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**Quiz Description** :

**Name: **Rocket Propulsion mcq test

**Subject: **Physics

**Topic: **Rocket Propulsion

**Questions: **8 Objective type

**Time Allowed: **15 min

**Important for: **Aerospace, Avionics Engineering students.

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- Question 1 of 8
##### 1. Question

1 pointsRocket propulsion works on the principle of conservation of

CorrectWe know that when the fuel in a rocket undergoes combustion, the burnt gases leave the body of rocket with a large speed and provide up-thrust to the rocket. If we assume that the fuel is burnt at constant rate, then the rate of change of momentum of the rocket will be constant.

As more and more fuel is burnt, mass of the rocket goes on decreasing and it increases velocity of the rocket more and more.

Therefore rocket propulsion works on the principle of conservation of linear momentum.IncorrectWe know that when the fuel in a rocket undergoes combustion, the burnt gases leave the body of rocket with a large speed and provide up-thrust to the rocket. If we assume that the fuel is burnt at constant rate, then the rate of change of momentum of the rocket will be constant.

As more and more fuel is burnt, mass of the rocket goes on decreasing and it increases velocity of the rocket more and more.

Therefore rocket propulsion works on the principle of conservation of linear momentum. - Question 2 of 8
##### 2. Question

1 pointsA rocket is moving with a velocity of v

_{1}, while the velocity of gases ejecting from rocket is v_{2}. Then velocity of gases with respect to rocket isCorrectGiven velocity of rocket (v

_{R}) = v_{1}and velocity of gases ejected from rocket (v_{G}) = – v_{2}(Minus sign due to opposite direction).

We know that as the rocket and goes ejecting from the rocket are moving in opposite directions, therefore velocity of gases with respect to rocket = v_{R}+ v_{G}= v_{1}+ v_{2}.IncorrectGiven velocity of rocket (v

_{R}) = v_{1}and velocity of gases ejected from rocket (v_{G}) = – v_{2}(Minus sign due to opposite direction).

We know that as the rocket and goes ejecting from the rocket are moving in opposite directions, therefore velocity of gases with respect to rocket = v_{R}+ v_{G}= v_{1}+ v_{2}. - Question 3 of 8
##### 3. Question

1 pointsIf the force on a rocket moving with a velocity of 300 m-s

^{-1}is 210 N, then the rate of fuel combustion isCorrectGiven that the velocity of rocket (v) = 300 m-s

^{-1}and force (F) = 210 N.

We know that the rate of fuel combustion (dm/dt) = F/v = 210/300 = 0.7 kg-s^{-1}IncorrectGiven that the velocity of rocket (v) = 300 m-s

^{-1}and force (F) = 210 N.

We know that the rate of fuel combustion (dm/dt) = F/v = 210/300 = 0.7 kg-s^{-1} - Question 4 of 8
##### 4. Question

1 pointsA truck, moving on a smooth horizontal surface with a uniform speed v, carries stone dust. If a mass (dm) of the stone dust leaks from the truck in a time (dt), then force needed to keep the truck moving with uniform speed in horizontal direction, is

CorrectGiven: Uniform speed of truck = v and rate of leak of dust stone = dm/∆t

We know that force exerted on the truck in vertical upward direction due to leaking of stone dust

(F) = v × (dm /∆t)

Since the surface is smooth, therefore there is no friction. Thus no force is needed to keep the truck moving with a constant speed in horizontal direction.IncorrectGiven: Uniform speed of truck = v and rate of leak of dust stone = dm/∆t

We know that force exerted on the truck in vertical upward direction due to leaking of stone dust

(F) = v × (dm /∆t)

Since the surface is smooth, therefore there is no friction. Thus no force is needed to keep the truck moving with a constant speed in horizontal direction. - Question 5 of 8
##### 5. Question

1 pointsA rocket of mass 100kg consumes fuel at a rate of 40 kg-s

^{-1}. If the velocity of gases ejected from rocket is 5×10^{4}m-s^{-1}, then thrust exerted on the rocket isCorrectGiven: Mass of the rocket (m) = 1000kg; Rate of fuel consumption (dm/dt) = 40 kg-s

^{-1}and velocity of ejected gases (v) = 5×10^{4}m-s^{-1}.

We know that thrust exerted on the rocket (F) = v × (dm/dt) = (5×10^{4}) × 40 = 2 × 10^{6}N.IncorrectGiven: Mass of the rocket (m) = 1000kg; Rate of fuel consumption (dm/dt) = 40 kg-s

^{-1}and velocity of ejected gases (v) = 5×10^{4}m-s^{-1}.

We know that thrust exerted on the rocket (F) = v × (dm/dt) = (5×10^{4}) × 40 = 2 × 10^{6}N. - Question 6 of 8
##### 6. Question

1 pointsA rocket of mass 1000 kg is to be projected vertically upwards. The gases are exhausted vertically downwards with velocity 100 m-s

^{-1}with respect to the rocket. What is the minimum rate of burning of fuel, so as to just life the rocket upwards against the gravitational attraction? (Take g = 10m-s^{-2}).CorrectGiven that Mass of rocket (m) = 1000 kg; Exhaust speed of gases with respect to rocket (v) = 1000 kg-s

^{-1}and acceleration due to gravity = 10 m-s^{-1}.

We know that minimum force on the rocket to lift it against gravitational attraction (F_{min}) = mg = 1000 × 10 = 10000 N.

Therefore minimum rate of burning of fuel (dm/dt) = F_{min}/v = 10000/100 = 100 kg-s^{-1}.IncorrectGiven that Mass of rocket (m) = 1000 kg; Exhaust speed of gases with respect to rocket (v) = 1000 kg-s

^{-1}and acceleration due to gravity = 10 m-s^{-1}.

We know that minimum force on the rocket to lift it against gravitational attraction (F_{min}) = mg = 1000 × 10 = 10000 N.

Therefore minimum rate of burning of fuel (dm/dt) = F_{min}/v = 10000/100 = 100 kg-s^{-1}. - Question 7 of 8
##### 7. Question

1 pointsIf m is initial mass of rocket, r is rate of ejection of gases and u is velocity of gases with respect to rocket, then acceleration of the rocket dv/dt at time t after blast is

CorrectGiven: Initial mass of rocket = m; Rate of ejection of gases (dm/dt) = r;

Velocity of gases with respect to the rocket = u and time = t.

We know that force on the rocket (F) = u × (dm/dt) = ur.

And final mass of rocket after t seconds (m’) = m – (dm/dt) × t = m – rt.

Therefore acceleration of rocket at time t after blast (dv/dt) = (F/m’) = ur/(m – rt).IncorrectGiven: Initial mass of rocket = m; Rate of ejection of gases (dm/dt) = r;

Velocity of gases with respect to the rocket = u and time = t.

We know that force on the rocket (F) = u × (dm/dt) = ur.

And final mass of rocket after t seconds (m’) = m – (dm/dt) × t = m – rt.

Therefore acceleration of rocket at time t after blast (dv/dt) = (F/m’) = ur/(m – rt). - Question 8 of 8
##### 8. Question

1 pointsA rocket of initial mass 6000 kg, ejects gases at constant rate 16 kg-s

^{-1}with constant relative speed of 11 km-s^{-1}. What is acceleration of rocket one minute after the blast?CorrectGiven: Initial mass of rocket (m) = 6000 kg and rate of gases ejected (dm/dt) = 16 kg-s

^{-1};

Relative speed of rocket (v) = 11 km-s^{-1}= 11 × 10^{3}m-s^{-1}and time (t) = 60s.

We know that force on rocket (F) = v × (dm/dt) = (11 × 10^{3}) × 16 = 176 × 10^{3}N.

And final mass of rocket after 1 minute (m’) = m – (dm/dt) × t = m – rt

= 6000 – (16 × 60) = 6000 – 960 = 5040kg.

Therefore acceleration of rocket (a) = (F/m’) = 176 × 10^{3}/5040 = 35 m-s^{-2}.IncorrectGiven: Initial mass of rocket (m) = 6000 kg and rate of gases ejected (dm/dt) = 16 kg-s

^{-1};

Relative speed of rocket (v) = 11 km-s^{-1}= 11 × 10^{3}m-s^{-1}and time (t) = 60s.

We know that force on rocket (F) = v × (dm/dt) = (11 × 10^{3}) × 16 = 176 × 10^{3}N.

And final mass of rocket after 1 minute (m’) = m – (dm/dt) × t = m – rt

= 6000 – (16 × 60) = 6000 – 960 = 5040kg.

Therefore acceleration of rocket (a) = (F/m’) = 176 × 10^{3}/5040 = 35 m-s^{-2}.