Question

In the equation $\int \frac{dv}{v^{\frac{3}{2}}}=BC{e}^{-2Ct}$, where $v$ is velocity, $t$ is time, $e$ is Euler's number. The dimensions of $B$ are

• A. $\left[L^{-\frac{1}{2}}{T}^{\frac{3}{2}}\right]$
• B. $\left[L^2{T}^{-\frac{3}{2}}\right]$
• C. $\left[L^{-\frac{1}{2}}{T}^{\frac{5}{2}}\right]$
• D. $\left[L^{-\frac{1}{2}}{T}^{-\frac{5}{2}}\right]$

Answer 1

The correct option is A $\left[L^{-\frac{1}{2}}{T}^{\frac{3}{2}}\right]$

Applying the principle of homogeneity, we have
$\frac{\left[dv\right]}{[v{]}^{\frac{3}{2}}}=\left[BC\right](\because [e^{-2Ct}]=1)$
$\Rightarrow \left[BC\right]=\frac{\left[L{T}^{-1}\right]}{[L{T}^{-1}{]}^{\frac{3}{2}}}$
$\left[BC\right]=\left[L^{-\frac{1}{2}}{T}^{\frac{1}{2}}\right]....\left(1\right)$

Since powers of exponentials must also be dimensionless,
$\left[2Ct\right]=1(\because [2]=1)$
$\Rightarrow \left[C\right]=\frac{1}{\left[t\right]}$
$\left[C\right]=\left[T^{-1}\right]....\left(2\right)$

$\left(2\right)$ in $\left(1\right)$ gives,
$\left[B\right]=\frac{\left[L^{-\frac{1}{2}}{T}^{\frac{1}{2}}\right]}{\left[T^{-1}\right]}$
$\left[B\right]=\left[L^{-\frac{1}{2}}{T}^{\frac{3}{2}}\right]$