Import Question JSON

Current Question (ID: 17310)

Question:

$\text{Alternating voltage } V = V_0 \sin \omega t \text{ is applied to a circuit containing some elements in series.}$ $\text{Match the values of current in Column I with the possible elements connected in the circuit in Column II.}$ $\begin{array}{|c|c|} \hline \text{Column-I} & \text{Column-II} \\ \hline (A) \ I = I_0 \sin \omega t & (P) \ \text{only inductor circuit} \\ (B) \ I = -I_0 \cos \omega t & (Q) \ \text{may be } C-R \text{ circuit} \\ (C) \ I = I_0 \sin \left( \omega t + \frac{\pi}{4} \right) & (R) \ \text{may be } L-R \text{ circuit} \\ (D) \ I = I_0 \sin \left( \omega t - \frac{\pi}{4} \right) & (S) \ \text{only resistance circuit} \\ \hline \end{array}$

Options:

1. $A \rightarrow S, \ B \rightarrow Q, \ C \rightarrow P, \ D \rightarrow R$
2. $A \rightarrow P, \ B \rightarrow S, \ C \rightarrow R, \ D \rightarrow Q$
3. $A \rightarrow S, \ B \rightarrow P, \ C \rightarrow Q, \ D \rightarrow R$
4. $A \rightarrow S, \ B \rightarrow P, \ C \rightarrow R, \ D \rightarrow Q$

Solution:

$\text{Hint: Recall the different types of AC circuits.}$ $\text{Step: Find the correct match.}$ $(A) \ \text{In a purely resistive circuit the phase difference between the current and voltage is zero. Therefore, } V = V_0 \sin \omega t \text{ and } I = I_0 \sin \omega t \text{ represents the purely resistive circuit.}$ $(B) \ \text{In a purely inductive circuit the current lags the voltage by } \frac{\pi}{2}. \text{ Therefore, } V = V_0 \sin \omega t \text{ and } I = -I_0 \cos \omega t \text{ represents the purely inductive circuit.}$ $(C) \ \text{For } V = V_0 \sin \omega t \text{ and } I = I_0 \sin \left( \omega t + \frac{\pi}{4} \right), \text{ the current is leading but the phase is less than } \frac{\pi}{2}. \text{ Therefore, the given circuit may be } C-R \text{ circuit.}$ $(D) \ \text{For } V = V_0 \sin \omega t \text{ and } I = I_0 \sin \left( \omega t - \frac{\pi}{4} \right), \text{ the current is lagging but the phase is less than } \frac{\pi}{2}. \text{ Therefore, the given circuit may be } L-R \text{ circuit.}$ $\text{Therefore, the correct match is } A \rightarrow S; \ B \rightarrow P; \ C \rightarrow Q; \ D \rightarrow R. \text{ Hence, option (3) is the correct answer.}$

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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
}