21. The following figure shows a mercury barometer tu
in the upper part of this tube. Calculate (i) the volume
the trapped air (ii) pressure of the trapped air if nomm
atmospheric pressure is 76 cm of Hg and (iii) the volum
of cross-section area 1.2 cm². There is some air
trapp
of air at NTP. (Figure 4.30)
16) 12 cm; (ii) 27.2 If/cm²; (iii) 10.3 4
Answers
Answer:
i ) The volume of the trapped air will be (atmospheric pressure = 70 cm of Hg and density of Hg = 13.6 g/cm3 )
ii) Atmospheric pressure falls linearly with ascent.
Here atmospheric pressure at sea level =76cm of Hg
For every 10mm=1cm decrease there is a increase of 120m so, at 70cm of Hg
height=76−70=6×120=720m
iii) By Pascal's law, the pressure increased using one piston is communicated everywhere in the fluid. Thus, the same pressure acts on the other piston.
Cross-sectional area of piston 1 is A
1
=100 cm
2
=0.01 m
2
Force on piston 1 is F
1
=10
7
dyne=100 N
Force on piston 2 is F
2
=2000 kg wt=2000×10=2×10
4
N
Let the cross-sectional area of Piston 2 be A
2
=A.
By Pascal's law, P
1
=P
2
⇒
A
1
F
1
=
A
2
F
2
⇒A
2
=
F
1
F
2
A
1
⇒A=
100
2×10
4
×0.01
=2 m
2
=2×10
4
cm
2