Q. A river is 300 m wide. If a person swims across the river at a speed of 2 m/s and the river flows at 1 m/s, how long will it take to cross the river? (2020)
A.150 s
B.100 s
C.200 s
D.300 s
Solution
Time = Distance / Speed. The effective speed across the river is 2 m/s. Time = 300 m / 2 m/s = 150 s.
Q. A river meanders and has a total length of 150 km. If the straight-line distance from the source to the mouth is 100 km, what is the river's meandering ratio? (2023)
A.1.5
B.1.2
C.1.3
D.1.4
Solution
Meandering ratio = Total length / Straight-line distance = 150 km / 100 km = 1.5.
Q. A river's depth increases by 2 meters due to rainfall. If the river's width is 50 meters, what is the increase in volume per kilometer of the river? (2020)
A.100,000 m³
B.200,000 m³
C.300,000 m³
D.400,000 m³
Solution
Increase in volume = Width × Depth × Length = 50 m × 2 m × 1000 m = 100,000 m³.
Q. A river's sediment load is measured at 200 kg per cubic meter of water. If the river has a flow of 100 m³/s, what is the sediment load per second? (2023)
A.200 kg/s
B.400 kg/s
C.600 kg/s
D.800 kg/s
Solution
Sediment load per second = Sediment load per m³ x Flow rate = 200 kg/m³ x 100 m³/s = 20,000 kg/s.
Q. A roller coaster at the top of a hill has a potential energy of 5000 J. If it descends to a height of 10 m, what is its speed at the bottom? (g = 9.8 m/s²)
A.10 m/s
B.20 m/s
C.30 m/s
D.40 m/s
Solution
Using conservation of energy, initial PE + KE = final PE + KE. 5000 J = mgh + 0.5mv². Solving gives v = √(2(5000 - mgh)/m) = 30 m/s.
Q. A roller coaster starts from rest at a height of 30 m. What is its speed at the lowest point? (g = 9.8 m/s²)
A.10 m/s
B.15 m/s
C.20 m/s
D.25 m/s
Solution
Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv². Solving gives v = √(2gh) = √(2 * 9.8 * 30) = 24.5 m/s.
Q. A roller coaster starts from rest at a height of 50 m. What is its speed at the lowest point?
A.10 m/s
B.20 m/s
C.30 m/s
D.40 m/s
Solution
Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv^2. Solving gives v = sqrt(2gh) = sqrt(2*9.8*50) = 31.3 m/s.
Q. A rolling object has both translational and rotational motion. Which of the following quantities remains constant for a rolling object on a flat surface?
A.Linear velocity
B.Angular velocity
C.Total energy
D.Kinetic energy
Solution
The total energy remains constant for a rolling object on a flat surface, assuming no external work is done.
Q. A rotating body has an angular momentum L. If its moment of inertia is doubled and angular velocity is halved, what will be the new angular momentum? (2021)
Q. A rotating disc has an angular velocity of ω. If the radius of the disc is doubled while keeping the mass constant, what happens to the angular momentum?
A.It doubles
B.It remains the same
C.It quadruples
D.It halves
Solution
Angular momentum L = Iω, where I is the moment of inertia. If radius is doubled, I increases by a factor of 4, but ω decreases by a factor of 2, so L remains the same.
