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Question: State and prove Pascal’s Law of transmission of fluid pressure....

State and prove Pascal’s Law of transmission of fluid pressure.

Explanation

Solution

In this question, the concept of Pascal's law is used, that is it states that if a static pressure is applied from outside of a confined fluid, the pressure is evenly distributed throughout every portion of fluid.

Complete step by step solution:
As we know that Pascal's law states that if a static pressure is applied from outside of a confined fluid, the pressure is evenly distributed throughout every portion of fluid. According to the law, if FF is the applied force, PPis the transmitted pressure, and AA is the area of applied pressure then from Pascal’s law relation of these three will be
F=P×A\Rightarrow F = P \times A

Fig: Pressure and Force distribution in triangular prism

Now, we assume a right angled prismatic triangle containing a fluid of densityρ\rho .the prism being very small, we consider every point at the same depth, and gravitational effect on prism is also the same in those points.
Let psps, qsqsand rsrs be the area of surfaces PSURPSUR, PSTQPSTQ and TURQTURQ of the prism.
And also, pressure on those surfaces be, Pb{P_b} and Pc{P_c} respectively.
A force is exerted by the pressure in surfaces, in PRUSPRUS face the force be Fa{F_a}, in PSTQPSTQ the force be Fb{F_b} and in TURQTURQ the force be Fc{F_c}.
Now, we write the expression for the forces at different surfaces as,
Fa=Pa×ps\Rightarrow {F_a} = {P_a} \times ps
Fb=Pb×qs\Rightarrow {F_b} = {P_b} \times qs
Fc=Pc×rs\Rightarrow {F_c} = {P_c} \times rs
As it is right angled triangular prism,
sinθ=qp\Rightarrow \sin \theta = \dfrac{q}{p}
cosθ=rp\Rightarrow \cos \theta = \dfrac{r}{p}
As we know that the total force inside the prism will be zero, as equilibrium is attained.
Breaking the Fa{F_a} force into horizontal and vertical directions as Fasinθ{F_a}\sin \theta and Facosθ{F_a}\cos \theta , we get
Fasinθ=Fb......(1)\Rightarrow {F_a}\sin \theta = {F_b}......(1)
Facosθ=Fc......(2)\Rightarrow {F_a}\cos \theta = {F_c}......(2)
Replacing Fa{F_a} with Pa×ps{P_a} \times ps and the value of Fb{F_b} with Pb×qs{P_b} \times qs putting values of sinθ\sin \theta in equation(1)(1)
Pa×ps×qp=Pb×qs.....(3)\Rightarrow {P_a} \times ps \times \dfrac{q}{p} = {P_b} \times qs.....(3)
Similarly, from equation (2)(2) putting the values we get
Pa×ps×rp=Pc×rs......(4)\Rightarrow {P_a} \times ps \times \dfrac{r}{p} = {P_c} \times rs......(4)
Now, from equation (3)(3) and (4)(4) we get,
Pa=Pb{P_a} = {P_b} And Pa=Pc{P_a} = {P_c}
Pa=Pb=Pc\therefore {P_a} = {P_b} = {P_c}

So, the pressure distributed in the fluid is equal in every direction.

Note: As we know that the Pascal Law has wide application for example it is used in the hydraulic press, which is the great achievement of Pascal’s Law application, in modern planes it is used for brakes, landing and various purposes.