Chapter 1, Problem 10Q

Summary Introduction

To calculate: The average size of a brain cell with the assumption that each cell is entirely filled with water. 1 cm³ of water weighs 1g.

Explanation of Solution

Calculation:

As given in the problem, the average weight of a human brain is $\text{1}\text{kg}$ and the average number of cells in an average sized brain is ${10}^{12}$ cells.

$\begin{array}{c}\text{Weight}\text{of}\text{an}\text{average}\text{brain}\text{cell}\text{=}\frac{\text{Total weight}\text{of}\text{the}\text{brain}}{\text{Total}\text{number}\text{of}\text{cells}\text{in}\text{the}\text{brain}}\\ =\frac{1000}{{10}^{12}}\\ =\frac{{10}^3}{{10}^{12}}\\ ={10}^{-9}g\end{array}$

Hence, the weight of an average brain cell is ${\text{10}}^{\text{-9}}\text{g}$.

Summary Introduction

To calculate: The length of one side of an average sized brain cell when the cell is a simple cube.

Explanation of Solution

Calculation:

$\text{1}\text{g}$ of ${\text{H}}_{\text{2}}\text{O}$ occupies $\text{1 ml}{\text{= 1 cm}}^{\text{3}}$

${\text{1 cm}}^{\text{3}}{\text{=10}}^{\text{-6}}{\text{m}}^{\text{3}}$

$\begin{array}{c}\text{Volume of 1 brain cell},\\ a^3={10}^{-9}g\times {10}^{-6}{m}^3/g\\ ={10}^{-15}{m}^3\\ \end{array}$

$\begin{array}{c}\text{Length}\text{of}\text{one}\text{side}\text{of}\text{the}\text{average}\text{sized}\text{brain}\text{cell},\\ a=3\sqrt{{10}^{-15}}\\ =0.000094\\ =0.00001\\ ={10}^{-5}m\end{array}$

Hence, the length of one side of the average sized brain cell is ${\text{10}}^{\text{-5}}\text{m}$.

Summary Introduction

To calculate: Number of pages the cells would cover in the book, if the cells were spread out as a thin layer that is only a single cell thick.

Explanation of Solution

Calculation:

$\begin{array}{c}\text{Length}\text{of}\text{the}\text{book},l=21cm\\ \text{Breadth}\text{of}\text{the}\text{book},b=27.5cm\\ \text{Surface}\text{of}\text{a}\text{page}\text{in}\text{the}\text{book}=l\times b\\ =21\times 27.5cm\\ =577.5c{m}^2\\ =0.057m^2\end{array}$

$\begin{array}{c}\text{Footprint}\text{of}\text{each}\text{cell}=a\times a\\ ={10}^{-5}\times {10}^{-5}\\ ={10}^{-10}{m}^2\end{array}$

$\begin{array}{c}\text{No}\text{of}\text{cells}\text{which}\text{can}\text{fit}\text{one}\text{page}\text{when}\text{spread}\text{out}\text{in}\text{a}\text{single}\text{layer}\\ \text{=}\frac{\text{Surface}\text{area}\text{of}\text{a}\text{page}\text{in}\text{the}\text{book}}{\text{Footprint}\text{of}\text{each}\text{cell}}\\ =\frac{0.057m^2}{{10}^{-10}{m}^2}\\ =57\times {10}^{7}cells\end{array}$

The number of cells that can fit one page when spread out in a single layer is ${\text{57×10}}^{\text{7}}\text{cells}$.

$\begin{array}{c}\text{No}\text{of}\text{pages}\text{the total}\text{number}\text{of}\text{brain}\text{cells}\text{would}\text{occupy is calculated below}\\ \text{=}\frac{\text{Total}\text{number}\text{of}\text{brain}\text{cells}}{\text{No}\text{of}\text{cells}\text{for}\text{a}\text{single}\text{page}}\\ =\frac{{10}^{12}}{57\times {10}^7}\\ =\frac{{10}^5}{57}\\ =1754.38\\ \simeq 1754\text{pages}.\end{array}$

Hence, the number of pages the cells would cover in the book, if the cells were spread out as a thin layer that is only a single cell thick would be $1754\text{pages}$