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(D) The dimensional formula for time period $T$ is $[M^0 L^0 T^1]$.The dimensional formula for pressure $p$ is $[M^1 L^{-1} T^{-2}]$.The dimensional f

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

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(D) The dimensional formula for time period $T$ is $[M^0 L^0 T^1]$.The dimensional formula for pressure $p$ is $[M^1 L^{-1} T^{-2}]$.The dimensional formula for density $D$ is $[M^1 L^{-3} T^0]$.The dimensional formula for surface tension $S$ is $[M^1 L^0 T^{-2}]$.Given the relation $T = p^a D^b S^c$,we equate the dimensions on both sides:$[M^0 L^0 T^1] = [M^1 L^{-1} T^{-2}]^a [M^1 L^{-3} T^0]^b [M^1 L^0 T^{-2}]^c$$[M^0 L^0 T^1] = [M^{a+b+c} L^{-a-3b} T^{-2a-2c}]$Comparing the powers of $M, L,$ and $T$ on both sides:$1$) $a + b + c = 0$$2$) $-a - 3b = 0 \implies a = -3b$$3$) $-2a - 2c = 1$Substitute $a = -3b$ into equation $(1)$:$-3b + b + c = 0 \implies c = 2b$Substitute $a = -3b$ and $c = 2b$ into equation $(3)$:$-2(-3b) - 2(2b) = 1$$6b - 4b = 1 \implies 2b = 1 \implies b = \frac{1}{2}$Now,find $a$ and $c$:$a = -3b = -3(\frac{1}{2}) = -\frac{3}{2}$$c = 2b = 2(\frac{1}{2}) = 1$Thus,the values are $a = -\frac{3}{2}, b = \frac{1}{2}, c = 1$.

The time period of a body undergoing simple harmonic motion is given by $T=p^{a} D^{b} S^{c}$,where $p$ is the pressure,$D$ is density,and $S$ is surface tension. The values of $a, b,$ and $c$ respectively are

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