. A water filled cylinder of height 50 cm and base area 20 cm is placed on a table with
the base on the table. The thrust offered by water on the table is
Answers
The thrust offered by water on the table is 9.8 N.
• Given,
Height (h) of the cylinder = 50 cm
Area (a) of the base of the cylinder = 20 cm²
• Thrust offered by water means force exerted by water on the table.
Force = Pressure × Area
• Pressure exerted by water on table = height of the cylinder (h) × density of water (d) × acceleration due to gravity (g)
Density of water = 1 g/cm³
Acceleration due to gravity =
980 cm/s²
Pressure = 50 cm × 1 g/cm³ × 980 cm/s² = 49000 g/cm/s² = 49 × 10³ g/cm/s²
• Now, thrust (F) offered by water on the table = P × A
F = 49 × 10³ g/cm/s² × 20 cm²
=> F = 980 × 10³ g/cm/s²
=> F = 9.8 × 10² × 10³ dynes ( 1 g/cm/s = 1 dyne) = 9.8 × 10^5 dynes
=> F = 9.8 N ( 10^5 dynes = 1 N)
• Thus, the thrust offered by water on the table is 9.8 N.
The answer is 10N
step by step explanation is provided in the image above
