A) \[1,\frac{1}{2}\]
B) \[{{M}^{0}}{{L}^{2}}{{T}^{-2}}\]
C) \[\frac{1}{2},\,1\]
D) \[1,\,1\]
Correct Answer: B
Solution :
\[v\propto {{g}^{p}}{{h}^{q}}\] (given) By substituting the dimension of each quantity and comparing the powers in both sides we get \[[L{{T}^{-1}}]={{[L{{T}^{-2}}]}^{p}}{{[L]}^{q}}\] \[\Rightarrow \] \[p+q=1,\,\,-2p=-1,\,\therefore \,\,p=\frac{1}{2},\,q=\frac{1}{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
