# Farm Machinery Questions and Answers – Numericals on Estimation of Tractor Power

This set of Farm Machinery online test focuses on “Numericals on Estimation of Tractor Power”.

1. What will be the interest per hour when the cost of 35hp tractor is ₨ 3,00,000/-, life of tractor is 10 years, rate of interest 10% and working hours per year is 1000 hours?
a) Rs 16.50
b) Rs 17.34
c) Rs 15.30
d) Rs 14.28

Explanation: Interest = $$\frac{C+S}{2}*\frac{i}{h}=\frac{[300000+(300000*0.1)]}{2}*\frac{10}{100*100}=\frac{330000}{2}*\frac{1}{10000}$$ = Rs 16.50.

2. A motor car has a wheel base of 2.73 m and pivot centre of 1.065 m. the front and rear wheel track is 1.217 m. What will be angle of outside lock and turning circle radius of the outer front and inner rear wheels when the angle of inside lock is 40°?
a) 34.6°
b) 42°
c) 32.4°
d) 29°

Explanation: cot φ – cot θ = $$\frac{c}{b}$$
cot φ = $$\frac{c}{b}$$ + cot θ
cot φ = $$\frac{1.065}{2.743}$$ + cot⁡ 40°
cot φ = 0.388 + 1.19175 = 1.579
cot φ = 1.58m
φ = cot-1 (1.58) = 32.4°.

3. A straight tooth gear having involute profile with 20° pressure angle transmits 20Nm torque. Pitch diameter of the gear is 24 cm. If only one pair of tooth is engaged at any instant of time, the force experienced by the tooth will be ______
a) 78.30 N
b) 88.70 N
c) 156.60 N
d) 177.36 N

Explanation: Torque = Ft * r
20 = Ft * 0.12
Ft = 20/0.12 = 166.67 N
Ft = F cos φ
F = $$\frac{Ft}{cosφ} = \frac{166.67}{cos20°}$$ = 177.36 N.

4. A flange mounted shear pin is used on a shaft as a safety device. The steel shear pin has a diameter of 2.38 mm and is to be mounted on the flange of shaft rotating at 650 rpm. The maximum power transmitted by the shaft is 4.5 KW. If the shear strength of material of pin is 30 MPa, the radial distance of its mounting is ____
a) 5.02 mm
b) 11.98 mm
c) 47.94 mm
d) 301.20 mm

Explanation: Kw = 8.225 * N * rs * d2 * Ss * 10-8
4.5 = 8.225 * 650 * rs * (2.38)2 * 310
rs = 47.94 mm.

5. A simple band brake has the tight side of the band attached to a fixed pivot. The angle of wrap is 280° about a 450-mm diameter drum. A torque of 170 mm is sustained at 900 revs min-1. If the co-efficient of friction is 0.2 the required belt tension at tight side of the belt will be _____
a) 453.04N
b) 549.12N
c) 605.35N
d) 1211.53N

Explanation: $$\frac{T1}{T2}$$ = eμθ
$$\frac{T1}{T2}$$ = e0.2*280 * π/180
$$\frac{T1}{T2}$$ = 2.657
Power = (T1-T2)v
$$\frac{2\pi NT}{60}$$ = (T1-T2)*$$(\frac{\pi DN}{60})$$
2*170 = (T1-T2) * 0.45
T1-T2 = 340/0.45
T1 – $$\frac{T2}{2.657} = \frac{340}{0.45}$$
T1 = 1211.53 N.

6. In an epicyclic gear train, an arm carries two wheels ‘A’ and ‘B’ having 24 teeth and 30 teeth respectively. If the arm rotates at 100 rpm in the clockwise direction about the centre of the wheel. ‘A’ which is fixed, then the speed of wheel on its own axis is _____
a) 20 rpm anticlockwise
b) 25 rpm anticlockwise
c) 180 rpm clockwise
d) 225 rpm clockwise

Explanation: y=-100
X + y = 0
X = -y = -100
Speed of wheel B = Y – X $$\frac{Ta}{Tb}$$
= -100 * 100 * $$\frac{24}{30}$$
= -100-80 = -180
180 rpm clockwise.

7. A hydraulic circuit uses 25 litres of fluid per min. The fluid is supplied by a pump having a fixed displacement of 12.5 cm3 per revolution driven at 3000 rpm. The pump has a volumetric efficiency of 0.85 and torque efficiency of 0.88. If the system pressure is set at 18 MPa by the relief valve. What will be the power required to drive the pump?
a) 13.48 KW
b) 12.20 KW
c) 11.09 KW
d) 12.78 KW

Explanation: np = $$\frac{Dp* ∆p}{2πTp}$$
Tp = $$\frac{12.6*10^{-6}*18*10^6}{2π*0.88}$$=40.69 Nm
P required = 2π*Tp*Np
P = 2π*40.69*3000/60
P = 12.78 KW.

8. What will be the torque required to drive a 0.1 litre/min pump at 2000 rpm of delta pressure across pump is 240 bars?
a) 0.87 Nm
b) 0.992 Nm
c) 0.191 Nm
d) 0.367 Nm

Explanation: Ph = Qπ
= $$\frac{0.1*10^{-3}}{60}$$*240*106 = 40 W = 0.040 Kw
P = $$\frac{2πNT}{60000}$$
T = $$\frac{0.040*60000}{2π*2000}$$ = 0.191 KW.

Sanfoundry Global Education & Learning Series – Farm Machinery.