Import Question JSON

Current Question (ID: 10192)

Question:

\text{Given below are two statements:} \text{Assertion (A): For a body under translatory as well as rotational equilibrium, net torque about any axis is zero.} \text{Reason (R): Together } \sum \vec{F}_i = 0 \text{ and } \sum (\vec{r}_i \times \vec{F}_i) = 0 \text{ implies that } \sum (\vec{r}_i - \vec{r}_0) \times \vec{F}_i = 0.

Options:

1. $\text{Both (A) and (R) are True and (R) is the correct explanation of (A).}$
2. $\text{Both (A) and (R) are True but (R) is not the correct explanation of (A).}$
3. $\text{(A) is True but (R) is False.}$
4. $\text{Both (A) and (R) are False.}$

Solution:

$\text{Hint: The net torque must be zero for rotational equilibrium.}$ $\text{Explanation: For a body under translatory as well as rotational equilibrium, net torque about any axis is zero. This equivalence can be demonstrated mathematically as follows;}$ $\text{For translation and rotation equilibrium;}$ $\sum \vec{F}_i = 0 \text{ and } \sum \left(\vec{r}_i \times \vec{F}_i\right) = 0$ $\text{The torque about any point}(\vec{r}_0) \text{ is given by;}$ $\vec{\tau} = \sum \left(\vec{r}_i - \vec{r}_0\right) \times \vec{F}$ $= \sum \vec{r}_i \times \vec{F} - \sum \vec{r}_0 \times \vec{F}$ $\Rightarrow \sum \vec{r}_i \times \vec{F} - \vec{r}_0 \times \sum \vec{F}$ $\Rightarrow 0$ $\text{Therefore, Both (A) and (R) are true, and (R) is the correct explanation of (A).}$ $\text{Hence, option (1) is the correct answer.}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}