Question

A thin tube sealed at both ends is 100 cm long . It lies horizontally,the middle 20cm containing mercury and 2 equal ends containing air at std. atm. pressur. If the tube is now turned to a vertical position, by what amount will the mercury be displaced? (given cross scetion is uniform)

Answer 1

@ParthKohli @ganeshie8

Answer 2

oh god why do you hate me so much :(

Answer 3

hint: boyle's law

Answer 4

Lolxcc @pk

Answer 5

Okk @priyar
Did u get it ??

Answer 6

no...

Answer 7

hey I'm going out for a walk but\[P\ell = P_1 \ell_1 = P_2 \ell_2\]\[P_2 = P_1 + \rho_{mercury}\cdot 10 ~\rm cm \cdot g\]

Answer 8

Better write as l only because we can assume pressure equal to h...

Answer 9

I mean l2=l1+10

Answer 10

why 10?

Answer 11

@samigupta8?

Answer 12

why h=10?

Answer 13

pls explain properly..

Answer 14

why is h=10??? why not h=20? anyone?

Answer 15

Sorry i wrote it wrong ...it should be what u told just now 20

Answer 16

I just saw parth's post and thought that the mercury column was 10cm long..

Answer 17

how did you both make the same mistake?!!

Answer 18

I didn't see the value in ur question for mercury column length

Answer 19

and..density of mercury is not given!

Answer 20

It doesn't matter ...bcoz the pressure are being taken in term of length so we can assume that l2=l1+20

Answer 21

Sorry @priyar ..i m wrong ...
I were considering a very diff case from ur question..
Wait lemme correct it in my notebook and then i should tell you how to move with this problem

Answer 22

hmm..okay.. thats why i got confused..

Answer 23

What are the options pls..?

Answer 24

sorry guys I just came back
that 10 cm was a typo

Answer 25

ok.. but density of mercury isn't given..

Answer 26

that's ok... actually the thing is that pressure is measured as length of mercury too. 1 atm = 76 cm mercury.

Answer 27

i just came to ask if we were to use that..

Answer 28

yes, we have to use that.
in the upper column of gas, the pressure is the atmospheric pressure (76 cm). and in the lower one, it is atmospheric pressure + 20 cm = 96 cm mercury.

Answer 29

U don't need it even...
U can do it in this fashion...
P2=P1+rho *20*g
Then P2= 40*76/80-h
And P1=40*P/h
And now write P as 76 cm of Hg
Now eq iz this
40*76/80-h -40*76/h=20

Answer 30

Here i assumed p2 is pressure in upper column of the tube
And p1 in lower column

Answer 31

how P2= 40*76/80-h?

Answer 32

Sorry i wrote it wrong p2 is pressure in lower column

Answer 33

still explain how you got P2..

Answer 34

P*40=p1*h
P*40=p2(80-h)

Answer 35

isn't that (60 - h) ?.. and the answer is not matching.....

Answer 36

Why would it be 60-h?

Answer 37

coz 40+20 = 60..right.. and 2 is at 60cm from the top.. @samigupta8

Answer 38

@IrishBoy123 @imqwerty

Answer 39

@ParthKohli @Michele_Laino

Answer 40

The answer should be a displacement of 5.18 cm.

Answer 41

@Vincent-Lyon.Fr that's correct! pls explain how to solve..

Answer 42

Since section is constant, PV = cst is the same as \(P.l=cst\)
If pressures are expressed in cm Hg, and lengths in cm, then \(\rho_{Hg}.g=1\)
It makes equations to solve easier when expressed numerically.

You have to solve:

\(P_o.l_o=P_1.l_1=P_2.l_2\) (Boyle's law)
\(P_o=76\) and \(l_o=40\)
\(l_1+l_2=80\)
\(P_2-P_1=20\)

Answer 43

how ρHg.g=1 ?

Answer 44

pressure = ρHg.g.h right?

Answer 45

@Vincent-Lyon.Fr

Answer 46

If h = 1 cm, then P = \(\rho_{Hg}.g.h=1\; cm Hg\)

Answer 47

yes...okay

Answer 48

are these what you have taken as l1, l2