
A wide hose pipe is held horizontally by a fireman. It delivers water through a nozzle at one litres per second. On increasing the pressure, this increases two litres per second. The fireman has now to
A. Push forward twice as hard.
B. Push forward four times as hard.
C. Push backward four times as hard.
D. Push backward twice as hard.
Answer
575.7k+ views
Hint: Here we have given mass flow rate which is rate of change of mass with respect to time. We can write the second law of motion in the form of rate of change of mass. Applying the second law of motion we can find the direction in which the fireman should move and with how much force.
Formula used:
\[\begin{align}
& \dfrac{dm}{dt}=Av\rho \\
& V=Av \\
& F=v\dfrac{dm}{dt} \\
\end{align}\]
Complete answer:
Water is flowing through the pipe which means its mass is changing over time and the mass flow rate is given as
\[\dfrac{dm}{dt}=Av\rho \]
Where A is the cross-sectional area, v is the flow velocity and ρ is the density of the liquid (here water). Here we can see the rate of flow is directly proportional to velocity. So when the flow rate is doubled velocity is doubled too.
Now the Newton’s second law can be expressed in term of \[\dfrac{dm}{dt}\] as follows
\[F=v\dfrac{dm}{dt}\]
Where F is the force. Substituting the mass flow rate, force becomes
\[F=A{{v}^{2}}\rho \]
From the above formula I get that the force is proportional to the second power of velocity.
When pressure is increases the velocity is doubled and the as
\[F\propto {{v}^{2}}\]
Therefore, force is increased four times.
As the flow of water is in forward direction, the hose pipe will try to oppose the force and tends to come backward and therefore the fireman should push forward four times as hard when pressure is increased.
So, the correct answer is “Option B”.
Note:
We can also compare this situation by third Newton’s law that every action is equal and opposite. The hose pipe is opposing the flow of water and hence the force with which it is flowing and this is the reason that firemen should oppose the force of the hose pipe and move forward in direction of flow.
Formula used:
\[\begin{align}
& \dfrac{dm}{dt}=Av\rho \\
& V=Av \\
& F=v\dfrac{dm}{dt} \\
\end{align}\]
Complete answer:
Water is flowing through the pipe which means its mass is changing over time and the mass flow rate is given as
\[\dfrac{dm}{dt}=Av\rho \]
Where A is the cross-sectional area, v is the flow velocity and ρ is the density of the liquid (here water). Here we can see the rate of flow is directly proportional to velocity. So when the flow rate is doubled velocity is doubled too.
Now the Newton’s second law can be expressed in term of \[\dfrac{dm}{dt}\] as follows
\[F=v\dfrac{dm}{dt}\]
Where F is the force. Substituting the mass flow rate, force becomes
\[F=A{{v}^{2}}\rho \]
From the above formula I get that the force is proportional to the second power of velocity.
When pressure is increases the velocity is doubled and the as
\[F\propto {{v}^{2}}\]
Therefore, force is increased four times.
As the flow of water is in forward direction, the hose pipe will try to oppose the force and tends to come backward and therefore the fireman should push forward four times as hard when pressure is increased.
So, the correct answer is “Option B”.
Note:
We can also compare this situation by third Newton’s law that every action is equal and opposite. The hose pipe is opposing the flow of water and hence the force with which it is flowing and this is the reason that firemen should oppose the force of the hose pipe and move forward in direction of flow.
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