A $U$ -tube is of non uniform cross-section. The area of cross-sections of two sides of tube are $A$ and $2 A$ (see fig.). It contains non-viscous liquid of mass $m$ . The liquid is displaced slightly and free to oscillate. Its time period of oscillations is

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$T = 2 π \sqrt{\frac{m}{3 ρgA}}$

$T = 2 π \sqrt{\frac{m}{2 ρgA}}$

$T = 2 π \sqrt{\frac{m}{ρgA}}$

none of these

answer is A.

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Detailed Solution

Suppose the liquid in left side limb is displaced slightly by $y$ , the liquid in right limb will increase by $y / 2$ . The restoring force

$F = - P A = ρ g (\frac{3 y}{2}) \times 2 A = 3 ρ g A (- y) a = \frac{F}{m} = 3 ρ g A (- y) / m$

On comparing with, $a = - ω^{2} y$ , we get

$ω = \sqrt{\frac{3 ρ g A}{M}}$ and $T = 2 π \sqrt{\frac{m}{3 ρgA}}$

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