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State and prove Pascal's law in fluids.

Answer» <html><body><p></p>Solution :<img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PRE_GRG_PHY_XI_V02_C07_E01_049_S01.png" width="80%"/> <br/>Pascal's law states that if the effect of <a href="https://interviewquestions.tuteehub.com/tag/gravity-18707" style="font-weight:bold;" target="_blank" title="Click to know more about GRAVITY">GRAVITY</a> can be generated then the pressure in a fluid in equilibrium is the same <a href="https://interviewquestions.tuteehub.com/tag/everywhere-2065767" style="font-weight:bold;" target="_blank" title="Click to know more about EVERYWHERE">EVERYWHERE</a>.<br/> Let us consider any two points A and B inside the fluid imagined. A cylinder is such that points A and B lie at the centre of the <a href="https://interviewquestions.tuteehub.com/tag/circular-916697" style="font-weight:bold;" target="_blank" title="Click to know more about CIRCULAR">CIRCULAR</a> surface at the top and bottom of the cylinder. Let the fluid inside this cylinder be in equilibrium under the action of <a href="https://interviewquestions.tuteehub.com/tag/forces-16875" style="font-weight:bold;" target="_blank" title="Click to know more about FORCES">FORCES</a> from <a href="https://interviewquestions.tuteehub.com/tag/outside-1143075" style="font-weight:bold;" target="_blank" title="Click to know more about OUTSIDE">OUTSIDE</a> the fluid. The forces acting on the circular, top and bottom surfaces are perpendicular to the forces acting on the cylindrical surface. Therefore the forces acting on the faces at A and B are equal and opposite and hence add to zero.<br/>As the areas of these two faces are equal, pressure at A = pressure at B.<br/>This is the proof of Pasccal's law when the effect of gravity is not taken into account.</body></html>


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