Dimensions and Dimensional Formula Questions in English

(C) The dimensions of a physical quantity are given as $[L^{a} M^{b} T^{c}]$.
Let us check the dimensions of the given options:
$1$. Velocity: $[M^{0} L^{1} T^{-1}]$. Here,$a=1, b=0, c=-1$. Option $A$ is incorrect.
$2$. Force: $[M^{1} L^{1} T^{-2}]$. Here,$a=1, b=1, c=-2$. Option $B$ is incorrect.
$3$. Pressure: Pressure is defined as $\text{Force} / \text{Area} = [M^{1} L^{1} T^{-2}] / [L^{2}] = [M^{1} L^{-1} T^{-2}]$. Here,$a=-1, b=1, c=-2$. This matches option $C$.
$4$. Acceleration: $[M^{0} L^{1} T^{-2}]$. Here,$a=1, b=0, c=-2$. Option $D$ is incorrect.
Therefore,the correct option is $C$.

(C) The correct option is $C$.
From the relation $E = h \nu$,where $E$ is energy,$h$ is Planck's constant,and $\nu$ is frequency.
$h = \frac{E}{\nu}$.
The dimension of energy $E$ is $[ML^2 T^{-2}]$.
The dimension of frequency $\nu$ is $[T^{-1}]$.
Therefore,the dimension of $h = \frac{[ML^2 T^{-2}]}{[T^{-1}]} = [ML^2 T^{-1}]$.
Now,let us check the dimensions of the product of force,displacement,and time:
Dimension of force $F = [MLT^{-2}]$.
Dimension of displacement $d = [L]$.
Dimension of time $t = [T]$.
Product dimension $= [MLT^{-2}] \times [L] \times [T] = [ML^2 T^{-1}]$.
This matches the dimension of Planck's constant.

(A) The time constant of an $LR$ circuit is given by $\tau = \frac{L}{R}$.
Dimensional formula of inductance $L$ is $[M^{1} L^{2} T^{-2} A^{-2}]$.
Dimensional formula of resistance $R$ is $[M^{1} L^{2} T^{-3} A^{-2}]$.
Therefore,the dimensions of $\frac{L}{R}$ are:
$\frac{[M^{1} L^{2} T^{-2} A^{-2}]}{[M^{1} L^{2} T^{-3} A^{-2}]} = M^{1-1} L^{2-2} T^{-2-(-3)} A^{-2-(-2)} = M^{0} L^{0} T^{1} A^{0}$.
Thus,the dimensions of $\left(\frac{L}{R}\right)$ are $[L^{0} M^{0} T^{1}]$.

(B) In the given equation $F = P \cos(Ax) + Q \sin(Bt)$,the arguments of trigonometric functions must be dimensionless.
Therefore,the dimensions of $(Ax)$ must be $[M^0 L^0 T^0]$.
Since $[x] = [L]$,we have $[A][L] = [M^0 L^0 T^0]$,which implies $[A] = [L^{-1}]$.
Similarly,the dimensions of $(Bt)$ must be $[M^0 L^0 T^0]$.
Since $[t] = [T]$,we have $[B][T] = [M^0 L^0 T^0]$,which implies $[B] = [T^{-1}]$.
Now,calculating the dimensions of the ratio $\left[\frac{B}{A}\right]$:
$\left[\frac{B}{A}\right] = \frac{[T^{-1}]}{[L^{-1}]} = [L T^{-1}]$.
The dimension $[L T^{-1}]$ corresponds to the physical quantity of velocity.

(A) The dimension of torque is $[M^1 L^2 T^{-2}]$.
Torque is defined as the product of force and the perpendicular distance from the axis of rotation,given by $\tau = F \times r$.
The dimensional formula for force $(F)$ is $[M^1 L^1 T^{-2}]$.
The dimensional formula for distance $(r)$ is $[L^1]$.
Therefore,the dimensional formula for torque $(\tau)$ is $[M^1 L^1 T^{-2}] \times [L^1] = [M^1 L^2 T^{-2}]$.

(B) The gyromagnetic ratio $(\gamma)$ is defined as the ratio of the magnetic dipole moment $(M)$ to the angular momentum $(L)$ of a system.
$\gamma = \frac{M}{L}$
Since the magnetic moment $M = I \cdot A$ (where $I$ is current and $A$ is area),its dimensions are $[I^1 L^2]$.
Since angular momentum $L = mvr$,its dimensions are $[M^1 L^2 T^{-1}]$.
Therefore,the dimensions of the gyromagnetic ratio are:
$\text{Dimension} = \frac{[I^1 L^2]}{[M^1 L^2 T^{-1}]} = [M^{-1} L^0 T^1 I^1]$.

(A) The dimensional formula for torque is given by $\tau = r \times F$.
Since the dimensions of force $F$ are $[M^1L^1T^{-2}]$ and the dimensions of distance $r$ are $[L^1]$,the dimensions of torque are $[M^1L^1T^{-2}] \times [L^1] = [M^1L^2T^{-2}]$.
The moment of force is defined as the product of force and the perpendicular distance from the axis of rotation,which is identical to the definition of torque.
Therefore,the dimensions of the moment of force are also $[M^1L^2T^{-2}]$.
Thus,the dimensions of torque are the same as those of the moment of force.

(A) Given the equation $x = \pi R \left( \frac{P^2 - Q^2}{2} \right)$.
Here,$P, Q,$ and $R$ are lengths,so their dimensions are $[L]$.
The term $(P^2 - Q^2)$ has dimensions $[L^2] - [L^2] = [L^2]$.
The constant $\pi$ and the factor $1/2$ are dimensionless.
Therefore,the dimension of $x$ is $[L] \times [L^2] = [L^3]$.
Physical quantities with dimensions of $[L^3]$ represent volume.

(C) The dimensions of Planck's constant $h$ are given by $[h] = [M L^2 T^{-1}]$.
The dimensions of the moment of inertia $I$ are given by $[I] = [M L^2]$.
Taking the ratio of the dimensions of Planck's constant to the moment of inertia:
$\frac{[h]}{[I]} = \frac{[M L^2 T^{-1}]}{[M L^2]} = [T^{-1}]$.
The dimension $[T^{-1}]$ corresponds to the dimension of frequency.

(D) The refractive index $n$ of a medium is given by the ratio of the speed of light in vacuum $c$ to the speed of light in the medium $v$,i.e.,$n = \frac{c}{v}$.
We know that $c = \frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$ and $v = \frac{1}{\sqrt{\mu \varepsilon}} = \frac{1}{\sqrt{\mu_{0} \mu_{r} \varepsilon_{0} \varepsilon_{r}}}$.
Substituting these values,we get $n = \frac{1/\sqrt{\mu_{0} \varepsilon_{0}}}{1/\sqrt{\mu_{0} \mu_{r} \varepsilon_{0} \varepsilon_{r}}} = \sqrt{\mu_{r} \varepsilon_{r}}$.
Since the refractive index $n$ is a ratio of two speeds,it is a dimensionless quantity.
Therefore,the dimensional formula is $M^{0} L^{0} T^{0} A^{0}$,which is not listed in the options.
Thus,the correct option is $D$.

(A) The intensity of radiation $(I)$ is defined as the energy $(E)$ incident per unit area $(A)$ per unit time $(t)$.
Mathematically,$I = \frac{E}{A \times t}$.
The dimensional formula for energy $(E)$ is $[M^{1} L^{2} T^{-2}]$.
The dimensional formula for area $(A)$ is $[L^{2}]$.
The dimensional formula for time $(t)$ is $[T^{1}]$.
Substituting these into the formula:
$I = \frac{[M^{1} L^{2} T^{-2}]}{[L^{2}] \times [T^{1}]}$
$I = [M^{1} L^{2-2} T^{-2-1}]$
$I = [M^{1} L^{0} T^{-3}]$
Therefore,the correct option is $A$.

(A) Polarization $P$ is defined as the dipole moment per unit volume.
$P = \frac{p}{V}$
Since $p = q \times d$ (charge $\times$ distance) and $V = L^3$ (volume),
$P = \frac{q \times d}{L^3} = \frac{q}{L^2}$
The unit of charge $q$ is $A \cdot T$ (Ampere-second) and the unit of area $L^2$ is $m^2$.
Therefore,the unit of polarization is $A \cdot T \cdot m^{-2}$.
The dimensional formula is $[A^1 T^1 L^{-2}]$.
Thus,the correct option is $A$.

(D) The formula for capacitance is $C = \frac{Q}{V}$.
Since potential $V = \frac{W}{Q}$,where $W$ is work and $Q$ is charge,we substitute this into the capacitance formula:
$C = \frac{Q^2}{W}$.
The dimensional formula for work $W$ is $[M^1 L^2 T^{-2}]$.
Substituting the dimensions:
$C = \frac{Q^2}{[M^1 L^2 T^{-2}]}$.
$C = M^{-1} L^{-2} T^2 Q^2$.

(B) The torsional constant $(k)$ for a spring or a wire is defined by the relation $\tau = k\theta$,where $\tau$ is the torque and $\theta$ is the angular displacement in radians.
Since $\theta$ is dimensionless,the dimensions of the torsional constant $(k)$ are the same as the dimensions of torque $(\tau)$.
Torque is defined as the product of force and perpendicular distance: $\tau = F \times d$.
The dimensional formula for force $(F)$ is $[M^1 L^1 T^{-2}]$.
The dimensional formula for distance $(d)$ is $[L^1]$.
Therefore,the dimensional formula for torque $(\tau)$ is $[M^1 L^1 T^{-2}] \times [L^1] = [M^1 L^2 T^{-2}]$.
Thus,the dimensional formula of the effective torsional constant is $[M^1 L^2 T^{-2}]$.

(C) The activity of a radioactive substance is defined as the rate of decay,which is the number of disintegrations per unit time.
Mathematically,Activity $A = -\frac{dN}{dt}$.
Since $N$ (number of nuclei) is a dimensionless quantity and $t$ represents time,the dimensions of activity are $[T^{-1}]$.
Therefore,the dimensional formula is $[M^0 L^0 T^{-1}]$.

(B) $1$. Thermal conductivity $(K)$: From $Q = \frac{KA(T_2 - T_1)t}{d}$,we get $[K] = [M^{1} L^{1} T^{-3} K^{-1}]$. This is correct.
$2$. Potential $(V)$: $V = \frac{W}{q}$. $[V] = \frac{[M^{1} L^{2} T^{-2}]}{[A^{1} T^{1}]} = [M^{1} L^{2} T^{-3} A^{-1}]$. The given option $B$ states $[M^{1} L^{2} T^{3} A^{-1}]$,which is incorrect.
$3$. Permeability of free space $(\mu_{0})$: From $F = \frac{\mu_{0} I_1 I_2 L}{2\pi d}$,we get $[\mu_{0}] = [M^{1} L^{1} T^{-2} A^{-2}]$. This is correct.
Therefore,option $B$ is the incorrect statement.

(D) The dimensional formula for angular frequency is $[M^{0} L^{0} T^{-1}]$.
Comparing this with $[M^{a} L^{b} T^{c}]$,we get $a=0, b=0, c=-1$.
For spring constant,the formula is $[M^{1} L^{0} T^{-2}]$.
For surface tension,the formula is $[M^{1} L^{0} T^{-2}]$.
For force,the formula is $[M^{1} L^{1} T^{-2}]$.
Thus,option $D$ is the correct match.

(C) The dimensional formula for resistance $R$ is derived from $R = \frac{V}{I} = \frac{W}{qI}$.
Substituting the dimensions: $R = \frac{[ML^2 T^{-2}]}{[AT][A]} = [ML^2 T^{-3} A^{-2}]$.
Now,let us check the dimensions of $\frac{h}{e^2}$.
The dimensions of Planck's constant $h$ are $[ML^2 T^{-1}]$.
The dimensions of charge $e$ are $[AT]$.
Therefore,the dimensions of $\frac{h}{e^2} = \frac{[ML^2 T^{-1}]}{[AT]^2} = \frac{[ML^2 T^{-1}]}{[A^2 T^2]} = [ML^2 T^{-3} A^{-2}]$.
Since the dimensions of $R$ and $\frac{h}{e^2}$ are identical,the correct option is $C$.

(A) Impulse is defined as the product of force and time interval.
Impulse = Force $\times$ Time
Dimensional formula of Force = $[MLT^{-2}]$
Dimensional formula of Time = $[T]$
Therefore,the dimensional formula of impulse = $[MLT^{-2}] \times [T] = [MLT^{-1}]$.

(A) According to Newton's law of gravitation,the force $F$ between two masses $M_1$ and $M_2$ separated by a distance $R$ is given by:
$F = \frac{G M_1 M_2}{R^2}$
Rearranging the formula to solve for the universal gravitational constant $G$:
$G = \frac{F R^2}{M_1 M_2}$
Substituting the dimensional formulas for force $[F] = [MLT^{-2}]$,distance $[R] = [L]$,and mass $[M] = [M]$:
$[G] = \frac{[MLT^{-2}] [L^2]}{[M] [M]} = \frac{[ML^3 T^{-2}]}{[M^2]} = [M^{-1} L^3 T^{-2}]$

(C) Planck's constant $(h)$ is related to the energy $(E)$ of a photon and its frequency $(
u)$ by the equation: $E = h
u$.
From this,we can write $h = \frac{E}{\nu}$.
The dimensional formula of energy $(E)$ is $[ML^2 T^{-2}]$.
The dimensional formula of frequency $(
u)$ is $[T^{-1}]$.
Substituting these into the equation for $h$:
$h = \frac{[ML^2 T^{-2}]}{[T^{-1}]} = [ML^2 T^{-2+1}] = [ML^2 T^{-1}]$.
Therefore,the correct dimensional formula for Planck's constant is $[ML^2 T^{-1}]$.

(C) To determine which pair does not have the same dimensional formula,we analyze each option:
$1$. Work and Torque: Both have the dimensional formula $[ML^2T^{-2}]$.
$2$. Angular momentum and Planck's constant: Both have the dimensional formula $[ML^2T^{-1}]$.
$3$. Stress and Linear momentum: Stress is force per unit area,$[ML^{-1}T^{-2}]$. Linear momentum is mass times velocity,$[MLT^{-1}]$. These are not the same.
$4$. Surface tension and Force constant: Both have the dimensional formula $[MT^{-2}]$.
Therefore,the pair that does not have the same dimensional formula is stress and linear momentum.

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

(A) The dimensions of kinetic energy $(K.E.)$ are $[M L^2 T^{-2}]$.
The dimensions of surface tension $(S)$ are $[M T^{-2}]$.
Let the required quantity be $X = \sqrt{\frac{K.E.}{S}}$.
Substituting the dimensions: $X = \sqrt{\frac{[M L^2 T^{-2}]}{[M T^{-2}]}}$.
Simplifying the expression: $X = \sqrt{[L^2]} = [L]$.
The dimension $[L]$ corresponds to length or distance.
Therefore,the correct option is $A$.

(B) The expression given is $\frac{e^2}{4 \pi \varepsilon_0 \hbar}$.
We know that the Coulomb force is $F = \frac{1}{4 \pi \varepsilon_0} \frac{e^2}{r^2}$,so $\frac{e^2}{4 \pi \varepsilon_0} = F r^2$.
The dimensions of $\frac{e^2}{4 \pi \varepsilon_0}$ are $[M L T^{-2}] [L^2] = [M L^3 T^{-2}]$.
The dimensions of Planck's constant $h$ (or $\hbar$) are $[M L^2 T^{-1}]$.
Therefore,the dimensions of the expression are $\frac{[M L^3 T^{-2}]}{[M L^2 T^{-1}]} = [L^1 T^{-1}]$.
This represents the dimensions of velocity (speed of light $c$).

(A) The $SI$ unit of the Boltzmann constant $(k_B)$ is Joule per Kelvin $(J/K)$.
Energy $(J)$ has the dimensional formula $\left[ML^2 T^{-2}\right]$.
Temperature $(K)$ has the dimensional formula $\left[K^1\right]$.
Therefore,the dimensional formula for the Boltzmann constant is $\left[ML^2 T^{-2}\right] / \left[K^1\right] = \left[ML^2 T^{-2} K^{-1}\right]$.

(A) Stress is defined as the restoring force per unit area of the material. It is denoted by the Greek letter $\sigma$.
$\text{Stress} = \frac{\text{Force}}{\text{Area}}$
Substituting the dimensional formulas for force and area:
$\text{Force} = [M L T^{-2}]$
$\text{Area} = [L^2]$
$\text{Stress} = \frac{[M L T^{-2}]}{[L^2]} = [M L^{1-2} T^{-2}] = [M L^{-1} T^{-2}]$

(A) We know that the electrostatic force $F$ between two charges $q_1$ and $q_2$ separated by a distance $r$ is given by Coulomb's Law:
$F = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r^2}$
Rearranging the formula to solve for permittivity $\varepsilon_0$:
$\varepsilon_0 = \frac{q_1 q_2}{4 \pi F r^2}$
Now,substituting the dimensional formulas for each physical quantity:
$[q] = [AT]$
$[F] = [MLT^{-2}]$
$[r] = [L]$
Substituting these into the expression for $\varepsilon_0$:
$[\varepsilon_0] = \frac{[AT][AT]}{[MLT^{-2}][L^2]} = \frac{[A^2 T^2]}{[ML^3 T^{-2}]}$
$[\varepsilon_0] = [M^{-1} L^{-3} T^4 A^2]$

(C) The unit $kg m s^{-2}$ represents a newton $(N)$,which is the unit of force.
The unit $kg m^2 s^{-3}$ represents a watt $(W)$,which is the unit of power.
The unit $kg m^2 s^{-2}$ represents a joule $(J)$,which is the unit of work or energy. Work is defined as force multiplied by displacement $(W = F \times d)$. Since force is $kg m s^{-2}$ and displacement is $m$,the unit of work is $(kg m s^{-2}) \times m = kg m^2 s^{-2}$.
The unit $kg m^{-1} s^{-2}$ represents a pascal $(Pa)$,which is the unit of pressure. Pressure is defined as force divided by area $(P = F / A)$. Thus,the unit is $(kg m s^{-2}) / m^2 = kg m^{-1} s^{-2}$.

(A) From Maxwell's relation for electromagnetic waves,the speed of light $c$ is given by the formula: $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$.
Squaring both sides,we get: $c^2 = \frac{1}{\mu_0 \varepsilon_0}$.
Therefore,the product $\mu_0 \varepsilon_0 = \frac{1}{c^2}$.
The dimensions of the speed of light $c$ are $[LT^{-1}]$.
Substituting the dimensions,we get: $[\mu_0 \varepsilon_0] = [LT^{-1}]^{-2}$.
Simplifying this,we obtain: $[\mu_0 \varepsilon_0] = [L^{-2} T^2]$.
In terms of mass,length,and time,this is written as $[M^0 L^{-2} T^2]$.

(C) The dimension of the spring constant $k$ is given by $k = \frac{F}{x}$,so its unit is $\frac{N}{m} = \frac{kg \cdot m/s^2}{m} = kg \cdot s^{-2}$. The dimensional formula is $[M L^0 T^{-2}]$.
Surface energy is defined as energy per unit area: $\text{Surface energy} = \frac{\text{Energy}}{\text{Area}} = \frac{J}{m^2} = \frac{N \cdot m}{m^2} = \frac{N}{m}$.
Thus,the unit of surface energy is also $\frac{N}{m}$,which corresponds to the dimensional formula $[M L^0 T^{-2}]$.
Since both have the same dimensional formula,option $C$ is correct.

(C) Angular momentum $(L)$ for a particle is defined as the product of linear momentum $(p)$ and the perpendicular distance $(r)$ from the axis of rotation.
Mathematically,$L = p \times r = m \times v \times r$.
The dimensions are:
Mass $(m)$ = $[M]$
Velocity $(v)$ = $[LT^{-1}]$
Distance $(r)$ = $[L]$
Therefore,the dimension of angular momentum = $[M] \times [LT^{-1}] \times [L] = [ML^2 T^{-1}]$.

(B) The dimension of the electric field $E$ is given by:
$[E] = \frac{[F]}{[q]} = \frac{[M^1 L^1 T^{-2}]}{[A^1 T^1]} = [M^1 L^1 T^{-3} A^{-1}] \quad ... (i)$
The dimension of the permeability of free space $\mu_0$ is derived from the relation $B = \frac{\mu_0 I}{2 \pi r}$ for a long straight wire,or more fundamentally from the energy density of a magnetic field $u_B = \frac{B^2}{2 \mu_0}$. Using the force law $F = qvB$,we find:
$[\mu_0] = [M^1 L^1 T^{-2} A^{-2}] \quad ... (ii)$
Now,we calculate the dimension of $\frac{E^2}{\mu_0}$:
$\left[\frac{E^2}{\mu_0}\right] = \frac{[E]^2}{[\mu_0]} = \frac{[M^1 L^1 T^{-3} A^{-1}]^2}{[M^1 L^1 T^{-2} A^{-2}]}$
$= \frac{[M^2 L^2 T^{-6} A^{-2}]}{[M^1 L^1 T^{-2} A^{-2}]}$
$= [M^{2-1} L^{2-1} T^{-6-(-2)} A^{-2-(-2)}]$
$= [M^1 L^1 T^{-4}] = [MLT^{-4}]$

(A) Given the relation $P V^{5/3} = K$.
Pressure $P$ has dimensions $[M L^{-1} T^{-2}]$.
Volume $V$ has dimensions $[L^3]$.
Substituting these into the equation:
$[K] = [P] [V]^{5/3} = [M L^{-1} T^{-2}] ([L^3])^{5/3}$.
$[K] = [M L^{-1} T^{-2}] [L^5]$.
$[K] = [M L^{4} T^{-2}]$.
Thus,the dimensions of $K$ are $ML^4T^{-2}$.

(D) In the equation $x(t) = \frac{v_0}{A}(1 - e^{-At})$,the exponent of the exponential function must be dimensionless. Therefore,the dimensions of $At$ must be $[M^0 L^0 T^0]$.
Since $[t] = [T]$,we have $[A][T] = [1]$,which implies $[A] = [T^{-1}] = [M^0 L^0 T^{-1}]$.
Next,the term $(1 - e^{-At})$ is dimensionless. Thus,the dimensions of $x$ must be equal to the dimensions of $\frac{v_0}{A}$.
$[x] = \frac{[v_0]}{[A]} \implies [L] = \frac{[v_0]}{[T^{-1}]}$.
Therefore,$[v_0] = [L][T^{-1}] = [M^0 LT^{-1}]$.
Thus,the dimensions of $v_0$ and $A$ are $[M^0 LT^{-1}]$ and $[M^0 L^0 T^{-1}]$ respectively.

(B) Coefficient of viscosity $\eta = \frac{F dr}{A dv} = \frac{[MLT^{-2}][L]}{[L^2][LT^{-1}]} = [ML^{-1}T^{-1}]$. Thus, $A-IV$.
$(B)$ Surface tension $S = \frac{F}{L} = \frac{[MLT^{-2}]}{[L]} = [MT^{-2}]$ or $[ML^0T^{-2}]$. Thus, $B-III$.
$(C)$ Pressure $P = \frac{F}{A} = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$. Thus, $C-I$.
$(D)$ Surface energy $E = S \times A = [MT^{-2}][L^2] = [ML^2T^{-2}]$. Thus, $D-II$.
Therefore, the correct matching is $A-IV, B-III, C-I, D-II$.

(A) The quantity $\frac{B^2}{2\mu_0}$ represents the magnetic energy density $(u_B)$,which is defined as the energy stored per unit volume in a magnetic field.
Energy $(E)$ has the dimensional formula $[ML^2T^{-2}]$.
Volume $(V)$ has the dimensional formula $[L^3]$.
Therefore,the dimensional formula for magnetic energy density is $\frac{[ML^2T^{-2}]}{[L^3]} = [ML^{-1}T^{-2}]$.
Thus,the correct option is $A$.

(C) Planck's constant $(h)$ has the dimensions of action,which is defined as energy multiplied by time $(E \times t)$.
The $SI$ unit of energy is Joule $(J)$ and time is second $(s)$,so the unit of Planck's constant is $J \cdot s$.
Angular momentum $(L)$ is defined as the product of moment of inertia $(I)$ and angular velocity $(\omega)$,or $L = mvr$.
The dimensions of angular momentum are $[M L^2 T^{-1}]$,which are identical to the dimensions of Planck's constant.
Therefore,both Planck's constant and angular momentum have the same $SI$ unit,which is $kg \cdot m^2/s$ or $J \cdot s$.
Thus,the correct option is $C$.

(B) The expression $\frac{1}{2}\epsilon_0 E^2$ represents the energy density of an electric field,which is defined as energy per unit volume.
Dimensions of energy = $[ML^2 T^{-2}]$.
Dimensions of volume = $[L^3]$.
Therefore,dimensions of energy density = $\frac{[ML^2 T^{-2}]}{[L^3]} = [ML^{-1} T^{-2}]$.
Comparing this with $[M^a L^b T^c]$,we get $a = 1$,$b = -1$,and $c = -2$.
Now,calculating the value of $2a - b + c$:
$2(1) - (-1) + (-2) = 2 + 1 - 2 = 1$.

(B) The dimensional formula for the universal gravitational constant $G$ is $[M^{-1}L^3T^{-2}]$.
The dimensional formula for Planck's constant $h$ is $[ML^2T^{-1}]$.
We need to express $G$ in terms of $h, L, M, T$. Let $G = h^a L^b M^c T^d$.
Substituting the dimensions: $[M^{-1}L^3T^{-2}] = [ML^2T^{-1}]^a [L]^b [M]^c [T]^d$.
$[M^{-1}L^3T^{-2}] = M^a L^{2a} T^{-a} \cdot L^b \cdot M^c \cdot T^d = M^{a+c} L^{2a+b} T^{-a+d}$.
Comparing the powers of $M, L, T$ on both sides:
$a + c = -1$ $(1)$
$2a + b = 3$ $(2)$
$-a + d = -2$ $(3)$
Testing option $(B)$: $a = 1, b = 1, c = -2, d = -1$.
From $(1)$: $1 + (-2) = -1$ (Correct).
From $(2)$: $2(1) + 1 = 3$ (Correct).
From $(3)$: $-1 + (-1) = -2$ (Correct).
Therefore, $G = h^1 L^1 M^{-2} T^{-1}$, which is $[hT^{-1}LM^{-2}]$.