The force of interaction between two atoms is | vedclass

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(B) The argument of the exponential function must be dimensionless.Therefore,$[\frac{x^2}{\alpha kt}] = [M^0L^0T^0]$.Since $[x^2] = L^2$ and $[kt] = [

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$M^2LT^{-4}$

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(B) The argument of the exponential function must be dimensionless.Therefore,$[\frac{x^2}{\alpha kt}] = [M^0L^0T^0]$.Since $[x^2] = L^2$ and $[kt] = [Energy] = ML^2T^{-2}$,we have:$[\alpha] = \frac{[x^2]}{[kt]} = \frac{L^2}{ML^2T^{-2}} = M^{-1}T^2$.The force $F$ is given by $F = \alpha \beta \exp(\dots)$. Since the exponential term is dimensionless,the dimensions of $F$ must be equal to the dimensions of $\alpha \beta$.$[F] = [\alpha][\beta]$$MLT^{-2} = (M^{-1}T^2) [\beta]$$[\beta] = \frac{MLT^{-2}}{M^{-1}T^2} = M^2LT^{-4}$.

Column-$I$	Column-$II$
$(1)$ Torque	$(a)$ $M^1L^1T^{-1}$
$(2)$ Angular momentum	$(b)$ $M^1L^2T^{-1}$
$(3)$ Linear momentum	$(c)$ $M^1L^2T^{-2}$

The force of interaction between two atoms is given by $F = \alpha \beta \exp \left( - \frac{x^2}{\alpha kt} \right)$,where $x$ is the distance,$k$ is the Boltzmann constant,$T$ is temperature,and $\alpha$ and $\beta$ are two constants. The dimension of $\beta$ is:

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Match the physical quantities in Column-$I$ with their corresponding dimensional formulas in Column-$II$.

Column-$I$ Column-$II$

$(1)$ Torque $(a)$ $M^1L^1T^{-1}$

$(2)$ Angular momentum $(b)$ $M^1L^2T^{-1}$

$(3)$ Linear momentum $(c)$ $M^1L^2T^{-2}$

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