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Q.

Suppose, the torque acting on a body is given by $τ = KL + XI / (ω)$ Where L = angular momentum, I= moment of inertia, $ω$ =angular speed. The dimensional formula for KX is same as that for

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a

time²

b

time⁴

c

time^-2

d

time^-4

answer is D.

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Detailed Solution

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$\begin{array}{l} τ^{1} = KL + \frac{XI}{W} \Rightarrow τ^{1} = [{ML}^{2} T^{- 2}] L = [{ML}^{2} T^{- 1}] \\ I = [{ML}^{2}]; W = [T^{- 1}] \\ KL = [{ML}^{2} T^{- 2}] \\ \Rightarrow K [{ML}^{2} T^{- 1}] = [{ML}^{2} T^{- 2}] \\ K = \frac{T^{- 2}}{T^{- 1}} = T^{- 1} \Rightarrow \frac{XI}{W} = [{ML}^{2} T^{- 2}] \\ \frac{X [{ML}^{2}]}{[T^{- 1}]} = [{ML}^{2} T^{- 2}] \Rightarrow X = T^{- 1} T^{- 2} = [T^{- 3}] \\ KX = T^{- 1} T^{- 3} = T^{- 4} \end{array}$

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