Velocity of a freely falling body changes as {g^p}{h^q} where g is acceleration due to gravity and h

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1.Units, Dimensions and Measurement

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

A

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

B

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

C

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

D

$1,\,1$

Solution

(b) $v \propto {g^p}{h^q}$ (given)

By substituting the dimension of each quantity and comparing the powers in both sides we get $[L{T^{ – 1}}] = {[L{T^{ – 2}}]^p}{[L]^q}$

$ \Rightarrow $ $p + q = 1,\,\, – 2p = – 1,\,$

$\therefore \,\,p = \frac{1}{2},\,q = \frac{1}{2}$

Standard 11

Physics

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