UPRVUNL’21 ME: Daily Practices Quiz. 2-Sep-2021

Quiz: Mechanical Engineering
Exam: UPRVUNL
Topic: Miscellaneous

Each question carries 1 mark
Negative marking: 1/4 mark
Time: 8 Minutes

Q1. The law which measure the temperature of the systems is?
(a) Zeroth law of thermodynamics
(b) First law of thermodynamics
(c) Second law of thermodynamics
(d) Third law of thermodynamics

Q2. A Carnot engine operates between 27℃ and 327℃. If the engine produces 300 kJ of work, what is the entropy change during heat addition?
(a) 0.5 kJ/K
(b) 1.0 kJ/K
(c) 1.5 kJ/K
(d) 2.0 kJ/K

Q3. Air injection in IC engine refers to injection of
(a) Air only
(b) Liquid fuel only
(c) Liquid fuel and air
(d) Supercharging air

Q4. For a system to be in thermodynamic equilibrium, the system and its surroundings
are to be in
(a) Thermal equilibrium
(b) Chemical equilibrium
(c) Mechanical equilibrium
(d) Thermal, chemical and mechanical equilibrium

Q5. A man is climbing up a ladder which is resting against a vertical wall. When he was exactly halfway up, the ladder started slipping. The path traced by the man is:
(a) Parabola
(b) Hyperbola
(c) Ellipse
(d) Circle

Q6. The static friction
(a) bears a constant ratio to the normal reaction between the two surfaces
(b) is independent of the area of contact, between the two surfaces
(c) always acts in a direction, opposite to that in which the body tends to move
(d) all of the above

Solutions
S1. Ans. (a)
Sol. Measurement of temperature is based on Zeroth law of thermodynamics.

S2. Ans. (b)
Sol. Given data
T2=27℃=27+273K
=300 K
T1=327℃=327+273K
= 600 K
W=300 kJ

S3. Ans. (c)
Sol. Air injection in IC engines refers to the injection of liquid fuel and air.

S4. Ans. (d)
Sol. A thermodynamic system will be in thermodynamic equilibrium with its surrounding when it is in thermal equilibrium (temperature is equal to surrounding), chemical equilibrium (chemical potential of system with respect to surrounding is zero) and mechanical equilibrium (Pressure of system is equal to that of surrounding) simultaneously.

S5. Ans.(d)
Sol.

P is mid-point on the ladder.
BP = AP=L/2,OC=CB=x
When x = 0, then y = L/2
(Point P lie on OA)
When y = 0, then x = L/2
This condition satisfies the circle equation,
(x)^2+ (y)^2=(L/2)^2
So, the path traced by point is circular.

S6. Ans. (d)
Sol. In case of static friction it is the ratio of Normal reaction between the surface to the friction force.
It is always opposing the motion or tending to motion of the body so always act in opposite direction.

Sharing is caring!