Messages posted by Ramandeep Singh

Suppose you have 2 objects with masses M and m that exert a force on each other as shown. 

A summary of the mass, force & acceleration is given in the table below

                  Quantity       Mass M       mass m     
       Mass       M       m     
       Force       -F       F     
       Acceleration       aM         = -F/M       am         = F/m     
    

Newton's laws of action/reaction have been used.  Calculating quantities such as distances between them looks complex, but there is a nifty tool.  View the system in the reference frame of one of the objects (say mass M).

Now mass M is stationary, but mass m is still free to move under the force F, with one catch.  When one views a situation in the frame of an object with acceleration a, all objects have as there acceleration the acceleration due to the forces on it minus this acceleration a. For the case of mass m

a_m´ = F/m - a_M = F·(1/m + 1/M) = F/(m·M/(m + M))

In other words, mass m behaves as if it instead has a mass

m = m·M/(m + M)

This is called the reduced mass.

I have tried my best to expliain the concept!

pls rate is crrect....................

i think u shud go 4 morrison & boyd.
its the best book 4 organic

it will disintegrate by

-decay.

look in ₁₃Al²⁹ it has 2 extra neutrons than in ₁₃Al²⁷ which is stable.
so in order to become stable its neutron will disintegrate into a

-particle and a proton to reduce the energy caused due to the higher no. of neutrons.

the actual reaction is shown below

₁₃Al²⁹ -------------------> ₁₄Si²⁹ + ₀e^-1

the reaction process may continue depending on the stability of ₁₄Si²⁹.

hope u have understood

ok mr obaid u r forgiven but on the condition u must not repeat the same thing again.

kool article.
did u write dat urself

i think the function is not integrable because its not specified wrt what we have to integrate it.
if u think it should be integrated wrt x then dx shud be in numerator.

look let me illustrate more specifically. suppose we differentiate y wrt x then dx is always in denominator. in dy/dx u cannot say x is differentiated wrt y. its convention which is disturbed in this case.

similarly the convention is not valid in the question provided by u.
pls re-check it from the source!

dan brown coz da vinci code rocks

lawn-tennis or table-tennis

Hey dude!

I have tried to post almost all the structures in our syllabus.
after that i think u shud have no problem

Simple lattices and their unit cells

Simple Cubic (SC) - There is one host atom ("lattice point") at each corner of a cubic unit cell. The unit cell is described by three edge lengths a = b = c = 2r (r is the host atom radius), and the angles between the edges, alpha = beta = gamma = 90 degrees. There is one atom wholly inside the cube (Z = 1). Unit cells in which there are host atoms (or lattice points) only at the eight corners are called primitive.

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Body Centered Cubic (BCC) - There is one host atom at each corner of the cubic unit cell and one atom in the cell center. Each atom touches eight other host atoms along the body diagonal of the cube (a = 2.3094r, Z = 2).

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Face Centered Cubic (FCC) - There is one host atom at each corner, one host atom in each face, and the host atoms touch along the face diagonal (a = 2.8284r, Z = 4). This lattice is "closest packed", because spheres of equal size occupy the maximum amount of space in this arrangment (74.05%); since this closest packing is based on a cubic array, it is called "cubic closest packing": CCP = FCC.

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FCC Primitive - It is also possible to choose a primitive unit cell to describe the FCC lattice. The cell is a rhombohedron, with a = b = c = 2r, and alpha = beta = gamma = 60 degrees. [A cube is a rhombohedron with alpha = beta = gamma = 90 degrees!]

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Simple Hexagonal (SH) - Spheres of equal size are most densely packed (with the least amount of empty space) in a plane when each sphere touches six other spheres arranged in the form of a regular hexagon. When these hexagonally closest packed planes (the plane through the centers of all spheres) are stacked directly on top of one another, a simple hexagonal array results; this is not, however, a three-dimensional closest packed arrangement. The unit cell, outlined in black, is composed of one atom at each corner of a primitive unit cell (Z = 1), the edges of which are: a = b = c = 2r, where cell edges a and b lie in the hexagonal plane with angle a-b = gamma = 120 degrees, and edge c is the vertical stacking distance.

Closest Packing

Hexagonal Closest Packing (HCP) - To form a three-dimensional closest packed structure, the hexagonal closest packed planes must be stacked such that atoms in successive planes nestle in the triangular "grooves" of the preceeding plane. Note that there are six of these "grooves" surrounding each atom in the hexagonal plane, but only three of them can be covered by atoms in the adjacent plane. The first plane is labeled "A" and the second plane is labeled "B", and the perpendicular interplanar spacing between plane A and plane B is 1.633r (compared to 2.000r for simple hexagonal). If the third plane is again in the "A" orientation and succeeding planes are stacked in the repeating pattern ABABA... = (AB), the resulting closest packed structure is HCP.

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HCP Coordination - Each host atom in an HCP lattice is surrounded by and touches 12 nearest neighbors, each at a distance of 2r: six are in the planar hexagonal array (B layer), and six (three in the A layer above and three in the A layer below) form a hexagonal prism around the central atom.

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Cubic Closest Packing (CCP) - If the atoms in the third layer lie over the three grooves in the A layer which were not covered by the atoms in the B layer, then the third layer is different from either A or B and is labeled "C". If a fourth layer then repeats the A layer orientation, and succeeding layers repeat the pattern ABCABCA... = (ABC), the resulting closest packed structure is CCP = FCC. Again, the perpendicular spacing between any two successive layers is 1.633r.

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CCP Coordination - Each host atom in a CCP lattice is surrounded by and touches 12 nearest neighbors, each at a distance of 2r: six are in the planar hexagonal (B) plane, and six (three in the C layer above and three in the A layer below) form a hexagonal anti-prism (also known as a distorted octahedron) around the central atom.

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Rhombohedral (R) lattice - If, in the (ABC) layered lattice, the interplanar spacing is not the closest packed value (1.633r), then the primitive (Z = 1) unit cell is a rhombohedron with a = b = c <> 2r and alpha = beta = gamma <> 60 degrees. The non-primitive hexagonal unit cell (Z = 3).may also be chosen.

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2- & 3-layer repeats - There is only one way to produce a repeat pattern (crystal lattice) in two layers of hexagonally closest packed planes: (AB) = HCP. Likewise, there is only one way to produce a repeat pattern in three layers of hexagonally closest packed planes: (ABC) = CCP.

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4-layer repeats - However, there are two ways to produce a closest packed lattice in four layers: (ABAC) and (ABCB). By extension, there are increasing numbers of ways to produce closest packed lattices in five layers, six layers, etc., up to and including non-repeating random stacking. Thus, there are many closest (and pseudo-closest) packings in natural and artificial materials.

Holes ("Interstices") in Closest Packed Arrays

Tetrahedral Hole - Consider any two successive planes in a closest packed lattice. One atom in the A layer nestles in the triangular groove formed by three adjacent atoms in the B layer, and the four atoms touch along the edges (of length 2r) of a regular tetrahedron; the center of the tetrahedron is a cavity called the Tetrahedral (or Td) hole; a guest sphere will just fill this cavity (and touch the four host spheres) if its radius is 0.2247r.

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Octahedral Hole - Adjacent to the Td hole, three atoms in the B layer touch three atoms in the A layer such that a trigonal antiprismatic polyhedron (a regular octahedron) is formed; the center of the octahedron is a cavity called the Octahedral (or Oh) hole. A guest sphere will just fill this cavity (and touch the six host spheres) if its radius is 0.4142r. It can be shown that there are twice as many Td as Oh holes in any closest packed bilayer.

Simple Crystal Structures

CsCl Structure - Each ion resides on a separate, interpenetrating SC lattice such that the cation is in the center of the anion unit cell and visa versa. The two lattices have the same unit cell dimension.

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NaCl Structure - Each ion resides on a separate, interpenetrating FCC lattice. The two lattices have the same unit cell dimension.

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Halite Structure - The sodium chloride structure may also be viewed as a CCP lattice of anions (Z = 4), with smaller cations occupying all Oh cavities (Z = 4).

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Fluorite Structure - The structure of the mineral fluorite (calcium fluoride) may be viewed as a CCP lattice of cations (Z = 4), with the smaller anion occupying all of the Td holes (Z = 8). The Td cavities reside on a SC lattice which is half the dimension of the CCP lattice.

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Zinc Blende Structure - The structure of cubic ZnS (mineral name "zinc blende") may be viewed as a CCP lattice of anions (Z = 4), with the smaller cations occupying every other Td hole (Z = 4). [Note: the other ZnS mineral, wurtzite, can be described as a HCP lattice of anions with cations in every other Td hole.]

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Zinc Blende lattices - The lattice of cations in zinc blende is a FCC lattice of the same dimension as the anion lattice, so the structure can be described as interpenetrating FCC lattices of the same unit cell dimension. Note that the only difference between the halite and zinc blende structures is a simple shift in relative position of the two FCC lattices.

Source:http://members.kr.inter.net/joo/physics/curri-sub/crystal/lattice.html

i think i have got one more for english.
its mp3000.net

Hydrogen Gas Thermometer

In 1780, J. A. C. Charles, a French physician, showed that for the same increase in temperature, all gases exhibited the same increase in volume. Because the expansion coefficient of gases is so very nearly the same, it is possible to establish a temperature scale based on a single fixed point rather than the two fixed- point scales, such as the Fahrenheit and Celsius scales. This brings us back to a thermometer that uses a gas as the thermometric medium.

In a constant volume gas thermometer a large bulb B of gas, hydrogen for example, under a set pressure connects with a mercury-filled "manometer" by means of a tube of very small volume. (The Bulb B is the temperature-sensing portion and should contain almost all of the hydrogen). The level of mercury at C may be adjusted by raising or lowering the mercury reservoir R. The pressure of the hydrogen gas, which is the "x" variable in the linear relation with temperature, is the difference between the levels D and C plus the pressure above D.

P. Chappuis in 1887 conducted extensive studies of gas thermometers with constant pressure or with constant volume using hydrogen, nitrogen, and carbon dioxide as the thermometric medium. Based on his results, the Comité International des Poids et Mesures adopted the constant-volume hydrogen scale based on fixed points at the ice point (0° C) and the steam point (100° C) as the practical scale for international meteorology.

Experiments with gas thermometers have shown that there is very little difference in the temperature scale for different gases. Thus, it is possible to set up a temperature scale that is independent of the thermometric medium if it is a gas at low pressure. In this case, all gases behave like an "Ideal Gas" and have a very simple relation between their pressure, volume, and temperature: pV= (constant)T.

Source:www.cartage.org.lb

Platinum Resistance Thermometers (RTDs)

Resistance thermometers are slowly replacing thermocouples in many lower temperature industrial applications (below 600°C). Resistance thermometers come in a number of construction forms and offer greater stability, accuracy and repeatability. The resistance tends to be almost linear with temperature. A small power source is required.

No special extension cables or cold junction compensations are required The resistance of a conductor is related to its temperature. Platinum is usually used due to its stability with temperature. The Platinum detecting wire needs to be kept free of contamination to remain stable. A Platinum wire or film is created and supported on a former in such a way that it gets minimal differential expansion or other strains from its former, yet is reasonably resistant to vibration.

Commercial platinum grades are produced which exhibit a change of resistance of 0.385 Ohms/°C (European Fundamental Interval) The sensor is usually made to have 100 Ohms at 0 °C. This is defined in BS EN 60751:1996. The American Fundamental Interval is 0.392 Ohms/°C.

Resistance thermometers require a small current to be passed through in order to determine the resistance. This can cause self heating and manufacturers limits should always be followed along with heat path considerations in design. Care should also be taken to avoid any strains on the resistance thermometer in its application.

Lead wire resistance should be considered and adopting three and four wire connection strategies can result in eliminating connection lead resistance effects from measurements.

Resistance thermometers elements are available in a number of forms. The most common are:

Wire Wound in a ceramic insulator - High temperatures to 850 °C
Wires encapsulated in glass - Resists the highest vibration and offers most protection to the Pt
Thin film with Pt film on a ceramic substrate - Inexpensive mass production

Practical Construction

These elements will nearly always require insulated leads attached. At low temperatures PVC, Silicon rubber or PTFE insulators are common to 250 °C. Above this Glass fibre or ceramic are used. The measuring point and usually most of the leads require a housing or protection sleeve. This is often a metal alloy which is inert to a particular process.

Often more consideration goes in to selecting and designing protection sheaths than sensors as this is the layer that must withstand chemical or physical attack along with offering convenient process attachment features.

Standard Resistance Thermometer Data

Temperature sensors are usually supplied with Thin film Elements These are rated as:

Continuous operation	-70 to +500 °C
Tolerance class B	-70 to +500 °C
Tolerance class A (1/2B)	-30 to +350 °C
Tolerance class 1/3B	0 to +100 °C

Resistance Thermometer elements can be supplied which function up to 850 °C Sensor Tolerances are calculated as:

Class B	change in t=+/- (0.3+0.005\|t\|)
Class A	change in t=+/- (0.15+0.002\|t\|)
1/3 Class B	change in t=+/- 1/3 x (0.3+0.005\|t\|)
1/5 Class B	change in t=+/- 1/5 x (0.3+0.005\|t\|)
1/10 Class B	change in t=+/- 1/10 x (0.3+0.005\|t\|)

Where |t| = temperature in °C. Where elements have a resistance of n x 100 Ohms then the basic values and tolerances also have to be multiplied by n

Resistance Thermometer Wiring Configurations

Two Wire Configuration

The simplest resistance thermometer configuration uses two wires. It is only used when high accuracy is not required as the resistance of the connecting wires is always included with that of the sensor leading to errors in the signal. Using this configuration you will be able to use 100 metres of cable. This applies equally to balanced bridge and fixed bridge systems. The values of the lead resistance can only be determined in a separate measurement without the resistance thermometer sensor and therefore a continuous correction during the temperature measurement is not possible.

Three Wire Configuration

In order to minimise the effects of the lead resistances a three wire configuration can be used. Using this method the two leads to the sensor are on adjoining arms, there is a lead resistance in each arm of the bridge and therefore the lead resistance is cancelled out. High quality connection cables should be used for this type of configuration because an assumption is made that the two lead resistances are the same. This configuration allows for up to 600 metres of cable.

Four Wire Configuration

Four Wire Resistance Thermometer Diagram

The four wire resistance thermometer configuration even further increases the accuracy and reliability of the resistance being measured. In the diagram above a standard two terminal RTD is used with another pair of wires to form an additional loop that cancels out the lead resistance. The above wheatstone bridge method uses a little more copper wire and is not a perfect solution. Below is an better alternative configuration that we use in all our RTD's. It provides full cancellation of spurious effects and cable resistance of up to 15 ohms can be handled.

Source:www.peaksensors.co.uk

hey dude!

first for balloon
using

s = ut + 1/2at²
s = 0 + 1/2 x 1.25 x 64
s = 40m

and using v = u + at
v = 0 + 1.25 x 8
v = 10m/s

now coming back to stone-

h = ut - 1/2gt²
h = 10t - 5t².............................{1}

also v = u - gt
0 = 10 - 10t
t = 1s
ie stone ascends for 1s.

putting in {1}, h = 5m..........................{2}

at this instant height of stone from ground,
H = 5 + 40 = 45m

now while descending u = 0
using H = ut - 1/2gt²-45 = 0 - 1/2 x 10 x t²t² = 9
t = 3s

so
i)dist covered by stone = 5m(upwards)+45m(downwards) = 50m
ii)time taken by stone = 1s(upwards) + 3s(downwards) = 4s

pls rate if crrect

ye bacchon ka khel nahi

Hello dude!
Check this out below-
hope it helps

Balancing Redox Reactions Using the Half Reaction Method

Many redox reactions occur in aqueous solutions or suspensions. In this medium most of the reactants and products exist as charged species (ions) and their reaction is often affected by the pH of the medium. The following provides examples of how these equations may be balanced systematically. The method that is used is called the ion-electron or "half-reaction" method.

Example 1 -- Balancing Redox Reactions Which Occur in Acidic Solution

Organic compounds, called alcohols, are readily oxidized by acidic solutions of dichromate ions. The following reaction, written in net ionic form, records this change. The oxidation states of each atom in each compound is listed in order to identify the species that are oxidized and reduced, respectively.

An examination of the oxidation states, indicates that carbon is being oxidized, and chromium, is being reduced. To balance the equation, use the following steps:

First, divide the equation into two halves; an oxidation half-reaction and reduction half- reaction by grouping appropriate species.
```
(red.)   (Cr₂O₇)^-2   ---->   Cr⁺³
 
(ox.)    C₂H₆O  ---->   C₂H₄O 
```
Second, if necessary, balance both equations by inspection. In doing this ignore any oxygen and hydrogen atoms in the formula units. In other words, balance the non-hydrogen and non-oxygen atoms only. By following this guideline in the example above, only the reduction half-reaction needs to be balanced by placing the coefficient, 2 , in front of Cr⁺³ as shown below.
```
(red.)    (Cr₂O₇)^-2  ---->  2 Cr⁺³
 
(ox.) C₂H₆O     ---->  C₂H₄O  
```
(as there are equal numbers of carbon atoms on both sides of this equation, skip this step for this half-reaction. Remember, in this step, one concentrates on balancing only non-hydrogen and non-oxygen atoms)
The third step involves balancing oxygen atoms. To do this, one must use water (H₂O) molecules. Use 1 molecule of water for each oxygen atom that needs to be balanced. Add the appropriate number of water molecules to that side of the equation required to balance the oxygen atoms as shown below.
```
(red.)   (Cr₂O₇)^-2   ---->  2 Cr⁺³  +  7 H₂O
 
(ox.)    C₂H₆O      ---->  C₂H₄O 
```
(as there are equal numbers of oxygen atoms, skip this step for this half-reaction)
The fourth step involves balancing the hydrogen atoms. To do this one must use hydrogen ions (H⁺). Use one (1) H⁺ ion for every hydrogen atom that needs to balanced. Add the appropriate number of hydrogen ions to that side of the equation required to balance the hydrogen atoms as shown below
```
(red.)   14 H⁺ + (Cr₂O₇)^-2   --->   2 Cr⁺³ + 7 H₂O
```
(as there are 14 hydrogen atoms in 7 water molecules, 14 H⁺ ions must be added to the opposite side of the equation)
```
(ox.)    C₂H₆O ---> C₂H₄O + 2 H⁺
```
(2 hydrogen ions must be added to the "product" side ofthe equation to obtain a balance)
The fifth step involves the balancing of positive and negative charges. This is done by adding electrons (e-). Each electron has a charge equal to (-1). To determine the number of electrons required, find the net charge of each side the equation.

The electrons must always be added to that side which has the greater positive charge as shown below.

note: the net charge on each side of the equation does not have to equal zero.

The same step is repeated for the oxidation half-reaction.

As there is a net chargae of +2 on the product side, two electrons must be added to that side of the equation as shown below.

At this point the two half-reactions appear as:
```
(red)   6e^-  +  14 H⁺  +  (Cr₂O₇)^-2  ------->  2 Cr⁺³ + 7 H₂O
 
(ox)    C₂H₆O  ------>  C₂H₄O  +  2 H⁺  +  2e^-
```
The reduction half-reaction requires 6 e-, while the oxidation half-reaction produces 2 e-.
The sixth step involves multiplying each half-reaction by the smallest whole number that is required to equalize the number of electrons gained by reduction with the number of electrons produced by oxidation. Using this guideline, the oxidation half reaction must be multiplied by "3" to give the 6 electrons required by the reduction half-reaction.
```
(ox.)   3 C₂H₆O ---> 3 C₂H₄O + 6 H⁺ + 6e^-
```
The seventh and last step involves adding the two half reactions and reducing to the smallest whole number by cancelling species which on both sides of the arrow.
```
6e^-  +  14 H⁺  +   (Cr₂O₇)^-2  ----->   2 Cr⁺³ + 7 H₂O
                    3 C₂H₆O   ----->   3 C₂H₄O + 6 H⁺ + 6e^-
```
adding the two half-reactions above gives the following:

6e^-+ 14H⁺ + (Cr₂O₇)^-2 + 3C₂H₆O ---> 2Cr⁺³ + 7H₂O + 3C₂H₄O + 6H⁺ + 6e^-

Note that the above equation can be further simplified by subtracting out 6 e- and 6 H⁺ ions from both sides of the equation to give the final equation.

Note: the equation above is completely balanced in terms of having an equal number of atoms as well as charges.

Example 2 - Balancing Redox Reactions in Basic Solutions

The active ingredient in bleach is the hypochlorite (OCl^-) ion. This ion is a powerful oxidizing agent which oxidizes many substances under basic conditions. A typical reaction is its behavior with iodide (I^-) ions as shown below in net ionic form.

I^- (aq) + OCl^-(aq) ------> I₂ + Cl^- + H₂O

Balancing redox equations in basic solutions is identical to that of acidic solutions except for the last few steps as shown below.

First, divide the equation into two halves; an oxidation half-reaction and reduction-reaction by grouping appropriate species.
```
(ox)    I^-   ----> I₂
 
(red)   OCl^- ----> Cl^- + H₂O
```
Second, if needed, balance both equations, by inspection ignoring any oxygen and hydrogen atoms. (The non-hydrogen and non-oxygen atoms are already balanced,hence skip this step)
Third, balance the oxygen atoms using water molecules . (The hydrogen and oxygen atoms are already balanced; hence, skip this step also.
Fourth, balance any hydrogen atoms by using an (H+) for each hydrogen atom
```
(ox)    2 I^- ----> I₂
```
(as no hydrogens are present, skip this step for this half-reaction)
```
(red)   2 H⁺ + OCl^- -----> Cl^- + H₂O
```
(two hydrogen ions must be added to balance the hydrogens in the water molecule).
Fifth, use electrons (e-) to equalize the net charge on both sides of the equation. Note; each electron (e-) represents a charge of (-1).
Sixth, equalize the number of electrons lost with the number of electrons gained by multiplying by an appropriate small whole number.
```
(ox)   2 I^-  ---->  I₂  +  2e^-
(red)  2e^-  +  2 H⁺  +  OCl^-  ---->  Cl^-  +  H₂O
```
(as the number of electrons lost equals the number of electrons gained, skip this step)
Add the two equations, as shown below.
```
2 e^-  +  2 I^-  +  2 H⁺  +  OCl^-  ----> I₂  +  Cl^-  +  H₂O  +  2e^-
```
and subtract "like" terms from both sides of the equation. Subtracting "2e-" from both sides of the equation gives the net equation:
To indicate the fact that the reaction takes place in a basic solution, one must now add one (OH-) unit for every (H+) present in the equation. The OH- ions must be added to both sides of the equation as shown below.
```
2 OH^- + 2 I^- + 2 H⁺ + OCl^-  ----->  I₂ + Cl^- + H₂O + 2 OH^-
```
Then, on that side of the equation which contains both (OH^-) and (H⁺) ions, combine them to form H₂O. Note, combining the 2 OH^- with the 2 H⁺ ions above gives 2 HOH or 2 H₂O molecules as written below.
```
2 H₂O + 2 I^- + OCl^-  ---->  I₂ + Cl^- + H₂O + 2 OH^-
```
Simplify the equation by subtracting out water molecules, to obtain the final, balanced equation.

Note that both the atoms and charges are equal on both sides of the equation, and the presence of hydroxide ions (OH^-) indicates that the reaction occurs in basic solution.

pls rate is useful

hum aapke hain kaun....bathroom mein

i think the answer is B.
light is scattered due to diffraction.

my vote-

boys - himanshu

gals - manasi

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thank u sir

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Simple lattices and their unit cells

Closest Packing

Holes ("Interstices") in Closest Packed Arrays

Simple Crystal Structures

Platinum Resistance Thermometers (RTDs)

Practical Construction

Standard Resistance Thermometer Data

Resistance Thermometer Wiring Configurations

Two Wire Configuration

Three Wire Configuration

Four Wire Configuration

Balancing Redox Reactions Using the Half Reaction Method

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