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 \[\times \] velocity or \[p=m\times v\] or \[[p]=[m]\times [v]=[M]\,[L{{T}^{-1}}]\,=\,[ML{{T}^{-1}}]\] (b) Angular momentum = moment of inertia \[\times \] 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 \[\times \] 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.
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