A) \[\left[ {{M}^{0}}L{{T}^{-1}} \right]\]
B) \[\left[ M{{L}^{0}}{{T}^{-1}} \right]\]
C) \[\left[ M{{L}^{-1}}{{T}^{0}} \right]\]
D) \[\left[ {{M}^{0}}{{L}^{0}}{{T}^{0}} \right]\]
Correct Answer: C
Solution :
Key Idea: Put the dimensions for each physical quantity in the given relation. Given, \[v=\frac{p}{2l}{{\left[ \frac{F}{m} \right]}^{1/2}}\] Squaring the equation on either side, we have \[{{v}^{2}}=\frac{{{p}^{2}}}{4{{l}^{2}}}\left[ \frac{F}{m} \right]\] \[\Rightarrow \] \[m=\frac{{{p}^{2}}\,F}{4{{l}^{2}}\,v}\] Putting the dimensions of equations on RHS, we get \[F=\left[ ML{{T}^{-2}} \right]\], \[l=\left[ L \right]\], \[v=\left[ {{T}^{-1}} \right]\], \[p\] being a number is dimensionless, we have \[\left[ m \right]=\frac{\left[ ML{{T}^{-2}} \right]}{\left[ {{L}^{2}} \right]{{\left[ {{T}^{-1}} \right]}^{2}}}\] \[=\left[ M{{L}^{-1}}{{T}^{0}} \right]\]
You need to login to perform this action.
You will be redirected in
3 sec
