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The angular speed of a flywheel is increased from $600 \text{ rpm}$ to $1200 \text{ rpm}$ in $10 \text{ s}$. The number of revolutions completed by the flywheel during this time is :

This question was previously asked in
NEET UG Re-Exam 2026 Question Paper (21-Jun-2026)
The correct answer is
150

Flywheel Revolutions Calculation

The problem asks for the total number of revolutions a flywheel completes while its angular speed increases over a specific time period.

Initial Data Conversion

First, convert the initial and final angular speeds from revolutions per minute (rpm) to radians per second (rad/s), as physics calculations require SI units.

  • Conversion factor: $1 \text{ rpm} = \frac{2\pi}{60} \text{ rad/s}$
  • Initial angular speed ($\omega_0$): $600 \text{ rpm} = 600 \times \frac{2\pi}{60} \text{ rad/s} = 20\pi \text{ rad/s}$
  • Final angular speed ($\omega_f$): $1200 \text{ rpm} = 1200 \times \frac{2\pi}{60} \text{ rad/s} = 40\pi \text{ rad/s}$
  • Time ($t$): $10 \text{ s}$

Calculating Angular Acceleration

Determine the angular acceleration ($\alpha$) using the formula $\alpha = \frac{\omega_f - \omega_0}{t}$.

  • $\alpha = \frac{40\pi \text{ rad/s} - 20\pi \text{ rad/s}}{10 \text{ s}}$
  • $\alpha = \frac{20\pi \text{ rad/s}}{10 \text{ s}} = 2\pi \text{ rad/s}^2$

Calculating Angular Displacement

Use the kinematic equation for angular displacement: $\theta = \omega_0 t + \frac{1}{2} \alpha t^2$.

  • $\theta = (20\pi \text{ rad/s})(10 \text{ s}) + \frac{1}{2} (2\pi \text{ rad/s}^2)(10 \text{ s})^2$
  • $\theta = 200\pi \text{ rad} + \frac{1}{2} (2\pi \text{ rad/s}^2)(100 \text{ s}^2)$
  • $\theta = 200\pi \text{ rad} + 100\pi \text{ rad}$
  • $\theta = 300\pi \text{ radians}$

Determining Number of Revolutions

Convert the total angular displacement ($\theta$) from radians to revolutions. Since $1 \text{ revolution} = 2\pi \text{ radians}$.

  • Number of revolutions = $\frac{\theta}{2\pi}$
  • Number of revolutions = $\frac{300\pi \text{ radians}}{2\pi \text{ radians/revolution}}$
  • Number of revolutions = $150$

Therefore, the flywheel completes 150 revolutions.

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Important Questions from Mechanics

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  3. In case of vertical circular motion of a particle by a thread of length $r$ if the tension in the thread is zero at an angle $30^\circ$ shown in figure, the velocity at the bottom point ($A$) of the circular path is
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  4. The fifth harmonic of a closed organ pipe is found to be in unison with the first harmonic of an open pipe. The ratio of lengths of closed pipe to that of the open pipe is $5/x$. The value of $x$ is _______.
  5. In a vernier callipers, 50 vernier scale divisions are equal to 48 main scale divisions. If one main scale division $= 0.05 \text{ mm}$, then the least count of the vernier callipers is _______ mm.
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