The polar modulus for a hollow shaft of outer diameter (D) and inner diameter (d) is

A $$\frac{\pi }{4}\left( {\frac{{{{\text{D}}^2} - {{\text{d}}^2}}}{{\text{D}}}} \right)$$
B $$\frac{\pi }{{16}}\left( {\frac{{{{\text{D}}^3} - {{\text{d}}^3}}}{{\text{D}}}} \right)$$
C $$\frac{\pi }{{16}}\left( {\frac{{{{\text{D}}^4} - {{\text{d}}^4}}}{{\text{D}}}} \right)$$
D $$\frac{\pi }{{32}}\left( {\frac{{{{\text{D}}^4} - {{\text{d}}^4}}}{{\text{D}}}} \right)$$

Correct Answer: $$\frac{\pi }{{16}}\left( {\frac{{{{\text{D}}^4} - {{\text{d}}^4}}}{{\text{D}}}} \right)$$

Bissoy MCQ
#Mechanical Engineering #Strength Of Materials In ME

Related Questions

The polar moment of inertia of a hollow shaft of outer diameter (D) and inner diameter (d) is
A $$\frac{\pi }{{16}}\left( {{{\text{D}}^3} - {{\text{d}}^3}} \right)$$
B $$\frac{\pi }{{16}}\left( {{{\text{D}}^4} - {{\text{d}}^4}} \right)$$
C $$\frac{\pi }{{32}}\left( {{{\text{D}}^4} - {{\text{d}}^4}} \right)$$
D $$\frac{\pi }{{64}}\left( {{{\text{D}}^4} - {{\text{d}}^4}} \right)$$
The strain energy stored in a hollow circular shaft of outer diameter (D) and inner diameter (d) subjected to shear stress is
A $$\frac{{{\tau ^2}}}{{2{\text{C}}}}\left( {\frac{{{{\text{D}}^2} - {{\text{d}}^2}}}{{\text{D}}}} \right) \times {\text{Volume of shaft}}$$
B $$\frac{{{\tau ^2}}}{{2{\text{C}}}}\left( {\frac{{{{\text{D}}^2} + {{\text{d}}^2}}}{{\text{D}}}} \right) \times {\text{Volume of shaft}}$$
C $$\frac{{{\tau ^2}}}{{4{\text{C}}}}\left( {\frac{{{{\text{D}}^2} - {{\text{d}}^2}}}{{\text{D}}}} \right) \times {\text{Volume of shaft}}$$
D $$\frac{{{\tau ^2}}}{{4{\text{C}}}}\left( {\frac{{{{\text{D}}^2} + {{\text{d}}^2}}}{{\text{D}}}} \right) \times {\text{Volume of shaft}}$$
The torque transmitted by a hollow shaft of outer diameter (D) and inner diameter (d) is
A $$\frac{\pi }{4} \times \tau \left( {\frac{{{{\text{D}}^2} - {{\text{d}}^2}}}{{\text{D}}}} \right)$$
B $$\frac{\pi }{{16}} \times \tau \left( {\frac{{{{\text{D}}^3} - {{\text{d}}^3}}}{{\text{D}}}} \right)$$
C $$\frac{\pi }{{16}} \times \tau \left( {\frac{{{{\text{D}}^4} - {{\text{d}}^4}}}{{\text{D}}}} \right)$$
D $$\frac{\pi }{{32}} \times \tau \left( {\frac{{{{\text{D}}^4} - {{\text{d}}^4}}}{{\text{D}}}} \right)$$
Two shafts A and B are made of the same material. The diameter of shaft A is twice as that of shaft B. The power transmitted by the shaft A will be __________ of shaft B.
A Twice
B Four times
C Eight times
D Sixteen times
The polar modulus for a solid shaft of diameter (D) is
A $$\frac{{\pi {{\text{D}}^2}}}{4}$$
B $$\frac{{\pi {{\text{D}}^3}}}{{16}}$$
C $$\frac{{\pi {{\text{D}}^3}}}{{32}}$$
D $$\frac{{\pi {{\text{D}}^4}}}{{64}}$$
Two shafts ‘A’ and ‘B’ under pure torsion are of identical length and identical weight and are made of the same material. The shaft ‘A’ is solid and the shaft ‘B’ is hollow. We can say that
A Shaft B is better than shaft A
B Shaft A is better than shaft B
C Both the shafts are equally good
D None of these
The relation between Young's modulus (E), shear modulus (C) and bulk modulus (K) is given by
A $${\text{E}} = \frac{{3{\text{KC}}}}{{3{\text{K}} + {\text{C}}}}$$
B $${\text{E}} = \frac{{6{\text{KC}}}}{{3{\text{K}} + {\text{C}}}}$$
C $${\text{E}} = \frac{{9{\text{KC}}}}{{3{\text{K}} + {\text{C}}}}$$
D $${\text{E}} = \frac{{12{\text{KC}}}}{{3{\text{K}} + {\text{C}}}}$$
If the modulus of elasticity for a given material is twice its modulus of rigidity, then bulk modulus is equal to
A 2C
B 3C
C $$\frac{{2{\text{C}}}}{3}$$
D $$\frac{{3{\text{C}}}}{2}$$
Which of the following relationships is correct for relating the three elastic constants of an isotropic elastic material (where, E = Young's modulus, G = Modulus of rigidity or shear modulus v = Poisson's ratio)?
A E = 2G (1 + v)
B E = G (1 + v)
C $${\text{E}} = \frac{{{\text{G}}\left( {1 + {\text{v}}} \right)}}{2}$$
D E = 2G (1 + 2v)
A hollow shaft will transmit a __________ torque than a solid shaft of the same mass and same material.
A Smaller
B Greater
C Medium
D None of these

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