A U-tube of base length L filled with the same volume of two liquids of densities ρ and 2ρ is moving with acceleration a on the horizontal plane. If the height difference between the two surfaces (open to atmosphere) becomes zero, then the height h is given by

Question

A U-tube of base length L filled with the same volume of two liquids of densities ρ and 2ρ is moving with acceleration a on the horizontal plane. If the height difference between the two surfaces (open to atmosphere) becomes zero, then the height h is given by <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-m1u9spgd6baj.png" /> (A) (aL/2 g) (B) (3 aL/2 g) (C) (aL/g) (D) (2 aL/3 g)

Tardigrade Admin · Accepted Answer

The pressure at the left bottom due to left vertical pipe

(= P)_{L} = P_{a t m} + h (2 ρ) g

The pressure at the right bottom due to right vertical pipe

= P_{R} = P_{a t m} + h (ρ) g

and Pressure difference in the horizontal tube due to acceleration

P_{L R} = (\frac{L}{2}) ρ a + (\frac{L}{2}) 2 ρ a

P_{L R} = \frac{3 L ρ a}{2}

The pressure at the left bottom due to right vertical pipe and horizontal pipe

P_{L} = P_{R} + P_{L R}

P_{L} = P_{a t m} + h ρ g + \frac{3 L ρ a}{2}

The pressure at the left bottom due to left vertical pipe=Pressure at the left bottom due to right vertical pipe and horizontal pipe

P_{a t m} + h (2 ρ) g = P_{a t m} + h ρ g + \frac{3 L ρ a}{2}

\Rightarrow h = \frac{3 a L}{2 g}

Q. A U-tube of base length $L$ filled with the same volume of two liquids of densities $ρ$ and $2 ρ$ is moving with acceleration $a$ on the horizontal plane. If the height difference between the two surfaces (open to atmosphere) becomes zero, then the height $h$ is given by

$\frac{a L}{2 g}$

$\frac{3 a L}{2 g}$

$\frac{a L}{g}$

$\frac{2 a L}{3 g}$

Q. A U-tube of base length L filled with the same volume of two liquids of densities ρ and 2ρ is moving with acceleration a on the horizontal plane. If the height difference between the two surfaces (open to atmosphere) becomes zero, then the height h is given by

2gaL​

2g3aL​

gaL​

3g2aL​

Q. A U-tube of base length $L$ filled with the same volume of two liquids of densities $ρ$ and $2 ρ$ is moving with acceleration $a$ on the horizontal plane. If the height difference between the two surfaces (open to atmosphere) becomes zero, then the height $h$ is given by

$\frac{a L}{2 g}$

$\frac{3 a L}{2 g}$

$\frac{a L}{g}$

$\frac{2 a L}{3 g}$