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If the time period (t) of vibration of a liquid drop depends on density (rho) of the liquid, radius (r) of the drop and surface tension (S), then the expression of t is where k is a dimensionless constant.

Answer» <html><body><p>`t= ksqrt((rhor^(3))/(S))`<br/>surface tension<br/>`t= ksqrt((rhor^(3))/(S^(1//2)))`<br/>`t= ksqrt((rhor)/(S))`</p>Solution :Let `t= krho^(x)r^(y)S^(z)` <br/> where k is a dimensionless constant <br/> Equating dimensions on both sides, we get <br/> `[T]= [<a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a>^(-3)]^(x)[<a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>]^(y)[ML^(0)T^(-2)]^(z)` or `[M^(0)L^(0)T]= [M^(x+z)L^(-3x+z)T^(-<a href="https://interviewquestions.tuteehub.com/tag/2z-301338" style="font-weight:bold;" target="_blank" title="Click to know more about 2Z">2Z</a>)]` <br/> According to <a href="https://interviewquestions.tuteehub.com/tag/principle-1166255" style="font-weight:bold;" target="_blank" title="Click to know more about PRINCIPLE">PRINCIPLE</a> of homogeneity of dimensions we get <br/> x+z= 0...(i) <br/> -3x+y=0 ..(ii) <br/> -2z = 1 ...(iii) <br/><a href="https://interviewquestions.tuteehub.com/tag/solving-1217196" style="font-weight:bold;" target="_blank" title="Click to know more about SOLVING">SOLVING</a> equations (i), (ii), and (iii), we get <br/> `x= (1)/(2), y= (3)/(2), z= (-1)/(2) :. t= krho^(1//2)r^(3//2)S^(-1//2)` or `t= ksqrt((rhor^(3))/(S))`</body></html>


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