Import Question JSON

Current Question (ID: 8789)

Question:

$\text{The pitch of a screw gauge is 1 mm and there are 100 divisions on the circular scale. While measuring the diameter of a wire, the linear scale reads 1 mm and } 47^{\text{th}} \text{ division on the circular scale coincides with the reference line. The length of the wire is 5.6 cm. Curved surface area (in cm}^2\text{) of the wire in appropriate number of significant figures will be:}$

Options:

1. $2.4 \text{ cm}^2$
2. $2.56 \text{ cm}^2$
3. $2.6 \text{ cm}^2$
4. $2.8 \text{ cm}^2$

Solution:

\text{Hint: Diameter of wire} = \text{MSR} + \text{LC} \times \text{CSD} \text{Step 1: Find the least count of the screw gauge} \text{L.C.} = \frac{\text{Pitch}}{\text{Number of circular scale divisions}} \text{L.C.} = \frac{1\text{ mm}}{100} = 0.01\text{ mm} \text{Step 2: Find the diameter of the wire.} d = \text{MSR} + \text{LC} \times n d = 1\text{ mm} + 0.01\text{ mm} \times 47 d = 1.47\text{ mm} \text{Step 3: Find the curved surface area of the wire.} A = 2\pi rl A = 2 \times 3.14 \times \frac{1.47}{2} \times 10^{-1} \times 5.6 A = 2.584\text{ cm}^2 \text{Step 4: Report the answer according to the correct significant figures.} \text{For multiplication, the final number should have the least number of significant figures as in the given numbers} \text{Here, the least significant figure in the given numbers} = 2 \text{The final number should have 2 significant figures.} \text{Therefore, the curved surface area (in cm}^2\text{) of the wire is } A = 2.6 \text{Hence, option (3) is the correct answer.}

Import JSON File

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
}