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The torque and the work both have the same dimensions because they are given as the product of force and the distance.

$[Momentum]=[MLT_{−1}]$

$[h]=vE =ML_{2}T_{−2}T=[ML_{2}T_{−1}]$.

On the other hand, the young's modulus is the ratio of stress and strain. The strain is a dimensionless quantity. Therefore, the unit of stress and young's modulus will be the same.

The speed of light is given by:

$c=μ_{0}ϵ_{0} 1 $

Therefore, both will have the same dimensions.

So, option $(B)$ is correct.

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$αt_{2}$ must be dimension less in $e_{−αt_{2}}$

$⇒dim(αt_{2})=M_{0}L_{0}T_{0}⇒dim(α)=T_{2}M_{0}L_{0}T_{0} $

$∴$ $dim(α)=M_{0}L_{0}T_{−2}$

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Given, $U=x_{2}+BAx_{1/2} $

$x_{2}+B$ should have some dimension $⇒$ $L_{2}=dim(B)$

$dim(A)=dim(x_{1/2}U×(x_{2}+B) )=L_{1/2}[ML_{2}T_{−2}][L_{2}] $

$=ML_{7/2}T_{−2}$

$dim(A×B)=[ML_{7/2}T_{−2}][L_{2}]=[ML_{11/2}T_{−2}]$

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$h=vE =[T_{−1}][ML_{2}T_{−2}] =[ML_{2}T_{−1}]$

Angular momentum $=$ Moment of inertia $×$ Angular velocity (Angular momentum)

$=[ML_{2}][T_{−1}]=[ML_{2}T_{−1}]$

So, here we can see that both the Planck's constant and Angular momentum have the same dimensions.

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$h=vE =[T_{−1}][ML_{2}T_{−2}] =[ML_{2}T_{−1}]$

Angular momentum $=$ Moment of inertia $×$ Angular velocity (Angular momentum)

$=[ML_{2}][T_{−1}]=[ML_{2}T_{−1}]$

So, here we can see that both the Planck's constant and Angular momentum have the same dimensions.

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The dimensional formula of u is $[M_{0}LT_{−1}]$.

The dimensional formula of v is $[M_{0}LT_{−1}]$.

The dimensional formula of at is $[M_{0}LT_{−2}][T]=[M_{0}LT_{−1}]$.

Here the dimensions of every term in the given physical relation are the same, hence the given physical relation is dimensionally correct.

$(μ_{0}ε_{0})_{−1/2}$ are:

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We know that the speed of an electromagnetic wave is given by :

$c=μ_{0}ε_{0} 1 $Therefore, $(μ_{0}ε_{0})_{−1/2}$ has the dimension the same as that of the speed.

$∴dimensionLT_{−1}$

$∴dimensionLT_{−1}$

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$Latentheat(L)=MassHeat $ . . . . . . (1)

The dimensional formula of mass $=[M_{1}L_{0}T_{0}]$. . . . (2)

Also, the dimensions of heat $=$ dimensions of energy $=$ dimensions of work

Since, $work=force×displacement$ . . . . (3)

And, the dimensional formula of,

Displacement $=[M_{0}L_{1}T_{0}]$. . . (4)

Force $=m×a=[M_{1}L_{1}T_{−2}]$ . . . (5)

On substituting equation (4) and (5) in equation (3) we get,

Work $=[M_{1}L_{1}T_{−2}]×[L_{1}]$

Therefore, the dimensions of work or heat $=[M_{1}L_{2}T_{−2}]$ . . . . (6)

On substituting equation (2) and (6) in equation (1) we get,

Or, $L=[M_{1}L_{0}T_{0}][M_{1}L_{2}T_{−2}] =[M_{0}L_{2}T_{−2}]$

Therefore, latent heat is dimensionally represented as $[M_{0}L_{2}T_{−2}]$

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$N=$ no. of particle per unit area per unit time

$n_{1}$ & $n_{2}$ $=$ no. of particle per unit volume

$x_{1}$ & $x_{2}$ are distance from some reference point.

Dimension of $_{′}N_{′}=[L_{−2}T_{−1}]$

Dimensional formula of $n_{1}$ & $n_{2}$ $=[L_{−3}]$

Dimensional formula of $x_{1}$ & $x_{2}$ $=$ $[L]$

$N=−Dx_{2}−x_{1}n_{2}−n_{1} $

$D=−N[n_{2}−n_{1}x_{2}−x_{1} ]$

$=[L_{−3}][L_{−2}T_{−1}][L] $

$D=[M_{0}L_{2}T_{−1}] $

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