You just have to tell whther I'm right or Wrong and Why..No need to do any calculations :

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 ...

5 Answers

1
The Enlightened One - jsg ·

None ? ! [11]...

1
nish ·

I 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

1
The Enlightened One - jsg ·

You 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...

1
buddyboyyash ·

@jsg

well i havent done fluids yet!!!...bt if u r right...be ready..u r about to prove archimedes wrong!!!

1
The Enlightened One - jsg ·

[16][17]

Your Answer

Close [X]