
A tube of length L is filled completely with an incompressible liquid of mass M and closed at both ends. The tube is then rotated in a horizontal plane about one of its ends with uniform angular velocity $\omega $. The force exerted by the liquid at the other end is
A) $\dfrac{1}{2}M{\omega ^2}{L^2}$
B) $M{\omega ^2}L$
C) $\dfrac{1}{4}M{\omega ^2}L$
D) $\dfrac{1}{2}M{\omega ^2}L$
Answer
511.5k+ views
Hint: According to the question tube is rotated in a horizontal plane about one of its ends with uniform angular velocity $\omega$. Now, when force is applied at one end, then force acting on the other end is the centripetal force.
Formula used:
To solve this type of question we use the following formula.
Centrifugal force= $m{\omega ^2}r$, where $\omega$ is the angular frequency, $m$ is mass and $r$ is the distance the axes to the point of rotation where force is applied.
Complete step by step answer:
The formula of centrifugal force is $m{\omega ^2}r$.
Now, we know that the centre of mass will be at $\dfrac{L}{2}$.
Let us substitute the value in $m{\omega ^2}r$.
Centrifugal force=$m{\omega ^2} \times \dfrac{L}{2} = \dfrac{{mL{\omega ^2}}}{2}$
This is the force that will act on the other end of the tube.
$\therefore $ Hence, option (D) $\dfrac{{mL{\omega ^2}}}{2}$ is the correct option.
Additional information:
The centripetal force always acts perpendicular to the displacement direction. Consider an object accelerating with a change in velocity, the object will either change the direction of motion or the speed. This happens only in the centripetal force. If an object moves in a circular path and experiences an outward force then the object is said to be in the centrifugal force.
Note:
-A body performing circular motion is acted upon by a force which is always directed towards the center of the circle. This is called the centripetal force.
-Centrifugal force is a pseudo force. There are certain situations in which we feel that a body is acted upon by a force, but actually there is no force on the body.
-An effect that seems to cause an object moving in a curve to be pushed away from the curve’s center. This is called the centrifugal force.
Formula used:
To solve this type of question we use the following formula.
Centrifugal force= $m{\omega ^2}r$, where $\omega$ is the angular frequency, $m$ is mass and $r$ is the distance the axes to the point of rotation where force is applied.
Complete step by step answer:

The formula of centrifugal force is $m{\omega ^2}r$.
Now, we know that the centre of mass will be at $\dfrac{L}{2}$.
Let us substitute the value in $m{\omega ^2}r$.
Centrifugal force=$m{\omega ^2} \times \dfrac{L}{2} = \dfrac{{mL{\omega ^2}}}{2}$
This is the force that will act on the other end of the tube.
$\therefore $ Hence, option (D) $\dfrac{{mL{\omega ^2}}}{2}$ is the correct option.
Additional information:
The centripetal force always acts perpendicular to the displacement direction. Consider an object accelerating with a change in velocity, the object will either change the direction of motion or the speed. This happens only in the centripetal force. If an object moves in a circular path and experiences an outward force then the object is said to be in the centrifugal force.
Note:
-A body performing circular motion is acted upon by a force which is always directed towards the center of the circle. This is called the centripetal force.
-Centrifugal force is a pseudo force. There are certain situations in which we feel that a body is acted upon by a force, but actually there is no force on the body.
-An effect that seems to cause an object moving in a curve to be pushed away from the curve’s center. This is called the centrifugal force.
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