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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

$T h e f i g u r e s h o w s a m e r c u r y$ $b a r o m e t e r .$ $f i n d r e a d i n g o f t h e w e i g h i n g$ $m a c h i n e . D e n s i t y o f m e r c u r y$ $= 13.6 g / {c m}^{3} . A r e a o f c r o s s e c t i o n$ $o f g l a s s t u b e = 20 {c m}^{2} . N e g l e c t$ $w e i g h t o f g l a s s t u b e .$
Commented by math solver last updated on 14/Nov/17

$is the answer 136 N$
Commented by math solver last updated on 14/Nov/17

$lol, how is the area of cross section of$ $glass tube 20 cm possible!!$
Commented by Physics lover last updated on 14/Nov/17

$y e a h, 136 N i s t h e r i g h t a n s w e r .$
Commented by Physics lover last updated on 14/Nov/17

$c a n y o u p l z p r o v i d e s o l u t i o n .$
Commented by Physics lover last updated on 14/Nov/17

$a n d a l s o f i n d r e a d i n g i f t h e$ $t u b e i s 1 m l o n g .$
Commented by Physics lover last updated on 14/Nov/17

$w e l l i d o n t . : p$
Commented by Physics lover last updated on 14/Nov/17

$i j u s t w a n t t o s h o w i t t o s o m e o n e .$
Commented by Physics lover last updated on 14/Nov/17

$w h o b e l i e v e s t h a t t h e g l a s s t u b e$ $w i l l j u m p u p w a r d s . s m h .$
Commented by Physics lover last updated on 14/Nov/17

$w h i c h i s c o m p l e t e v i o l a t i o n$ $o f l a w o f c o n s e r v a t i o n$ $o f e n e r g y . I s n t i t ?$
Commented by Physics lover last updated on 14/Nov/17

$n v m, d o n t g i v e t h e s o l u t i o n .$ $i j u s t w a n t t o c o n f i r m m y a n s w e r .$ $C a n y o u p l s t e l l t h e a n s w e r f o r$ $s e c o n d p a r t .$ $m y a n s w e r i s 206.72 N i s$ $m y a n s w e r .$
Commented by math solver last updated on 14/Nov/17

$it should be 272 N . .$
Commented by Physics lover last updated on 14/Nov/17

$l o l, t h e m e r c u r y w o n t r i s e$ $u p t o 100 c m . i t w i l l s t o p a t 76 c m .$
	Answered by math solver last updated on 14/Nov/17

	$reading of the weighing machine is$ $given by : V ρ g$ $v = volume ρ = density g = acc . due to gravity$ $so lets put the value now,$ $v = area \times height$ $\Rightarrow (20 \times 10^{- 4}) m^{2} \times (50 \times 10^{-^{} 2} m) .$ $ρ = 13.6 \times 10^{3} kg / m^{3} .$ $g = 10 m / s^{2} .$ $so, reading = 136 N .$ $Note : we have not taken atmospheric$ $pressure in account as it is closed .$
	Commented by Physics lover last updated on 14/Nov/17

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

	$i {didn}^{'} t understand why the upper$ $force is not taken .$ $i mean A (P - h ρ g) ?$ $when it was 50 cm and when it [is 100$ $in both cases it is closed . right ?$
	Commented by Physics lover last updated on 14/Nov/17

	$n o s i r t h a t s n o t t h e r i g h t w a y$ $t h o u g h t t h e a n s w e r i s r i g h t .$
	Commented by Physics lover last updated on 14/Nov/17

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

	$t h a t r e a d i n g i s n o t d i r e c t l y d u e$ $t o w e i g h t o f t h e m e r c u r y b u t$ $d u e t o a t m o s p h e r i c p r e s s u r e . S o$ $y o u c a n n o t i g n o r e i t .$
	Commented by Physics lover last updated on 14/Nov/17

	$a n d r e a d t h e f i r s t l i n e o f t h e$ $q u e s t i o n . i t s a m e r c u r y b a r o m e t e r$ $t h e t u b e i s n o t c l o s e d . i t i s o p e n a t$ $t h e l o w e r e n d .$
	Commented by math solver last updated on 14/Nov/17

	$well, i think i am correct .$ $you can just say i have skip 1 step .$ $coz whenever it is closed you will$ $always notice the atmospheric pressure$ $is not taken in account in finding the$ $final ans .$
	Commented by Physics lover last updated on 14/Nov/17

	$w e l l y e a h, t h a t s r i g h t .$
	Commented by Physics lover last updated on 14/Nov/17

	$w h a t a b o u t t h e s e c o n d p a r t,$ $c o n s i d e r i n g t h e g l a s s t u b e t o$ $b e o p e n a t t h e l o w e r e n d .$
	Commented by math solver last updated on 14/Nov/17

	$barometer is open at lower end and$ $not upper! .$ $if it is open from upper end then only$ $you have to take atmopheric pressure$ $in account .$ $hey, can you show F . B . D when$ $tube is open as you have already$ $taken atmospheric pressure in account ?$
	Commented by math solver last updated on 15/Nov/17

	$hmmm, finally we have got correct$ $result :)$
	Commented by Physics lover last updated on 15/Nov/17

	$: D$
	Commented by Physics lover last updated on 15/Nov/17

	$b e c a u s e t h e m e r c u r y i s n o t$ $i n c o n t a c t w i t h t h e u p p e r e n d$ $o f t h e g l a s s t u b e a n d a l i q u i d$ $c a n n o t e x e r t a n y t a n g e n t i a l$ $f o r c e o n t h e s u r f a c e o f t h e i n n e r$ $g l a s s t u b e .$ $i n o t h e r w o r d b e l o w t h e u p p e r$ $c i r c u l a r d i s c, t h e r e i s t o r r c e l l i a n$ $v a c u u m o r t h e p r e s s u r e i s z e r o$ $\Rightarrow A (p) = A (0) = 0$
	Commented by Physics lover last updated on 15/Nov/17

	$a n d b e t t e r c h e c k o u t t h e c o n s t r u c t i o n$ $o f a m e r c u r y b a r o m e t e r, i t$ $w i l l b e m o r e c l e a r t h e n . I t s i s a l w a y s$ $o p e n a t t h e l o w e r e n d a n d c l o s e d$ $a t t h e u p p e r e n d . i n b o t h c a s e s$ $i t i s a n o p e n g l a s s t u b e .$
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