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## Class 11 Physics (India)

# Solved example: percentage change in density

Let's explore how to figure out the percentage change in density of a material when it is subjected to bulk stress in this problem-solving video. Created by Mahesh Shenoy.

## Want to join the conversation?

- At10:40,if we assume that V and V' are pretty much the same,then V-V' should be equal to zero. I didn't understand how V-V' = -ΔV?(2 votes)
- 'ΔV' itself means that change in volume which is (V-V') and Minus indicated it is compression...final volume is less than initial volume...and 'pretty much same' is different from 'exactly same', if they are exactly same and then V-V' is zero and it also means the material is not compressed(2 votes)

- How is delta v upon v = 1 the same as 100% compression? (At2:03), shouldn't it be 0% compression?(2 votes)
- When I tried to solve Δρ/ρ , I got the answer as V/ΔV which does not seem correct because I got the % change as 400000% which is just crazy. Why isn't this correct?

Δρ/ρ = V/ΔV(2 votes)

## Video transcript

diamond is being compressed under a pressure of 1000 atmospheres find the percentage change in its density given that 1 atmosphere equals 10 to the power 5 Pascal's and the bulk modulus of diamond is 400 Giga Pascal's alright the first thing we'll do is gather the data and see what's given to us we are given pressure and we've been given the bulk modulus so from this we can pretty much guess we're dealing with bulk stress in bulk strain over here so the first thing we'll do is write the expression that's Hookes law let's write that down Hookes law tells us that bulk stress which is the same as pressure is proportional to bulk strain bulk strain is is the change in volume per unit volume and that proportionality constant is the bulk modulus B and we've talked about this in previous videos so if you're not familiar with this or you require refresher it would be a great idea to go back watch it those videos first and then come back over here but anyways a good way to think about bulk modulus is we could say that bulk modulus equals the pressure when when this number goes to 1 this number becomes 1 and what does it mean that this number equals 1 well Delta V is the change in the volume and over here under pressure the volume decreases the material shrinks so really the change in the volume is negative so there should be a negative sign over here but let's not worry about those signs don't worry about that but anyways Delta V is the change in the volume and V is the initial volume so if we say Delta V or V equals 1 we're saying changing the volume should be equal to initial volume meaning if the initial volume was say v with V compress it by v if the initial volume is 10 we compress it by 10 in other words we are doing a hundred percent compression Delta V over V equals 1 means hundred percent compression 100% compression so we can think of bulk modulus as the amount of pressure needed for 100% compression all right as if you think of it this way we can now understand for Diamond 400 Giga Pascal's is needed to compress diamond by 100 percent so if the pressure is thousand atmospheres how much will be the compression or we can easily calculate that right we can easily calculate that by doing some cross multiplication maybe but the problem is we are not asked how much compression this this pressure produces we are asked how much change in the density how much percentage change in the density it produces so density we're talk a little bit about density density tells us how crowded a material is how much mass is concentrated in a unit volume so we usually write density as the mass or volume how much mass is constant in unit volume and since the volume is changing density will also change right and that's what we need to calculate how much is the density changing so instead of asking this they're asking us what is the percentage change in density that means what is Delta Rho over Rho this is what we need to find out okay and the way I see it is by knowing by using the knowledge of bulk modulus we can calculate this and by using this expression we can actually relate changes in density to changes in volume we should be able to do that right that's actually not physics that's mathematics if you think about it right so we have been we know it's the move it's connection in density and volume we just have to figure out the connection between changes in density and changes in volume and then we are done all right so here's what we'll do instead of trying to derive this right now over here and you know losing focus from the physics to maths who I will vote directly see what the relationship over here is I'll just write the relationship over here we'll do this problem solve the problem and then towards the end we'll we'll do the math all right so we'll do it that way so if you do the math over here if you try to figure out the changes and density change more him it turns out that the change in the density per unit density in fact turns out to be this is a very nice relationship turns out to be equal to equal to the change in volume per unit volume so the percentage change in density turns out to be exactly equal to percentage change in volume and this is true for small changes in volume all right so for small change in volume only so for small small changes in volume changes in volume and we'll prove this later okay as if not we'll accept this and of course there will be a negative sign over here and the reason why there's a negative sign is because it's telling us that if the Delta is positive that means if volume increases then the density will decrease right I mean there's an inverse relationship so that's so if one increases the other one should decrease and that's the only reason is a negative sign but as far as the amount goes the person to change magnitude of the percentage changes are exactly the same so that makes our life very easy because we just need to calculate this and then we then then we know what this is all right all right so let's go ahead and do that so from our definition of bulk modulus we just learned that 400 giga pascals so here it is 400-year Pascal's of pressure is needed to compress diamond by 100% so this much pressure compresses diamond by hundred percent hundred percent compression so the pressure of thousand atmospheres thousand atmospheres and one atmosphere is a 10 to the power 5 Pascal so instead of an atmosphere we can just say 10 to the power 5 Pascal produces how much compression that's the question produces how much compression and we can solve this by a cross multiplication we could say that the compression that it produces would be this would be 1,000 times 10 to the power five times 100 divided by 400 gigapascals 400 gigapascal is it there's a Pascal on the top as well it's a Pascal's over here this Pascal's in this Pascal gets cancelled and Giga is 10 to the power 9 and on the top let's see how many zeroes we have 1 2 3 4 5 we have 10 zeros so what we could do is you can cancel 10 zeros 9 zeros from here and 1 zero over here all right so what remains now if you see carefully I hope I'm not skipping too many steps over here but if you just cut a lot of zeros or you'll end up with 1 or 40 all right you can just pause and just confirm everything is going going fine anyways we'll get 1 or 40 and that will be you 1 over 4 is 0.25 1 over 40 is point 0 to 5 and that is the amount of compression so what we have calculated is is this number Delta V or V that's the amount of compression that we're getting but since we now saw that we just saw that change in compression is the same as the change in the density of percentage change in the density we could also say that the percentage change in density is also so much in fact since volume decreased by so much amount the density would have increased by so much amount so I'm just gonna box that I'm gonna say that is our final answer all right so that the density is going to increase by this much amount so yeah we are done let's end the video don't worry I remember we have one last thing to do we'd have to prove this is to be true and it would be a great idea to pause the video and see if you can try this yourself first so you I'll give a little bit of clue what we could do is so let me do that in the next hey let me just yeah let's do that on a fresh pitch we just saw the physics problem so we need to connect between the densities and volume so let's start with the connection we know that density is mass or volume so we can think of this as the initial density and this is the initial volume once we compress it the volume changes so the new density row - will be the new mass divided by the new volume I'll notice that the mass doesn't change so M remains the same alright so again this could be the clues now just try and figure out what Delta remember that's what we want to calculate what is Delta oops use the same color what is Delta Rho over Rho this is what we need to find out remember Delta Rho is change in density final density - initial density just plug in and see what you get all right let's do it so Delta Rho would be no let's write that Delta Rho over Rho is what we want this would be final density minus Delta Rho is final minus initial right so that would be final density minus initial density divided by initial density final density we'll have it over here that's masked you have a final volume - mastered by initial volume divided by mass divided by initial volume we can take em common and we can take the common denominator here we get V - times V and on the numerator we get V - V - divided by the denominator we get M divided by V that is M cancels and we cancels so what we are left with is we minus V dash divided by V - alright so let's see what's that equal to what is we - we - oh that's Delta V that's the change in volume right oh yeah but no change in volume is final minus initial right changes final - this is initial minus final so that's negative of Delta V that's where the minus sign comes from we already seen there's a minus sign and in the denominator you might denominator is a V - we need a V right oh that's where we could just say look if the changes are very tiny then we - and we are pretty much the same right we can just say that you might be like whoa whoa whoa how can that be it's let me just give you an example what I'm saying is if you're dividing say okay some some number divided by a hundred and you're dividing some number by a hundred point one well that's pretty much the same right I'm not gonna have a huge difference that's why this only works when you are dealing with a very small quantity and so notice that what you have found now we just proved that the change in density per unit density is indeed indeed the same as change in volume per unit volume provided provided now volume doesn't change too much all right so there we have it