Question

The area of base of a cylindrical vessel is 200 cm 2 . Water (density = 1000 kg m -3 ) is poured into it up to a depth of 8 cm. Calculate (a) the pressure and (b) the thrust of water on the base. (g = 10 ms -2 ) (3) c) State how does the

Answer 1

Answer:1) 80kg/sq.m

2) 1.6 kg

Explanation:

Pressure - Mass/ area, mass- volume x density

Mass - 0.02m2x 0.08 cm - 0.0016m3

Density - 1000 kg/m3

Pressure - 80 Kg/m2

Trust- pressure x area

- 80kg/m2 x 0.02m2

= 1.6 Kg

Answer 2

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

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

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