1@jsg
well i havent done fluids yet!!!...bt if u r right...be ready..u r about to prove archimedes wrong!!!
To all folks in TargetIIT , I'm weak in mechanics especially Fluid Mechanics and I don't know even the basic concepts [2] .
In the quest for more concepts..I'm posting an analysis of a fluid statics concept regarding Archimedes Principle , JUST TELL ME WHETHER I'M WRONG OR RIGHT AND IF I'M WRONG THEN WHY..
Q: A wooden block[M] with a coin[m] placed on its top , floats in water as as shown in the figure [given below] .The distance L,x and h are shown these.After sometime the coin falls into the water.Then :
[A] x decreases and h increases
[B] x increases and h decreases
[C] both increase
[D] both decrease
Answer given : D
My Mathematical solution :
I assumed the following :
A : Base area of container
d : Density of block
dcoin : Density of coin [I think it is always greater than d ,rt ? ]
x1 and x2 :Different values of x before and after coin falls
h1 and h2 :Different values of h before and after coin falls.
Initially [ before coin falls ] :
Height of water displaced = h1 = x1L(M+m)d.A------------------------(1)
Finally [after coin falls ] :
Height of water displaced = h2 = x2LMd.A+mdcoin.A---------------(2)
-> Now as coin falls from the block , it is obvious that x2 < x1 [so x decreases]
Now h1-h2 = M(x1-x2)L.d.A +mA(x1L.d - 1dcoin)
Doubt 1 : Isn't the answer [D] , only when x1=L ( I mean , h2 <h1 only when it is initially x1=L right ? ) ?
Doubt 2 : From (1) and (2) ,we can see that the term d is in the denominator so: I conclude that , as the density of the floating material ( d ) increases , the level of the liquid in which it floats ( h )[here it is water] decreases , AM I RIGHT / WRONG..if I 'm wrong , then please give the reason ...
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5 Answers
1I still dont know how u got the eq for height of water displaced
let area of cross-section of box be A,P be density of liquid
h1 x P x A = M + m ............
I tried it without much math...
In the first case,
let total volume of water displaced be P
P1 = volume to support M + volume required to support m
volume required to support m will be greater than volume of m(assuming m is more dense than the liquid)
but when coin falls
P2 = volume to support M + volume of m
P2<P1 since volume to support m > volume of m
therefore h decreases
it is easy to see that x also decreases
1You know what..I did the same itself..just check..
SO I SUPPOSE WE CAN SAY THAT WHEN DENSITY OF A FLOATING SUBSTANCES IS INCREASED [ I MEAN , A NEW SUBSTANCE WITH LARGER DENSITY IS USED ] , the level of water / any liquid recedes...
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