Govt. Exams
Advertisement
Topics in JEE Physics
A block slides down a frictionless incline of height 5 m. What is its speed at the bottom? (g = 10 m/s²)
Correct Answer:
B. 10 m/s
EXPLANATION
Using energy conservation: mgh = ½mv². v = √(2gh) = √(2×10×5) = √100 = 10 m/s
What is the moment of inertia of a uniform disc of mass M and radius R about an axis perpendicular to the disc passing through its center?
Correct Answer:
B. ½MR²
EXPLANATION
The moment of inertia of a uniform disc about its central perpendicular axis is I = ½MR²
A particle moves with constant velocity v = 5 m/s in a circular path of radius 10 m. What is its centripetal acceleration?
Correct Answer:
C. 2.5 m/s²
EXPLANATION
Centripetal acceleration = v²/r = 25/10 = 2.5 m/s²
A block of mass 5 kg is placed on a frictionless inclined plane at 30°. What is the acceleration down the plane? (g = 10 m/s²)
Correct Answer:
A. 5 m/s²
EXPLANATION
a = g sin(30°) = 10 × 0.5 = 5 m/s²
A body is projected vertically upward with velocity 40 m/s. If g = 10 m/s², what is the maximum height reached?
Correct Answer:
B. 80 m
EXPLANATION
Using v² = u² - 2gh, at maximum height v = 0. So 0 = 1600 - 2(10)h, h = 80 m
A uniform rod of mass M and length L is pivoted at one end. The moment of inertia about the pivot is:
Correct Answer:
A. ML²/3
EXPLANATION
For a uniform rod about one end, I = ML²/3 (standard formula)
A particle experiences a force F = 2i + 3j (N) and displacement s = 4i + 6j (m). The work done is:
Correct Answer:
C. 26 J
EXPLANATION
W = F·s = (2×4) + (3×6) = 8 + 18 = 26 J
A rotating wheel has moment of inertia 2 kg·m² and angular acceleration 3 rad/s². The torque required is:
Correct Answer:
D. 6 N·m
EXPLANATION
τ = Iα = 2 × 3 = 6 N·m
A man of mass 60 kg stands on a weighing scale in an elevator moving downward with acceleration 2 m/s². The scale reading is (g = 10 m/s²):
Correct Answer:
A. 480 N
EXPLANATION
Normal force N = m(g - a) = 60(10 - 2) = 60 × 8 = 480 N
A disc of radius 0.5 m rolls without slipping on a horizontal surface with angular velocity 4 rad/s. The velocity of the center of mass is:
Correct Answer:
B. 2 m/s
EXPLANATION
For rolling without slipping, v = ωr = 4 × 0.5 = 2 m/s
