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Current Question (ID: 20355)

Question:

$\text{Which of the following will NOT be observed when a multimeter (operating in resistance measuring mode) probes connected across a component, are just reversed?}$

Options:

1. $\text{Multimeter shows an equal deflection in both cases i.e. before and after reversing the probes if the chosen component is resistor.}$
2. $\text{Multimeter shows NO deflection in both cases i.e. before and after reversing the probes if the chosen component is metal wire.}$
3. $\text{Multimeter shows a deflection, accompanied by a splash of light out of connected component in one direction and NO deflection on reversing the probes if the chosen component is LED.}$
4. $\text{Multimeter shows NO deflection in both cases i.e. before and after reversing the probes if the chosen component is capacitor.}$

Solution:

$\text{Hint: When a multimeter is used in resistance mode across a capacitor it shows some deflection.}$ $\text{(2) If we assume that LED has negligible resistance then multimeter shows no deflection for the forward bias but when it connects in reverse direction, it break down occurs so splash of light out.}$ $\text{(3) The resistance of metal wire may be taken zero, so no deflection in multimeter.}$ $\text{(4) No matter, how we connect the resistance across multimeter. It shows same deflection}$

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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
}