An unknown quantity \displaystyle "\alpha" is expressed \displaystyle \alpha =\frac{2ma}{\beta }\log \left ( 1+\frac{2\beta \varphi }{ma} \right ) where m = mass, a = acceleration, \displaystyle \varphi = length The unit of \displaystyle \alpha should be

\alpha =\dfrac { 2ma }{ \beta } \log { 1+\dfrac { 2\beta \psi }{ ma } } First find unit of \beta .The value of log is dimensionless so dimension of: \beta \psi =ma \beta (m)=(kg)(m/sec^2) Unit of \beta =\dfrac { kg }{ { sec }^{ 2 } } Unit of \alpha =\dfrac { kg.m/{ sec }^{ 2 } }{ kg/{ sec }^{ 2 } } =m= meter

The velocity of a body moving viscous medium is given by V=\dfrac { A }{ B } [1-{ e }^{ { -t }/{ B } }] where t is time, A and B are constants. Then the dimensional formula of A is:

In e^\frac{-t}{B} is the exponential term. So, the \dfrac{t}{B} is constant.So, B has the dimensions as: B=[T] On the other hand, the dimension of \dfrac AB will be the same as that of the velocity. \dfrac{A}{B} has dimensions as velocity [LT^{-1}]=\dfrac{a}{[T]} dimensional formula of A is M^0L^1 T^0

Find the dimensional formula for {\mu}_{o} and { \varepsilon }_{ o } :

Since, k=\dfrac{1}{4\pi\epsilon_0} , dimension of [\epsilon_0]=[k^{-1}]=[ML^3T^{-4}A^{-2}]^{-1}=[M^{-1}L^{-3}T^4A^2] Also we know that , c=\dfrac{1}{\sqrt(\mu_0\epsilon_0)} this gives , dimension of [\mu_0]=[c^2\epsilon_0]^{-1}=[MLT^{-2}A^{2}]

\displaystyle \alpha =\frac{F}{v^{2}}\sin (\beta t) (here v = velocity, F = force, t = time)Find the dimension of \displaystyle \alpha and \displaystyle \beta

\displaystyle \alpha =\frac{F}{v^{2}}\sin (\beta t) dimensionless dimensionless \Rightarrow \displaystyle [\beta ][t]=1=[\beta ]=[T^{-1}] So \displaystyle [\alpha ]=\frac{[F]}{V^{2}}=\frac{[M^{1}L^{1}T^{-2}]}{[L^{1}T^{-1}]^{2}}=M^{1}L^{-1}T^{0}

In the relation, \displaystyle P=\frac{\alpha }{\beta }e^{{\alpha z}/{k\theta }} P is pressure, Z is distance, K is Boltzmann constant and \displaystyle \theta is the temperature. The dimensions of \displaystyle \beta will be

P=\dfrac { \alpha }{ \beta } { e }^{ \dfrac { \alpha z }{ k\theta } } The dimension of power of e is zero. \dfrac { \alpha z }{ k\theta } =\quad \left[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 } \right] Thus unit of: \alpha =\dfrac { k\theta }{ z } Unit of boltzman's constant K is K=\left[ { M }^{ 1 }{ L }^{ 2 }{ T }^{ -2 }K^{ -1 } \right] By putting thes values we get,Unit of \alpha =\dfrac { \left[ { M }^{ 1 }{ L }^{ 2 }{ T }^{ -2 }K^{ -1 } \right] \left[ { K }^{ 1 } \right] }{ \left[ { L }^{ 1 } \right] } =\left[ { M }^{ 1 }{ L }^{ 1 }{ T }^{ -2 }K^{ 0 } \right] Therefore unit of P=\dfrac { \alpha }{ \beta } Unit of \beta =\dfrac { \alpha }{ P } \beta =\dfrac { \left[ { M }^{ 1 }{ L }^{ 1 }{ T }^{ -2 }K^{ 0 } \right] }{ \left[ { { M }^{ 1 }{ L }^{ -1 }{ T }^{ -2 } } \right] } =\left[ { { M }^{ 0 }{ L }^{ 2 }{ T }^{ 0 } } \right]

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Important questions on Dimensions And Dimensional Analysis

Q: Match List I with List II and select the correct answer using the codes given below the lists:

\mathrm{A}.\ \rightarrow(4) ; \mathrm{B}.\ \rightarrow(2) ; \mathrm{C}.\ \rightarrow(1) ; \mathrm{D}.\ \rightarrow(3) \mathrm{A}.\ \displaystyle \mathrm{K}\mathrm{E}=\frac{3}{2}\mathrm{K}\mathrm{T} \Rightarrow [\mathrm{M}\mathrm{L}^{2}\mathrm{T}^{-2}]=\mathrm{K}'[\mathrm{K}]\Rightarrow \mathrm{K}'=[\mathrm{M}\mathrm{L}^{2}\mathrm{T}^{-2}\mathrm{K}^{-1}] \mathrm{B}.\ \mathrm{F}=6\pi\eta \mathrm{r}\mathrm{v} \Rightarrow [\mathrm{M}\mathrm{L}\mathrm{T}^{-2}]=\eta[\mathrm{L}][\mathrm{L}\mathrm{T}^{-1}]\Rightarrow \eta =[\mathrm{M}\mathrm{L}^{-1}\mathrm{T}^{-1}] \mathrm{C}.\ \mathrm{E}= \mathrm{h}\mathrm{f} \Rightarrow [\displaystyle \mathrm{M}\mathrm{L}^{2}\mathrm{T}^{-2}]=\frac{\mathrm{h}}{[\mathrm{T}]} \Rightarrow \mathrm{h}=[\mathrm{M}\mathrm{L}^{2}\mathrm{T}^{-1}] \mathrm{D}.\ \displaystyle \frac{\mathrm{d}\mathrm{Q}}{\mathrm{d}\mathrm{t}}=\frac{\mathrm{K}'\mathrm{A}(\Delta \mathrm{T})}{\Delta \mathrm{x}}\Rightarrow \frac{[\mathrm{M}\mathrm{L}^{2}\mathrm{T}^{-2}]}{[\mathrm{T}]}=\frac{\mathrm{k}[\mathrm{L}^{2}][\mathrm{K}']}{[\mathrm{L}]} \mathrm{K}'=[\mathrm{M}\mathrm{L}\mathrm{T}^{-3}\mathrm{K}^{-1}]

Q: Which of the following does not have the same dimension?

B) Pressure: \dfrac { force }{ Area } Stress : \dfrac { force }{ Area } Youngs modulus : \dfrac { stress }{ strain } , here strain is dimension less So, all three has same dimension.C) Electromotive force: \dfrac { Energy }{ charge } Potential difference: \dfrac { W}{ Q } , with is same as Electromotive Electric voltage: If you see in a capacitor, then Voltage is given by \dfrac { Energy }{ charge } D) Heat: ML^2T^{-2} Potential energy: mgh \Rightarrow ML^2T^{-2} Work done =f.d\Rightarrow ML^2T^{-2} A) Electric flux: M^{-2}SA Electric field: M^2S^{-3}A^{-1} Hence, it has quantities with different units.

Q: The Van der Waal's equation of 'n' moles of a real gas is \displaystyle \left( P+\frac { a }{ { V }^{ 2 } } \right) \left( V-b \right) =nRT Where P is pressure, V is volume, T is absolute temperature, R is molar gas constant and a, b, c are Van der Waal constants. The dimensional formula for ab is:

Since, the dimensions of P must be same as \dfrac { a }{ { V }^{ 2 } } , Hence, \dfrac { [a] }{ L^{ 6 } } =[M{ L }^{ -1 }{ T }^{ -2 }] [a]=[M{ L }^{ 5 }{ T }^{ -2 }] Also, dimension of b must be same as that of V . Hence, [b]={ L }^{ 3 } .Thus, [ab]=[M{ L }^{ 8 }{ T }^{ -2 }]

Q: Pressure \displaystyle P_{r}=\frac{3Fv^{2}}{\pi ^{2}t^{2}x} (where \displaystyle P_{r} = Pressure F = force, v = velocity, t = time, x = distance)State whether true or false: the above equation is dimentionally correct

Pressure \displaystyle P_{r}=\frac{3Fv^{2}}{\pi ^{2}t^{2}x} (where \displaystyle P_{r} = Pressure F = force, v = velocity, t = time, x = distance)Dimension of L.H.S = \displaystyle [P_{r}]=M^{1}L^{-1}T^{-2} Dimension of R.H.S. = \displaystyle \frac{[3][F][V^{2}]}{[\pi ][t^{2}][x]}=\frac{[M^{1}L^{1}T^{-2}][L^{2}T^{2}]}{[T^{2}][L]}=M^{1}L^{2}T^{-6} Dimension of L.H.S. and R.H.S are not same So the relation cannot be correct Sometimes a question is asked which is beyond our syllabus, then certainly it must be the question of dimensional analyses

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easy117 Questionsmedium298 Questionshard25 Questions

An unknown quantity $" α "$ is expressed $α = \frac{2 m a}{β} lo g (1 + \frac{2 β φ}{m a})$
where $m$ = mass, $a$ = acceleration, $φ$ = length
The unit of $α$ should be

meter

$m / s$

$m / s$ $^{2}$

$s$ $^{- 1}$

View Answer

$α = \frac{2 m a}{β} lo g 1 + \frac{2 β ψ}{m a}$

First find unit of $β$ .

The value of log is dimensionless so dimension of:

$β ψ = m a$

$β (m) = (k g) (m / s e c^{2})$

Unit of $β = \frac{k g}{s e c ^{2}}$

Unit of $α = \frac{k g . m / s e c ^{2}}{k g / s e c ^{2}} = m = m e t e r$

Match List I with List II and select the correct answer using the codes given below the lists:

View Answer

$A . \to (4)$ ; $B . \to (2)$ ;

$C . \to (1)$ ; $D . \to (3)$

$A . K E = \frac{3}{2} K T \Rightarrow [M L^{2} T^{- 2}] = K^{'} [K] \Rightarrow K^{'} = [M L^{2} T^{- 2} K^{- 1}]$

$B . F = 6 π η r v \Rightarrow [M L T^{- 2}] = η [L] [L T^{- 1}] \Rightarrow η = [M L^{- 1} T^{- 1}]$

$C . E = h f \Rightarrow [M L^{2} T^{- 2}] = \frac{h}{[ T ]} \Rightarrow h = [M L^{2} T^{- 1}]$

$D . \frac{d Q}{d t} = \frac{K ^{'} A ( Δ T )}{Δ x} \Rightarrow \frac{[ M L ^{2} T ^{- 2} ]}{[ T ]} = \frac{k [ L ^{2} ] [ K ^{'} ]}{[ L ]}$

$K^{'} = [M L T^{- 3} K^{- 1}]$

The velocity of a body moving viscous medium is given by $V = \frac{A}{B} [1 - e^{- t / B}]$ where $t$ is time, $A$ and $B$ are constants. Then the dimensional formula of $A$ is:

$M^{0} L^{0} T^{0}$

$M^{0} L^{1} T^{0}$

$M^{0} L^{1} T^{- 1}$

$M^{1} L^{1} T^{- 1}$

View Answer

In $e^{\frac{- t}{B}}$ is the exponential term. So, the $\frac{t}{B}$ is constant.
So, B has the dimensions as:
$B = [T]$

On the other hand, the dimension of $\frac{A}{B}$ will be the same as that of the velocity.
$\frac{A}{B}$ has dimensions as velocity

$[L T^{- 1}] = \frac{a}{[ T ]}$

dimensional formula of A is $M^{0} L^{1} T^{0}$

Find the dimensional formula for $μ_{o}$ and $ε_{o}$ :

View Answer

Since, $k = \frac{1}{4 π ϵ _{0}}$ ,
dimension of $[ϵ_{0}] = [k^{- 1}] = [M L^{3} T^{- 4} A^{- 2}]^{- 1} = [M^{- 1} L^{- 3} T^{4} A^{2}]$
Also we know that , $c = \frac{1}{( μ _{0} ϵ _{0} )}$
this gives , dimension of $[μ_{0}] = [c^{2} ϵ_{0}]^{- 1} = [M L T^{- 2} A^{2}]$

Which of the following does not have the same dimension?

Electric flux, Electric field, Electric dipole moment

Pressure, stress, Youngs modulus

Electromotive force, Potential difference, Electric voltage

Heat, Potential energy, Work done

View Answer

B) Pressure: $\frac{f o r c e}{A r e a}$
Stress : $\frac{f o r c e}{A r e a}$
Youngs modulus : $\frac{s t r e s s}{s t r a i n}$ , here strain is dimension less So, all three has same dimension.

C) Electromotive force: $\frac{E n e r g y}{c h a r g e}$
Potential difference: $\frac{W}{Q}$ , with is same as Electromotive
Electric voltage: If you see in a capacitor, then Voltage is given by $\frac{E n e r g y}{c h a r g e}$

D) Heat: $M L^{2} T^{- 2}$
Potential energy: $m g h \Rightarrow M L^{2} T^{- 2}$
Work done $= f . d \Rightarrow M L^{2} T^{- 2}$

A) Electric flux: $M^{- 2} S A$
Electric field: $M^{2} S^{- 3} A^{- 1}$

Hence, it has quantities with different units.

$α = \frac{F}{v ^{2}} sin (β t)$ (here v = velocity, F = force, t = time)
Find the dimension of $α$ and $β$

View Answer

$α = \frac{F}{v ^{2}} sin (β t)$
dimensionless dimensionless $\Rightarrow$ $[β] [t] = 1 = [β] = [T^{- 1}]$
So $[α] = \frac{[ F ]}{V ^{2}} = \frac{[ M ^{1} L ^{1} T ^{- 2} ]}{[ L ^{1} T ^{- 1} ] ^{2}} = M^{1} L^{- 1} T^{0}$

The Van der Waal's equation of $^{'} n^{'}$ moles of a real gas is
$(P + \frac{a}{V ^{2}}) (V - b) = n R T$
Where $P$ is pressure, $V$ is volume, $T$ is absolute temperature, $R$ is molar gas constant and $a, b, c$ are Van der Waal constants. The dimensional formula for $a b$ is:

[ $M L^{8} T^{- 2}$ ]

[ $M L^{6} T^{- 2}$ ]

[ $M L^{4} T^{- 2}$ ]

[ $M L^{2} T^{- 2}$ ]

View Answer

Since, the dimensions of $P$ must be same as $\frac{a}{V ^{2}}$ ,
Hence,
$\frac{[ a ]}{L ^{6}} = [M L^{- 1} T^{- 2}]$

$[a] = [M L^{5} T^{- 2}]$

Also, dimension of $b$ must be same as that of $V$ . Hence, $[b] = L^{3}$ .

Thus, $[a b] = [M L^{8} T^{- 2}]$

Two forces $P$ and $Q$ act at a point and have resultant $R$ . If $Q$ is replaced by $\frac{( R ^{2} - P ^{2} )}{Q}$ acting in the direction opposite to that of $Q$ , the resultant:

remains same

becomes half

becomes twice

none of these

View Answer

Pressure $P_{r} = \frac{3 F v ^{2}}{π ^{2} t ^{2} x}$ (where $P_{r}$ = Pressure F = force, v = velocity, t = time, x = distance)
State whether true or false: the above equation is dimentionally correct

True

False

View Answer

Pressure $P_{r} = \frac{3 F v ^{2}}{π ^{2} t ^{2} x}$ (where $P_{r}$ = Pressure F = force, v = velocity, t = time, x = distance)

Dimension of L.H.S = $[P_{r}] = M^{1} L^{- 1} T^{- 2}$

Dimension of R.H.S. = $\frac{[ 3 ] [ F ] [ V ^{2} ]}{[ π ] [ t ^{2} ] [ x ]} = \frac{[ M ^{1} L ^{1} T ^{- 2} ] [ L ^{2} T ^{2} ]}{[ T ^{2} ] [ L ]} = M^{1} L^{2} T^{- 6}$

Dimension of L.H.S. and R.H.S are not same So the relation cannot be correct Sometimes a question is asked which is beyond our syllabus, then certainly it must be the question of dimensional analyses

In the relation, $P = \frac{α}{β} e^{α z / k θ}$ $P$ is pressure, $Z$ is distance, $K$ is Boltzmann constant and $θ$ is the temperature. The dimensions of $β$ will be

$[M^{\circ} L^{2} T^{0}]$

$[M L^{2} T]$

$[M L^{\circ} T^{1}]$

$[M^{\circ} L^{2} T^{- 1}]$

View Answer

$P = \frac{α}{β} e^{\frac{α z}{k θ}}$

The dimension of power of e is zero.
$\frac{α z}{k θ} = [M^{0} L^{0} T^{0}]$

Thus unit of: $α = \frac{k θ}{z}$
Unit of boltzman's constant K is $K = [M^{1} L^{2} T^{- 2} K^{- 1}]$

By putting thes values we get,
Unit of $α = \frac{[ M ^{1} L ^{2} T ^{- 2} K ^{- 1} ] [ K ^{1} ]}{[ L ^{1} ]} = [M^{1} L^{1} T^{- 2} K^{0}]$

Therefore unit of $P = \frac{α}{β}$

Unit of $β = \frac{α}{P}$

$β = \frac{[ M ^{1} L ^{1} T ^{- 2} K ^{0} ]}{[ M ^{1} L ^{- 1} T ^{- 2} ]} = [M^{0} L^{2} T^{0}]$

Important questions on Dimensions And Dimensional Analysis

An unknown quantity "α" is expressed α=β2ma​log(1+ma2βφ​) where m = mass, a = acceleration, φ = length The unit of α should be

meter

m/s

m/s2

s−1

α=β2ma​log1+ma2βψ​ First find unit of β. The value of log is dimensionless so dimension of: βψ=ma β(m)=(kg)(m/sec2) Unit of β=sec2kg​ Unit of α=kg/sec2kg.m/sec2​=m=meter

Match List I with List II and select the correct answer using the codes given below the lists:

A. →(4) ; B. →(2) ; C. →(1) ; D. →(3) A. KE=23​KT⇒[ML2T−2]=K′[K]⇒K′=[ML2T−2K−1] B. F=6πηrv⇒[MLT−2]=η[L][LT−1]⇒η=[ML−1T−1] C. E=hf⇒[ML2T−2]=[T]h​⇒h=[ML2T−1] D. dtdQ​=ΔxK′A(ΔT)​⇒[T][ML2T−2]​=[L]k[L2][K′]​ K′=[MLT−3K−1]

The velocity of a body moving viscous medium is given by V=BA​[1−e−t/B] where t is time, A and B are constants. Then the dimensional formula of A is:

M0L0T0

M0L1T0

M0L1T−1

M1L1T−1

In eB−t​ is the exponential term. So, the Bt​ is constant.So, B has the dimensions as:B=[T] On the other hand, the dimension of BA​ will be the same as that of the velocity.BA​ has dimensions as velocity [LT−1]=[T]a​ dimensional formula of A is M0L1T0

Find the dimensional formula for μo​ and εo​:

Since, k=4πϵ0​1​, dimension of [ϵ0​]=[k−1]=[ML3T−4A−2]−1=[M−1L−3T4A2]Also we know that , c=(​μ0​ϵ0​)1​this gives , dimension of [μ0​]=[c2ϵ0​]−1=[MLT−2A2]

Which of the following does not have the same dimension?

Electric flux, Electric field, Electric dipole moment

Pressure, stress, Youngs modulus

Electromotive force, Potential difference, Electric voltage

Heat, Potential energy, Work done

α=v2F​sin(βt) (here v = velocity, F = force, t = time) Find the dimension of α and β

α=v2F​sin(βt) dimensionless dimensionless ⇒ [β][t]=1=[β]=[T−1] So [α]=V2[F]​=[L1T−1]2[M1L1T−2]​=M1L−1T0

The Van der Waal's equation of ′n′ moles of a real gas is (P+V2a​)(V−b)=nRT Where P is pressure, V is volume, T is absolute temperature, R is molar gas constant and a,b,c are Van der Waal constants. The dimensional formula for ab is:

[ML8T−2]

[ML6T−2]

[ML4T−2]

[ML2T−2]

Since, the dimensions of P must be same as V2a​, Hence,L6[a]​=[ML−1T−2] [a]=[ML5T−2] Also, dimension of b must be same as that of V. Hence, [b]=L3. Thus, [ab]=[ML8T−2]

Two forces P and Q act at a point and have resultant R. If Q is replaced by Q(R2−P2)​ acting in the direction opposite to that of Q, the resultant:

remains same

becomes half

becomes twice

none of these

Pressure Pr​=π2t2x3Fv2​ (where Pr​ = Pressure F = force, v = velocity, t = time, x = distance)State whether true or false: the above equation is dimentionally correct

True

False

In the relation, P=βα​eαz/kθ P is pressure, Z is distance, K is Boltzmann constant and θ is the temperature. The dimensions of β will be

[M∘L2T0]

[ML2T]

[ML∘T 1]

[M∘L2T−1]

More important questions

An unknown quantity $" α "$ is expressed $α = \frac{2 m a}{β} lo g (1 + \frac{2 β φ}{m a})$
where $m$ = mass, $a$ = acceleration, $φ$ = length
The unit of $α$ should be

$m / s$

$m / s$ $^{2}$

$s$ $^{- 1}$

$α = \frac{2 m a}{β} lo g 1 + \frac{2 β ψ}{m a}$

First find unit of $β$ .

The value of log is dimensionless so dimension of:

$β ψ = m a$

$β (m) = (k g) (m / s e c^{2})$

Unit of $β = \frac{k g}{s e c ^{2}}$

Unit of $α = \frac{k g . m / s e c ^{2}}{k g / s e c ^{2}} = m = m e t e r$

The velocity of a body moving viscous medium is given by $V = \frac{A}{B} [1 - e^{- t / B}]$ where $t$ is time, $A$ and $B$ are constants. Then the dimensional formula of $A$ is:

$M^{0} L^{0} T^{0}$

$M^{0} L^{1} T^{0}$

$M^{0} L^{1} T^{- 1}$

$M^{1} L^{1} T^{- 1}$

Find the dimensional formula for $μ_{o}$ and $ε_{o}$ :

Since, $k = \frac{1}{4 π ϵ _{0}}$ ,
dimension of $[ϵ_{0}] = [k^{- 1}] = [M L^{3} T^{- 4} A^{- 2}]^{- 1} = [M^{- 1} L^{- 3} T^{4} A^{2}]$
Also we know that , $c = \frac{1}{( μ _{0} ϵ _{0} )}$
this gives , dimension of $[μ_{0}] = [c^{2} ϵ_{0}]^{- 1} = [M L T^{- 2} A^{2}]$

$α = \frac{F}{v ^{2}} sin (β t)$ (here v = velocity, F = force, t = time)
Find the dimension of $α$ and $β$

$α = \frac{F}{v ^{2}} sin (β t)$
dimensionless dimensionless $\Rightarrow$ $[β] [t] = 1 = [β] = [T^{- 1}]$
So $[α] = \frac{[ F ]}{V ^{2}} = \frac{[ M ^{1} L ^{1} T ^{- 2} ]}{[ L ^{1} T ^{- 1} ] ^{2}} = M^{1} L^{- 1} T^{0}$

The Van der Waal's equation of $^{'} n^{'}$ moles of a real gas is
$(P + \frac{a}{V ^{2}}) (V - b) = n R T$
Where $P$ is pressure, $V$ is volume, $T$ is absolute temperature, $R$ is molar gas constant and $a, b, c$ are Van der Waal constants. The dimensional formula for $a b$ is:

[ $M L^{8} T^{- 2}$ ]

[ $M L^{6} T^{- 2}$ ]

[ $M L^{4} T^{- 2}$ ]

[ $M L^{2} T^{- 2}$ ]

Since, the dimensions of $P$ must be same as $\frac{a}{V ^{2}}$ ,
Hence,
$\frac{[ a ]}{L ^{6}} = [M L^{- 1} T^{- 2}]$

$[a] = [M L^{5} T^{- 2}]$

Also, dimension of $b$ must be same as that of $V$ . Hence, $[b] = L^{3}$ .

Thus, $[a b] = [M L^{8} T^{- 2}]$

Two forces $P$ and $Q$ act at a point and have resultant $R$ . If $Q$ is replaced by $\frac{( R ^{2} - P ^{2} )}{Q}$ acting in the direction opposite to that of $Q$ , the resultant:

Pressure $P_{r} = \frac{3 F v ^{2}}{π ^{2} t ^{2} x}$ (where $P_{r}$ = Pressure F = force, v = velocity, t = time, x = distance)
State whether true or false: the above equation is dimentionally correct

In the relation, $P = \frac{α}{β} e^{α z / k θ}$ $P$ is pressure, $Z$ is distance, $K$ is Boltzmann constant and $θ$ is the temperature. The dimensions of $β$ will be

$[M^{\circ} L^{2} T^{0}]$

$[M L^{2} T]$

$[M L^{\circ} T^{1}]$

$[M^{\circ} L^{2} T^{- 1}]$