Engineering Questions with Answers - Multiple Choice Questions

# Mechanical Operations MCQ’s – Power

1 - Question

Using Rittinger’s law, calculate the power required to crush 12 ton/hr of rocks, if 10 ton/hr rocks were crushed at a power of 9 hp. The diameter of the rocks was reduced from 3 cm to 0.5 cm in each case.
a) 10.7 hp
b) 11 hp
c) 10.8 hp
d) 12 hp

Explanation: Given,
Power required to crush 9 ton/hr of rocks = 9 hp
Initial diameter of the rocks (d) = 3cm
Final diameter of the crushed product (D) = 0.5 cm
To find: Power required to crush 12 ton/hr of rocks
Mathematically, Rittinger’s law can be written as: Pm = Kr (1D1d)
We first calculate the Rittinger’s constant, substituting the following
P = 9 hp
m = 10 ton/hr
d = 3 cm
D = 0.5 cm
Therefore, Kr = 0.54 hpcmhrton
Substituting this value of Kr along with m = 12 ton/hr
d = 3 cm
D = 0.5 cm
We get P = 10.8 hp

2 - Question

Work index is the gross energy required to reduce very large feed to a size such that, 65 percent of the product passes through a 100 – µm mesh screen.
a) True
b) False

Explanation: Work index is the gross energy required to reduce very large feed to a size such that, 80 percent of the product passes through a 100 – µm mesh screen. Work index is represented by Wi, which can be mathematically written as, Wi = kb100×103 where Kb is the Bond’s constant.

3 - Question

Calculate the power required in Watts, if the work done can be given as w = 3t4 – 16t at t = 5 seconds.
a) 1795 W
b) 1420 W
c) 1484 W
d) 1675 W

Explanation: Power is defined as the rate of doing work. If a function of work, is given in terms of time, then power can be obtained by differentiating the function with respect to time.
w = 3t4 – 16t
dwdt = 12t3 – 16
At t = 5 seconds P = dwdt = 1484 W

4 - Question

What is the unit of Kick’s constant?
a) kg / kW
b) kW * s/kg
c) kW * kg/s
d) kg * s/kW

Explanation: Kick’s law can represented as: Pm = kkln⁡(DsaDsb) where kk is Kick’s constant. Dsa and Dsb are initial and final diameters respectively. The ln() function always gives a constant, and is therefore dimensionless. The units of kk are therefore the same as that of P/m.
Pm=kWkgs. Therefore, the units of kk are kW * s/kg

5 - Question

What are the dimensions of power?
a) [ML2T-3]
b) [ML2T-1]
c) [ML1T-3]
d) [ML3T-3]

Explanation: The dimensions can be understood by simplifying the formula of power.
Power = WorkTime
Power = ForcedisplacementTime
Power = MassaccelerationdisplacementTime
Substituting the dimensions, Mass = [M]
Acceleration = [LT-2]
Displacement = [L]
Time = [T]
We get, Power = [ML2T-3]

6 - Question

Which of the following is the correct formula of Bond’s law?
a) Pm = kb(1Dsb+1Dsa)
b) P = kb(1Dsb1Dsa)
c) Pm = kb(1Dsb1Dsa)
d) P = kb(1Dsb+1Dsa)

Explanation: Bond’s law states that, the work required for crushing is proportional to the difference of reciprocal of feed diameter and reciprocal of product diameter. Here, P is the power required, m is the mass flow rate, Dsb is the product diameter, Dsa is the feed diameter and Kb is Bond’s constant.

7 - Question

What is the unit of Bond’s constant?
a) kW mm1/2 s/kg
b) kW mm s/kg
c) kW mm1/2 kg/s
d) kW mm kg/s

Explanation: Mathematically, Kick’s law can be written as Pm = kb(1Dsb1Dsa).
Now, when we substitute the units of Power ‘P’, mass flow rate ‘m’, diameters Dsa and Dsb, we get the units of Bond’s constant as = kWmmkgs.

8 - Question

The force applied on a body can be given as F = 4i + 2j + 6k. The displacement of the body is given by s = 3i + 9k. Calculate the power at t = 2 seconds.
a) 66 W
b) 33 W
c) 30 W
d) 15 W

Explanation: Power = Work/Time
Work is given by the dot product of force and displacement. W = F . s
W = (4i + 2j + 6k) . (3i + 9k)
Therefore, W = (4 × 3) + (6 × 9)
W = 66 Joules
Power = W/t
P = 66/2
P = 33 W

9 - Question

Which of the following is not a unit of power?
a) erg/second
b) kgm2s-3
c) Horsepower
d) Joule

Explanation: erg/second is the unit of power in the CGS system of units. kgm2s-3 which can also be written as Watts, is the unit of power in the SI system of units. Horsepower is the unit of power which is used usually in reference to output of engines. However, Joule is the unit of work.

10 - Question

Power is the time derivative of which quantity?
a) Energy
b) Acceleration
c) Velocity
d) Position