A) \[0{}^\circ \]
B) \[180{}^\circ \]
C) \[90{}^\circ \]
D) \[45{}^\circ \]
Correct Answer: C
Solution :
Here, \[\overrightarrow{P}=\hat{i}2\cos t+\hat{j}2\sin t\] \[\overrightarrow{F}=\frac{d\overrightarrow{P}}{dt}=-2\hat{i}\sin t+2\hat{j}\cos t\] \[\overrightarrow{F}.\overrightarrow{P}=(-2\hat{i}\sin t+2\hat{j}\cos t).(2\hat{i}\cos +2\hat{j}\sin t)\] \[=-4\sin t\cos t+4\sin t\cos t\] Therefore, \[\theta =90{}^\circ \]
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