Question

Energy due to the position of a particle is given by, $U=\frac{\alpha \sqrt{y}}{y+\beta }$, where $\alpha$ and $\beta$ are constants, $y$ is distance. The dimensions of $(\alpha \times \beta )$ is:

• A. $\left[M^{0}L{T}^0\right]$
• B. $\left[M^{1/2}{L}^{3/2}{T}^{-2}\right]$
• C. $\left[M^0{L}^{-7/2}{T}^0\right]$
• D. $\left[M{L}^{7/2}{T}^{-2}\right]$

Answer 1

The correct option is D $\left[M{L}^{7/2}{T}^{-2}\right]$

$U=\frac{\alpha \sqrt{y}}{y+\beta }$ ... (1)
Dimension of energy $=\left[M^1{L}^2{T}^{-2}\right]$ ... (2)
Dimension of $\beta =\left[L^1\right]$ ... (3)
$\therefore$ From equation (1)
$\left[M^1{L}^2{T}^{-2}\right]=\frac{\alpha [L{]}^{1/2}}{\left[L\right]}$
$\alpha =\frac{\left[M^1{L}^2{T}^{-2}\right]\left[L^1\right]}{[L{]}^{1/2}}$
$\alpha =\left[M^1{L}^{5/2}{T}^{-2}\right]$
$\therefore$ Dimension of $[\alpha \times \beta ]=\left[M^1{L}^{5/2}{T}^{-2}\right]\left[L^1\right]$
$=\left[M^1{L}^{7/2}{T}^{-2}\right]$
$\therefore$ Option (D) is correct.