MHT CET · Physics · Rotational Motion

When a 12000 joule of work is done on a flywheel, its frequency of rotation increases from 10 Hz to 20 Hz. The moment of inertia of flywheel about its axis of rotation is $(π^{2} = 10)$

A $1 k g m^{2}$
B $2 k g m^{2}$
C $1.688 k g m^{2}$
D $1.5 k g m^{2}$

Verified Solution

Answer & Solution

Correct Answer

(B) $2 k g m^{2}$

Step-by-step Solution

Detailed explanation

Given, work done, W = 12000 J,
Initial frequency,

f_{1} = 10 H z

Angular velocity for rotational motion is given by

ω = 2 π f

∴ ω_{1} = 2 π f_{1} = 2 π \times 10 = 20 π \frac{r a d}{s}

a n d ω_{2} = 2 π f_{2} = 2 π \times 20 = \frac{40 π r a d}{s}

According to work-energy theorem,
Work done in rotation = change in rotational kinetic energy

\Rightarrow W = \frac{\frac{1}{2}}{ω_{2}^{2}} - \frac{\frac{1}{2}}{ω_{1}^{2}} [∵ \frac{K E_{r o t a t i o n a l} \frac{1}{2}}{ω^{2}}]

\frac{1}{2} / (ω_{2}^{2} - ω_{1}^{2})

…(i)
Where, l = moment of inertia of the flywheel.
Substituting given values in Eq. (i), we get

\frac{12000 \frac{1}{2}}{1600 π^{2} - 400 π^{2}}

\Rightarrow = \frac{1}{1} (1200 \times 10) [∵ π^{2} = 10]

\Rightarrow l = \frac{12000 \times 2}{12000} = 2 k g m^{2}

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From MHT CET

Explore more questions on app

When a 12000 joule of work is done on a flywheel, its frequency of rotation increases from 10 Hz to 20 Hz. The moment of inertia of flywheel about its axis of rotation is π2=10

Step-by-step Solution

When a 12000 joule of work is done on a flywheel, its frequency of rotation increases from 10 Hz to 20 Hz. The moment of inertia of flywheel about its axis of rotation is $(π^{2} = 10)$