Solution to Question 5
1. Analysis of Liquid Distribution
The U-tube accelerates horizontally with acceleration $a = g$. The liquid surface tilts, creating a height difference between the two arms. Based on the diagram, the liquid rises in the right arm, implying the acceleration was directed to the left.
Let $h_1$ be the height in the left arm and $h_2$ be the height in the right arm. The pressure difference due to horizontal acceleration balances the hydrostatic pressure difference: $$ P_{right} – P_{left} = \rho a l $$ Substituting $P = \rho g h$ and $a=g$: $$ \rho g h_2 – \rho g h_1 = \rho g l $$ $$ h_2 – h_1 = l \quad \text{…(i)} $$
The total length of the liquid column is given as $4l$. The bottom horizontal arm has length $l$, so the liquid in the vertical arms must sum to: $$ h_1 + h_2 = 4l – l = 3l \quad \text{…(ii)} $$
Solving (i) and (ii): $$ (h_2 – h_1) + (h_2 + h_1) = l + 3l \implies 2h_2 = 4l \implies h_2 = 2l $$ $$ h_1 = 3l – 2l = l $$ State at $t=0$: Left arm height is $l$, Right arm height is $2l$.
Figure: Liquid distribution at the moment acceleration stops. The unbalanced head ($2l – l$) drives fluid acceleration $a_f$ from right to left.
2. Dynamics After Acceleration Stops
When the external acceleration stops, the liquid column is momentarily at rest but unbalanced. The extra height in the right arm exerts a net force driving the fluid to the left.
Calculating Fluid Acceleration ($a_f$):
Net driving force = Weight of the unbalanced column of height $(h_2 – h_1)$.
$$ F_{net} = (\rho A l) g $$
Total mass being accelerated = Mass of total liquid length $4l$.
$$ m_{total} = \rho A (4l) $$
Using Newton’s Second Law ($F = ma_f$):
$$ (\rho A l) g = (\rho A 4l) a_f $$
$$ a_f = \frac{g}{4} $$
The fluid accelerates **down** in the right arm, **left** in the horizontal arm, and **up** in the left arm.
3. Calculation of Pressure at Midpoint
We can find the pressure at midpoint $M$ by starting from the open surface of the left arm (where $P=0$ gauge) and moving to $M$.
Path: Surface of Left Arm $\rightarrow$ Bottom Left Corner $\rightarrow$ Midpoint M
Step A: Vertical Descent (Left Arm)
We move down by depth $h_1 = l$. The fluid in this arm is accelerating **upwards** with $a_f = g/4$.
The effective gravity is $g_{eff} = g + a_f$.
$$ P_{bottom\_left} = P_{surface} + \rho (g + a_f) h_1 $$
$$ P_{bottom\_left} = 0 + \rho \left(g + \frac{g}{4}\right) (l) $$
$$ P_{bottom\_left} = \rho \left(\frac{5g}{4}\right) l = 1.25 \rho g l $$
Step B: Horizontal Movement (Bottom Arm)
We move horizontally to the right by distance $x = l/2$ to reach $M$.
The fluid acceleration is to the **left** ($a_f = g/4$).
Since we are moving to the right (opposite to the acceleration vector), pressure **increases**.
$$ P_M = P_{bottom\_left} + \rho a_f x $$
$$ P_M = 1.25 \rho g l + \rho \left(\frac{g}{4}\right) \left(\frac{l}{2}\right) $$
$$ P_M = 1.25 \rho g l + 0.125 \rho g l $$
$$ P_M = 1.375 \rho g l $$
Start at Right Surface ($h_2=2l$). Fluid accelerates down. $g_{eff} = g – a_f$. $$ P_{bottom\_right} = \rho (g – \frac{g}{4}) (2l) = \rho (\frac{3g}{4})(2l) = 1.5 \rho g l $$ Move Left to M ($x=l/2$). Moving **with** acceleration direction $\rightarrow$ Pressure decreases. $$ P_M = 1.5 \rho g l – \rho (\frac{g}{4}) (\frac{l}{2}) = 1.5 \rho g l – 0.125 \rho g l = 1.375 \rho g l $$
Answer
The gauge pressure at the midpoint is $1.375 \rho g l$.
Correct Option: (d)