Question

Due to an explosion underneath water, a bubble stalled 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 $Y$

• A. $\alpha=-\frac{3}{2}, \beta=\frac{1}{3}, \gamma=-\frac{5}{6}$
• 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. $\alpha=-\frac{1}{3}, \beta=\frac{3}{2}, \gamma=-\frac{4}{3}$

Answer 1

Answer: (B)

Given, time-period of oscillation $T$ is
$T \,\propto \,p^{\alpha} \,S^{\beta} \,E^{\gamma}$
or $T=k p^{\alpha} \,S^{\beta} \,E^{\gamma}$
Now, substituting dimensions of $T, p, S$ and $E$, we have $
\left[M^{0} \, L^{0} \, T^{l}\right]=k\left[M \, L^{-1} \,T^{-2}\right]^{\alpha}\left[M \,L^{-3}\right]^{\beta}\left[M \,L^{2} \, T^{-2}\right]^{\gamma}$
Equating powers of similar terms, we have
$\alpha+\beta+\gamma=0 \ldots$ (i)
$-\alpha-3 \beta+2 \gamma=0 \ldots$.(ii)
$-2 \alpha-2 \gamma=1 \ldots$ (ii)
Solving, we get $\alpha=\frac{-5}{6}, \beta=\frac{1}{2}, \gamma=\frac{1}{3}$