A soap film is formed on a parabolic, vertical, rigid, fixed frame $(y = - 4 x^{2})$ . A sufficient long thin rod is hanging with frame with support of film in uniform gravity as shown in diagram. Now rod is slightly displaced in vertical direction and released. ‘T’ is surface tension. Take complete wetting condition. Mass of rod is ‘m’

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

Initially equilibrium of rod is unstable

Initially equilibrium of rod is stable

For small oscillations time period of oscillation will be $2 π \sqrt{\frac{m \sqrt{H}}{2 T}}$

For small oscillations time period of oscillation will be $2 π \sqrt{\frac{m \sqrt{H}}{T}}$

answer is A, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Equilibrium condition at $y = - H$
$m g = 2 T \sqrt{H} - - (1)$
When displaced slightly down by ‘ $y_{0}$ ’
$2 T [1 + \frac{Y_{0}}{2 H}] \sqrt{H} - m g = m a$
$m a = \frac{T Y_{0}}{\sqrt{H}}$ (or) $ω = \sqrt{\frac{T}{m \sqrt{H}}}$
Timer period $= 2 π \sqrt{\frac{m \sqrt{H}}{T}}$