Entrance Exams
Govt. Exams
Elastic PE = ½kx² = ½ × 100 × (0.1)² = ½ × 100 × 0.01 = 0.5 J.
On a frictionless incline, a = g sin(θ) = 10 × sin(30°) = 10 × 0.5 = 5 m/s².
Initial KE = ½m(v)² + ½m(2v)² = 2.5mv². Final velocity = 3v/2 (by momentum conservation). Final KE = ½(2m)(3v/2)² = 2.25mv²/4 = 0.84mv². Loss = 2.5 - 0.84 = 1.66, which is 2/3 of initial KE.
Electric field at surface: E = kQ/r² = (9 × 10⁹ × 8 × 10⁻⁶)/(0.02)² = (72)/(4 × 10⁻⁴) = 1.8 × 10⁷ N/C.
Wavelength λ = v/f = 1500/(50 × 10³) = 1500/50000 = 0.03 m = 3 cm. Higher frequencies in denser mediums produce shorter wavelengths.
Maximum velocity in SHM: v_max = ωA = 2πfA = 2π × 2 × 5 = 20π cm/s ≈ 62.8 cm/s.
After 30 years = 3 half-lives: N = N₀ × (1/2)³ = N₀/8. The sample reduces by half three times successively.
Height = L × sin(60°) = 10 × (√3/2) = 5√3 m ≈ 8.66 m. The vertical component of the ladder's length is found using trigonometry.
Object at 2f (40 cm = 2 × 20 cm) forms image at 2f. Using mirror equation confirms v = 40 cm, m = -1 (real, inverted, same size as object).
Each half has resistance R/2. When connected in parallel: 1/R_new = 1/(R/2) + 1/(R/2) = 4/R, so R_new = R/4.