In fluid mechanics, which of the following statements about Pascal's law is correct?

The correct answer is A: Pressure applied to a confined fluid is transmitted equally in all directions. Pascal's law states that pressure applied to a confined fluid is transmitted undiminished in all directions. This principle is fundamental to hydraulic systems used in Indian industries.

A V-belt drive operates with a belt velocity of 20 m/s and transmits 30 kW. If the belt has a mass of 0.5 kg/m, what is the centrifugal tension in the belt?

The correct answer is D: 200 N. Centrifugal tension T_c = m × v² = 0.5 × 20² = 200 N

In a diesel engine, if the cutoff ratio is 1.5 and compression ratio is 16, calculate the thermal efficiency. (Assume γ = 1.4)

The correct answer is B: 63.5%. Diesel cycle efficiency = 1 - (1/r^(γ-1)) × [(r_c^γ - 1)/(γ(r_c - 1))] where r_c = cutoff ratio. With r=16, r_c=1.5, γ=1.4, efficiency ≈ 63.5%

A heat engine operates between two thermal reservoirs at temperatures 600 K and 300 K. What is the maximum possible efficiency of this engine?

The correct answer is B: 50%. Maximum efficiency of a heat engine = 1 - (T_cold/T_hot) = 1 - (300/600) = 0.50 or 50% (Carnot efficiency)

A reciprocating compressor compresses air from 1 bar, 300 K to 10 bar. If the process follows PVⁿ = constant with n=1.25, what is the specific work required per kg of air? (R=287 J/kg·K)

The correct answer is B: 220 kJ/kg. W = [nRT₁/(n-1)][(P₂/P₁)^((n-1)/n) - 1] = [1.25×287×300/0.25][10^0.2 - 1] ≈ 220 kJ/kg

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Mechanical Engineering
Thermodynamics

Thermodynamics, hydraulics, machine design

36 Q 3 Topics Take Mock Test

Difficulty: All Easy Medium Hard 21–30 of 36

Topics in Mechanical Engineering

All Thermodynamics 100 Fluid Mechanics 79 Machine Design 80

Q.21 Easy Thermodynamics

A system undergoes a cyclic process. The net work done by the system is 150 kJ and the net heat absorbed is 150 kJ. Which statement is correct?

A The process violates the first law of thermodynamics

B The process obeys the first law; ΔU_cycle = 0

C The surroundings have done negative work on the system

D The internal energy of the system has increased by 300 kJ

Correct Answer: B. The process obeys the first law; ΔU_cycle = 0

EXPLANATION

For any cycle: ΔU_cycle = 0. First law: Q_net = W_net confirms energy conservation. 150 = 150 ✓

Test

Q.22 Easy Thermodynamics

If the specific heat at constant volume (Cv) for a diatomic ideal gas is 5R/2, what is its specific heat at constant pressure (Cp)?

A 7R/2

B 5R/2

C 3R/2

D 9R/2

Correct Answer: A. 7R/2

EXPLANATION

Using Mayer's relation: Cp - Cv = R. Therefore, Cp = 5R/2 + R = 7R/2

Test

Q.23 Easy Thermodynamics

What is the SI unit of entropy?

A J/K

B J·K

C kJ/mol

D cal/K

Correct Answer: A. J/K

EXPLANATION

Entropy is defined as dS = dQ_rev/T. Units: Joules/Kelvin = J/K. This is the fundamental SI unit for entropy.

Test

Q.24 Easy Thermodynamics

In a constant volume process, 300 J of heat is removed from a gas. Calculate the work done and change in internal energy.

A W = 0, ΔU = -300 J

B W = -300 J, ΔU = 0

C W = 300 J, ΔU = -300 J

D W = 0, ΔU = 300 J

Correct Answer: A. W = 0, ΔU = -300 J

EXPLANATION

In constant volume process: W = ∫PdV = 0. From first law: ΔU = Q - W = -300 - 0 = -300 J

Test

Q.25 Easy Thermodynamics

Which of the following has the highest specific heat capacity among common substances at room temperature?

A Water

B Iron

C Air

D Copper

Correct Answer: A. Water

EXPLANATION

Water has exceptionally high specific heat (~4.18 kJ/kg·K), much higher than metals (iron ~0.46) and air (~1.01). This is due to hydrogen bonding in water.

Test

Q.26 Easy Thermodynamics

A gas undergoes an isobaric process. If 500 J of heat is added and the gas expands such that work done by the gas is 200 J, what is the change in internal energy?

A 300 J

B 700 J

C 250 J

D -300 J

Correct Answer: A. 300 J

EXPLANATION

From first law: ΔU = Q - W = 500 - 200 = 300 J. Internal energy increases by 300 J.

Test

Q.27 Easy Thermodynamics

Which thermodynamic property is NOT a state function?

A Internal energy

B Enthalpy

C Heat

D Entropy

Correct Answer: C. Heat

EXPLANATION

Heat (Q) and work (W) are path-dependent quantities, not state functions. Internal energy, enthalpy, and entropy are state functions depending only on initial and final states.

Test

Q.28 Easy Thermodynamics

A reversible adiabatic process for an ideal gas is also known as:

A Isothermal process

B Isobaric process

C Isentropic process

D Isochoric process

Correct Answer: C. Isentropic process

EXPLANATION

A reversible adiabatic process has constant entropy (dS = 0), making it isentropic. This is a key assumption in many thermodynamic analysis for ideal processes.

Test

Q.29 Easy Thermodynamics

What is the relationship between specific heats C_p and C_v for an ideal gas?

A C_p = C_v

B C_p - C_v = R

C C_p + C_v = R

D C_p × C_v = R

Correct Answer: B. C_p - C_v = R

EXPLANATION

The Mayer relation: C_p - C_v = R, where R is the specific gas constant. This relation holds for all ideal gases.

Test

Q.30 Easy Thermodynamics

What is the first law of thermodynamics in differential form?

A dU = δQ - δW

B dU = δQ + δW

C dU = δQ × δW

D δQ = dU + δW

Correct Answer: A. dU = δQ - δW

EXPLANATION

The correct form is dU = δQ - δW, where δW = PdV for expansion work. This represents energy conservation in thermodynamic systems.

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