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Current Question (ID: 19897)

Question:

$\text{A simple pendulum has a time period } T \text{ in air. The bob is then completely immersed and continues to oscillate freely in a non-viscous liquid whose density is } \left( \frac{1}{16} \right) \text{th of that of the bob.}$ $\text{Assuming no damping and only the effect of buoyancy, what is the new time period of oscillation?}$

Options:

1. $2T \sqrt{\frac{1}{14}}$
2. $2T \sqrt{\frac{1}{10}}$
3. $4T \sqrt{\frac{1}{15}}$
4. $4T \sqrt{\frac{1}{14}}$

Solution:

$\text{Hint: Tension decreases when immersed in liquid.}$ $\text{Step 1: Find the effective acceleration due to gravity.}$ $\text{Effective weight} = mg - F_B$ $\Rightarrow \text{Effective weight} = mg - \rho V g \quad [\rho = \text{Density of liquid}]$ $\Rightarrow \text{Effective weight} = mg - \frac{1}{16} \rho_{\text{bob}} V g$ $\Rightarrow \text{Effective weight} = mg - \frac{1}{16} mg$ $\Rightarrow m g_{\text{eff}} = \frac{15mg}{16}$ $\Rightarrow g_{\text{eff}} = \frac{15g}{16}$ $\text{Step 2: Find a new period of oscillation.}$ $T = 2\pi \sqrt{\frac{L}{g}}$ $\Rightarrow \frac{T'}{T} = \sqrt{\frac{g}{g_{\text{eff}}}}$ $\Rightarrow \frac{T'}{T} = \sqrt{\frac{g}{\left(\frac{15}{16}\right)g}}$ $\Rightarrow \frac{T'}{T} = \sqrt{\frac{1}{15/16}}$ $\Rightarrow \frac{T'}{T} = \frac{4}{\sqrt{15}}$ $\Rightarrow T' = \frac{4T}{\sqrt{15}}$ $\text{Hence, option (3) is the correct answer.}$

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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
}