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Current Question (ID: 10167)
Question:
\text{The figure shows a lamina in XY-plane. Two axes } z \text{ and } z' \text{ pass}
\text{perpendicular to its plane. A force } \vec{F} \text{ acts in the plane of the lamina}
\text{at point P as shown. (The point P is closer to the } z'\text{-axis than the } z\text{-axis.)}
\text{(a) torque } \vec{\tau} \text{ caused by } \vec{F} \text{ about } z\text{-axis is along } -\hat{k}
\text{(b) torque } \vec{\tau}' \text{ caused by } \vec{F} \text{ about } z'\text{-axis is along } -\hat{k}
\text{(c) torque caused by } \vec{F} \text{ about the } z\text{-axis is greater in}
\text{magnitude than that about the } z'\text{-axis}
\text{(d) total torque is given by } \vec{\tau}_{\text{net}} = \vec{\tau} + \vec{\tau}'
Options:
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1. $(c, d)$
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2. $(a, c)$
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3. $(b, c)$
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4. $(a, b)$
Solution:
\text{Hint: } \tau = \vec{r} \times \vec{F}
\text{Step 1: Find the direction of the torque about the two axes.}
\text{Consider the adjacent diagram, where } r > r'
\text{Torque } \vec{\tau} \text{ about z-axis } = \vec{r} \times \vec{F} \text{ which is along } \hat{k}
\text{Torque } \vec{\tau}' \text{ about } z'\text{-axis } = \vec{r}' \times \vec{F} \text{ which is along } -\hat{k}
\text{Step 2: Find the magnitude of the torque about the two axes.}
\text{(c) } |\vec{\tau}|_z = Fr_{\perp} = \text{the magnitude of the torque about the z-axis}
\text{where } r_{\perp} \text{ is the perpendicular distance between F and z-axis.}
\text{Similarly, } |\vec{\tau}'|_{z'} = Fr'_{\perp}
\text{Clearly, } r_{\perp} > r'_{\perp} \Rightarrow |\vec{\tau}|_z > |\vec{\tau}'|_{z'}
\text{(d) We are always calculating resultant torque about a common axis.}
\text{Hence, total torque } \vec{\tau}_{\text{net}} \neq \vec{\tau} + \vec{\tau}'
\text{because } \vec{\tau} \text{ and } \vec{\tau}' \text{ are not about a common axis.}
\text{Therefore, the correct statements are (b) and (c).}
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