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Current Question (ID: 16704)

Question:

$12$ $\text{cells each having the same emf are connected in series with some cells wrongly connected. The arrangement is connected in series with an ammeter and two similar cells which are in series. Current is } 3 \text{ A when cells and battery aid each other and is } 2 \text{ A when cells and battery oppose each other. The number of cells wrongly connected is/are:}$

Options:

1. $4$
2. $1$
3. $3$
4. $2$

Solution:

$\text{Hint: Apply series and parallel combinations of cells.}$ $\text{Step: Find the net EMF and total resistance of the cells.}$ $\text{Let } n \text{ be the number of wrongly connected cells.}$ $\text{Number of cells helping one another } = (12 - n)$ $\text{Total e.m.f. of such cells } = (12 - n)E$ $\text{Total e.m.f. of cells opposing } = nE$ $\text{Resultant e.m.f. of battery } = (12 - n)E - nE = (12 - 2n)E$ $\text{Total resistance of cells } = 12r$ $\text{(since resistance remains same irrespective of connections of cells)}$ $\text{With additional cells}$ $(a) \text{ Total e.m.f. of cells when additional cells help battery } = (12 - 2n)E + 2E$ $\text{Total resistance } = 12r + 2r = 14r$ $\frac{(12 - 2n)E + 2E}{14r} = 3 \quad \ldots (i)$ $(b) \text{ Similarly when additional cells oppose the battery}$ $\frac{(12 - 2n)E - 2E}{14r} = 2 \quad \ldots (ii)$ $\text{Solving (i) and (ii), } n = 1$ $\text{Therefore, the number of wrongly connected cells are } 1.$ $\text{Hence, option } (2) \text{ is the correct answer.}$

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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
}