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The dimension of kinematic viscosity is

1. L / (M T)
2. L / T^2
3. L^2 / T
4. M L / T

Correct answer: L^2 / T

Solution

Kinematic viscosity \(\nu\) is defined as \(\nu=\mu/\rho\). Since \([\mu]=M L^{-1} T^{-1}\) and \([\rho]=M L^{-3}\), we get \([\nu]=L^2 T^{-1}\). Therefore, the correct dimension is \(L^2/T\).

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