At what depth below the surface of water will pressure be equal to twice the atmospheric pressure? Atmospheric pressure = 10⁵ N/m², density of water is 10³ kg m–3 and g = 9.8 ms–2.
Answers
Answer:
If your ears have ever popped on a plane flight or ached during a deep dive in a swimming pool, you have experienced the effect of depth on pressure in a fluid. At the Earth’s surface, the air pressure exerted on you is a result of the weight of air above you. This pressure is reduced as you climb up in altitude and the weight of air above you decreases. Under water, the pressure exerted on you increases with increasing depth. In this case, the pressure being exerted upon you is a result of both the weight of water above you and that of the atmosphere above you. You may notice an air pressure change on an elevator ride that transports you many stories, but you need only dive a meter or so below the surface of a pool to feel a pressure increase. The difference is that water is much denser than air, about 775 times as dense. Consider the container in Figure 1.
Explanation:
Sol: Pressure at a depth h below the surface of water = atmospheric pressure + pressure due to water column of height h.
If atmospheric pressure is P, then 2P = P + hρg or hρg = P
But P = 10 N cm–2
= 10N/(10–2 m)2
= 105 N m–2
Hence h = P/ρg
= 105/(103 × 9.8)
= 10.2 m
Thus a water column of height 10.2 m exerts a pressure equal to the atmospheric pressure.
Given:
The atmospheric pressure, P° = 10⁵ N/m²
The density of water, ρ = 10³ kg/m3
g = 9.8 m/s2
The pressure below the surface of the water, P = 2 P°
To Find:
The depth 'h' at which P = 2 P°.
Calculation:
- The pressure below the surface of the water is given as:
P = P° + hρg
⇒ 2 P° = P° + hρg
⇒ 2 P°-P° = hρg
⇒ P° = hρg
⇒ h = P°/ρg
⇒ h = 10⁵/(10³ × 9.8)