A loud sound can be distinguished from a faint sound by its MCQ with Answer and Explanation

Explanation:
Distance covered in nth second: sₙ = u + a(2n-1)/2. With u=0, sₙ ∝ (2n-1). For n=1: 2(1)-1=1; n=2: 3; n=3:5. Thus ratio 1:3:5. This result holds for any uniformly accelerated motion starting from rest. The distances covered in successive equal time intervals follow odd number ratio. Memory aid: This is a standard result derivable from s = ut + ½at² by calculating s at t=n and t=n-1. Frequently appears in competitive exams testing equation of motion applications.

Explanation:
Weight on surface W = mg = 72 N. Acceleration due to gravity at height h is g' = g [R / (R+h)]². Here h = R/2. So, g' = g [R / (R + R/2)]² = g [1 / (3/2)]² = g (2/3)² = 4g/9. The new weight W' = mg' = m(4g/9) = (4/9)mg = (4/9) × 72 = 32 N.

Explanation:
ΔX/X = 2(Δa/a) + 3(Δb/b) + 2(Δc/c) = 2×2% + 3×2% + 2×2% = 4% + 6% + 4% = 14%. Always add absolute relative errors.