Two discs of same mass and different radii are made of different materials such that their thicknesses are 1 cm and 0.5 cm respectively. The densities of materials are in the ratio 3: 5. The moment of inertia of these discs respectively about their diameters will be in the ratio of x/6. The value of x is ________.

Solution

Physics sir · 3 steps

Same mass fixes the radii, and with equal masses the ratio of moments of inertia is just the ratio of R².

1. Ratio of I. About a diameter, a disc has I = 1/4mR². The masses are equal, so I₁/I₂ = R₁²/R₂².
2. Use equal mass. The mass of a disc is density times volume, m = ρ π R² t. Setting the two masses equal gives ρ₁ R₁² t₁ = ρ₂ R₂² t₂.

   R₁²/R₂² = (ρ₂ t₂)/(ρ₁ t₁) = (5 × 0.5)/(3 × 1) = 2.5/3 = 5/6

3. So. I₁/I₂ = 5/6 = x/6, so x = 5.

The common slip: 7.2. Writing R₁²/R₂² = (ρ₁ t₁)/(ρ₂ t₂) inverts the ratio to 6/5 and gives x = 7.2. The denser, thicker disc must have the smaller radius.

The idea to keep. When masses are equal, I depends only on the size. Use m = ρ × volume to turn density and thickness data into a radius ratio, and remember that a denser or thicker disc must be smaller.

Answer from the NTA final answer key. Solution by PrepWiser.