A metallic cube is floating inside a beaker. The temperature of the system is increased by a small amount ΔT. It is found that the depth of the submerged portion of the cube does not change. Then the ratio of the coefficient of linear expansion of the cube and the coefficient of volume expansion of the liquid is given by b a , where a and b are coprime positive integers.
What is a + b ?
Note: Ignore the expansion of the beaker.
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Could you please explain what is wrong with the solution given below :
s = initial length of side of cube x = increase in cube length after expansion Initial Volume of fluid = V y = Increase in volume of water M = Mass of Body Mw = Mass of water
I am assuming that the cube represents an inanimate object and that it is submerged symmetrically, that is the proportion o volume submerged stays the same.
Total Volume of floating body * density of body = Volume of submerged portion * density of water
Volume of submerged portion / Total volume of body = density of body / density of water....(1)
Initial density of body = M / s^3 final density of body = M/(s+x)^3
Initial Density of water = Mw / V Final Density of water = Mw / (V+y)
Since submerged portion stays same (because of (1))
(M /s^3 / Mw / V) = (M / (s+x)^3 / Mw / (V+y))
MV / (s^3* Mw) = M (V+y) / ((s+x)^3 Mw
V (s+x)^3 = (V+y) s^3 ....(2)
What is needed is the ratio (x/s) / (y/V)
Let a=(x/s) and b= y/V
(2) gives V [s(1+x/s)]^3 = V(1+y/V) s^3
(1+x/s)^3 = (1+y/V)
(1+ a^3 + 3a(a+1)) = 1 + b
a^3+3a^2+3a = b
a(a^2+3a + 3) = b
We are given that a and b are relatively prime. So the only possible values are a = 1 and b = 1^2+3(1)+3 = 7
Therefore, a + b is 8 but it seemed to be wrong.. Presumably, your assumption is that the submerged height remains the same and that the orientation remains the same but i still can't understand how area could increase only by a factor off 2 since it should increase by (s+x)^2 - s^2 = 2xs + x^2
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This is wrong: (M /s^3 / Mw / V) = (M / (s+x)^3 / Mw / (V+y))
What you've essentially written is ratio of densities initially = ratio of densities finally. But this is not so. I think you misunderstood "submerged portion stays same".
Remember, only the depth of the submerged portion remained same not the volume. So with the depth constant, volume of submerged portion actually increased (because length & breadth of cube are also changing).
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Okay, got it, keeping the same variables and adding h = submerged depth and using the principle that a floating body displaces its own weight,, we get :
h s^2 (Mw / V) = Mass of cube * g
where h*s^2 is the volume of water displaced and Mw / V is the initial density of water and g is the gravitational constant
After expansion, the weight of water displaced =
h (s+x)^2 (Mw)/(V+y) where (s+x) is the new dimension of the cube and V+y is the increased volume of the body of water
Now since weight of the cube remains the same
(h s^2 Mw ) / V = h (s+x)^2 Mw / (V+y) h s^2 Mw * (V+y) = h (s+x)^2 Mw V s^2(1+x/s)^2 * V = s^2 (V+y) (1+X/s)^2 = (V+y)/V = (1+y/V)
What we need is the ratio (x/s)/ (y/v)
let a = x/s and b = y/V
we get
(1+a)^2 = (1+b) a^2+2a+1 = 1 + b b = a^2+2a = a(a+2)
We are given that a and b are relatively prime. Given this, the above can occur only if a=1 and b = 3
giving a/b = 1/3
and a + b = 4
But again, this does not agree with what is given.
C'ant help being a little perplexed
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@Sundar R – You've got every reason to be perplexed.
I think the answer you are pointing to, "4" is more accurate than "3".
Here's how we can arrive at "3" from your method.. (1+a)^2 is approximated using binomial approximation. So it will become 1+2a
Hence
(1+a)^2 = (1+b)
1+2a = 1+b
2a=b
But binomial approximation is only right when a is very close to 0. Usually thermal expansion coefficients ARE very small numbers so this should hold true. But in this case picking numbers like 1 or 2 or 3 the approximation fails.. So your approach points to a more accurate answer..
I suggest you talk about this with the author of the question.. Let me know what happens..
I'm sure it has a perfectly good explanation, but just take a look at mine.
Let the height of the cube be h, the density of the cube d, the density of the liquid D an the coeff of linear expansion of cube and liquid a and A.
So the height of cube submerged initially is h*d/D. After the temperature rise, d=>d/(1+3aT).......where T represents rise in temperature. D=>D/(1+3AT) h=>h(1+aT)
Equating the two, we get (1+3AT)(1+aT)=1+3aT Neglecting 3aA*T^2, 3A=2a.
Thus a/A=3/2 Hence the sum is 5.
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Your answer is absolutely right .. but you made a slight error. Firstly a liquid does not have a coefficient of linear expansion so writing gamma as 3A was not necessary. Plus the question says coefficient of volume expansion of the liquid not linear expansion. So the required ratio is a/gamma and not a/A. Answer is still 3. You found a/A.
You are making the same assumption that i intially made, that the submerged proportion of height remains the same, whereas only the height remains the same as pointed out to me.. Also., are you approximating the volume by (1+3xT) term ?
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It is given in the question that the height submerged remains constant, which I have used.
The basic formula for volume expansion is V(1+3xT). I don't think this is an approximation, at least as long as we assume x to be constant with T.
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@Ketan Kanishka – Since this is not a dynamic problem, we d'ont need to bring in calculus and we can just use the plain formula of Volume = (side)^3 or Height * area^2 and then look at changes
Should you not consider the change in density of the cube??
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ρ l ′ = ρ l ( 1 − γ Δ T ) This equation is considering change in density.
If γ is coefficient of volume expansion, then V ′ = V ( 1 + γ Δ T )
Since density is inversely proportional to volume (mass remains same), we have a negative sign for new density.
it past year iit ques. i myself was thinking to upload it
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Let A be the area of the cube, L be the length of the cube. Let x be the length of the cube inside the liquid.
Let ρ c and ρ l denote the density of the cube and liquid respectively.
Initially, for the equilibrium of the cube, A L ρ c g = A x ρ l g
After the temperature has risen by Δ T ,
We know that,
A ′ = A ( 1 + 2 α Δ T )
ρ l ′ = ρ l ( 1 − γ Δ T )
For the equilibrium of the cube,with an approximation,
A ( 1 + 2 α Δ T ) x ρ l ( 1 − γ Δ T ) = A L ρ c
1 + 2 α Δ T − γ Δ T = 1
or γ = 2 α .