## Sunday, June 7, 2015

### Circuit Solution Methods (By Examples)

To find the equivalent resistance between A  and B.

After marking the junction

Hence equivalent resistance between A and B is $url=http://latex.codecogs.com/gif.latex?%7B%7BR%7D_%7BAB%7D%7D%3D%5Cfrac%7B3R%7D%7B5%7D&container=blogger&gadget=a&rewriteMime=image/*$

Nodal analysis:
To find the current distribution in arms

To solve this one must assume the potential of one node equal to zero and then corresponding other potentials as marked in following diagram

The solving equation for the unknown V is

$url=http://latex.codecogs.com/gif.latex?%5Cfrac%7BV-%7B%7BE%7D_%7B1%7D%7D%7D%7B%7B%7BR%7D_%7B1%7D%7D%7D+%5Cfrac%7BV-%7B%7BE%7D_%7B2%7D%7D%7D%7B%7B%7BR%7D_%7B2%7D%7D%7D+%5Cfrac%7BV%7D%7B%7B%7BR%7D_%7B3%7D%7D%7D%3D0&container=blogger&gadget=a&rewriteMime=image/*$
Symmetry method:

(a) Folding symmetry

(b) Input output symmetry

In such cases the current distribution follows the symmetry as shown below

Example: A battery of 10 V and negligible internal resistance is connected across the
diagonally opposite corners of a cubical network consisting of 12 resistors
each of resistance  as shown in the figure. Determine the equivalent
resistance of the network and the current along each edge of the cube.
Solution:        The network is not reducible to a simple series and parallel combinations of resistors. There is, however, a clear symmetry in the problem which we can exploit to obtain the equivalent resistance of the network.

The paths, AD and AB are obviously symmetrically placed in the network. Thus, the current in each must be the same, say, I. Further, at the corners, B and D, the incoming current I must split equally into the two outgoing branches. In this manner, the current in all the 12 edges of the cube are easily written down in terms of I, using Kirchhoff’s first rule and the symmetry in the problem. Next take a closed loop, say, and apply Kirchhoff’s second rule:

$-IR-(1/2)IR-IR+\varepsilon =0$
Where R is the resistance of each edge and  the emf  $\varepsilon$ of battery.
Thus,
$\varepsilon =\frac{15}{2}IR$    ${{R}_{eq}}=\frac{\varepsilon }{31}=\frac{5}{6}R$
The equivalent resistance  of the network is

For  $R=1\,\Omega ,\,\,{{R}_{eq}}=(5/6)\,\Omega$   and for  $\varepsilon =10V,$

the total current (= 3I) in the network is

$3I=10V/(5/6)\,\Omega =12A,\,i.e.,\,I=4\,A$

The current  in each edge can now be read off from the figure.

## Thursday, May 28, 2015

### Number Series & Letter Series

Hello MyRank Viewers,Here are some series.Try it........

1.     1, 2, 4, 7, 11, 16, 22, 29, ___
A.      27
B.      29
C.      30
D.     26
E.      None of these
2.    1, 2, 6, 24, 120, 720, __
_                 A.      4320
B.      5040
C.      5760
D.     6480
E.      None of these
3.      5      3      4     ?      38
A.      5
B.      6
C.      7.5
D.     8
E.      None of these

4.     5      11       ?          55       117
A.      22
B.      27
C.      23
D.     25
E.      None of these

5.    2, 10, 30, 68, 130, 222, 350, ___
A.      520
B.      524
C.      504
D.     666
E.      none of these

6.     A, F, K, ___________

A.       P                           B. Q                             C. R                              D. S
E. None of these

7.     DIL, GLO, JOR, ___________

A. XAD            B. GJM             C. MRU                        D. PSV
E. None of these

8.     BAT, DCV, FEX, _________

A. HGI             B. HGZ            C. HIJ               D. HGY
E. None of these

9.     MN, LO, KP, ___________

A. RS                           B. JQ                            C. IJ                              D. QR
E. None of these

10. BFZ, IMQ, PTX, _________

A. XBF             B. XAE             C. WAE                        D. WDF
E. None of these

## Wednesday, May 27, 2015

### Explanation Of S-Block(Group IIA Elements)

Group IIA Elements:
Beryllium (Atomic number - 4)
Magnesium (Atomic number - 12)
Calcium (Atomic number - 20)
Strontium (Atomic number - 38)
Barium (Atomic number - 56)
•  Have 2 electrons in the outermost orbit.
•  By loosing two electrons they form stable + 2 ions.
•  They are divalent and show + 2 as the stable oxidation state.
General trends in Periodic Table:
• As we move from top to bottom, atomic radius increases.
• Order of atomic radius: Be < Mg < Ca < Sr < Ba < Ra
• Ionization potential decreases down the group.
• Density increases with increase in atomic mass.
• Electronegativity decreases from top to bottom.
• Stability of thermal decomposition of metal carbonates increases down the group.
• Solubility of carbonates in water decreases down the group.
• Hydration enthalpy decreases down the group.
• Reactivity towards Halogens
All the alkaline earth metals combine with halogen forming their halides.
M + X₂ → M X₂ (X = F, Cl, Br, I)

Flame Test: Be and Mg give negative flame test as the electrons are strongly bound.

Hydration Enthalpy: Like alkali metals ions, the hydration enthalpies of alkaline earth metals ion decrease with increase in ionic size down the group.
General characteristics of compounds of alkaline earth metals:-

1. Oxides and Hydroxides: The alkaline earth metals burn in oxygen to form monoxide of the form MO. BeO is Amphoteric and rest are ionic in nature.
MO + H₂O → M (OH)₂ (hydroxide)
The solubility, thermal stability and the basic character of hydroxides increases with increase in atomic number.
2. Halides:  Except for Beryllium halides, all halides of alkaline earth metals are ionic in nature.
3. Salts of Oxoacids:
(i) Carbonates:
• Carbonates of alkaline earth metals are insoluble in water.
• Metal carbonates on heating give metal oxide and carbon dioxide gas on heating.
• Beryllium carbonate in highly unstable hence are kept in CO₂ atmosphere.
MCO₃ (s) → MO (s) + CO₂ (g).
(ii) Nitrates:
• On heating give metal oxides and Nitrogen dioxide.
• Tendency of formation of hydration decreases with increase in size.
M (NO₃)₂ (s) → 2MO (s) + 4NO₂ (g) + O₂ (g).

## Sunday, May 24, 2015

### Different Types Of Thermodynamic Process At A Glance

S.No.Change or name
of the process
1.DefinitionP=ConstantV=ConstantT=Constanta) Q=Constant
b) Entropy
S = Constant
2.dQ1. For solids
dQ = mcpdT
For gases ncvDt
2. For change in state
dQ = mL
1. For solids
dQ = mcvdT
For gases ncvdT
dQ = dWZero
3.dU1. dQ-pdV
2. dQ-nRdT
dQZero-dW
4.dW1. PdV
2. nRdT
Zero$2.303nRt\log \frac{v_2}{v_1}$
$2.303p_1v_1\log \frac{v_2}{v_1}$
$2.303p_1v_1\log \frac{p_1}{p_2}$
$\frac{R(T_2-T_1)}{1-\gamma}$
$\frac{(P_2V_2-P_1V_1)}{1-\gamma}$
5.Equation of state $\frac{V}{T}= \text{constant}$
$\frac{V_1}{T_1} = \frac{V_2}{T_2}$
$\frac{P}{T} = \text{constant}$
$\frac{P_1}{T_1} = \frac{P_2}{T_2}$
$PV = \text{constant}$
$P_1V_1 = P_2V_2$
PVγ=constant
TVγ-1=constant
P1-γTγ=constant
6.Slope of
p-v curve
Zero$\infty$$-\frac{p}{v}$$-\frac{\gamma p}{v}$
7.LawCharle's lawGay-lussac's lawBoyle's lawPoisson's law
8.Form of First lawsdQ=dU+dW
=ncpdT+PdV
dQ=dU=ncvdTdQ=dW=dV-dU=dW
9.Bulk modulusZeroInfinity-p-γp
10.Result of
maximum work
MaximumZeroLess from isobaric
process but greater
Minimum but not
zero

Example :
A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having  a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?

Ans: As no heat is allowed to be exchanged, the process is adiabatic.

$\therefore P_2V_2^{\gamma} = P_1V_1^{\gamma} \,\, (or) \,\, \frac{P_2}{P_1} = \left( \frac{V_1}{V_2} \right )$
$\text{As } V_2 = \frac{V_1}{2}$

$\therefore \frac{P_2}{P_1} = \left( \frac{V_1}{\frac12 V_2} \right)^{1.4} = 2^{1.4} = 2.64$

## Saturday, May 23, 2015

### PHYLLOTAXY

Wondering how leaves are arranged on the stem! Then Check it out...............

Phyllotaxis or phyllotaxy is the arrangement of leaves on a plant stem. Which originates from a Greek word   phýllon "leaf" and  taxis "arrangement".

The basic arrangements of leaves on a stem are opposite, or alternate = spiral. Leaves may also be whorled if several leaves arise, or appear to arise, from the same level (at the same node) on a stem. This arrangement is fairly unusual on plants except for those with particularly short internodes.

With an opposite leaf arrangement, two leaves arise from the stem at the same level (at the same node), on opposite sides of the stem. An opposite leaf pair can be thought of as a whorl of two leaves. In an opposite pattern, if successive leaf pairs are perpendicular, this is called decussate.

With an alternate (spiral) pattern, each leaf arises at a different point (node) on the stem.Distichous phyllotaxis, also called "two-ranked leaf arrangement" is a special case of either opposite or alternate leaf arrangement where the leaves on a stem are arranged in two vertical columns on opposite sides of the stem. Examples include various bulbous plants such as Boophone, Aloe seedlings, and also mature Aloe plicatilis.

A whorl can occur as a basal structure where all the leaves are attached at the base of the shoot and the internodes are small or nonexistent. A basal whorl with a large number of leaves spread out in a circle is called a rosette. Example: Alstonia.

Try observing phyllotaxy of plants in and around your vicinity. And name few of them.