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प्रश्न
A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom?
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उत्तर १
When the tube is held horizontally, the mercury thread of length 76 cm traps a length of air = 15 cm. A length of 9 cm of the tube will be left at the open end. The pressure of air enclosed in tube will be atmospheric pressure. Let area of cross-section of the tube be 1 sq. cm.
.’. P1 = 76 cm and V1 = 15 cm3
When the tube is held vertically, 15 cm air gets another 9 cm of air (filled in the right handside in the horizontal position) and let h cm of mercury flows out to balance the atmospheric pressure. Then the heights of air column and mercury column are (24 + h) cm and (76 – h) cm respectively.
The pressure of air = 76 - (76 - h) = h cm of mercury
`:. V_2 = (24 + h) cm^3` and `P_2 = h cm`
if we assume that temperature remain constant then
`P_1V_1 = P_2V_2 or 76 xx 15 = h xx (24 + h ) or h^2 + 24h - 1140 = 0`
or `h = (-24 +- sqrt((24)^2 + 4xx 1140))/2` = 23.8 cm or - 47.8 cm
Since h cannot be negative (because more mercury cannot flow into tube), there fore h = 23.8 cm
Thus in the verticle position of the tube , 23.8 cm of mercury flow out
उत्तर २
Length of the narrow bore, L = 1 m = 100 cm
Length of the mercury thread, l = 76 cm
Length of the air column between mercury and the closed end, la = 15 cm
Since the bore is held vertically in air with the open end at the bottom, the mercury length that occupies the air space is: 100 – (76 + 15) = 9 cm
Hence, the total length of the air column = 15 + 9 = 24 cm
Let h cm of mercury flow out as a result of atmospheric pressure.
∴Length of the air column in the bore = 24 + h cm
And, length of the mercury column = 76 – h cm
Initial pressure, P1 = 76 cm of mercury
Initial volume, V1 = 15 cm3
Final pressure, P2 = 76 – (76 – h) = h cm of mercury
Final volume, V2 = (24 + h) cm3
Temperature remains constant throughout the process.
∴P1V1 = P2V2
76 × 15 = h (24 + h)
h2 + 24h – 1140 = 0
`:. h = (-24+-sqrt((24)^2 + 4xx 1xx 1140))/(2xx1)`
= 23.8 cm or –47.8 cm
Height cannot be negative. Hence, 23.8 cm of mercury will flow out from the bore and 52.2 cm of mercury will remain in it. The length of the air column will be 24 + 23.8 = 47.8 cm.
संबंधित प्रश्न
Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms:
| Substance | Atomic Mass (u) | Density (103 Kg m-3) |
| Carbon (diamond) | 12.01 | 2.22 |
| Gold | 197.00 | 19.32 |
| Nitrogen (liquid) | 14.01 | 1.00 |
| Lithium | 6.94 | 0.53 |
| Fluorine (liquid) | 19.00 | 1.14 |
[Hint: Assume the atoms to be ‘tightly packed’ in a solid or liquid phase, and use the known value of Avogadro’s number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few Å].
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