Home › NEET UG › physics › c11phy_lom_friction_circ › Per NCERT Example 4.8 (inclined plane), if a mas…
Per NCERT Example 4.8 (inclined plane), if a mass just starts to slide at angle θ = 15°, the coefficient of static friction is which?
Acos 15° ≈ 0.97
Btan 15° ≈ 0.27
Csin 15° ≈ 0.26
D1.00
Answer & Solution
Correct answer: B. tan 15° ≈ 0.27
NCERT Ex 4.8: at impending slip, mg sin θ = μ_s mg cos θ ⇒ μ_s = tan θ = tan 15° ≈ 0.27.
Related questions
Per NCERT Example 4.11, a R=300 m banked-at-15° racetrack with μ = 0.2 has optimum speed aPer NCERT Example 4.10, a cyclist taking a R=3 m turn at 18 km/h (μ_s = 0.1) is in which cPer NCERT §4.10 eqn 4.21, the MAXIMUM speed on a banked road (with friction μ_s) is which Per NCERT §4.10 eqn 4.22, the OPTIMUM speed (no friction needed) on a banked road of anglePer NCERT §4.10 eqn 4.18, the maximum speed of a car on a LEVEL circular road of radius R Per NCERT §4.10 eqn 4.16, the CENTRIPETAL force needed for a body of mass m moving at speePer NCERT §4.9.1 (rolling friction note), among rolling, sliding (kinetic), and static friPer NCERT Example 4.7, the MAXIMUM acceleration of a train such that a box on its floor (μ