Question

the area of base of a cylindrical vessel is 300 cm^2 water (density 1000kg m^-3) is poured into it upto a depth of 6cm calculate the pressure and the thrust at the base of vessel

Answer 1

pressure=force/ area
here force= weight of water=mg=300×10^-4×6×10^-2×1000×10
=18N
Area=300×10^-4 m^2
-------> pressure= 18/(3×10^-2)
=600 atmospheres.

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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