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(A) The time period $T$ is given by $T \propto P^a d^b E^c$. Writing the dimensional formulas for each quantity: $[T] = [M^0 L^0 T^1]$ $[P] = [M L^{-1

Vedclass · Accepted Answer

$ - \frac{5}{6}, \frac{1}{2}, \frac{1}{3}$

Vedclass · Answer

(A) The time period $T$ is given by $T \propto P^a d^b E^c$. Writing the dimensional formulas for each quantity: $[T] = [M^0 L^0 T^1]$ $[P] = [M L^{-1} T^{-2}]$ $[d] = [M L^{-3} T^0]$ $[E] = [M L^2 T^{-2}]$ Substituting these into the 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 exponents of $M, L,$ and $T$ on both sides: $1) a + b + c = 0$ $2) -a - 3b + 2c = 0$ $3) -2a - 2c = 1$ From equation $(3)$,$a + c = -1/2$,so $c = -1/2 - a$. Substitute $c$ into equation $(1)$: $a + b + (-1/2 - a) = 0 \implies b = 1/2$. Substitute $b = 1/2$ and $c = -1/2 - a$ into equation $(2)$: $-a - 3(1/2) + 2(-1/2 - a) = 0$ $-a - 3/2 - 1 - 2a = 0$ $-3a = 5/2 \implies a = -5/6$. Finally,$c = -1/2 - (-5/6) = -3/6 + 5/6 = 2/6 = 1/3$. Thus,$a = -5/6, b = 1/2, c = 1/3$.

$A$ gas bubble from an explosion under water oscillates with a period proportional to $P^a d^b E^c$,where $P$ is the static pressure,$d$ is the density of water,and $E$ is the energy of the explosion. Then $a, b,$ and $c$ are:

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