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Current Question (ID: 19538)

Question:

$\text{The length of a metallic wire is } l_1, \text{ when the tension in it is } T_1 \text{ and is } l_2 \text{ when the tension is } T_2. \text{ The original (natural) length of the wire is:}$

Options:

1. $\frac{l_1 + l_2}{2}$
2. $\frac{T_2 l_1 + T_1 l_2}{T_1 + T_2}$
3. $\frac{T_2 l_1 - T_1 l_2}{T_2 - T_1}$
4. $\frac{T_1 l_1 - T_2 l_2}{T_2 - T_1}$

Solution:

$\text{Hint: } T \propto \Delta l$ $\text{Step: Find the original length of the wire.}$ $\text{According to Hooke's law, change in length is directly proportional to tension, i.e., } T \propto \Delta l$ $\text{Let the natural length (original length) of the wire is } l_0.$ $\Rightarrow T \propto \Delta l$ $\Rightarrow T = k \Delta l = k(l - l_0)$ $\text{The tension in the wire is written as;}$ $\Rightarrow T_1 = k(l_1 - l_0) \quad \cdots (1)$ $\Rightarrow T_2 = k(l_2 - l_0) \quad \cdots (2)$ $\text{Dividing equation (1) by (2) we get;}$ $\frac{T_1}{T_2} = \frac{(l_1 - l_0)}{(l_2 - l_0)}$ $\Rightarrow T_1(l_2 - l_0) = T_2(l_1 - l_0) \Rightarrow T_1 l_2 - T_1 l_0 = T_2 l_1 - T_2 l_0$ $\Rightarrow l_0 = \frac{T_2 l_1 - T_1 l_2}{T_2 - T_1}$ $\text{Therefore, the original length of the wire is } \frac{T_2 l_1 - T_1 l_2}{T_2 - T_1}.$ $\text{Hence, option (3) is the correct answer.}$

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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
}