Respuesta :
We have that the surface of the cylinder is given by 2prh. The diameter is 0.1 μm, so the radius is 0.05 μm. Hence, substituting we get that the total surface area is 0.314 μm^2=2*0.05*1*3.14.
The surface of a hemisphere is half the surface of a sphere. The surface of a sphere is: 4pr^2. Hence, the hemisphere has an area of 2pr^2. Substituting now again, we get:
[tex]2\pi r^2=2\pi *(0.05)^2=2*3.14*0.0025=0.016[/tex] μ[tex]m^2[/tex].
This is the area of the top.
We can add these two areas to find the total area.
The surface of a hemisphere is half the surface of a sphere. The surface of a sphere is: 4pr^2. Hence, the hemisphere has an area of 2pr^2. Substituting now again, we get:
[tex]2\pi r^2=2\pi *(0.05)^2=2*3.14*0.0025=0.016[/tex] μ[tex]m^2[/tex].
This is the area of the top.
We can add these two areas to find the total area.
The surface area of the cylinder that forms the sides of the microvillus rod is [tex]\fbox{\begin\\0.314 \times {10^{ - 12}}\,{{\text{m}}^{\text{2}}}\end{minispace}}[/tex] or [tex]\fbox{\begin\\31.4\times {10^{ - 14}}\,{{\text{m}}^{\text{2}}}\end{minispace}}[/tex] and the surface area of the hemispherical top of the microvillus rod is [tex]\boxed{1.57 \times {10^{ - 14}}\,{{\text{m}}^{\text{2}}}}[/tex].
Further explanation:
The surface area of an object is defined as the measure of the total area occupied by the surface of the object.
The surface are is classified into two categories:
• Curved surface area- It is the surface area of the surfaces of an object or the area of curved surfaces.
• Lateral surface area- It is the surface area of an object around the sides excluding top and bottom of the objects like prism etc.
Given:
The height of rod like structure of microvillus is [tex]1\mu \text{m}[/tex].
The microvillus has the hemispherical surface of the diameter [tex]0.1\mu \text{m}[/tex].
The value of [tex]\pi[/tex] is [tex]3.14[/tex].
Concept:
Radius is calculated by using the expression:
[tex]r=\frac{d}{2}[/tex]
Here, [tex]r[/tex] is the radius of the hemispherical surface of the microvillus and [tex]d[/tex] is the diameter of the hemispherical surface of the microvillus.
Substitute [tex]0.1\mu \text{m}[/tex] for [tex]d[/tex] in the expression of radius we get
[tex]\begin{aligned}r&=\frac{0.1\mu \text{m}}{2}\\ &=0.05\mu \text{m}\end{aligned}[/tex]
The microvillus is a rod like surface which is surmounted by a hemispherical top.
Let [tex]{A_1}[/tex] represent the surface area of the cylinder that forms the sides of a microvillus and [tex]{A_2}[/tex] represents the surface area of the hemispherical top of the microvillus rod.
The surface area of the curved part of a cylinder which forms the sides of the microvillus is given by the expression:
[tex]\fbox{\begin\\{A_1}=2\pi rh\end{minispace}}[/tex]
Here, [tex]h[/tex] is the height of the cylindrical part of microvillus.
Substitute [tex]3.14[/tex] for [tex]\pi[/tex], [tex]0.05\mu \text{m}[/tex] for [tex]r[/tex] and [tex]1\mu \text{m}[/tex] for [tex]h[/tex] in the above expression.
[tex]\begin{aligned}{A_1}&=2\times3.14\times0.05\times10^{-6}\times1\times10^{-6}\\&=31.4\times10^{-14}\text{m}^2\end{aligned}[/tex]
The surface area of the hemispherical top of the microvillus rod is given by the expression:
[tex]\fbox{\begin\\{A_2} = 2\pi {r^2}\end{minispace}}[/tex]
Substitute [tex]3.14[/tex] for [tex]\pi[/tex] and [tex]0.05\mu \text{m}[/tex] for r in the above expression
[tex]\begin{aligned}\\{A_2}&=2\times3.14\times(\left 0.05\times10^{-6}\right)^2\\&=1.57\times10^{-14}\text{m}^2\end{aligned}[/tex]
Thus, the surface area of the cylinder that forms the sides of the microvillus rod is [tex]\fbox{\begin\\0.314 \times {10^{ - 12}}\,{{\text{m}}^{\text{2}}}\end{minispace}}[/tex] or [tex]\fbox{\begin\\31.4\times {10^{ - 14}}\,{{\text{m}}^{\text{2}}}\end{minispace}}[/tex] and the surface area of the hemispherical top of the microvillus rod is [tex]\boxed{1.57 \times {10^{ - 14}}\,{{\text{m}}^{\text{2}}}}[/tex].
Learn more:
1. The motion of a body under friction brainly.com/question/4033012
2. A ball falling under the acceleration due to gravity brainly.com/question/10934170
3. Conservation of energy brainly.com/question/3943029
Answer Details:
Grade: College
Subject: Physics
Chapter: Units and Measurements
Keywords:
Microvillus, cylinder, hemisphere, rod like structure, hemispherical top, area, surface area, curved surface area, 1.57x10^-14 meter square, 1.57*10-14m2, 1.57x10-14m2, 31.4x10^-14 meter square, 31.4*10-14m2, 31.4x10-14m2.
