longitudinal wave

Both transverse and longitudinal waves may be represented by sine wave forms. Both types of waves can(a) be reflected,(b) be refracted,(c) interfere,(d) be diffracted.In general, what applies to one set of waves applies equally well to the other. Transverse waves can be polarised whereas longitudinal waves cannot, but this is outside the scope of this book.

inductance of a coil

In 1820, Oersted discovered the magnetic effect of electric current, according to which an electric current produces a magnetic field. The converse effect was discovered and demonstrated by Michael Faraday in UK and by Joseph Henry in USA in the year 1831. They showed that electric current can be produced in a coil of wire with the help of a magnetic field. The phenomenon is known as electromagnetic induction. Many vital devices like generators and transformers work on the principle of electromagnetic induction. The laws of physics governing the phenomenon of electromagnetic induction are known as Faraday's law of electromagnetic induction. The discovery of the phenomenon of electromagnetic induction lent support to the belief of physicists that 'symmetry is beauty in physics'.

congruent figures

Let PQRS be a parallelogram and PR be one of its diagonals. What can you say about triangles PQRC andRSP?Solution:Step 1: In a four-sided figure, opposite sides are congruent. Hence sides Step 2: QR and PS are congruent, and also sides PQ and RS are congruent.Step 3: In a four-sided figure opposite angles are congruent. Step 4: Hence angles PQR and RSP are four side figure.Step 5: Two sides and an included angle of triangle PQR are congruent to two corresponding sides and an included angle in triangle RSP. According to the above postulate the two triangles PQR and RSP are congruent.

1 man 1 cup

Sexes, male and female are distinct in all higher animals including human. Hereditary differences are quite pronounced in male and female individuals.1. Sex of an organism is associated with certain chromosomes, known as sex chromosomes. In human the sex chromosomes are XY in male and XX in female.2. The male which is heterozygous, produces two types of gametes or sperm X and Y. The homozygous female produces only one type of gamete or ovum X.3. An ovum fertilized with Y bearing sperm gives rise to a male offspring (XY) while an ovum fertilized with X bearing sperm produces a female (XX)4. The maleness and femaleness of the offsprings is dependent on the type of sex chromosomes in the zygote.5. The only difference between male and female sex is that the male has a haploid set and the female a diploid set of sex chromosomes.*

biology grade 10

Self Evaluation1. Draw and label the parts of transverse section of monocot root.2. Draw the transverse section of dicot root and label the parts.3. Distinguish the anatomy of dicot roots from monocot roots. Differences between monocot and dicot roots.Monocot rootsDicot roots(0 Xylem is polyarch.(ii) Pith is usually large at the centre.(iii) Metaxylem vessels are generally circular in cross section.(/V) Conjunctive tissue is sclerenchy-matous in monocot roots like maize, (v) There is no secondary growth.Xylem is usually tetrarch. Pith is usually absent. Metaxylem vessels are generally polygonal in cross section. Conjunctive tissue is usually • parenchymatous. Secondary growth is present.Other Important Question4. What are the three tissue systems in the root of a matured plant?The embryo develops into an adult plant with roots, stem and leaves due to the activity of the apical meristem. A mature plant has three kinds of tissue systems — (0 the dermal, (ii) the fundamental and (iii) the vascular system.(/') The dermal system includes the epidermis, which is the primary outer protective covering of the plant body. The periderm is another protective tissue that supplants the epidermis in the roots and stems that undergo secondary growth.(ii) The fundamental tissue system includes tissues that form the ground substance of the plant in which other permanent tissues are found embedded. Parenchyma, collenchyma and sclerenchyma are the main ground tissues.(iii) The vascular system contains the two conducting tissues, the phloem and xylem. In different parts of the plants, the various tissues are distributed in characteristic patterns.

cubic yard calculator

1. Explain the concept of reciprocal lattice. Discuss its properties. What is its importance?2. Derive the Bragg conditions in terms of the reciprocal lattice vectors. (Raj., 1974)3. Explain the concept of Brillouin zones. Give the construction of the first Brillouin zone of an oblique lattice in two dimensions.4. Show that the reciprocal lattice of a simple cubic lattice is another simple cubic lattice. Hence deduce the first Brillouin zone of the simple cubic crystal lattice.5. Show that the reciprocal lattice for a body centred cubic crystal is face centred cubic.(Raj., 1974; Jodhpur, 1968)6. What do you mean by reciprocal lattice? Calculate the reciprocal lattice for a direct fee lattice. What is meant by first Brillouin Zone? (Delhi, 1983)7. Discuss the origin of the geometrical structure factor and the atomic scattering factor in x-ray diffraction and derive suitable mathematical expressions for them. What is the importance of the geometrical structure factor in the analysis of the crystal structures? (Raj., 1977)8. Explain the origin of van der Waals forces in molecular crystal. Show that for van der Waals forces the interaction energy varies as 1 / Rf\ where R is the separation of two interacting atoms.(Raj., 1984)9. Define Cohesive energy and determine its value for crystals of inert gases.10. Obtain the binding energy of an ionic crystal and derive the expression for the Madelung constant. Calculate the value of this constant for a linear ionic lattice. (Raj., 1970)11. Obtain an expression for the total cohesive energy of an ionic crystal in terms of Madelung constant and other parameters.12. State and prove Bloch theorem.13. Discuss the Kronig-Penny model for a linear lattice. How does it lead to formation of energy bands in solid? (Raj., 1983)14. Give Kronig -Penney model of electron in a periodic potential. What are its consequences?(Rohilkand, 1993, 92, 91; Meerut, 1993)

aluminum hydroxide formula

INTRODUCTIONEthers may be considered as the anhydrides of alcohols, as they may be prepared by elimination of one molecule of water from two molecules of any alcohol.e-g-,C2H5OH + HOC2H5—» C2H5OC2H5 + H2OEthyl alcohol (2 mols) Diethyl etherSince alcohols are also alkyl hydroxides, likewise ethers may be called alkyl oxides by analogy to sodium oxide which is also prepared by elimination of the molecule of water from two molecules of sodium hydroxide.Na|OH + H]QNa—» Na —O—Na + H20Sod. hydroxide (2 mols) Sodium oxideR|OH + H|OR—*R — O — R + H20Alkyl hydroxide (2 mols) Alkyl oxideAs the alcohols may be considered as the monoalkyl derivatives of water, so the ethers may be considered as the dialkyl derivatives of water.H—O—H R—O—H R —O —RWater Alcohol EtherEthers also form a homologous series with the general molecular formula C^H^ + 20 where the value of n is always more than one. Ethers possess the general functional group — O — and are in general represented as R — O — R'.

big lots hours

Water power has been used since ancient times by diverting water from natural streams or rivers over various kinds of paddle wheels or turbines. The power output from waterwheels being low people started building high dams from the last century to obtain a substantial head of hydrostatic pressure. Thus, the water under high pressure, flows through the base of the dam and drives turbo-generators producing hydroelectric power. In U.S., about 300 large dams generate 9.5% of its total electrical power production.Although hydroelectric power is basically a non-polluting renewable energy source, it is still associated with serious problems :1. Dams have drowned out beautiful stretch of rivers, wildlife habitat, forests, productive farmlands, and areas of historic, archeological, and geological significance. The construction of big dams have also rendered several farmers and tribals homeless, and without any livelihood.2."The reservoir behind the Aswan High Dam in Egypt has caused the spread of a parasitic worm which caused a debilitating disease. Further, the increase in humidity over a large area because of the reservoir is causing rapid deterioration of ancient monuments and artefacts which were existing over many centuries.3. Since water flow from the dam is regulated as per the requirement of power, dams play havoc downstream because water levels may change from extremes of near flood levels to virtual dryness and back to flood even in a single day. Other ecological factors are also affected because sediments rich in nutrients settle in the reservoir and only small amounts reach the river's mouth.4. Devastating earthquakes, observed near Koyana in India, are attributed to the Koyana dam (Maharashtra) by some Scientists.Many developing countries have great potential for large hydel power projects but due to the above problems, there is lot of opposition from people as well as from Environmental protection organisations.

what is surface area

The surface area of sphere is 4 * `Pi` * r2 square units.Let us see sample problems for the surface area of sphere,Example 1:Find the surface area of the sphere which is of radius equal to 7 cm.solution:Given the radius of the sphere: (r) = 7 cm.we need to find surface area of the sphere.we know surface area of sphere is given by 4 * `Pi` * r2 cubic units.substitute the radius, r = 7 cm in the formula,volume of the sphere = 4 * 3.14 * 72 = 4 * 3.14 * 7 * 7volume of the sphere = 615.44 cm2Example 2:Find radius of sphere which is of surface area equal to 615.44 cm2.solution:Given the surface area of the sphere: 615.44 cm2To find radius of sphere.we know volume of sphere is given by 4*`Pi` * r2 cubic units.equate the given volume of sphere with the formula,volume of the sphere = 4 * 3.14 * r2 615.44 cm2 = 4 * 3.14 * r2 r2 = 615.44 * `(1/4)` * `(1/(3.14)) ` cm2 r2 = 49 r = 7 cm.

electrons and protons

The proton has a mass of approximately 1 a.m.u. and a charge of +e. It is the hydrogen nucleus. The charge is equal and opposite to that on the electron, and its mass is approximately two thousand times as large.When a-particles bombard nitrogen, protons are emitted. Since the proton is smaller than the a-particle it has greater penetrating power. It can be deflected by both magnetic and electric fields, because it is electrically charged.

rounding off numbers

1) 665.3682 becomes:(round to nearest whole number) Solution: In the first step to make a nearest thousand 665.368 The 2 is neglected, because less than 5 Then the second step to rounding the value of 8, it is greater than 5, 665.37 Again in the third step tenth place it is rounded to 4 665.4 And the final result is 665

integration formula

Formulas on Exponential integration1) `int e^(x)dx` = `e^x`2) `int e^(cx)dx` = `(e^(cx))/(c)` 3) `int a^(cx)dx` =`(a^(cx))/(c.ln a)` [ a > 0, a ≠1 ]4) `int_0^oo e^(-ax) sin bx dx` = `b/(a^2 + b^2)` [ a > 0 ] 5) `int_0^oo e^(-ax) cos bx dx` = `a/(a^2 + b^2)` [ a > 0 ]

properties of gases

Groups: A vertical column in the periodic table is termed as a group or family. They are considered to be the most important method of classifying the elements. In some groups, the elements have same properties and show a clear pattern in properties down the group. These groups are to be given trivial names, like the alkali metals, alkaline earth metals, halogens, and noble gases. Some groups in the periodic table show less similarities and these have no trivial names and are simply recognized by their group numbers.Periods: A horizontal row in the periodic table can be defined as a period. Although groups are the common way of classifying elements, there are some regions of the periodic table where the periods are more significant than groups. This can be true in case of d-block or "transition metals", and especially for the f-block, where the lanthanides and actinides form two horizontal series of elements.Blocks: Due to the importance of the outermost shell, blocks can be refered by the different regions of the periodic table, and are named according to the subshell in which the "last" electron is present. The s-block comprises the alkali metals and alkaline earth metals as well as hydrogen and helium. The p-block comprises the groups 13 through 18 and contains all of the semimetals. The d-block comprises groups 3 through 12 and has all of the transition metals. The f-block is comprised of the rare earth metals.Other: The chemical elements are also grouped together in many different ways. Some of these groupings are often stated on the periodic table, like transition metals, poor metals, and metalloids. Other groupings may also be present, which are informal like the platinum group and the noble metals.

energy stored in a capacitor

Q. 51. Find the equivalent capacitance between points A and B in the following diagrams (Fig. 3 • 26).Solution: (i) Capacitors of 8|iFand 8 |xF are in series, then 1112 1— = — + — = — = — => equivalent capacitance q = 4(.iF q 8 8 8 4 Similarly for other branch11-13 + 25 . . t— =--1--=-= — => equivalent capacitance C, = — = 6 uFC2 10 15 30 30 5Now two capacitances Q andC2areinparallel/henceequivalentcnpncitance isC =CX +C2 =4+ 6=10nF (ii) Do yourself like part (i). [Ans. 2(.iF]Q. 52. Three capacitors are connected to a battery of 20 volt as shown in the figure 3 - 27. Calculate:(i) equivalent capacitance of the combination,(ii) charge stored on the capacitor of 3|iF. (1999)Solution : (i) Capacitors of 3(iF and 6 (iF are in series, henceJ__i 1 - 2 + 1Q ~ 3 + 6 ~ 6equivalent capacity C| = = 2 |iFNow capacitors of 2(.iFand 8|.iFare in parallel, hence equivalent capacitance C = 2 + 8 = 10|.iF(ii) Since capacitors of 3|.iFand 6(iFare connected in series with 20 vol t source. Charge on 3|iFcapacitor=charge on series combination of 3 and 6(.iF = Cj x 20 = 2|iFx 20 volt = 40 |iC Q. 53. The combination of four condensers of equal capacity is shown in the fig 3 • 28. If the resultant capacity between the points P and Q is l|iF, find the capacity of each condenser.Solution : Let the capacity of each condenser be C. These condensers are arranged in series in between the points P and Q. Therefore,111114 „ . _- = — + — + — + — = — or C= 4liF1 c c c c cQ. 54. Find the equivalent capacitance between points A and B in the adjoining figure3-29.Solution : Capacitance 2(.iF and 2j.iF are connected in parallel, hence their resultant capacitance Cj = 2 + 2 = 4(iF.Now 8|iF and Cj are in series, theirresultant capacitance^ 8x 4 32 8 _C2 =-= — = —uF8 + 4 12 3Again, 12 (iF and 6|iF are in series, hence their resultant„ 12 x.6 72 . _C, =-= -i— = 4|iF12+6 18» ,ii i „ c,c2Note : — = — + — => C = —C Cj C2 Cx + C2Now C 3 and 4p.F are in parallel, their resultant C4 = C3 + 4|iF = 4^F + 4|iF = 8(iFAgain,C4 and l|iF are in series, their resultant_ C4xl 8x1 8 _Cs = —-=-= - uF' C4+l 8 + 1 9Now C2, the resultant capacitance between C and £ and Cs, the resultant capacitance between C and D are in parallel and their resultant r 8 8 32 .Q. 55. The equivalent capacitance between points A and B in the adjoining figure 3 ■30, is 1 • 0 (iF. Find the capacity of the capacitor C. (1995)Solution : As in last question, it can be shown that capacitance between points P and Q is Q = 2(j.FNow Cj = 2nF is in series with C and their equivalent capacitance is l|iF, hence1=1 I1 ~2 + C1 i 1 1 ' r i n=> —=1--=-/. C = 2 uFC 2 2Q. 56. In the adjoining figure (3 -31) of a circuit, B is earthed and A is kept at 1500 volts. Calculate the potential at point 1Solution: Capacitors of 5 and 5|iFare in series, their equivalent capacitance is 5/2 |iF. Now 5/2 (iF is in parallel with l|iF capacitor, their equivalent capacitance is7/2 (iF. Capacitors of ^and 3-5^= ^j(iFare in series, hence equivalent capacitance isP.D.between/4 and B = 1500- 0=1500 volts7Charge on the combination q=CV Fx 1500 volt = 2625 (iC.-. Charge on capacitor of 3 -5 |.iF = 2625 (.iCPotential difference between A and P= -i_=^C =750 volt. 3-5(iF 3-5nCi.e. 1500 - VP = 750 /.Potential at P, VP = 1500 - 750 = 750 volt.Q. 57. Find the equivalent capacitance between points A and B in the adjoining circuit by drawing its simple equivalent circuit. (2000)Solution : In the given circuit first plate of each capacitor is connected to A and second plate of each is connected to B, hence it is a parallel combination.Equivalent circuit can be drawn as in fig. 3.33Now do yourself. [Ans. 6|xF]Q. 58. Find equivalent capacity between the points A and B in the adjoining diagram (fig.3 -34).Solution : The given circuit consists of two closed mesh, first made of 2,2 and 5|iF and other made of 6,6 and 5nF. This circuit can be rearranged as shown in fig 3-35.According to fig. 3 ■35 given circuit is Wheatstone bridge.2 _ 6 2 6i.e. Ratio of capacitances in sides AC and AD is equal to that of capacitances in sides BC and BD. Hence this is a balanced bridge.There will be no charge on the capacitor of 5|iF. Hence this capacitor can be removed. (Note). Then reduced circuit will be as shown in the fig 3 • 36.Now do yourself. [Ans. 3jiF]Q. 59. Find the energy stored in the capacitor shown in the adjoining figure 3-37. (1995)Solution : Once capacitor is charged „ it will not allow the current to flow.Resistances 2 fl, 3 Q and 5 Q are in series, Their equivalent resistanceR = 2+ 3 + 5= 10 fi This resistance is in parallel with capacitor hence P.D. of capacitorV=iR =2x10 =20 volt /. Energy stored in capacitorU = - CV2 = - x 2 x 10"6 x (20)2 2 2= 4x 10-4 jouleQ. 60. In seady state, find the charges stored and potential difference of two capacitors, shown in adjoining figures-38. (1997,2002)Solution : In steady state, nc current will flow through capacitoi branch, hence equivalent resistance ol circuitR = 4 + 5 + 1= 10 Q ^ 10V .Current i =-= 1 amp.lOfi rv Capacitor branch is in parallel with branch of resistances 4 Q and 5 Q. .•. P.D. of capacitor branchV = ix R' = 1 x (4 + 5) = 9 volt Equivalent capacitanceCj + C2 2+3 Charge stored on capacitor combinationq = CV = 1 • 2nF x 9 volt = 10 • 8 |*C Charge on each capacitor,<h=l2= 10 • 8 |iC P.D. of first capacitor V1=^- = 1C>28^C = 5-4 voltP.D. of second capacitor V2=— =^ ^^ =3-6 voltC2 3nFPotential difference between A and P= -i_=^C =750 volt. 3-5(iF 3-5nCi.e. 1500 - VP = 750 /.Potential at P, VP = 1500 - 750 = 750 volt.

free form amino acids

The following are chemical properties of sulphuric acid:-It is acidic only in water solution when hydronium and sulphate ions are formed. Hydronium ions result in acidic property. It is diabasic as it ionises in two stages- 1 H2SO4 + H2O ↔ H3O+ + HSO4- 2 HSO4- + H2O ↔ H3O+ +SO42-1) It react with metals and form metallic sulphate and hydrogen at ordinary temperature. Mg + H2SO4→ MgSO4 + H22) It neutralises bases to form slats and water. CuO + H2SO4→ CuSO4 + H2O3) It liberates carbondioxide from metallic carbonate (ZnCO3) and bicarbonate(2NaHCO3). ZnCO3 + H2SO4 → ZnSO4 + H2O + CO24)It evolves hydrogen from metal sulphides. F2S + H2SO4 → FeSO45)It evolves sulphurdioxide and sulphites and hydrogen sulphites Na2SO3 + H2SO4→ Na2SO4 + H2O + SO2

science working models

240 volts is the home electricity voltage in India which is supplied to our homes. On voltage of 240 maximum appliances work but some devices like: hair dryer, battery chargers, laptops, radios etc, do not accept 240 volts for them use of voltage converter is must. Three types of voltage converters are there:-Resistors networks TransformersCombination When our appliances are turned on, a force of 240 volts is passed through appliance. The energy is used in completing the works like heating, running motor, lighting, etc it is analogous to the water is forced through the pipe using a mono block pump.

primary resources maths

Ecological efficiency is the percentage of energy transferred from one trophic level to the next. Alternatively, it is the ratio of the net productivity, i.e., the biomasS, at one trophic level to the net productivity (biomass) at the level below. Ecological efficiency varies among organisms. Usually it is 10%. This means that 90% of the energy available at onp trophic level never transfers to the next. (There are number of ratios used to express the efficiency with which organisms exploit their food resources and convert the food into biomass. Important efficiency measures for producers are —1. Photosynthetic efficiency. It is the percentage ratio between gross primary productivity and incident total solar radiation. It generally varies from 1 to 5 per cent.Photosynthetic efficiency_ Gross primary productivity ^ Incident total solar radiation2. Net production efficiency. It is the percentage ratio between net primary productivity and gross primary productivity. It is around 50%.Net production efficiency_ Net primary productivity Gross primary productivityImportant efficiency measures for consumers include —1. Assimilation efficiency. It is the percentage ratio between food energy assimilated and food energy ingested at one trophic level.Assimilation efficiency_ Food energy assimilated ^ ^^ Food energy ingested2. Ecological efficiency. It is the percentage ratio between energy in biomass production at one trophic level and energy in biomass production at previous trophic level. It is also called trophic level efficiency.Ecological efficiency = Energy in biomass production at one trophic level Energy in biomass production at previous trophic level

biology study guide

The data gathered in the experimental study of artificial transmutations are so vast and varied that it is not possible to give here a detailed account of them ; nor can this serve any useful purpose to the general student. It is more profitable to indicate the main guiding principles used in the classification and interpretation of experimental data.

life of pi summary

1. Experiments show that if a discharge tube is exhausted to a low pressure (0.01 mm of Hg), cathode rays are produced.2. Cathode rays are fast moving electrons having a negative charge. They contain the smallest quantity of electricity from nature i.e., 1.602 x 10'19 coulombs.An electron has a mass 9.1 x 10~31 kg and hence is about —— limes lighter 1840than the hydrogen atom.3. Prof. J.J. Thomson proposed the hypothesis that cathode rays were streams ofnegatively charged particles. He was the pioneer who devised an experiment £by which the specific charge — of such particles was determined. m4. Milikan's oil drop experiment was the first direct experimental proof of the atomic nature of electric charge.5. The photoelectric effect is a process whereby electrons can be ejected from a metallic surface when light is incident on that surface. Einstein provided a successful explanation of this effect by extending Planck's quantum hypothesis to the electromagnetic field, in this model, light is viewed as a stream of particles called photons, each with energy E = hv, where v is the frequency and h is Planck's constant The kinetic energy of the ejected photoelectron is given by (hv - wj, where wo is the work function of the metal.6. X-rays from an incident beam are scattered at various angles by elections in a target such as carbon. In such a scattering event, a shift in wavelength is observed for the scattered X-rays, and the phenomenon is known cs the Compton effect. Classical physics does not explain this effect. If the X-rays is treated as a photonconservation of energy and momentum applied to die photon-election collisions yields the following expression for the Comptonshift:-(1-cos a), mcwhere m is the mass of the electron, c is the speed of light and a is the scattering angle.7. All matter exhibits both particle and wave character. The dualistic nature of matter was proposed by de-Broglie. The de-Broglie wavelength of any particle of mass m and velocity v is given by x = t.-±. p mv£and the frequency of matter waves obeys the Einstein relation i) = —, where hE is the total energy of the particle. Subsequent experiments that confirmed the concept of matter waves included the observation of electron diffraction by Davisson and Germer and independently by Thomson.8. The wavelengths of visible objects are far too small for their wavelike nature to be apparent in everyday life. The wavelike nature of electrons becomes visible when they are reflected from single crystals or diffracted by ultra-thin materials.

pa school requirements

Find the co-ordinates of a point which is equidistant from the points (-2,9), (10,-7) and (12,-5) Solution Let the co-ordinates of the required point P be (x,y) which is equidistant from the given points A(-2,9),B(10,-7) and C(12,-5) Now; $PA=\sqrt{(x+2)^{2}+(y-9)^{2}}$ $PB=\sqrt{(x-10)^{2}+(y+7)^{2}}$ $PC=\sqrt{(x-12)^{2}+(y+5)^{2}}$ Since PA=PB $\therefore PA^{2}=PB^{2}$ $\Rightarrow(x+2)^{2}+(y-9)^{2}=(x-10)^{2}+(y+7)^{2}$ $\Rightarrow x^{2}+4x+4+y^{2}-18y+81=x^{2}-20x+100+y^{2}+14y+49$ $\Rightarrow24x-32y=64$ $\Rightarrow3x-4y=8$.......(i) Also, since PA=PC $PA^{2}=PC^{2}$ $\Rightarrow(x+2)^{2}+(y-9)^{2}=(x-12)^{2}+(y+5)^{2}$ $\Rightarrow x^{2}+4x+4+y^{2}-18y+81=x^{2}-24x+144+y^{2}+10y+25$ $\Rightarrow28x-28y=84$ $\Rightarrow x-y=3$....(ii) Solving equation (i) & (ii); x=4;y=1 Hence the co-ordinates of the required point P are (4,1) Ans.