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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:
  • 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'\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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{
  "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
}