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Question: If the time period \((T)\) of vibration of a liquid drop depends on surface tension \((S)\), radius ...

If the time period (T)(T) of vibration of a liquid drop depends on surface tension (S)(S), radius (r)(r) of the drop and density (ρ)(\rho) of the liquid, then the expression of TT is

A

T=kρr3/ST = k\sqrt{\rho r^{3}/S}

B

T=kρ1/2r3/ST = k\sqrt{\rho^{1/2}r^{3}/S}

C

T=kρr3/S1/2T = k\sqrt{\rho r^{3}/S^{1/2}}

D

None of these

Answer

T=kρr3/ST = k\sqrt{\rho r^{3}/S}

Explanation

Solution

Let T∝SxryρzT \propto S^{x}r^{y}\rho^{z}

by substituting the dimension of [T]=[T]\lbrack T\rbrack = \lbrack T\rbrack

[S]=[MT−2],[r]=[L],[ρ]=[ML−3]\lbrack S\rbrack = \lbrack MT^{- 2}\rbrack,\lbrack r\rbrack = \lbrack L\rbrack,\lbrack\rho\rbrack = \lbrack ML^{- 3}\rbrack

and by comparing the power of both the sides

x=−1/2,y=3/2,z=1/2x = - 1/2,y = 3/2,z = 1/2

so T∝ρr3/S⇒T=kρr3ST \propto \sqrt{\rho r^{3}/S} \Rightarrow T = k\sqrt{\frac{\rho r^{3}}{S}}

T∝SxryρzT \propto S^{x}r^{y}\rho^{z}

[T]=[T][S]=[MT−2],[r]=[L],[ρ]=[ML−3]\lbrack T\rbrack = \lbrack T\rbrack\lbrack S\rbrack = \lbrack MT^{- 2}\rbrack,\lbrack r\rbrack = \lbrack L\rbrack,\lbrack\rho\rbrack = \lbrack ML^{- 3}\rbrack

T∝ρr3/S⇒T=kρr3ST \propto \sqrt{\rho r^{3}/S} \Rightarrow T = k\sqrt{\frac{\rho r^{3}}{S}})