The potential energy of a particle that varies with distance ${x}$ is ${V}(x)=-\alpha x+\beta \sin \frac{x}{\gamma}$. A dimensionless combination of constants $\alpha, \beta, \gamma$ is

$\frac{\alpha^2}{\beta \alpha}$

JEE PHYSICS MCQ: UNITS & MEASUREMENTS

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Application +4 | -1

The potential energy of a particle that varies with distance

{x}

{V}(x)=-\alpha x+\beta \sin \frac{x}{\gamma}

. A dimensionless combination of constants

\alpha, \beta, \gamma

View answer and explanation

Correct option: D

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

The correct answer is

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

because the dimensions of each term in the potential energy equation must be the same, allowing us to find a dimensionless combination.

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