 All right, friends, so I am Dhiraj from Centrum Academy. So we are live again, and we are taking up a question on nuclei chapter, all right? So this is a different kind of question which is based on some scenario as such, fine? So let us see what this question is about. So as you can see on your screen, there is a huge question, all right? So what we'll do, we'll solve it in a step-by-step manner. So there are multiple parts to this question, fine? So we'll take one by one all of these. So there is a town which has a population of 1 million. So there is a town, fine? Which has a population of 10 to the power six people, all right? The average electric power needed per person is 300 watt. So every person requirement is 300 watt. Fine, so the requirement of town is what? So it must be about the town which is asked later on. So we are just readily finding out some of the obvious parameters, like per person energy consumption is given. So I'll find out the consumption by the entire town. So that will be what? 300 into 10 is for six. So town will require three into 10 to the power eight watt. Okay? The reactor is designed to supply power to this town. The efficiency with which thermal power is converted into electrical power is 25% fine? So thermal power, if it is 100, then 25 will be converted into electrical power, fine? So if the requirement of town is this much electrical energy, so how much thermal energy is required to create this much electrical energy? Assume that to be x, fine? So 25% of x, which is x into 0.25 should be equal to three into 10 to the power eight watts. Fine? So x will be equal to what? Three divided by 0.25, which is 12 into 10 to the power eight joules per second, fine? So this much amount of heat energy must be supplied to create that much amount of power in the town, okay? So these are some of the initial analysis we did. Now let us see what is asked. This I'll write here. This is the thermal energy required, okay? Now part A of the question. Assuming 200 MeV of thermal energy to come from each fission event, find the number of events that should take place every day. Now we know that one fission will give 200 MeV, okay? If you convert it into joules, you'll get here as 200 into 10 to the power six into charge of electron. This much joules you'll get, right? So one fission gives this much joule and the requirement is 12 into 10 to the power eight joules every second, fine? So every second, how many reactions should take place? Let's call that as N and that will be equal to 1.2 into 10 ratio power nine, which is X, okay? That divided by this energy, okay? That is equal to two into 1.6 into 10 ratio power. This is eight and that is minus 11, okay? So this many events should happen per second, okay? Now how many events should happen per day? That will be equal to N into 24 hours, 60 minutes into 60 seconds, right? So this is the answer for part A. This many events per second. So per day, if I have to find out, I have to multiply that with number of seconds one day has, okay? So like this, you can do part A of the question, all right? Let us see what is part B. In part B, assuming that the fission to take place largely through uranium 235, at what rate will the amount of uranium 235 decrease, okay? We know that every second, this many fission reaction should happen, fine? And one fission reaction, one fission will consume one nucleus of uranium 235, fine? So how many nucleases will get consumed of uranium 235? It will be equal to this only number of events per second will be equal to number of nucleus that should get consumed every second, isn't it? Now, if you want to find out what is the mass rate of consumption, all you have to do is divide this N with the Avogadro number, fine? If you divide it with Avogadro number, you'll get number of moles that should get consumed every second. This divided by 6.023 into 10 raise to power 23, fine? So this is number of moles of uranium 235 that should get consumed every second, fine? Now, what is a consumption rate in terms of mass? You can multiply this with the molecular mass for the molar mass, sorry. So if you multiply this with 235, you'll get grams per second that should get consumed, gram per second that should get consumed, okay? So if you want to find in terms of kg per second, you just divide it with 1000, fine? So like this you can do the B part of the question, okay? Now it says that you should also express in kg per day. So this is gram per second, so I'm sure you will be able to convert gram per second in kg per day, all right? So we can move to the C part of the question. Assuming that uranium enriched to 3% of 235 uranium, that will be used. You need to find how much uranium is needed per month, all right? So in B part, we have got how many kgs of uranium 235 is required per day, fine? Let's call that as Y. So this much uranium 235 suppose is required per day, fine? So for 30 days, what is the requirement of uranium 235? That is 30 into Y, okay? This much uranium 235 is required per month, kg per month, all right? Now this much uranium 235 is required, but uranium doesn't exist in pure 235. It is written that only 3% of naturally occurring uranium has uranium 235, okay? So if that is a case, we can say that if X kg of uranium, naturally occurring uranium is required, then this into 3% which is 0.03 will be equal to 13 to Y, okay? So if you solve this equation, you'll get the value of X and X is the amount of naturally occurring uranium that you require to conduct all this, okay? So I hope you have learned something today. In case you have any doubts, feel free to get in touch with us and we'll be clearing your doubts.