Question

If vibration frequency $f$ of the star is given by the following equation: $f=K{R}^a{P}^b{G}^c$ where $R$ is the radius of star, $\rho$ is the density of star, $G$ is the universal gravitational constant and $k$is a constant, then $(b+c)$is equal to

Answer 1

Given:$f=K{R}^a{P}^b{G}^c\cdot \cdot \cdot \cdot \left(i\right)$

Where,

$\left[T^{-1}\right]=f=$ vibration frequency
$\left[L\right]=R$ is the radius of star,
$\left[M{L}^{-3}\right]=\rho$ is the density of star,

$\left[M^{-1}{L}^3{T}^{-2}\right]=\frac{F{r}^2}{m_1{m}_2}=G$
is the universal gravitational constant,
$k$ is a constant.
by putting all the dimensions in equation $\left(i\right)$ we get

$\left[M^0{L}^0{T}^{-1}\right]={\left[L\right]}^a{\left[M{L}^{-3}\right]}^b{\left[M^{-1}{L}^3{T}^{-2}\right]}^c$

By comparing both sides equation we get,

$b-c=0$
$a-3b+3c=0$
$-2c=-1$
$c=\frac{1}{2}$
$b=\frac{1}{2}$
$a=0$
Then $b+c=\frac{1}{2}+\frac{1}{2}=1$
Final Answer: $1$