Question

A vessel contains water up to a height of 6 cm. Taking the density of water 1000 kg m-3 and acceleration due to gravity 10 ms-2  calculate the pressure at the base of vessel​

Answer 1

Explanation:

Pressure =ρgh

where ρ=density of water

g=acceleration due to gravity

P=10

3

×9.8×1.5=1.47×10

4

Nm

2

Upthrust = Force exerted = Pressure \times Area

⇒Upthrust=1.47×10

4

Nm

2

×100×10

−4

m=147N

Answer 2

$\bold{(a) \: Pressure} \\ \bold{Height = 1.5m} \\ \bold{Density = 1000 kg/m³} \\ \bold{Acceleration \: due \: to \: gravity \: = \: 9.8m/s²} \\ \bold \green{Pressure \: = \: hpg} \\ \bold{ = \: 1.5 \times 1000 \times 9.8} \\ \bold{ = \: 14700 Pa} \\ \bold \red{ = \: 1.47 \times {10}^{4} N/m² \: \: ....ans}$

$\bold{(b) \: Thrust \: at \: the \: base \: of \: vessel} \\ \bold{Pressure \: = \: \frac{force}{area} } \\ \bold \green{force \: = \: Pressure \times area} \\ \bold{ = \: 1.47 \times {10}^{4} \times 100} \\ \bold \red{ = \: 147 N \: \: ....ans}$

Hope it helps you...

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