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Current Question (ID: 18916)

Question:

$\text{The least count of the main scale of a vernier callipers is } 1 \text{ mm. Its vernier scale is divided into } 10 \text{ divisions and coincide with } 9 \text{ divisions of the main scale.}$ $\text{When jaws are touching each other, the } 7^{\text{th}} \text{ division of vernier scale coincides with a division of main scale and the zero of vernier scale is lying right side of the zero of main scale.}$ $\text{When this vernier is used to measure length of a cylinder the zero of the vernier scale between } 3.1 \text{ cm and } 3.2 \text{ cm and } 4^{\text{th}} \text{ VSD coincides with a main scale division.}$ $\text{The length of the cylinder is: (VSD is vernier scale division)}$

Options:

1. $3.21 \text{ cm}$
2. $2.99 \text{ cm}$
3. $3.2 \text{ cm}$
4. $3.07 \text{ cm}$

Solution:

$\text{Hint: Zero error is positive.}$ $\text{Step 1: Find the least count of the instrument.}$ $10 \text{ VSD} = 9 \text{ MSD}$ $\Rightarrow 1 \text{ VSD} = 0.9 \text{ MSD}$ $\text{The least count is given by:}$ $\text{LC} = \text{MSD} - \text{VSD}$ $\Rightarrow \text{LC} = \text{MSD} - 0.9 \text{ MSD}$ $\Rightarrow \text{LC} = 0.1 \text{ MSD}$ $\Rightarrow \text{LC} = 0.1 \text{ mm} \quad [\because 1 \text{ MSD} = 1 \text{ mm}]$ $\text{Step 2: Find the zero error.}$ $\text{zero error} = +7 \times 0.1 \text{ mm}$ $\Rightarrow \text{zero error} = +0.7 \text{ mm}$ $\Rightarrow \text{zero error} = +0.07 \text{ cm}$ $\text{Step 3: Find the length of the cylinder.}$ $\text{CR} = \text{MSR} + \text{VSR} - \text{zero error}$ $\Rightarrow \text{CR} = 3.1 \text{ cm} + 4 \times 0.01 \text{ cm} - 0.07 \text{ cm}$ $\Rightarrow \text{CR} = 3.14 \text{ cm} - 0.07 \text{ cm}$ $\Rightarrow \text{CR} = 3.07 \text{ cm}$ $\text{Hence, option (4) is the correct answer.}$

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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
}