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Current Question (ID: 9729)

Question:

$\text{Sand is being dropped on a conveyor belt at a rate of } M \text{ kg/s. The force necessary to keep the belt moving with at a constant velocity of } v \text{ m/s will be:}$

Options:

1. $Mv \text{ newton}$
2. $2Mv \text{ newton}$
3. $\frac{Mv}{2} \text{ newton}$
4. $\text{zero}$

Solution:

$\text{Hint: The force is given by the rate of change of momentum.}$ $\text{Step 1: Find the rate of change in momentum.}$ $\text{The rate of change in momentum is given by,}$ $\frac{dp}{dt} = \frac{d(mv)}{dt} = \frac{mdv}{dt} + v\frac{dm}{dt}$ $\text{Here v is a constant, so, } \frac{dp}{dt} = v\left(\frac{dm}{dt}\right) = vM$ $\text{Step 2: Find the force required.}$ $F = \frac{dp}{dt} = vM \text{ Newton}$ $\text{Explanation: When sand is dropped onto the moving conveyor belt, each grain of sand initially has zero horizontal velocity. The belt must accelerate this sand to its velocity } v\text{. The rate at which momentum is being added to the system is } vM\text{, where } M \text{ is the mass rate of sand being added. By Newton's second law, the force required to maintain constant belt velocity equals this rate of momentum change.}$

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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
}