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Question Number 24211 by Physics lover last updated on 14/Nov/17

Commented by Physics lover last updated on 14/Nov/17

The figure shows a mercury barometer. find reading of the weighing machine. Density of mercury = 13.6 g/cm^3 .Area of crossection of glass tube = 20 cm^2 .Neglect weight of glass tube.

$${The}\:{figure}\:{shows}\:{a}\:{mercury} \\ $$$${barometer}. \\ $$$${find}\:{reading}\:{of}\:{the}\:{weighing}\: \\ $$$${machine}.\:{Density}\:{of}\:{mercury} \\ $$$$=\:\mathrm{13}.\mathrm{6}\:{g}/{cm}^{\mathrm{3}} \:.{Area}\:{of}\:{crossection} \\ $$$${of}\:{glass}\:{tube}\:=\:\mathrm{20}\:{cm}^{\mathrm{2}} .{Neglect} \\ $$$${weight}\:{of}\:{glass}\:{tube}. \\ $$$$ \\ $$

Commented by math solver last updated on 14/Nov/17

is the answer 136 N

$$\mathrm{is}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{136}\:\mathrm{N} \\ $$

Commented by math solver last updated on 14/Nov/17

lol , how is the area of cross section of glass tube 20 cm possible !!

$$\mathrm{lol}\:,\:\mathrm{how}\:\mathrm{is}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{cross}\:\mathrm{section}\:\mathrm{of}\: \\ $$$$\mathrm{glass}\:\mathrm{tube}\:\mathrm{20}\:\mathrm{cm}\:\mathrm{possible}\:!! \\ $$

Commented by Physics lover last updated on 14/Nov/17

yeah,136 N is the right answer.

$${yeah},\mathrm{136}\:{N}\:{is}\:{the}\:{right}\:{answer}. \\ $$

Commented by Physics lover last updated on 14/Nov/17

can you plz provide solution.

$${can}\:{you}\:{plz}\:{provide}\:{solution}. \\ $$

Commented by Physics lover last updated on 14/Nov/17

and also find reading if the tube is 1 m long.

$${and}\:{also}\:{find}\:{reading}\:{if}\:{the} \\ $$$${tube}\:{is}\:\mathrm{1}\:{m}\:{long}. \\ $$

Commented by Physics lover last updated on 14/Nov/17

well i dont.:p

$${well}\:{i}\:{dont}.:{p} \\ $$

Commented by Physics lover last updated on 14/Nov/17

i just want to show it to someone.

$${i}\:{just}\:{want}\:{to}\:{show}\:{it}\:{to}\:{someone}. \\ $$

Commented by Physics lover last updated on 14/Nov/17

who believes that the glass tube will jump upwards. smh.

$${who}\:{believes}\:{that}\:{the}\:{glass}\:{tube} \\ $$$${will}\:{jump}\:{upwards}.\:{smh}. \\ $$$$ \\ $$

Commented by Physics lover last updated on 14/Nov/17

which is complete violation of law of conservation of energy.Isnt it?

$${which}\:{is}\:{complete}\:{violation} \\ $$$${of}\:{law}\:{of}\:{conservation} \\ $$$${of}\:{energy}.{Isnt}\:{it}? \\ $$

Commented by Physics lover last updated on 14/Nov/17

nvm,dont give the solution. i just want to confirm my answer. Can you pls tell the answer for second part. my answer is 206.72 N is my answer.

$${nvm},{dont}\:{give}\:{the}\:{solution}. \\ $$$${i}\:{just}\:{want}\:{to}\:{confirm}\:{my}\:{answer}. \\ $$$${Can}\:{you}\:{pls}\:{tell}\:{the}\:{answer}\:{for} \\ $$$${second}\:{part}. \\ $$$${my}\:{answer}\:{is}\:\mathrm{206}.\mathrm{72}\:{N}\:{is} \\ $$$${my}\:{answer}. \\ $$

Commented by math solver last updated on 14/Nov/17

it should be 272 N..

$$\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:\mathrm{272}\:\mathrm{N}.. \\ $$

Commented by Physics lover last updated on 14/Nov/17

lol,the mercury wont rise upto 100 cm.it will stop at 76 cm.

$${lol},{the}\:{mercury}\:{wont}\:{rise}\: \\ $$$${upto}\:\mathrm{100}\:{cm}.{it}\:{will}\:{stop}\:{at}\:\mathrm{76}\:{cm}. \\ $$$$ \\ $$

Answered by math solver last updated on 14/Nov/17

reading of the weighing machine is given by : Vρ g v = volume ρ=density g= acc. due to gravity so lets put the value now, v= area × height ⇒( 20 ×10^(−4) )m^(2 ) × (50×10^(−^ 2) m). ρ = 13.6 ×10^(3 ) kg/m^3 . g=10m/s^2 . so, reading = 136N. Note: we have not taken atmospheric pressure in account as it is closed.

$$\mathrm{reading}\:\mathrm{of}\:\mathrm{the}\:\mathrm{weighing}\:\mathrm{machine}\:\mathrm{is}\: \\ $$$$\mathrm{given}\:\mathrm{by}\::\:\mathrm{V}\rho\:\mathrm{g} \\ $$$$ \\ $$$$\mathrm{v}\:=\:\mathrm{volume}\:\rho=\mathrm{density}\:\mathrm{g}=\:\mathrm{acc}.\:\mathrm{due}\:\mathrm{to}\:\mathrm{gravity} \\ $$$$\mathrm{so}\:\mathrm{lets}\:\mathrm{put}\:\mathrm{the}\:\mathrm{value}\:\mathrm{now},\: \\ $$$$\mathrm{v}=\:\mathrm{area}\:×\:\mathrm{height}\: \\ $$$$\Rightarrow\left(\:\mathrm{20}\:×\mathrm{10}^{−\mathrm{4}} \:\right)\mathrm{m}^{\mathrm{2}\:\:} \:×\:\left(\mathrm{50}×\mathrm{10}^{−^{} \mathrm{2}} \:\mathrm{m}\right). \\ $$$$\rho\:=\:\mathrm{13}.\mathrm{6}\:×\mathrm{10}^{\mathrm{3}\:} \:\mathrm{kg}/\mathrm{m}^{\mathrm{3}} . \\ $$$$\mathrm{g}=\mathrm{10m}/\mathrm{s}^{\mathrm{2}} . \\ $$$$\mathrm{so},\:\mathrm{reading}\:=\:\mathrm{136N}. \\ $$$$\mathrm{Note}:\:\mathrm{we}\:\mathrm{have}\:\mathrm{not}\:\mathrm{taken}\:\mathrm{atmospheric}\: \\ $$$$\mathrm{pressure}\:\mathrm{in}\:\mathrm{account}\:\mathrm{as}\:\mathrm{it}\:\mathrm{is}\:\mathrm{closed}. \\ $$$$ \\ $$

Commented by Physics lover last updated on 14/Nov/17

Commented by math solver last updated on 14/Nov/17

i didn′t understand why the upper force is not taken. i mean A(P−hρg)? when it was 50 cm and when it[is 100 in both cases it is closed . right?

$$\mathrm{i}\:\mathrm{didn}'\mathrm{t}\:\mathrm{understand}\:\mathrm{why}\:\mathrm{the}\:\mathrm{upper}\: \\ $$$$\mathrm{force}\:\mathrm{is}\:\mathrm{not}\:\mathrm{taken}. \\ $$$$\mathrm{i}\:\mathrm{mean}\:\mathrm{A}\left(\mathrm{P}−\mathrm{h}\rho\mathrm{g}\right)? \\ $$$$\mathrm{when}\:\mathrm{it}\:\mathrm{was}\:\mathrm{50}\:\mathrm{cm}\:\mathrm{and}\:\mathrm{when}\:\mathrm{it}\left[\mathrm{is}\:\mathrm{100}\right. \\ $$$$\mathrm{in}\:\mathrm{both}\:\mathrm{cases}\:\mathrm{it}\:\mathrm{is}\:\mathrm{closed}\:.\:\mathrm{right}? \\ $$

Commented by Physics lover last updated on 14/Nov/17

no sir thats not the right way thought the answer is right.

$${no}\:{sir}\:{thats}\:{not}\:{the}\:{right}\:{way} \\ $$$${thought}\:{the}\:{answer}\:{is}\:{right}. \\ $$

Commented by Physics lover last updated on 14/Nov/17

Commented by Physics lover last updated on 14/Nov/17

that reading is not directly due to weight of the mercury but due to atmospheric pressure.So you cannot ignore it.

$${that}\:{reading}\:{is}\:{not}\:{directly}\:{due} \\ $$$${to}\:{weight}\:{of}\:{the}\:{mercury}\:{but} \\ $$$${due}\:{to}\:{atmospheric}\:{pressure}.{So} \\ $$$${you}\:{cannot}\:{ignore}\:{it}. \\ $$

Commented by Physics lover last updated on 14/Nov/17

and read the first line of the question.its a mercury barometer the tube is not closed. it is open at the lower end.

$${and}\:{read}\:{the}\:{first}\:{line}\:{of}\:{the}\: \\ $$$${question}.{its}\:{a}\:{mercury}\:{barometer} \\ $$$${the}\:{tube}\:{is}\:{not}\:{closed}.\:{it}\:{is}\:{open}\:{at} \\ $$$${the}\:{lower}\:{end}. \\ $$

Commented by math solver last updated on 14/Nov/17

well , i think i am correct . you can just say i have skip 1 step. coz whenever it is closed you will always notice the atmospheric pressure is not taken in account in finding the final ans.

$$\mathrm{well}\:,\:\mathrm{i}\:\mathrm{think}\:\mathrm{i}\:\mathrm{am}\:\mathrm{correct}\:. \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{just}\:\mathrm{say}\:\mathrm{i}\:\mathrm{have}\:\mathrm{skip}\:\mathrm{1}\:\mathrm{step}. \\ $$$$\mathrm{coz}\:\mathrm{whenever}\:\mathrm{it}\:\mathrm{is}\:\mathrm{closed}\:\mathrm{you}\:\mathrm{will}\: \\ $$$$\mathrm{always}\:\mathrm{notice}\:\mathrm{the}\:\mathrm{atmospheric}\:\mathrm{pressure} \\ $$$$\mathrm{is}\:\mathrm{not}\:\mathrm{taken}\:\mathrm{in}\:\mathrm{account}\:\mathrm{in}\:\mathrm{finding}\:\mathrm{the}\: \\ $$$$\mathrm{final}\:\mathrm{ans}. \\ $$$$ \\ $$

Commented by Physics lover last updated on 14/Nov/17

well yeah, thats right.

$${well}\:{yeah},\:{thats}\:{right}.\: \\ $$

Commented by Physics lover last updated on 14/Nov/17

what about the second part, considering the glass tube to be open at the lower end.

$${what}\:{about}\:{the}\:{second}\:{part}, \\ $$$${considering}\:{the}\:{glass}\:{tube}\:{to} \\ $$$${be}\:{open}\:{at}\:{the}\:{lower}\:{end}. \\ $$

Commented by math solver last updated on 14/Nov/17

barometer is open at lower end and not upper !. if it is open from upper end then only you have to take atmopheric pressure in account . hey, can you show F.B.D when tube is open as you have already taken atmospheric pressure in account?

$$\mathrm{barometer}\:\mathrm{is}\:\mathrm{open}\:\mathrm{at}\:\mathrm{lower}\:\mathrm{end}\:\mathrm{and}\: \\ $$$$\mathrm{not}\:\mathrm{upper}\:!. \\ $$$$\mathrm{if}\:\mathrm{it}\:\mathrm{is}\:\mathrm{open}\:\mathrm{from}\:\mathrm{upper}\:\mathrm{end}\:\mathrm{then}\:\mathrm{only} \\ $$$$\mathrm{you}\:\mathrm{have}\:\mathrm{to}\:\mathrm{take}\:\mathrm{atmopheric}\:\mathrm{pressure}\: \\ $$$$\mathrm{in}\:\mathrm{account}\:.\: \\ $$$$\mathrm{hey},\:\mathrm{can}\:\mathrm{you}\:\mathrm{show}\:\:\mathrm{F}.\mathrm{B}.\mathrm{D}\:\mathrm{when}\: \\ $$$$\mathrm{tube}\:\mathrm{is}\:\mathrm{open}\:\mathrm{as}\:\mathrm{you}\:\mathrm{have}\:\mathrm{already}\: \\ $$$$\mathrm{taken}\:\mathrm{atmospheric}\:\mathrm{pressure}\:\mathrm{in}\:\mathrm{account}? \\ $$$$ \\ $$

Commented by math solver last updated on 15/Nov/17

hmmm, finally we have got correct result :)

$$\mathrm{hmmm},\:\mathrm{finally}\:\mathrm{we}\:\mathrm{have}\:\mathrm{got}\:\mathrm{correct} \\ $$$$\left.\mathrm{result}\::\right) \\ $$

Commented by Physics lover last updated on 15/Nov/17

:D

$$\:\:\::\boldsymbol{\mathrm{D}} \\ $$

Commented by Physics lover last updated on 15/Nov/17

because the mercury is not in contact with the upper end of the glass tube and a liquid cannot exert any tangential force on the surface of the inner glass tube. in other word below the upper circular disc,there is torrcellian vacuum or the pressure is zero ⇒A(p) = A(0) = 0

$${because}\:{the}\:{mercury}\:{is}\:{not} \\ $$$${in}\:{contact}\:{with}\:{the}\:{upper}\:{end} \\ $$$${of}\:{the}\:{glass}\:{tube}\:{and}\:{a}\:{liquid} \\ $$$${cannot}\:{exert}\:{any}\:{tangential} \\ $$$${force}\:{on}\:{the}\:{surface}\:{of}\:{the}\:{inner} \\ $$$${glass}\:{tube}. \\ $$$$\:{in}\:{other}\:{word}\:{below}\:{the}\:{upper} \\ $$$${circular}\:{disc},{there}\:{is}\:{torrcellian} \\ $$$${vacuum}\:{or}\:{the}\:{pressure}\:{is}\:{zero} \\ $$$$\Rightarrow{A}\left({p}\right)\:=\:{A}\left(\mathrm{0}\right)\:=\:\mathrm{0} \\ $$

Commented by Physics lover last updated on 15/Nov/17

and better check out the construction of a mercury barometer, it will be more clear then.Its is always open at the lower end and closed at the upper end.in both cases it is an open glass tube.

$${and}\:{better}\:{check}\:{out}\:{the}\:{construction} \\ $$$${of}\:{a}\:{mercury}\:{barometer},\:{it} \\ $$$${will}\:{be}\:{more}\:{clear}\:{then}.{Its}\:{is}\:{always} \\ $$$${open}\:{at}\:{the}\:{lower}\:{end}\:{and}\:{closed} \\ $$$${at}\:{the}\:{upper}\:{end}.{in}\:{both}\:{cases} \\ $$$${it}\:{is}\:{an}\:{open}\:{glass}\:{tube}. \\ $$

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