Question

A gas bubble forms an explosion under water which oscillate with a period $ T\propto {{p}^{a}}{{\rho }^{b}}{{E}^{c}}, $ where p is the static pressure, p is the density of water and E is the total energy of explosion. Then, the values of q, b and c will be

• A. $ -\frac{5}{6},\frac{1}{2},\frac{1}{3} $
• B. $ \frac{2}{5},\frac{1}{3},\frac{1}{4} $
• C. $ \frac{4}{5},\frac{9}{8},\frac{1}{4} $
• D. $ \frac{3}{2},\frac{1}{4},\frac{5}{6} $

Answer 1

Answer: (A)

As $ T\propto {{P}^{a}}{{\rho }^{b}}{{E}^{c}} $ From principle of homogeneity $ [T]={{[M{{L}^{-1}}{{T}^{-2}}]}^{a}}{{[M{{L}^{-3}}]}^{b}}{{[M{{L}^{2}}{{T}^{-2}}]}^{c}} $ $ [{{M}^{0}}{{L}^{0}}{{T}^{1}}]=[{{M}^{a+b+c}}{{L}^{-a-3b-2c}}{{T}^{-2a-3b-2c}}] $ Equating the exponents $ a+b+c=0, $ $ -a-3b-2c=0 $ and $ -2a-3b-2c=1 $ Solving these equations, we get $ a=-\frac{5}{6},b=\frac{1}{2} $ and $ c=\frac{1}{3} $