$\alpha=\frac{F}{v^2}sin(\beta t)$.Then dimensions of $\alpha$ is
Answer & explanation
Correct answer: option 2
$sin(\beta t)$ is dimensionless. So dimension of $\alpha$ is same as dimension of $\frac{F}{V^2}$ = $\frac{[MLT^{-2}]}{[L^2T^{-2}]}$ = $[ML^{-1}T^0]$