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Standard XII

Physics

Accelerating Containers

A 'U' - tube ...

Question

A ‘U’ - tube containing a liquid is accelerated horizontally with a constant acceleration $a_{0}$ . If the separation between the vertical limbs is l, the difference in the heights of the liquid in the two arms is

\frac{a_{0} g}{l}

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\frac{a_{0} l}{g}

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\frac{g l}{a_{0}}

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\frac{l}{a_{0} g}

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Solution

The correct option is B $\frac{a_{0} l}{g}$
If $^{'} P_{a}^{'}$ is the atmospheric pressure
‘A’ is the area of cross section
‘h’ is the increase in height
$∴ P_{a} A + A ρ a_{0} = P_{a} A + (h ρ g) A$
$\Rightarrow h g = a_{0} l$
$\Rightarrow h = \frac{a_{0} l}{g}$

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A ‘U’ - tube containing a liquid is accelerated horizontally with a constant acceleration a0. If the separation between the vertical limbs is l, the difference in the heights of the liquid in the two arms is

The correct option is B a0lgIf ′P′a is the atmospheric pressure ‘A’ is the area of cross section ‘h’ is the increase in height ∴PaA+Aρa0=PaA+(hρg)A ⇒hg=a0l ⇒h=a0lg

A ‘U’ - tube containing a liquid is accelerated horizontally with a constant acceleration $a_{0}$ . If the separation between the vertical limbs is l, the difference in the heights of the liquid in the two arms is

The correct option is B $\frac{a_{0} l}{g}$
If $^{'} P_{a}^{'}$ is the atmospheric pressure
‘A’ is the area of cross section
‘h’ is the increase in height
$∴ P_{a} A + A ρ a_{0} = P_{a} A + (h ρ g) A$
$\Rightarrow h g = a_{0} l$
$\Rightarrow h = \frac{a_{0} l}{g}$