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The time period of a body under S.H.M. is represented by: T = P^a D^b S^c where P is pressure, D is density and S is surface tension, then values of a, b and c are

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$- \frac{3}{2}, \frac{1}{2}, 1$

$- 1, - 2, 3$

$\frac{1}{2}, - \frac{3}{2}, - \frac{1}{2}$

$1, 2, \frac{1}{3}$

answer is A.

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Detailed Solution

$T = P^{a} D^{b} S^{c}$

$\begin{array}{l} M^{0} L^{0} T^{1} = {({ML}^{- 1} T^{- 2})}^{a} {({ML}^{- 3})}^{b} {({MT}^{- 2})}^{c} \\ = M^{a + b + c} L^{- a - 3 b} T^{- 2 a - 2 c} \end{array}$

Applying principle of homogeneity

$a + b + c = 0; - a - 3 b = 0; - 2 a - 2 c = 1$

on solving, we get $a = - 3 / 2, b = 1 / 2, c = 1$

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