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 $<br/>\left[M^0{L}^0{T}^l\right]=k{\left[M{L}^{-1}{T}^{-2}\right]}^{\alpha }{\left[M{L}^{-3}\right]}^{\beta }{\left[M{L}^2{T}^{-2}\right]}^{\gamma }$
Equating powers of similar terms, we have
$\alpha +\beta +\gamma =0\dots$ (i)
$-\alpha -3\beta +2\gamma =0\dots$.(ii)
$-2\alpha -2\gamma =1\dots$ (ii)
Solving, we get $\alpha =\frac{-5}{6},\beta =\frac{1}{2},\gamma =\frac{1}{3}$