A) \[127.5\,\,kg\,{{s}^{-1}}\]
B) \[187.\,5\,kg\,{{s}^{-1}}\]
C) \[185.5\,kg\,{{s}^{-1}}\]
D) \[137.5\,kg\,{{s}^{-1}}\]
Correct Answer: B
Solution :
Key idea: Thrust force on the rocket balances the weight of the rocket. Thrust force on the rocket. \[{{F}_{t}}={{v}_{r}}\left( -\frac{dm}{dt} \right)(upward)\] Weight of the rocket w = mg (downward) Net force on the rocket \[{{F}_{net}}={{F}_{t}}-w\] \[\Rightarrow ma={{v}_{r}}\left( \frac{-dm}{dt} \right)-mg\] \[\Rightarrow \left( \frac{-dm}{dt} \right)=\frac{m\,(g+a)}{{{v}_{r}}}\] \[\therefore \] Rate of gas ejected per second \[=\frac{500\,(10+20)}{800}=\frac{5000\times 30}{800}\] \[=187.5\,kg\,{{s}^{-1}}\] Note: Problems related to variable mass can be solved in following three steps: 1. Make a list of all the forces acting on the main mass and apply them on it. 2. Apply an additional thrust force \[{{\vec{F}}_{t}}\] on the mass, the magnitude of which is \[\left| {{{\vec{v}}}_{r}}\left( \pm \frac{dm}{dt} \right) \right|\] and direction is given by the direction of \[{{\vec{v}}_{r}}\] in case the mass is increasing and otherwise the direction of \[-{{\vec{v}}_{r}}\] if it is decreasing. 3. Find net force on the mass and apply \[{{\vec{F}}_{net}}=m\frac{d\,\vec{v}}{dt}\] (\[m=\] mass at that particular instat)
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