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Laws of Motion question

2021 · 26 Aug · Shift 1 · Q59
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  5. /2021 · 26 Aug · Shift 1 · Q59

Laws of Motion question

2021 · 26 Aug · Shift 1 · Q59

JEE MainPhysicsLaws of MotionMCQ+4 / −1
The initial mass of a rocket is 1000 kg. Calculate at what rate the fuel should be burnt so that the rocket is given an acceleration of 20 ms-2. The gases come out at a relative speed of 500 ms −-− 1 with respect to the rocket : [Use g = 10 m/s2]
  1. A
    6.0 ×\times× 102 kg s −-− 1
  2. B
    500 kg s −-− 1
  3. C
    10 kg s −-− 1
  4. D
    60 kg s −-− 1
View written solutionFree

Correct answer: D

  1. Use the rocket thrust equation

For a vertically upward moving rocket,

T−mg=maT - mg = maT−mg=ma

where thrust

T=u dmdtT = u\,\frac{dm}{dt}T=udtdm​

Here,

  • initial mass of rocket: m=1000 kgm = 1000\,\text{kg}m=1000kg
  • required upward acceleration: a=20 m s−2a = 20\,\text{m s}^{-2}a=20m s−2
  • exhaust speed relative to rocket: u=500 m s−1u = 500\,\text{m s}^{-1}u=500m s−1
  • g=10 m s−2g = 10\,\text{m s}^{-2}g=10m s−2

So,

udmdt−mg=mau \frac{dm}{dt} - mg = maudtdm​−mg=ma 500dmdt−1000×10=1000×20500\frac{dm}{dt} - 1000\times 10 = 1000\times 20500dtdm​−1000×10=1000×20
  1. Substitute values
500dmdt−10000=20000500\frac{dm}{dt} - 10000 = 20000500dtdm​−10000=20000 500dmdt=30000500\frac{dm}{dt} = 30000500dtdm​=30000 dmdt=30000500=60 kg s−1\frac{dm}{dt} = \frac{30000}{500} = 60\,\text{kg s}^{-1}dtdm​=50030000​=60kg s−1
  1. Match with options
60 kg s−1\boxed{60\,\text{kg s}^{-1}}60kg s−1​

This corresponds to Option D.

  1. Verification with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.

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