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A liquid drop placed on a horizontal pla...

A liquid drop placed on a horizontal plane has a near spherical shape (slightly flattened due to gravity). Let R be the radius of its largest horizontal section. A small disturbance causes the drop to vibrate with frequency v about its equilibrium shape. By dimensional analysis the ratio `(v)/(sqrt(sigma//rho R^(3)))` can be (Here `sigma` is surface tension, `rho` is density, g is acceleration due to gravity, and k is arbitrary dimensionless constant)–

A

`(krhogR^2)/sigma`

B

`(krhogR^3)/(gsigma)`

C

`(krhogR^2)/(gsigma)`

D

`(krho)/(gsigma)`

Text Solution

Verified by Experts

The correct Answer is:
A
`(upsilon)/(sqrt(sigma//rhoR^3))=upsilonsqrt((rhoR^3)/(sigma))=T^(-1)[(ML^(-3)L^(3))/(MT^(-2))]^(1//2)=T^(-1)T =1`
`(upsilon)/(sqrt(sigma//rhoR^3))=K implies upsilon^2(rhoR^3)/sigma=K^2`
`=K^2 =(rho"R^3)/sigmaT^(-2)=(rhoR^(2))/(sigma)RT^(-2)impliesK=(rhoR^2)/(sigma)g`
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