It is a common observation that rain clouds can be at about a kilomete

1.	It is a common observation that rain clouds can be at about a kilometer altitude above the grounds.(a) If a rain drop falls from such a height freely under gravity, what will be its speed? Also calculate in km/h. (g = 10 m/s2 )(b) A typical rain drop is about 4 mm diameter. Momentum is mass × speed in magnitude. Estimate its momentum when it hits ground.(c) Estimate the time required to flatten the drop.(d) Rate of change of momentum is force. Estimate how much force such a drop would exert on you.(e) Estimate the order of magnitude force on umbrella. Typical lateral separation between two rain drops is 5 cm.(Assume that umbrella is circular and has a diameter of 1 m and cloth is not pierced through !!)
Answer» (a) v = \(\sqrt {2gh}\) = \(\sqrt{2\times10\times1000}\) = 141 m/s = 507.6 km/h. (b) \(m=\frac{4\pi}{3}r^3ρ=\frac{4\pi}{3}{(2\times10^{-3})}^3(10^3)\) = 3.4 x 10^-5kg. ρ = mv ≈ 4.7 × 10^-3 kg m/s ≈ 5 × 10^-3 kg m/s. (c) Diameter ≈ 4 mm ∆t ≈ \(\frac {d}{v}\) = 28 μs ≈ 30 μs (d) F = \(\frac{\Delta p}{\Delta t}\) = \(\frac{4.7\times 10^{-3}}{28\times10^{-4}}\) ≈ 168 N ≈ 1.7 × 10² N. (e) Area of cross-section =\(\frac{\pi d^2}{4}\) ≈ 0.8 m². With average separation of 5 cm, no. of drops that will fall almost simultaneously is \(\frac{0.8\, m^2}{{(5\times10{-2})}^2}\) ≈ 320. Net force ≈ 54000 N (Practically drops are damped by air viscosity).

It is a common observation that rain clouds can be at about a kilometer altitude above the grounds.(a) If a rain drop falls from such a height freely under gravity, what will be its speed? Also calculate in km/h. (g = 10 m/s2 )(b) A typical rain drop is about 4 mm diameter. Momentum is mass × speed in magnitude. Estimate its momentum when it hits ground.(c) Estimate the time required to flatten the drop.(d) Rate of change of momentum is force. Estimate how much force such a drop would exert on you.(e) Estimate the order of magnitude force on umbrella. Typical lateral separation between two rain drops is 5 cm.(Assume that umbrella is circular and has a diameter of 1 m and cloth is not pierced through !!)

Answer»

(a) v = \(\sqrt {2gh}\) = \(\sqrt{2\times10\times1000}\) = 141 m/s = 507.6 km/h.

(b) \(m=\frac{4\pi}{3}r^3ρ=\frac{4\pi}{3}{(2\times10^{-3})}^3(10^3)\) = 3.4 x 10^-5kg.

ρ = mv ≈ 4.7 × 10^-3 kg m/s ≈ 5 × 10^-3 kg m/s.

(c) Diameter ≈ 4 mm

∆t ≈ \(\frac {d}{v}\) = 28 μs ≈ 30 μs

(d) F = \(\frac{\Delta p}{\Delta t}\) = \(\frac{4.7\times 10^{-3}}{28\times10^{-4}}\) ≈ 168 N ≈ 1.7 × 10² N.

(e) Area of cross-section =\(\frac{\pi d^2}{4}\) ≈ 0.8 m².

With average separation of 5 cm, no. of drops that will fall almost simultaneously is

\(\frac{0.8\, m^2}{{(5\times10{-2})}^2}\) ≈ 320.

Net force ≈ 54000 N

(Practically drops are damped by air viscosity).