The velocity of a freely falling body changes | vedclass

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(B) Given the relation: $v \propto g^p h^q$.The dimensions of the quantities are:Velocity $(v)$: $[L T^{-1}]$Acceleration due to gravity $(g)$: $[L T^

Vedclass · Accepted Answer

$\frac{1}{2}, \frac{1}{2}$

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(B) Given the relation: $v \propto g^p h^q$.The dimensions of the quantities are:Velocity $(v)$: $[L T^{-1}]$Acceleration due to gravity $(g)$: $[L T^{-2}]$Height $(h)$: $[L]$Substituting these dimensions into the given relation:$[L T^{-1}] = [L T^{-2}]^p [L]^q$$[L T^{-1}] = L^p T^{-2p} L^q$$[L T^{-1}] = L^{p+q} T^{-2p}$Comparing the powers of $L$ and $T$ on both sides:For $T$: $-2p = -1 \Rightarrow p = \frac{1}{2}$For $L$: $p + q = 1 \Rightarrow \frac{1}{2} + q = 1 \Rightarrow q = \frac{1}{2}$Thus,the values are $p = \frac{1}{2}$ and $q = \frac{1}{2}$.

The velocity of a freely falling body changes as ${g^p}{h^q}$ where $g$ is the acceleration due to gravity and $h$ is the height. The values of $p$ and $q$ are:

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