Question

No spam!!

Content quality required!!

Thanks!​

Answer 1

Hey mate, the question is very interesting.

Solving step by step,

Moment of inertia of the complete docs about a perpendicular axis passing through the centre O, shown in the figure is,

I1 = 1/2(9M).R^2 = 9/2(MR^2) ...... [ ¶1]

m= [9M/πR^2] • π • R^2 = M { as the rest of the terms get cancelled}

Now, the inertia of the remaining part...

I2 = 1/2 • M (R/3)^2 + M(2R/3)^2 = 1/2 • M • R^2 ..... [ ¶2]

∆ I = I1 - I2 as I1 is greater than I2

∆I = 4 • M • R^2

This is the answer.

The answer is (4)

Hope this helps you.

PLZ MARK MY ANSWER AS BRAINLIEST

Answer 2

Answer:

Total Moment of Inertia of the whole

Disc

= 9MR²/2

where M is the mass of total disc, R is the total radius.

Now, the part which is cut off will be considered as Negative mass and negative Moment of Inertia.

Mass of the cut portion

= {(Total mass)/(Total Area)} × (Cut area)

= (9M/πR²) × {π(R/3)²}

= M

Moment of Inertia of the cut portion with axis at the centre

= [{(M) × (R/3)²}/2] + (M) × (2R/3)²

= MR²/18 + 4MR²/9

= MR²/2

Net Moment of Inertia

= 9MR²/2 - MR²/2

= (9 - 1)MR²/2

= 4MR².

So the answer is option 4)