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To travel straight north, the swimmer must aim upstream (westward) to cancel river drift.
If $\phi$ is the angle west of north:
$v_{\text{swim}}\sin\phi = v_{\text{river}} \Rightarrow 20\sin\phi = 10 \Rightarrow \sin\phi = \dfrac{1}{2} \Rightarrow \phi = \mathbf{30ยฐ}$ west of north
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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