First slide

Dimensions of physical quantities

Question

$\frac{α}{t^{2}} = F v + \frac{β}{x^{2}}$ . Find the dimension formula for $[α] and [β] (here t = time, F = force, v = velocity, x = distance)$

Moderate

A

${α} = [{ML}^{2} T^{- 1}], [β] = [{ML}^{4} T^{- 3}]$
B

${α} = [{ML}^{- 2} T^{- 1}], [β] = [{ML}^{4} T^{- 3}]$
C

${α} = [{ML}^{2} T^{- 1}], [β] = [{ML}^{4} T^{3}]$
D

${α} = [{ML}^{2} T^{2}], [β] = [{ML}^{4} T^{- 3}]$

Solution

Since, dimension of F v
$= [Fv] = [{MLT}^{- 2}] [{LT}^{- 1}] = [{ML}^{2} T^{- 3}]$
So, $[\frac{β}{x^{2}}]$ should also be $[{ML}^{2} T^{- 3}]$
$\frac{[β]}{[x^{2}]} = [{ML}^{2} T^{- 3}] [β] = [{ML}^{4} T^{- 3}]$
And $[Fv + \frac{β}{x^{2}}]$ will also have dimension $[{ML}^{2} T^{- 3}]$ so LHS should also have the same dimension $[{ML}^{2} T^{- 3}]$
So, $\frac{[α]}{[t^{2}]} = [{ML}^{2} T^{- 3}] {α} = [{ML}^{2} T^{- 1}]$

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