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For a given solvent: elevation in boiling point for a 1 molal glucose solution is 2 K, and depression in freezing point for a 2 molal glucose solution is 2 K. What is the relation between $K_b$ and $K_f$?
Boiling point elevation: $\Delta T_b = K_b \cdot m$
$2 = K_b \times 1 \Rightarrow K_b = 2\ \text{K kg mol}^{-1}$
Freezing point depression: $\Delta T_f = K_f \cdot m$
$2 = K_f \times 2 \Rightarrow K_f = 1\ \text{K kg mol}^{-1}$
Therefore $K_b = 2 = 2K_f$, i.e., $\mathbf{K_b = 2\,K_f}$
Wait — that gives option D. $K_b=2$ and $K_f=1$, so $K_b=2K_f$. But let me recheck: $\Delta T_b=K_b\cdot m_1$: $2=K_b\cdot1$, $K_b=2$. $\Delta T_f=K_f\cdot m_2$: $2=K_f\cdot2$, $K_f=1$. So $K_b=2K_f$. Official JEE answer is $K_b=K_f$ — this would require $K_b=K_f=2$. In that case $\Delta T_f=2\times2=4\neq2$. The correct computation gives $K_b=2K_f$ (index 3).
For decades, economists believed that increased productivity would naturally lead to higher wages. However, recent studies indicate that while productivity has surged, wage growth has stagnated, suggesting that other structural factors — such as labor market policies, globalization, and technological automation — may be influencing income distribution.
Sub-Questions:
The passage emphasizes that higher productivity no longer ensures wage increases because of other structural factors — exactly what option C states.
We integrate $\frac{dP}{dx} = 100 - 12\sqrt{x}$ from $x = 0$ to $x = 25$:
$P = \int_0^{25}(100 - 12\sqrt{x})\,dx = \left[100x - 12 \cdot \frac{2}{3}x^{3/2}\right]_0^{25}$
$= \left[100x - 8x^{3/2}\right]_0^{25} = 100(25) - 8(25)^{3/2} = 2500 - 8(125) = 2500 - 1000 = 1500$
New production $= 2000 + 1500 = \mathbf{3500}$
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