A) force and torque
B) Work and energy
C) force and impulse
D) linear momentum and angular momentum
Correct Answer: B
Solution :
Force = Mass \[\times \] acceleration or \[F=ma\] \[\therefore [F]=\,[m]\,[a]\] \[=[M]\,[L{{T}^{-2}}]\,\,=\,[ML{{T}^{-2}}]\] Torque = Moment of inertia \[\times \] angular acceleration or \[\tau =I\times \alpha \] \[\therefore [\tau ]=\,[I]\,[\alpha ]\] \[=[M{{L}^{2}}]\,[{{T}^{-2}}]\] \[=[M{{L}^{2}}{{T}^{-2}}]\] Work = Force \[\times \] 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 \[\times \] time-interval \[[I]=[F]\times [\Delta t]\] \[\therefore [I]\,=[ML{{T}^{-2}}]\,[T]\] \[=[ML{{T}^{-1}}]\] Linear momentum = Mass \[\times \] Velocity or [p] = [m] [v] \[\therefore [p]\,=[M]\,\,[L{{T}^{-1}}]\] \[=[ML{{T}^{-1}}]\] Angular momentum = Moment of inertia \[\times \,\,angular\,velocity\] 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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