Question

Consider an e. coli to be a cylinder with a diameter of 1 micrometer (um) and with a length of 2 micrometer (um).

Answer 1

Answer:

$A=6.28\ \mu m^2$

Part 1

$V=1.57 \ \mu m^3$

Part 2

$A=6.28\ \mu m^2$

Explanation:

Given that

Diameter,d=1 μm

Length ,l=2 μm

As we know that volume of cylinder given as

$V=\pi r^2l$

$V=\pi \times 0.5^2\times 2 \ \mu m^3$

$V=1.57 \ \mu m^3$

Surface area,A

A=π d l

$A=\pi \times 1 \times 2\ \mu m^2$

$A=6.28\ \mu m^2$

Part 1

$V=1.57 \ \mu m^3$

Part 2

$A=6.28\ \mu m^2$