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Given that $r > s > 0$, compare the two quantities below.
Column A: $\dfrac{rs}{r}$
Column B: $\dfrac{rs}{s}$
Simplify each expression:
- Column A: $\dfrac{rs}{r} = s$
- Column B: $\dfrac{rs}{s} = r$
Since we are given $r > s > 0$, it follows that $r > s$.
Therefore Column B ($r$) is greater than Column A ($s$).
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).
F₂ is the strongest oxidising agent because:
- Highest electronegativity
- Lowest bond dissociation energy among halogens (easy to break F–F)
- Highest electron affinity when bond breaking is factored in
- F has no d-orbitals, preventing back bonding
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