Question

The area of base of a cylindrical vessel is 300 cm^2. Water (density = 1000 kg/m^3) is poured into it up to a depth of 6 cm. Calculate the pressure and the thrust of water on the base (g = 10 m/s^2)

Answer 1

Explanation:

static pressure exerted by water is

= density*depth*accln due to gravity

= (1000*0.06*10) = 600 N/m^2

thrust force = pressure*area

= 600*300*0.000=18 newton

Answer 2

Question -

The area of base of a cylindrical vessel is 300 cm^2. Water (density = 1000 kg/m^3) is poured into it up to a depth of 6 cm. Calculate the pressure and the thrust of water on the base (g = 10 m/s^2).

Given -

• Area of the base of a cylindrical vessel = 300 cm².
• Density of water = 1000 kg/m³.
• Depth or height = 6 cm
• Acceleration due to gravity = 10 m/s².

To find -

• The pressure.
• The thrust of water on the base.

Solution -

Height = 6 cm to m

$\bold{➺ \: \dfrac{6}{100}m } \\ \\ \bold{➺ \: 0.06 \: m}$

Density = 1000 kg/m³

Acceleration due to gravity = 10 m/s²

∴ Pressure = hpg

➺ 0.06 m × 1000 kg/m³ × 10 m/s²

➺ 600 Pa ........ (ans)

To calculate the thrust of water on the base,

$\bold{Pressure = \dfrac{ Force }{Area} } \\ \\ \bold{Force = Pressure \times Area }$

➺ 600 Pa × 300 cm²

➺ 180000 to N

➺ 18 N ........(ans)