Explanation

Given that, Time period, $T \propto p^{a}d^{b}E^{c}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$ The dimensions of these quantities are given as $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ p = \left\lbrack ML^{- 1}T^{- 2} \right\rbrack$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ d = \left\lbrack ML^{- 3} \right\rbrack$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E = \left\lbrack ML^{2}T^{- 2} \right\rbrack$ In Eq. (i), on writing the dimensions on both sides. $\left\lbrack M^{0}L^{0}T \right\rbrack\ \propto \left\lbrack {ML}^{- 1}T^{- 2} \right\rbrack^{a}\left\lbrack {ML}^{- 3} \right\rbrack^{b}\left\lbrack ML^{2}T^{- 2} \right\rbrack^{c}$ $\Rightarrow \left\lbrack M^{0}L^{0}T \right\rbrack \propto \left\lbrack M^{a\ + \ b\ + \ c}L^{- a - 3b + 2c}T^{- 2a - 2c} \right\rbrack$ On comparing the powers of M, L, T on both sides. $\Rightarrow \ \ \ \ \ \ \ \ \ \ \ a + b + c = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$ $\ \ \ \ \ \ \ \ - a - 3b + 2c = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(iii)\$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - 2a - 2c = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(iv)$ Solving, we get value of $a,\ b\ and\ c,\ - \frac{5}{6},\frac{1}{2}and\frac{1}{3}\ repectively.\$