Question

The speed of a freely falling body from rest is proportional to $g^p{h}^q$, where $g$ is acceleration due to gravity and $h$ is the distance travelled. The values of $p$ and $q$ are

• A. $\frac{1}{3},\frac{1}{3}$
• B. $\frac{1}{2},\frac{1}{2}$
• C. $\frac{1}{4},\frac{1}{4}$
• D. $\frac{1}{5},\frac{1}{5}$

Answer 1

The correct option is B $\frac{1}{2},\frac{1}{2}$

Dimensional formula of speed $v$,
$\left[v\right]=\left[M^{0}L{T}^{-1}\right]$
Dimensional formula of acceleration due to gravity $g$,
$\left[g\right]=\left[M^{0}L{T}^{-2}\right]$
Dimensional formula of distance travelled $h$,
$\left[h\right]=\left[M^{0}L{T}^0\right]$
As, $v\propto g^p{h}^q$
$\Rightarrow v=k{g}^p{h}^q$
where $k$ is proportionality constant
$\Rightarrow \left[M^{0}L{T}^{-1}\right]=\left[\right(M^{0}L{T}^{-2}{]}^p\times [M^{0}L{T}^0{]}^q$
$\Rightarrow \left[M^{0}L{T}^{-1}\right]=\left[M^{(0+0)}{L}^{(p+q)}{T}^{(-2p+0)}\right]$
On comparing, we get,
$p+q=1$ and $-2p=-1$
So, $p=\frac{1}{2}$, putting this in $p+q=1$, we get $q=\frac{1}{2}$.