The velocity of a body which falls under gravity varies as $g^{a} h^{b}$ , where $g$ is acceleration due to gravity and $h$ is the height. The values of $a$ and $b$ are

a = 1, b = 1 / 2

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a = b = 1

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a = 1 / 2, b = 1

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a = 1 / 2, b = 1 / 2

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Solution

The correct option is D $a = 1 / 2, b = 1 / 2$
Let velocity $v = g^{a} h^{b}$
so, $[v] = [M^{0} L T^{- 1}] = (L T^{- 2})^{a} L^{b} = [L^{a + b} T^{- 2 a}]$
Applying principle of homogeneity we get, $a + b = 1, - 2 a = - 1$
On solving we get $a = 1 / 2, b = 1 / 2$

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The velocity of a body which falls under gravity varies as gahb, where g is acceleration due to gravity and h is the height. The values of a and b are

The correct option is D a=1/2,b=1/2Let velocity v=gahbso, [v]=[M0LT−1]=(LT−2)aLb=[La+bT−2a]Applying principle of homogeneity we get, a+b=1,−2a=−1On solving we get a=1/2,b=1/2

The velocity of a body which falls under gravity varies as $g^{a} h^{b}$ , where $g$ is acceleration due to gravity and $h$ is the height. The values of $a$ and $b$ are

The correct option is D $a = 1 / 2, b = 1 / 2$
Let velocity $v = g^{a} h^{b}$
so, $[v] = [M^{0} L T^{- 1}] = (L T^{- 2})^{a} L^{b} = [L^{a + b} T^{- 2 a}]$
Applying principle of homogeneity we get, $a + b = 1, - 2 a = - 1$
On solving we get $a = 1 / 2, b = 1 / 2$