A) \[\eta \frac{S\Delta \theta }{h}\]
B) \[\eta \left( \frac{S\Delta \theta }{h} \right)\left( \frac{1}{\rho g} \right)\]
C) \[\frac{S\Delta \theta }{\eta h}\]
D) \[\left( \frac{S\Delta \theta }{\eta h} \right)\left( \frac{1}{\rho g} \right)\]
Correct Answer: A
Solution :
Let \[\left( \frac{\theta }{A} \right)\]is derived quantity which is derived by three fundamental quantities \[\eta ,\left( \frac{S\Delta \theta }{h} \right)\]and\[\left( \frac{1}{eg} \right)\]By using property of homogeneity. \[\left[ \frac{\theta }{A} \right]={{\left[ \eta \right]}^{x}}{{\left[ \frac{S\Delta \theta }{h} \right]}^{y}}{{\left[ \frac{1}{eg} \right]}^{z}}\] \[\left[ \frac{\theta }{A} \right]=\left[ {{m}^{1}}{{T}^{-3}} \right]\] \[\left[ \eta \right]=\left[ {{m}^{1}}{{L}^{-1}}{{T}^{-1}} \right]\] \[\left[ \frac{S\Delta \theta }{h} \right]=\left[ {{L}^{-1}}{{T}^{-2}} \right]\] \[\left[ \frac{1}{eg} \right]=\left[ {{m}^{-1}}{{L}^{2}}{{T}^{+2}} \right]\] \[\left[ {{m}^{1}}{{L}^{0}}{{T}^{-3}} \right]={{\left[ {{m}^{1}}{{L}^{-1}}{{T}^{-1}} \right]}^{x}}{{\left[ {{m}^{0}}{{L}^{1}}{{T}^{-2}} \right]}^{y}}{{\left[ {{m}^{1}}{{L}^{2}}{{T}^{+2}} \right]}^{z}}\]\[x+0-z=1,-x+y+2z=0\And -x-2y+2z=-3\]\[-x+y+2z=0\] \[-x-2y+2z=-3\] \[\frac{+\,\,+-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,}{3y=3\Rightarrow y=1,}x=1,z=0\] So, \[\frac{\theta }{A}=\eta .\frac{S\Delta \theta }{h}\]
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