An open and wide glass tube is immersed vertically in mercury in such a way that length 0.05 m extends above mercury level. The open end of the tube is closed and the tube is raised further by 0.43 m. The length of air column above mercury level in the tube will be: Take Pa t m=76 cm of mercury

Question

An open and wide glass tube is immersed vertically in mercury in such a way that length 0.05 m extends above mercury level. The open end of the tube is closed and the tube is raised further by 0.43 m. The length of air column above mercury level in the tube will be: Take Pa t m=76 cm of mercury (A) 0.215 m (B) 0.2 m (C) 0.1 m (D) 0.4 m

Tardigrade Admin · Accepted Answer

In the final condition. Let atmospheric pressure is P and ht of liquid column is h. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/a89e6a37527643e669ecc0f46c1c4772-.png" /> 380=(76- h )(43- h ) h =38 cm So, 48- h =10 cm =0.1 m

Q. An open and wide glass tube is immersed vertically in mercury in such a way that length 0.05m extends above mercury level. The open end of the tube is closed and the tube is raised further by 0.43m. The length of air column above mercury level in the tube will be : Take Patm​=76cm of mercury

0.215 m

0.2 m

0.1 m

0.4 m

In the final condition. Let atmospheric pressure is P and ht of liquid column is h. 380=(76−h)(43−h) h=38cm So, 48−h=10cm=0.1m

In the final condition.
Let atmospheric pressure is $P$ and $h t$ of liquid column is $h$ .

$380 = (76 - h) (43 - h)$
$h = 38 c m$
So, $48 - h = 10 c m = 0.1 m$