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I just recently wrote this function to load in files containing level information for a 2-D side scroller game I'm working on, where anything with an m_ in front of it is a member variable of the mm9Core class, and the first character following the underscore is the variable type:
MM9ErrorType mm9Core::LoadLevelInfo()
{
char*		szLevelFileName = NULL;
char*		szBossFileName = NULL;
char*		szBossNameFileName = NULL;
int		iNameWidth, iNameHeight;
fstream		kLevelFile;

//Choose appropriate level file to load with respect to chosen boss.
switch(m_iSCursorBossPos)
{
case 0:
szLevelFileName = "LevelInfo/SampleLevel2.txt";
break;
default:
return LEVELFILENOTFOUND;
}

kLevelFile.open(szLevelFileName, ios::in);

//clear all of the arrays so that they can store new level data

if(m_szTileFiles != NULL)
delete [] m_szTileFiles;

if(m_iTileMap != NULL)
delete [] m_iTileMap;

if(m_szNoCollide != NULL)
delete [] m_szNoCollide;

if(m_szWaterZone != NULL)
delete [] m_szWaterZone;

if(m_szKillZone != NULL)
delete [] m_szKillZone;

if(kLevelFile.is_open())
{
//read level width and height in tiles
kLevelFile>>m_iLvlWidth>>m_iLvlHeight;

//read in the number of distinct tiles
kLevelFile>>m_iNumTiles;

m_szTileFiles = new char*[m_iNumTiles];

for(int i = 0; i < m_iNumTiles; i++)
{
m_szTileFiles[i]       = new char[MAXFILENAMELEN];
}

for(i = 0; i < m_iNumTiles; i++)
{
kLevelFile>>m_szTileFiles[i]    ;
}

kLevelFile>>m_iNumNoCollide;

if(m_iNumNoCollide != 0)
m_szNoCollide = new char[m_iNumNoCollide*2];

for(i = 0; i < m_iNumNoCollide; i+=2)
{
kLevelFile>>m_szNoCollide[i]     ;
m_szNoCollide[i+1] = ' ';
}

kLevelFile>>m_iNumWaterZone;

if(m_iNumWaterZone != 0)
m_szWaterZone = new char[m_iNumWaterZone*2];

for(i = 0; i < m_iNumWaterZone; i+=2)
{
kLevelFile>>m_szWaterZone[i]     ;
m_szWaterZone[i+1] = ' ';
}

kLevelFile>>m_iNumKillZone;

if(m_iNumKillZone != 0)
m_szKillZone = new char[m_iNumKillZone*2];

for(i = 0; i < m_iNumKillZone; i++)
{
kLevelFile>>m_szKillZone[i]     ;
m_szKillZone[i+1] = ' ';
}

m_iTileMap = new int* [m_iLvlHeight];

for (i = 0; i < m_iLvlHeight; i++)
{
m_iTileMap[i]      = new int [m_iLvlWidth];
}

for(i = 0; i < m_iLvlHeight; i++)
{
for(int j = 0; j < m_iLvlWidth; j++)
{
kLevelFile>>m_iTileMap[i][j]    ;
}
}

kLevelFile>>m_iStartX>>m_iStartY;

//!!!//
//INFORMATION FROM THE FILE NEVER GETS READ INTO THESE VARIABLES!!//
//IT IS AS IF THE FILE ENDED HERE//
kLevelFile>>szBossFileName;
kLevelFile>>szBossNameFileName;
kLevelFile>>iNameWidth>>iNameHeight;

//end of file-reading portion part of function
}

The level file I'm attempting to read in is this: -------------------------------------------------
10 8
5
Graphics/SquareMan/tile0.bmp
Graphics/SquareMan/tile1.bmp
Graphics/SquareMan/tile2.bmp
Graphics/SquareMan/tile3.bmp
Graphics/SquareMan/tile4.bmp
1
0
0
0
0 1 1 1 1 1 1 1 1 0
4 0 0 0 0 0 0 0 0 2
4 0 0 0 0 0 0 0 0 2
4 0 0 0 0 0 0 0 0 2
4 0 0 0 0 0 0 0 0 2
4 0 0 0 0 0 0 0 0 2
4 0 0 0 0 0 0 0 0 2
0 3 3 3 3 3 3 3 3 0
0 0
ABOs/mm9SquareManABO.txt
ABOs/mm9NameSquareManABO.txt
181 16

------------------------------------------------- The last 4 variables (the section of the code I marked with //!!!//) never receive any values; as if there is no more data left in the file to read. However, I investigated the file reading routine using peek() and after reading in the pair of 0s after that big matrix of ints, peek() returns ASCII character code 013, which is a carriage return. Once kLevelFile>>szBossFileName; is executed, peek() returns ASCII character code -1, which is garbage I guess. Also I tested for EOF after the pair of 0s, and also tested is_open() again and both returned that the end of the file was not reached, and that it was still open. Why can't I read in those two file names and the pair of integers at the end of the file? Am I missing something? I appreciate all help. **PLEASE NOTE** the forum form formatting (that's a mouthful) isn't too conducive to code; so anything regarding loop control variables (i in particular since, [i] is the code for italics) that looks incorrect is most likely just a formatting error _________________ ~Firestorm Fenix~ [edited by - FirestormFenix on August 13, 2003 12:45:47 AM]

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not too sure what the problem is, but i noticed that you opened the file in binary mode, why is that? you should process the file in ascii mode, seeing as it is purely text that you are reading in.

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Oh man, forgot about that. I was working with bitmaps earlier and I just typed that in by accident. I did however remove ios::binary from it just now, and I'm still having the same problem......

I'm going to fix the first post so it doesn't show that anymore.

[edited by - FirestormFenix on August 13, 2003 12:45:10 AM]

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hint.. use $source$ $/source$ (edit - minus the \)tags to post large chunks of code. I am only familiar with using FILE and fopen for text files, but have you tried reading back the data as you gather it to make sure you are where you think you are in the file? I know it sucks but sometimes I miss things and it is the only way I will find them, especially if the file gets much bigger than what youve already got. I use a file class I wrote to handle reading everything and I know it helps that if I get something unexpected it gives me the value and line number it occured on. I use a similar map file, heres what one of mine looks like: 20x15x2
MAPFILE NewMap.map~ NewMap0~ 20 15 2TILESET_BMP tileset0.bmp~ 32 32MAP_DATA   -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1     4    -1    -1    -1     4    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1     4    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1    -1 END_OF_FILE

[edited by - evillive2 on August 14, 2003 1:51:34 AM]

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My code would look nothing like that for the same function=)
I use fopen/fclose/sprintf for most of these files reading projects.

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ok, i looked again at the code and saw this

char* szBossFileName = NULL;
char* szBossNameFileName = NULL;

you are creating char pointers, but not allocating memory for them. so you can either:

char szBossFileName[100];

or better yet, seeing as you are using c++ streams...

String szBossFileName;

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Okay.

Kuladus - I tried what you said, and that solved the problem. I was thinking about that solution before, but I dismissed the idea because I was able to do this:
char*		szLevelFileName = NULL;//...szLevelFileName = "LevelInfo/SampleLevel2.txt";

I'm supposing it makes a difference when the data is coming from a file?

evillive2 - What do you mean by reading back the data that I gather, do you mean doing something such as outputting it to a console to check? I've mainly just been walking through with the debugger in VC++ to see what's going on. Actually, is there a way (I'm sure there's got to be) to output to the console in a Win32 project? I tried using std::cout but that didn't work. I'm a newb when it comes to Win32 stuff, so pardon my ignorance.
Oh, on a side note, I used $code$$/code$ in the first post since I've seen it used in phpBB forums to do the same thing. Using the source tags here is way better though, thanks.

chbfiv - What don't you like about what I've done? I'm just curious...improving my code is also something I'm always interested in doing.

_________________
~Firestorm Fenix~

[edited by - FirestormFenix on August 17, 2003 8:51:04 PM]

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char*		szLevelFileName = NULL;szLevelFileName = "LevelInfo/SampleLevel2.txt";

When people learn C, they need to first read a book about C, then read a book about pointers. then another book on pointers, then another book on pointers

repeat ad infinitum until they figure out why that works, and why the reading from a file does not work.

also, you need a .ignore(80 /*reasonable number*/, ''\n''); when switching from cin>>int to cin>>string.

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# Multiplying properties

Discussion in 'Physics & Math' started by Magical Realist, Sep 18, 2013.

1. ### Magical RealistValued Senior Member

Messages:
12,926
Who was the first person to decide properties like mass, energy, speed, and force can be multiplied and divided and even squared? What does it actually mean to use a property like it was a number? Isn't a quality something fundamentally different from a quantity? Or are we talking magnitudes used simply as variables?

3. ### originIn a democracy you deserve the leaders you elect.Valued Senior Member

Messages:
11,056
The answer seems so obvious that I assume I must not understand your question.

5. ### RJBeeryNatural PhilosopherValued Senior Member

Messages:
4,215
I agree with Origin that the answer seems obvious...at first. But a property must be measurable and quantifiable before we can apply math to it, yes? We can measure colors, for example, but until we understood wavelengths of light we had no objective way to apply math to those values. What the hell is red squared? And is blue > yellow?

7. ### Magical RealistValued Senior Member

Messages:
12,926
Precisely. And what does it even mean to multiply say mass by acceleration to get force? By "multiply" I assume we mean the arithmetical sense of the number of times a value is added to itself. So why is force equal to mass added to itself "acceleration" number of times? Or why does energy equal mass added to itself the speed of light times squared? What is speed of light squared anyway? I can see how a certain QUANTITY of a property can be multiplied or squared. But the property or constant itself? What does that even mean?

8. ### RJBeeryNatural PhilosopherValued Senior Member

Messages:
4,215
This...I believe I can answer. Physics is all about dimensions and how they relate to one another. I have this "thing" called position and this "thing" called time; well, let's say I can associate the position of a particular object with time; I can relate the dimensions to each other mathematically. With that I can calculate things like velocity, acceleration, etc. Basically all of our equations are really just taking a handful of these fundamental dimensions and making various sandwiches out of them (e.g. Force, etc) to give us higher levels of understanding about their complexities.

Of course it all relies on the assumption that there is a mathematical relationship between, for example, an object's position and time in the first place. If objects had non-discrete and/or non-continuous positions with regard to time (such as quantum systems) then applying particular physics treatments to them would be futile...

9. ### wellwisherBannedBanned

Messages:
5,160
One math operation that is useful but raises a yellow flag is division by a fraction. If I have 1 kg of mass, and divide this by 1/2, I now have 2 kg of mass. What exactly did I do here, in that I was able to create mass. The process of dividing by 1/2 did not even require the energy =mc2, where m-1 kg. I was able to do it with much less energy.

In the miracle of the multitudes, Jesus feeds thousands with a few fish and few loaves of bread. in terms of math, he used division by a fraction. ALl he had to do was divide by 1/1000.

Messages:
5,051
11. ### Russ_WattersNot a Trump supporter...Valued Senior Member

Messages:
5,051
Thus proving that math, like any other language can say things that aren't true.

12. ### RJBeeryNatural PhilosopherValued Senior Member

Messages:
4,215
It's as if millions of brain cells all cried out in terror and were suddenly silenced...

13. ### brucepValued Senior Member

Messages:
4,098
You can't divide 2 kg with a number [1/2] and wind up with 4 kg. Really dude? How many apples do you get if you divide 2 apples by 1/2? How many 1/2 apples do you get when divide 2 apples by 1/2 apple? Ever hear of dimensional analysis?

14. ### arfa branecall me arfValued Senior Member

Messages:
6,930
One way to see a disconnection between physics and the mathematics used to describe it, is that derivatives like dx/dt are axiomatic; mathematically dx/dt is not zero if x is some function of t. It seems obvious that a change in position 'includes' an interval of time.

15. ### RJBeeryNatural PhilosopherValued Senior Member

Messages:
4,215
Yes, time has been given a special treatment in Physics and unraveling that point was difficult for me, as a non-Physicist, because it was never explained. We could just as easily study, say, temperature with respect to position* of an object and then analyze the derivatives of this temperature/position relationship...it was simply a matter of convention and convenience that we try to associate properties to change in time.

* Newton's law of cooling, for example, could be thought of as relating the temperature of an object with the position of the hands of a clocking device.

16. ### hansdaValued Senior Member

Messages:
2,424
I think your basic question is about "multiplying units" and whether it makes sense or not.

If the unit is "length", multiplication will make geometrical sense. For other units, it may make some equivalent sense which can be abstract also.

17. ### Magical RealistValued Senior Member

Messages:
12,926
Excellent insight. That makes it much clearer. But re: these dimensions..To what extent are they just artificial constructs made of certain ways of measuring a certain magnitude? The mathematical relationships between the dimensions imparts to them a certain objective reality as if they are just out there on their own waiting to be discovered. But aren't the dimensions still just conceptual abstractions dependent on certain arbitrary standards and criteria we are applying to our experience. Like your color example. We have a pretty much continuous "analog" property there. We could quantify it into units of color measurement. Color suddenly equals these units times another variable say like wavelength. But how much of this mathematical structure comes from our own quantitative model of color? Aren't we really just finding equalities between dimensions we have already just mathematically projected upon the world?

18. ### RJBeeryNatural PhilosopherValued Senior Member

Messages:
4,215
I'd say it all comes down to practicality. Declaring that "the position of this train has a very close approximation to the mathematical formula f(x) = 3t" then it lets us make predictions based on that declaration. If our declaration is not accurate then our predictions will break down. Squaring colors doesn't really provide any practical use that I can immediately think of, but squaring length might be useful when, say, ordering fabric or some such...

19. ### AlphaNumericFully ionizedRegistered Senior Member

Messages:
6,702
It is really that nothing with units like mass, length etc has fundamental physical meaning, since our choice of units is artibrary. Why pick metres over yards over furlongs? Something which has no units is better, hence why things like the fine structure constant are unitless.

Likewise you cannot take the sin, cos, tan, exp or log of a quantity with units, they must be dimensionless. This can be seen from their Taylor expansion, $\exp(x) = \sum_{n=0}^{\infty} \frac{x^{n}}{n!}$. If x has units of (for instance) length, then $x^{2}$ has units of length squared, etc. and adding them is meaningless. Given that any function can be expressed as a Taylor series (pipe down those of you thinking "non-analytic functions!") of the form $f(x) = \sum_{n=0}^{\infty}f^{(n)}(y)\frac{(x-y)^{n}}{n!}$ then it follows that ANY function of a quantity must only be applied to dimensionless quantities.

The expressions used in physical formulae must satisfy this underlying principle if they are rearranged in the right way. In reality equations are often written in the form of most convenience (particularly in engineering domains) so something like F=ma having units on both sides is not troubling to anyone.

In terms of historical development it comes from experiments. If the force a spring applies is twice as much when it is compressed by twice the amount that tells you something about how $F \propto x$. If a ball dropped from 4 times as high only takes twice as long to hit the ground then you have $d \propto t^{2}$. Once you have got all of these proportionality relationships then you pick your preferred units and put in the appropriate proportionality constants. For example, if you have worked out that the gravitational force between two objects of masses $M_{1},M_{2}$ a distance r apart goes like $F \propto M_{1}$, $F \propto M_{2}$, $F \propto \frac{1}{r^{2}}$ then you have $F = G\frac{M_{1}M_{2}}{r^{2}}$ and you determine G in your preferred units.

Of interesting note is that you can deduce $F = G\frac{M_{1}M_{2}}{r^{2}}$ using a few reasonable assumptions without ever having to do an experiment. It follows from dimensional analysis.

20. ### RJBeeryNatural PhilosopherValued Senior Member

Messages:
4,215
This bewildered me a bit. I suppose, since the dimensions in an equation must cancel each other out by definition of equation, that any mathematical analysis could be done without dimensions...but does this mean we can't use dimensions in mathematical functions?

Could you elaborate on this? I'd love to see the minimal list of assumptions to deduce that equality.

21. ### AlphaNumericFully ionizedRegistered Senior Member

Messages:
6,702
Yes, you must always be able to rearrange an equation to make it dimensionless, though few people ever bother for practical purposes. If I have something which is twice as massive as another object I shouldn't care if its 2kg vs 1kg or 100 billion nano-grams vs 50 billion nano-grams or whatever, all that matters is their ratio, which is dimensionless, is 2 (or 1/2, depending on your perspective).

The book which does it properly (rather than me giving a less than clear 'off the top of my head' version) I've left in my office but I'll write it out sometime this week. In the mean time here's plenty of sites and pdfs which go through the application of the Buckingham Pi theorem (which is the underlying method we're talking about) for things like drag forces or pendulum oscillations.

22. ### PeteIt's not rocket surgeryRegistered Senior Member

Messages:
10,167
It means that if you and a friend stand close together on a slippery surface, then push each other apart, then the lighter of you will accelerate away faster. The ratio of your accelerations will be the same as the ratio of your masses.

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# Thermal Properties of Materials - Calorimetry, Thermal Expansion, Heat Transmission Practice Problems, FAQs

Have you ever seen that the shape of the railway tracks gets changed because of the expansion of the tracks due to increase in temperature. So to avoid this issue they often leave some gap between the two consecutive tracks so that if tracks expand it will get adjusted in that gap. This expansion of material under the application of heat is one of the thermal properties of the material. Thermal properties are exhibited by the material when heat is passed through it. In this article we will be looking at some important thermal properties.

• Heat
• Calorimetry
• Thermal Expansion
• Heat Transmission
• Practice Problems
• FAQs

## Heat

When two bodies with differing surface temperatures come into contact, heat is a type of energy that forms and travels from the body with a higher surface temperature to the one with a lower surface temperature. It is energy that is moving or in transit. Heat is not a system's property. A system does not contain heat, yet it can emit or absorb heat. It is the type of energy that controls how a body's thermal state changes and is characterised as the transfer of energy from one body to another as a result of the two bodies' different levels of heat (temperature). It transfers heat from one body's high temperature to another's low temperature.

Internal energy is the energy involved in the arrangement and haphazard movement of the atoms and molecules within a body. Heat is the portion of internal energy that is transferred from one body to another as a result of temperature differences.

Heat can be converted into other forms of energy and vice versa because it is a kind of energy.

For instance, a resistor transforms electrical energy into thermal energy, while a thermocouple does the opposite. Mechanical energy is transformed into heat energy through friction. Heat energy is transformed into mechanical energy by a heat engine.

## Calorimetry

A calorimeter is a device that is used for heat measurements, necessary for calorimetry.

The figure below shows a basic calorimeter.

It consists of an insulated chamber having ignition wires with a thermometer and a stirrer attached to it as shown. The sample placed in the vessel is burned by the aid of ignition wires and a passage through which oxygen is passed for burning. The amount of heat released by the sample is measured by measuring the temperature rise in the water and further using it along with water mass and the specific heat capacity of water. The heat loss is prevented by the insulated chamber. The stirrer ensures uniform heat distribution in the water.

Principle behind calorimetry is that in the case of an insulated system, the heat energy lost by a hot body is the same as the heat energy gained by the cold body.

The Calorimeter is based on the following formula.

We know that from the conservation of the energy at steady state situation,

Heat loss = Heat gain

q=mcΔt

Where q denotes the measure of the transferred heat into the system, m denotes the mass of substance, c denotes the specific heat capacity of the substance , Δt denotes the temperature change.

## Thermal Expansion

• Thermal expansion is the property of matter to change in shape, volume, area, and configuration in response to a change in temperature.
• It can be thought of as a photographic enlargement of the object, where the distance between any two points increases/decreases in accordance with the temperature.

The graphical representation of an object at two different temperatures T and T0 (T >T0) is shown in the figure. It can be observed that the photographic enlargement is not only restricted to an increase in length, breadth, and area, but also increases the distance between any two points chosen at random.

Thermal expansion at molecular level

We can understand thermal expansion qualitatively on the molecular level. The interatomic forces in a solid can be considered as springs, as in figure below. There is an analogous relationship between spring forces and interatomic forces. Each atom vibrates about its equilibrium position. As spring is easier to stretch as compared to compress, the interatomic spring behaves in a similar manner and these are not symmetrical about equilibrium position.

With the rise in temperature, the amplitude of vibrations at atomic level increases inside the matter. This results in the increase of energies and average distances between the atoms. Therefore, the matter as a whole expands.

Point to Remember

 With the rise in temperature, the amplitude of vibrations at atomic level increases inside the matter, which results in the increase of energies and average distances between the atoms. Therefore, the matter as a whole expands.

Thermal expansion in solids

Solids can expand in one, two, or three dimensions. These are known as linear expansion, areal expansion, and volume expansion, respectively.

Isotropic Solids

The solids whose physical properties are independent of the orientation of the system are known as isotropic solids. In other words, substances that expand linearly at the same rate in every direction are known as isotropic substances.

The coefficient of linear expansion (𝛼) of an isotropic solid is the same in x, y, and z directions.

Anisotropic Solids

The solids whose physical properties change in different orientations of the system are known as anisotropic or non-isotropic solids.

For an anisotropic solid, let 𝛼1, 𝛼2, and 𝛼3 be the coefficients of linear expansion in x, y, and z directions. Then, the coefficients of areal and volume expansion will be as follows:

𝛽 = 𝛼1 + 𝛼2

𝛾 = 𝛼1 + 𝛼2 + 𝛼3

Linear expansion

It refers to the change in the length of the body due to thermal expansion.

Consider a rod calibrated at length L0 initially at temperature T1. Heat has been supplied to the rod for some time. The temperature of the rod rises up to T2 and its length increases by amount ΔL.

According to the principle of thermal expansion, the change in length (ΔL) depends on the following two factors.

1. LT
2. LLo

Introducing a proportionality constant,

$\mathrm{\Delta L}=\alpha {L}_{o}\mathrm{\Delta T}$

Where is coefficient of linear expansion and it is property of a material.

Units of isoC-1 or K-1

${L}_{o}+\mathrm{\Delta L}={L}_{o}+\alpha {L}_{o}\mathrm{\Delta T}$

$={L}_{o}\left(1+\alpha \Delta T\right)$

For small increments in length,

$\alpha =\frac{1}{L}\frac{dL}{dT}$

${\int }_{{L}_{o}}^{L}\frac{dL}{L}=\alpha {\int }_{{T}_{1}}^{T}dT$

$ln\left[L\right]LoL=\alpha \Delta T$

$ln\frac{L}{{L}_{o}}=\alpha \Delta T$

$⇒L={L}_{o}exp\alpha \Delta T$

Even for temperature differences in the range of 102, the value of expression αΔT is of order 10−3.

Hence, for practical purposes, one can expand the exponential series.

$L\approx {L}_{o}\left[1+\alpha \Delta T\right]$

Linear expansion coefficients for some material

 Material (10-6 / oC) Ordinary Glass 9 Aluminium 23 Brass 19 Copper 17 Iron 12 Steel 13 Tungsten 4.3 Gold 17 Quartz 0.59

Areal expansion

Areal or superficial expansion refers to the expansion in the area of an object due to the increase in temperature.

The expression for areal thermal expansion is similar to that of linear expansion.

$\mathrm{\Delta A}=\beta {A}_{o}\mathrm{\Delta T}$

= coefficient of real expansion

Ao= Initial area

A and T represents change in area and temperature respectively.

Final area A=Ao+A

$A={A}_{o}+\beta {A}_{o}\mathrm{\Delta T}$

$A={A}_{o}\left(1+\beta \Delta T\right)$

Consider a case of square sheets of area A0 for simplicity. It undergoes expansion due to a rise in temperature.

L0 and L are the lengths of the sides of the sheet before and after the expansion takes place. The final value of the area of the sheet is A.

Therefore,

${A}_{o}={L}_{o}^{2}$

$A={L}^{2}={L}_{o}^{2}\left(1+\alpha \Delta T{\right)}^{2}$

$A\approx {L}_{o}^{2}\left(1+2\alpha \Delta T\right)$

$A={A}_{o}\left(1+2\alpha \Delta T\right)$

Since A=Ao(1+T)

$⇒\beta =2\alpha$

Thus the coefficient of areal expansion is twice as much as linear expansion.

Volume expansion

Volume expansion, also known as cubical expansion, is the increase in the volume of a solid due to the rise in temperature.

Similar to linear and areal expansions, the expression for volume expansion can be deduced as follows:

ΔV = 𝛾VoΔT

Thus,

V = Vo+ΔV

Or

V = Vo + 𝛾VoΔT

Or

V = Vo(1 + 𝛾ΔT)

Consider a cubical solid of side L0 initially at temperature T. The temperature of this solid has been raised by an amount ΔT. It goes under volume expansion and as a result, the length of its sides increases up to L.

From the expression of linear expansion, we can write it as:

${L}^{3}={L}_{o}^{3}\left(1+\alpha \Delta T{\right)}^{3}$

$⇒V={V}_{o}\left(1+\alpha \Delta T{\right)}^{3}$

Using binomial expansion as T<<1,

$V={V}_{o}\left(1+3\alpha \Delta T\right)$

$⇒\gamma =3\alpha$

Thus the coefficient of volume expansion is thrice as much as linear expansion.

It can be concluded that for solids,

$\alpha :\beta :\gamma =1:2:3$

## Heat Transmission

There are three major modes of heat transfer:

1. Conduction:
1. Heat conduction is the flow of internal energy from a region of higher temperature to that of a lower temperature by the interaction of adjacent particles (atoms, molecules, ions, electrons, etc.) in the intervening space.
2. Heat transfer happens only due to the collisions between the neighbouring molecules.
3. The rate of heat transfer through a medium via conduction is given by Fourier’s Law of Heat Conduction,

$Q=-kA\frac{dT}{dx}$

Where k = thermal conductivity of the material

A = cross-sectional area

dTdx= temperature gradient across the medium.

1. Convection:
1. Heat transfer by the bulk movement of molecules within fluids such as liquids and gases.
2. Convection heat transfer is further of two types i.e. natural and forced convection depending upon the involvement of external factors.
3. The rate of heat transfer by convection is given by,

$q=hA\left(T-{T}_{o}\right)$

h = Convection coefficient

A = Surface area of the body

T = Temperature of the body

T0 = Temperature of the surrounding

1. It is a type of heat transfer that does not require any intervening medium.
2. Radiation heat transfer uses electromagnetic waves to transfer heat from one body to another.
3. There are many laws that govern radiation heat transfer but the most important is Stefan Boltzmann law.
4. According to Stefan Boltzmann law, radiant energy emitted by a body is given by

E = ε𝜎AT4

Where ε= emissivity of the material.

A= surface area of the body.

𝜎= Stefan’s constant = 5.67 × 10-8 W m-2K-4

T= absolute temperature of the body

## Practice Problems

Q. 1 g ice is mixed with 1 g of steam. After thermal equilibrium is reached, find the temperature of the mixture.

(Latent heat of fusion of ice; Lice=80 cal/g, latent heat of condensation of vapour; Lv=540 cal/g, Specific heat capacity of water; c=1 cal/g oC)

A.

We know,

Latent heat of fusion of ice; Lice=80 cal/g

Latent heat of condensation of vapour; Lv=540 cal/g

Specific heat capacity of water; c=1 cal/g oC

Heat supplied by steam if it is to condense totally and convert into water at 100 oC.

Heat required by ice to convert totally into water at 0 oC

${Q}_{2}=1×{L}_{ice}$

Heat required by ice to convert totally into water at 100 oC

=80+100

=180 cal

Since Q1>Q3, entire steam will not condense and the final temperature of mixture of water and steam will be 100 oC.

Q. If the temperature of an iron rod of length 1 m changes by 100oC, what will be the percentage change in its length?

A.

Given,

Percentage change in length $=\left(\frac{\mathrm{\Delta L}}{L}×100\right)%$

$=\left(\frac{{\alpha }_{iron}{L}_{o}\mathrm{\Delta T}}{{L}_{o}}×100\right)%$

$=\left(12×{10}^{-5}×{10}^{2}×100\right)%=1.2%$

Q. Find the ratio of the lengths of an iron rod and an aluminium rod for which the difference in the lengths is independent of temperature.

A.

Temperatures of both the rods were raised from T0 to T, which results in the increment of their lengths.

Assuming that the lengths of aluminium and iron rods at temperature T0 are l1 and l2, respectively. In response to the rise in temperature, their lengths are increased to l1’ and l2’, respectively.

It is also given that the difference in the lengths of both the rods (l1 − l2) is independent of temperature.

Therefore,

${l}_{1}-{l}_{2}={l}_{1}\text{'}-{l}_{2}\text{'}$

$⇒{l}_{1}-{l}_{1}\text{'}={l}_{2}-{l}_{2}\text{'}$

$⇒\mathrm{\Delta }{l}_{1}=\mathrm{\Delta }{l}_{2}$

${\alpha }_{Al}{l}_{1}\mathrm{\Delta T}={\alpha }_{Fe}{l}_{2}\mathrm{\Delta T}$

$⇒{\alpha }_{Al}{l}_{1}={\alpha }_{Fe}{l}_{2}$ [Since T=T-To is same for both the rods]

$\frac{{l}_{2}}{{l}_{1}}=\frac{{\alpha }_{Al}}{{\alpha }_{Fe}}=\frac{23×{10}^{-5}}{12×{10}^{-5}}=\frac{23}{12}$

Hence, the ratio of the lengths of iron and aluminium rods is 23:12

Q. A thin brass sheet which is at a temperature of 10oC have same surface area as that of a thin steel sheet at 20oC. Find the common temperature at which both would have the same area.

Coefficient of areal expansion for brass and steel are 1010-6/oC and 2010-6/oC respectively.

A. We know ,

Final area A=Ao(1+T)

where Ao is the initial area.

Given:

Ao is the same for both brass and steel sheet at 10oC and 20oC respectively.

Let at temperature T, both have the same area A.

${A}_{o}\left(1+{\beta }_{b}\mathrm{\Delta }{T}_{1}\right)={A}_{o}\left(1+{\beta }_{s}\mathrm{\Delta }{T}_{2}\right)$

${A}_{o}\left(1+{\beta }_{b}\left(T-10\right)\right)={A}_{o}\left(1+{\beta }_{s}\left(T-20\right)\right)$

$⇒\frac{{\beta }_{b}}{{\beta }_{s}}=\frac{T-20}{T-10}$

$⇒\frac{10×{10}^{-6}}{20×{10}^{-6}}=\frac{T-20}{T-10}$

$⇒T={30}^{o}C$

## FAQs

Q. Why are metals used in calorimeters?
A.
Metals are usually good conductors of heat and have low specific heat. Thus they reach equilibrium temperature quickly.

Q. Can a process occur without any heat transfer across the system?
A.
Yes. Adiabatic process occurs without any heat transfer across the system.

Q. Is it possible that the temperature of a system decreases with the addition of heat?
A.
Yes, it is possible that the temperature of the system may decrease upon heat addition. This occurs in polytropic processes which have negative specific heat. Basically the system does more work than the amount of heat added and due to this the internal energy of the system decreases, thus resulting in decrease in temperature of the system.

Q. Two rods having the same material and initial temperature but different lengths are heated to the same temperature. For which rod change in length will be greater?
A.
Change in length will be more for longer rod as compared to shorter rod having same material and same initial temperature.

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text:
Acta Physico-Chimica Sinica ›› 2018, Vol. 34 ›› Issue (8): 904-911.

Special Issue: Green Chemistry

• ARTICLE •

### Investigation on the Thermal Stability of Deep Eutectic Solvents

Wenjun CHEN1,Zhimin XUE2,*(),Jinfang WANG1,Jingyun JIANG1,Xinhui ZHAO1,Tiancheng MU1,*()

1. 1 Department of Chemistry, Renmin University of China, Beijing 100872, P. R. China
2 College of Materials Science and Technology, Beijing Forestry University, Beijing 100083, P. R. China
• Received:2017-12-12 Accepted:2017-12-26 Published:2018-04-03
• Contact: Zhimin XUE,Tiancheng MU E-mail:zmxue@bjfu.edu.cn;tcmu@ruc.edu.cn
• Supported by:
The project was supported by the National Natural Science Foundation of China(21773307);The project was supported by the National Natural Science Foundation of China(21503016)

Abstract:

In recent years, deep eutectic solvents (DESs) have attracted considerable attention. They have been applied in many fields such as dissolution and separation, electrochemistry, materials preparation, reaction, and catalysis. The DESs are generally formed by the hydrogen bonding interactions between hydrogen-bond donors (HBDs) and acceptors (HBAs). Knowledge of the thermal stability of DESs is very important for their application at high temperatures. However, there have been relatively few studies on the thermal stability of DESs. Herein, a systematic investigation on the thermal stability of 40 DESs was carried out using thermal gravimetric analysis (TGA), and the onset decomposition temperatures (Tonset) of these solvents were obtained. The most important conclusion drawn from this work is that the thermal behavior of DESs is quite different from that of ionic liquids. The anions or cations of ionic liquids decompose first, followed by the decomposition of the opposite ion at elevated temperatures. On the other hand, the DESs generally first decompose to HBDs and HBAs at high temperatures through the weakening of the hydrogen bond interactions. Subsequently, the HBDs with relatively low boiling points or poor stabilities undergo volatilization or decomposition; the HBAs also undergo volatilization or decomposition but at a higher temperature. For example, the most commonly used HBA choline chloride (ChCl) begins to decompose at around 250 ℃. The hydrogen bond plays an important role in the thermal stability of DESs. It hinders the "escape" of molecules and requires greater energy to break than pure HBAs and HBDs, which causes the Tonset of DESs to shift to higher temperatures. Note that the thermal stability of HBDs has a crucial effect on the Tonset of DESs. The HBDs would decompose or volatilize first during TGA because of their relatively poor thermal stability or lower boiling points. The more stable the HBDs are, the greater would be the Tonset values of the corresponding DESs. Further, the effects of anions on HBAs, molar ratio of HBAs to HBDs, and heating rate in fast scan TGA have been discussed. As the heating rate increased, the TGA curves of DESs shifted to higher temperatures gradually, and the temperature hysteretic effect became prominent when the rate reached 10 ℃?min?1. From an industrial application point of view, there is an overestimation of the onset decomposition temperatures of DESs by Tonset, so the long-term stability of DESs was investigated at the end of the study. This study could help understand the thermal behavior of DESs (progressive decomposition) and provide guidance for designing DESs with appropriate thermal stability for practical applications.

MSC2000:

• O642


text:
# lower bound for probability distribution of a random variable

If $X$ is a random variable with finite mean $\mu$ and variance $\sigma^2$, how do I show that

$P[\mu − d\sigma < X < \mu + d\sigma] ≥ 1 − 1/d^2~~~~~~\forall d>1$

I found this in a book but unable to see the proof. Note that $X$ may not be normal.

-
$$P[\mu - d \sigma < X < \mu + d \sigma] \geq 1 - 1/d^2$$ is equivalent to $$P[|X - \mu| < d \sigma] \geq 1 - 1/d^2,$$ which is equivalent to $$P[|X - \mu| > d \sigma] \leq 1/d^2,$$


text:
# What is the most secure/reliable CSPRNG currently available? [closed]

In terms of period length, uniformity/distribution, what is the most secure/reliable CSPRNG currently available?

## closed as primarily opinion-based by kelalaka, AleksanderRas, Maarten Bodewes♦Sep 1 at 16:18

Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. If this question can be reworded to fit the rules in the help center, please edit the question.

• Try the Yarrow suite of Algorithms by Bruce Schneier – crytoknight Sep 1 at 13:50

• ChaCha and Salsa20 are 256-bit stream ciphers which can output $$2^{64} \times 512$$ bits of data per key.
• AES in CTR mode is secure, but becomes distinguishable after generating about $$2^{64} \times 128$$ bits.


text:
Problems on Average - Solved Examples
 1. In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the run rate in the remaining 40 overs to reach the target of 282 runs? A. 6.25 B. 5.5 C. 7.4 D. 5

Explanation:

Runs scored in the first 10 overs = 10 × 3.2 = 32
Total runs = 282

Remaining runs to be scored = 282 - 32 = 250
Remaining overs = 40

Run rate needed = $\dfrac{250}{40} = 6.25$

 2. A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs. 6500? A. 4800 B. 4991 C. 5004 D. 5000

Explanation:

Let the sale in the sixth month $=x$

Then $\dfrac{6435+6927+6855+7230+6562+x}{6}$ $=6500$

=> $6435+6927+6855+7230+6562+x$ $=6×6500$

=> $34009 + x = 39000$
=> $x = 39000 - 34009 = 4991$

 3. The average of 20 numbers is zero. Of them, How many of them may be greater than zero, at the most? A. 1 B. 20 C. 0 D. 19

Explanation:

Average of 20 numbers = 0

=> $\dfrac{\text{Sum of 20 numbers}}{20} = 0$

=> Sum of 20 numbers = 0

Hence at the most, there can be 19 positive numbers.
(Such that if the sum of these 19 positive numbers is x, 20th number will be -x)

 4. The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. Find out the average age of the team. A. 23 years B. 20 years C. 24 years D. 21 years

Explanation:

Number of members in the team = 11

Let the average age of of the team = $x$

=> $\dfrac{\text{Sum of ages of all 11 members}}{11} = x$

=> Sum of the ages of all 11 members = $11x$

Age of the captain = 26
Age of the wicket keeper = 26 + 3 = 29

Sum of the ages of 9 members of the team excluding captain and wicket keeper
$= 11x - 26 - 29 = 11x - 55$

Average age of 9 members of the team excluding captain and wicket keeper
$= \dfrac{11x - 55}{9}$

Given that $\dfrac{11x - 55}{9} = (x - 1)$

$\Rightarrow 11x - 55 = 9(x - 1)\\ \Rightarrow 11x - 55 = 9x - 9\\ \Rightarrow 2x = 46\\ \Rightarrow x = \dfrac{46}{2} = 23\text{ years}$

 5. The average monthly income of A and B is Rs. 5050. The average monthly income of B and C is Rs. 6250 and the average monthly income of A and C is Rs. 5200. What is the monthly income of A? A. 2000 B. 3000 C. 4000 D. 5000

Explanation:

Let monthly income of A = a
monthly income of B = b
monthly income of C = c

a + b = 2 × 5050 .... (Equation 1)
b + c = 2 × 6250 .... (Equation 2)
a + c = 2 × 5200 .... (Equation 3)

(Equation 1) + (Equation 3) - (Equation 2)
=> a + b + a + c - (b + c) = (2 × 5050) + (2 × 5200) - (2 × 6250)
=> 2a = 2(5050 + 5200 - 6250)
=> a = 4000

i.e., Monthly income of A = 4000

 6. A car owner buys diesel at Rs.7.50, Rs. 8 and Rs. 8.50 per litre for three successive years. What approximately is the average cost per litre of diesel if he spends Rs. 4000 each year? A. Rs. 8 B. Rs. 7.98 C. Rs. 6.2 D. Rs. 8.1

Explanation:

Total Cost = 4000 × 3

Total diesel used = $\dfrac{4000}{7.5}+\dfrac{4000}{8}+\dfrac{4000}{8.5}$

Average cost per litre of diesel
$=\dfrac{4000 \times 3}{\left(\dfrac{4000}{7.5} + \dfrac{4000}{8} + \dfrac{4000}{8.5}\right)}\\ = \dfrac{3}{\left(\dfrac{1}{7.5} + \dfrac{1}{8} + \dfrac{1}{8.5}\right)}$

It is important how you proceed from this stage. Remember time is very important and if we solve this in the normal method, it may take lot of time. Instead, we can find out the approximate value easily and select the right answer from the given choices.

$=\dfrac{3}{\left(\dfrac{1}{7.5} + \dfrac{1}{8} + \dfrac{1}{8.5}\right)}\\ \approx \dfrac{3}{\left(\dfrac{1}{8} + \dfrac{1}{8} + \dfrac{1}{8}\right)} \approx \dfrac{3}{\left(\dfrac{3}{8}\right)} \approx 8$

We got that answer as approximately equal to 8. From the given choices, the answer can be 8 or 7.98 or 8.1 . But which one from these?

It is easy to figure out. We approximated the denominator,$\left(\dfrac{1}{7.5}+\dfrac{1}{8}+\dfrac{1}{8.5}\right)$ to $\dfrac{3}{8}$. However
$\dfrac{1}{7.5}+\dfrac{1}{8.5}\\=\dfrac{1}{8-0.5}+\dfrac{1}{8+0.5}\\=\dfrac{8+0.5+8-0.5}{(8-0.5)(8+0.5)}\\ =\dfrac{16}{\left(8^2-0.5^2\right)} ~~\small{\left[∵ (a-b)(a+b)=a^2 - b^2\right]}\\ =\dfrac{16}{64-0.25}$

i.e., $\dfrac{1}{7.5} + \dfrac{1}{8.5} = \dfrac{16}{\left(64-0.25\right)}$
We know that $\dfrac{1}{8}+\dfrac{1}{8}=\dfrac{1}{4}=\dfrac{16}{64}$

=> $\dfrac{1}{7.5}+\dfrac{1}{8.5} > \dfrac{1}{8} + \dfrac{1}{8}$

Early we had approximated the denominator to $\dfrac{3}{8}$. However, from the above mentioned equations, now we know that actual denominator is slightly greater than $\dfrac{3}{8}$. It means answer is slightly lesser than 8. Hence we can pick the choice 7.98 as the answer

Try to remember such relations between numbers which can save lot of time in calculations.

 7. In Kiran's opinion, his weight is greater than 65 kg but less than 72 kg. His brother does not agree with Kiran and he thinks that Kiran's weight is greater than 60 kg but less than 70 kg. His mother's view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of Kiran? A. 70 kg B. 69 kg C. 61 kg D. 67 kg

Explanation:

Let Kiran's weight = x. Then

According to Kiran, 65 < x < 72 ----(equation 1)

According to brother, 60 < x < 70 ----(equation 2)

According to mother, x $\leq$ 68 ----(equation 3)

Given that equation 1,equation 2 and equation 3 are correct. By combining these equations,we can write as

$65 < x \leq 68$

i.e., x = 66 or 67 or 68

Average of different probable weights of Kiran = $\dfrac{66+67+68}{3}=67$

 8. The average weight of 16 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg. Find the average weights of all the boys in the class. A. 48.55 B. 42.25 C. 50 D. 51.25

Explanation:

Average weight of 16 boys = 50.25
Total weight of 16 boys = 50.25 × 16

Average weight of remaining 8 boys = 45.15
Total weight of remaining 8 boys = 45.15 × 8

Total weight of all boys in the class = (50.25 × 16)+ (45.15 × 8)
Total boys = 16 + 8 = 24

Average weight of all the boys = $\dfrac{\left( 50.25 \times 16 \right)+\left( 45.15 \times 8\right)}{24}$
$= \dfrac{(50.25 \times 2)+(45.15 \times 1)}{3}\\=(16.75 \times 2)+ 15.05\\= 33.5 + 15.05\\=48.55$

 9. A library has an average of 510 visitors on Sundays and 240 on other days. What is the average number of visitors per day in a month of 30 days beginning with a Sunday? A. 290 B. 304 C. 285 D. 270

Explanation:

In a month of 30 days beginning with a Sunday, there will be 4 complete weeks and another two days which will be Sunday and Monday.

Hence there will be 5 Sundays and 25 other days in a month of 30 days beginning with a Sunday

Average visitors on Sundays = 510
Total visitors of 5 Sundays = 510 × 5

Average visitors on other days = 240
Total visitors of other 25 days = 240 × 25

Total visitors = (510 × 5) + (240 × 25)
Total days = 30

Average number of visitors per day
$=\dfrac{(510 \times 5)+(240 \times 25)}{30}\\ =\dfrac{(51 \times 5)+(24 \times 25)}{3}\\=(17\times 5)+(8\times 25)\\=85+200\\=285$

 10. A student's mark was wrongly entered as 83 instead of 63. Due to that the average marks for the class got increased by $\dfrac{1}{2}$. What is the number of students in the class? A. 45 B. 40 C. 35 D. 30

Explanation:

Let the total number of students = $x$

The average marks increased by $\dfrac{1}{2}$ due to an increase of 83 - 63 = 20 marks.

But total increase in the marks = $\dfrac{1}{2}\times x = \dfrac{x}{2}$

Hence we can write as
$\dfrac{x}{2}=20\\ \Rightarrow x = 20 \times 2 = 40$

 11. A family consists of two grandparents, two parents and three grandchildren. The average age of the grandparents is 67 years, that of the parents is 35 years and that of the grandchildren is 6 years. The average age of the family is A. $32 \dfrac{2}{7}$ years B. $31 \dfrac{5}{7}$ years C. $28\dfrac{1}{7}$ years D. $30 \dfrac{5}{7}$ years

Explanation:

Total age of the grandparents = 67 × 2
Total age of the parents = 35 × 2
Total age of the grandchildren = 6 × 3

Average age of the family
$=\dfrac{(67 \times 2)+(35 \times 2) + (6 \times 3)}{7}\\ =\dfrac{134+70+18}{7}\\=\dfrac{222}{7}\\= 31\dfrac{5}{7}$

 12. The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, what is the weight of B? A. $31$ kg B. $28 \dfrac{1}{2}$ kg C. $32$ kg D. $30 \dfrac{1}{2}$ kg

Explanation:

Let the weight of A, B and C are a,b and c respectively.

Average weight of A,B and C = 45
a + b + c = 45 × 3 = 135 --- equation(1)

Average weight of A and B = 40
a + b = 40 × 2 = 80 --- equation(2)

Average weight of B and C = 43
b + c = 43 × 2 = 86 --- equation(3)

equation(2) + equation(3) - equation(1)
=> a + b + b + c - (a + b + c) = 80 + 86 - 135
=> b = 80 + 86 -135 = 166 - 135 = 31

Weight of B = 31 Kg

 13. If the average marks of three batches of 55, 60 and 45 students respectively is 50, 55, 60, what is the average marks of all the students? A. 53.23 B. 54.68 C. 51.33 D. 50

Explanation:

Average marks of batch1 = 50
Students in batch1 = 55
Total marks of batch1 = 55 × 50

Average marks of batch2 = 55
Students in batch2 = 60
Total marks of batch2 = 60 × 55

Average marks of batch3 = 60
Students in batch3 = 45
Total marks of batch3 = 45 × 60

Total students = 55 + 60 + 45 = 160

Average marks of all the students
$=\dfrac{(55 \times 50 )+ (60 \times 55)+(45 \times 60)}{160}\\ =\dfrac{275+ 330+ 270}{16}\\= \dfrac{875}{16}\\= 54.68$

 14. The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. What is the present age of the husband? A. 40 B. 32 C. 28 D. 30

Explanation:

Let the present age of the husband = h
Present age of the wife = w
Present age of the child = c

3 years ago, average age of husband, wife and their child = 27
=> Sum of age of husband, wife and their child before 3 years = 3 × 27 = 81
=> (h-3) + (w-3) + (c-3) = 81
=> h + w + c = 81 + 9 = 90 --- equation(1)

5 years ago, average age of wife and child = 20
=> Sum of age of wife and child before 5 years = 2 × 20 = 40
=> (w-5) + (c-5) = 40
=> w + c = 40 + 10 = 50 --- equation(2)

Substituting equation(2) in equation(1)
=> h + 50 = 90
=> h = 90 - 50 = 40

i.e., Present age of the husband = 40

 15. The average weight of 8 person's increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What is the weight of the new person? A. 75 Kg B. 50 Kg C. 85 Kg D. 80 Kg

Explanation:

Total increase in weight = 8 × 2.5 = 20

If $x$ is the weight of the new person, total increase in weight = $x - 65$
=> 20 = $x$ - 65
=> $x$ = 20 + 65 = 85

 16. There are two divisions A and B of a class, consisting of 36 and 44 students respectively. If the average weight of divisions A is 40 kg and that of division b is 35 kg. What is the average weight of the whole class? A. 38.25 B. 37.25 C. 38.5 D. 37

Explanation:

Total weight of students in division A = 36 × 40
Total weight of students in division B = 44 × 35

Total students = 36 + 44 = 80

Average weight of the whole class
$=\dfrac{(36 \times 40)+(44 \times 35)}{80}\\ =\dfrac{(9 \times 40)+(11\times 35)}{20}\\ =\dfrac{(9 \times 8)+(11\times 7)}{4}\\ =\dfrac{72+77}{4}\\ =\dfrac{149}{4}\\ =37.25$

 17. A batsman makes a score of 87 runs in the 17th inning and thus increases his averages by 3. What is his average after 17th inning? A. 39 B. 35 C. 42 D. 40.5

Explanation:

Let the average after 17 innings = x
Total runs scored in 17 innings = 17x

Average after 16 innings = (x-3)
Total runs scored in 16 innings = 16(x-3)

Total runs scored in 16 innings + 87 = Total runs scored in 17 innings
=> 16(x-3) + 87 = 17x
=> 16x - 48 + 87 = 17x
=> x = 39

 18. A student needed to find the arithmetic mean of the numbers 3, 11, 7, 9, 15, 13, 8, 19, 17, 21, 14 and x. He found the mean to be 12. What is the value of x? A. 12 B. 5 C. 7 D. 9

Explanation:

$\dfrac{\text{3+11+7+9+15+13+8+19+17+21+14+}x}{12}$ $= 12$

$\Rightarrow \dfrac{137 + x}{12} = 12\\~\\ \Rightarrow 137 + x = 144\\~\\ \Rightarrow x = 144 - 137 = 7$

 19. Arun obtained 76, 65, 82, 67 and 85 marks (out in 100) in English, Mathematics, Chemistry, Biology and Physics. What is his average mark? A. 53 B. 54 C. 72 D. 75

Explanation:

Average mark = $\dfrac{76 + 65 + 82 + 67 + 85}{5}=\dfrac{375}{5}=75$

 20. Distance between two stations A and B is 778 km. A train covers the journey from A to B at 84 km per hour and returns back to A with a uniform speed of 56km per hour. Find the average speed of the train during the whole journey? A. 69.0 km /hr B. 69.2 km /hr C. 67.2 km /hr D. 67.0 km /hr

Explanation:

-------------------------------------------
Solution 1
-------------------------------------------
If a car covers a certain distance at $x$ kmph and an equal distance at $y$ kmph. Then,
average speed of the whole journey = $\dfrac{2xy}{x+y}$ kmph.

By using the same formula, we can find out the average speed quickly.

Average speed
$=\dfrac{2 \times 84 \times 56}{84 + 56}=\dfrac{2\times84 \times 56}{140}\\ =\dfrac{2 \times 21 \times 56}{35}=\dfrac{2 \times 3 \times 56}{5}\\ =\dfrac{336}{5}=67.2$

-------------------------------------------
Solution 2
-------------------------------------------
Though it is a good idea to solve the problems quickly using formulas, you should know the fundamentals too. Let's see how we can solve this problems using basics.

Let the distance between A and B = $x$

Train travels from A to B at 84 km per hour
Total time taken for traveling from A to B = $\dfrac{\text{distance}}{\text{speed}}=\dfrac{x}{84}$

Train travels from B to A at 56 km per hour
Total time taken for traveling from B to A = $\dfrac{\text{distance}}{\text{speed}}=\dfrac{x}{56}$

Total distance travelled = $x+x=2x$
Total time taken = $\dfrac{x}{84}+\dfrac{x}{56}$

Average speed $=\dfrac{\text{Total distance traveled}}{\text{Total time taken}}$

$=\dfrac{2x}{\dfrac{x}{84}+\dfrac{x}{56}}=\dfrac{2}{\dfrac{1}{84}+\dfrac{1}{56}}\\~\\ =\dfrac{2 \times 84 \times 56}{56 + 84}=\dfrac{2\times84 \times 56}{140}\\~\\ =\dfrac{2 \times 21 \times 56}{35}=\dfrac{2 \times 3 \times 56}{5}\\ =\dfrac{336}{5}= 67.2$

 21. The average age of boys in a class is 16 years and that of the girls is 15 years. What is the average age for the whole class? A. 15 B. 16 C. 15.5 D. Insufficient Data

Explanation:

We do not have the number of boys and girls. Hence we cannot find out the answer.

 22. The average age of 36 students in a group is 14 years. When teacher's age is included to it, the average increases by one. Find out the teacher's age in years? A. 51 years B. 49 years C. 53 years D. 50 years

Explanation:

average age of 36 students in a group is 14
Sum of the ages of 36 students = 36 × 14

When teacher's age is included to it, the average increases by one
=> average = 15
Sum of the ages of 36 students and the teacher = 37 × 15

Hence teachers age
= 37 × 15 - 36 × 14
= 37 × 15 - 14(37 - 1)
= 37 × 15 - 37 × 14 + 14
= 37(15 - 14) + 14
= 37 + 14
= 51

 23. The average of five numbers id 27. If one number is excluded, the average becomes 25. What is the excluded number? A. 30 B. 40 C. 32.5 D. 35

Explanation:

Sum of 5 numbers = 5 × 27
Sum of 4 numbers after excluding one number = 4 × 25

Excluded number
= 5 × 27 - 4 × 25
= 135 - 100 = 35

 24. The batting average for 40 innings of a cricket player is 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, the average of the remaining 38 innings is 48 runs. Find out the highest score of the player. A. 150 B. 174 C. 180 D. 166

Explanation:

Total runs scored by the player in 40 innings = 40 × 50
Total runs scored by the player in 38 innings after excluding two innings = 38 × 48
Sum of the scores of the excluded innings = 40 × 50 - 38 × 48 = 2000 - 1824 = 176

Given that the scores of the excluded innings differ by 172. Hence let's take
the highest score as x + 172 and lowest score as x

Now x + 172 + x = 176
=> 2x = 4
=> x $=\dfrac{4}{2}$ = 2

Highest score = x + 172 = 2 + 172 = 174

 25. The average score of a cricketer for ten matches is 38.9 runs. If the average for the first six matches is 42, what is the average for the last four matches? A. 34.25 B. 36.4 C. 40.2 D. 32.25

Explanation:

Total runs scored in 10 matches = 10 × 38.9

Total runs scored in first 6 matches = 6 × 42

Total runs scored in the last 4 matches = 10 × 38.9 - 6 × 42

Average of the runs scored in the last 4 matches = $\dfrac{10 \times 38.9 - 6 \times 42}{4}$
$=\dfrac{389 - 252}{4}=\dfrac{137}{4}= 34.25$

 26. The average of six numbers is x and the average of three of these is y. If the average of the remaining three is z, then A. None of these B. x = y + z C. 2x = y + z D. x = 2y + 2z

Explanation:

Average of 6 numbers = x
=> Sum of 6 numbers = 6x

Average of the 3 numbers = y
=> Sum of these 3 numbers = 3y

Average of the remaining 3 numbers = z
=> Sum of the remaining 3 numbers = 3z

Now we know that 6x = 3y + 3z
=> 2x = y + z

 27. Suresh drives his car to a place 150 km away at an average speed of 50 km/hr and returns at 30 km/hr. What is his average speed for the whole journey ? A. 32.5 km/hr. B. 35 km/hr. C. 37.5 km/hr D. 40 km/hr

Explanation:

-------------------------------------------
Solution 1
--------------------------------------------
If a car covers a certain distance at x kmph and an equal distance at y kmph. Then,
average speed of the whole journey = $\dfrac{2xy}{x+y}$ kmph.

Therefore, average speed
$=\dfrac{2 \times 50\times 30}{50 + 30} = \dfrac{2\times 50 \times 30}{80} \\ = \dfrac{2 \times 50 \times 3}{8}= \dfrac{50 \times 3}{4}\\ = \dfrac{25\times 3}{2} = \dfrac{75}{2} = 37.5$
-------------------------------------------
Solution 2
--------------------------------------------
Though it is a good idea to solve the problems quickly using formulas, you should know the fundamentals too. Let's see how we can solve this problems using basics.

Total time taken for traveling one side = $\dfrac{\text{distance}}{\text{speed}}=\dfrac{150}{50}$
Total time taken for return journey = $\dfrac{\text{distance}}{\text{speed}}=\dfrac{150}{30}$
Total time taken = $\dfrac{150}{50}+\dfrac{150}{30}$

Total distance travelled $= 150 + 150 = 2 × 150$

Average speed = $\dfrac{\text{Total distance traveled}}{\text{Total time taken}}$

$=\dfrac{2 \times 150}{\dfrac{150}{50}+\dfrac{150}{30}}=\dfrac{2}{\dfrac{1}{50}+\dfrac{1}{30}}\\~\\ =\dfrac{2 \times 50 \times 30}{30+ 50}=\dfrac{2\times 50 \times 30}{80}\\ =\dfrac{2 \times 50 \times 3}{8}=\dfrac{50 \times 3}{4}\\ =\dfrac{25\times 3}{2} = \dfrac{75}{2} = 37.5$

 28. The average age of a husband and his wife was 23 years at the time of their marriage. After five years they have a one year old child. What is the average age of the family ? A. 21 years B. 20 years C. 18 years D. 19 years

Explanation:

Total age of husband and wife (at the time of their marriage) = 2 × 23 = 46

Total age of husband and wife after 5 years + Age of the 1 year old child
= 46 + 5 + 5 + 1 = 57

Average age of the family = $\dfrac{57}{3}$ = 19

 29. In an examination, a student's average marks were 63. If he had obtained 20 more marks for his Geography and 2 more marks for his history, his average would have been 65. How many subjects were there in the examination? A. 12 B. 11 C. 13 D. 14

Explanation:

Let the number of subjects = x
Then, total marks he scored for all subjects = 63x

If he had obtained 20 more marks for his Geography and 2 more marks for his history, his average would have been 65
=> Total marks he would have scored for all subjects = 65x

Now we can form the equation as 65x - 63x = additional marks of the student = 20 + 2 = 22
=> 2x = 22
=> x = $\dfrac{22}{2}$ = 11

 30. The average salary of all the workers in a workshop is Rs.8000. The average salary of 7 technicians is Rs.12000 and the average salary of the rest is Rs.6000. How many workers are there in the workshop? A. 21 B. 22 C. 23 D. 24

Explanation:

Let the number of workers = x

Given that average salary of all the workers = Rs.8000
Then, total salary of all workers = 8000x

Given that average salary of 7 technicians is Rs.12000
=> Total salary of 7 technicians = 7 × 12000 = 84000

Count of the rest of the employees = (x - 7)
Average salary of the rest of the employees = Rs.6000
Total salary of the rest of the employees = (x - 7)(6000)

8000x = 84000 + (x - 7)(6000)
=> 8x = 84 + (x - 7)(6)
=> 8x = 84 + 6x - 42
=> 2x = 42
=> x = $\dfrac{42}{2}$ = 21

Swati
2015-03-22 05:29:32
A batsman in his 20th inning makes a score of 110 and thereby increases his average by 4. What is his average before 20th inning?
Pankaj Maheshwari
2015-04-24 06:28:12
20*4 = 80.....
110-80 = 30
thats all
Dev
2015-04-05 17:32:24
Let the initial average be x
Then the total score in the first 19 innings = 19x

He scores 110 runs in the 20th inning.
Total score becomes 19x+110

average of the 20 innings is (x+4).
Total score of the 20 innings = 20(x+4)

19x+110 = 20(x+4)
x = 30
his average before 20th inning = 30
mahender
2015-03-17 05:11:48
A pool is there. In order to fill that pool, it takes 28 days.how many days does it take to fill half of the pool.day by day it will goes like twice
shyam
2015-09-25 11:05:47
since everyday pool it fills up twice, so it will fill to half in 27 days.
yavanika sharma
2015-03-12 07:11:49
At the end of a soccer season every player had a prime number of goals and the average of the 11 players was also a prime number. No player's individual tally was the same as of anyone else's or as the average. nobody has scored more than 45 goals.

Q1 what was the average of their goals scored?
Q2 what was the maximum goals scored by a single person?
Q3 what was the minimum number of goals scored by a single person?
Raj
2015-03-14 08:52:12
Consider all prime numbers less than 45 which is a possible super set of the 11 unique scores
{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43}

Total of these numbers = 281
if we remove 3,5,7 (lowest three),  average will slightly above 24
if we remove
37,41,43, average will be slightly above 14
So, Possible values of the average are 17, 19 and 23 (because average is also a prime)

possible set is                                {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43}
reminder (when divided by by 11)   {2, 3, 5, 7,  0,    2,  6,   8,   1,  7,   9,   4,   8,  10
}

if the average is 17,
possible set is                                  {2, 3, 5, 7, 11, 13, 19, 23, 29, 31, 37, 41, 43}

Total of these 13 numbers = 264 .
Total of the actual 11 numbers = 17 * 11 = 187 (because average is 17)

Remainder of
264 when divided by 11 is 0.  So the two numbers which are subtracted from 264 should have the remainder sum of 11 (ie, sum of the numbers as 11, 22 etc) so that the net sum will be divisible by 11.

Even when we subtract any two numbers from 264 whose reminder sum is 11, we will not get 187. So average cannot be 17.

if the average is 19,
possible set is                                  {2, 3, 5, 7, 11, 13, 17, 23, 29, 31, 37, 41, 43}

Total of these 13 numbers = 262
Total of the actual 11 numbers = 19 * 11 = 209

Remainder of 262 when divided by 11 is 9. so the two numbers which are subtracted from 262 should have the remainder sum of 2
(ie, sum of the numbers as 2, 13, 24 etc) so that the net sum will be divisible by 11)

Even when we subtract any two numbers from 262 whose reminder sum is 2, we will not get 209 and hence, average cannot be 19

if the average is 23,
possible set is                                  {2, 3, 5, 7, 11, 13, 17, 19, 29, 31, 37, 41, 43}
remainder (when divided by by 11)    {2, 3, 5, 7,  0,    2,  6,  8,  7,   9,   4,   8,  10
}

Total of these 13 numbers = 258(remainder when divided by 11 is 5)
Total of the actual 11 numbers = 23 * 11 = 243
we can subtract 2 and 3 (
whose reminder sum is 5) such that is sum becomes 253 which is divisible by 11

So the average is 23
The numbers are
{5, 7, 11, 13, 17, 19, 29, 31, 37, 41, 43}

Now the questions can be easily answered
1. average of their goals scored : 23
2.
maximum goals scored by a single person : 43
3. minimum number of goals scored by a single person : 5
hyma
2015-03-11 10:48:18
there are 27 students in a class whose average weight is 36kg, among these 27 students 15 are boys and 12 are girls, the average weight of 15 boys is 6.75kg more than the average weight of 12 girls. what is the average weight of 12 girls(in kg)?
Dev
2015-03-11 17:41:36
Total weight of all students = 27*36 = 972

Let average weight of the 12 girls be x kg.
Then, the average weight of 15 boys = (x+6.75)

Total weight of the 12 girls = 12x
Total weight of the 15 boys = 15(x+6.75)

12x+15(x+6.75)=972
x = 32.25 kg
average weight of 12 girls = 32.25 kg
ishan
2015-01-20 15:15:13
the length of a rectangular plot is $4\dfrac{1}{2}$ times that of its breadth if the area of plot is 200 square metres then, then what is its length ?
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# Finding Summation with Binomial Expansion

Having trouble figuring out this question. Any help would be appreciated! Thanks.

$\sum_{k=2}^n k(k-1)\binom{n}{k}$

• What have you tried? Why don't you try a different form for $\binom{n}{k}$? The $k=2$ is a big hint. – michaelrccurtis Sep 7 '15 at 1:31
• I simplified the original to $\sum_{k=2}^n \cfrac {n!}{(n-k!)/(k-2)!}$ which simplifies to $\sum_{k=2}^n \cfrac{n(n-1)*...*(n-k+1)}{(k-2)!}$ but do not know how to proceed. How does the k=2 factor in? Thank you. – Adam Taché Sep 7 '15 at 1:42
• If you take your first simplification, can you think of a way to make that look like $\binom{n-2}{m}$ for some m, say? – michaelrccurtis Sep 7 '15 at 1:44

First, we can substitute in the factorial form for the binomial coefficient and simplify to:

$$\sum_{k=2}^n \frac{n!}{(k-2)!(n-k)!}$$

If we then make the substitution $m = k-2$:

$$\sum_{m=0}^{n-2} \frac{n(n-1)(n-2)!}{m!((n-2)-m)!}$$

We can then bring out constant factors:

$$n(n-1)\sum_{m=0}^{n-2} \binom{n-2}{m}$$

Finally, we can note that the sum part is the expansion for $(1+1)^{n-2}$ (or from Pascals Triangle), which means that the result is:

$$n(n-1)2^{n-2}$$

Note that these types of simplification problem often appear, and the strategy is frequently to manipulate them into a form that looks like a binomial expansion of some kind.

• Well explained :-) – user220382 Sep 8 '15 at 7:18

Consider the series $$S(t) = \sum_{k=0}^{n} \binom{n}{k} \, t^{k} = (1 + t)^{n}$$ then $$S''(t) = n(n-1) \, (1+t)^{n-2} = \sum_{k=0}^{n} k(k-1) \, \binom{n}{k} \, t^{k-2} = \sum_{k=2}^{n} k(k-1) \, \binom{n}{k} \, t^{k-2}$$ Now, by letting $t=1$ the identity $$\sum_{k=2}^{n} k(k-1) \, \binom{n}{k} = 2^{n-2} \, n(n-1)$$ is obtained.