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A tube's radius is doubled at constant wall thickness. Its torsion constant is multiplied by

  • A.8
  • B.2
  • C.4
  • D.16

Answer: A

For a thin circular tube \( J = 2\pi r^3 t \), so the cube on the radius gives a factor of \( 2^3 = 8 \). Enlarging the tube is therefore far more effective than thickening its wall, which would give only a factor of two.

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