Question

A gas bubble formed from an explosion under water oscillates with a period $T$ proportional to $P^{a} d^{b} E^{c}$, where $P$ is pressure, $d$ is the density of water and $E$ is the total energy of explosion. The value of $a, b, c$ are

• A. $ a=1\,b=1,\,c=2 $
• B. $ a=1,\,b=2,\,c=1 $
• C. $ a=\frac{5}{6}, b=\frac{1}{2},c=\frac{1}{3} $
• D. $ a=-\frac{5}{6},b=\frac{1}{2},c=\frac{1}{3} $

Answer 1

Answer: (D)

Every equation relating Physical quantities should be in dimensional balance.
Given $T \propto p^{a} d^{b} E^{c}$
Since, only similar quantities can be equated, therefore dimensions of the terms on both sides of the equation must be same.
Hence, we have Dimensions of $=\left[M^{o} L^{o} T^{1}\right]$
Dimensions of pressure $P=\left[M L^{-1} T^{-2}\right]$
Dimensions of density $d=\left[M L^{-3}\right]$
Dimensions of energy $E=\left[M L^{2} T^{-2}\right]$
$ \therefore $ We have
$\left[M^{o} L^{o} T^{1}\right]=k\left[M L^{-1} T^{-2}\right]^{a}\left[M L^{-3}\right]^{b}\left[M L^{2} T^{-2}\right]^{c}$
where, $k$ is a constant.
Comparing dimensions of similar terms, we have
$\left[M^{o} L^{o} T^{1}\right]=k\left[M^{a+b+c} L^{-a-3 b+2 c} T^{-2 a-2 c}\right]$
Comparing powers of $M$, we have $0=a+b+c \,\,\,\ldots$(i)
Comparing powers of $L$, we have $0=-a-3 b+2 c \,\,\,\ldots$(ii)
Comparing powers of $T$, we have $1=-2 a-2 c \,\,\,\ldots$(iii)
Solving Eqs. (i), (ii) and (iii), we have
$a=-\frac{5}{6}, b=\frac{1}{2}, c=\frac{1}{3}$