A) \[1,\,\,-1,\,\,~-1\]
B) \[-1,\,\,-1,\,\,~1\]
C) \[-1,\,\,-1,\,\,~-1\]
D) \[1,\,\,1,\,\,1\]
Correct Answer: A
Solution :
| [a] Key Concept According to principle of homogeneity of dimension states that, a physical quantity equation will be dimensionally correct, if the dimensions of all the terms occurring on both sides of the equations are same. |
| Given critical velocity of liquid flowing through a tube are expressed as |
| \[{{v}_{c}}\propto {{\eta }^{n}}{{\rho }^{y}}{{r}^{z}}\] |
| Coefficient of viscosity of liquid, \[\eta =[M{{L}^{-1}}{{T}^{-1}}]\] |
| Density of liquid, \[\rho =[M{{L}^{-3}}]\] |
| Radius of a tube, \[r=[L]\] |
| Critical velocity of liquid \[{{v}_{c}}=[{{M}^{0}}L{{T}^{-1}}]\] |
| \[\Rightarrow \] \[[{{M}^{0}}{{L}^{-1}}{{T}^{-1}}]={{[M{{L}^{-1}}{{T}^{-1}}]}^{x}}.{{[M{{L}^{-3}}]}^{y}}.{{[L]}^{z}}\] |
| \[[{{M}^{0}}{{L}^{1}}{{T}^{-1}}]=[{{M}^{x+y}}.{{L}^{x-3y+z}}{{T}^{x}}]\] |
| Comparing exponents of M, L and L, we get |
| \[x+y=0,-x-3y+z=1,-x=-1\] |
| \[\Rightarrow \] \[z=-1,x=1,y=-1\] |
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