Question

Frequency is the function of density $\left(\rho \right)$, length $\left(a\right)$ and surface tension $\left(T\right)$. Then its value is ($k$ is a constant)

• A. $k{\rho }^{-1/2}{a}^{-3/2}{T}^{1/2}$
• B. $k{\rho }^{1/2}{a}^{3/2}{T}^{-1/2}$
• C. $k{\rho }^{1/2}{a}^{-3/2}{T}^{1/2}$
• D. $k{\rho }^{-1/2}{a}^{3/2}{T}^{-1/2}$

Answer 1

The correct option is A $k{\rho }^{-1/2}{a}^{-3/2}{T}^{1/2}$

Frequency $\left(n\right)=k{\rho }^a{a}^b{T}^c,k=$ constant
Where density $\left(\rho \right)=\frac{\text{Mass}}{\text{volume}}$
$=\frac{\left[M\right]}{\left[L^3\right]}=\left[M^1{L}^{-3}\right]$
$a=$ length $=\left[L^1\right]$
Surface Tension $\left(T\right)=\frac{\text{Force}}{\text{length}}$
$=\frac{M^1{L}^1{T}^{-2}}{L^1}=\left[M^1{T}^{-2}\right]$
$\because$ Frequency $=\frac{1}{\text{time}}=\left[T^{-1}\right]$
Now from equation (1)
$\left[T^{-1}\right]={\left[M^1{L}^{-3}\right]}^a[L^1{]}^b{\left[M^1{T}^{-2}\right]}^c$
$\left[M^0{L}^0{T}^{-1}\right]=\left[M^{a+c}{L}^{-3a+b}{T}^{-2c}\right]$
Comparing the powers of LHS and RHS
$0=a+c\Rightarrow c=-a$ ... (1)
$-3a+b=0$ ... (2)
$-2c=-1\Rightarrow c=\frac{1}{2}$ ... (3)
$\therefore$ From equation (1) and (3)
$a=-\frac{1}{2}$
Now from equation (2)
$-3(-\frac{1}{2})+b=0$
$b=-\frac{3}{2}$
$b=-\frac{3}{2}$
Now substituting the value of $a,b$ and $c$ in equation (1)
$n=k{\rho }^{-1/2}{a}^{-3/2}{T}^{1/2}$