SECTION-5

MACHINE DESIGN MCQ

##### Q21. The power transmitted by a belt or rope is given by

(a) (T1 – T2) × v/1000
(b) (T1 + T2) × v/1000
(c) T1 × v/1000
(d)  T2 × v/1000
where T1 = Tension in the tight side, T2 = Tension in the slack side, and
v = Linear speed of the belt in m/sec.

Ans:(b) (T1 + T2) × v/1000

##### Q22. The relation between the tension (T1) in the belt on the tight side to the tension (T2) on the slack side is given by

(a) T1/T2= μθ
(b) T1/T2= -μθ
(c) T1/T2= e– μθ
(d) T1/T2= eμθ.
where μ = Co-efficient of friction between the bolt and the pulley, and
θ = Angle of lap in radians.

Ans:(d) T1/T2= eμθ.

##### Q23. The centrifugal tension caused by centrifugal force is given by

(a) 1/2  mv2
(b) mv2/2g
(c) mv2 /g
(d) 2wv2/g
where v = Linear velocity of the belt, w = Weight per unit length of the belt, and
m = Mass of the belt.

Ans:(c) mv2 /g

##### Q24. If α is the angle of groove of a grooved pulley in which a rope is running, then ratio of tensions of the two sides of the rope is given by

(a) T1/T2= μθ
(b) T1/T2= eμα
(c)T1/T2= eμ1×θ.
(d) T1/T2= μ × θ × α.
where μ1 = μ cosec(α/2), μ = Co-efficient of friction between the rope and the grooved pulley, and θ = Angle of lap in radians.

Ans:(c)T1/T2= eμ1×θ.

##### Q25. The initial tension in the perfectly elastic material of the belt neglecting centrifugal tension is given by

(a) T1-T2/2
(b) T1+T2/2
(c)  2T1-T2/2
(d)  T1-2T2/2
where T1 = Tension on tight side, and T2 = Tension on slack side.

Ans:(b) T1+T2/2

##### Q26. The initial tension in the perfectly elastic material of the belt considering the centrifugal  tension is given by

(a) T1-T2-Tc/2
(b) T1+T2+Tc/2
(c) T1+T2+2Tc/2
(d) T1+T2-2Tc/2
where Tc = Centrifugal tension in the belt.

Ans:(c) T1+T2+2Tc/2

##### Q27. The mean width (b) of the collar in case of cotter joint is given by

(a) P/2t × τ
(b) 2P/t × τ
(c) P/t × τ
(d) 4P/t × τ
where P = Axial load, t = Thickness of cotter, and
τ = Permissible shear stress in the cotter material.

Ans:(a) P/2t × τ

##### Q28. The diameter of the spigot (or inside diameter of the socket) of a cotter joint is obtained from

$\text { (a) } P=\frac{\pi}{4} d_{1}^{2} \times \sigma_{t}$

(b) P = d1 × t × σt

$\text { (c) } P=\left(\frac{\pi}{4} d_{1}^{2}-d_{1} \times t\right) \sigma_{t}$

(d) P = 2d1 × t × τ
where P = Axial load, t = Thickness of cotter,
d1 = Diameter of spigot, σt = Permissible tensile stress, and
τ = Permissible shear stress.

Ans:$\text { (c) } P=\left(\frac{\pi}{4} d_{1}^{2}-d_{1} \times t\right) \sigma_{t}$

(a)  p ×D/2t
(b) p×D/4×t
(c)  p×D/8×t
(d)  2p ×D/t

Ans: (b) p×D/4×t

##### Q30.  In the above question, the two stresses developed in the material of the cylinder are called longitudinal stress and circumferential stress is given by

(a)  p ×D/2t
(b) p×D/4×t
(c)  p×D/8×t
(d)  2p ×D/t

Ans:(a)  p ×D/2t

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##### Read More Sections of Machine Design

Each section contains maximum 80 Questions. To practice more questions visit other sections.

Machine Design MCQ – Section-1

Machine Design MCQ – Section-2

Machine Design MCQ – Section-3

Machine Design MCQ – Section-4

Machine Design MCQ – Section-5

Machine Design MCQ – Section-6

Machine Design MCQ – Section-7