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# A cylinder of mass 5 kg is held in vertical position. If the height of the cylinder is 6 cm and radius of cross section is 4 cm then find the pressure acting on its bottom surface.

Last updated date: 21st Jun 2024
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Hint: When a person is standing in a vertical position on his feet his entire weight will be borne by his feet. Feet are the bottom most part of the body which are in contact with the floor. Similarly in case of a cylinder the entire weight will be borne by its base and we can find pressure from it.

Formula used:
$P = \dfrac{F}{A}$

Let us assume the mass of the cylinder is ‘m’ and the radius of the cross section of the cylinder is ‘r’ and the height of the cylinder is ‘h’. Entire mass of the cylinder will be concentrated at the center of mass of the cylinder. Which is nothing but the geometric center of the cylinder if the mass distribution is uniform. If it is not uniform then the position might be varied. But whatever may be the mass distribution the entire weight of the cylinder is balanced by the base of the cylinder.
Base of the cylinder is a circle and it’s area will be
$A = \pi {r^2}$
Radius is given as 4cm and mass is given as 5kg and height is given as 6cm.
\eqalign{ & A = \pi {r^2} \cr & \Rightarrow A = \pi {(4 \times {10^{ - 2}}m)^2} \cr & \Rightarrow A = 16\pi \times {10^{ - 4}}{m^2} \cr}
Weight is given as $W = mg$
$W = mg$
\eqalign{ & \Rightarrow W = 5 \times 10 \cr & \Rightarrow W = 50N \cr}
Where ‘g’ is the acceleration due to gravity whose value is equal to $10m/{s^2}$
Now the total pressure will be
$P = \dfrac{F}{A}$
\eqalign{ & \Rightarrow P = \dfrac{W}{A} \cr & \Rightarrow P = \dfrac{{50}}{{16\pi \times {{10}^{ - 4}}}} \cr & \Rightarrow P = 0.97 \times {10^4}N{m^{ - 2}} \cr}
Hence the pressure on the bottom surface is $0.97 \times {10^4}N{m^{ - 2}}$

Note: If the cylinder is hollow and if it is filled tightly with some fluid then the fluid will apply force on the curved surface of the cylinder too. Then due to that force pressure also will be applied on the curved face of the cylinder.