Question

A cylindrical block of weight 314 N is placed on horizontal surface. If it exert pressure of 200 Pa on the surface, then find the circumference of circular part of cylinder.​

Answer 1

Answer:

T

his chapter deals with forces applied by fluids at rest or in rigid-body

motion. The fluid property responsible for those forces is pressure,

which is a normal force exerted by a fluid per unit area. We start this

chapter with a detailed discussion of pressure, including absolute and gage

pressures, the pressure at a point, the variation of pressure with depth in a

gravitational field, the manometer, the barometer, and pressure measure-

ment devices. This is followed by a discussion of the hydrostatic forces

applied on submerged bodies with plane or curved surfaces. We then con-

sider the buoyant force applied by fluids on submerged or floating bodies,

and discuss the stability of such bodies. Finally, we apply Newton’s second

law of motion to a body of fluid in motion that acts as a rigid body and ana-

lyze the variation of pressure in fluids that undergo linear acceleration and

in rotating containers. This chapter makes extensive use of force balances

for bodies in static equilibrium, and it will be helpful if the relevant topics

Answer 2

Given,

Weight of block, $\[F = 314N\]$

Pressure, $\[P = 200Pa\]$

Solution,

Know that,

$\[\begin{array}{l}area\left( A \right) = \frac{{force}}{{pressure}}\\ \Rightarrow A = \frac{{314}}{{200}}\\ \Rightarrow A = 1.57{m^2}\end{array}\]$

Calculate the radius of circular cross section of cylindrical block,

$\[\begin{array}{l}A = 1.57\\ \Rightarrow \pi {r^2} = 1.57\\ \Rightarrow r = 0.7069m\end{array}\]$

Calculate the circumference of circular part of cylinder.

$\[\begin{array}{l} = 2\pi r\\ = 2 \times \pi \times 0.7069\\ = 4.44m\end{array}\]$

Hence the circumference is $\[4.44m\]$