Assertion (A): Time period of oscillation of a liquid drop depends on surface tension(S), if density of the liquid is $ρ$ and radius of the drop is r, then $T = k \sqrt{\frac{{ρr}^{3}}{S^{3 / 2}}}$ is dimensionally correct, where K is dimensionless.
Reason (R): Using dimensional analysis we get R.H.S having different dimension than that of time period. In the light of above statements, choose the correct answer from the options given below.

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(A) is true but (R) is false

(A) is false but (R) is true

Both (A) and (R) are true but (R) is not the correct explanation of (A)

Both (A) and (R) are true and (R) is the correct explanation of (A)

answer is D.

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Detailed Solution

$T = k \sqrt{\frac{{ρr}^{3}}{S^{3 / 2}}}$
Dimensions of RHS $= \frac{[M^{1 / 2} L^{- 3 / 2}] [L^{3 / 2}]}{{[{MT}^{- 2}]}^{3 / 4}} = M^{- 1 / 4} L^{0} T^{3 / 2}$
Dimensions of L.H.S $\neq$ Dimensions of R.H.S

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