For the equation $F \propto A^{a}v^{b}d^{c}$, where $F$ is the force, $A$ is the area $v$ is the velocity and $d$ is the density, the value of $a,b$ and $c$ are respectively

Question

Dr. Abubacker · Accepted Answer

$\left\lbrack MLT^{- 2} \right\rbrack = \left\lbrack L^{2a} \right\rbrack\left\lbrack L^{b}T^{b} \right\rbrack\left\lbrack M^{c}L^{- 3c} \right\rbrack = \lbrack M^{c}L^{2a + c - 3c}T^{- b}\rbrack$ Comparing powers of M, L and T, we get $c = 1,2a + b - 3c = 1, - b = - 2$ or $b = 2$ $2a + 2 - 3(1) = 1\ \ \ \ \Rightarrow \ \ \ 2a = 2$ or $a = 1$.

Q. For the equation $F \propto A^{a}v^{b}d^{c}$, where $F$ is the force, $A$ is the area $v$ is the velocity and $d$ is the density, the value of $a,b$ and $c$ are respectively

Explanation

Reviewed by Dr. Abubacker

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