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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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