Back to Directory
NEET PHYSICSMedium

A 5000 kg5000 \text{ kg} rocket is set for vertical firing. The exhaust speed is 800 ms−1800 \text{ ms}^{-1}. To give an initial upward acceleration of 20 ms−220 \text{ ms}^{-2}, the amount of gas ejected per second to supply the needed thrust will be (g=10 ms−2g = 10 \text{ ms}^{-2})

A

127.5 kg s−1127.5 \text{ kg s}^{-1}

B

187.5 kg s−1187.5 \text{ kg s}^{-1}

C

185.5 kg s−1185.5 \text{ kg s}^{-1}

D

137.5 kg s−1137.5 \text{ kg s}^{-1}

Step-by-Step Solution

  1. Identify Forces: The rocket experiences two main forces: the downward force of gravity (mgmg) and the upward thrust force (FtF_t) generated by the ejected gas. The net acceleration (aa) is upwards.
  2. Apply Newton's Second Law: Fnet=Ft−mg=maF_{net} = F_t - mg = ma Ft=m(g+a)F_t = m(g + a)
  3. Relate Thrust to Mass Ejection: The thrust is given by the product of the exhaust speed (vv) and the rate of mass ejection (dm/dtdm/dt). Ft=vdmdtF_t = v \frac{dm}{dt}
  4. Combine Equations: vdmdt=m(g+a)v \frac{dm}{dt} = m(g + a) dmdt=m(g+a)v\frac{dm}{dt} = \frac{m(g + a)}{v}
  5. Substitute Values: Given: m=5000 kgm = 5000 \text{ kg}, v=800 m/sv = 800 \text{ m/s}, a=20 m/s2a = 20 \text{ m/s}^2, g=10 m/s2g = 10 \text{ m/s}^2. dmdt=5000(10+20)800\frac{dm}{dt} = \frac{5000 (10 + 20)}{800} dmdt=5000×30800=150000800=15008\frac{dm}{dt} = \frac{5000 \times 30}{800} = \frac{150000}{800} = \frac{1500}{8} dmdt=187.5 kg s−1\frac{dm}{dt} = 187.5 \text{ kg s}^{-1} (Reference: NCERT Class 11, Physics Part I, Chapter 5: Laws of Motion, Exercise 5.9).
Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started