Dimensions and Dimensional Formula Questions in English

(A) According to the principle of homogeneity of dimensions,only physical quantities with the same dimensions can be added or subtracted.
Since $v$ is velocity and $v = a + bt + ct^2$,the dimensions of each term on the right-hand side must be equal to the dimensions of velocity $(v)$.
Therefore,the dimension of $a$ must be equal to the dimension of $v$.
Since the unit of velocity $v$ is $m/s$,the unit of $a$ must also be $m/s$.

(D) Planck's constant $(h)$ relates the energy of a photon $(E)$ to its frequency $(
u)$ by the equation $E = h\nu$.
Rearranging for $h$,we get $h = E / \nu$.
The unit of energy $(E)$ is Joule $(J)$ and the unit of frequency $(
u)$ is Hertz $(Hz)$,which is equivalent to $s^{-1}$.
Therefore,the unit of $h$ is $J / s^{-1} = J \cdot s$ (Joule-second).

(A) Pressure is defined as force per unit area,so its dimensional formula is $[M L^{-1} T^{-2}]$.
Stress is defined as restoring force per unit area,so its dimensional formula is also $[M L^{-1} T^{-2}]$.
Since both have the same dimensional formula,the correct pair is Pressure and stress.

(C) The dimensional formula for $Young's modulus$,$Stress$,and $Pressure$ is given by $\frac{Force}{Area} = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$.
$Strain$ is defined as the ratio of change in dimension to the original dimension,i.e.,$Strain = \frac{\Delta L}{L}$.
Since $Strain$ is a ratio of two similar physical quantities,it is a dimensionless quantity,represented as $[M^0 L^0 T^0]$.
Therefore,the dimensional formula $ML^{-1}T^{-2}$ does not represent $Strain$.

(B) The dimensional formula for Power is given by the ratio of Work to Time.
Power = $\frac{\text{Work}}{\text{Time}}$.
The dimensional formula for Work is $M{L^2}{T^{ - 2}}$.
The dimensional formula for Time is $T$.
Therefore,Power = $\frac{M{L^2}{T^{ - 2}}}{T} = M{L^2}{T^{ - 3}}$.
Thus,the correct option is $B$.

(A) Calorie is a unit of heat,which is a form of energy.
Since energy is defined as the capacity to do work,its dimensional formula is the same as that of work.
Work = Force $\times$ Displacement.
Dimensional formula of Force = $[M L T^{-2}]$.
Dimensional formula of Displacement = $[L]$.
Therefore,the dimensional formula of energy (and thus calorie) = $[M L T^{-2}] \times [L] = [M L^2 T^{-2}]$.

(B) The dimensional formula for angular momentum $(L)$ is given by the product of linear momentum $(p)$ and radius $(r)$.
$L = p \times r = (mv) \times r$
Substituting the dimensions of mass $(M)$,velocity $(L{T^{-1}})$,and length $(L)$:
$L = [M] \times [L{T^{-1}}] \times [L] = M{L^2}{T^{-1}}$
Therefore,the correct option is $(b)$.

(C) The dimension of inductance $L$ is given by the formula $E = \frac{1}{2} L i^2$,where $E$ is energy and $i$ is current. Thus,$L = \frac{2E}{i^2}$.
Substituting dimensions: $[L] = \frac{[ML^2T^{-2}]}{[A^2]} = [ML^2T^{-2}A^{-2}]$.
The dimension of resistance $R$ is given by $V = iR$,so $R = \frac{V}{i} = \frac{W}{qi} = \frac{W}{i^2t}$.
Substituting dimensions: $[R] = \frac{[ML^2T^{-2}]}{[A^2T]} = [ML^2T^{-3}A^{-2}]$.
Now,the dimension of $\frac{L}{R}$ is $\frac{[ML^2T^{-2}A^{-2}]}{[ML^2T^{-3}A^{-2}]} = [T]$.
Therefore,the dimensions of $\frac{L}{R}$ are $[M^0 L^0 T^1]$.

(C) The impulse is defined as the change in momentum,according to the impulse-momentum theorem: $J = \Delta p$. Since impulse is the change in momentum,both quantities share the same physical dimensions. The dimensional formula for both is $[M^1 L^1 T^{-1}]$. Therefore,the correct pair is momentum and impulse.

(B) The time constant of an $RC$ circuit is given by the product of resistance $R$ and capacitance $C$.
Dimensional formula of resistance $R$ is $[R] = [M L^2 T^{-3} A^{-2}]$.
Dimensional formula of capacitance $C$ is $[C] = [M^{-1} L^{-2} T^4 A^2]$.
Multiplying these dimensions: $[RC] = [M L^2 T^{-3} A^{-2}] \times [M^{-1} L^{-2} T^4 A^2]$.
$[RC] = [M^{1-1} L^{2-2} T^{-3+4} A^{-2+2}] = [M^0 L^0 T^1 A^0]$.
Thus,the dimensions of $RC$ are equivalent to time $T$.

(A) The dimensional formula for Torque is $[ML^2T^{-2}]$.
The dimensional formula for Work is $[ML^2T^{-2}]$.
Since both have the same dimensional formula,the correct pair is Torque and Work.
Therefore,the correct option is $A$.

(A) The formula for heat energy $Q$ required to change the state of a substance of mass $m$ is given by $Q = mL$,where $L$ is the latent heat.
Therefore,the latent heat $L$ is given by $L = \frac{Q}{m}$.
Since heat $Q$ is a form of energy,its dimensional formula is $[M^1 L^2 T^{-2}]$.
The dimensional formula for mass $m$ is $[M^1]$.
Substituting these into the formula for $L$:
$L = \frac{[M^1 L^2 T^{-2}]}{[M^1]} = [M^0 L^2 T^{-2}]$.
Thus,the correct option is $A$.

(D) Volume elasticity (Bulk Modulus) is defined as the ratio of stress to volumetric strain.
Volume elasticity = $\frac{\text{Stress}}{\text{Volumetric strain}} = \frac{\text{Force/Area}}{\text{Change in volume/Original volume}}$.
Since strain is a ratio of two similar quantities (change in volume/original volume),it is dimensionless.
Therefore,the dimensions of volume elasticity are the same as the dimensions of stress (Force/Area).
Dimensional formula of Force = $[M^1 L^1 T^{-2}]$.
Dimensional formula of Area = $[L^2]$.
Dimensional formula of Volume elasticity = $\frac{[M^1 L^1 T^{-2}]}{[L^2]} = [M^1 L^{-1} T^{-2}]$.
Thus,the correct option is $D$.

(B) According to Newton's law of universal gravitation,the force $F$ between two masses $m_1$ and $m_2$ separated by a distance $d$ is given by $F = \frac{G m_1 m_2}{d^2}$.
Rearranging the formula to solve for the gravitational constant $G$,we get $G = \frac{F d^2}{m_1 m_2}$.
The dimensional formula for force $F$ is $[MLT^{-2}]$,for distance $d$ is $[L]$,and for mass $m$ is $[M]$.
Substituting these into the expression for $G$: $[G] = \frac{[MLT^{-2}][L^2]}{[M][M]} = \frac{[ML^3T^{-2}]}{[M^2]} = [M^{-1}L^3T^{-2}]$.

(A) Angular velocity $(\omega)$ is defined as the rate of change of angular displacement $( \theta)$ with respect to time $(t)$.
Mathematically,$\omega = \frac{\theta}{t}$.
The dimensional formula for angular displacement $(\theta)$ is dimensionless,represented as $[M^0L^0T^0]$.
The dimensional formula for time $(t)$ is $[T]$.
Therefore,the dimensional formula for angular velocity is $[\omega] = \frac{[M^0L^0T^0]}{[T]} = [M^0L^0T^{-1}]$.

(A) Power is defined as the rate of doing work,which is given by the formula: $P = \frac{W}{t}$.
Work done $(W)$ has the dimensional formula $[M^1 L^2 T^{-2}]$.
Time $(t)$ has the dimensional formula $[T^1]$.
Therefore,the dimensions of power are: $[P] = \frac{[M^1 L^2 T^{-2}]}{[T^1]} = [M^1 L^2 T^{-3}]$.
Thus,the correct option is $A$.

(A) couple is defined as the product of a force and the perpendicular distance between the two forces (arm length).
Mathematically,$\text{Couple} = \text{Force} \times \text{Arm length}$.
The dimensional formula of force is $[MLT^{-2}]$.
The dimensional formula of length is $[L]$.
Therefore,the dimensions of a couple are $[MLT^{-2}] \times [L] = [ML^2T^{-2}]$.

(B) Angular momentum $(L)$ is defined as the product of linear momentum $(p)$ and the perpendicular distance $(r)$ from the axis of rotation.
Mathematically,$L = p \times r$.
Since linear momentum $p = m \times v$,where $m$ is mass and $v$ is velocity.
The dimensional formula for mass $m$ is $[M]$.
The dimensional formula for velocity $v$ is $[L{T^{ - 1}}]$.
The dimensional formula for distance $r$ is $[L]$.
Therefore,the dimensional formula for angular momentum is $[M] \times [L{T^{ - 1}}] \times [L] = [M{L^2}{T^{ - 1}}]$.

(B) 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) The $r.m.s.$ (root mean square) velocity is a type of velocity,and all velocities have the same dimensions as speed.
Velocity is defined as displacement divided by time,which is $\frac{\text{Length}}{\text{Time}}$.
The dimensional formula for length is $[L]$ and for time is $[T]$.
Therefore,the dimensional formula for velocity is $[L T^{-1}]$.
In terms of mass,length,and time,this is written as $[M^0 L T^{-1}]$.
Thus,the correct option is $A$.

(C) The energy of a photon is given by the equation $E = h \nu$,where $E$ is energy,$h$ is Planck's constant,and $\nu$ is frequency.
Rearranging for $h$,we get $h = \frac{E}{\nu}$.
The dimensional formula for energy $E$ is $[M L^2 T^{-2}]$.
The dimensional formula for frequency $\nu$ is $[T^{-1}]$.
Substituting these into the equation: $[h] = \frac{[M L^2 T^{-2}]}{[T^{-1}]} = [M L^2 T^{-1}]$.
Therefore,the correct option is $C$.

(D) The energy of a photon is given by $E = h\nu$,where $h$ is Planck's constant and $\nu$ is the frequency.
Thus,the dimension of $h$ is $[h] = [E] / [\nu]$.
$[E] = [M L^2 T^{-2}]$ and $[\nu] = [T^{-1}]$.
So,$[h] = [M L^2 T^{-2}] / [T^{-1}] = [M L^2 T^{-1}]$.
Angular momentum $L$ is defined as $L = mvr$,where $m$ is mass,$v$ is velocity,and $r$ is radius.
$[L] = [M] [L T^{-1}] [L] = [M L^2 T^{-1}]$.
Therefore,the dimension of Planck's constant is equal to that of angular momentum.

(A) The energy stored in an inductor is given by the formula $U = \frac{1}{2}Li^2$.
Since energy has the dimensions of work,its dimensional formula is $[M{L^2}{T^{-2}}]$.
Because $\frac{1}{2}$ is a dimensionless constant,the dimensions of $Li^2$ are the same as the dimensions of energy.
Therefore,the dimensions of $Li^2$ are $[M{L^2}{T^{-2}}]$.

(C) Given the relation $m = A/B$,we can express $B$ as $B = A/m$.
Linear density $m$ is defined as mass per unit length,so its dimensions are $[M L^{-1}]$.
Force $A$ has dimensions $[M L T^{-2}]$.
Substituting these into the expression for $B$:
$[B] = \frac{[M L T^{-2}]}{[M L^{-1}]} = [L^2 T^{-2}]$.
Latent heat $L$ is defined as energy per unit mass,so its dimensions are $[M L^2 T^{-2}] / [M] = [L^2 T^{-2}]$.
Thus,the dimensions of $B$ are the same as those of latent heat.

(C) The velocity of sound is a type of speed or velocity,which is defined as the rate of change of displacement with respect to time.
Dimensional formula of velocity = $\frac{[\text{Displacement}]}{[\text{Time}]}$.
Since displacement has the dimension of length $[L]$ and time has the dimension $[T]$,the dimensional formula is $\frac{[L]}{[T]} = [L{T^{ - 1}}]$.
Expressing this in terms of mass,length,and time,we get $[{M^0}L{T^{ - 1}}]$.

(A) The capacitance $C$ is defined as the ratio of charge $Q$ to potential difference $V$,given by $C = \frac{Q}{V}$.
Since $V = \frac{W}{Q}$,where $W$ is work done,we can write $C = \frac{Q^2}{W}$.
The dimensional formula for charge $Q$ is $[A^1T^1]$.
The dimensional formula for work $W$ is $[M^1L^2T^{-2}]$.
Substituting these into the formula: $C = \frac{[A^1T^1]^2}{[M^1L^2T^{-2}]} = \frac{[A^2T^2]}{[M^1L^2T^{-2}]}$.
Simplifying the expression,we get $[C] = [M^{-1}L^{-2}T^4A^2]$.

(B) The dimensional formula for momentum is calculated as follows:
Momentum $(p)$ = mass $(m)$ $\times$ velocity $(v)$.
The dimensions of mass are $[M]$.
The dimensions of velocity are $[LT^{-1}]$.
Therefore,the dimensional formula for momentum is $[M] \times [LT^{-1}] = [MLT^{-1}]$.
Thus,option $(b)$ is correct.

(A) Heat energy is a form of energy. The dimensional formula for any form of energy (kinetic,potential,heat,work,etc.) is derived from the formula $Work = Force \times Displacement$.
Since $Force = Mass \times Acceleration$,the dimensions of force are $[M^1L^1T^{-2}]$.
Therefore,the dimensions of energy are $[M^1L^1T^{-2}] \times [L^1] = [M^1L^2T^{-2}]$.
Thus,the correct option is $A$.

(D) The dimension of energy is $[ML^2T^{-2}]$.
Work is defined as the product of force and displacement,$W = F \cdot d$.
The dimensions of work are $[MLT^{-2}] \times [L] = [ML^2T^{-2}]$.
Since energy and work have the same dimensional formula $[ML^2T^{-2}]$,the correct option is $D$.

(D) The dimensional formula for Work is $[ML^2T^{-2}]$.
The dimensional formula for Energy is $[ML^2T^{-2}]$.
The dimensional formula for Power is $[ML^2T^{-3}]$.
The dimensional formula for Momentum is $[MLT^{-1}]$.
The dimensional formula for Force is $[MLT^{-2}]$.
Since Work and Energy both have the dimensional formula $[ML^2T^{-2}]$,they have the same dimensions.
Therefore,option $D$ is correct.

(D) The dimensions of the given pairs are as follows:
$(a)$ Work and energy both have dimensions $[M L^2 T^{-2}]$.
$(b)$ Angle (arc/radius) and strain (change in length/original length) are both dimensionless quantities,i.e.,$[M^0 L^0 T^0]$.
$(c)$ Relative density (density of substance/density of water) and refractive index (speed of light in vacuum/speed of light in medium) are both ratios of similar physical quantities,hence they are dimensionless,i.e.,$[M^0 L^0 T^0]$.
$(d)$ Planck constant $(h)$ has dimensions $[M L^2 T^{-1}]$,while energy $(E)$ has dimensions $[M L^2 T^{-2}]$.
Since their dimensions are different,option $(d)$ is the correct answer.

(B) Frequency is defined as the number of oscillations or cycles per unit time.
Mathematically,$\text{Frequency} = \frac{1}{\text{Time period}}$.
The dimensional formula for time is $[T]$.
Therefore,the dimensions of frequency are $\frac{1}{[T]} = [T^{-1}]$.
In terms of mass,length,and time,this is expressed as $[M^0 L^0 T^{-1}]$.
Thus,the correct option is $B$.

(B) The dimensional formula of energy (kinetic or potential) is given by the product of force and displacement.
Force $F = [M L T^{-2}]$
Displacement $d = [L]$
Energy $E = F \times d = [M L T^{-2}] \times [L] = [M L^2 T^{-2}]$
Pressure is $[M L^{-1} T^{-2}]$,Momentum is $[M L T^{-1}]$,and Power is $[M L^2 T^{-3}]$.
Therefore,the expression $[M L^2 T^{-2}]$ represents Kinetic energy.

(A) The speed of light in vacuum is given by the relation $c = \frac{1}{\sqrt{\varepsilon_0 \mu_0}}$.
Here,$\varepsilon_0$ is the permittivity of free space and $\mu_0$ is the permeability of free space.
Since $c$ represents the speed of light,its dimension is the same as that of velocity,which is $[LT^{-1}]$.
Therefore,the dimension of $\frac{1}{\sqrt{\varepsilon_0 \mu_0}}$ is that of velocity.

(A) According to the problem,$\text{muscle} \times \text{speed} = \text{power}$.
Therefore,$\text{muscle} = \frac{\text{power}}{\text{speed}}$.
The dimensional formula for power is $[ML^2T^{-3}]$.
The dimensional formula for speed is $[LT^{-1}]$.
Substituting these values: $\text{muscle} = \frac{[ML^2T^{-3}]}{[LT^{-1}]} = [MLT^{-2}]$.

(B) The wave number is defined as the reciprocal of the wavelength $(\lambda)$.
Wave number = $\frac{1}{\lambda}$.
Since the wavelength $(\lambda)$ is a measure of length,its dimensional formula is $[M^0 L^1 T^0]$.
Therefore,the dimensional formula of the wave number is $\frac{1}{[M^0 L^1 T^0]} = [M^0 L^{-1} T^0]$.
Thus,the correct option is $B$.

(C) Pressure is defined as the force applied per unit area.
$P = \frac{F}{A}$
The dimensional formula for force $(F)$ is $[MLT^{-2}]$.
The dimensional formula for area $(A)$ is $[L^2]$.
Therefore,the dimensions of pressure are:
$[P] = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$.

(A) The force per unit length between two parallel current-carrying wires is given by the formula: $F/l = \frac{\mu_0 I_1 I_2}{2\pi r}$.
Rearranging for $\mu_0$,we get: $\mu_0 = \frac{2\pi r F}{l I_1 I_2}$.
The dimensional formula for force $F$ is $[M^1 L^1 T^{-2}]$.
The dimensional formula for length $l$ is $[L^1]$.
The dimensional formula for current $I$ is $[A^1]$.
The dimensional formula for distance $r$ is $[L^1]$.
Substituting these into the formula: $[\mu_0] = \frac{[L^1] [M^1 L^1 T^{-2}]}{[L^1] [A^1] [A^1]}$.
Simplifying the expression: $[\mu_0] = [M^1 L^1 T^{-2} A^{-2}]$.

(A) The energy $E$ stored in an inductor is given by the formula $E = \frac{1}{2} L i^2$.
Rearranging for $L$,we get $L = \frac{2E}{i^2}$.
The dimensional formula for energy $E$ is $[M L^2 T^{-2}]$.
The dimensional formula for current $i$ is $[A]$.
Substituting these into the expression for $L$:
$L = \frac{[M L^2 T^{-2}]}{[A]^2} = [M L^2 T^{-2} A^{-2}]$.
Thus,the correct option is $A$.

(C) Power $(P)$ is defined as the rate of doing work or the rate of energy transfer.
$P = \frac{\text{Work}}{\text{Time}} = \frac{W}{t}$
The dimensional formula for work $(W)$ is $[ML^2T^{-2}]$.
The dimensional formula for time $(t)$ is $[T]$.
Therefore,the dimensional formula for power is:
$P = \frac{[ML^2T^{-2}]}{[T]} = [ML^2T^{-3}]$
Comparing this with the general dimensional formula $[M^aL^bT^c]$,the dimension of time is $c = -3$.
Thus,the correct option is $C$.

(A) The formula for kinetic energy is $K.E. = \frac{1}{2}mv^2$.
Here,$m$ is mass,which has dimensions $[M]$.
$v$ is velocity,which has dimensions $[LT^{-1}]$.
Substituting these into the formula,we get:
$[K.E.] = [M] \times [LT^{-1}]^2 = [M] \times [L^2T^{-2}] = [ML^2T^{-2}]$.
Therefore,the dimensions of kinetic energy are $[ML^2T^{-2}]$.

(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 $A$.