Question

A large fluid star oscillates in shape under the influence of its own gravitational field. Using the method of dimensions, obtain a relation for the period of oscillation (T) in terms of a radius (R) of the star, its density ($\rho$) and universal gravitational constant (G) is :

• A. $T=\frac{KR}{\sqrt{pG}}$
• B. $T=\frac{K}{\sqrt{pG}}$
• C. $T=\frac{K}{\sqrt{pRG}}$
• D. $T=\frac{K}{p^2{G}^2}$

Answer 1

Answer: (B)

$T=\frac{K}{\sqrt{pG}}$

Let $T=K{R}^a{\rho }^b{G}^c$, where $k$ is a constant.
Thus, equating the dimensions for both sides, using dimensions for $G$ as $L^3{M}^{-1}{T}^{-2}$.
$T=L^a{\left(M{L}^{-3}\right)}^b{\left(L^3{M}^{-1}{T}^{-2}\right)}^c$
Or, $-2c=1$, giving $c=\frac{-1}{2}$
$b-c=0$ or $b=c=\frac{-1}{2}$
$a-3b+3c=0$ or $a=0$
Thus we get, $T=K{(\rho .G)}^{-1/2}=\frac{K}{\sqrt{\rho G}}$