The mass per unit area of a circular disc (of negligible thickness) varies with distance from centre r as ρ=ρ₀(1+r), where ρ₀ is a constant. Its moment of inertia about an axis passing through center perpendicular plane is [πρ₀R⁴/10](Y+4R). Then Y=________ [take the radius of the disc as R]
Physics · Mixed Chapter
JEE Mains

Question

The mass per unit area of a circular disc (of negligible thickness) varies with distance from centre r as ρ=ρ₀(1+r), where ρ₀ is a constant. Its moment of inertia about an axis passing through center perpendicular plane is [πρ₀R⁴/10](Y+4R). Then Y=________ [take the radius of the disc as R]

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: A ring of radius r and thickness dr has area dA=2πr dr and mass dm=ρ₀(1+r)2πr dr. Step 2: Its moment of inertia about the disc axis is dI=r²dm=2πρ₀(r³+r⁴)dr. Step 3: Integrate 0→R: I=2πρ₀(R⁴/4+R⁵/5)=[πρ₀R⁴/10](5+4R). Step 4: Compare with [πρ₀R⁴/10](Y+4R). Hence Y=5.