Q. A conical pendulum swings in a horizontal circle. If the angle of the string with the vertical is θ, what is the expression for the tension in the string?
A.T = mg
B.T = mg/cos(θ)
C.T = mg/sin(θ)
D.T = mg tan(θ)
Solution
The vertical component of tension balances the weight: T cos(θ) = mg, thus T = mg/cos(θ).
Q. A conical pendulum swings in a horizontal circle. If the angle of the string with the vertical is θ, what is the relationship between the tension and the gravitational force acting on the pendulum bob?
A.T = mg
B.T = mg cos(θ)
C.T = mg sin(θ)
D.T = mg tan(θ)
Solution
The vertical component of tension balances the weight: T cos(θ) = mg.
Q. A conical pendulum swings with a constant speed. If the angle of the string with the vertical is θ, what is the expression for the tension in the string?
A.mg/cos(θ)
B.mg/sin(θ)
C.mg/tan(θ)
D.mg
Solution
Tension T = mg/cos(θ) to balance the vertical component of weight.
Q. A convex lens has a focal length of 20 cm. If an object is placed at a distance of 60 cm from the lens, what is the distance of the image from the lens?
A.15 cm
B.30 cm
C.45 cm
D.60 cm
Solution
Using the lens formula 1/f = 1/v - 1/u, we find v = 30 cm.
Q. A convex lens has a focal length of 20 cm. If an object is placed at a distance of 40 cm from the lens, what is the distance of the image from the lens?
A.20 cm
B.40 cm
C.60 cm
D.80 cm
Solution
Using the lens formula 1/f = 1/v - 1/u, where f = 20 cm and u = -40 cm, we find v = 20 cm. The image is formed at 20 cm on the opposite side.
Q. A convex lens has a focal length of 20 cm. If an object is placed at a distance of 30 cm from the lens, what is the distance of the image from the lens?
A.60 cm
B.15 cm
C.30 cm
D.10 cm
Solution
Using the lens formula 1/f = 1/v - 1/u, we find v = 60 cm.
Q. A cyclist accelerates from rest to a speed of 15 m/s. If the mass of the cyclist and the bicycle is 75 kg, what is the kinetic energy at that speed?
A.500 J
B.750 J
C.1000 J
D.1250 J
Solution
Kinetic Energy = 0.5 × mass × velocity² = 0.5 × 75 kg × (15 m/s)² = 8437.5 J.
Q. A cyclist is moving at 15 m/s and a pedestrian is walking at 5 m/s in the same direction. What is the relative speed of the pedestrian with respect to the cyclist?
A.10 m/s
B.5 m/s
C.20 m/s
D.15 m/s
Solution
Relative speed = Speed of pedestrian - Speed of cyclist = 5 m/s - 15 m/s = -10 m/s (10 m/s behind).
Q. A cyclist is moving at 15 m/s and a pedestrian is walking at 5 m/s in the same direction. What is the speed of the cyclist relative to the pedestrian?
A.10 m/s
B.15 m/s
C.5 m/s
D.20 m/s
Solution
Relative speed = Speed of cyclist - Speed of pedestrian = 15 m/s - 5 m/s = 10 m/s.
Q. A cyclist is moving at 15 m/s and passes a stationary observer. If the observer starts moving at 5 m/s in the same direction, what is the speed of the cyclist relative to the observer?
A.10 m/s
B.15 m/s
C.20 m/s
D.5 m/s
Solution
Relative speed = Speed of cyclist - Speed of observer = 15 m/s - 5 m/s = 10 m/s.
Q. A cyclist is moving at 15 m/s towards the east while a car is moving at 25 m/s towards the west. What is the relative speed of the cyclist with respect to the car?
A.10 m/s
B.15 m/s
C.40 m/s
D.25 m/s
Solution
Relative speed = Speed of cyclist + Speed of car = 15 m/s + 25 m/s = 40 m/s.