A) \[{{M}^{0}}L{{T}^{2}}\]
B) \[{{M}^{0}}{{L}^{2}}{{T}^{-4}}\]
C) \[{{M}^{0}}L{{T}^{-3}}\]
D) \[{{M}^{0}}{{L}^{2}}{{T}^{-1}}\]
Correct Answer: D
Solution :
[n] = Number of particles crossing a unit area in unit time = \[[{{L}^{-2}}{{T}^{-1}}]\] \[\left[ {{n}_{2}} \right]=\left[ {{n}_{1}} \right]=\]number of particles per unit volume = [L?3] \[[{{x}_{2}}]=[{{x}_{1}}]\]= positions \ \[D=\frac{[n]\ \left[ {{x}_{2}}-{{x}_{1}} \right]}{\left[ {{n}_{2}}-{{n}_{1}} \right]}=\frac{\left[ {{L}^{-2}}{{T}^{-1}} \right]\times [L]}{[{{L}^{-3}}]}\]=\[\left[ {{L}^{2}}{{T}^{-1}} \right]\]
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