question_answer

A large fluid star oscillates in shape under the influence of its own gravitational field. Using dimensional analysis, find the expression for period of oscillation (T) in terms of radius of star (R), mean density of fluid \[\left( \rho  \right)\] and universal gravitational constant (G).

Answer:

                Let                                \[\text{T}=\text{K}{{\text{R}}^{\text{a}}}{{\rho }^{\text{b}}}{{\text{G}}^{\text{c}}}\]                                                             ..(i) \[\left[ {{\text{M}}^{0}}{{\text{L}}^{0}}{{\text{T}}^{\text{1}}} \right]={{\left[ \text{L} \right]}^{\text{a}}}{{\left[ \text{M}{{\text{L}}^{-\text{3}}} \right]}^{\text{b}}}{{\left[ {{\text{M}}^{-\text{1}}}{{\text{L}}^{\text{3}}}{{\text{T}}^{-\text{2}}} \right]}^{\text{c}}}={{\text{M}}^{\text{b}-\text{c}}}{{\text{L}}^{\text{a}-\text{3b}+\text{3c}}}{{\text{T}}^{-\text{2c}}}\] Applying principle of homogeneity ef dimensions, we get                         \[\text{b}\text{c}=0,\text{ a}\text{3b}+\text{3c}=0,\text{2c}=\text{l},\text{ c}=\frac{1}{2}\]                                                                                    \[\text{b}=\text{c }=\frac{1}{2}\] and                          \[\text{a}\text{3}\left( -\frac{1}{2} \right)+\text{3}\left( -\frac{1}{2} \right)=0,\text{ a}=0\] Putting in (i), we get \[\text{T}=\text{KR}{}^\circ {{\rho }^{-\text{1}/\text{2}}}{{\text{G}}^{-\text{1}/\text{2}}}=\text{K}{{(\rho \text{G})}^{-\text{1}/\text{2}}}\]