Pressure in Fluids and Atmospheric Pressure Exe-4B Atmospheric Pressure and it’s Measurement Numericals Answer Type for Class-9 ICSE Concise Physics. There is the solutions of Numericals Answer type Questions of your latest textbook which is applicable in 2023-24 academic session. Visit official Website CISCE for detail information about ICSE Board Class-9.
Pressure in Fluids Exe-4B Atmospheric Pressure and it’s Measurement Numericals Answer
(ICSE Class – 9 Physics Concise Selina Publishers)
| Board | ICSE |
| Class | 9th |
| Subject | Physics |
| Writer / Publication | Concise selina Publishers |
| Chapter-4 | Pressure in Fluids and Atmospheric Pressure |
| Exe – 4B | Atmospheric Pressure and it’s Measurement |
| Topics | Solution of Exe-4(B) Numericals Answer Type |
| Academic Session | 2023-2024 |
Exe-4B Atmospheric Pressure and it’s Measurement Numericals Answer Type
Ch-4 Pressure in Fluids and Atmospheric Pressure Physics Class-9 ICSE Concise
Page 113
Question 1. Convert 1 mm of Hg into pascal. Take density of Hg = 13.6 x 103 kg m-3 and g = 9.8 m s-2 .
Answer:
We know that, the pressure is given by the formula,
δP = hρg
Putting the given values in above equation, we get,
δP = 0.001 x 13.6 x 1000 x 9.8
δP = 133.28 Pa
Therefore, 1 mm of Hg is equal to 133.28 pascals.
Question 2. At a given place, a mercury barometer records a pressure of 0.70 m of Hg. What would be the height of water column if mercury in barometer is replaced by water? Take density of mercury to be = 13.6 x 103
Answer:
Relative Density of Hg = 1.36 = kgm 13.6 × 103 kgm-3
Acceleration due to gravity , g = 9.8 msms-2
Height of mercury column = 0.70 m
∴ Pressure , P = hρg
or , P = (0.7)(1.36 × 103)(9.8) pascal
or , P = 93.3 × 103 Pa
Let h be the height of water column
Then , P = h (density of water) g
or , 93.3× 103 = h × 103 ×9.8
or , h = 9.52 m
Question 3. At sea level, the atmospheric pressure is 76 cm of Hg. If air pressure falls by 10 mm of Hg per 120m of ascent, what is the height of a hill where the barometer reads 70 cm Hg. State the assumption made by you.
Answer:
Atmospheric pressure , P = 76 cm Hg
Rate at which pressure falls = 10 mm of Hg per 120 m of ascent = 1 cm of Hg per 120 m of ascent
Let h be the height of the hill.
Pressure at hill , P’ = 70 cm Hg
Total fall in pressure = P – P’ = (76 – 70) cm Hg = 6 cm Hg
Now , fall in pressure is 1 cm Hg for every 120 m increase in height
Thus , if the fall in pressure is 6 cm Hg , increase in height shall be (6 × 120)m = 720 m
∴ Height of the hill = 720 m
Assumption: Atmospheric pressure falls linearly with ascent.
(Pressure in Fluids and Atmospheric Pressure Exe-4B Numericals Class-9 ICSE)
Question 4. At sea level, the atmospheric pressure is 1.04 x 105 Pa. Assuming g = 10 m s-2 and density of air to be uniform and equal to 1.3 kg m-3, find the height of the atmosphere.
Answer:
Pressure = hdg
h = Pressure / (dg)
= (1.04 × 105 Pa) / (1.3 kg/m³ × 10 m/s²)
= 8000 m
= 8 kilometers
Height of the atmosphere is 8 kilometers
Question 5. Assuming the density of air to be 1.295 kg m-3, find the fall in barometric height in mm of Hg at a height of 107 m above the sea level. Take density of mercury = 13.6 × 103 kg m-3.
Answer:
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