If the time period T propto P^a d^b E^c,where | vedclass

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(C) The dimensional formula for time period $T$ is $[M^0 L^0 T^1]$.The dimensional formulas for the given physical quantities are:Pressure $P = [M L^{

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$\frac{1}{3}$

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(C) The dimensional formula for time period $T$ is $[M^0 L^0 T^1]$.The dimensional formulas for the given physical quantities are:Pressure $P = [M L^{-1} T^{-2}]$Density $d = [M L^{-3}]$Energy $E = [M L^2 T^{-2}]$Given the relation $T = k P^a d^b E^c$,we write the dimensional equation:$[M^0 L^0 T^1] = [M L^{-1} T^{-2}]^a [M L^{-3}]^b [M L^2 T^{-2}]^c$$[M^0 L^0 T^1] = M^{a+b+c} L^{-a-3b+2c} T^{-2a-2c}$Comparing the powers of $M, L,$ and $T$ on both sides:$1$) $a + b + c = 0$$2$) $-a - 3b + 2c = 0$$3$) $-2a - 2c = 1$From equation $(3)$,$-2(a + c) = 1$,so $a + c = -1/2$.Substitute $a + c = -1/2$ into equation $(1)$: $(-1/2) + b = 0$,which gives $b = 1/2$.Substitute $b = 1/2$ into equation $(2)$: $-a - 3(1/2) + 2c = 0 \Rightarrow -a + 2c = 3/2$.Now we have a system of two equations:$a + c = -1/2$$-a + 2c = 3/2$Adding these two equations: $3c = 1$,so $c = 1/3$.

If the time period $T \propto P^a d^b E^c$,where $P$ is pressure,$d$ is density,and $E$ is energy,then the value of $c$ is:

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