Question

what are the dimensions of a and c in the relation x=a+c/m where x is the displacement and m is the mass of the body

Answer 1

Topic :- Units and Dimensions

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Given relation :

$\longrightarrow\:\:\sf x = a + \dfrac{c}{m}$

Where, x is the displacement. Therefore the dimensions of 'x' is :

$\longrightarrow\:\:\sf [x] = [M^0L^1T^0]$

We're also given that, 'm' is the mass of the body. Therefore dimensions of 'm' is :

$\longrightarrow\:\:\sf [m] = [M^1L^0T^0]$

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Now, according to the principle of homogeneity we have :

$\dashrightarrow\:\:\sf [x] = [a]$

By rearranging the above equation we have :

$\dashrightarrow\:\:\sf [a] = [x]$

$\dashrightarrow\:\: \underline{ \boxed{\sf [a] = [M^0L^1T^0]}}$

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Now, let's find the Dimensions of 'c'.

Again we have to apply the principle of homogeneity :

$: \implies\:\:\sf [x] = \dfrac{[c]}{[m]}$

$: \implies\:\:\sf [M^0L^1T^0] = \dfrac{[c]}{[M^1L^0T^0]}$

$: \implies\:\:\sf [M^0L^1T^0].[M^1L^0T^0] = [c]$

By rearranging the above equation we have :

$: \implies\:\:\sf [c] = [M^0L^1T^0].[M^1L^0T^0]$

$: \implies\: \: \underline{ \boxed{\sf [c] =[M^1L^1T^0]}}$

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