00:01
We have efficiency eta equals to 40%, that is 0 .40 and uranium to 35 fission produced at 200 mega electron volt.
00:12
So the power will be equal to 100 by 40 multiplied by 10 to the power 9, that is equal to 25 multiplied by 10 to the power 8 watt.
00:24
So the energy production in a day will be equal to power multiplicity.
00:30
By time and power is 25 multiplied by 10 to the power 8 watt and in a day there is 24 mudflared by 300 360 00 seconds so from here energy e comes out to be 216 mudplared by 10 to the power 12 jule okay so the mass for the part a of the question the mass of the coal required will be m equals to energy 216 divided 10 to the power 12 jule divided by 3 .3 multiplied by 10 to the power 7.
01:04
So from here after solving we get mass m equals to 65 .45 multiplied by 10 to the power 5 kg.
01:14
So this is the mass required of the code.
01:18
So this is mass required of the core.
01:21
Now moving to the part b in which we have to determine the ratio of mass of the coal to uranium so as we know that one mole that will be equals to 6 .0 .2 multiplied by 10 to the power 23 of uranium atoms so the energy produced in 235 gram of uranium will be 2 multiplied by 10 to the power 8 multiplied by 1 .6 multiplied by 10 to the power minus 19 this is 200 mega electron volt this energy okay multiplied by 1 .m.
02:00
That is 6 .022 multiplied by 10 to the power 23 and this comes out to be 19 .27 multiplied by 10 to the power 12 june...