Q. A concave mirror has a focal length of 10 cm. An object is placed 30 cm in front of the mirror. Where will the image be formed?
A.10 cm
B.15 cm
C.20 cm
D.30 cm
Solution
Using the mirror formula, 1/f = 1/v + 1/u, where f = -10 cm (concave mirror), u = -30 cm. Solving gives v = -15 cm, which means the image is formed 15 cm in front of the mirror.
Q. A concave mirror produces a virtual image of an object placed 10 cm in front of it. If the focal length of the mirror is 5 cm, what is the distance of the image from the mirror?
A.5 cm
B.10 cm
C.15 cm
D.20 cm
Solution
Using the mirror formula, 1/f = 1/v + 1/u. Here, f = -5 cm (concave mirror), u = -10 cm. Solving gives v = -10 cm.
Q. A conical pendulum consists of a mass attached to a string that 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.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 conical pendulum consists of a mass m attached to a string of length L, swinging in a horizontal circle. 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 = m(v²/r)
Solution
In a conical pendulum, T = mg/cos(θ) where θ is the angle with the vertical.
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 in a horizontal circle. If the angle of the string with the vertical increases, what happens to the tension in the string?
A.Increases
B.Decreases
C.Remains the same
D.Becomes zero
Solution
As the angle increases, the vertical component of tension must increase to balance the weight.
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 in the string and the gravitational force?
A.T = mg
B.T = mg/cos(θ)
C.T = mg/sin(θ)
D.T = mg/tan(θ)
Solution
Tension T provides the centripetal force and balances the weight, 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 in the string 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
Tension provides the vertical component to balance 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 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 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 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.