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A hollow circular tube has a mean radius of 8 cm and a wall thickness of 0.04 cm. It is melted and remade into a solid cylindrical rod of the same length. What is the ratio of the torsional rigidities of the tube to the rod?

1. ((8.02)⁴-(7.98)⁴)/(0.8)⁴
2. ((8.02)²-(7.98)²)/(0.8)²
3. ((8.02)⁴-(7.98)⁴)/(0.8)²
4. ((8.02)²-(7.98)²)/(0.8)⁴

Correct answer: ((8.02)⁴-(7.98)⁴)/(0.8)⁴

Solution

Tube J ~ (Ro^4 - Ri^4) = (8.02^4 - 7.98^4). Equal-volume solid rod: 2*pi*R*t = pi*r^2 -> r^2 = 2*8*0.04 = 0.64 -> r = 0.8, so J_rod ~ 0.8^4. Ratio = ((8.02)^4-(7.98)^4)/(0.8)^4.

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