The torque acting on a body about a point is | vedclass

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(D) Given the torque $\overrightarrow{	au} = \overrightarrow{A} 	imes \overrightarrow{L}$. Since $\overrightarrow{	au} = \frac{d\overrightarrow{L}}

Vedclass · Accepted Answer

All of the above.

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(D) Given the torque $\overrightarrow{	au} = \overrightarrow{A} 	imes \overrightarrow{L}$. Since $\overrightarrow{	au} = \frac{d\overrightarrow{L}}{dt}$,we have $\frac{d\overrightarrow{L}}{dt} = \overrightarrow{A} 	imes \overrightarrow{L}$.$1$. By the definition of the cross product,$\frac{d\overrightarrow{L}}{dt}$ is perpendicular to both $\overrightarrow{A}$ and $\overrightarrow{L}$. Thus,$\frac{d\overrightarrow{L}}{dt} \perp \overrightarrow{L}$,making option $(a)$ correct.$2$. To check the magnitude of $\overrightarrow{L}$,consider $L^2 = \overrightarrow{L} \cdot \overrightarrow{L}$. Differentiating with respect to time $t$: $\frac{d}{dt}(L^2) = 2\overrightarrow{L} \cdot \frac{d\overrightarrow{L}}{dt}$. Since $\frac{d\overrightarrow{L}}{dt} \perp \overrightarrow{L}$,the dot product $\overrightarrow{L} \cdot \frac{d\overrightarrow{L}}{dt} = 0$. Therefore,$\frac{d}{dt}(L^2) = 0$,which means the magnitude $L$ is constant. Thus,option $(c)$ is correct.$3$. Since $\frac{d\overrightarrow{L}}{dt} = \overrightarrow{A} 	imes \overrightarrow{L}$,the rate of change of $\overrightarrow{L}$ is always perpendicular to $\overrightarrow{A}$. The component of $\overrightarrow{L}$ along $\overrightarrow{A}$ is given by $L_A = \overrightarrow{L} \cdot \hat{A}$. Differentiating with respect to time: $\frac{dL_A}{dt} = \frac{d\overrightarrow{L}}{dt} \cdot \hat{A} = (\overrightarrow{A} 	imes \overrightarrow{L}) \cdot \hat{A}$. Since the cross product $(\overrightarrow{A} 	imes \overrightarrow{L})$ is perpendicular to $\overrightarrow{A}$,the dot product with $\hat{A}$ is zero. Thus,$\frac{dL_A}{dt} = 0$,meaning the component of $\overrightarrow{L}$ along $\overrightarrow{A}$ is constant. Thus,option $(b)$ is correct.Since all statements are correct,the answer is $(d)$.

The torque acting on a body about a point is equal to $\overrightarrow{A} \times \overrightarrow{L}$,where $\overrightarrow{A}$ is a constant vector and $\overrightarrow{L}$ is the angular momentum about that point. This implies that:

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