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Q.
If equation $\int\frac{dt}{\sqrt{3at-2t^{2}}}=a^{x} sin ^{-1}\left(\frac{t^{2}}{a^{2}}-1\right)$ the value of $x$ is
Physical World, Units and Measurements
Solution:
$\left(\frac{t^{2}}{a^{2}}-1\right)$ is dimensionless
$\therefore \left[a\right]=\left[t\right]$
As,$\sqrt{\left(3at-2t^{2}\right)}=\left[t\right]$
$\therefore \left[\frac{dt}{\sqrt{3at-t^{2}}}\right]=\frac{\left[t\right]}{\left[t\right]}=\left[M^{0}L^{0}T^{0}\right]$
$a^{x}$ should be dimensionless, so $x = 0$