A rocket of mass 1000 kg is to be projected vertically upwards. The gases are exhausted vertically downwards with velocity 100 ms−1 with respect to the rocket. What is the minimum rate of burning of fuel, so as to just lift the rocket upwards against, the gravitational attraction? (Take g = 10ms−2)

Question

A rocket of mass 1000 kg is to be projected vertically upwards. The gases are exhausted vertically downwards with velocity 100 ms−1 with respect to the rocket. What is the minimum rate of burning of fuel, so as to just lift the rocket upwards against, the gravitational attraction? (Take g = 10ms−2) (A) 50 kgs−1 (B) 100 kgs−1 (C) 200 kgs−1 (D) 400 kgs−1

Tardigrade Admin · Accepted Answer

Given, the velocity of exhaust gases with respect to rocket

= 100 m s^{- 1}

The minimum force on the rocket to lift it

F_{m i n} = m g = 1000 \times 10 = 10000 N

Hence, minimum rate of burning of fuel is given by'

\frac{d m}{d t} = \frac{F _{m i n}}{v} = \frac{10000}{100}

= 100 k g s^{- 1}

$50 k g s^{- 1}$

$100 k g s^{- 1}$

$200 k g s^{- 1}$

$400 k g s^{- 1}$

Given, the velocity of exhaust gases with respect to rocket $= 100 m s^{- 1}$
The minimum force on the rocket to lift it
$F_{m i n} = m g = 1000 \times 10 = 10000 N$
Hence, minimum rate of burning of fuel is given by'
$\frac{d m}{d t} = \frac{F _{m i n}}{v} = \frac{10000}{100}$
$= 100 k g s^{- 1}$

50kgs−1

100kgs−1

200kgs−1

400kgs−1

Given, the velocity of exhaust gases with respect to rocket =100ms−1 The minimum force on the rocket to lift it Fmin​=mg=1000×10=10000N Hence, minimum rate of burning of fuel is given by' dtdm​=vFmin​​=10010000​ =100kgs−1

$50 k g s^{- 1}$

$100 k g s^{- 1}$

$200 k g s^{- 1}$

$400 k g s^{- 1}$

Given, the velocity of exhaust gases with respect to rocket $= 100 m s^{- 1}$
The minimum force on the rocket to lift it
$F_{m i n} = m g = 1000 \times 10 = 10000 N$
Hence, minimum rate of burning of fuel is given by'
$\frac{d m}{d t} = \frac{F _{m i n}}{v} = \frac{10000}{100}$
$= 100 k g s^{- 1}$