No emission below threshold frequency (ν < ν₀).

Inside ideal solenoid, field is uniform and parallel to axis.

The reactions are as follows:

• Acidic Medium: $2[Fe(CN)_{6}]^{4-} + H_{2}O_{2} + 2H^{+} \rightarrow 2[Fe(CN)_{6}]^{3-} + 2H_{2}O$. Here, $H_{2}O_{2}$ is reduced to $H_{2}O$.
• Alkaline Medium: $2[Fe(CN)_{6}]^{3-} + H_{2}O_{2} + 2OH^{-} \rightarrow 2[Fe(CN)_{6}]^{4-} + 2H_{2}O + O_{2}$. Here, $H_{2}O_{2}$ is oxidized to $O_{2}$.

Thus, the products are $H_{2}O$ and $(H_{2}O + O_{2})$.

The National Education Policy 1978, formulated during General Zia ul-Haq’s government, emphasized both Islamic values and the ideology of Pakistan as the ideological foundation of the national education system. Islamization of curriculum and promotion of national ideology were its defining features.

Correct Answer: C — Both A and B.

Work done = $Q \times V_{origin}$, where $V_{origin}$ is the potential at origin due to the four charges.

Distances from origin:

• $(0,2)$: $r_1 = 2$
• $(0,-2)$: $r_2 = 2$
• $(4,2)$: $r_3 = \sqrt{16+4} = \sqrt{20} = 2\sqrt{5}$
• $(4,-2)$: $r_4 = \sqrt{20} = 2\sqrt{5}$

$V = \frac{Q}{4\pi\varepsilon_0}\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{2\sqrt{5}}+\frac{1}{2\sqrt{5}}\right) = \frac{Q}{4\pi\varepsilon_0}\left(1 + \frac{1}{\sqrt{5}}\right)$

Work $= QV = \dfrac{Q^2}{4\pi\varepsilon_0}\left(1+\dfrac{1}{\sqrt{5}}\right)$