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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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Civil Engineering (SSC JE)

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Question 1 · Electrical Engineering (SSC JE) · Basic Electrical Engineering
A 12 V battery is connected across a 4 Ω resistor. The current drawn is
Choose your answer for question 1
Answer and explanation

Correct answer: A

By Ohm’s law I = V/R = 12/4 = 3 A.

Question 2 · Electrical Engineering (SSC JE) · Basic Electrical Engineering
A heater rated 230 V, 1150 W has a resistance of
Choose your answer for question 2
Answer and explanation

Correct answer: D

R = V²/P = 230² / 1150 = 52 900 / 1150 = 46 Ω.

Question 3 · Electrical Engineering (SSC JE) · Basic Electrical Engineering
The length of a wire is doubled and its diameter is halved. Its resistance becomes
Choose your answer for question 3
Answer and explanation

Correct answer: D

R = ρl/A. Length ×2 and area ÷4 (area ∝ d²) give R × 2 × 4 = 8 times.

Question 4 · Electrical Engineering (SSC JE) · Basic Electrical Engineering
Two resistors of 10 Ω and 15 Ω are connected in parallel. The equivalent resistance is
Choose your answer for question 4
Answer and explanation

Correct answer: C

R = (10 × 15)/(10 + 15) = 150/25 = 6 Ω.

Question 5 · Electrical Engineering (SSC JE) · Basic Electrical Engineering
Kirchhoff’s current law is based on the law of conservation of
Choose your answer for question 5
Answer and explanation

Correct answer: A

At a node, the charge entering equals the charge leaving, so the algebraic sum of currents is zero.

Question 6 · Electrical Engineering (SSC JE) · Basic Electrical Engineering
One kilowatt-hour is equal to
Choose your answer for question 6
Answer and explanation

Correct answer: D

1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J.

Question 7 · Electrical Engineering (SSC JE) · Basic Electrical Engineering
The energy stored in a 100 μF capacitor charged to 100 V is
Choose your answer for question 7
Answer and explanation

Correct answer: D

E = ½CV² = 0.5 × 100 × 10⁻⁶ × 10⁴ = 0.5 J.

Question 8 · Electrical Engineering (SSC JE) · Basic Electrical Engineering
The temperature coefficient of resistance of a semiconductor is
Choose your answer for question 8
Answer and explanation

Correct answer: D

Heating frees more charge carriers in a semiconductor, so its resistance falls as temperature rises. In metals it rises.

Question 9 · Electrical Engineering (SSC JE) · Basic Electrical Engineering
The energy stored in an inductor of 2 H carrying 3 A is
Choose your answer for question 9
Answer and explanation

Correct answer: B

E = ½LI² = 0.5 × 2 × 9 = 9 J.

Question 10 · Electrical Engineering (SSC JE) · Basic Electrical Engineering
The time constant of a series RL circuit with R = 10 Ω and L = 0.5 H is
Choose your answer for question 10
Answer and explanation

Correct answer: A

τ = L/R = 0.5/10 = 0.05 s.

Question 11 · Electrical Engineering (SSC JE) · Basic Electrical Engineering
The SI unit of conductance is
Choose your answer for question 11
Answer and explanation

Correct answer: A

Conductance G = 1/R and is measured in siemens (formerly mho).

Question 12 · Electrical Engineering (SSC JE) · Basic Electrical Engineering
The magnitude of the charge on an electron is approximately
Choose your answer for question 12
Answer and explanation

Correct answer: D

1.6 × 10⁻¹⁹ C is the elementary charge. 9.1 × 10⁻³¹ kg is the mass of an electron.

Question 13 · Electrical Engineering (SSC JE) · Basic Electrical Engineering
A 6 Ω and a 3 Ω resistor in parallel are connected in series with a 4 Ω resistor across a 12 V supply. The current drawn from the supply is
Choose your answer for question 13
Answer and explanation

Correct answer: D

6 Ω ∥ 3 Ω = 2 Ω. Total = 2 + 4 = 6 Ω. I = 12/6 = 2 A.

Question 14 · Electrical Engineering (SSC JE) · Basic Electrical Engineering
Resistors of 20 kΩ and 30 kΩ are connected in series across a 100 V supply. The voltage across the 30 kΩ resistor is
Choose your answer for question 14
Answer and explanation

Correct answer: D

V = 100 × 30 / (20 + 30) = 60 V.

Question 15 · Electrical Engineering (SSC JE) · Basic Electrical Engineering
A 2 kW heater runs for 3 hours every day for 30 days. At ₹6 per unit, the monthly cost of the energy is
Choose your answer for question 15
Answer and explanation

Correct answer: C

Energy = 2 × 3 × 30 = 180 kWh. Cost = 180 × 6 = ₹1,080.

Question 16 · Electrical Engineering (SSC JE) · Basic Electrical Engineering
A battery of emf 12 V and internal resistance 0.5 Ω supplies a 5.5 Ω load. The terminal voltage is
Choose your answer for question 16
Answer and explanation

Correct answer: B

I = 12/(0.5 + 5.5) = 2 A. Terminal voltage = 12 − 2 × 0.5 = 11 V.

Question 17 · Electrical Engineering (SSC JE) · Basic Electrical Engineering
Capacitors of 6 μF and 3 μF are connected in series. The equivalent capacitance is
Choose your answer for question 17
Answer and explanation

Correct answer: B

C = (6 × 3)/(6 + 3) = 2 μF.

Question 18 · Electrical Engineering (SSC JE) · Basic Electrical Engineering
If the distance between the plates of a parallel plate capacitor is doubled, keeping everything else the same, the capacitance
Choose your answer for question 18
Answer and explanation

Correct answer: A

C = εA/d, so C is inversely proportional to the plate separation d.

Question 19 · Electrical Engineering (SSC JE) · Basic Electrical Engineering
A resistor has a resistance of 10 Ω at 20 °C and a temperature coefficient of 0.004 per °C at 20 °C. Its resistance at 70 °C is
Choose your answer for question 19
Answer and explanation

Correct answer: A

Rt = R0 [1 + α (T − T0)] = 10 × [1 + 0.004 × 50] = 12 Ω.

Question 20 · Electrical Engineering (SSC JE) · Basic Electrical Engineering
A current of 5 A flows for 2 minutes through a 10 Ω resistor. The heat produced is
Choose your answer for question 20
Answer and explanation

Correct answer: B

H = I²Rt = 25 × 10 × 120 = 30 000 J.

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