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Current Question (ID: 9739)

Question:

$\text{A small ball is suspended from a thread. If it is lifted up with an acceleration of } 4.9 \text{ ms}^{-2} \text{ and lowered with an acceleration of } 4.9 \text{ ms}^{-2}\text{, then the ratio of the tension in the thread in both cases will be:}$

Options:

1. $1 : 3$
2. $3 : 1$
3. $1 : 1$
4. $1 : 5$

Solution:

$\text{When the ball is lifted up with acceleration } a = 4.9 \text{ ms}^{-2}\text{:}$ $T_1 - mg = ma$ $T_1 = m(g + a)$ $\text{When the ball is lowered with acceleration } a = 4.9 \text{ ms}^{-2}\text{:}$ $mg - T_2 = ma$ $T_2 = m(g - a)$ $\text{The ratio of tensions:}$ $\frac{T_1}{T_2} = \frac{g + a}{g - a} = \frac{9.8 + 4.9}{9.8 - 4.9} = \frac{14.7}{4.9} = \frac{3}{1}$

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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
}