This set of Mechanical Operations Multiple Choice Questions & Answers (MCQs) focuses on “Power”.

1. 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

View Answer

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: \(\frac {P}{m}\) = Kr (\(\frac {1}{D} – \frac {1}{d}\))

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 \(\frac {hp * cm * hr}{ton}\)

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. 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

View Answer

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 W

_{i}, which can be mathematically written as, W

_{i}= \(\frac {k_{b}}{\sqrt {100×10^{-3}}}\) where K

_{b}is the Bond’s constant.

3. Calculate the power required in Watts, if the work done can be given as w = 3t^{4} – 16t at t = 5 seconds.

a) 1795 W

b) 1420 W

c) 1484 W

d) 1675 W

View Answer

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 = 3t

^{4}– 16t

\(\frac {dw}{dt}\) = 12t

^{3}– 16

At t = 5 seconds P = \(\frac {dw}{dt}\) = 1484 W

4. What is the unit of Kick’s constant?

a) kg / kW

b) kW * s/kg

c) kW * kg/s

d) kg * s/kW

View Answer

Explanation: Kick’s law can represented as: \(\frac {P}{m}\) = k

_{k}ln\(( \frac {D_{sa}}{D_{sb}} )\) where k

_{k}is Kick’s constant. D

_{sa}and D

_{sb}are initial and final diameters respectively. The ln() function always gives a constant, and is therefore dimensionless. The units of k

_{k}are therefore the same as that of P/m.

\(\frac {P}{m} = \frac {kW}{\frac {kg}{s}}\). Therefore, the units of k

_{k}are kW * s/kg

5. What are the dimensions of power?

a) [ML^{2}T^{-3}]

b) [ML^{2}T^{-1}]

c) [ML^{1}T^{-3}]

d) [ML^{3}T^{-3}]

View Answer

Explanation: The dimensions can be understood by simplifying the formula of power.

Power = \(\frac {Work}{Time}\)

Power = \(\frac {Force * displacement}{Time}\)

Power = \(\frac {Mass * acceleration * displacement}{Time}\)

Substituting the dimensions, Mass = [M]

Acceleration = [LT

^{-2}]

Displacement = [L]

Time = [T]

We get, Power = [ML

^{2}T

^{-3}]

6. Which of the following is the correct formula of Bond’s law?

a) \(\frac {P}{m}\) = k_{b}\((\frac {1}{\sqrt {D_{sb}}} + \frac {1}{\sqrt {D_{sa}}}) \)

b) P = k_{b}\((\frac {1}{\sqrt {D_{sb}}} – \frac {1}{\sqrt {D_{sa}}}) \)

c) \(\frac {P}{m}\) = k_{b}\((\frac {1}{\sqrt {D_{sb}}} – \frac {1}{\sqrt {D_{sa}}}) \)

d) P = k_{b}\((\frac {1}{\sqrt {D_{sb}}} + \frac {1}{\sqrt {D_{sa}}}) \)

View Answer

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, D

_{sb}is the product diameter, D

_{sa}is the feed diameter and K

_{b}is Bond’s constant.

7. What is the unit of Bond’s constant?

a) kW mm^{1/2} s/kg

b) kW mm s/kg

c) kW mm^{1/2} kg/s

d) kW mm kg/s

View Answer

Explanation: Mathematically, Kick’s law can be written as \(\frac {P}{m}\) = k

_{b}\((\frac {1}{\sqrt {D_{sb}}} – \frac {1}{\sqrt {D_{sa}}}) \).

Now, when we substitute the units of Power ‘P’, mass flow rate ‘m’, diameters D

_{sa}and D

_{sb}, we get the units of Bond’s constant as = \(\frac {kW\sqrt {mm}}{\frac {kg}{s}}\).

8. 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

View Answer

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. Which of the following is not a unit of power?

a) erg/second

b) kgm^{2}s^{-3}

c) Horsepower

d) Joule

View Answer

Explanation: erg/second is the unit of power in the CGS system of units. kgm

^{2}s

^{-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. Power is the time derivative of which quantity?

a) Energy

b) Acceleration

c) Velocity

d) Position

View Answer

Explanation: The time derivative of acceleration is called jerk. Acceleration is the time derivative of velocity. When position is differentiated with respect to time, we get velocity. Power is the time derivative of energy and can be mathematically represented as P = dE/dT.

**Sanfoundry Global Education & Learning Series – Mechanical Operations.**

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