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Current Question (ID: 18906)

Question:

$\text{The current-voltage relation of the diode is given by}$ $I = \left(e^{\frac{1000V}{T}} - 1\right) \text{ mA},$ $\text{where the applied voltage } V \text{ is in volts and the temperature } T \text{ is in degree Kelvin.}$ $\text{If a student makes an error measuring } \pm 0.01 \text{ V while measuring the current of } 5 \text{ mA at } 300 \text{ K,}$ $\text{what will be the error in the value of current in mA?}$

Options:

1. 0.02 \text{ mA}
2. 0.5 \text{ mA}
3. 0.05 \text{ mA}
4. 0.2 \text{ mA}

Solution:

$I = \left[e^{\frac{1000V}{T}} - 1\right] \text{ mA}$ $\text{differentiating on both sides}$ $\text{d}I = \left[e^{\frac{1000V}{T}}\right]\left[\frac{1000}{T}\right]\text{d}V$ $\text{d}I = \left(I + 1\right)\left(\frac{1000}{T}\right)\left(\text{d}V\right)$ $\text{d}I = \left(5 + 1\right)\left(\frac{1000}{300}\right)\times (0.01)$ $\text{d}I = 6 \times \frac{10}{3} \times 0.01 = 0.2 \text{ mA}$

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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
}