The angular speed of a flywheel is increased from 600 ~rpm$600 \mathrm{~rpm}$ to 1200 ~rpm$1200 \mathrm{~rpm}$ in 10 ~s$10 \mathrm{~s}$. The number of revolutions completed by the flywheel during this time is :
We can solve this directly using average rotational speed since the acceleration is constant.
Initial frequency N₁ = 600 ~rpm = (600)/(60) = 10 ~rev/s$N_1 = 600 \mathrm{~rpm} = \frac{600}{60} = 10 \mathrm{~rev/s}$.
Final frequency N₂ = 1200 ~rpm = (1200)/(60) = 20 ~rev/s$N_2 = 1200 \mathrm{~rpm} = \frac{1200}{60} = 20 \mathrm{~rev/s}$.
Time t = 10 ~s$t = 10 \mathrm{~s}$.
Step 1: Calculate Revolutions via Average Frequency
Since angular acceleration is constant, the average frequency (revolutions per second) is:
Total number of revolutions = Navg × t$N_{\text{avg}} \times t$
Total revolutions = 15 × 10 = 150 revolutions$15 \times 10 = 150 \text{ revolutions}$.
Pattern Recognition
Converting strictly into radians per second is technically correct but unnecessarily tedious. Staying in "revolutions per second (Hz)" allows directly multiplying average Hz by time to get total revolutions instantly.
Chapter Mix
Class 11 Physics: Systems of Particles and Rotational Motion
Keywords:#number of revolutions completed by the flywheel#NEET 2026 Code 12 Q38#Systems of Particles and Rotational Motion NEET 2026#Kinematics of Rotational Motion about a Fixed Axis NEET 2026
More Systems of Particles and Rotational Motion Previous-Year Questions
Q20neet_2026_03_may_morningMoment of Inertia
A thin wire of length 'L$L$' and linear mass density 'm$m$' is bent into a circular ring (in x - y plane) with centre 'C$C$' as shown in figure. The moment of inertia of the ring about an axis yy' will be :
Ring mapped on x-y axis showing yy' axis tangent to the boundary of the ring.
A.3 ~mL³8π²$\frac{3 \mathrm{~mL}^3}{8\pi^2}$
B.3 ~mL³8π$\frac{3 \mathrm{~mL}^3}{8\pi}$
C.3 ~mL²8π²$\frac{3 \mathrm{~mL}^2}{8\pi^2}$
D.3 ~mL²8π$\frac{3 \mathrm{~mL}^2}{8\pi}$
Solution
Related Formula
M = m × L$M = m \times L$
L = 2π r r = (L)/(2π)$$L = 2\pi r \implies r = \frac{L}{2\pi}$$Idiametric = (M r²)/(2)$$I_{\text{diametric}} = \frac{M r^2}{2}$$Itangent = ICM + M r²$$I_{\text{tangent}} = I_{\text{CM}} + M r^2$$
Core Logic
Total mass of the ring M = m × L$M = m \times L$, where m$m$ is the linear mass density.
The radius r$r$ of the ring is derived from the circumference: r = (L)/(2π)$r = \frac{L}{2\pi}$.
The required axis yy' is a tangent to the ring in its own plane. By the parallel axis theorem, the moment of inertia about this tangential axis is:
Iyy' = Idiametric + M r²$$I_{yy'} = I_{\text{diametric}} + M r^2$$Iyy' = (M r²)/(2) + M r² = (3)/(2) M r²$$I_{yy'} = \frac{M r^2}{2} + M r^2 = \frac{3}{2} M r^2$$
Step 1: Substitute Length and Mass
Substitute M = mL$M = mL$ and r = (L)/(2π)$r = \frac{L}{2\pi}$ into the equation:
Whenever finding tangential MoI, verify if the tangent is "in-plane" (I = (3)/(2)MR²$I = \frac{3}{2}MR^2$) or "perpendicular to plane" (I = 2MR²$I = 2MR^2$). The figure strictly places the tangent in the x-y plane.
Chapter Mix
Class 11 Physics: Systems of Particles and Rotational Motion
More Systems of Particles and Rotational Motion Questions — neet_2026_03_may_morning
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
