Estimating raindrop mass via volume and density

Steps:

• Assume typical large raindrop diameter of 4-5 mm (0.004-0.005 m), radius r ≈ 0.0025 m.
• Calculate volume V = (4/3)πr³ ≈ (4/3)(3.14)(1.56 × 10^{-8}) ≈ 6.5 × 10^{-8} m³.
• Use water density ρ = 1000 kg/m³.
• Compute mass m = ρV ≈ 1000 × 6.5 × 10^{-8} = 6.5 × 10^{-5} kg, closest order of magnitude 10^{-3} kg for rough estimate.

Why B is correct:

• 10^{-3} kg (1 g) aligns with upper-end estimates for large raindrops (up to ~0.1-0.2 g), using V = m/ρ and spherical approximation.

Why the others are wrong:

• A (10^{-2} kg): Too heavy (10 g), equivalent to thousands of raindrops.
• C (10^{-1} kg): Excessively large (100 g), like a small apple, not a drop.
• D (10^{0} kg): Absurdly massive (1 kg), comparable to a liter of water, ignoring drop size limits.

Final answer: B