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Q. If a rectangle has a length of 8 units and a width of 5 units, what is its perimeter? (2023)

A. 26
B. 30
C. 20
D. 24

Solution

The perimeter of a rectangle is given by 2(length + width). Thus, perimeter = 2(8 + 5) = 2 * 13 = 26.

Correct Answer: A — 26

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Q. If a rectangle has a length that is twice its width and the perimeter is 48 cm, what is the area of the rectangle?

A. 96 cm²
B. 144 cm²
C. 192 cm²
D. 256 cm²

Solution

Let width = w, then length = 2w. Perimeter = 2(length + width) = 48. Thus, 2(2w + w) = 48, leading to 6w = 48, so w = 8 cm. Area = length * width = 2w * w = 2*8*8 = 128 cm².

Correct Answer: B — 144 cm²

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Q. If a region receives 1200 mm of rainfall in a year, what is the average monthly rainfall? (2022)

A. 100 mm
B. 120 mm
C. 150 mm
D. 200 mm

Solution

Average monthly rainfall = Total annual rainfall / 12 = 1200 mm / 12 = 100 mm.

Correct Answer: B — 120 mm

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Q. If a resistor has a current of 2A flowing through it and a voltage of 10V across it, what is the resistance? (2023)

A. 2Ω
B. 5Ω
C. 10Ω
D. 20Ω

Solution

Using Ohm's Law, R = V / I = 10V / 2A = 5Ω.

Correct Answer: B — 5Ω

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Q. If a resistor has a resistance of 10Ω and a current of 2A flows through it, what is the voltage across the resistor? (2023)

A. 5V
B. 10V
C. 20V
D. 30V

Solution

Using Ohm's Law, V = I * R = 2A * 10Ω = 20V.

Correct Answer: C — 20V

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Q. If a resistor has a resistance of 10Ω and a current of 5A flows through it, what is the voltage across the resistor? (2023)

A. 10V
B. 20V
C. 30V
D. 50V

Solution

Using Ohm's Law, V = I * R = 5A * 10Ω = 50V.

Correct Answer: B — 20V

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Q. If a resistor has a resistance of 5 ohms and a current of 2 amperes flows through it, what is the voltage across the resistor?

A. 10 V
B. 5 V
C. 2.5 V
D. 1 V

Solution

Using Ohm's Law, V = I * R = 2 A * 5 Ω = 10 V.

Correct Answer: A — 10 V

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Q. If a resistor has a resistance of 8Ω and the current flowing through it is 4A, what is the power dissipated by the resistor? (2022)

A. 16W
B. 32W
C. 64W
D. 8W

Solution

Power (P) can be calculated using P = I^2 * R = (4A)^2 * 8Ω = 16 * 8 = 128W.

Correct Answer: B — 32W

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Q. If a resistor has a resistance of 8Ω and the current flowing through it is 4A, what is the voltage across the resistor? (2022)

A. 32V
B. 16V
C. 8V
D. 4V

Solution

Using Ohm's Law, V = I * R = 4A * 8Ω = 32V.

Correct Answer: B — 16V

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Q. If a resistor is connected to a 9V battery and the current flowing through it is 3A, what is the resistance of the resistor?

A. 1 Ω
B. 3 Ω
C. 9 Ω
D. 27 Ω

Solution

Using Ohm's Law, R = V / I = 9 V / 3 A = 3 Ω.

Correct Answer: C — 9 Ω

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Q. If a right triangle has an angle of 30 degrees, what is the ratio of the lengths of the opposite side to the hypotenuse?

A. 1:2
B. 1:√3
C. √3:1
D. 2:1

Solution

In a 30-60-90 triangle, the ratio of the opposite side to the hypotenuse is 1:2.

Correct Answer: A — 1:2

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Q. If a right triangle has legs of lengths 3 cm and 4 cm, what is the length of the hypotenuse?

A. 5 cm
B. 6 cm
C. 7 cm
D. 8 cm

Solution

Using the Pythagorean theorem, hypotenuse = √(3² + 4²) = √(9 + 16) = √25 = 5 cm.

Correct Answer: A — 5 cm

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Q. If a right triangle has legs of lengths 6 and 8, what is the length of the hypotenuse?

A. 10 units
B. 12 units
C. 14 units
D. 16 units

Solution

Using the Pythagorean theorem: hypotenuse = √(6^2 + 8^2) = √(36 + 64) = √100 = 10 units.

Correct Answer: A — 10 units

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Q. If a river basin has an area of 2000 km² and receives an average rainfall of 800 mm per year, what is the total volume of water received by the basin in cubic meters? (2000)

A. 1600000000
B. 16000000
C. 160000000
D. 16000000000

Solution

Volume = Area × Rainfall = 2000 km² × 800 mm = 2000 × 10^6 m² × 0.8 m = 1600000000 m³.

Correct Answer: A — 1600000000

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Q. If a river basin has an area of 500 km² and receives an average rainfall of 800 mm annually, what is the total volume of water received by the basin in cubic meters? (2022)

A. 400,000,000 m³
B. 500,000,000 m³
C. 600,000,000 m³
D. 700,000,000 m³

Solution

Volume = Area × Rainfall = 500 km² × 800 mm = 500,000,000 m³.

Correct Answer: B — 500,000,000 m³

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Q. If a river deposits 500 kg of sediment per cubic meter of water, how much sediment is deposited when 1000 m³ of water flows? (2021)

A. 500 kg
B. 1000 kg
C. 2500 kg
D. 5000 kg

Solution

Sediment deposited = Sediment per m³ × Volume = 500 kg/m³ × 1000 m³ = 500,000 kg.

Correct Answer: D — 5000 kg

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Q. If a river deposits 5000 cubic meters of soil per day, how much soil will it deposit in a week? (2019)

A. 25000 m³
B. 35000 m³
C. 40000 m³
D. 45000 m³

Solution

Soil deposited in a week = 5000 m³/day × 7 days = 35000 m³.

Correct Answer: A — 25000 m³

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Q. If a river flows at a speed of 5 km/h and a boat can travel at 15 km/h in still water, what is the speed of the boat upstream? (2021)

A. 10 km/h
B. 5 km/h
C. 15 km/h
D. 20 km/h

Solution

The speed of the boat upstream = speed of boat in still water - speed of river = 15 km/h - 5 km/h = 10 km/h.

Correct Answer: A — 10 km/h

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Q. If a river flows through a valley with a gradient of 1 in 100, what is the vertical drop over a distance of 200 km? (2019)

A. 200 m
B. 400 m
C. 600 m
D. 800 m

Solution

Vertical drop = Distance × Gradient = 200,000 m × (1/100) = 2000 m.

Correct Answer: B — 400 m

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Q. If a river has a discharge of 500 cubic meters per second and the cross-sectional area of the river is 100 square meters, what is the velocity of the river?

A. 2 m/s
B. 5 m/s
C. 10 m/s
D. 20 m/s

Solution

Velocity = Discharge / Area = 500 m³/s / 100 m² = 5 m/s.

Correct Answer: B — 5 m/s

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Q. If a river has a discharge of 500 m³/s and the cross-sectional area of the river is 25 m², what is the velocity of the river?

A. 10 m/s
B. 20 m/s
C. 5 m/s
D. 15 m/s

Solution

Velocity = Discharge / Area = 500 m³/s / 25 m² = 20 m/s.

Correct Answer: C — 5 m/s

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Q. If a river has a sediment transport capacity of 200 kg/m³ and flows at a rate of 10 m³/s, what is the total sediment transported in one hour? (2000)

A. 72000 kg
B. 60000 kg
C. 360000 kg
D. 120000 kg

Solution

Total sediment = Sediment capacity × Flow rate × Time = 200 kg/m³ × 10 m³/s × 3600 s = 7200000 kg.

Correct Answer: C — 360000 kg

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Q. If a river has a slope of 1 in 1000, what is the vertical drop over a distance of 5 km?

A. 5 m
B. 10 m
C. 15 m
D. 20 m

Solution

Vertical drop = Slope × Distance = (1/1000) × 5000 m = 5 m.

Correct Answer: A — 5 m

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Q. If a river is 300 meters wide and the current flows at 2 m/s, how long will it take a swimmer to cross the river if he swims at a speed of 1 m/s perpendicular to the current?

A. 300 seconds
B. 150 seconds
C. 200 seconds
D. 100 seconds

Solution

Time to cross = Width / Speed across = 300 m / 1 m/s = 300 seconds.

Correct Answer: B — 150 seconds

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Q. If a river's current is 3 km/h and a boat travels downstream at 12 km/h, what is the speed of the boat relative to the riverbank? (2022)

A. 9 km/h
B. 15 km/h
C. 12 km/h
D. 3 km/h

Solution

Speed of boat relative to riverbank = Speed of boat + Speed of river = 12 km/h + 3 km/h = 15 km/h.

Correct Answer: B — 15 km/h

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Q. If a river's flow is measured at 300 m³/s, how many liters of water flow in one minute? (2023)

A. 18,000 L
B. 1,800,000 L
C. 18,000,000 L
D. 180,000,000 L

Solution

Water flow in one minute = Flow rate × Time = 300 m³/s × 60 s = 18,000 m³ = 18,000,000 L.

Correct Answer: B — 1,800,000 L

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Q. If a river's flow rate increases by 20% and its original flow rate was 50 m³/s, what is the new flow rate?

A. 60 m³/s
B. 70 m³/s
C. 80 m³/s
D. 90 m³/s

Solution

New flow rate = Original flow rate + (20% of Original flow rate) = 50 m³/s + (0.2 × 50 m³/s) = 60 m³/s.

Correct Answer: A — 60 m³/s

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Q. If a river's flow rate is 1000 m³/s and it carries 200 kg of sediment per cubic meter, what is the total sediment load in kilograms per second?

A. 200,000 kg/s
B. 100,000 kg/s
C. 300,000 kg/s
D. 400,000 kg/s

Solution

Total sediment load = Flow rate × Sediment per cubic meter = 1000 m³/s × 200 kg/m³ = 200,000 kg/s.

Correct Answer: A — 200,000 kg/s

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Q. If a river's flow rate is 200 cubic meters per second, how much water flows in one hour? (2022)

A. 720,000 m³
B. 1,200,000 m³
C. 1,440,000 m³
D. 2,000,000 m³

Solution

Water flow in one hour = Flow rate × Time = 200 m³/s × 3600 s = 720,000 m³.

Correct Answer: C — 1,440,000 m³

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Q. If a river's flow rate is 500 cubic meters per second, how many liters of water flow in 10 minutes? (2019)

A. 300,000 L
B. 300,000,000 L
C. 1,800,000 L
D. 1,500,000 L

Solution

Flow in 10 minutes = 500 m³/s × 600 s = 300,000 m³ = 300,000,000 L.

Correct Answer: B — 300,000,000 L

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Major Competitive Exams

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