Question

The time period of a body in $SHM$ is represented by $T_o=P^a{\rho }^b{S}^c$, where $P$ is pressure, $\rho$ is density and $S$ is surface tension. The value of $a,b$ and $c$ respectively for the relation to be dimensionally correct is

• A. $-\frac{3}{2},\frac{1}{2},1$
• B. $-1,-2,3$
• C. $\frac{1}{2},-\frac{3}{2},-\frac{1}{2}$
• D. $1,2,\frac{1}{3}$

Answer 1

The correct option is A $-\frac{3}{2},\frac{1}{2},1$

Dimensional formula of,
Time period, $T_o$
$\left[T_o\right]=\left[M^0{L}^0{T}^1\right]$
Pressure, $P$
$\left[P\right]=\left[M^1{L}^{-1}{T}^{-2}\right]$
Density, $\rho$
$\left[\rho \right]=\left[M^1{L}^{-3}{T}^0\right]$
Surface Tension, $S$
$\left[S\right]=\left[M^1{L}^0{T}^{-2}\right]$
Now,
$\left[T_o\right]=\left[P^a{\rho }^b{S}^c\right]$
$\Rightarrow \left[M^0{L}^0{T}^1\right]=[M^1{L}^{-1}{T}^{-2}{]}^a\times [M^1{L}^{-3}{T}^0{]}^b\times [M^1{L}^0{T}^{-2}{]}^c$
$\Rightarrow \left[M^0{L}^0{T}^1\right]=\left[M^{(a+b+c)}{L}^{(-a-3b+0)}{T}^{(-2a+0-2c)}\right]$
On comparing we get,
$a+b+c=0$, .............(1)
$-a-3b=0$, .............(2)
$-2a-2c=1$ .............(3)
On putting the value of $c$ from (1) in (3), we get, $b=\frac{1}{2}$
On putting the value of $b$ in (2), we get,
$a=\frac{-3}{2}$
On putting the value of $a$ in (3), we get,
$c=1$
Hence, the value of $a,b$ and $c$ is $\frac{-3}{2},\frac{1}{2}$ and $1$