A) \[{{c}^{2}}{{g}^{0}}{{p}^{-2}}\]
B) \[{{c}^{0}}{{g}^{2}}{{p}^{-1}}\]
C) \[c{{g}^{3}}{{p}^{-2}}\]
D) \[c{{g}^{0}}{{p}^{-1}}\]
Correct Answer: B
Solution :
: \[[c]=[L{{T}^{-1}}]\] \[[g]=[L{{T}^{-2}}]\] \[[P]=[M{{L}^{-1}}{{T}^{-2}}]\] \[\frac{c}{g}=\frac{L{{T}^{-1}}}{L{{T}^{-1}}}=T\] \[\therefore \] \[T=\frac{c}{g}\] ?..(i) Again \[\frac{{{c}^{2}}}{g}=\frac{{{L}^{2}}{{T}^{-2}}}{L{{T}^{-2}}}=L\] \[\therefore \] \[L=\frac{{{c}^{2}}}{g}\] ?.(ii) \[P=M{{L}^{-1}}{{T}^{-2}}\to P=M{{\left[ \frac{{{c}^{2}}}{g} \right]}^{-1}}{{\left[ \frac{c}{g} \right]}^{-2}}\] \[P\left[ \frac{{{c}^{2}}}{g} \right]{{\left[ \frac{c}{g} \right]}^{-2}}=M\] Or \[M=\frac{P{{c}^{4}}}{{{g}^{3}}}\] ?..(iii) Now \[G={{M}^{-1}}{{L}^{3}}{{T}^{-2}}\] \[\therefore \] \[[G]={{\left[ \frac{P{{c}^{4}}}{{{g}^{3}}} \right]}^{-1}}{{\left[ \frac{{{c}^{2}}}{g} \right]}^{3}}{{\left[ \frac{c}{g} \right]}^{-2}}\] \[[G]={{P}^{-1}}{{c}^{0}}{{g}^{2}}\] \[[G]={{c}^{0}}{{g}^{2}}{{P}^{-1}}\]
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