Question

The area of base of a cylindrical vessel is 300cm water (density = 1000kgm) is poured into it upto a depth od 6cm calculate the pressure and the thrust of water on the base.

Answer 1

Explanation:

Density (ρ) = 1000kg/m³

acceleration due to gravity (a) = 10m/s²

height (h) = 6cm = 0.06m

Area = 300 × 10-⁴

Pressure = ρgh

= 1000 × 10 × 0.06

= 600kg/m²

= 600 Pa

Thrust = Pressure × Area

= 600 × 300 × 10-⁴

= 18N

Answer 2

$\bold{(a) \: The \: pressure} \\ \bold{Height \: = \: 6cm \: to \: m} \\ \bold{ = \: \frac{6}{100} } \\ \bold{ = \: 0.06 \: m} \\ \bold{Density \: = \: 1000 \: kgm^-3} \\ \bold{Acceleration \: due \: to \: gravity \: = \: 10ms^-2} \\ \bold \green{pressure \: = \: hpg} \\ \bold{ = \: 0.06 \: m \times 1000 \: kgm^-3 \times 10 \: ms^-2} \\ \bold \red{ = \: 600 \: Pa \: \: ....ans}$

$\bold{(b) \: The \: thrust \: of \: water \: on \: the \: base} \\ \bold{Pressure \: = \: \frac{force}{area} } \\ \bold \green{force \: = \: Pressure \times area} \\ \bold{ = \: 600 \: Pa \times 300 \: {cm}^{2} } \\ \bold \red{ = \: 18 \: N \: \: ....ans}$

hope it helps you....

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