A) \[[{{L}^{2}}{{T}^{2}}]\]
B) \[[{{L}^{1/2}}\,{{T}^{2}}]\]
C) \[[{{L}^{-1/2}}\,{{T}^{2}}]\]
D) \[[ML{{T}^{-\,2}}]\]
Correct Answer: C
Solution :
Here, \[M\sqrt{x}\] has dimension of force F \[\therefore \,\,\,\,\,\,\,\,M=\frac{F}{\sqrt{x}}=\frac{\left[ ML{{T}^{-\,2}} \right]}{\left[ {{L}^{1/2}} \right]}=\left[ M{{L}^{1/2}}\,{{T}^{-\,2}} \right]\] Similarly, \[N=\frac{F}{{{t}^{2}}}=\frac{\left[ ML{{T}^{-\,2}} \right]}{\left[ {{T}^{2}} \right]}=\left[ ML{{T}^{-\,4}} \right]\] \[\therefore \,\,\,\,\frac{M}{N}\,=\frac{\left[ M{{L}^{1/2}}\,{{T}^{-\,2}} \right]}{\left[ ML{{T}^{-\,4}} \right]}=\left[ {{L}^{-1/2}}\,{{T}^{2}} \right]\]
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