A) torque
B) work
C) energy
D) coefficient of viscosity.
Correct Answer: D
Solution :
Torque is defined as \[\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}\] \[\therefore \] Dimensions of torque \[[\tau ]=[r]\times [F]\] \[=[L]\times [ML{{T}^{-2}}]\] \[=[M{{L}^{2}}{{T}^{-2}}]\] Similarly for work and energy the dimensions are same as that of torque i.e. \[=[M{{L}^{2}}{{T}^{-2}}]\] Now for viscosity, we know that \[F=6\pi \eta rv\] \[\Rightarrow \] \[\eta =\frac{F}{6\pi rv}\] \[\therefore \] \[[\eta ]=\frac{F}{[r][v]}=\frac{[ML{{T}^{-2}}]}{[L][L{{T}^{-1}}]}\] \[=[M{{L}^{-1}}{{T}^{-1}}]\] \[\therefore \] The given dimensions are related to the coefficient of viscosity.
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