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Current Question (ID: 16774)

Question:

$\text{A particle of charge } +q \text{ and mass } m \text{ moving under the influence of a uniform electric field } \vec{E}\hat{i} \text{ and a uniform magnetic field } B\hat{k} \text{ follows a trajectory from } P \text{ to } Q \text{ as shown in the figure.}$ $\text{The velocities at } P \text{ and } Q \text{ are } v\hat{i} \text{ and } -2v\hat{j} \text{ respectively.}$ $\text{Which of the following statement(s) is/are correct?}$

Options:

1. $E = \frac{3}{4} \frac{mv^2}{qa}.$
2. $\text{Rate of work done by electric field at } P \text{ is } \frac{3}{4} \frac{mv^3}{a}.$
3. $\text{Rate of work done by both fields at } Q \text{ is zero.}$
4. $\text{All of the above.}$

Solution:

$\text{Kinetic energy of the particle at point } P = \frac{1}{2} mv^2$ $\text{K.E. of the particle at point } Q = \frac{1}{2} m (2v)^2$ $\text{Increase in K.E. } = \frac{3}{2} mv^2$ $\text{It comes from the work done by the electric force } qE \text{ on the particle as it covers a distance } 2a \text{ along the } x\text{-axis.}$ $\text{Thus } \frac{3}{2} mv^2 = qE \times 2a \Rightarrow E = \frac{3}{4} \frac{mv^2}{qa}. \text{ The rate of work done by the electric field at}$ $P = \vec{F} \times \vec{v} = qE \times v = \frac{3mv^3}{4a}$ $\text{At } Q \; \vec{F} = q\vec{E} \text{ is along } x\text{-axis while velocity is along negative } y\text{-axis.}$ $\text{Hence rate of work done by electric field } = \vec{F}_e \cdot \vec{v} = 0 \; (\because \theta = 90^\circ) \text{ Similarly,}$ $\text{according to equation } \vec{F}_m = q \left( \vec{v} \times \vec{B} \right)$ $\text{Force } \vec{F}_m \text{ is also perpendicular to velocity vector } \vec{v}.$ $\text{Hence, the rate of work done by the magnetic field } = 0$

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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
}