1. The infinitesimal amount of work done on changing the length of a wire with tension T from L to L+dL is
a) -FdL
b) FdL
c) -2FdL
d) 2FdL
Explanation: It is the work done in stretching a wire.
2. Work done in stretching a wire is given by
a) -∫FdL
b) -∫EɛALdɛ
c) -AEL∫ɛdɛ
d) all of the mentioned
Explanation: Above formulae arrive when we limit the problem within the elastic limit.
3. The surface tension acts to make the surface area of the liquid
a) maximum
b) minimum
c) zero
d) none of the mentioned
Explanation: Characteristic of surface tension.
4. The work done per unit volume on a magnetic material is dW=-Hdl where H is
a) field strength
b) magnetization field
c) induced current
d) none of the mentioned
Explanation: H is the field strength of the magnetic field.
5. The work is equal to the integral of the product of an intensive property.
a) true
b) false
Explanation: Along with it, change in work is related to the extensive property.
6. A 1200 hp engine has a drive shaft rotating at 2000 RPM. Find the torque on the shaft?
a) 2214 Nm
b) 3214 Nm
c) 4214 Nm
d) 5214 Nm
Explanation: Power, rate of work = T ω and 1 hp = 0.7355 kW = 735.5 W
ω = RPM × (2π)/(60 s) = 209.44 rad/s
T = power/ω = (1200 hp × 735.5 W/hp)/ 209.44 rad/s
= 4214 Ws = 4214 Nm.
7. A 1200 hp engine drives a car with a speed of 100 km/h. Find the force between the tires and the road?
a) 11.8 kN
b) 21.8 kN
c) 31.8 kN
d) 41.8 kN
Explanation: Power = F V and 1 hp = 0.7355 kW = 735.5 W
velocity in m/s: V = 100 × 1000 / 3600 = 27.78 m/s
F = (1200 × 735.5)/(27.78) = 31 771 N
= 31.8 kN.
8. A work of 2.5 kJ is delivered on a rod from a piston/cylinder where the air pressure is 500 kPa. What should be the diameter of cylinder to restrict the rod motion to maximum 0.5 m?
a) 0.013 m
b) 0.113 m
c) 0.213 m
d) 0.313 m
Explanation: W = ⌠P dV = ⌠PA dx = PA ∆x = P(π/4)(D^2)∆x
Putting the values in above equation we get
D = 0.113 m.
9. A force of 1.2 kN moves a car with 60 km/h up a hill. Find the power
a) 20 kW
b) 30 kW
c) 40 kW
d) 50 kW
Explanation: Power = F V = 1.2 kN × 60 (km/h)
= ( 1.2 × 10^3 × 60 )*(1000/3600)
= 20 000 W = 20 kW.
10. When a battery(12 V) is charged with 6 amp for 3 hours how much energy is delivered?
a) 477.6 kJ
b) 577.6 kJ
c) 677.6 kJ
d) 777.6 kJ
Explanation: W = ⌠(power)dt = V i ∆t
= 12 V × 6 Amp × 3 × 3600 s
= 777 600 J = 777.6 kJ.