$\displaystyle\int_0^{\pi/3}\dfrac{\tan\theta}{\sqrt{2k}\cdot\sqrt{\sec\theta}}\,d\theta = \dfrac{1}{\sqrt{2k}}\int_0^{\pi/3}\dfrac{\sin\theta}{\cos\theta}\cdot\sqrt{\cos\theta}\,d\theta=\dfrac{1}{\sqrt{2k}}\int_0^{\pi/3}\sin\theta\cdot(\cos\theta)^{-1/2}d\theta$

Let $u=\cos\theta$, $du=-\sin\theta\,d\theta$; limits: $1$ to $1/2$:

$=\dfrac{1}{\sqrt{2k}}\int_1^{1/2}(-u^{-1/2})du=\dfrac{1}{\sqrt{2k}}\left[2\sqrt{u}\right]_1^{1/2}... $ Wait: $\int_1^{1/2}(-u^{-1/2})du=\int_{1/2}^1 u^{-1/2}du=[2\sqrt{u}]_{1/2}^1=2-\sqrt{2}=2(1-1/\sqrt{2})$

$\dfrac{2(1-1/\sqrt{2})}{\sqrt{2k}}=1-\dfrac{1}{\sqrt{2}} \Rightarrow \dfrac{2}{\sqrt{2k}}=1 \Rightarrow \sqrt{2k}=2 \Rightarrow 2k=4 \Rightarrow k=\mathbf{2}$

The circumference formula is $C = \pi d$, where $d$ is the diameter.

$18\pi = \pi d \Rightarrow d = 18$

Column A = 18, Column B = 9.

Since $18 > 9$, Column A is greater.

Article 19 of the Constitution of Pakistan (1973) reads: “Every citizen shall have the right of freedom of speech and expression, and there shall be freedom of the press, subject to any reasonable restrictions imposed by law in the interest of the glory of Islam or the integrity, security, or defence of Pakistan or any part thereof, friendly relations with foreign states, public order, decency or morality or in relation to the contempt of court, defamation or incitement to an offence.” This is a comprehensive list of permissible restrictions. Critics of Pakistani press law note that the breadth of these restrictions — especially “glory of Islam” and “public order” — gives governments considerable room to restrict media freedom. The Freedom of Information Ordinance 2002, while acknowledging citizens' right to know, was also criticised for its restrictive list of disclosable records and 21-day response timeframe, which were seen as undermining the public's true right of access to government information.

• Molar mass of $NH_4SH = 51\text{ g/mol}$. Initial moles $= 5.1/51 = 0.1\text{ mol}$.
• At equilibrium, moles of $NH_3 = H_2S = 0.1 \times 0.3 = 0.03\text{ mol}$.
• Using $PV = nRT$: $P_{NH_3} = P_{H_2S} = \frac{0.03 \times 0.082 \times 600}{3} = 0.492\text{ atm}$.
• $K_p = P_{NH_3} \times P_{H_2S} = (0.492)^2 \approx 0.242\text{ atm}^2$.