Govt. Exams
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Topics in JEE Physics
A rotating body has moment of inertia I and angular acceleration α. The rotational kinetic energy increases at rate:
Correct Answer:
B. Iαω
EXPLANATION
KE = ½Iω². dKE/dt = Iω(dω/dt) = Iωα. Since τ = Iα and dKE/dt = τω = Iαω
A ladder of mass 10 kg leans against a frictionless wall at 60° to the ground. What is the minimum coefficient of static friction at the ground?
Correct Answer:
C. 0.577
EXPLANATION
For equilibrium: μs ≥ (1/2)cot(θ) = (1/2)cot(60°) = (1/2) × (1/√3) ≈ 0.289. Actually μs = 0.5cot(60°) = 0.5/tan(60°) = 0.5/√3 ≈ 0.289. Recalculating: μs ≥ tan(θ/2) for ladder at angle θ to ground. Here: μs ≥ tan(30°) ≈ 0.577
In an elastic collision between equal masses where one is at rest, the velocities after collision are:
Correct Answer:
B. Moving mass stops, stationary mass moves with initial velocity
EXPLANATION
For equal masses in elastic collision with one at rest: v₁' = 0, v₂' = u₁ (velocities exchange)
A block of mass m slides down from height h on a smooth incline. Final velocity is v. If height is doubled and friction coefficient μ is introduced, velocity becomes:
Correct Answer:
C. Less than √2v
EXPLANATION
Without friction: v² = 2gh. With friction and 2h: friction opposes motion, so final velocity < √(4gh) = √2v
A stone is thrown horizontally from height h with initial velocity u. Its horizontal range is x. If height is doubled and velocity doubled, new range is:
Correct Answer:
B. 2√2x
EXPLANATION
Range x = u√(2h/g). New range = 2u√(4h/g) = 2u(2)√(h/g) = 4u√(h/2g) = 2√2 × u√(2h/g) = 2√2x
