# In the process of nuclear fission of 1 gram uranium, the mass lost is 0.92 milligram. The efficiency of powder house run by it is 10%. To obtain 400 megawatt powder from the powder house, how much uranium will be required per hour ?

Q: In the process of nuclear fission of 1 gram uranium, the mass lost is 0.92 milligram. The efficiency of powder house run by it is 10%. To obtain 400 megawatt powder from the powder house, how much uranium will be required per hour ? (c = 3 x 108 m/s )

Sol: Powder to be obtained from powder house = 400 mega watt/ Energy obtained per hour

= 400 megawatt x 1 hour = (400 x 106 watt) x 3600 second

= 144 x 1010 joule.

Here only 10% of input is utilized .

In order to obtain 144 x 1010 joule of useful energy, the output energy from the powder house

(10 E)/100 = 144 x 1010 J

E = 144 x 1011 joule

Let, this energy is obtained form a mass- loss of Δm kg.

Then (Δm)c2 = 144 × 1011 joule

Δm  = (144 × 1011)/(3 × 108)2 = 16 × 10-5 kg = 0.16 g

Since 0.92 milligram (= 0.92 x 10-3 g) mass is lost in 1 g uranium, hence for mass loss of 0.16 g, the

uranium required is = (1 × 0.16 )/(0.92 × 10-3 )= 174 g

Thus to rum the powder house, 174 gm uranium is required per hour.