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 patm=76cmof mercury

Answer 1

Hey dear,

◆ Answer- 0.1 m

◆ Explanation-
# Given-
l1 = 0.05 m
l2 = 0.43 m
Patm = 760 mm Hg = 0.76 m Hg

# Solution-
When glass tube is raised by 0.05 m,
Length of mercury column P1 = Patm = 0.76 m
Volume of air V2 = 0.05 × Area = 0.05A

Consider d be length of air column above mercury.
When glass tube is raised by 0.43 m,
Length of mercury column P2 = 0.76-(0.05+0.43-d) = d + 0.28 m
Volume of air V2 = d × area = dA

We know, P1V1 = P2V2
0.76 × 0.05A = (d+0.28) × dA
d^2 + 0.28d - 0.0380 = 0

Solving quadratic eqn.
d = 0.1 0r d = -0.38

Length can't be negative, d = 0.1 m.

Therefore length of air column above mercury is 0.1 m.

Hope this is helpful...