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Current Question (ID: 10287)

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

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}