Let the moment of inertia of a hollow cylinder of length 0cm (inner radius 10cm and outer radius 20cm), about its axis, be I. The radius of a thin cylinder of the same mass such that its moment of inertia about its axis is also I, is____?

The moment of inertia of a hollow cylinder about its axis is given by I = ½m(R² + r²), where m is the mass, R is the outer radius, and r is the inner radius. In this case, R = 20cm and r = 10cm. So, I = ½m(20² + 10²) = ½m(400 + 100) = ½m(500) = 250m.

For a thin cylinder (r ≈ 0), the moment of inertia about its axis is I = ½mk², where k is the radius of the cylinder. We are given that this moment of inertia is equal to I = 250m.

Therefore, 250m = ½mk²
500m = mk²
k² = 500
k = √500 = 10√5 ≈ 22.36 cm

However, the given options don't include this value. Let's re-examine the formula for the moment of inertia of a hollow cylinder. It's actually I = m(R² + r²)/2. For a solid cylinder, it's I = mr²/2. Let's use the correct formula for a hollow cylinder.

I_hollow = m(R² + r²)/2 = m(20² + 10²)/2 = m(500)/2 = 250m

For a thin cylinder (which is approximately a solid cylinder), the moment of inertia is:
I_thin = mk²/2

We set these equal since they have the same moment of inertia:
250m = mk²/2
500 = k²
k = √500 ≈ 22.36 cm

Again, this is not in the options. There appears to be an error in the question or provided solution. The provided solution 'm202+1022=mk2k√400+1002k=√250k=5√10k=16' is not mathematically sound or clearly explained and does not lead to a correct answer.
The correct radius for the thin cylinder, given the conditions, is approximately 22.36 cm.