A) linear momentum
B) angular momentum
C) energy
D) power
Correct Answer: B
Solution :
| \[E=hv\] |
| \[\Rightarrow \] \[h=Planck's\,\text{constant}=\frac{E}{v}\] |
| \[\therefore \] \[[h]=\frac{[E]}{[v]}=\frac{[M{{L}^{2}}{{T}^{-2}}]}{[{{T}^{-1}}]}=[M{{L}^{2}}{{T}^{-1}}]\] |
| [a] Linear momentum = mass × velocity |
| or \[p=m\times v\] |
| or \[[p]=[m]\times [v]=[M]\,[L{{T}^{-1}}]\,=\,[ML{{T}^{-1}}]\] |
| [b] Angular momentum = moment of inertia × angular velocity |
| or \[L=I\times \omega =m{{r}^{2}}\omega \] \[[\because \,I=m{{r}^{2}}]\] |
| \[\therefore \] \[[L]=[M]\,[{{L}^{2}}]\,[{{T}^{-1}}]\,=[M{{L}^{2}}{{T}^{-1}}]\] |
| [c] Energy \[[E]=[M{{L}^{2}}{{T}^{-2}}]\] |
| [d] Power = force × velocity |
| or \[P=F\times v\] |
| \[\therefore \] \[[P]=[ML{{T}^{-2}}]\,[L{{T}^{-1}}]\,=[M{{L}^{2}}{{T}^{-3}}]\] |
| Hence, option [b] is correct. |
| Note: According to homogeneity of dimensions, the dimensions of all the terms in a physical expression should be same. For example, in the physical expression \[s=ut+\frac{1}{2}a{{t}^{2}}\], |
| the dimensions of \[s,\,ut\] and \[\frac{1}{2}a{{t}^{2}}\] all are same. |
You need to login to perform this action.
You will be redirected in
3 sec
