Question

Energy $(E)$ is expressed in terms of mass $(m)$, distance $(x)$ and time $(t)$ as $E=a x+\frac{t^{2} \sqrt{b}}{m}$. If the dimensions of $\frac{b}{a}$ is $\left[ M ^{p} L ^{q} T ^{r}\right]$, then $p+q+r=$________.

Answer 1

From principle of homogeneity,
$[ a ]=\frac{[ E ]}{[ x ]}=\frac{\left[ M ^{1} L ^{2} T ^{-2}\right]}{ L ^{1}}=\left[ M ^{1} L ^{1} T ^{-2}\right]$
Also $[\sqrt{ b }] =\frac{[ m ][ E ]}{[ t ]^{2}}$
$=\frac{\left[ M ^{1}\right]\left[ M ^{1} L ^{2} T ^{-2}\right]}{\left[ T ^{2}\right]}$
$=\left[ M ^{2} L ^{2} T ^{-4}\right]$
$\because [b] =\left[ M ^{4} L ^{4} T ^{-8}\right]$
$\therefore \frac{[b]}{[a]}=\frac{\left[M^{4} L^{4} T^{-8}\right]}{\left[M^{1} L^{L} T^{-2}\right]}$
$=\left[M^{3} L^{3} T^{-6}\right]=\left[M^{P} L^{q} T^{r}\right]$
$\therefore p +q+ r=3+3-6=0$