Bottom arm of a U-tube of uniform section has length ℓ and the vertical arms have length 3 ℓ each. The tube is held with its parallel arms vertical and filled with a liquid of density ρ to occupy a to

Question

Bottom arm of a U-tube of uniform section has length $ℓ$ and the vertical arms have length $3 ℓ$ each. The tube is held with its parallel arms vertical and filled with a liquid of density $ρ$ to occupy a total length of $4 ℓ$ of the tube and then it is made to accelerate horizontally in the plane of the tube with an acceleration g that is modulus of acceleration of free fall. What will gauge pressure at the midpoint of the bottom arm be immediately after acceleration of the tube is made abruptly zero?

TestHub · Accepted Answer

While accelerating at g, level difference in arms: $ρ g ℓ = ρ g (h_{L} - h_{R})$ ⇒ heights 2ℓ and ℓ.

Immediately after acceleration becomes zero, liquid column accelerates along tube, say a'.

Then $g ℓ = a^{'} (2 ℓ + ℓ + ℓ) \Rightarrow a^{'} = \frac{g}{4}$

Pressure at left bottom corner: $P_{L} = ρ g (2 ℓ) - ρ \frac{g}{4} (2 ℓ) = \frac{3 ρ g ℓ}{2}$

Midpoint M: $P_{M} = P_{L} - ρ \frac{g}{4} \frac{ℓ}{2} = \frac{3 ρ g ℓ}{2} - \frac{ρ g ℓ}{8} = \frac{11}{8} ρ g ℓ = 1.375 ρ g ℓ$ Hence (D).

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