
A rocket consumes 20 kg fuel per second. The exhaust gases escape at a speed at $1000\;m{s^{ - 1}}$ relative to the rocket. Calculate the velocity acquired by the rocket, when its mass reduces to $1/100$ of its initial mass.
A) 4.5 km/sec
B) 4.6 km/sec
C) 4.06 km/sec
D) 4.05 km/sec
Answer
133.5k+ views
Hint: The rocket propels upwards due to the thrust produced by the burning of the fuel. This thrust onto the ground causes a force upwards which propels the rocket upwards. Due to the force acting upwards the rocket will be accelerated upwards. This up thrust is produced by the burning of fuel in the rocket.
Complete step by step answer:
The velocity V acquired when the final mass is $1/100$ of the initial mass is :
$V = u\ln \left( {\dfrac{M}{m}} \right)$
Here u is the initial velocity, m is the final mass and M is the initial mass.
$\begin{array}{l}
V = 1000\;\ln \left( {100} \right)\\
= 4606\;m/s\\
= 4.6\;km/s
\end{array}$
Therefore, the correct option is (B).
Additional Information: To understand the science behind the Rocket, you can take an example of a gunshot. Shooting a gun also demonstrates the application of conservation of momentum. As we pull the trigger, the bullet comes out at a very high speed, but we also observe a recoil of the gun. This happens to conserve momentum. The momentum gained by the bullet is equal to and also the reason for the recoil of the gun. Same as this, the gases inside a rocket are made to propel out of the Rocket at a very high speed. This, in turn, gives a push to the rocket in the opposite direction to conserve the momentum. Thus a lot of fuel needs to be burned to provide the rocket a sufficient amount of force to escape the earth's atmosphere.
Note: The force that propels the rocket upwards is the consequence of Newton’s Third Law. The force of thrust on to the ground and the force on the rocket is a Newtonian pair of forces which are equal and opposite. The force acting as thrust is due to the mass dissipation of the fuel per time which with the velocity of the exhaust produces a thrust force is responsible for the propulsion.
Complete step by step answer:
The velocity V acquired when the final mass is $1/100$ of the initial mass is :
$V = u\ln \left( {\dfrac{M}{m}} \right)$
Here u is the initial velocity, m is the final mass and M is the initial mass.
$\begin{array}{l}
V = 1000\;\ln \left( {100} \right)\\
= 4606\;m/s\\
= 4.6\;km/s
\end{array}$
Therefore, the correct option is (B).
Additional Information: To understand the science behind the Rocket, you can take an example of a gunshot. Shooting a gun also demonstrates the application of conservation of momentum. As we pull the trigger, the bullet comes out at a very high speed, but we also observe a recoil of the gun. This happens to conserve momentum. The momentum gained by the bullet is equal to and also the reason for the recoil of the gun. Same as this, the gases inside a rocket are made to propel out of the Rocket at a very high speed. This, in turn, gives a push to the rocket in the opposite direction to conserve the momentum. Thus a lot of fuel needs to be burned to provide the rocket a sufficient amount of force to escape the earth's atmosphere.
Note: The force that propels the rocket upwards is the consequence of Newton’s Third Law. The force of thrust on to the ground and the force on the rocket is a Newtonian pair of forces which are equal and opposite. The force acting as thrust is due to the mass dissipation of the fuel per time which with the velocity of the exhaust produces a thrust force is responsible for the propulsion.
Recently Updated Pages
JEE Main 2025 Session 2 Form Correction (Closed) – What Can Be Edited

Sign up for JEE Main 2025 Live Classes - Vedantu

JEE Main Books 2023-24: Best JEE Main Books for Physics, Chemistry and Maths

JEE Main 2023 April 13 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 11 Shift 2 Question Paper with Answer Key

JEE Main 2023 April 10 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Class 11 JEE Main Physics Mock Test 2025

Current Loop as Magnetic Dipole and Its Derivation for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

Motion In A Plane: Line Class 11 Notes: CBSE Physics Chapter 3

Waves Class 11 Notes: CBSE Physics Chapter 14
