Question

If equation $\int \frac{dt}{\sqrt{3at-2t^2}}=a^{x}si{n}^{-1}\left(\frac{t^2}{a^2}-1\right)$ the value of $x$ is

• A. $\frac{3}{2}$
• B. $0$
• C. $\frac{1}{2}$
• D. $-\frac{1}{2}$

Answer 1

Answer: (B)

$\left(\frac{t^2}{a^2}-1\right)$ 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 $x=0$