Compare boiling point of given solutions:
(i) \( 10^{-4} \, \text{M NaCl} \)>(ii) \( 10^{-3} \, \text{M NaCl} \)>(iii) \( 10^{-2} \, \text{M NaCl} \)>(iv) \( 10^{-4} \, \text{M urea} \)
(iii) \( 10^{-2} \, \text{M NaCl} \)>(ii) \( 10^{-3} \, \text{M NaCl} \)>(i) \( 10^{-4} \, \text{M NaCl} \)>(iv) \( 10^{-4} \, \text{M urea} \)
(ii) \( 10^{-3} \, \text{M NaCl} \)>(i) \( 10^{-4} \, \text{M NaCl} \)>(iii) \( 10^{-2} \, \text{M NaCl} \)>(iv) \( 10^{-4} \, \text{M urea} \)
(iii) \( 10^{-2} \, \text{M NaCl} \)>(ii) \( 10^{-3} \, \text{M NaCl} \)>(i) \( 10^{-4} \, \text{M NaCl} \)>(iv) \( 10^{-4} \, \text{M urea} \)
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.