Answer :

Given,

The radius of the weapon, r=0.5 m

Therefore, Surface area, A=4𝜋r^{2} = 𝜋

Temperature after detonation, T=10^{6} K

Stefan’s constant, σ = 5.67 × 10^{-8} W/ m^{2}K^{4}

(a) Therefore, power radiated (P) is given by

(ANS)

(b) Here, 10% of the energy(E) is to evaporate water which is at 30°C. This means the energy heats the water up to 100°C and then evaporates.

So, 0.1E = msΔT +mL_{v}

Here, the first term is due to the contribution of energy in heating the liquid to 100°C and the second term is due to vaporization and m is the mass of the water used.

Energy,

Here, t is the time which is 1 sec.

Given, Heat capacity, S_{w} = 4186.0 J/ kgK

Latent heat of vaporization, L_{v} = 22.6 X 10^{5} J / kg

Temperature difference, ΔT = 100°C - 30°C = 70°C

So, 7x10^{9} kg of water would be evaporated.

(c) Here, energy,

Therefore, momentum,

Now, momentum at a distance 1km (per unit time) is

Therefore, the momentum imparted at 1km is 47.7 N/m^{2}

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