Question

The area of base of a cylindrical vessel is 300 cm². Water (density = 1000 kg/m³) is poured into it up to a depth of 6 cm. Calculate : (a) the pressure and (b) the thrust of water on the base. (g = 10m/s).​

Answer 1

Answer:

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Explanation:

The pressure at the base will be

P = ρgh

here

ρ = density of water = 1000 kg/m3

g = acceleration due to gravity = 9.81 m/s2

h = height of the water column = 6 cm = 0.06 m

so,

P = 1000 x 9.81 x 0.06

thus, we get the pressure as

P = 588.6 Pa

the thrust (force) exerted will be

F = pressure x area of the piston = P X A

here

A = 300 cm2 = 300 x 10-4 m2

thus, we get

F = 588.6 x 300x10-4

or

the thrust will be

F = 17.658 N

Answer 2

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

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

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