Question

When at rest, a liquid stands at the same level in the tubes as shown in the figure. But as indicated, a height difference h occurs when the system is given acceleration $a$ towards the right. Then h is equal to:

Answer 1

Hint: Using the fluid mechanics formula to find the force which is applied by the weight of the water on the walls of the container when water is kept without any external pressure. The formula for such is written as:
\[F\text{ }=\text{ }pgAh\]
where p is the density of the liquid, g is the gravity, A is the area of the cross section of the water contained and h is the height of the water in two tubes.

Complete step by step solution:
First let us take the area of the horizontal tube as $A$.
To find the volume of water that is contained within the tube from 1 to 2 and with the length of the tube from 1 to 2 as $L$ and we take the area of the water in the tube from 1 to 2 as $A$. Hence, the volume of the water from 1 to 2 is \[LA\].
With volume found the mass of the water from 1 to 2 is density multiplied by volume and is written as:
\[m=pV\]
According to the question given the whole contraption is moving from left to right hence, the force of moving the contraption from left to right is given as F = ma.

Placing the value of m as pV in the above formula, we get the final value as mpV and then equate it with the force formula of the fluid, the force formula of the fluid is given as:
\[F\text{ }=\text{ }pgAh\]
Now equating the value of the fluid force with the force due to the motion of the contraption is:
\[pgAh\text{ }=\text{ }pV\] with \[V\text{ }=\text{ }LA\]
\[h\text{ }=\text{ }\dfrac{aL}{g}\]

Therefore, the extra height of the liquid is given as \[h\text{ }=\text{ }\dfrac{aL}{g}\].

Note: Student may go wrong if they try to take any extra variable for the unknown part of length and try to implement it into the equation if we take the unknown part of length below height h as x or any other variable and then solve the question we are left with an extra unsolvable variable hence, the we don't need to solve the unknown part of the tube to get the answer.