Q. A circular loop of radius R carries a current I. What is the magnetic field at the center of the loop?
A.μ₀I/(2R)
B.μ₀I/R
C.μ₀I/(4R)
D.μ₀I/(8R)
Solution
The magnetic field at the center of a circular loop carrying current I is given by the formula B = (μ₀I)/(2R), where μ₀ is the permeability of free space.
Q. A circular loop of radius r is placed in a uniform magnetic field B. If the magnetic field is perpendicular to the plane of the loop, what is the magnetic flux through the loop?
A.0
B.πr²B
C.2πrB
D.B/r
Solution
The magnetic flux Φ through a surface is given by Φ = B * A, where A is the area. For a circular loop, A = πr², so Φ = B * πr².
Q. A circular loop of wire is placed in a uniform magnetic field. What happens to the induced EMF if the magnetic field strength is doubled?
A.Induced EMF is halved
B.Induced EMF remains the same
C.Induced EMF is doubled
D.Induced EMF is quadrupled
Solution
According to Faraday's law of electromagnetic induction, the induced EMF is directly proportional to the rate of change of magnetic flux. If the magnetic field strength is doubled, the induced EMF will also double.
Q. A circular loop of wire is placed in a uniform magnetic field. What happens to the induced EMF if the area of the loop is increased?
A.Increases
B.Decreases
C.Remains the same
D.Depends on the magnetic field strength
Solution
According to Faraday's law of electromagnetic induction, the induced EMF is proportional to the rate of change of magnetic flux. Increasing the area increases the flux, thus increasing the induced EMF.
Q. A coil of wire is placed in a magnetic field. If the magnetic field strength is increased, what happens to the induced EMF in the coil?
A.It increases
B.It decreases
C.It remains the same
D.It becomes zero
Solution
According to Faraday's law of electromagnetic induction, the induced EMF in a coil is directly proportional to the rate of change of magnetic flux. Increasing the magnetic field strength increases the magnetic flux, thus increasing the induced EMF.
Q. A composite body consists of a solid cylinder and a solid sphere, both of mass M and radius R. What is the total moment of inertia about the same axis?
A.(7/10) MR^2
B.(9/10) MR^2
C.(11/10) MR^2
D.(13/10) MR^2
Solution
The total moment of inertia is I_cylinder + I_sphere = (1/2 MR^2) + (2/5 MR^2) = (7/10) MR^2.
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 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.
