\[ \begin{array}{|c|c|} \hline \textbf{LIST-I (Redox Reaction)} & \textbf{LIST-II (Type of Redox Reaction)} \\ \hline A. \, CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) & I. \, \text{Disproportionation reaction} \\ B. \, 2NaH(s) \rightarrow 2Na(s) + H_2(g) & II. \, \text{Combination reaction} \\ C. \, V_2O_5(s) + 5Ca(s) \rightarrow 2V(s) + 5CaO(s) & III. \, \text{Decomposition reaction} \\ D. \, 2H_2O(aq) \rightarrow 2H_2(g) + O_2(g) & IV. \, \text{Displacement reaction} \\ \hline \end{array} \]
Step 1: The reaction \( CH_4 + O_2 \) is a combination reaction (A-II).
Step 2: The reaction \( 2NaH \rightarrow 2Na + H_2 \) is a decomposition reaction (B-III).
Step 3: The reaction \( V_2O_5 + Ca \) is a redox displacement reaction (C-I).
Step 4: The reaction \( 2H_2O \rightarrow 2H_2 + O_2 \) is a displacement reaction (D-IV).
Final Conclusion: The correct matching is Option (3), A-II, B-III, C-I, D-IV.
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.