Question

please answer, question in attachments

Answer 1

Given :-

Radius of small disc = R

Radius of large disc = 2R

As the density of disc is constant.

Mass of larger disc = M = σ π(2R)²

M = 4σπR²

Similarly,

Mass of smaller disc = m = σ πR²

m = σ πR²

Now,

Moment of Inertia of Larger disc = I = MR²/2

I = 4σπR²(2R)²/2

I = 16σπR⁴/2

I = 8σπR⁴

Again,

Moment of Inertia of smaller disc about Perpendicular to the plane = I' = σπR²(R)²

I' = σπR⁴

Using Parallel axes theorem we will find the Centre of Mass of smaller disc along the tangential.

I* = I + mR²

I* = mR² + mR²/2

I* = 3mR²/2

I* = 3(σπR²)(R²)/2

I* = 3σπR⁴/2

Now,

For Moment of Inertia of Remaining disc.

Let the Moment of Inertia of Remaining disc = M

M = I - I*

M = 8σπR⁴ - 3σπR⁴/2

M = (16σπR⁴ - 3σπR⁴)/2

M = 13σπR⁴/2

Hence,

Option (B) is correct.