Import Question JSON

Current Question (ID: 9800)

Question:
$\text{Two stones of masses } m \text{ and } 2m \text{ are whirled in horizontal circles, the heavier one in a radius } \frac{r}{2} \text{ and the lighter one in the radius } r\text{. The tangential speed of lighter stone is } n \text{ times that of heavier stone when they experience the same centripetal forces. The value of } n \text{ is:}$
Options:
  • 1. $2$
  • 2. $3$
  • 3. $4$
  • 4. $1$
Solution:
$\text{Hint: } F = \frac{mv^2}{r}$ $\text{Step: Find the tangential speed and value of } n$ $\text{The centripetal force for a mass } m \text{ moving with speed } v \text{ in a circle of radius } r \text{ is given by:}$ $F = \frac{mv^2}{r}$ $\text{Since the centripetal forces on both stones are the same:}$ $\frac{mv_1^2}{r} = \frac{2mv_2^2}{r/2}$ $\text{Simplify the equation we get:}$ $\frac{mv_1^2}{r} = \frac{4mv_2^2}{r}$ $v_1^2 = 4v_2^2$ $\text{Taking the square root of both sides we get:}$ $v_1 = 2v_2$ $\text{Thus, the tangential speed of the lighter stone is 2 times that of the heavier stone.}$ $\text{Hence, option (1) is the correct answer.}$

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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
}