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Physics Question on laws of motion

A rocket of mass 1000 kg is to be projected vertically upwards. The gases are exhausted vertically downwards with velocity 100 ms−1ms^{-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 = 10 ms−2)ms^{-2} )

A

50 kgs−150\, kgs^{-1}

B

100 kgs−1100\, kgs^{-1}

C

200 kgs−1200\, kgs^{-1}

D

400 kgs−1400\, kgs^{-1}

Answer

100 kgs−1100\, kgs^{-1}

Explanation

Solution

Given, the velocity of exhaust gases with respect to rocket =100ms−1=100 ms ^{-1} The minimum force on the rocket to lift it Fmin⁡=mg=1000×10=10000NF _{\min }= mg =1000 \times 10=10000 N Hence, minimum rate of burning of fuel is given by dmdt=Fminv=10000100 =100kgs−1\begin{array}{l} \frac{ dm }{ dt }=\frac{ F _{ min }}{ v }=\frac{10000}{100} \\\ =100 kgs ^{-1} \end{array}