A soap film is formed on a vertical equilateral triangular frame ABC. Side BC can move vertically always remaining horizontal. If ‘m’ is mass of rod (slider) BC and ‘S’ is surface tension. If BC is slightly displaced and released, then the time period of small oscillations of BC rod about equilibrium position is $π \sqrt{\frac{\sqrt{x} m}{S}}$ . The value of $x$ is

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answer is 3.

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Detailed Solution

If rod(slider) is displaced through ‘x’
Then length of film in contact becomes $(l + \frac{2 x}{\sqrt{3}})$

Restoring force $| F | = 2 S . (l + \frac{2 x}{\sqrt{3}}) - m g$
$| F | = 2 S l + \frac{4 x S}{\sqrt{3}} - m g$
(as S.2 $l$ =mg)
$| F | = \frac{4 S}{\sqrt{3}} . x$
$m . a = \frac{4 S}{\sqrt{3}} . x$
$|a| = \frac{4 S}{\sqrt{3} m} . x$
Similar to |a|= $ω^{2}$ x
$ω^{2} = \frac{4 S}{\sqrt{3} m} ω = \frac{4 S}{\sqrt{3} m}$
$T = \frac{2 π}{ω} = π \sqrt{\frac{\sqrt{3} m}{S}}$