This set of Farm Machinery Questions & Answers for Exams focuses on “Power Transmission System – 2”.

1. In a belt drive the mass of the belt is 1 kg/m and its speed is 6m/s. The drive transmits 9.6 KW of power. Determine the strength of the belt when the coefficient of friction is 0.25 and wrap angle is 220°.

a) 2629.78 N

b) 2404.73 N

c) 2003.09 N

d) 1829.78 N

View Answer

Explanation: T

_{c}= mv

^{2}= 1*6

^{2}= 36 N

\(\frac{T1}{T2}\)=e

^{μθ}

\(\frac{T1}{T2}=e^{0.25*220*\frac{\pi}{180}}\) = 2.6102

P = (T1-T2) * V

9.6*10

^{3}= (T1-T2) * 6

T1-T2 = 1600

T1 = 2593.78 N

T2 = 993.78 N

Strength = T1 + T

_{c}= 2629.78 N

2. What power is transmitted by belt and pulley arrangement, if tension on tight side of belt is 60 kg and tension on slack side is 25kg. The diameter and speed of driving pulley is 0.95 m and 250 rpm respectively.

a) 5.239 KW

b) 9.624 KW

c) 2.264 KW

d) 4.269 KW

View Answer

Explanation: peripheral velocity = \(\frac{πDN}{60}=\frac{π*0.95*250}{60}\) = 12.4354 m/s

P = (T1-T2) * V

P = (60-25) * 9.81 * 12.4354 = 4.269 KW

3. A counter shaft of engineering workshop is being driven by electric motor with the help of canvas belt and pulley arrangement such that the speed of belt is 600 m/min and it transmits 50 KW. What will be the difference between tensions on either side of the belt?

a) 10000

b) 100

c) 5000

d) 50

View Answer

Explanation: V = 600*10/60 = 10 m/s

P = (T1-T2) * V

50000 = (T1-T2) * 10

T1 – T2 = 5000

4. If a driven pulley is made to rotate at 350 rpm by a driving pulley of 0.3 m diameter, running at 1400 rpm, what will be the length of open belt drive when the pulleys are 2 m apart?

a) 7.14 m

b) 6.45 m

c) 8.88 m

d) 10 m

View Answer

Explanation: N

_{1}D

_{1}= N

_{2}D

_{2}

1400 * 0.3 = 350 * D

_{2}

D

_{2}= 1.2 m

Length of open belt drive = 2C + \(\frac{[π(D1+D2)]}{2}+\frac{(D1-D2)^2}{4C} \)

= 2*2 + \(\frac{[π(1.5)]}{2}+\frac{(0.9)^2}{8} \)

= 6.45 m

5. Two pulleys are connected by a belt. The sum of the diameter of two pulleys is 90 cm and while one makes 50 rpm, the other makes 20 rpm. What will be the diameters of the other pulley?

a) 14.1234 cm, 21.9876 cm

b) 34.357 cm, 24.0473 cm

c) 18.426 cm, 15.921 cm

d) 64.2857 cm, 25.7142 cm

View Answer

Explanation: D1 + D2 = 90 cm

\(\frac{D1}{D2}=\frac{50}{20}\) = 2.5

3.5 D2 = 90

D2 = 25.7142 cm

D1 = 64.2857 cm

6. While testing a wheat thresher at recommended throughout, 80 Nm torque is recorded at 750 rpm at the main shaft of the threshing cylinder, which is operated by a 200-mm diameter V pulley. The overload factor and unit mass of V-belt are 1.2 and 0.9 kg/m respectively. At condition of maximum power transfer the maximum tension in V-belt in N is ________

a) 166.55 N

b) 233.78 N

c) 120.06 N

d) 211.23 N

View Answer

Explanation: N = 750 rpm

V = \(\frac{π*0.2*750}{60}\) = 7.854 m/s

M = 0.9 kg/m

T

_{c}= mv

^{2}= 0.9 * (7.854)

^{2}

T

_{c}= 55.5165 N

At maximum power transmission

T

_{max}= 3 T

_{c}

T

_{max}= 3 * 55.5165 = 166.55 N

7. A V-belt operates at 23.56 m/s peripheral speed. The mass of the belt per unit length is 0.9 kg/m. The value of centrifugal tension is _______

a) 514 N

b) 523 N

c) 500 N

d) 599 N

View Answer

Explanation: V = 23. 56 m/s

M=0.9 kg/m

T

_{c}= mv

^{2}= 0.9 * (23.56)

^{2}

T

_{c}= 500 (approx.)

8. Determine the length of cross-belt of connecting two pulleys 4 m apart. The diameters of driving and driven pulleys are 1.25 m and 0.75 m respectively.

a) 12.36 m

b) 11.35 m

c) 14.28 m

d) 13.27 m

View Answer

Explanation: L = 2C + \(\frac{[π(D1+D2)]}{2}+\frac{(D1+D2)^2}{4C} \)

= 2*4 + \(\frac{[π(1.25+0.75)]}{2}+\frac{(2)^2}{4*4}\) = 11.35 m

9. A leather belt 100 mm wide and 10 mm thick with a safe permissible stress of 1.5 MPa is used for transmitting a maximum power of 15 KW. If the density of belt material is 1 g/cm^{3}, the velocity of belt for maximum power transmission will be ________

a) 22.36 m/s

b) 34.13 m/s

c) 24.98 m/s

d) 12.12 m/s

View Answer

Explanation: Area = 100*10*20

^{-6}= 10

^{-3}m

^{2}

Tmax = Pmax * Area

= 1.5 * 10 * 10

^{-3}

Tmax = 1500 N

M = ρ*area

= 1000*10

^{-3}= 1 kg/m

For maximum transmission

Tmax = 3T

_{c}

Tmax= 3mv

^{2}

V = \(\sqrt{\frac{Tmax}{3m}} = \sqrt{\frac{1500}{3}} \)

V = 22.36 m/s

**Sanfoundry Global Education & Learning Series – Farm Machinery.**

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