# Question #b47e0

Sep 6, 2016

$a = \frac{1}{2} , b = - \frac{1}{2}$

#### Explanation:

Given equation is

$P \propto {L}^{a} {g}^{b}$
where $P \text{ has dimension of time} \implies {T}^{1}$, $L \text{ has dimension of Length} \implies {L}^{1}$ and
$g \text{ acceleration due to gravity has units as } m {s}^{-} 2 \implies {L}^{1} {T}^{-} 2$

Now writing given equation in dimensional form we get
${T}^{1} \propto {\left({L}^{1}\right)}^{a} . {\left({L}^{1} {T}^{-} 2\right)}^{b}$

To include both dimension on either side of the equation it can be rewritten as
${L}^{0} {T}^{1} \propto {\left({L}^{1}\right)}^{a} . {\left({L}^{1} {T}^{-} 2\right)}^{b}$
$\implies {L}^{0} {T}^{1} \propto {L}^{a + b} . {T}^{- 2 b}$

Comparing exponents of each dimension on LHS and RHS we get
$0 = a + b$ ........(1)
$1 = - 2 b$ ..........(2)
Solving (2) for $b$, we get
$b = - \frac{1}{2}$
Inserting this values in (1) we get
$a = \frac{1}{2}$