Question

A vessel contains water up to height of 2 m. Taking the density of water 103 kg.m-3

acceleration due to gravity 10 ms-2

and area of base of vessel 100 cm2
Calculate the
pressure.​

Answer 1

To FinD :-

• The pressure at the base .

$\red{\frak{ Given}}\begin{cases}\textsf{A vessel contains water up to height of 2 m.}\\\textsf{</p><p>The acceleration due to gravity is 10m/s ^{\sf 2} .}\\\textsf{</p><p>Density of water is 1000 kg / m ^{\sf 3} .}\end{cases}$

We know that if we have a vessel filled with a liquid of density p , having a mass m , and volume as v , then the density is ,

$\sf \to Density (\rho) =\dfrac{Mass(m)}{Volume (V)}$

Here the area of base is 100cm² . On converting it into SI unit we have , 100 / 10⁴ m² = 1/100 m² = 0.01 m² . Let's find out Volume and then mass ! .

Volume will be Area of base × height , That will be 2m × 0.01m² = 0.02 m³ .

$\sf:\implies \pink{ Density (\rho) =\dfrac{Mass(m)}{Volume (V)} }\\\\\sf:\implies 10^4 \ kg/m^3 = \dfrac{mass}{0.02 \ m^3} \\\\\sf:\implies mass = 10^4 \times 0.02 kg \\\\\sf:\implies \boxed{\pink{\frak{ Mass = 200 kg }}}$

Now the force exterted at the base will be mg . And we know Pressure as Force / Area . So that ,

$\sf:\implies \pink{ Pressure (P) =\dfrac{Force}{Area}} \\\\\sf:\implies Pressure = \dfrac{mg}{0.01 m^2}\\\\\sf:\implies Pressure =\dfrac{ (200kg)(10 m/s^2)}{0.01 m^2 } \\\\\sf:\implies\underset{\blue{\sf Required \ Pressure }}{\underbrace{ \boxed{\pink{\frak{Pressure = 2\times 10^5 \ Pa }}}}}$