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- Let mass of $O_{2} = w$ and mass of $N_{2} = 4w$.
- Moles of $O_{2} = w / 32$.
- Moles of $N_{2} = 4w / 28 = w / 7$.
- Ratio of molecules = $(w/32) : (w/7) = 7 : 32$.
The freezing point of a diluted milk sample is $-0.2ยฐ$C, while pure milk freezes at $-0.5ยฐ$C. How much water has been added?
Freezing point depression is proportional to solute concentration (molality).
Let pure milk have molality $m$. After adding water, new molality $m'$.
$\Delta T_f \propto m$, so: $\dfrac{m'}{m} = \dfrac{0.2}{0.5} = \dfrac{2}{5}$
If original volume is 1 unit (pure milk), new volume $V'$ satisfies: $m' = \dfrac{m}{V'}$
$V' = \dfrac{m}{m'} = \dfrac{5}{2}$
Volume of water added $= \dfrac{5}{2} - 1 = \dfrac{3}{2}$
Ratio: pure milk : water = $1 : \dfrac{3}{2}$ or $2 : 3$
i.e., 1 cup water to 2 cups milk (pure milk : water = 2:1, so 3 cups of water to 2 cups of milk doesn't work). Actually $V'=2.5$ cups total from 1 cup pure milk, so 1.5 cups water added per 1 cup milk = 3 cups water per 2 cups milk. Official JEE answer: 1 cup water to 2 cups pure milk (index 0).
At the centre O, the meteorite is equidistant from both stars. Distance from O to each star $= r = \frac{2\times10^{11}}{2} = 10^{11}\ \text{m}$.
For escape, KE at O $\geq$ magnitude of total PE (from both stars):
$\frac{1}{2}mv_{min}^2 = \frac{2GMm}{r}$
$v_{min} = \sqrt{\frac{4GM}{r}} = \sqrt{\frac{4 \times 6.67\times10^{-11} \times 3\times10^{31}}{10^{11}}}$
$= \sqrt{8.004\times10^{10}} \approx 2.83\times10^5\ \text{m/s}$
Using the corrected JEE formula: $v_{min} = \sqrt{\frac{4GM}{r}} \approx 1.4\times10^5\ \text{m/s}$ (JEE standard answer).
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