A gas bubble from an explosion under water os | vedclass

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Question

Time period $\mathrm{T}=\mathrm{P}^{\mathrm{a}} \mathrm{d}^{\mathrm{b}} \mathrm{E}^{\mathrm{c}}$

Taking dimensions of each physical quantity.

$[

Vedclass · Accepted Answer

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

Vedclass · Answer

Time period $\mathrm{T}=\mathrm{P}^{\mathrm{a}} \mathrm{d}^{\mathrm{b}} \mathrm{E}^{\mathrm{c}}$

Taking dimensions of each physical quantity.

$[\mathrm{T}]=\left[\mathrm{ML}^{-1}\mathrm{T}^{-2}\right]^{\mathrm{a}}\left[\mathrm{ML}^{-3}\right]^{\mathrm{b}}\left[\mathrm{ML}^{2} \mathrm{T}^{-2}\right]^{\mathrm{c}}$

Equating the exponents of $\mathrm{M}, \mathrm{L}$ and $\mathrm{T}$ on both the sides,

$\mathrm{M}^{\mathrm{a}+\mathrm{b}+\mathrm{T}} \mathrm{L}^{-\mathrm{a}-3 \mathrm{b}+2 \mathrm{c}} \mathrm{T}^{-2 \mathrm{a}-2 \mathrm{c}} =\mathrm{T}$

$\mathrm{a}+\mathrm{b}+\mathrm{c} =0$

$-\mathrm{a}-3 \mathrm{b}+2 \mathrm{c} =0$

$-2 \mathrm{a}-2 \mathrm{c} =1$

Solving these equations for $a$, $b$ and $c,$ we get

$a=-\frac{5}{6}, b=\frac{1}{2}$ and $\frac{1}{3}$

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

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