A liquid of density ρ is filled in a U-tube is accelerated with an acceleration a so that the height of liquid in its two vertical arms are h1 and h2 as shown in the figure. If l is the length of horizontal arm of the tube, the acceleration a is

Question

A liquid of density ρ is filled in a U-tube is accelerated with an acceleration a so that the height of liquid in its two vertical arms are h1 and h2 as shown in the figure. If l is the length of horizontal arm of the tube, the acceleration a is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-yuciyufack6z.png" /> (A) (g (. h1 - h2 .)/2 l) towards right (B) (g (. h1 - h2 .)/2 l) towards left (C) (g (. h1 - h2 .)/l) towards right (D) (g (. h1 - h2 .)/l) towards left

Tardigrade Admin · Accepted Answer

Pressure on left end of horizontal tube,

p_{1} = p_{0} + h_{1} ρ g

Pressure on right end of horizontal tube,

p_{2} = p_{0} + h_{2} ρ g

As

p_{1} > p_{2}

, so acceleration should be towards right hand side. If

A

is the area of cross-section of the tube in the horizontal portion of U-tube, then

p_{1} A - p_{2} A = (l A ρ) a

Or

(h_{1} - h_{2}) ρ g A = l A ρ a

a = g \frac{( h _{1} - h _{2} )}{l}

Q. A liquid of density $ρ$ is filled in a U-tube is accelerated with an acceleration $a$ so that the height of liquid in its two vertical arms are $h_{1}$ and $h_{2}$ as shown in the figure. If $l$ is the length of horizontal arm of the tube, the acceleration $a$ is

$\frac{g ( h _{1} - h _{2} )}{2 l}$ towards right

$\frac{g ( h _{1} - h _{2} )}{2 l}$ towards left

$\frac{g ( h _{1} - h _{2} )}{l}$ towards right

$\frac{g ( h _{1} - h _{2} )}{l}$ towards left

Q. A liquid of density ρ is filled in a U-tube is accelerated with an acceleration a so that the height of liquid in its two vertical arms are h1​ and h2​ as shown in the figure. If l is the length of horizontal arm of the tube, the acceleration a is

2lg(h1​−h2​)​ towards right

2lg(h1​−h2​)​ towards left

lg(h1​−h2​)​ towards right

lg(h1​−h2​)​ towards left

Q. A liquid of density $ρ$ is filled in a U-tube is accelerated with an acceleration $a$ so that the height of liquid in its two vertical arms are $h_{1}$ and $h_{2}$ as shown in the figure. If $l$ is the length of horizontal arm of the tube, the acceleration $a$ is

$\frac{g ( h _{1} - h _{2} )}{2 l}$ towards right

$\frac{g ( h _{1} - h _{2} )}{2 l}$ towards left

$\frac{g ( h _{1} - h _{2} )}{l}$ towards right

$\frac{g ( h _{1} - h _{2} )}{l}$ towards left