List-I | List-II |
---|---|
A. Melting point [K] | I. Tl > In > Ga > Al > B |
B. Ionic Radius [M3/pm] | II. B > Tl > Al ≈ Ga ≈ In |
C. ΔiH1 [kJ mol-1] | III. Tl > In > Al > Ga > B |
D. Atomic Radius [pm] | IV. B > Al > Tl > In > Ga |
To match the items from List I with List II:
A. Melting point [K]: Based on the trend, Tl > In > Ga > Al > B. (Matches I).
B. Ionic Radius [M$^{+3}$/pm]: Boron has the smallest ionic size, followed by Tl > Al $\approx$ Ga > In (Matches II).
C. $\Delta_i H_1$ [kJ mol$^{-1}$]: Ionization enthalpy follows the trend Tl > In > Al > Ga > B.(Matches III).
D. Atomic Radius [pm]: Atomic radius increases as B > Al > Tl > In > Ga.(Matches IV).
Correct Answer: (3)
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.