Due to an explosion underneath water, a bubble started oscillating. If this oscillation has time period $T$, which is proportional to $p^\alpha S^\beta E^\gamma$, where $p$ is static pressure, $S$ is density of water and $E$ is total energy of explosion. Determine $\alpha, \beta$ and $\gamma$.

• A

  $\alpha=-\frac{3}{2}, \beta=\frac{1}{3}, \gamma=-\frac{5}{6}$
• Correct answer
  B

  $\alpha=-\frac{5}{6}, \beta=\frac{1}{2}, \gamma=\frac{1}{3}$
• C

  $\alpha=\frac{1}{2}, \beta=-\frac{5}{6}, \gamma=\frac{7}{4}$
• D

  $\alpha=\frac{1}{3}, \beta=\frac{3}{2}, \gamma=\frac{4}{3}$

Solution

Given, time-period of oscillation $T$ is $T \propto p^\alpha S^\beta E^\gamma \text { or } T=k p^\alpha S^\beta E^\gamma$ Now, subtituting dimensions of $T, p, S$ and $E$, we have $\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^{\mathrm{l}}\right]=k\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]^\alpha\left[\mathrm{ML}^{-3}\right]^\beta\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]^\gamma$ Equating powers of similar terms, we have
Solving, we get $\alpha=\frac{-5}{6}, \beta=\frac{1}{2}, \gamma=\frac{1}{3}$

Hence, the correct answer is option: 'B'.