Dimensions and Dimensional Formula Questions in English

(A) Torque $( \tau)$ is defined as the product of force and the perpendicular distance from the axis of rotation.
Mathematically,$\tau = \text{Force} \times \text{Distance}$.
The dimensional formula for force is $[M L T^{-2}]$.
The dimensional formula for distance is $[L]$.
Therefore,the dimensional formula for torque is $[M L T^{-2}] \times [L] = [M L^2 T^{-2}]$.
Thus,the correct option is $A$.

(C) The viscous force $F$ acting on a fluid layer is given by Newton's law of viscosity: $F = -\eta A \frac{dv}{dx}$.
Here,$F$ is the force,$\eta$ is the coefficient of viscosity,$A$ is the area,and $\frac{dv}{dx}$ is the velocity gradient.
Rearranging for $\eta$: $\eta = \frac{F}{A (dv/dx)}$.
Substituting the dimensions: $[F] = [M L T^{-2}]$,$[A] = [L^2]$,$[dv] = [L T^{-1}]$,and $[dx] = [L]$.
Thus,$[\eta] = \frac{[M L T^{-2}]}{[L^2] [L T^{-1} / L]} = \frac{[M L T^{-2}]}{[L^2] [T^{-1}]} = [M L^{-1} T^{-1}]$.
Therefore,the correct option is $C$.

(A) The formula for angular momentum is $L = mvr$,where $m$ is mass,$v$ is velocity,and $r$ is the radius (distance).
The formula for linear momentum is $p = mv$.
The ratio of angular momentum to linear momentum is $\frac{L}{p} = \frac{mvr}{mv}$.
By canceling the common terms $m$ and $v$,we get $\frac{L}{p} = r$.
Since $r$ represents a distance or length,its dimensional formula is $[M^0 L^1 T^0]$.

(D) Surface tension is defined as the force per unit length acting on the surface of a liquid.
Mathematically,$\text{Surface tension} = \frac{\text{Force}}{\text{Length}}$.
The dimensional formula for force is $[M L T^{-2}]$.
The dimensional formula for length is $[L]$.
Therefore,the dimensions of surface tension are $\frac{[M L T^{-2}]}{[L]} = [M T^{-2}]$.
Thus,the correct option is $D$.

(B) According to Coulomb's Law,the force $F$ between two charges $q_1$ and $q_2$ separated by a distance $r$ is given by:
$F = \frac{1}{{4\pi {\varepsilon _0}}}\,\frac{{{q_1}{q_2}}}{{{r^2}}}$
Rearranging the formula to solve for permittivity ${\varepsilon _0}$:
${\varepsilon _0} = \frac{{{q_1}{q_2}}}{{4\pi F{r^2}}}$
Substituting the dimensions of each quantity:
$[q] = [AT]$
$[F] = [MLT^{-2}]$
$[r] = [L]$
Therefore,the dimensions of ${\varepsilon _0}$ are:
$[{\varepsilon _0}] = \frac{[AT][AT]}{[MLT^{-2}][L^2]}$
$[{\varepsilon _0}] = \frac{[A^2T^2]}{[ML^3T^{-2}]}$
$[{\varepsilon _0}] = [A^2T^4M^{-1}L^{-3}]$
Thus,the correct option is $B$.

(D) The dimension of pressure is defined as force per unit area: $[P] = [F]/[A] = [MLT^{-2}]/[L^2] = [ML^{-1}T^{-2}]$.
Stress is also defined as restoring force per unit area: $[Stress] = [F]/[A] = [ML^{-1}T^{-2}]$.
Coefficient of elasticity is defined as stress divided by strain. Since strain is dimensionless,the dimension of the coefficient of elasticity is the same as stress: $[E] = [Stress]/[Strain] = [ML^{-1}T^{-2}] / [1] = [ML^{-1}T^{-2}]$.
Therefore,all three quantities have the same dimensions: $[ML^{-1}T^{-2}]$.

(B) Planck's constant $(h)$ is related to energy $(E)$ and frequency $(f)$ by the equation $E = hf$. Therefore,$h = E/f$. The dimension of energy is $[M L^2 T^{-2}]$ and the dimension of frequency is $[T^{-1}]$. Thus,the dimension of $h$ is $[M L^2 T^{-2}] / [T^{-1}] = [M L^2 T^{-1}]$.
Angular momentum $(L)$ is given by the formula $L = mvr$,where $m$ is mass,$v$ is velocity,and $r$ is radius. The dimensions are $[M] [L T^{-1}] [L] = [M L^2 T^{-1}]$.
Since both quantities have the same dimensions $[M L^2 T^{-1}]$,the correct option is $B$.

(B) $1$. Dimensional formula of permittivity of free space $(\varepsilon_0)$:
From Coulomb's Law,$F = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2}$.
Rearranging,$\varepsilon_0 = \frac{q_1 q_2}{4\pi F r^2}$.
Dimensions: $[q] = [IT]$,$[F] = [MLT^{-2}]$,$[r] = [L]$.
Substituting: $[\varepsilon_0] = \frac{[IT][IT]}{[MLT^{-2}][L^2]} = \frac{I^2 T^2}{ML^3 T^{-2}} = M^{-1} L^{-3} T^4 I^2$.
$2$. Dimensional formula of permeability of free space $(\mu_0)$:
From the force between two parallel wires,$F = \frac{\mu_0 I_1 I_2 l}{2\pi r}$.
Rearranging,$\mu_0 = \frac{F \cdot 2\pi r}{I_1 I_2 l}$.
Dimensions: $[F] = [MLT^{-2}]$,$[r] = [L]$,$[I] = [I]$,$[l] = [L]$.
Substituting: $[\mu_0] = \frac{[MLT^{-2}][L]}{[I][I][L]} = MLT^{-2}I^{-2}$.
Comparing with the options,option $B$ is correct.

(D) Strain is defined as the ratio of change in dimension to the original dimension. Since it is a ratio of two similar physical quantities,it is a dimensionless quantity.
Angular velocity has dimensions of $[T^{-1}]$.
Linear momentum has dimensions of $[MLT^{-1}]$.
Angular momentum has dimensions of $[ML^2T^{-1}]$.
Therefore,the correct option is $(d)$.

(C) The magnetic force $F$ on a current-carrying wire is given by $F = I L B \sin \theta$.
Rearranging for the magnetic field intensity $B$,we get $B = \frac{F}{I L}$.
The dimensional formula for force $F$ is $[M L T^{-2}]$.
The dimensional formula for current $I$ is $[A]$.
The dimensional formula for length $L$ is $[L]$.
Substituting these into the formula: $B = \frac{[M L T^{-2}]}{[A] [L]} = [M L^0 T^{-2} A^{-1}]$.
Thus,the correct option is $C$.

(A) The interatomic force constant $K$ is related to Young's modulus $Y$ and the interatomic distance $r_0$ by the relation $K = Y \times r_0$.
The dimensional formula for Young's modulus $Y$ is $[M L^{-1} T^{-2}]$.
The dimensional formula for distance $r_0$ is $[L]$.
Therefore,the dimensions of $K$ are $[M L^{-1} T^{-2}] \times [L] = [M T^{-2}]$.
Thus,the correct option is $A$.

(A) The formula for resistivity is $\rho = \frac{R \cdot A}{l}$.
Here,the dimension of resistance $R$ is $[M{L^2}{T^{ - 1}}{Q^{ - 2}}]$.
The dimension of area $A$ is $[L^2]$.
The dimension of length $l$ is $[L]$.
Substituting these into the formula:
$[\rho] = \frac{[M{L^2}{T^{ - 1}}{Q^{ - 2}}] \cdot [L^2]}{[L]} = [M{L^3}{T^{ - 1}}{Q^{ - 2}}]$.
Thus,the given dimension corresponds to resistivity.

(A) Electric current $(I)$ is defined as the rate of flow of electric charge $(Q)$ with respect to time $(t)$.
Mathematically,$I = \frac{Q}{t}$.
In terms of dimensional analysis,the dimension of charge is $[Q]$ and the dimension of time is $[T]$.
Therefore,the dimension of electric current is $[I] = \frac{[Q]}{[T]} = [M^0L^0T^{-1}Q]$.
Thus,the correct option is $A$.

(C) The dimensional formula for torque $( au)$ is given by: $[M^1 L^2 T^{-2}]$.
The dimensional formula for angular momentum $(L)$ is given by: $[M^1 L^2 T^{-1}]$.
Comparing the two,the dimensions of mass $(M)$ and length $(L)$ are the same in both formulas ($M^1$ and $L^2$ respectively).
Therefore,the correct option is $C$.

(D) The dimensional formula of Energy is $[M L^2 T^{-2}]$.
The dimensional formula of Mass is $[M]$.
The dimensional formula of Length is $[L]$.
Therefore,the dimensional formula of the given expression is:
$\frac{[M L^2 T^{-2}]}{[M] \times [L]} = \frac{[M L^2 T^{-2}]}{[M L]} = [L T^{-2}]$
$[L T^{-2}]$ is the dimensional formula for Acceleration.
Thus,the correct option is $D$.

(B) Luminous flux is defined as the rate of flow of light energy,which is equivalent to radiant power.
The $SI$ unit of luminous flux is the lumen $(lm)$,which is a unit of power.
Therefore,the dimensions of luminous flux are the same as the dimensions of power.
Power $(P)$ is defined as work done $(W)$ per unit time $(t)$: $P = W / t$.
The dimensions of work are $[M L^2 T^{-2}]$ and the dimensions of time are $[T]$.
Thus,the dimensions of power are $[M L^2 T^{-2}] / [T] = [M L^2 T^{-3}]$.

(D) The dimensions of the given quantities are as follows:
$1$. Planck's constant $(h)$ = $[M{L^2}{T^{ - 1}}]$ and Angular momentum $(L)$ = $[M{L^2}{T^{ - 1}}]$. They have the same dimensions.
$2$. Work $(W)$ = $[M{L^2}{T^{ - 2}}]$ and Energy $(E)$ = $[M{L^2}{T^{ - 2}}]$. They have the same dimensions.
$3$. Pressure $(P)$ = $[M{L^{ - 1}}{T^{ - 2}}]$ and Young's modulus $(Y)$ = $[M{L^{ - 1}}{T^{ - 2}}]$. They have the same dimensions.
$4$. Torque $( au)$ = $[M{L^2}{T^{ - 2}}]$ and Moment of inertia $(I)$ = $[M{L^2}]$. They do not have the same dimensions.
Therefore,the correct option is $D$.

(C) The dimensional formula for energy is $[M{L^2}{T^{ - 2}}]$.
Latent heat $(L)$ is defined as the heat energy $(Q)$ per unit mass $(m)$,given by $L = \frac{Q}{m}$.
Substituting the dimensions: $L = \frac{[M{L^2}{T^{ - 2}}]}{[M]} = [M^0{L^2}{T^{ - 2}}]$.
Therefore,the dimensional formula ${M^0}{L^2}{T^{ - 2}}$ represents latent heat.

(A) The capacitance $C$ is defined as the ratio of charge $Q$ to potential difference $V$,given by $C = \frac{Q}{V}$.
The dimensional formula for charge $Q$ is $[AT]$.
The dimensional formula for potential difference $V$ is $[ML^2T^{-3}A^{-1}]$.
Substituting these into the formula for $C$:
$C = \frac{[AT]}{[ML^2T^{-3}A^{-1}]} = [M^{-1}L^{-2}T^4A^2]$.
Thus,the correct option is $A$.

(D) The formula for electric charge $(Q)$ is given by the product of electric current $(I)$ and time $(t)$.
$Q = I \times t$
Since the dimensional formula for current $(I)$ is $[A]$ and the dimensional formula for time $(t)$ is $[T]$,the dimensional formula for charge is $[AT]$.
Thus,the correct option is $D$.

(B) The formula for viscous force is $F = - \eta A \frac{\Delta v}{\Delta z}$.
Rearranging for $\eta$,we get $\eta = \frac{F}{A (\Delta v / \Delta z)}$.
Substituting the dimensions of each quantity:
Force $F = [M L T^{-2}]$
Area $A = [L^2]$
Velocity gradient $\frac{\Delta v}{\Delta z} = \frac{[L T^{-1}]}{[L]} = [T^{-1}]$
Now,calculating the dimensions of $\eta$:
$[\eta] = \frac{[M L T^{-2}]}{[L^2] [T^{-1}]} = [M L^{1-2} T^{-2+1}] = [M L^{-1} T^{-1}]$.
Thus,the correct option is $B$.

(D) The viscous damping force $F$ is proportional to the velocity $v$,so we can write $F = kv$,where $k$ is the constant of proportionality.
Rearranging for $k$,we get $k = \frac{F}{v}$.
The dimensional formula for force $F$ is $[M L T^{-2}]$.
The dimensional formula for velocity $v$ is $[L T^{-1}]$.
Substituting these into the equation for $k$:
$[k] = \frac{[M L T^{-2}]}{[L T^{-1}]} = [M L^0 T^{-1}]$.
Therefore,the correct dimension is $M L^0 T^{-1}$.

(D) $emf$ (electromotive force) is defined as the work done per unit charge.
$emf = \frac{W}{Q}$
Since the dimensions of work $(W)$ are $[M{L^2}{T^{ - 2}}]$ and the dimension of charge is $[Q]$,
$[emf] = \frac{[M{L^2}{T^{ - 2}}]}{[Q]} = [M{L^2}{T^{ - 2}}{Q^{ - 1}}]$

(D) The dimensional formula for the Gravitational constant $(G)$ is $[M^{-1}L^3T^{-2}]$.
The dimensional formula for Planck's constant $(h)$ is $[ML^2T^{-1}]$.
The Power of a convex lens is defined as the reciprocal of focal length,so its dimensional formula is $[L^{-1}]$.
Since all the given quantities have dimensions,none of them are dimensionless.

$(A)$ Boltzmann's constant $(k)$ is related to the ideal gas constant $(R)$ and Avogadro's number $(N_A)$ by the relation $k = R / N_A$.
The ideal gas equation is $PV = nRT$,where $R = PV / (nT)$.
The dimensions of pressure $(P)$ are $[M L^{-1} T^{-2}]$,volume $(V)$ are $[L^3]$,amount of substance $(n)$ is $[\text{mol}]$,and temperature $(T)$ is $[\theta]$.
Thus,the dimensions of $R$ are $[M L^{-1} T^{-2}] \cdot [L^3] / ([\text{mol}] \cdot [\theta]) = [M L^2 T^{-2} \theta^{-1} \text{mol}^{-1}]$.
Since Avogadro's number $(N_A)$ is a dimensionless quantity (number of particles per mole),the dimensions of Boltzmann's constant $(k)$ are the same as the gas constant per mole,which is $[M L^2 T^{-2} \theta^{-1}]$.

(A) Given the equation $W = \frac{1}{2}K{x^2}$,where $W$ is work and $x$ is displacement.
Dimensional formula of work $W$ is $[M^1 L^2 T^{-2}]$.
Dimensional formula of displacement $x$ is $[L^1]$.
Rearranging the equation for $K$,we get $K = \frac{2W}{x^2}$.
Substituting the dimensions: $[K] = \frac{[M^1 L^2 T^{-2}]}{[L^1]^2} = \frac{[M^1 L^2 T^{-2}]}{[L^2]} = [M^1 L^0 T^{-2}]$.
Thus,the dimensions of $K$ are $[M^1 L^0 T^{-2}]$.

(C) The dimensions of the given quantities are as follows:
$1$. Speed: $[LT^{-1}]$ and ${({\mu _0}{\varepsilon _0})^{-1/2}} = c$ (speed of light): $[LT^{-1}]$. They have the same dimensions.
$2$. Torque: $[ML^2T^{-2}]$ and Work: $[ML^2T^{-2}]$. They have the same dimensions.
$3$. Momentum: $[MLT^{-1}]$ and Planck's constant: $[ML^2T^{-1}]$. They do not have the same dimensions.
$4$. Stress: $[ML^{-1}T^{-2}]$ and Young's modulus: $[ML^{-1}T^{-2}]$. They have the same dimensions.
Therefore,the correct option is $C$.

(B) According to Ohm's law,$R = \frac{V}{I}$.
First,find the dimensions of potential difference $(V)$: $V = \frac{W}{q} = \frac{[M L^2 T^{-2}]}{[A T]} = [M L^2 T^{-3} A^{-1}]$.
Now,substitute the dimensions of $V$ and $I$ into the formula for $R$:
$R = \frac{[M L^2 T^{-3} A^{-1}]}{[A]} = [M L^2 T^{-3} A^{-2}]$.
Therefore,the correct option is $B$.

(D) Relative density is defined as the ratio of the density of a substance to the density of water at $4^{\circ}C$.
$\text{Relative density} = \frac{\text{Density of substance}}{\text{Density of water}}$
Since both the numerator and the denominator have the same dimensions $([M L^{-3}])$,they cancel each other out.
Therefore,relative density is a dimensionless quantity,represented as $[M^0 L^0 T^0]$.

(B) Potential energy is defined as the work done against a conservative force,given by the formula $U = mgh$.
The dimensions of mass $m$ are $[M^1]$.
The dimensions of acceleration due to gravity $g$ are $[LT^{-2}]$.
The dimensions of height $h$ are $[L^1]$.
Therefore,the dimensional formula for potential energy is $[M^1] \times [LT^{-2}] \times [L^1] = [ML^2T^{-2}]$.

(D) The dimension of pressure gradient is calculated as: $\frac{P}{x} = \frac{[M L^{-1} T^{-2}]}{[L]} = [M L^{-2} T^{-2}]$.
$1$. Velocity gradient: $\frac{v}{x} = \frac{[L T^{-1}]}{[L]} = [T^{-1}]$.
$2$. Potential gradient: $\frac{V}{x} = \frac{[M L^2 T^{-3} A^{-1}]}{[L]} = [M L T^{-3} A^{-1}]$.
$3$. Energy gradient: $\frac{E}{x} = \frac{[M L^2 T^{-2}]}{[L]} = [M L T^{-2}]$.
Comparing these,the dimension of pressure gradient does not match any of the given options. Therefore,the correct option is $(d)$.

(A) The ideal gas equation is given by $PV = nRT$,where $R$ is the universal gas constant.
Rearranging for $R$,we get $R = \frac{PV}{nT}$.
The dimensions of pressure $P$ are $[M L^{-1} T^{-2}]$.
The dimensions of volume $V$ are $[L^3]$.
The dimensions of amount of substance $n$ are $[mol]$ (or often represented as dimensionless in basic dimensional analysis if $\theta$ represents temperature).
The dimensions of temperature $T$ are $[\theta]$.
Substituting these into the formula: $R = \frac{[M L^{-1} T^{-2}] \times [L^3]}{[\theta]} = [M L^2 T^{-2} \theta^{-1}]$.
Thus,the correct option is $A$.

(A) To determine which pair does not have identical dimensions,we analyze the dimensional formulas of each:
$1$. Moment of inertia $(I = MR^2)$ has dimensions $[ML^2T^0]$. Moment of force (Torque,$\tau = r \times F$) has dimensions $[ML^2T^{-2}]$. These are not identical.
$2$. Work $(W = F \cdot d)$ has dimensions $[ML^2T^{-2}]$. Torque $(\tau = r \times F)$ has dimensions $[ML^2T^{-2}]$. These are identical.
$3$. Angular momentum $(L = mvr)$ has dimensions $[ML^2T^{-1}]$. Planck's constant $(h = E/f)$ has dimensions $[ML^2T^{-1}]$. These are identical.
$4$. Impulse $(J = F \Delta t)$ has dimensions $[MLT^{-1}]$. Momentum $(p = mv)$ has dimensions $[MLT^{-1}]$. These are identical.
Therefore,the pair that does not have identical dimensions is Moment of inertia and moment of force.

(D) dimensional constant is a physical quantity that has dimensions and a constant value.
$1$. Acceleration due to gravity $(g)$ varies with location on the Earth's surface.
$2$. Surface tension of water varies with temperature.
$3$. The weight of a standard kilogram mass depends on the local gravitational field.
$4$. The velocity of light in vacuum $(c)$ is a universal constant with a value of approximately $3 \times 10^{8} \ m/s$,and it possesses dimensions of $[LT^{-1}]$.
Therefore,the velocity of light in vacuum is a dimensional constant.

(C) Surface tension $(T)$ is defined as the force $(F)$ per unit length $(l)$.
Mathematically,$T = \frac{F}{l}$.
The dimensional formula for force $(F)$ is $[MLT^{-2}]$.
The dimensional formula for length $(l)$ is $[L]$.
Therefore,the dimensions of surface tension are $T = \frac{[MLT^{-2}]}{[L]} = [ML^0T^{-2}]$.

(D) The relative permittivity $\varepsilon_r$ (also known as the dielectric constant $K$) is defined as the ratio of the permittivity of a medium $\varepsilon$ to the permittivity of free space $\varepsilon_0$.
Mathematically,$\varepsilon_r = \frac{\varepsilon}{\varepsilon_0}$.
Since both $\varepsilon$ and $\varepsilon_0$ have the same dimensions,their ratio is a dimensionless quantity.
Therefore,$\varepsilon_r$ has no dimensions,i.e.,$[M^0 L^0 T^0 A^0]$.

(B) According to Coulomb's law,the force between two charges is given by $F = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q_1 q_2}{r^2}$.
Rearranging for permittivity,we get $\epsilon_0 = \frac{q_1 q_2}{4 \pi F r^2}$.
The dimensional formula for charge $(q)$ is $[AT]$,for force $(F)$ is $[MLT^{-2}]$,and for distance $(r)$ is $[L]$.
Substituting these dimensions: $\epsilon_0 = \frac{[AT][AT]}{[MLT^{-2}][L^2]} = \frac{[A^2T^2]}{[ML^3T^{-2}]}$.
Simplifying the expression,we get $[M^{-1} L^{-3} T^4 A^2]$.

(B) According to Coulomb's law,the force $F$ between two charges $q_1$ and $q_2$ separated by a distance $r$ is given by:
$F = \frac{1}{4\pi \varepsilon_0} \frac{q_1 q_2}{r^2}$
Rearranging for $\varepsilon_0$:
$\varepsilon_0 = \frac{q_1 q_2}{4\pi F r^2}$
Substituting the dimensions:
$[q] = [AT]$,$[F] = [MLT^{-2}]$,$[r] = [L]$
$[\varepsilon_0] = \frac{[AT][AT]}{[MLT^{-2}][L^2]} = \frac{[A^2 T^2]}{[ML^3 T^{-2}]}$
$[\varepsilon_0] = [M^{-1} L^{-3} T^4 A^2]$

(A) According to Ohm's law,$R = \frac{V}{I}$.
Since potential difference $V = \frac{W}{q}$,where $W$ is work and $q$ is charge,we have $R = \frac{W}{qI}$.
We know that $q = I \times T$.
Substituting this into the equation,we get $R = \frac{W}{(I \times T) \times I} = \frac{W}{I^2 T}$.
The dimensional formula for work $W$ is $[M^1L^2T^{-2}]$,for current $I$ is $[I^1]$,and for time $T$ is $[T^1]$.
Therefore,the dimensional formula for $R$ is $\frac{[M^1L^2T^{-2}]}{[I^2][T^1]} = [M^1L^2T^{-3}I^{-2}]$.

(D) Linear density $m$ has dimensions of $\frac{\text{Mass}}{\text{Length}} = [ML^{-1}]$.
Force $A$ has dimensions $[MLT^{-2}]$.
Given the relation $A/B = m$,we can write $B = A/m$.
Substituting the dimensions: $B = \frac{[MLT^{-2}]}{[ML^{-1}]} = [L^2T^{-2}]$.
Latent heat is defined as energy per unit mass: $\frac{[ML^2T^{-2}]}{[M]} = [L^2T^{-2}]$.
Thus,the dimensions of $B$ are the same as those of latent heat.

(C) dimensionless quantity is a quantity to which no physical dimension is assigned.
However,a dimensionless quantity can still possess a unit of measurement.
For example,the plane angle is measured in radians $(rad)$ and the solid angle is measured in steradians $(sr)$,yet both are dimensionless quantities because they are defined as the ratio of two lengths or areas,respectively.

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

(B) According to Ohm's Law,$V = IR$,where $V$ is the potential difference,$I$ is the current,and $R$ is the resistance.
Therefore,$R = \frac{V}{I}$.
We know that potential difference $V = \frac{W}{q}$,where $W$ is work done and $q$ is charge.
The dimensional formula for work $W$ is $[ML^{2}T^{-2}]$ and for charge $q$ is $[AT]$.
Thus,the dimensional formula for potential $V$ is $[V] = \frac{[ML^{2}T^{-2}]}{[AT]} = [ML^{2}T^{-3}A^{-1}]$.
Substituting this into the expression for resistance:
$[R] = \frac{[ML^{2}T^{-3}A^{-1}]}{[A]} = [ML^{2}T^{-3}A^{-2}]$.
Hence,the correct option is $B$.

(C) The gradient of a physical quantity $Q$ is defined as the rate of change of that quantity with respect to distance,given by $\frac{dQ}{dx}$.
For a gradient to be unitless,the dimensions of the numerator $Q$ must be the same as the dimensions of the denominator $x$ (which is length,$L$).
$1$. Velocity gradient: $\frac{[LT^{-1}]}{[L]} = [T^{-1}]$,which has units of $s^{-1}$.
$2$. Pressure gradient: $\frac{[ML^{-1}T^{-2}]}{[L]} = [ML^{-2}T^{-2}]$,which has units of $Pa/m$.
$3$. Displacement gradient: $\frac{[L]}{[L]} = [M^0L^0T^0]$,which is dimensionless and unitless.
$4$. Force gradient: $\frac{[MLT^{-2}]}{[L]} = [MT^{-2}]$,which has units of $N/m$.
Therefore,the displacement gradient is the unitless quantity.

(B) In the given wave equation $Y = A \sin \omega \left( \frac{x}{v} - k \right)$,the argument of the sine function must be dimensionless.
Furthermore,terms inside the parentheses must have the same dimensions to be subtracted.
Therefore,the dimension of $k$ must be equal to the dimension of $\frac{x}{v}$.
Dimension of $x$ (length) is $[L]$.
Dimension of $v$ (velocity) is $[LT^{-1}]$.
Thus,the dimension of $k = \frac{[L]}{[LT^{-1}]} = [T]$.

(C) Dimensions of a physical quantity are the powers to which the fundamental units must be raised to represent the given physical quantity.
By definition,a unitless quantity is one that does not have any units of measurement.
Since units are derived from fundamental quantities,if a quantity has no units,it implies that the powers of all fundamental quantities (like $M, L, T$) are zero.
Therefore,a unitless quantity is always dimensionless.
Examples include angle (ratio of arc length to radius),elastic strain (ratio of change in length to original length),and Poisson's ratio.