Question

The mass of a liquid contained in a cylindrical vessel of cross sectional area A is 800 kg. If the<br />pressure at the bottom ofthe vessel is 4 x 10 power 4 Pascal, then A is equal to (g = 10 ms )<br />a) 1 m*m<br />b) 2 m*m<br />c) 0.1 m*m<br />d) 0.2 m*m​

Answer 1

Given:

The mass of liquid contained in a cylindrical vessel of cross sectional area A is 800 kg. Pressure at the bottom of the vessel is 4 × 10⁴ Pascal.

To find:

Value of area A

Concept:

Force in this question is better termed as Thrust , Thrust is the perpendicular force acting on the bottom surface of the vessel.

Pressure is a scalar quantity denoting the distribution of Thrust over a specified surface area.

Mathematically , pressure is calculated by the ratio of the force and the the area upon which it acts.

Calculation:

$\sf{pressure = \dfrac{force}{area} }$

$= > \sf{pressure = \dfrac{(mass \times gravity)}{area} }$

$= > \sf{4 \times {10}^{4} = \dfrac{(800 \times 10)}{A } }$

$= > \sf{ A= \dfrac{(800 \times 10)}{4 \times {10}^{4} } }$

$= > \sf{ A= 2 \times {10}^{(3 - 4)} }$

$= > \sf{ A= 2 \times {10}^{ - 1 } }$

$= > \sf{ A= 0.2 \: {m}^{2} }$

So final answer is :

Area of cross-section of vessel : 0.2 m²