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Current Question (ID: 11294)

Question:

$\text{A spring-block system oscillates with a time period } T \text{ on the earth's surface. When the system is brought into a deep mine, the time period of oscillation becomes } T'. \text{ Then, one can conclude that:}$

Options:

1. $T' > T$
2. $T' < T$
3. $T' = T$ (Correct)
4. $T' = 2T$

Solution:

$\text{Hint: } T = 2\pi \sqrt{\frac{m}{k}}.\\ \text{Explanation: The time period of the spring-block system does not depend on the effective acceleration. The time period of a spring-block system does not depend on gravity } (g) \text{ because the restoring force is provided solely by the spring, not by gravity. When the system is moved from the Earth's surface into a deep mine: The value of gravity } (g) \text{ decreases, but this does not affect the time period } T. \text{ The time period remains the same. } \Rightarrow T' = T. \text{ Hence, option (3) is the correct answer.}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}