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 2000 kg rocket in free space expels 0.5 kg of gas per second at an exhaust velocity of 400 m s-1 for 5 seconds. The increase in the speed of the rocket in this time is (Assume time interval is too small to consider variation in acceleration)
PhysicsCenter of Mass Momentum and CollisionNEET
Options:
  • A 2000 m s-1
  • B 200 m s-1
  • C 0.5 m s-1
  • D zero
Solution:
2392 Upvotes Verified Answer
The correct answer is: 0.5 m s-1
dmdt×u=Mdvdt

0.5×400=2000×Δv5     Δv=0.5 m s-1

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.