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(D) The potential energy of the particle is given by $V(x) = -\alpha x + \beta \sin(x / \gamma)$.For the equation to be dimensionally consistent,the d

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$\frac{\alpha \gamma}{\beta}$

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(D) The potential energy of the particle is given by $V(x) = -\alpha x + \beta \sin(x / \gamma)$.For the equation to be dimensionally consistent,the dimensions of each term must be the same as the dimensions of potential energy $[V] = [ML^2T^{-2}]$.$1$. For the term $\alpha x$: $[\alpha][x] = [V] \implies [\alpha][L] = [ML^2T^{-2}] \implies [\alpha] = [MLT^{-2}]$.$2$. For the term $\beta \sin(x / \gamma)$: The argument of the sine function must be dimensionless,so $[x / \gamma] = [M^0L^0T^0] \implies [L] / [\gamma] = 1 \implies [\gamma] = [L]$.$3$. Also,$[\beta] = [V] = [ML^2T^{-2}]$.Now,check the dimensions of the given options:For option $(d)$: $\frac{[\alpha][\gamma]}{[\beta]} = \frac{[MLT^{-2}][L]}{[ML^2T^{-2}]} = \frac{[ML^2T^{-2}]}{[ML^2T^{-2}]} = [M^0L^0T^0]$.Since this combination is dimensionless,the correct option is $(d)$.

The potential energy of a point particle is given by the expression $V(x) = -\alpha x + \beta \sin(x / \gamma)$. $A$ dimensionless combination of the constants $\alpha, \beta$ and $\gamma$ is

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