Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
A $5000 \mathrm{~kg}$ rocket is set for vertical firing. The exhaust speed is $800 \mathrm{~m} / \mathrm{s}$. To give an initial upward acceleration of $20 \mathrm{~m} / \mathrm{s}^{2}$, the amount of gas ejected per second to supply the needed thrust will be (Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ )
PhysicsCenter of Mass Momentum and CollisionVITEEEVITEEE 2017
Options:
  • A $127.5 \mathrm{~kg} / \mathrm{s}$
  • B $137.5 \mathrm{~kg} / \mathrm{s}$
  • C $155.5 \mathrm{~kg} / \mathrm{s}$
  • D $187.5 \mathrm{~kg} / \mathrm{s}$
Solution:
1356 Upvotes Verified Answer
The correct answer is: $187.5 \mathrm{~kg} / \mathrm{s}$
Given : Mass of rocket $(\mathrm{m})=5000 \mathrm{Kg}$ Exhaust speed $(v)=800 \mathrm{~m} / \mathrm{s}$
Acceleration of rocket $(\mathrm{a})=20 \mathrm{~m} / \mathrm{s}^{2}$
Gravitational acceleration $(\mathrm{g})=10 \mathrm{~m} / \mathrm{s}^{2}$
We know that upward force
$$
\begin{aligned}
\mathrm{F} &=\mathrm{m}(\mathrm{g}+\mathrm{a})=5000(10+20) \\
&=5000 \times 30 &=150000 \mathrm{~N}
\end{aligned}
$$
We also know that amount of gas ejected
$$
\left(\frac{\mathrm{dm}}{\mathrm{dt}}\right)=\frac{\mathrm{F}}{\mathrm{v}}=\frac{150000}{800}=187.5 \mathrm{~kg} / \mathrm{s}
$$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.