Question:
A fan is rotating with an angular speed \(300\ \text{rpm}\). The fan is switched off, and it takes \(80\ \text{s}\) to come to rest. Assuming constant angular deceleration, the number of revolutions made by the fan before it comes to rest is:
A fan is rotating with an angular speed \(300\ \text{rpm}\). The fan is switched off, and it takes \(80\ \text{s}\) to come to rest. Assuming constant angular deceleration, the number of revolutions made by the fan before it comes to rest is:
Show Hint
For constant angular deceleration,
\[
\omega_{\text{avg}}=\frac{\omega_0+\omega}{2}
\]
and
\[
\text{Number of revolutions}=\omega_{\text{avg}}\times t
\]
when angular speed is measured in revolutions per second.
Updated On: Jun 25, 2026
- \(400\)
- \(200\)
- \(300\)
- \(314\)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Convert angular speed into revolutions per second.
Given angular speed: \[ 300\ \text{rpm} \] This means \[ 300\ \text{revolutions per minute} \] Since \[ 1\ \text{minute}=60\ \text{seconds}, \] we get \[ 300\ \text{rpm}=\frac{300}{60}\ \text{rps} \] \[ =5\ \text{rps} \] So, the initial angular speed is \[ \omega_0=5\ \text{rev/s} \] Final angular speed is \[ \omega=0 \] Time taken is \[ t=80\ \text{s} \]
Step 2: Use average angular speed.
Since the fan comes to rest with constant angular deceleration, the average angular speed is \[ \omega_{\text{avg}}=\frac{\omega_0+\omega}{2} \] Substituting values, \[ \omega_{\text{avg}}=\frac{5+0}{2} \] \[ =\frac{5}{2} \] \[ =2.5\ \text{rev/s} \]
Step 3: Calculate total number of revolutions.
Number of revolutions is \[ N=\omega_{\text{avg}}t \] Substituting values, \[ N=2.5\times 80 \] \[ N=200 \]
Step 4: Final conclusion.
Hence, the number of revolutions made before coming to rest is \[ \boxed{200} \]
Given angular speed: \[ 300\ \text{rpm} \] This means \[ 300\ \text{revolutions per minute} \] Since \[ 1\ \text{minute}=60\ \text{seconds}, \] we get \[ 300\ \text{rpm}=\frac{300}{60}\ \text{rps} \] \[ =5\ \text{rps} \] So, the initial angular speed is \[ \omega_0=5\ \text{rev/s} \] Final angular speed is \[ \omega=0 \] Time taken is \[ t=80\ \text{s} \]
Step 2: Use average angular speed.
Since the fan comes to rest with constant angular deceleration, the average angular speed is \[ \omega_{\text{avg}}=\frac{\omega_0+\omega}{2} \] Substituting values, \[ \omega_{\text{avg}}=\frac{5+0}{2} \] \[ =\frac{5}{2} \] \[ =2.5\ \text{rev/s} \]
Step 3: Calculate total number of revolutions.
Number of revolutions is \[ N=\omega_{\text{avg}}t \] Substituting values, \[ N=2.5\times 80 \] \[ N=200 \]
Step 4: Final conclusion.
Hence, the number of revolutions made before coming to rest is \[ \boxed{200} \]
Was this answer helpful?
0
0
Top AP EAPCET Physics Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The radii of the two columns in a U - tube are r1 and r2. When a liquid of density p (angle of contact is $ 0{}^\circ $ ) is filled in it, the level difference of the liquid in the two arms is h. The surface tension of the liquid is: (g = acceleration due to gravity)
- A rigid metallic sphere is spinning around its own axis in the absence of external torque. If the temperature is raised, its volume increases by $9 \%$. The change in its angular speed is
View More Questions
Top AP EAPCET Rotational motion Questions
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- A planet is revolving around the Sun as shown in the figure. The radius vectors joining the Sun and the planet at points $A$ and $B$ are $90 \times 10^{6} km$ and $60 \times 10^{6} km$, respectively. The ratio of velocities of the planet at the points $A$ and $B$ when its velocities make angle $30^{\circ}$ and $60^{\circ}$ with major-axis of the orbit is
- The kinetic energy of a body rotating at 300 revolutions per minute is 62.8 J. Its angular momentum $ (\text{in kg}\,{{m}^{2}}{{s}^{-1}}) $ is approximately:
- The moment of inertia of a solid sphere of mass 20 kg and diameter 20 cm about the tangent to the sphere is:
- A disc starts rotating from rest with constant acceleration and attains angular velocity of 20 rad/s in 5 seconds. The total angular displacement during this interval is
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions