A1 and A2 represent area of cross-section of bottom and top of tank which is completely filled with water. If A1 = 10 cm square and A2 = 30 cm square and the volume is 1 Litre. The force exerted by the water on the bottom is?
Answers
The force exerted by the water on the bottom is 102 N.
Explanation:
=> It is given that,
Area of cross-section of bottom, A1 = 10 cm² = 10⁻³ m²
Area of cross-section of the top of tank, A2 = 30 cm²
Height of the tank, h = 10cm = 10* 10⁻² = 10⁻¹ m
Atmospheric pressure = Patm = 1.01 * 10⁵
Density of water, ρ = 1000 kg/m³
=> Since lid of tank is open, total pressure :
Ptotal = Patm + ρgh
Ptotal = 1.01 * 10⁵ + 1000 * 10 * 10⁻¹
= 101 * 10³ + 10³
= 10³ (101 + 1)
= 10³ (102)
=> The force exerted by the water on the bottom, F:
F = pressure * area
= 10³ (102) * A₁
= 10³ (102) * 10⁻³
= 102 N
Thus, The force exerted by the water on the bottom is 102 N and it will always act downwards.
Learn more:
Q:1 A glass full of water has a bottom of area 20 cm², top of area 20 cm², height 20 cm and volume half a liter. (a) Find the force exerted by the water on the bottom. (b) Considering the equilibrium of the water, find the resultant force exerted by the sides of the glass on the water. Atmospheric pressure = 1.0 x 10⁵ N/m². Density of water = 1000 kg/m³ and g = 10 m/s². Take all numbers to be exact.
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Q:2 Water is filled in a rectangular tank of size 3 m x 2 m x 1 m. (a) Find the total force exerted by the water on the bottom surface of the tank. (b) Consider a vertical side of area 2 m x 1 m. Take a horizontal strip of width ẟx meter in this side, situated at a depth of x meter from the surface of the water. Find the force by the water on this strip. (c) Find the torque of the force calculated in part (b) about the bottom of edge of this side. (d) Find the total force by the water on this side. (e) Find the total torque by the water on the side about the bottom edge. Neglect the atmospheric pressure and take g = 10 m/s².
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Answer:102N
Explanation:
