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Current Question (ID: 9869)

Question:

$\text{If two springs, A and B } (K_A = 2 K_B) \text{, are stretched by the same suspended weights, then the ratio of work done in stretching is equal to:}$

Options:

1. $1 : 2$
2. $2 : 1$
3. $1 : 1$
4. $1 : 4$

Solution:

$\text{Given formulas:}$ $W = \frac{1}{2}Kx^2, \quad F = -Kx$ $\text{From these, we can derive:}$ $W = \frac{1}{2}K \cdot \frac{F^2}{K^2} = \frac{F^2}{2K}$ $\text{Since the springs are stretched by the same suspended weights, the force F is the same for both springs.}$ $\text{Therefore, the work is inversely proportional to the spring constant:}$ $W \propto \frac{1}{K} = \frac{W_A}{W_B} = \frac{K_B}{K_A} = \frac{K_B}{2K_B} = \frac{1}{2}$ $\text{Hence, the ratio of work done } W_A : W_B = 1 : 2$

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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
}