A) force and torque
B) work and energy
C) force and impulse
D) linear momentum and angular momentum
Correct Answer: B
Solution :
| Force = Mass × acceleration |
| or \[F=ma\] |
| \[\therefore \] \[[F]=\,[m]\,[a]\] |
| \[=[M]\,[L{{T}^{-2}}]\,\,=\,[ML{{T}^{-2}}]\] |
| Torque = Moment of inertia × angular acceleration |
| or \[\tau =I\times \alpha \] |
| \[\therefore \] \[[\tau ]=\,[I]\,[\alpha ]\] |
| \[=[M{{L}^{2}}]\,[{{T}^{-2}}]\] |
| \[=[M{{L}^{2}}{{T}^{-2}}]\] |
| [b] Work = Force × displacement |
| or \[W=F\times d\] |
| \[\therefore \] \[[W]\,=[F]\,[d]\] |
| \[=\,[ML{{T}^{-2}}]\,[L]\] |
| \[[M{{L}^{2}}{{T}^{-2}}]\] |
| Energy \[=\frac{1}{2}\times Mass\times {{(Velocity)}^{2}}\] |
| or \[K=\frac{1}{2}m{{v}^{2}}\] |
| \[\therefore \] \[[K]=[m]\,[{{v}^{2}}]\] |
| \[=[M]\,[L{{T}^{-1}}]\,\,=[M{{L}^{2}}{{T}^{-2}}]\] |
| Force as discussed above |
| \[[F]\,=[ML{{T}^{-2}}]\] |
| Impulse = Force × time-interval |
| \[[I]=[F]\times [\Delta t]\] |
| \[\therefore \] \[[I]\,=[ML{{T}^{-2}}]\,[T]\] |
| \[=[ML{{T}^{-1}}]\] |
| [d] Linear momentum = Mass × Velocity |
| or \[\left[ p \right]=\left[ m \right]\text{ }\left[ v \right]\] |
| \[\therefore \] \[[p]\,=[M]\,\,[L{{T}^{-1}}]\] |
| \[=[ML{{T}^{-1}}]\] |
| Angular momentum = Moment of inertia \[\text{ }\!\!\times\!\!\text{ }\,\,\text{angular}\,\text{velocity}\] |
| or \[[L]=[I]\times [\omega ]\] |
| \[\therefore \] \[[L]=[M{{L}^{2}}]\,[{{T}^{-1}}]\] |
| \[=[M{{L}^{2}}{{T}^{-1}}]\] |
| Hence, we observe that choice [b] is correct. |
| Note: In this problem, the momentum of inertia and impulse are given same symbol \[l\]. |
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