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English SSC CGL Tier I papers. Free 60-minute mocks, 100 questions, 200 marks; +2 correct and −0.5 incorrect. Source: SSC response sheets preserved by SSC Portal. The marked archived keys have not been independently revalidated against final SSC keys; worked explanations are pending review. A/B/C/D correspond to source options 1/2/3/4.

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Question 1 · Mechanical Engineering (SSC JE) · Thermodynamics
The zeroth law of thermodynamics is the basis for the measurement of
Choose your answer for question 1
Answer and explanation

Correct answer: D

If two bodies are each in thermal equilibrium with a third, they are in equilibrium with each other. This makes a thermometer meaningful.

Question 2 · Mechanical Engineering (SSC JE) · Thermodynamics
A Carnot engine works between reservoirs at 800 K and 300 K. Its efficiency is
Choose your answer for question 2
Answer and explanation

Correct answer: D

η = 1 − T2/T1 = 1 − 300/800 = 0.625.

Question 3 · Mechanical Engineering (SSC JE) · Thermodynamics
The value of the universal gas constant is
Choose your answer for question 3
Answer and explanation

Correct answer: C

R̄ = 8.314 kJ/kmol·K. The gas constant of air, 0.287 kJ/kg·K, is R̄ divided by the molar mass 28.97.

Question 4 · Mechanical Engineering (SSC JE) · Thermodynamics
For a closed system undergoing a complete cycle, the net heat transfer is equal to
Choose your answer for question 4
Answer and explanation

Correct answer: D

Over a cycle ΔU = 0, so by the first law ∮δQ = ∮δW.

Question 5 · Mechanical Engineering (SSC JE) · Thermodynamics
The ratio of specific heats γ for air at ordinary temperatures is about
Choose your answer for question 5
Answer and explanation

Correct answer: D

Air is mostly diatomic, with γ = Cp/Cv ≈ 1.4. Monatomic gases give 1.67.

Question 6 · Mechanical Engineering (SSC JE) · Thermodynamics
In a reversible adiabatic process, the entropy of the system
Choose your answer for question 6
Answer and explanation

Correct answer: C

With δQ = 0 and no irreversibility, dS = δQrev/T = 0. Such a process is called isentropic.

Question 7 · Mechanical Engineering (SSC JE) · Thermodynamics
For an ideal gas, Cp − Cv is equal to
Choose your answer for question 7
Answer and explanation

Correct answer: A

This is Mayer’s relation, Cp − Cv = R, where R is the specific gas constant of that gas.

Question 8 · Mechanical Engineering (SSC JE) · Thermodynamics
A throttling process is
Choose your answer for question 8
Answer and explanation

Correct answer: B

In throttling through a valve or porous plug, h1 = h2 (adiabatic, no work, negligible change in kinetic energy), but entropy increases.

Question 9 · Mechanical Engineering (SSC JE) · Thermodynamics
The air standard efficiency of an Otto cycle with compression ratio 8 and γ = 1.4 is about
Choose your answer for question 9
Answer and explanation

Correct answer: B

η = 1 − 1/r^(γ−1) = 1 − 1/8^0.4 = 1 − 1/2.297 = 0.565.

Question 10 · Mechanical Engineering (SSC JE) · Thermodynamics
The critical temperature of water is about
Choose your answer for question 10
Answer and explanation

Correct answer: A

At 374 °C and 221 bar the liquid and vapour phases become indistinguishable.

Question 11 · Mechanical Engineering (SSC JE) · Thermodynamics
The Kelvin-Planck statement of the second law deals with
Choose your answer for question 11
Answer and explanation

Correct answer: A

No cyclic heat engine can convert all the heat from a single reservoir into work. The Clausius statement deals with refrigerators.

Question 12 · Mechanical Engineering (SSC JE) · Thermodynamics
Reheating the steam in a Rankine cycle mainly
Choose your answer for question 12
Answer and explanation

Correct answer: A

Passing the partly expanded steam through the boiler again reduces moisture in the last stages, which protects the blades.

Question 13 · Mechanical Engineering (SSC JE) · Thermodynamics
A rigid vessel of 0.5 m³ contains air at 2 bar and 300 K. Taking R = 0.287 kJ/kg·K, the mass of air is
Choose your answer for question 13
Answer and explanation

Correct answer: C

m = pV/RT = (200 × 0.5) / (0.287 × 300) = 100 / 86.1 = 1.16 kg.

Question 14 · Mechanical Engineering (SSC JE) · Thermodynamics
One kg of air at 300 K is compressed isothermally from 1 bar to 4 bar (R = 0.287 kJ/kg·K). The work done on the air is about
Choose your answer for question 14
Answer and explanation

Correct answer: C

W = RT ln(p2/p1) = 0.287 × 300 × ln 4 = 86.1 × 1.386 = 119.4 kJ.

Question 15 · Mechanical Engineering (SSC JE) · Thermodynamics
A heat engine receives 500 kJ of heat and rejects 300 kJ. Its thermal efficiency is
Choose your answer for question 15
Answer and explanation

Correct answer: D

W = 500 − 300 = 200 kJ. η = 200/500 = 0.4.

Question 16 · Mechanical Engineering (SSC JE) · Thermodynamics
A refrigerator has a COP of 4. When the same machine is used as a heat pump, its COP is
Choose your answer for question 16
Answer and explanation

Correct answer: A

COP(heat pump) = COP(refrigerator) + 1, because Q_H = Q_L + W.

Question 17 · Mechanical Engineering (SSC JE) · Thermodynamics
Heat of 600 kJ is added reversibly to a system at a constant temperature of 300 K. The change of entropy of the system is
Choose your answer for question 17
Answer and explanation

Correct answer: D

ΔS = Q/T = 600/300 = 2 kJ/K.

Question 18 · Mechanical Engineering (SSC JE) · Thermodynamics
Wet steam has a dryness fraction of 0.9. If hf = 500 kJ/kg and hfg = 2000 kJ/kg, its enthalpy is
Choose your answer for question 18
Answer and explanation

Correct answer: D

h = hf + x hfg = 500 + 0.9 × 2000 = 2300 kJ/kg.

Question 19 · Mechanical Engineering (SSC JE) · Thermodynamics
Among the Otto, Diesel and Dual cycles with the same compression ratio and the same heat input, the air standard efficiency is highest for the
Choose your answer for question 19
Answer and explanation

Correct answer: A

For the same compression ratio, Otto > Dual > Diesel, because all the heat in the Otto cycle is added at constant volume (at the highest mean temperature).

Question 20 · Mechanical Engineering (SSC JE) · Thermodynamics
According to Dalton’s law of partial pressures, the total pressure of a gas mixture is
Choose your answer for question 20
Answer and explanation

Correct answer: A

Each ideal gas in the mixture exerts the pressure it would exert if it alone occupied the volume.

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