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 Diwali cracker releases 25 gram gas per second, with a speed of $400 \mathrm{~ms}^{-1}$ after
explosion. The force exerted by gas on the cracker is
PhysicsLaws of MotionMHT CETMHT CET 2020 (13 Oct Shift 2)
Options:
  • A 100 dyne
  • B 16 newton
  • C 10 newton
  • D 10,000 dyne
Solution:
2240 Upvotes Verified Answer
The correct answer is: 10 newton
$\mathrm{F}=$ rate of change of momentum
$=25 \times 10^{-3} \mathrm{~kg} / \mathrm{s} \times 400 \mathrm{~m} / \mathrm{s}$ $=10 \mathrm{~N}$

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.