Initial fuel in a rocket is 1000 kg- If it was given an acceleration of 201exm- -1exs2- the velocity of fuel with respect to the rocket is 5001exm- -1exs- Find the rate of consumption of fuel-

Solution:Step 1: Given informationInitial fuel=1000 kg.Acceleration=201exm/ 1exs^2.Fuel's velocity=5001exm/ 1exs.Step 2: Calculate the rate of consumption of ...

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Standard XII

Physics

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Initial fuel ...

Question

Initial fuel in a rocket is $1000 k g .$ If it was given an acceleration of $20 \frac{m}{s^{2}}$ the velocity of fuel with respect to the rocket is $500 \frac{m}{s}$ . Find the rate of consumption of fuel.

$30 \frac{k g}{s}$

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$60 \frac{k g}{s}$

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$900 \frac{k g}{s}$

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$80 \frac{k g}{s}$

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Solution

The correct option is B
$60 \frac{k g}{s}$

Solution:
Step 1: Given information
Initial fuel $= 1000 k g .$
Acceleration $= 20 \frac{m}{s^{2}}$
Fuel's velocity $= 500 \frac{m}{s}$
Step 2: Calculate the rate of consumption of fuel
Know that, given a system with variable mass,
$T o t a l F o r c e (B y r o c k e t) - G r a v i t a t i o n a l F o r c e = A c t i n g F o r c e$ ,
$f_{T} - m g = m a$ …..(i)
Where,
$f_{T} =$ Total force
$m =$ Mass of fuel
$g =$ Acceleration due to gravity
$a =$ Acceleration produced by rocket
Therefore, from equation (i),
$m \frac{v_{r}}{t} - m g = m a$ …..(ii)
Where, $(f_{T} = m \times a)$
Step 3: Solving by putting the given values in the equation (ii)
Therefore, from equation (ii),
$(\frac{d m}{d t}) v_{r} = m g + m a$
$(\frac{d m}{d t}) 500 = 1000 (10 + 20)$
$(\frac{d m}{d t}) = 60 k g / s$
Hence, the correct option is (B).

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Initial fuel in a rocket is 1000kg. If it was given an acceleration of 20ms2 the velocity of fuel with respect to the rocket is 500ms. Find the rate of consumption of fuel.

Initial fuel in a rocket is $1000 k g .$ If it was given an acceleration of $20 \frac{m}{s^{2}}$ the velocity of fuel with respect to the rocket is $500 \frac{m}{s}$ . Find the rate of consumption of fuel.