Polar modulus of a circular shaft of diameter D is 

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UKPSC JE Civil 8 May 2022 Official Paper-I
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  1. \(\rm\frac{\pi}{16}D^3\)
  2. \(\rm\frac{\pi}{32}D^4\)
  3. \(\rm\frac{D^2}{8}\)
  4. \(\rm\frac{\pi D^3}{8}\)

Answer (Detailed Solution Below)

Option 1 : \(\rm\frac{\pi}{16}D^3\)
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Detailed Solution

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Concept:

Torsion Equation of a shaft is given by,

\(\frac{T}{J} = \frac{τ }{r} = \frac{{Gθ }}{L}\)

Polar modulus is defined as the ratio of the polar moment of inertia to the radius of the shaft. It is also called as torsional section modulus. It is denoted by Zp.

Polar section modulus of shaft is given by,

\({Z_p} = \frac{J}{r} = \frac{{\frac{{\pi {D^4}}}{{32}}}}{{\frac{D}{2}}} = \frac{{\pi {D^3}}}{{16}}\)

Where, T = Torque, J = Polar moment of inertia, τ = Shear stress, r = Radius of shaft, G = Shear modulus, θ = Angle of twist and L = Length of shaft

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