Mensuration Formulas

Nuclear fission and fusion

The nuclear fission may be defined as the splitting of a heavy nucleus into nearly two equal parts by the bombardment of slow neutron with the release of energy. Otton Hahn and Strassmann (1939) carried out first nuclear fission reaction as:

Nuclear fission and fusion

In this process the emitting neutrons bombard again on $^{235}_{92}U$ and a chain reaction may be set up. In this way a huge amount of energy is liberated. This process is used in atom bomb and in atomic reactors.

The process in which lighter nuclei fuse together to produce heavier nucleus is called nuclear fusion.

e.g.

$^2_1H + ^2_1H \to ^4_2He + ^1_0n + 17.6 Mev \\ ^7_3Li + ^1_1H \to2 ^4_2He + 17.6 MeV \\ ^6_3Li + ^2_1H \to 2 ^4_2He + 22.4 MeV \\ ^2_1H + ^2_1H \to ^4_2He + 24.9 Mev$

Since in these processes, energy is to be given to the bombarding particle to overcome the repulsion between the target and bombarding particle, hence such reactions are also known as thermo nuclear reactions. Such reactions are used in hydrogen bomb. Such type of reactions are responsible for the luminosity of stars and source of energy in the sun.

Such reactions are possible when temp is $> 10^6K$ . Bethe (1939) proposed ‘Carbon-Nitrogen Cycle’ for the energy source of stars including the sun. the cule is as follow

$^{12}_6C + ^1_1H \to ^{13}_7N + \gamma \\ ^{13}_7N \to ^{13}_6C + ^0_{+1}e \\ ^{13}_6C + 61_1H \to ^{14}_7N + \gamma \\ ^{14}_7N + ^1_1H \to ^{15}_8O \gamma \\ ^{15}_8O \to ^{15}_7N + ^0_{+1}e\\ \line(1,0){250}\\ ^{15}_7N + ^1_1H \to ^{12}_6C + ^4_2He \\ \line(1,0){250}\\ 4 ^1_1H \to ^4_2He + 2 _{+1}^0e + 26 MeV$

But according to modern view, proton-proton interaction leading to helium takes place as follows

$^1_1H + ^1_1H \to ^2_1H + ^0_{+1}e \\ ^2_1H + ^1_1H \to ^3_2He + \gamma \\ ^3_2He + ^1_1H \to ^1_1He + \gamma \\ \line(1,0){250} \\ 4 ^1_1H \to ^4_2He + 2 ^0_{+1}e + 24.7 MeV \\ \line(1,0){250}$

Spallation: according to Seaborg a nucleus can be splitted into several nuclei when hit by a high energy charged particles like $_1H^2$ of 200 MeV energy. Here (10 – 20) neutrons are emitted.

$^{75}_{33}As + _1H^2 \to ^{56}_{25}Mn + 9 _1H^1 + 12_0 n^1$

Nuclear binding energy

It may be defined as the energy released when the given number of protons and neutrons combine to form its nucleus. In other words, it is the energy required to disrupt its nucleus into its constituents. This released energy corresponds to mass defect $(\Delta m)$ , e.e., the difference between the expected mass and the actual mass of the atoms. It may be written as:

Binding energy = [mass of neutrons + mass of protons + mass of electron] of the atom – actual mass of the atom

= mass defect a.m.u.

931.5 × Mass defect (MeV)

(since 1 a.m.u. = 931.5 MeV)
Aston expressed the variation of isotopic mass from mass number in terms of packing fraction for each isotope as:

$\text{pacing fraction} = \dfrac{\text{Isotopic mass- Mass number}}{\text{Mass number}} \times 10^4$
It is evident that the smaller value of packing fraction shows grater stability of the atom and vice versa.

Example 1: the fission fragments in the thermal neutron fission of $^{325}_{92}U$ found to be $^{98}_{42}Mo \text{and} ^{136}_{54}Xe$ what are the elementary particles and energy, released in the reaction? (Isotopic mass of $^{325}U ^{98}Mo, ^{136}Xe \text{and} ^1_0n$ are 235.044, 97.906, 135.907 and 1.0086 amu, respectively).

Solution:

According to the example,

$^{325}_{92}U + ^1_0n \to ^{98}_{42}Mo + ^{136}_{54}Xe$

On balancing the equation by elementary particles the above equation many be written as:

$^{235}_{92}U + ^1_0n \to ^{98}_{42}Mo + ^{136}_{54}Xe + 2 ^1_0n + 4 ^1_{-1}e\\ \therefore$ elementary particles released are neutrons and electrons.
Mass of the reactants = 235.044 + 236.0526

Mass of the products

= 97.906 + 135.907 + 2(1.0086) + 4(0.0005486) = 235.8324

Therefore mass defect $(\Delta m)$ = 235.8324 – 236.0526 = -0.2202 amu

Since the value of $(\Delta m)$ is negative, it means energy is released during the process.

Energy = 0.2202 × 931 MeV

Example 5: calculate the loss in mass accompanying combustion of one mole of fuel to release 1800KJ heat energy

Solution: since $E = \Delta m c^2$

Where E = released energy, $\Delta m$ and c = velocity of ligh

Hence

$\Delta m = \dfrac{E}{c^2} = \dfrac{1800 \times 10^3 J}{(3 \times 10^8 m s^{-1})^2} \\ 2 \times 10^{-11} kg (\text{since} J = kg m^2s^{-2})$

Medicinal use of radioactive isotopes

Henry Danlos use Ra for the first time in the treatment of Tuber Culons Skin Lesin disease.
CO – 60 $\to$ in treatment of Cancer.
I-131 $\to$ in treatment of Thyroid Gland.
P-32 $\to$ in treatment of Leukemia.
Na-24 $\to$ to trace flow of blood
I-123 $\to$ for imaging the brain.
Au-198 $\to$ blood cancer treatment.
Technetium-99M $\to$ for scanning bones.

Other uses

$^3_1H$ is used as a Tracer in absorption of water by plant roots.

P-32 $\to$ TO Truce plant growth etc.

O-18 $\to$ in photosynthesis.

$6CO_2 + 12H_2O^{18} \to C_6H_{12}O_6 + O_2^{18} + 6 H_2O$

Na-24 to find leakage pipes.

Carbon dating

W.F. Libby (1950)has shown that by measuring the radioactivity of $^{14}C$ it is possible to calculate the date at which a plant was cut or an animal died. The principle of this technique is that $^{14}C$ is produced in the upper atmosphere as:

$^{14}_7N + ^1_0n o ^{14}_6C + ^1_1H$

There is one atom of $^{14}C$ for every $7.49 imes 10^{11}$ carbon atom in the $CO_2$ of the atmosphere and in all living animals and plants. This ratio always remains constant in nature. when the animal or plant dies $^{14}C$ starts disintegration as:

$^{14}C o ^{14}_7N + ^0_{-1}e [3mm] t = dfrac{2.303}{lambda} log_{10} dfrac{ ext{Ratio of} C_{14}/ C_{12} ext{in living organism}}{ ext{Ratio of} C_{14}/C_{12} ext{in dead organism}}[3mm] ext{or} t = dfrac{2.303}{lambda} log_{10} dfrac{ ext{activity in living organism}}{ ext{activity in dead organism}}$

Thus by measuring the radioactivity of the substance under consideration and knowing the $t_{1/2}$ of the $^{14}_6C$ is is possible to calculate when the sample was removed from the carbon cycle.

Rock dating: use to find age of earth crust, rocks etc. it is based on the fact that U-238 decays to give Pb-206. [ $t_{1/2}$ of $U-238 = 4.46 imes 10^9$ years]

$dfrac{2.303}{lambda}log_{10}dfrac{(U-238 + Pb-206)}{(U-238)}$

Formulae

$ext{Radius of nucleus (r)} = (1.33 imes 10^{-15})( ext{mass number})^{1/3}m [3mm] ext{Density of nucleus} = dfrac{ ext{Mass No.}}{6.023 imes 10^{23}} div dfrac{4}{3}pi r^2$

Mass Defect $(delta m)$ = Expected mass – Actual mass

Binding energy, $E = Delta m. c^2 = Delta m imes 921 MeV [3mm] dfrac{-dN}{dt} = lambda N [3mm] lambda = dfrac{2.303}{t}log dfrac{N_0}{N_t} [3mm] lambda = dfrac{0.693}{t_{1/2}} t = 1.44 t_{1/2} = dfrac{1}{lambda} [3mm] N_t = N_0 left(dfrac{1}{2}ight)^n [3mm] left( ext{where} n = dfrac{ ext{total time}}{ ext{Half life period}}ight)$

Example 1: A piece of charcoal from the ruins of a settlement in japan was found to have a $^{14}C^/{12}C$ ratio that was 0.617 times that found in living organisms. How old is this piece of charcoal? The half-life period of $^{14}C$ is 5770 years.

Solution:

From eq.3,

$lambda = dfrac{0.693}{t_{1/2}} = dfrac{0.693}{5770 ext{yr}} = 1.201 imes 10^{-4} yr^{-1}$

From eq.2,

$t = \dfrac{2.303}{\lambda} \log \dfrac{N_0}{N} = \dfrac{2.303}{1.2 \times 10^{-4}} \log \dfrac{1}{0.617}$

http://regentsprep.org/Regents/math/algtrig/ATP2/SequenceWordpractice.htm

The five highest cards in each suit (A- K - Q - J - 10) are called the honours or honour cards.

Mensuration Formulas

Mensuration is the branch of mathematics which deals with the study of Geometric shapes , Their area , Volume and different parameters in geometric objects.

Some important mensuration formulas are:

1. Area of rectangle (A) = length(l) * Breath(b) $A = l \times b$

2. Perimeter of a rectangle (P) = 2 * (Length(l) + Breath(b)) $P = 2 \times(l + b)$

3. Area of a square (A) = Length (l) * Length (l) $A = l \times l$

4. Perimeter of a square (P) = 4 * Length (l) $P = 4 \times l$

5. Area of a parallelogram(A) = Length(l) * Height(h) $A = l \times h$

Parallelogram

6. Perimeter of a parallelogram (P) = 2 * (length(l) + Breadth(b)) $P = 2 \times (l + b)$

7. Area of a triangle (A) = (Base(b) * Height(b)) / 2 $A = \frac{1}{2} \times b \times h$

Triangle

And for a triangle with sides measuring “a” , “b” and “c” , Perimeter = a+b+c

and s = semi perimeter = perimeter / 2 = (a+b+c)/2

And also . Area of triangle = $A = \sqrt{s(s-a)(s-b)(s-c)}$

This formulas is also knows as “Heroe’s formula”.

8. Area of triangle(A) = $\frac{1}{2} a \times b \times \angle C = \frac{1}{2} b \times c \times \angle A = \frac{1}{2} a \times c \times \angle B$

Where , A , B and C are the vertex and angle A , B , C are respective angles of triangles and a , b , c are the respective opposite sides of the angles as shown in figure below:

area of triangle - mensuration

9. Area of isosceles triangle = $\frac{b}{4}\sqrt{4a^2 - b^2}$

Where , a = length of two equal side , b= length of base of isosceles triangle.

10. Area of trapezium (A) = $\frac{1}{2} (a+b) \times h$

Where , “a” and “b” are the length of parallel sides and “h” is the perpendicular distance between “a” and “b” .

Trapezium

11. Perimeter of a trapezium (P) = sum of all sides

12. Area f rhombus (A) = Product of diagonals / 2

13. Perimeter of a rhombus (P) = 4 * l

where l = length of a side

14. Area of quadrilateral (A) = 1/2 * Diagonal * (Sum of offsets)

quadrilateral

15. Area of a Kite (A) = 1/2 * product of it’s diagonals

16. Perimeter of a Kite (A) = 2 * Sum on non-adjacent sides

17. Area of a Circle (A) = $\pi r^2 = \frac{\pi d^2}{4}$

Where , r= radius of the circle and d= diameter of the circle.

18. Circumference of a Circle = $2 \pi r = \pi d$

r= radius of circle

d= diameter of circle

19. Total surface area of

cuboid = $2 (lb + bh + lh)$

where , l= length , b=breadth , h=height

20. Total surface area of

cuboid = $6 l^2$

where , l= length

21. length of diagonal of

cuboid = $\sqrt{l^2+b^2+h^2}$

22. length of diagonal of

cube = $\sqrt{3 l}$

23. Volume of cuboid = l * b * h

24. Volume of cube = l * l* l

25. Area of base of a cone = $\pi r^2$

26. Curved surface area of a cone =C

Where , r = radius of base , l = slanting height of cone

27. Total surface area of a cone = $\pi r (r+l)$

28. Volume of right circular cone = $\frac{1}{3} \pi r^2 h$

Where , r = radius of base of cone , h= height of the cone (perpendicular to base)

29. Surface area of triangular prism = (P * height) + (2 * area of triangle)

Where , p = perimeter of base

30. Surface area of polygonal prism = (Perimeter of base * height ) + (Area of polygonal base * 2)

31. Lateral surface area of prism = Perimeter of base * height

32. Volume of Triangular prism = Area of the triangular base * height

33. Curved surface area of a cylinder = $2 \pi r h$

Where , r = radius of base , h = height of cylinder

34. Total surface area of a cylinder = $2 \pi r(r + h)$

35. Volume of a cylinder = $\pi r^2 h$

36. Surface area of sphere = $4 \pi r^2 = \pi d^2$

where , r= radius of sphere , d= diameter of sphere

37. Volume of a sphere = $\frac{4}{3} \pi r^3 = \frac{1}{6} \pi d^3$

38. Volume of hollow cylinder = $\pi r h(R^2-r^2)$

where , R = radius of cylinder , r= radius of hollow , h = height of cylinder

39. Surface area of a right square pyramid = $a \sqrt{4b^2 - a^2}$

Where , a = length of base , b= length of equal side ; of the isosceles triangle forming the slanting face.

40. Volume of a right square pyramid = $\frac{1}{2} \times base \, \, area \times height$

41. Area of a regular hexagon = $\frac{3\sqrt{3}a^2}{2}$

42. area of equilateral triangle = $\frac{\sqrt{3}}{4} a^2$

43. Curved surface area of a Frustums = $\pi h (r_1 + r_2)$

44. Total surface area of a Frustums = $\pi (r_1^2 + h(r_1+r_2) + r_2^2)$

45. Curved surface area of a Hemisphere = $2 \pi r^2$

46. Total surface area of a Hemisphere = $3 \pi r^2$

47. Volume of a Hemisphere = $\frac{2}{3} \pi r^3 = \frac{1}{12} \pi d^3$

48. Area of sector of a circle = $\frac{\theta r^2 \pi}{360}$

where , $\theta$ = measure of angle of the sector , r= radius of the sector

http://questions.ascenteducation.com/iim_cat_mba_free_sample_questions_math_quant/probability/

http://www.analyzemath.com/statistics/probability_questions.html

http://www.analyzemath.com/statistics/discrete_pro_distribution.html

Thats right.

Some useful links are

http://regentsprep.org/Regents/math/algtrig/ATT10/trigequations2.htm

http://www.youtube.com/watch?v=CdZ2ntcY61g

http://www.math.uakron.edu/~tprice/Trig/Identities.pdf

Prime factorization of 1539 = 3 x 3 x 3 x 3 x 19

So cube root of 1539 =1539 ^1/3 = (3 x 3 x 3 x 3 x 19)^1/3= 3 * 57 ^1/3

Prime factorization of 1539 = 3 x 3 x 3 x 3 x 19

So cube root of 1539 =1539 ^1/3 = (3 x 3 x 3 x 3 x 19)^1/3= 3 * 57 ^1/3

Use VSEPR theory.http://en.wikipedia.org/wiki/VSEPR_theory

(A) Boiling point of H2O2 is higher than that of H2O : TRUE as BP of H2O2 is 150.2 °C, and that of H2O is 100 C

(B) D2O and H2O have same chemical properties. TRUE

(C) It is impossible to prepare 100 % pure ortho hydrogen. TRUE

(D) The simplest borane is BH3. TRUE

Polarization by Reflection

Since the reflection coefficient for light which has electric field parallel to the plane of incidence goes to zero at some angle between 0° and 90°, the reflected light at that angle is linearly polarized with its electric field vectors perpendicular to the plane of incidence and parallel to the plane of the surface from which it is reflecting. The angle at which this occurs is called the polarizing angle or the Brewster angle. At other angles the reflected light is partially polarized.

From Fresnel's equations it can be determined that the parallel reflection coefficient is zero when the incident and transmitted angles sum to 90°. The use of Snell's law gives an expression for the Brewster angle.

It is similar to the following problem

Problem: if a disc of radius R is cut from a disc of radius 2R then what will be the centre of mass of left plate. If the cut is made in such a way that edge of smaller disc touches the edge of original disc

Solution:

To solve this problem we can follow two different approaches:

1) Using Integral Calculus

2) Converting the problem in Two body system.

However, First approach is fundamental to solve all center of mass problems, but, here we follow second approach for its simplicity and to save time in problem solving.

Now we consider center of main disc as the origin and disc of radius R removed is lying in third and fourth quadrants.

Let us consider that the mass of Circular disc of radius 2R (say D) = M

After cutting the disc of radius R such that it touches the edge of main disc

Mass of disc of radius R (say D1) removed from main disc

= (M/4

R²) ×

R² = M/4

Therefore, mass of the remaining portion of the disc (say D2) = 3M/4

Now consider discs D, D1 and D2

Coordinates for the Center of mass of disc D = (0,0)

Coordinates for the Center of mass of disc D1 = (0,-R)

Let us assume that Coordinates for the Center of mass of disc D2 = (x,y)

Now, it can be considered that entire mass of discs D1 and D2 are concentrated at their center of mass coordinates, that is the problem is converted in two body which are point objects.

Therefore,

Coordinates of D =

[(M/4) × (coordinates of D1) + (3M/4) × (coordiantes of D2]/(M/4 + 3M/4) = 0

so, 0 = [(M/4) × (-R) + (3M/4) × y]/ M

or y = R/3,

similarly, x = 0 (as it should, well evident from symmetry of the problem)

Thus, coordinates of the remaining disc is given by (0, R/3)

It is essential concept and must be known for all types of engineering entrance related examinations.

Particle in a Box

When the momentum expression for the particle in a box :

is used to calculate the energy associated with the particle

For first excited state n = 2

a) mass of electron m= 9.1 * 10^-31 kg

L = 1cm = 0.01 m

Thus E = h² / 8mL²= 2.41 * 10^-33 J = 1.5 * 10^-14eV

c) for proton m= 1.67 * 10^-27 kg

L = 5 fm = 5 * 10^-15 m

Substitute and get the values

The phase lag between current and voltage in purely inductive circuit is pi/2. Unlike in purely resistive circuit where the change in voltage (ac) is synchronized with change in current (ac) and the two are in phase that is phase difference is ZERO. But in inductive circuit current (ac) is not able to chase the voltage and this results in a phase lag, certainly current can be non zero when voltage is zero as here there is a phase difference and the applied voltage is ac. Hope it is clear !!!

http://en.wikipedia.org/wiki/Spy_satellite

The statement goes like this

Rational number is a number that can be in the form p/q, where p and q are integers and q is not equal to ZERo.

Energy loss due to frictional force is converted into heat energy that is resposible for the melting of ice.

Energy converted into thermal or heat enegy = loss of kinetic energy = 0.5 m v² =0.5 * 50 * 5²= 625 J

Using,

Q = mL

where Q = 625 J

m = mass of ice melted

L = latent heat of fusion

Mensuration Formulas

Polarization by Reflection

Particle in a Box

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