A) \[u=\frac{hc}{{{e}^{2}}{{a}_{0}}}\]
B) \[u=\frac{{{e}^{2}}h}{c{{a}_{0}}}\]
C) \[u=\frac{{{e}^{2}}{{a}_{0}}}{hc}\]
D) \[u=\frac{{{e}^{2}}c}{h{{a}_{0}}}\]
Correct Answer: C
Solution :
| [c] \[C={{e}^{x}}a_{0}^{y}{{h}^{z}}{{c}^{a}}\] |
| Capacitance \[C=\left[ {{M}^{-1}}{{L}^{-2}}{{A}^{2}}{{T}^{4}} \right]\] |
| \[e=\left[ AT \right]\] \[{{a}_{0}}=\left[ L \right]\] |
| \[c=\left[ {{L}^{1}}{{T}^{-1}} \right]\] \[h=\left[ {{M}^{1}}{{L}^{2}}{{T}^{-1}} \right]\] |
| \[\left[ {{M}^{-1}}{{L}^{-2}}{{A}^{2}}{{T}^{4}} \right]={{\left[ AT \right]}^{x}}{{\left[ L \right]}^{y}}{{\left[ {{M}^{1}}{{L}^{2}}{{T}^{-1}} \right]}^{z}}{{\left[ {{L}^{1}}{{T}^{-1}} \right]}^{a}}\]Compare x = 2 |
| \[z=-1\] |
| \[\frac{{{M}^{-1}}{{L}^{-2}}{{A}^{2}}{{T}^{4}}}{{{A}^{2}}{{T}^{2}}{{M}^{-1}}{{L}^{-2}}{{T}^{1}}}={{\left[ L \right]}^{y}}{{\left[ {{L}^{1}}{{T}^{-1}} \right]}^{a}}\] |
| \[T=\left[ {{L}^{y}} \right]{{\left[ {{L}^{1}}{{T}^{-1}} \right]}^{a}}\]So \[u=\frac{{{e}^{2}}{{a}_{0}}}{hc}\] |
| \[\Rightarrow \] \[a=-1\]\[\Rightarrow \] \[y=1\] |
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