A) \[\text{ }\!\![\!\!\text{ M}{{\text{L}}^{\text{2}}}{{\text{T}}^{\text{-2}}}\text{ }\!\!]\!\!\text{ }\]
B) \[\text{ }\!\![\!\!\text{ M}{{\text{L}}^{\text{-1}}}{{\text{T}}^{\text{-1}}}\text{ }\!\!]\!\!\text{ }\]
C) \[\text{ }\!\![\!\!\text{ M}{{\text{L}}^{-2}}{{\text{T}}^{-2}}\text{ }\!\!]\!\!\text{ }\]
D) \[\text{ }\!\![\!\!\text{ }{{\text{M}}^{0}}{{\text{L}}^{0}}{{\text{T}}^{0}}\text{ }\!\!]\!\!\text{ }\]
Correct Answer: B
Solution :
Coefficient of viscosity \[\eta =\frac{\text{force}\,\text{ }\!\!\times\!\!\text{ }\,\text{length}}{\text{area}\,\text{ }\!\!\times\!\!\text{ velocity}}\] Dimensions of \[\eta \] \[=\frac{\text{Dimension of}\,\text{force}\,\,\text{ }\!\!\times\!\!\text{ }\,\,\text{dimension of length}}{\text{Dimension of area}\,\,\text{ }\!\!\times\!\!\text{ }\,\,\text{dimension of velocity}}\] \[=\frac{ML{{T}^{-2}}\times L}{{{L}^{2}}\times L{{T}^{-1}}}=[M{{L}^{-1}}{{T}^{-1}}]\]You need to login to perform this action.
You will be redirected in
3 sec