\( \dfrac{3\left(\dfrac{\sqrt{5}+1}{4}\right) + 5\left(\dfrac{\sqrt{5}-1}{4}\right)}{5\left(\dfrac{\sqrt{5}+1}{4}\right) - 3\left(\dfrac{\sqrt{5}-1}{4}\right)} = \dfrac{8\sqrt{5}-2}{2\sqrt{5}+8} \)
\( = \dfrac{4\sqrt{5}-1}{\sqrt{5}+4} \times \dfrac{\sqrt{5}-4}{\sqrt{5}-4} \)
\( = \dfrac{20 - 16\sqrt{5} - \sqrt{5} + 4}{-11} \)
\( = \dfrac{17\sqrt{5}-24}{11} \Rightarrow a = 17, \, b = 27, \, c = 11 \)
\( a + b + c = 52 \)
MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K.
MX(s) $\rightleftharpoons M^{+(aq) }+ X^{-}(aq)$; $K_{sp} = 10^{-10}$
If the standard reduction potential for $M^{+}(aq) + e^{-} \rightarrow M(s)$ is $(E^{\circ}_{M^{+}/M}) = 0.79$ V, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $E^{\circ}_{X^{-}/MX(s)/M}$ is ____________ mV. (nearest integer)
[Given : $\frac{2.303 RT}{F} = 0.059$ V]
An infinitely long straight wire carrying current $I$ is bent in a planar shape as shown in the diagram. The radius of the circular part is $r$. The magnetic field at the centre $O$ of the circular loop is :
