| (i) \[\vec{L}\] does not change with time |
| (ii) \[\frac{\overset{\to }{\mathop{dL}}\,}{dt}\] is perpendicular to \[\vec{L}\] at all instants of time |
| (iii) The magnitude of \[\vec{L}\] does not change with time |
| (iv) \[\frac{d\overset{\to }{\mathop{L}}\,}{dt}\] is parallel to \[\vec{L}\] at all instants of time |
| The correct statements are |
A) (i), (ii) and (iii)
B) (i), (ii) and (iv)
C) (iii) and (iv)
D) (ii) and (iii)
Correct Answer: D
Solution :
[d] \[\overset{\to }{\mathop{\tau }}\,=\overset{\to }{\mathop{C}}\,\times \overset{\to }{\mathop{L}}\,\] i.e., \[\frac{\overset{\to }{\mathop{d}}\,L}{dt}=\overset{\to }{\mathop{C}}\,\times \overset{\to }{\mathop{L}}\,\] \[\therefore \,\,\,\frac{\overset{\to }{\mathop{d}}\,L}{dt}\]is perpendicular to \[\overset{\to }{\mathop{C}}\,\] and \[\overset{\to }{\mathop{L}}\,\] both. Therefore option (ii) is correct. Further \[\overset{\to }{\mathop{L}}\,.\overset{\to }{\mathop{L}}\,={{L}^{2}}\] ... (1) Differentiating equation (1) with respect to time \[\overset{\to }{\mathop{L}}\,\frac{\overrightarrow{dL}}{dt}+\frac{\overrightarrow{dL}}{dt}.\vec{L}=2L\frac{dL}{dt}\] or \[2\overset{\to }{\mathop{L}}\,.\frac{d\overset{\to }{\mathop{L}}\,}{dt}=2L\frac{dL}{dt}\] or \[\overset{\to }{\mathop{L}}\,.\frac{\overset{\to }{\mathop{dL}}\,}{dt}=L\frac{dL}{dt}\] ??(2) But since \[\overset{\to }{\mathop{L}}\,\bot \frac{\overset{\to }{\mathop{dL}}\,}{dt}\] \[\therefore \,\,\,\,\,\,\overset{\to }{\mathop{L}}\,.\frac{\overset{\to }{\mathop{dL}}\,}{dt}=0\] or \[\frac{dL}{dt}=0\] (from equation (2)) \[\therefore \] Magnitude of \[\overset{\to }{\mathop{L}}\,\]or L does not change with time. Therefore, option (iii) is also correct.
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