A) \[\overset{\to }{\mathop{T}}\,=\,{{\omega }^{\to }}\times {{L}^{\to }}\]
B) \[{{L}^{\to }}=L\frac{\,\,\,{{\omega }^{\to }}}{dt}\]
C) \[{{T}^{\to }}={{\omega }^{\to }}\frac{\,\,\,d{{L}^{\to }}}{dt}\]
D) \[{{T}^{\to }}=\frac{\,\,\,d{{L}^{\to }}}{dt}\]
Correct Answer: D
Solution :
We know that \[\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}\] \[\therefore \] \[\frac{\overrightarrow{dL}}{dt}=\overrightarrow{r}\times \frac{\overrightarrow{dp}}{dt}+\frac{\overrightarrow{dr}}{dt}\times \overrightarrow{p}\] \[\Rightarrow \] \[\frac{\overrightarrow{dL}}{dt}=\overrightarrow{r}\times \overrightarrow{F}+\overrightarrow{v}\times \overrightarrow{p}\] \[\left[ \because \overrightarrow{F}=\frac{\overrightarrow{dp}}{\overrightarrow{dt}} \right]\] \[\frac{\overrightarrow{dL}}{ddt}=\overrightarrow{r}\times \overrightarrow{F}\] \[[\because \overrightarrow{v}||\overrightarrow{p}]\] Or \[\frac{\overrightarrow{dL}}{dt}=\overrightarrow{\tau }\] \[[\because \tau =\overrightarrow{r}\times \overrightarrow{F}]\]
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