A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom ?

Question

A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom ?,

MARKS App · Accepted Answer

In horizontal position, the bore will leave an air-column of 9 cm as shown in figure (a) so as to balance the atmospheric pressure ( = 76 cm of Hg ) $ ∴ P 1 ​ = 76 cm & V o l u m eo f ai rco l u mn =15 \mathrm{~cm}^3 a ss u min g a re a o f crosssec t i o n o f t u b e =1 \mathrm{~cm}^3 T o t a ll e n g t h o f ai rco l u mnin t h e t u b e =24 \mathrm{~cm} . im g src = " h ttp s : // c d n − q u es t i o n − p oo l . g e t ma r k s . a pp / m o d u l es / c b se / J o W y J 4 ltT o f w f i H f 823 u D k v E 4 o w ​ k I w b 7 g S Y u L P ​ SBM . or i g ina l . f u ll s i ze . p n g " > b r > I n v er t i c a lp os i t i o na ss h o w nin f i gu re ( b ) i f h^{\prime} \mathrm{cm} o f m erc u ry f l o w so u tt o ba l an ce t h e a t m os p h er i c p ress u re t h e nn e wp ress u reo nm erc u ryco l u mn =76-(76-h)= h \mathrm{~cm} o f \mathrm{Hg} T h e n e w v o l u m eo f ai rco l u mn =24+h . S in ce t h e t e m p er a t u rere main sco n s t an t d u r in g t h e t r an s i t i o n f ro mh or i zo n t a lt o v er t i c a lp os i t i o n . ​ ∴ P 1 ​ V 1 ​ = P 2 ​ V 2 ​ 76 × 15 = h × ( 24 + h ) or h 2 + 24 h − 1140 = 0 ⇒ h = 23.8 cm ​ He n ce , 23.8 \mathrm{~cm}$ of mercury flows out from the tube in vertical position.

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