A) \[{{T}^{2}}=\frac{4{{\pi }^{2}}{{r}^{3}}}{GM}\]
B) \[{{T}^{2}}=4{{\pi }^{2}}{{r}^{3}}\]
C) \[{{T}^{2}}=\frac{4{{\pi }^{2}}{{r}^{3}}}{G}\]
D) \[T=\frac{2{{\pi }^{2}}{{r}^{3}}}{G}\]
Correct Answer: A
Solution :
\[{{T}^{2}}=\frac{4{{\pi }^{2}}{{r}^{3}}}{GM}\] Taking dimensions on both sides, we get \[{{[T]}^{2}}=\,\,\frac{{{[L]}^{3}}}{[{{M}^{-1}}{{L}^{3}}{{T}^{-2}}\,M]}\,\,=\,\,\left[ {{M}^{0}}{{L}^{0}}{{T}^{2}} \right]\] \[\therefore \,\,\,\,LHS=RHS\] Now, \[{{T}^{2}}=4{{\pi }^{2}}{{r}^{2}}\] Taking dimensions on both sides \[{{\left[ T \right]}^{2}}=\left[ {{L}^{2}} \right]\,\,\,\,\,\,\therefore \,\,\,LHS\ne RHS\] Now, \[{{T}^{2}}=\frac{4{{\pi }^{2}}{{r}^{3}}}{G}\] Taking dimension on both sides \[{{\left[ T \right]}^{2}}=\frac{{{\left[ L \right]}^{3}}}{\left[ {{M}^{-1}}\,{{L}^{3}}\,{{T}^{2}} \right]}=\left[ {{M}^{1}}\,{{L}^{0}}\,{{T}^{-2}} \right]\,\,\,\,\,\,\therefore \,\,\,LHS\ne RHS\]Now, \[T=\frac{4{{\pi }^{2}}{{r}^{3}}}{G}\] \[\left[ T \right]=\frac{\left[ {{L}^{3}} \right]}{\left[ {{M}^{-1}}{{L}^{3}}{{T}^{2}} \right]}=\left[ {{M}^{1}}{{L}^{0}}{{T}^{2}} \right]\,\,\,\,\,\therefore \,LHS\ne RHS\]Taking dimensions on both sides
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