracted.
In the denominator,$x^2$ is added to $B$,so $[B] = [x^2] = [L^2]$.
The energy $U$ has dimensions $[M L^2 T^{-2}]$.
The equation is $U = \frac{A\sqrt{x}}{x^2 + B}$.
Substituting the dimensions: $[M L^2 T^{-2}] = \frac{[A][L^{1/2}]}{[L^2]}$.
Solving for $[A]$: $[A] = [M L^2 T^{-2}] \times \frac{[L^2]}{[L^{1/2}]} = [M L^{2+2-1/2} T^{-2}] = [M L^{7/2} T^{-2}]$.
Now,calculate the dimensions of $AB$: $[AB] = [A][B] = [M L^{7/2} T^{-2}] \times [L^2] = [M L^{7/2+2} T^{-2}] = [M L^{11/2} T^{-2}]$.

(D) In the equation $y = 2A \sin \left( \frac{2\pi ct}{\lambda} \right) \cos \left( \frac{2\pi x}{\lambda} \right)$,the arguments of the trigonometric functions must be dimensionless.
Therefore,$\left[ \frac{2\pi ct}{\lambda} \right] = M^0 L^0 T^0$ and $\left[ \frac{2\pi x}{\lambda} \right] = M^0 L^0 T^0$.
From this,we conclude that the units of $ct$ and $\lambda$ are the same,and the units of $x$ and $\lambda$ are the same.
Checking option $C$: The unit of $\frac{2\pi c}{\lambda}$ is $T^{-1}$ (frequency),while the unit of $\frac{2\pi x}{\lambda t}$ is $\frac{L}{L \cdot T} = T^{-1}$. Thus,they have the same units.
Checking option $D$: The unit of $\frac{c}{\lambda}$ is $T^{-1}$,while the unit of $\frac{x}{\lambda}$ is dimensionless $(L/L = 1)$. Since $T^{-1} \neq 1$,statement $D$ is not true.

(C) The dimensional formula for Young's modulus $Y$ is $[M L^{-1} T^{-2}]$.
In $CGS$ system,the unit is $g \cdot cm^{-1} \cdot s^{-2}$.
In $MKS$ system,the unit is $kg \cdot m^{-1} \cdot s^{-2}$.
Using the conversion formula $n_2 = n_1 [M_1/M_2]^1 [L_1/L_2]^{-1} [T_1/T_2]^{-2}$,where $n_1$ is the value in $CGS$ and $n_2$ is the value in $MKS$:
$n_2/n_1 = [1 \ g / 10^3 \ g]^1 \cdot [1 \ cm / 10^2 \ cm]^{-1} \cdot [1 \ s / 1 \ s]^{-2}$.
$n_2/n_1 = [10^{-3}]^1 \cdot [10^{-2}]^{-1} \cdot [1]^{-2}$.
$n_2/n_1 = 10^{-3} \cdot 10^2 = 10^{-1} = 0.1$.
Thus,to convert from $CGS$ to $MKS$,we multiply by $0.1$.

(A) The dimensional formula for power is $[P] = [M^1 L^2 T^{-3}]$.
We use the formula for conversion between two systems: $n_2 = n_1 [M_1/M_2]^a [L_1/L_2]^b [T_1/T_2]^c$.
Here,$n_1 = 10^6 \; W$,$M_1 = 1 \; kg$,$L_1 = 1 \; m$,$T_1 = 1 \; s$.
In the new system,$M_2 = 10 \; kg$,$L_2 = 1 \; dm = 0.1 \; m$,$T_2 = 1 \; \text{minute} = 60 \; s$.
Substituting the values: $n_2 = 10^6 \times [1 \; kg / 10 \; kg]^1 \times [1 \; m / 0.1 \; m]^2 \times [1 \; s / 60 \; s]^{-3}$.
$n_2 = 10^6 \times [1/10] \times [10]^2 \times [1/60]^{-3}$.
$n_2 = 10^6 \times 0.1 \times 100 \times 60^3$.
$n_2 = 10^7 \times 216000 = 2.16 \times 10^{12} \; \text{units}$.

(B) Given the relations:
$v_2 = v_1 \frac{\alpha^2}{\beta} \Rightarrow [L_2 T_2^{-1}] = [L_1 T_1^{-1}] \frac{\alpha^2}{\beta} \dots (i)$
$a_2 = a_1 \alpha \beta \Rightarrow [L_2 T_2^{-2}] = [L_1 T_1^{-2}] \alpha \beta \dots (ii)$
$F_2 = \frac{F_1}{\alpha \beta} \Rightarrow [M_2 L_2 T_2^{-2}] = [M_1 L_1 T_1^{-2}] \frac{1}{\alpha \beta} \dots (iii)$
Dividing equation $(iii)$ by equation $(ii)$:
$\frac{[M_2 L_2 T_2^{-2}]}{[L_2 T_2^{-2}]} = \frac{[M_1 L_1 T_1^{-2}]}{[L_1 T_1^{-2}]} \times \frac{1}{(\alpha \beta)(\alpha \beta)} \Rightarrow M_2 = \frac{M_1}{\alpha^2 \beta^2}$
Dividing equation $(ii)$ by equation $(i)$:
$\frac{[L_2 T_2^{-2}]}{[L_2 T_2^{-1}]} = \frac{[L_1 T_1^{-2}]}{[L_1 T_1^{-1}]} \times \frac{\alpha \beta}{\alpha^2 / \beta} \Rightarrow T_2^{-1} = T_1^{-1} \frac{\beta^2}{\alpha} \Rightarrow T_2 = T_1 \frac{\alpha}{\beta^2}$
Substituting $T_2$ into equation $(i)$:
$L_2 (T_1 \frac{\alpha}{\beta^2})^{-1} = L_1 T_1^{-1} \frac{\alpha^2}{\beta} \Rightarrow L_2 = L_1 \frac{\alpha^2}{\beta} \cdot \frac{\alpha}{\beta^2} = L_1 \frac{\alpha^3}{\beta^3}$
Thus,$M_2 = \frac{1}{\alpha^2 \beta^2} M_1, L_2 = \frac{\alpha^3}{\beta^3} L_1, T_2 = T_1 \frac{\alpha}{\beta^2}$.

(B) Let the original units be $L_1 = 1 \; m$,$T_1 = 1 \; s$,$M_1 = 1 \; kg$. The new units are $L_2 = 100 \; m$,$T_2 = 100 \; s$,$M_2 = 0.1 \; kg$.
$1$. Velocity $(v = L T^{-1})$: New unit $v_2 = L_2 T_2^{-1} = 100 \; m / 100 \; s = 1 \; m/s$. Thus,velocity remains the same.
$2$. Force $(F = M L T^{-2})$: New unit $F_2 = M_2 L_2 T_2^{-2} = (0.1 \; kg) \times (100 \; m) / (100 \; s)^2 = 0.1 \times 100 / 10000 = 0.001 \; N = \frac{1}{1000} \; N$. The new unit of force is $\frac{1}{1000}$ times the original.
$3$. Energy $(E = M L^2 T^{-2})$: New unit $E_2 = M_2 L_2^2 T_2^{-2} = (0.1 \; kg) \times (100 \; m)^2 / (100 \; s)^2 = 0.1 \times 10000 / 10000 = 0.1 \; J = \frac{1}{10} \; J$.
$4$. Pressure $(P = M L^{-1} T^{-2})$: New unit $P_2 = M_2 L_2^{-1} T_2^{-2} = (0.1 \; kg) / (100 \; m \times (100 \; s)^2) = 0.1 / (100 \times 10000) = 0.1 / 10^6 = 10^{-7} \; Pa$.

(D) Let the relation be $m = K \rho^a v^b g^c$,where $K$ is a dimensionless constant.
Dimensional formulas are:
$m = [M^1 L^0 T^0]$
$\rho = [M^1 L^{-3} T^0]$
$v = [M^0 L^1 T^{-1}]$
$g = [M^0 L^1 T^{-2}]$
Substituting these into the equation:
$[M^1 L^0 T^0] = [M^1 L^{-3} T^0]^a [M^0 L^1 T^{-1}]^b [M^0 L^1 T^{-2}]^c$
$[M^1 L^0 T^0] = [M^a L^{-3a+b+c} T^{-b-2c}]$
Comparing powers of $M, L,$ and $T$:
For $M$: $a = 1$
For $T$: $-b - 2c = 0 \Rightarrow b = -2c$
For $L$: $-3a + b + c = 0$
Substituting $a=1$ and $b=-2c$ into the $L$ equation:
$-3(1) + (-2c) + c = 0$
$-3 - c = 0 \Rightarrow c = -3$
Now,$b = -2(-3) = 6$
Thus,$m \propto \rho^1 v^6 g^{-3} = \frac{\rho v^6}{g^3}$.

(A) Given the formula $P = \frac{a^3 b^2}{cd}$.
To find the percentage error in $P$,we use the formula for propagation of errors:
$\frac{\Delta P}{P} \times 100 = \left( 3 \frac{\Delta a}{a} + 2 \frac{\Delta b}{b} + \frac{\Delta c}{c} + \frac{\Delta d}{d} \right) \times 100$
Substituting the given percentage errors:
$\frac{\Delta P}{P} \times 100 = [3(1\%) + 2(2\%) + 3\% + 4\%]$
$= [3\% + 4\% + 3\% + 4\%]$
$= 14\%$
Thus,the percentage error in $P$ is $14\%$.

(D) Let mass $m \propto F^a V^b T^c$.
Then $m = k F^a V^b T^c$,where $k$ is a dimensionless constant.
Writing dimensions on both sides:
$[M L^0 T^0] = [M L T^{-2}]^a [L T^{-1}]^b [T]^c$
$[M L^0 T^0] = [M^a L^{a+b} T^{-2a-b+c}]$
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 $T$: $-2a - b + c = 0 \implies -2(1) - (-1) + c = 0 \implies -2 + 1 + c = 0 \implies c = 1$
Substituting the values of $a, b,$ and $c$ in the equation,we get:
$[m] = [F^1 V^{-1} T^1] = [F V^{-1} T]$.

(C) Let the dimensional formula of surface tension $(S)$ be $S = E^x V^y T^z$.
The dimensions of surface tension are $[S] = [MT^{-2}]$.
The dimensions of the fundamental quantities are:
$[E] = [ML^2T^{-2}]$
$[V] = [LT^{-1}]$
$[T] = [T]$
Substituting these into the equation:
$[MT^{-2}] = [ML^2T^{-2}]^x [LT^{-1}]^y [T]^z$
$[MT^{-2}] = [M^x L^{2x+y} T^{-2x-y+z}]$
Comparing the powers of $M, L,$ and $T$ on both sides:
For $M$: $x = 1$
For $L$: $2x + y = 0 \Rightarrow 2(1) + y = 0 \Rightarrow y = -2$
For $T$: $-2x - y + z = -2 \Rightarrow -2(1) - (-2) + z = -2 \Rightarrow -2 + 2 + z = -2 \Rightarrow z = -2$
Thus,the dimensional formula is $[EV^{-2}T^{-2}]$.

(B) Given the dimensional relation: $[v_c] = [\eta^x \rho^y r^z]$ $(i)$
Writing the dimensions of the physical quantities:
$[v_c] = [M^0 L T^{-1}]$
$[\eta] = [M L^{-1} T^{-1}]$
$[\rho] = [M L^{-3} T^0]$
$[r] = [M^0 L T^0]$
Substituting these into equation $(i)$:
$[M^0 L T^{-1}] = [M L^{-1} T^{-1}]^x [M L^{-3} T^0]^y [M^0 L T^0]^z$
$[M^0 L T^{-1}] = [M^{x+y} L^{-x-3y+z} T^{-x}]$
Applying the principle of homogeneity of dimensions,we equate the powers of $M, L,$ and $T$ on both sides:
$x + y = 0$ $(ii)$
$-x - 3y + z = 1$ $(iii)$
$-x = -1$ $(iv)$
From $(iv)$,we get $x = 1$.
Substituting $x = 1$ into $(ii)$,we get $1 + y = 0$,so $y = -1$.
Substituting $x = 1$ and $y = -1$ into $(iii)$,we get $-1 - 3(-1) + z = 1 \implies -1 + 3 + z = 1 \implies 2 + z = 1 \implies z = -1$.
Thus,the values are $x = 1, y = -1, z = -1$.

(C) Let the dimension of length $l$ be expressed as $l \propto h^p c^q G^r$.
Substituting the dimensions of each constant:
$[L] = [ML^2T^{-1}]^p [LT^{-1}]^q [M^{-1}L^3T^{-2}]^r$
$[M^0 L^1 T^0] = M^{p-r} L^{2p+q+3r} T^{-p-q-2r}$
Comparing the powers of $M, L,$ and $T$ on both sides:
$p - r = 0 \implies p = r$
$2p + q + 3r = 1$
$-p - q - 2r = 0$
Substituting $p = r$ into the third equation: $-r - q - 2r = 0 \implies q = -3r$.
Substituting $p = r$ and $q = -3r$ into the second equation: $2r - 3r + 3r = 1 \implies 2r = 1 \implies r = 1/2$.
Thus,$p = 1/2$ and $q = -3/2$.
Therefore,$l \propto h^{1/2} c^{-3/2} G^{1/2} = \sqrt{\frac{hG}{c^3}} = \frac{\sqrt{hG}}{c^{3/2}}$.

(C) Let the physical quantity of length $l$ be expressed as $l = k \left( \frac{e^2}{4\pi \varepsilon_0} \right)^p G^q c^r$.
Dimensions of $\frac{e^2}{4\pi \varepsilon_0} = [F \cdot d^2] = [ML^3T^{-2}]$.
Dimensions of $G = [M^{-1}L^3T^{-2}]$.
Dimensions of $c = [LT^{-1}]$.
Equating dimensions: $[L^1] = [ML^3T^{-2}]^p [M^{-1}L^3T^{-2}]^q [LT^{-1}]^r$.
Comparing powers of $M$: $p - q = 0 \implies p = q$.
Comparing powers of $T$: $-2p - 2q - r = 0 \implies -4p = r$.
Comparing powers of $L$: $3p + 3q + r = 1 \implies 6p - 4p = 1 \implies 2p = 1 \implies p = 1/2$.
Thus,$q = 1/2$ and $r = -2$.
Therefore,$l = \frac{1}{c^2} \sqrt{\frac{Ge^2}{4\pi \varepsilon_0}}$.

(A) The dimensional formulas for the given quantities are:
$[E] = [ML^2T^{-2}]$
$[m] = [M]$
$[l] = [ML^2T^{-1}]$
$[G] = [M^{-1}L^3T^{-2}]$
Substituting these dimensions into the expression $\frac{El^2}{m^5G^2}$:
$\frac{[ML^2T^{-2}] \cdot [ML^2T^{-1}]^2}{[M]^5 \cdot [M^{-1}L^3T^{-2}]^2} = \frac{[ML^2T^{-2}] \cdot [M^2L^4T^{-2}]}{[M^5] \cdot [M^{-2}L^6T^{-4}]} = \frac{[M^3L^6T^{-4}]}{[M^3L^6T^{-4}]} = [M^0L^0T^0]$
Since the dimensions are $[M^0L^0T^0]$,the quantity is dimensionless. Among the given options,an angle is a dimensionless quantity.

(C) The dimension of $\tan \theta$ is $[M^0 L^0 T^0]$ as it is a dimensionless quantity.
For the right-hand side,the dimensions are: $[r] = [L]$,$[g] = [L T^{-2}]$,and $[v^2] = [L^2 T^{-2}]$.
Thus,the dimension of $\frac{rg}{v^2}$ is $\frac{[L] [L T^{-2}]}{[L^2 T^{-2}]} = [M^0 L^0 T^0]$.
Since both sides have the same dimensions,the equation is dimensionally correct.
However,the correct physical formula for the angle of banking is $\tan \theta = \frac{v^2}{rg}$.
Therefore,the given equation is numerically incorrect.

(C) According to the principle of dimensional homogeneity,the dimensions of each term in an equation must be the same.
$1$. For the term $c$: Since $c$ is added to $t$ (time),the dimension of $c$ must be the same as the dimension of $t$. Therefore,$[c] = [T]$.
$2$. For the term $at$: The dimension of $at$ must be equal to the dimension of velocity $v$. Thus,$[a][T] = [LT^{-1}]$,which gives $[a] = [LT^{-2}]$.
$3$. For the term $\frac{b}{t+c}$: The dimension of the entire term must be equal to the dimension of velocity $v$. Since $[t+c] = [T]$,we have $\frac{[b]}{[T]} = [LT^{-1}]$. Therefore,$[b] = [L]$.
Thus,the dimensions are $a = LT^{-2}, b = L, c = T$.

(A) To check the dimensional consistency of the equation $T = 2\pi \sqrt {\frac{{{R^3}}}{{GM}}} $:
$1$. The dimension of time period $T$ is $[T]$.
$2$. The dimension of radius $R$ is $[L]$. So,$R^3$ has dimensions $[L^3]$.
$3$. The gravitational constant $G$ has dimensions $[M^{-1}L^3T^{-2}]$.
$4$. The mass $M$ has dimensions $[M]$.
$5$. The product $GM$ has dimensions $[M^{-1}L^3T^{-2}] \times [M] = [L^3T^{-2}]$.
$6$. Substituting these into the expression under the square root: $\sqrt{\frac{R^3}{GM}} = \sqrt{\frac{L^3}{L^3T^{-2}}} = \sqrt{T^2} = [T]$.
$7$. Since the dimensions of the left-hand side and right-hand side match,the equation is dimensionally correct.

(A) The argument of an exponential function must be dimensionless. Therefore,the dimension of $\alpha t$ must be $[M^0 L^0 T^0]$.
Since $[t] = [T]$,we have $[\alpha] [T] = [1]$,which gives $[\alpha] = [T^{-1}]$.
In the expression $x(t) = (v_0 / \alpha) (1 - e^{-\alpha t})$,the term $(1 - e^{-\alpha t})$ is dimensionless.
Therefore,the dimension of $x(t)$ must be equal to the dimension of $(v_0 / \alpha)$.
$[x] = [L]$,so $[v_0 / \alpha] = [L]$.
$[v_0] = [L] \times [\alpha] = [L] \times [T^{-1}] = [L T^{-1}]$.
Thus,the dimensions of $v_0$ and $\alpha$ are $[M^0 L^1 T^{-1}]$ and $[T^{-1}]$ respectively.

(A) According to the principle of dimensional homogeneity,the dimensions of each term added or subtracted in an equation must be the same.
In the given equation $(P + \frac{a}{V^2}) = \frac{b\theta}{l}$,the term $P$ is added to $\frac{a}{V^2}$.
Therefore,$[P] = [\frac{a}{V^2}]$.
We know that the dimensional formula for pressure $P$ is $[ML^{-1}T^{-2}]$ and for volume $V$ is $[L^3]$.
Thus,$[a] = [P] \times [V^2]$.
Substituting the dimensions: $[a] = [ML^{-1}T^{-2}] \times [L^3]^2$.
$[a] = [ML^{-1}T^{-2}] \times [L^6]$.
$[a] = [ML^5T^{-2}]$.

(A) To determine the correct expression,we use dimensional analysis.
The dimension of speed $v$ is $[LT^{-1}]$.
For option $(A)$,the dimension of $\sqrt{g\lambda}$ is $\sqrt{[LT^{-2}] \cdot [L]} = \sqrt{[L^2T^{-2}]} = [LT^{-1}]$,which matches the dimension of speed.
For option $(B)$,the dimension of $\sqrt{g/h}$ is $\sqrt{[LT^{-2}] / [L]} = \sqrt{[T^{-2}]} = [T^{-1}]$,which is incorrect.
For option $(C)$,the dimension of $\sqrt{\rho gh}$ is $\sqrt{[ML^{-3}] \cdot [LT^{-2}] \cdot [L]} = \sqrt{[MT^{-2}]}$,which is incorrect.
For option $(D)$,the dimension of $\sqrt{g/\rho}$ is $\sqrt{[LT^{-2}] / [ML^{-3}]} = \sqrt{[M^{-1}L^4T^{-2}]}$,which is incorrect.
Therefore,the only dimensionally consistent expression is $\sqrt{g\lambda}$.

(D) The kinetic energy per unit mass $(KE/m)$ has the dimensions of velocity squared,which is $[L^2 T^{-2}]$.
We analyze the dimensions of the given options:
For option $D$,the expression is $\frac{\alpha \sigma}{\rho}$.
The dimensions of tensile strength $\sigma$ (pressure) are $[M L^{-1} T^{-2}]$.
The dimensions of density $\rho$ are $[M L^{-3}]$.
Therefore,the dimensions of $\frac{\sigma}{\rho}$ are $\frac{[M L^{-1} T^{-2}]}{[M L^{-3}]} = [L^2 T^{-2}]$.
Since $\alpha$ is a dimensionless constant,the expression $\frac{\alpha \sigma}{\rho}$ correctly represents the dimensions of kinetic energy per unit mass.

(D) In the expression $\cos \left( \frac{t}{p} - qx \right)$,the argument of the trigonometric function must be dimensionless.
Therefore,the dimensions of $\frac{t}{p}$ must be equal to the dimensions of $1$ (dimensionless).
This implies that the dimensions of $t$ must be equal to the dimensions of $p$.
Similarly,the dimensions of $qx$ must be equal to the dimensions of $1$,meaning $q$ has the dimensions of $1/x$ (inverse of length).
Thus,the unit of $t$ is the same as the unit of $p$.

(B) Given the equation: $\frac{dV}{dt} = At - BV$
According to the principle of homogeneity of dimensions,the dimensions of each term on both sides of the equation must be the same.
The dimension of $\frac{dV}{dt}$ (acceleration) is $[LT^{-2}]$.
For the term $At$:
$[At] = [LT^{-2}]$
$[A] [T] = [LT^{-2}]$
$[A] = [LT^{-2}] / [T] = [LT^{-3}]$
For the term $BV$:
$[BV] = [LT^{-2}]$
$[B] [LT^{-1}] = [LT^{-2}]$
$[B] = [LT^{-2}] / [LT^{-1}] = [T^{-1}]$
Therefore,the dimensions of $A$ and $B$ are $[LT^{-3}]$ and $[T^{-1}]$ respectively.

(C) The dimensional formula for time period $T$ is $[M^0 L^0 T^1]$.
The dimensional formulas for the given physical quantities are:
Pressure $P = [M L^{-1} T^{-2}]$
Density $d = [M L^{-3}]$
Energy $E = [M L^2 T^{-2}]$
Given the relation $T = k P^a d^b E^c$,we write the dimensional 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 powers of $M, L,$ and $T$ on both sides:
$1$) $a + b + c = 0$
$2$) $-a - 3b + 2c = 0$
$3$) $-2a - 2c = 1$
From equation $(3)$,$-2(a + c) = 1$,so $a + c = -1/2$.
Substitute $a + c = -1/2$ into equation $(1)$: $(-1/2) + b = 0$,which gives $b = 1/2$.
Substitute $b = 1/2$ into equation $(2)$: $-a - 3(1/2) + 2c = 0 \Rightarrow -a + 2c = 3/2$.
Now we have a system of two equations:
$a + c = -1/2$
$-a + 2c = 3/2$
Adding these two equations: $3c = 1$,so $c = 1/3$.

(A) According to the principle of homogeneity of dimensions,physical quantities can only be added or subtracted if they have the same dimensions.
Since $A$ and $B$ have unequal dimensional formulas,the operation $(A - B)$ is physically meaningless and not possible.
In the expression $log(A - B)$,the argument of the logarithm must be dimensionless,but the subtraction $(A - B)$ itself is invalid because the dimensions are unequal.
Therefore,the operation $log(A - B)$ is not possible.

(A) According to the principle of dimensional homogeneity,physical quantities can only be added or subtracted if they have the same dimensions.
However,multiplication and division of physical quantities are permissible regardless of whether their dimensions are the same or different.
Therefore,the operation $\frac{A}{B}$ can be performed even if $A$ and $B$ have different dimensions.

(B) The dimensional formulas are:
Force $F = [MLT^{-2}] = 10\, N \dots (i)$
Energy $E = [ML^2T^{-2}] = 100\, J \dots (ii)$
Velocity $v = [LT^{-1}] = 5\, m/s \dots (iii)$
Dividing equation $(ii)$ by equation $(i)$:
$\frac{[ML^2T^{-2}]}{[MLT^{-2}]} = \frac{100\, J}{10\, N} \implies [L] = 10\, m$
Substituting $[L] = 10\, m$ into equation $(iii)$:
$10\, m / [T] = 5\, m/s \implies [T] = \frac{10\, m}{5\, m/s} = 2\, s$
Substituting $[L] = 10\, m$ and $[T] = 2\, s$ into equation $(i)$:
$[M][10\, m][2\, s]^{-2} = 10\, N$
$[M][10\, m][0.25\, s^{-2}] = 10\, kg\cdot m/s^2$
$[M] = \frac{10}{2.5} = 4\, kg$
Thus,the units are length $= 10\, m$,mass $= 4\, kg$,and time $= 2\, s$.

(B) Energy $(E)$ is defined as the product of force $(F)$ and displacement $(L)$.
$E = F \times L$
We know that velocity $(V)$ is displacement divided by time $(T)$,so $V = L/T$,which implies $L = V \times T$.
Substituting this into the energy equation:
$E = F \times (V \times T)$
$E = [V^1 F^1 T^1]$
Therefore,the dimensions of energy in terms of velocity,force,and time are $[V^1 F^1 T^1]$.

(A) The dimensional formula for force is $[F] = [MLT^{-2}]$.
Given basic dimensions are density $[D] = [ML^{-3}]$,velocity $[V] = [LT^{-1}]$,and area $[A] = [L^2]$.
Let force be represented as $[F] = [A^x V^y D^z]$.
Substituting the dimensions:
$[MLT^{-2}] = [L^2]^x [LT^{-1}]^y [ML^{-3}]^z$
$[MLT^{-2}] = [M^z L^{2x+y-3z} T^{-y}]$
Comparing the powers of $M, L,$ and $T$ on both sides:
For $M$: $z = 1$
For $T$: $-y = -2 \implies y = 2$
For $L$: $2x + y - 3z = 1$
Substituting $y=2$ and $z=1$ into the equation for $L$:
$2x + 2 - 3(1) = 1$
$2x - 1 = 1$
$2x = 2 \implies x = 1$
Thus,the force is represented as $[A^1 V^2 D^1]$ or $[AV^2D]$.

(D) The formula for conversion between two systems of units is $n_2 = n_1 \left[ \frac{M_1}{M_2} \right]^a \left[ \frac{L_1}{L_2} \right]^b \left[ \frac{T_1}{T_2} \right]^c$.
Density has dimensions $[M^1 L^{-3} T^0]$,so $a=1, b=-3, c=0$.
Given: $n_1 = 4$,$M_1 = 1\,g$,$L_1 = 1\,cm$.
New system: $M_2 = 100\,g$,$L_2 = 10\,cm$.
Substituting the values: $n_2 = 4 \left[ \frac{1\,g}{100\,g} \right]^1 \left[ \frac{1\,cm}{10\,cm} \right]^{-3}$.
$n_2 = 4 \times \left( \frac{1}{100} \right) \times (10)^3$.
$n_2 = 4 \times 0.01 \times 1000 = 40$.
Thus,the value of density in the new system is $40$.

(D) The dimensional formula for energy is $[M L^2 T^{-2}]$.
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: $M_2 = 10 \, kg$,$L_2 = 10 \, m$,$T_2 = 1 \, minute = 60 \, s$.
Using the conversion formula $n_2 = n_1 [M_1/M_2]^1 [L_1/L_2]^2 [T_1/T_2]^{-2}$:
$n_2 = 1 \times [1 \, kg / 10 \, kg]^1 \times [1 \, m / 10 \, m]^2 \times [1 \, s / 60 \, s]^{-2}$
$n_2 = (1/10) \times (1/10)^2 \times (1/60)^{-2}$
$n_2 = (1/10) \times (1/100) \times (60)^2$
$n_2 = (1/1000) \times 3600 = 3.6 = 36 \times 10^{-1}$.

(B) The mass flow rate per unit area is given by $\frac{M}{A t} \propto P^{x} v^{y}$.
The dimensional formula for mass per unit area per unit time is $[M L^{-2} T^{-1}]$.
The dimensional formula for pressure $P$ is $[M L^{-1} T^{-2}]$ and for velocity $v$ is $[L T^{-1}]$.
Equating the dimensions on both sides:
$[M L^{-2} T^{-1}] = [M L^{-1} T^{-2}]^x [L T^{-1}]^y$
$[M L^{-2} T^{-1}] = M^x L^{-x+y} T^{-2x-y}$
Comparing the powers of $M, L,$ and $T$:
For $M$: $x = 1$
For $L$: $-x + y = -2$
For $T$: $-2x - y = -1$
Substituting $x = 1$ into the equation for $L$: $-1 + y = -2 \Rightarrow y = -1$.
Since $x = 1$ and $y = -1$,we have $x = -y$.

(D) The formula for pressure is $P = \frac{\text{Force}}{\text{Area}}$.
Since $\text{Force} = F$,we need to express $\text{Area}$ in terms of $F, V,$ and $T$.
We know that $\text{Velocity} (V) = \frac{\text{Distance}}{\text{Time}}$,so $\text{Distance} = V \times T$.
Therefore,$\text{Area} = (\text{Distance})^2 = (V \times T)^2 = V^2 T^2$.
Substituting these into the pressure formula: $P = \frac{F}{V^2 T^2} = F^1 V^{-2} T^{-2}$.
The dimension of force in pressure is the exponent of $F$,which is $1$.

(B) The argument of an exponential function must be dimensionless.
Therefore,the dimensions of the exponent $[\alpha t^2]$ must be equal to the dimensions of a dimensionless quantity,which is $1$.
$[\alpha t^2] = [M^0 L^0 T^0] = 1$.
Since $t$ represents time,its dimension is $[T]$.
Thus,$[\alpha] [T^2] = 1$.
$[\alpha] = \frac{1}{[T^2]} = [T^{-2}]$.
Therefore,the constant $\alpha$ has the dimensions of $T^{-2}$.

(A) dimensional constant is a physical quantity that has a constant value and possesses dimensions.
Gravitational constant $(G)$ is a universal constant with a value of $6.67 \times 10^{-11} \ N \ m^2 \ kg^{-2}$ and its dimensional formula is $[M^{-1} L^3 T^{-2}]$.
Relative density,refractive index,and Poisson ratio are ratios of similar physical quantities,which makes them dimensionless constants.

(D) The speed of light in vacuum is given by the relation $C = \frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$.
Squaring both sides,we get $C^2 = \frac{1}{\mu_{0} \varepsilon_{0}}$.
Therefore,the product $\mu_{0} \varepsilon_{0} = \frac{1}{C^2}$.
Since $C$ represents velocity,the dimensions of the product $\mu_{0} \varepsilon_{0}$ are related to velocity as $1/(velocity)^2$.