Histone proteins are positively charged and bind to negatively charged DNA, helping to compact and organize DNA into nucleosomes and ultimately into chromosomes.

Sulphur shows multiple oxidation states

Rotating distribution direction is an administration procedure specifically designed to ensure simultaneous, equitable starting conditions. If papers are distributed linearly, students at one end begin before others, gaining time advantage. Rotation also complicates organized cheating by preventing predictable paper flow. This supports fairness and the standardization needed for valid score interpretation.

Molar mass of Na$^+ = 23\ \text{g mol}^{-1}$

Moles of Na$^+ = \dfrac{92}{23} = 4\ \text{mol}$

Molality $= \dfrac{4\ \text{mol}}{1\ \text{kg water}} = \mathbf{4\ \text{mol kg}^{-1}}$

Wait — that's 4, which is option B. But checking: the question asks for molality of Na$^+$. Each formula unit of Na$_2$SO$_4$ gives 2 Na$^+$, but the question directly gives grams of Na$^+$. Moles $= 92/23 = 4$, per kg water → molality $= 4$. But the official JEE answer is 4 (option B, index 1). Let me correct the index.

Elongation $\Delta L = \frac{FL}{AY} = \frac{FL}{\pi(d/2)^2 Y} \propto \frac{L}{d^2}$.

• New elongation $\Delta L' \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{1}{2} \Delta L$
• $\Delta L' = \frac{1}{2} \times 0.04 = 0.02\text{ m}$.

Rule 42(b)(i) and (iii) together require that at least three quotations be obtained from GST registered firms, original equipment manufacturers, or authorized dealers when procurement value exceeds PKR 100,000 but does not exceed PKR 500,000.

Meningitis can be bacterial or viral and is a medical emergency.

Formula: \(\lambda = \dfrac{c}{\nu}\)

\(\lambda = \dfrac{3 \times 10^{17} \text{ nm s}^{-1}}{6 \times 10^{15} \text{ s}^{-1}} = 50\) nm

This wavelength (50 nm) lies in the extreme UV (EUV) region. Note: \(c = 3\times10^8\) m/s \(= 3\times10^{17}\) nm/s.

Compact states (roughly circular/hexagonal) minimize maximum distance from capital to borders, facilitating easier governance, efficient transport networks, cultural cohesion, and defense. Elongated states face challenges: distant regions are harder to govern and may develop separate regional identities.

Reliability requires consistency across administrations, scorers, and conditions. Portfolios are diverse in content (student-selected), judged subjectively by different assessors using loosely defined criteria, and lack uniform conditions. All of these factors introduce variability into scores, which by definition reduces reliability. This is listed as a known weakness of portfolios.

In a neutral or weakly alkaline medium, permanganate ($MnO_{4}^{-}$) oxidizes thiosulphate ($S_{2}O_{3}^{2-}$) to sulphate ($SO_{4}^{2-}$). The reaction is:

$8MnO_{4}^{-} + 3S_{2}O_{3}^{2-} + H_{2}O \rightarrow 8MnO_{2} + 6SO_{4}^{2-} + 2OH^{-}$

In the sulphate ion ($SO_{4}^{2-}$), the oxidation state of sulfur is $+6$.

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The Zimmermann Telegram was an intercepted diplomatic communication from German Foreign Secretary Arthur Zimmermann to the German ambassador in Mexico, proposing a military alliance: Germany would help Mexico recover Texas, New Mexico, and Arizona from the United States if Mexico would declare war on the U.S. The telegram was intercepted and decoded by British intelligence. The textbook describes its publication in 1917 as “undoubtedly the most important propaganda achievement of the British” — the British had been conducting covert propaganda in the U.S. to build support for the Allied cause, and the Zimmermann revelation dramatically shifted American public opinion. It helped persuade President Wilson and the U.S. Congress to declare war on Germany, bringing America into WWI on the Allied side. This is a classic example of intelligence-based propaganda — using real (but strategically timed and framed) information as a propaganda weapon.

Tetracycline chelates with divalent cations like $Ca^{2+}$ in milk, preventing its absorption.

The required plane contains the line of intersection of \(2x-5y+z=3\) and \(x+y+4z=5\), so it can be written as \((2x-5y+z-3) + \lambda(x+y+4z-5) = 0\), which is \((2+\lambda)x + (-5+\lambda)y + (1+4\lambda)z - (3+5\lambda) = 0\). For this to be parallel to \(x+3y+6z=1\), the normal vectors must be proportional: \(\frac{2+\lambda}{1} = \frac{-5+\lambda}{3} = \frac{1+4\lambda}{6}\). From the first two: \(3(2+\lambda) = -5+\lambda\), giving \(6+3\lambda = -5+\lambda\), so \(2\lambda = -11\) and \(\lambda = -\frac{11}{2}\). Substituting back: the plane is \((2-\frac{11}{2})x + (-5-\frac{11}{2})y + (1-22)z - (3-\frac{55}{2}) = 0\), which simplifies to \(-\frac{7}{2}x - \frac{21}{2}y - 21z + \frac{49}{2} = 0\). Multiplying by \(-\frac{2}{7}\): \(x + 3y + 6z = 7\).

Homogenization reduces fat globule diameter to below \(2\;\mu\text{m}\). By Stokes' law, creaming velocity \(v \propto r^2\); smaller globules rise negligibly slowly, preventing cream layer formation.

Laspeyres' index \(= \dfrac{\sum p_1 q_0}{\sum p_0 q_0} \times 100\) uses base period quantities \(q_0\) as fixed weights. Paasche's index \(= \dfrac{\sum p_1 q_1}{\sum p_0 q_1} \times 100\) uses current period quantities \(q_1\). Fisher's Ideal index is the geometric mean of both.

Dura mater (tough mother) is the thick, protective outer layer.

The myenteric and submucosal plexuses manage motility and secretion locally.

Region: $0 \leq x \leq 2$ and $\sqrt{x} \leq y \leq 2$. This equals $0 \leq y \leq 2$ and $0 \leq x \leq y^2$. New integral: $\int_0^2 \int_0^{y^2} f(x, y) \, dx \, dy$.

This is a standard chemical test for tannins.