A mercury barometer reads 75 cm. Now 3 cm3 of atmospheric air is introduced into the tube. The mercury falls to a height of 65 cm and the length of air column above the mercury is found to be 15 cm. Calculate the cross-sectional area of the barometer tube.
Answers
Answer:
Here there are given two separate cases,
We see that when 3cm3 of air is allowed inside the barometer, the mercury column falls to length of 65cm and 15cm is length of air column above the mercury column.
From here it can be inferred that total height of the mercury column = 65 + 15 = 80cm
In the first case , when height of the mercury is 75cm, the column of air above the mercury is
80-75 = 5cm
Let the volume of air be v
Height of air column = 5cm
Let the cross sectional area be c
Now volume = cross sectional area * height
Therefore v = c*5……………………………eqn1
Now in the next case,
Height of air column above mercury when the height of mercury reduces to 65cm is,
15cm as given.
Volume of air which is added is 3cm3
Hence the new volume = v+3
Hence v+3 = c * 15………………………….eqn 2
Taking ratio of eqn1 and eqn2 we get:
v/(v+3) =( c*5)/(c*15)
v/(v+3) = 1/3
3v = v+3
2v = 3
v = 3/2
using equation 1 we get:
c*5 = 3/2
c = 3/10cm2
Answer:
1.5cm^2
Explanation:
Initial volume of air v1=3cm^2 at a pressure p1=atmospheric pressure=75cm of mercury . the level of mercury falls to 65cm because the air in the tube exert a pressure on it. Therefore, pressure of air inside the tube, p2=75-65=10 cm of HG. Length of air column =15cm. If A cm^2 is the area of cross section of tube,then volume of the trapped air v2= 15×A cm^3...By Boyle.s law,p1v1=p2v2..75×3=10×(15×A)..So,A=75×3/10×15=1.5cm^2.. Mark as brainlieast...hope it helps you...