Import Question JSON

Current Question (ID: 10330)

Question:

$\text{A body of mass M is moving on a circular track of radius r in such a way that its kinetic energy K depends on the distance travelled by the body s according to relation } K = \beta s, \text{ where } \beta \text{ is a constant. The angular acceleration of the body is:}$

Options:

1. $\frac{\beta r}{M^2}$
2. $\sqrt{\frac{\beta r}{M}}$
3. $\frac{Mr^2}{\beta}$
4. $\frac{\beta}{Mr}$

Solution:

\text{1. Kinetic Energy and Linear Speed: The kinetic energy } K = \frac{1}{2}Mv^2\text{. We are given } K = \beta s\text{.} \text{So, } \frac{1}{2}Mv^2 = \beta s\text{.} \text{From this, we get } v^2 = \frac{2\beta s}{M}\text{.} \text{2. Tangential Acceleration: Differentiate } v^2 \text{ with respect to time (t):} 2v\frac{dv}{dt} = \frac{2\beta}{M}\frac{ds}{dt} \text{We know that } \frac{dv}{dt} \text{ is the tangential acceleration } (a_t) \text{ and } \frac{ds}{dt} \text{ is the linear speed } (v)\text{.} \text{So, } 2va_t = \frac{2\beta}{M}v\text{.} \text{Since the body is moving, } v \neq 0\text{, so we can divide by } 2v\text{:} a_t = \frac{\beta}{M} \text{3. Angular Acceleration: For circular motion, the tangential acceleration } (a_t) \text{ is related to the angular acceleration } (\alpha) \text{ by the formula } a_t = r\alpha\text{.} \text{Therefore, } \alpha = \frac{a_t}{r}\text{.} \text{Substitute the expression for } a_t\text{:} \alpha = \frac{(\beta/M)}{r} = \frac{\beta}{Mr}

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
}