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Current Question (ID: 19890)

Question:

$\text{A pendulum made of a uniform wire of cross-sectional area } A \text{ has time period } T. \text{ When an additional mass } M \text{ is added to its bob, the time period changes to } T_M. \text{ If the Young's modulus of the material of the wire is } Y \text{ then } \frac{1}{Y} \text{ is equal to:}$ $\left( g = \text{gravitational acceleration} \right)$

Options:

1. $\left( \frac{T_M}{T} \right)^2 - 1 \frac{Mg}{A}$
2. $1 - \left( \frac{T_M}{T} \right)^2 \frac{A}{Mg}$
3. $1 - \left( \frac{T}{T_M} \right)^2 \frac{A}{Mg}$
4. $\left( \frac{T_M}{T} \right)^2 - 1 \frac{A}{Mg}$

Solution:

$\text{Hint: } T = 2\pi \sqrt{\frac{l}{g}}$ $T = 2\pi \sqrt{\frac{l}{g}} \& T_m = 2\pi \sqrt{\frac{l+\Delta l}{g}}$ $Y = \frac{\left( \frac{F}{A} \right)}{\left( \frac{\Delta l}{l} \right)} \Rightarrow \Delta l = \frac{Fl}{AY}$ $T_m = 2\pi \sqrt{\frac{l+\frac{Fl}{AY}}{g}}$ $T_m = 2\pi \sqrt{\frac{l}{g}} \sqrt{\left( 1 + \frac{F}{AY} \right)}$ $\frac{T_m}{T} = \sqrt{1 + \frac{F}{AY}} \Rightarrow \left( \frac{T_m}{T} \right)^2 = 1 + \frac{F}{AY}$ $\left( \frac{T_m}{T} \right)^2 - 1 = \frac{F}{AY}$ $\frac{1}{Y} = \left( \frac{T_m}{T} \right)^2 - 1 \frac{A}{F}$ $\frac{1}{Y} = \left( \frac{T_m}{T} \right)^2 - 1 \frac{A}{Mg}$

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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
}