Let's determine the oxidation states of Mn in the reactants and products:
In MnO$_4^{2-}$: Let $x$ be the oxidation state of Mn.
Then $x + 4(-2) = -2$, so $x = +6$.
In MnO$_4^-$: $x + 4(-2) = -1$, so $x = +7$.
In MnO$_2$: $x + 2(-2) = 0$, so $x = +4$.
The oxidation states of Mn in the reaction are +6, +7, and +4. Therefore, the oxidation states not shown are +2 and +3.
A bob of heavy mass \(m\) is suspended by a light string of length \(l\). The bob is given a horizontal velocity \(v_0\) as shown in figure. If the string gets slack at some point P making an angle \( \theta \) from the horizontal, the ratio of the speed \(v\) of the bob at point P to its initial speed \(v_0\) is :