Q. A beam of light passes through a narrow slit and produces a diffraction pattern. What happens to the width of the central maximum if the slit width is decreased? (2019)
A.It increases
B.It decreases
C.It remains the same
D.It becomes zero
Solution
According to the diffraction principle, as the slit width decreases, the width of the central maximum increases.
Q. A block of mass 2 kg is sliding down a frictionless incline of height 5 m. What is its speed at the bottom?
A.5 m/s
B.10 m/s
C.15 m/s
D.20 m/s
Solution
Using conservation of energy, potential energy at the top (mgh) converts to kinetic energy at the bottom (1/2 mv²). Thus, v = sqrt(2gh) = sqrt(2 * 9.8 * 5) ≈ 10 m/s.
Q. A block of mass 2 kg is sliding down a frictionless incline of height 5 m. What is its speed at the bottom of the incline?
A.10 m/s
B.5 m/s
C.20 m/s
D.15 m/s
Solution
Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv². Thus, v = sqrt(2gh) = sqrt(2 * 9.81 * 5) = 10 m/s.
Q. A block of mass 5 kg is pulled on a horizontal surface with a force of 20 N. If the frictional force is 5 N, what is the acceleration of the block? (2020)
A.2 m/s²
B.3 m/s²
C.4 m/s²
D.5 m/s²
Solution
Net force = Applied force - Frictional force = 20 N - 5 N = 15 N. Using F = ma, a = F/m = 15 N / 5 kg = 3 m/s².
Q. A charge of +4 µC is placed in a uniform electric field of strength 500 N/C. What is the work done in moving the charge 0.2 m in the direction of the field?
A.400 J
B.200 J
C.100 J
D.50 J
Solution
W = F * d = (E * q) * d = (500 N/C * 4 × 10^-6 C) * 0.2 m = 0.4 J.
Q. A charged particle moves in a magnetic field. What is the path of the particle if it enters the field perpendicularly? (2023)
A.Straight line
B.Circular path
C.Elliptical path
D.Parabolic path
Solution
A charged particle moving perpendicularly to a magnetic field experiences a magnetic force that acts as a centripetal force, causing it to move in a circular path.
Q. A current-carrying conductor experiences a force in a magnetic field. What is the direction of this force? (2020)
A.Parallel to the field
B.Opposite to the field
C.Perpendicular to both current and field
D.In the direction of current
Solution
According to Fleming's left-hand rule, the force on a current-carrying conductor in a magnetic field is perpendicular to both the current and the magnetic field.
Correct Answer: C — Perpendicular to both current and field
