The polar moment of inertia of a hollow circular shaft of outer diameter (D) and inner diameter (d) is

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  1. \(\frac{\pi}{16}\times (D^3-d^3)\)
  2. \(\frac{\pi}{16}\times (D^4-d^4)\)
  3. \(\frac{\pi}{32}\times (D^4-d^4)\)
  4. \(\frac{\pi}{64}\times (D^4-d^4)\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{\pi}{32}\times (D^4-d^4)\)
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Explanation:

Polar moment of inertia of Solid cylinder:

\({I_Z} = J = \frac{{\pi {D^4}}}{{32}}\)

Polar moment of inertia of Hollow Cylinder:

\({I_Z} = J = \frac{{\pi \left( {{D^4} - {d^4}} \right)}}{{32}}\)

where outer diameter = D and inner diameter = d

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