Thursday, August 12, 2010
number pattern
number pattern
Proper Fractions: Fractions having numerator less than the denominator are called proper fractions. Example: `3/7, 7/13.`Improper Fractions: Fractions having numerator that are larger than or equal to their denominators are called improper fractions. Example: `12/5, 6/6, 21/11.` Mixed Fractions: Numbers having a whole number part and a fractional part are called mixed numbers. We denote a mixed number in the form a `b/c` . Example: `2 5/2, 1 1/3` . Decimal Fractions: Fractions having denominator as 10, 100 or 1000 or any other higher power of 10 are called decimal fractions. Example: `4/100` ,` 35/1000.` Simple Fractions: Fractions having both the numerator and denominator as whole numbers are called simple fractions. Example: 4/5, 6/95. Like and Unlike Fractions: Fractions having the same denominators are called like fractions, whereas fractions having different denominators are called unlike fractions. Example: `4/9, 1/9` are all like fractions. `3/5, 4/7` are all unlike fractions.pressure definition
pressure definition
Introduction to vapour pressure mixture:Consider a mixture of liquids like water and alcohol in a container which is closed.The vapor presure exertedby each component is partial vapor presure of that component.First we would consider water vapor presure.general knowledge facts
general knowledge facts
The zoogeography or the knowledge of the distribution of animals on the earth has led to the division of the world into a number of distinct zoogeographical (biogeo-graphical also) realms (Fig. 42.1). Originally these were based on the mammalian faunas of the various parts of the world, but their validity is general.writing chemical formulas
writing chemical formulas
We have to find the equation of the line using the following formula`(y-y1)/(y2-y1)`=`(x-x1)/(x2-x1)` Here (x1,y1) and (x2,y2) are the given two points. Substitute the given points in the above formula we have to get the equation of the line.Example:(1, 3) and (4, 8) write the equation of the line that passing through the given two points by Method - 1Solution:Here x1 = 1 and y1=3 and x2=4 and y2=8The formula for equation of the line is `(y-y1)/(y2-y1)`=`(x-x1)/(x2-x1)`Substitute the x1, y1, x2 and y2 values in the above equation we have to get, `(y-3)/(8-3)`=`(x-1)/(4-1)`Simplifying this we can get, 3y=5x+4This is the equation of the line that passes through the given two points.binomial formula
binomial formula
Q. 6. Explain Stefan's law of thermal radiation. (1996,2000,01,04)Ans. Stefan's Law : This law states that the radiant energy '£' emitted from the unit area of surface of a perfect black-body in unit time is proportional to the 4th powerof the absolute temperature of the body. -i.e. £ oc T4 or £ = oT4 _Where c is a constant which is called Stefan's constant and its value is 5 • 67 x 10~8 j/m2-s-K4.Suppose that a black-body having absolute temperature T\ is surrounded by a black enclosure at absolute temperature Tz- The body will emit a7|4 joule energy per second per unit area of its surface. At the same time it will absorb <rlf joule energy per second per unit area from the surrounding atmosphere. Thus net energy emitted by unit area of the body in unit time will be£1-£2=o(rl4-T24)The rate of emission of radiant energy from the area A of a body with emissive power e, at the absolute temperature T is given byE = oT4eAQ. 7. Establish Newton's law of cooling from Stefan's law. (1996,2000,01,04)Ans. Newton's Law of Cooling : This law states that if the temperature difference between a body and its surrounding is small then rate of cooling of the body (or rate of loss of heat by the body) is proportional to the temperature difference between the body and surroundings.i.e. rate of loss of heat « temperature differenceDerivation of Newton's Law From Stefan's Law: Suppose that a body is placed in air for cooling. Let the absolute temperature of the body and that of surroundings be Ti and T. Let e be the emissive power of the body. Then according to Stefan's law the net rate of loss of heat by the body is given by.A£ = ae(Ti4-T4)Let Ti-T + t wheret = 7} -T > 0is smallthen AE = ce[(T + t)i-T4]= aeT4 jl + 4—.......j -1 (Using Binomial Expansion)t < T, hence ^ is very small, hence higher degrees of ^ are neglected.= 4oeT3fv Atmosphere is very big as compared to the body, hence its temperature remains almost constant.Hence, rate of heat loss oc t (temperature difference)Q. 8. Draw spectral distribution curves of black-body radiation and write the effect of rise in temperature on it. (1996,2001)Or Draw the curves between energy and wavelength at different temperatures of a black-body radiations and write the conclusions drawn from these curves. (2003)Ans. Spectral Distribution of Black-Body Radiation : A perfect black-body, when heated upto high temperature, emits radiation of all possible wavelengths, that is why it is called full radiator. In 1899, two scientists Lummer and Pringsheim studied the spectral energy distribution in black-body radiation at differenl temperatures. They heated the black-body at differenl temperatures and at each temperature they plotted the radiant energy, emitted at different wavelengths againsl wavelength. The general shape of these curves are shown in the fig. 13 -2.Following conclusions were drawn from these graphs:(i) At a given temperature, the radiant energy 'E\ emitted by the black-body at wavelength 'A.' first increases with the wavelength and attains a maxima at a certain wavelength and then goes on decreasing.(ii) At a given temperature, maximum energy is emitted at a certain wavelength which is denoted by Xm.(iii) As the temperature is increased, the energy 'Ex emitted at wavelength X is also increased for each wavelength.(iv) The product of absolute temperature T of black-body and wavelength X,„ at which maximum energy is emitted, is a constant.i.e. XmT = b (constant)This law is called Wein's displacement law and the constant 'b' is called Wein's constant.From this law we can observe that, at low temperatures the maximum energy is emitted at larger wavelengths, but as the temperature is increased more energy is emitted in shorter wavelengths and maximum energy peak is shifted towards shorter wavelengths (see graphs).(iv) The area under these curves increases with increasing temperature directly as the 4th power of the absolute temperature of the body. That means the total radiant energy (measured by the area under the curve) emitted by the body is directly proportional to 4th power of the absolute temperature.i.e. ExT4Thus Stefan's law is also verified by these curves.Q. 9. Explain Planck's hypothesis of radiation. Discuss its importance in modern physics. (1997,98,99)Or State Planck's hypothesis. (2004)Ans. Planck's Hypothesis: According to this hypothesis, emission or absorption of radiant energy is not continuous process but is takes place in the form of small packets of energy. Each such packet is called a photon or quanta. The energy associated with a photon of radiation having frequency v is hv. Here 'h' is a constant called Planck's constant. Thus we can conclude that energies emitted by a body can be hv, 2hv,3hv....(integer multiple of hv) but not in between.Classical mechanics and thermo-dynamics could not explain the spectral energy distribution in black-body radiation. Planck, used his hypothesis and derive a formula to explain the spectral energy distribution in black-body radiation and found it in close agreement with the results of Lummer and Pringsheim experiments. Further Einstein successfully explained photo-electric effect on the basis of Planck's hypothesis. In this way Planck's hypothesis gained recogni tion.Q. 10. Find the expression for kinetic mass and momentum of photon using Planck's hypothesis. (1999,2004)Ans. Momentum of Photon: According to Planck's hypothesis, the rest mass mo of photon is zero and each photon travels with the speed of light c.If v is the frequency of photon, then its energy will be E = hv. Further if m is the kinetic mass of photon then by mass energy relation, we haveE = mc2.2 ' hv h f v 1)E = hv = mc => m = = — '•' ~ = -c2 cX V c X):. Kinetic mass of photon m = ~= —v c2cXIf p is the linear momentum of photon, thenp = kinetic mass x velocityor p=mc or p = (v velocity = c)h hvor p = — = —c/v c (■-• X = c/v)Momentum p = — = —y X cin situ conservation
in situ conservation
In situ conservation is the most appropriate method to maintain species of wild animals and plants in their natural habiats. This approach in-cludes protection of total ecosystems through a network of protected areasThe common natural habitats (protectedareas) that have been set for in situ conservation ofwild animals and plants include —1. National parks2. Wildlife sanctuaries3. Biosphere reserves4. Several wetlands, tnangrooves and coralreefs.5. Sacred grooves and Lakes.In situ conservation ilso includes the intro-duction of plants and ani mal species back intoagricultural, horticultural and animal husbandrypractices so that they are cultivated/reproduced fortheir reuse by the farmers/animal husbandrypeople. Farmers and horticulturists have beentraditionally maintaining large genetic diversity ofcrop plants/flowers by savir g seeds for next plant-ing season by a wide variety of indegenously developed practices. For example, tubers,rhizomes, bulbs and seeds of large variety of plant species ai e stored by the far ners/horticulturists fortheir cultivation in the next sieason. Similarly, native species of cattle, which are better adapted to dis-eases, drought and other adverse conditions, arebeing maintained by animal husbandry people.heat of neutralization
heat of neutralization
The heat of neutralization of all strong acids and bases are the same in dilute solution. It can be explained easily on the basis of this theory. According to this theory, the strong acids and bases are completely ionised in solution. They combine to form a highly ionised salt and feebly ionised water and since in all cases water is formed, the heat produced should be same. Heat of neutralization is, in fact, the heat of formation of water from H+ and OH" ions, which is always the same irrespective of the nature of acids and bases.H+ + CI" + Na+ + OH" Na+ + CF + H20 + 13-7k calsor H+ + OH"—»H20 + 13-7 k calsmatrix division
matrix division
A AEROBIC RESPIRATION oxidation of food in the presence of oxygen ANAEROBIC RESPIRATION oxidation of food in the absence of oxygep ANTIBIOTIC medicine that kills disease-causing bacteria ANUS opening at one end of the digestive tract through which undigested food leaves the body ATP (adenosine triphosphate) small molecule in a cell which can store, transfer or release energy AUTOTROPHS organisms that make their own food AXON long fibre thai extends from the cell body of a neuronB BALL AN D SOCKET JOINT joint in which end of one bone is rounded, which fits into the hollow part of the other bone BERI BERI condition caused due to deficiency of vitamin B BOWMAN'S CAPSULE cup-shaped structure at one end of a nephron BRONCHUS {plural bronchi) thin pipe that branches off from the trachea to enter a lung C CAMBIUM lateral meristematic tissue that produces new xylem and phloem cells in a stem CARNIVORES animals that eat other animalsCELL BODY central part of a neuron CELL WALL structure in plant cell Outside the cell membrane CENTRIOLE organelle in animal cells which helps in cell division CHLOROPLASTplastid containing chlorophyll CHROMOPLAST plastid containing coloured pigments CHROMOSOME small, thread-like structure in the nucleus of a cell CISTERNAE flattened sac-like vesicles of endoplasmic reticulum and golgi bodiesCONNECTIVE TISSUE tissue that joins and supports different body parts CRISTAE folds of the inner membrane of mitochondrion CUTANEOUS RESPIRATION exchange of gases through the skin of animals CYTOPLASM jelly-like'substance that fills most of the cell D DENDRITE short branched extension of a neuronDERMIS inner thick layer of the skin DIGESTION process of converting complex food into a simple, absorbable form E ENDOPLASMIC RETICULUM network of membranes in the cytoplasm EPIDERMIS outer layer of the skin EPITHELIAL TISSUE layer of cells that covers the external surface of the body and also forms the lining of the internal organs EXTEN50R muscle that contracts to straighten a part of the body F FLEXOR muscle that contracts and causes a part of the body to bend towards the bodyGGEOTROPiSM movement of plant parts in response to force of gravity GLAND organ that produces secretions GLIDING JOINT joint that allows sliding movement between two bonesH HAEMOGLOBIN red oxygen-carrying pigment in RBCs HERBIVORES animals that feed or plantsIHETEROTROPHS organisms that obtain food directly or indirectly from plantsINTERNAL RESPIRATION using oxygen to break down food and release energy in the bodyJJOINT place where two or more bones meet together in the skeletonkKIDNEY organ that filters wastes from the bloodL LACUNA tiny fluid-filled spaces in cartilagenous connective tissue LIGAMENT strong band of connective tissue that hold two or more bones at jointsLIVER organ that produces bile which helps in digestion LOCOMOTION movement of the entire body of the animal from one place to another LYMPH colourless fluid connective tissue; it surrounds cells of the bodyM MATRIX ground substance of connective tissue in which cells are presentMERISTEMATIC TISSUE plant tissue made up of cells that divide continuously MICROORGANISMS living organisms that can be seen only through a microscope MITOCHONDRION {plural mitochondria} rod-like organelle which helps get energy from food MUSCULAR SYSTEM system consisting of the muscles in the body MUSCULAR TISSUE animal tissue that forms the muscles in the body NNASTIC MOVEMENTS non-directional plant movements that occur in response to environmental stimuli NERVE bundle of nerve cells that carry messages throughout the body NEURON specialized cell that makes up the brain, spinal cord and nerves O OESOPHAGUS tube that leads from the mouth to stomach ORGAN part of an organism made up of one or more .types of tissues, which performs a specific function ORGAN SYSTEM group of organs that work togetherORGANELLE small structure found in the cytoplasm of a cellP PALISADE TISSUE parenchyma tissue in leaves containing chloroplastsPARENCHYMA plant tissue composed of simple cells with thin wailsPERMANENT TISSUE plant tissue made up of ceils that cannot dividePLASMA liquid part of blood in which the blood cells floatPROTOZOA group consisting of only single-celled animalsRRED BLOOD CELL cell that transports oxygenRESPIRATION process of taking in oxygen and using it,to release energy from foodRESPONSE reaction of an organism to a stimulus S SCAVENGERS animals that consume dead animals and dispose them SCLERENCHYMA hard and strong supporting tissue made up of dead and thick-walled cells SIEVE CELL elongated cell of phloem having minute pores SPECIES a group of living organisms which can breed among themselves SPIRILLA spiral-shaped bacteriaTTENDON connective tissue that joins skeletal muscles to bones TISSUE cells that have similar structure and perform similar functions TRACHEA tube passing through the neck that divides into two bronchi TRACHEID tube-like dead cell in xylem TRANSPIRATION loss of water from leaves through the stomataV VACUOLE sac-like organelle which stores food, water or wastesVASCULAR BUNDLE xylem and phloem tissues present togetherVASCULAR TISSUE permanent tissue in plants made up of xylem and phloemVERTEBRA smalt bone of a vertebral columnVESICLE small, round membranous structure in a cellVESSEL non-living part of xylem that, has tube-like ceils joined end to endW WARM BLOODED ANIMAL animal ' that has a constant body temperatureWHITE BLOOD CELL cell in blood that, fights disease-causing microorganisms xXYLEM complex vascular tissue that transports water and mineral saltsaldol condensation
aldol condensation
(i) With hydroxylamine : Aldehydes and ketones react with hydroxylamine to give corresponding oximes containing (> C = N - OH) group. The reaction is carried in the presence of a base like NaOH.(a) Acetaldehyde with hydroxylamine gives acetaldoxime. h hI ________ICH3 - C = Jo + _H2]nOH ) CH3 - cf NOH + HzO *acetaldehyde acetaldoximeCH3 CH3 ....CH3 - c =€f +~H21NOH ) CH3 - c = NOH + H20i____.Z'i ; - acetone acetoxime (ii) With hydrazine : Aldehydes and ketones react with hydrazine to give corresponding hydrazones. h h 1 r-___„ * IH3C-C = jO + H2|- N-NH2-> H3C-C =N-NH2 + H2O acetaldehyde acetaldehyde hydrazone ch3 ch3 I .1 h3c - c =jo_ + JH2_j- n - nh2—> h3c - c = n - nh2 + h2oacetone acetone hydrazone(iii) With phenylhydrazine : Aldehydes and ketones react with phenylhydrazine to give the corresponding phenylhydrazone.(a) Acetaldehyde with phenylhydrazine gives acetaldehyde phenylhydrazone.H HI Ich3 - c = jo + H2j nnhc6h5 -4 ch3 - c = nnhc6h5 + h2oacetaldehyde acetaldehyde phenylhydrazone(b) Acetone with phenylhydrazine gives acetone phenylhydrazone.CH3 CH3I ICH3- C=iO + H^N -NH-CgHs —» CH3-C = N-NH_C6H5 +H20acetone acetone phenylhydrazone(iv) Aldol condensation : Aldehydes or ketones containing active a-hydrogen atom, i.e. hydrogen on the carbon atom adjacent to carbonyl group (a-carbon atom) give P-hydroxy aldehyde, i.e. aldol or p-hydroxy ketone respectively on warming with dilute solution of sodium hydroxide (10%) or sodium carbonate. This reaction is known as aldol condensation. Formaldehyde and benzaldehyde do not give this reaction as they have no a-hydrogen atom. Aldol contains an aldehydic group (-cho) and also alcoholic group (-oh). (Aid for aldehyde group and -ol for alcohol, hence the name aid + ol = aldol.)(a) This is an addition reaction. This addition takes place in such a way that a-carbon atom of one aldehyde molecule is attached to the carbonyl carbon of the second molecule and a-hydrogen of one aldehyde is attached to carbonyl oxygen of the other to give P-hydroxy aldehvde.Aldol loses one water molecule on heating with an acid to form unsaturated aldehyde.When aldol condensation takes place between two molecules of different aldehydes, then it is known as cross aldol condensation. h h I „ Na2C03 I „ch3ch2c =o+hch2c = o-> ch3ch2-cp-ch2-c = o| solution ( |h oh hpropionaldehyde acetaldehyde propionaldol(b) Certain ketones like acetone in presence of barium hydroxide or sodium hydroxide give diacetone alcohol (p-hydroxy ketone). CH3 H O CH3 H O I I II Ba(OH)2 J J II CH3 -C + H-c«-c-CH3---^ cha-pc-aC -C-CH3Oh oh hacetone acetone diacetone alcohol(4-hydroxy, 4 -methyl, pentan-2-one)modern periodic table
modern periodic table
Learn table of Pythagorean related trigonometric identities:cos2θ + sin2θ = 1sin θ = ± `sqrt(1 - cos^2 theta)`cos θ = ± `sqrt(1 - sin^2 theta)`sin θ = `1/(csc theta)`cos θ = `1/(sec theta)` tan θ = `1/(cot theta)`csc θ = `1/(sin theta)` sec θ = `1/(cos theta)` cot θ = `1/(tan theta)`1 + tan2θ = sec2θ1 + cot2θ = csc2θsin θ = ± `(tan theta)/(sqrt(1 + tan^2theta))`cos θ = ± `1/(sqrt(1 + tan^2theta))`tan θ = ± `sqrt(sec^2 theta - 1)`csc θ = ± `(sqrt(1 + tan^2theta))/(tan theta)` sec θ = ± `(sqrt(1 + tan^2theta))`cot θ = ± `1/(sqrt(sec^2 theta - 1))`Learn table of symmetry related trigonometric identities:sin (-θ) = - sin θ cos (-θ) = + cos θ tan (-θ) = - tan θ csc (-θ) = - csc θ sec (-θ) = + sec θ cot (-θ) = - cot θ sin (∠- θ) = + sin θcos (∠- θ) = - cos θtan (∠- θ) = - tan θcsc (∠- θ) = + csc θsec (∠- θ) = - sec θcot (∠- θ) = - cot sin (`pi/2` - θ) = + cos θcos (`pi/2` - θ) = + sin θtan (`pi/2` - θ) = + cot θcsc (`pi/2` - θ) = + sec θsec (`pi/2` - θ) = + csc θcot (`pi/2` - θ) = + tan θLearn table of shifts and periodicity related trigonometric identities:sin (θ + `pi/2` ) = + cos θ cos (θ + `pi/2` ) = - sin θ tan (θ + `pi/2` ) = - cot θ csc (θ + `pi/2` ) = + sec θsec (θ + `pi/2` ) = - csc θ cot (θ + `pi/2` ) = - tan θ sin (θ + âˆ) = - sin θcos (θ + âˆ) = - cos θtan (θ + âˆ) = + tan θcsc (θ + âˆ) = - csc θsec (θ + âˆ) = - sec θcot (θ + âˆ) = + cot θsin (θ + 2âˆ) = + cos θcos (θ + 2âˆ) = + sin θtan (θ + 2âˆ) = + cot θcsc (θ + 2âˆ) = + sec θsec (θ + 2âˆ) = + csc θcot (θ + 2âˆ) = + tan θprocess mapping examples
process mapping examples
1)What is the addition of the following array in math. Solution The given array is as follows, Now we can add the numbers . This is the addition process.2)What is the difference of the following Array? Solution The given array is as follows, Now subtract the second group from the first group. This is the process of the subtraction.3) Multiply the following array. Solution Here we can multiply the first row of the first group into all columns of the second group. So the multiplication is as follows, This is the multiplication process.math test generator
math test generator
Example 1- Study online exponential generating functionDraw the graph for given exponential function `f(x)=(50/25)+e^x`Solution: We have to find the corresponding value of f(x) by substituting the value of x. After find out the f(x)-value for all x-value make them into table format as represented below.`f(x)=(50/25)+e^x``f(0)= (50/25)+e^0=3``f(1)= (50/25)+e^1=3.718``f(2)= (50/25)+e^2=9.38``f(3)= (50/25)+e^3=22.08`x0123f(x)34.7189.38922.08 When plotting the graph we have to assume scale as per our x and f(x) value. After find out the x-value and y-value ie f(x) we have to mark the points on a graph sheet then we have to join the points as coordinate respectively.Scale:In x-axis 1 unit=1 cmIn y-axis 1 unit=5 cmGraph for the given function is given belowequation for photosynthesis
equation for photosynthesis
Before 1930, it was considered by physiologists that one molecule each of CO2 and H2O form a molecule of formaldehyde (HCHO), of which 6 mols are polymerized to one molecule of glucose (a hexose sugar).LightCO2 + H2O -> HCHO + 02Chlorophyll (Formaldehyde) Polymerization6CH20 (or 6HCHO)-> C6Hi206(Formaldehyde) (Hexose sugar)However formaldehyde is a toxic substance which may kill the plants. Hence, formaldehyde hypothesis could not be accepted.On the basis of discovery of Nicolas de Saussure (1804) that "The amount of O2 released fromplants is equal to the amount of CO2 absorbed by plants", it was considered that O2 released inphotosynthesis comes from CO2, but later on it was proved wrong.In 1930, C.B. van Niel proved that, sulphur bacteria use H2S (in place of water) and CO2 tosynthesize carbohydrates as follows :6CO2 + 12H2S-> C6H1206 + 6H20 + 12SThis led van Niei to the postulation that in green plants, water (H2O) is utilized in place of H2Sand O2 is evolved in place of sulphur (S).6C02 + 12H20-> C6H1206 + 6H20 + 602This was confirmed by Ruben and Kamen in 1941 using chlorella a green alga. He used water(H20), having heavy isotope of oxygen (O18), oxygen of 18 atomic weight (normal oxygen is of 1618atomic weight) and found that oxygen released in the process of photosynthesis was of O 0 type (i.e., heavy oxygen). On the contrary when CO28 was used, the released oxygen was of normal type, i.e., O26. From this it can be concluded that, oxygen released by green plants comes from splitting of water (oxidation of water).Furthermore because 6 molecules of oxygen can be produced by 12 molecules of water only, the summary equation of photosynthesis is written as follows :Oxygenic and Anoxygenic Photosynthesis(i) Oxygenic photosynthesis (with evolution of O2) is found in eukaryotes and cyanobacteria. Water acts as H-donar to produce NADP.2H thus photolysis of water is essentialLightC02 + H2O-> Sugar + OxygenChlorophyll(ii) Anoxygenic photosynthesis (without the evolution of O2) is found in photosynthetic bacteria. H2S acts as H-donar and no photolysis of water is needed.LightCO2 + H2S-> Sugar + Sulphur or other oxidised compounds orChlorophyll some other inorganic compoundsAnoxygenic photosynthesizing forms are strictly anaerobic and thus O2 acts as a poison for them.alternating current generator
alternating current generator
An AC circuit consists of combinations of circuit elements like resistors, inductors and capacitors fed by an AC generator which provides sinusoidal voltage. To analyse an AC circuit is to find the current through the given circuit and the phase difference between the current and the applied voltage.grade 12 math
grade 12 math
Olfactory I: This sensory cranial nerve located in olfactory bulb of brain and has olfactory receptors for sense of smell.Optic Nerve II: This sensory cranial nerve leads from eyes to thalamus. It is responsible for sense of sight of retina.Oculomotor Nerve III: This motor nerve arise from midbrain and leads to eye muscles (including eyelids and lens) and pupil. This nerve is associated with eye movement and pupil constriction.Trochlear Nerve IV: This motor nerve arises from midbrain and leads to eye muscles. It works with oculomotor nerve to produce the eye movements.Trigeminal Nerve V: This cranial nerve, located in pons, has both sensory and motor nerve fibers. It leads to most of the face (including eyes and mouth) and carry somatosensory information to face, head and chewing muscles of jaws.Abducens Nerve VI: Another mixed cranial nerve, located in pons. This is associated with eye movement.Facial VII: This is mixed cranial nerve, located in pons. Facial nerve splits into several branches that control the muscles used for facial expressions (smiling, frowning etc). It also stimulates salivary glands to produce saliva.A branch of facial nerve carries taste sensation from the front 2/3 part of the tongue. The facial nerve comes out just below the ear and passes through the salivary gland. Part of facial nerves carries sensation from the outside of the ear.Vestibulocochlear VIII: This sensory cranial nerve leads from inner ear to pons and is associated with sense of hearing and balance.Glossopharyngeal IX: This is a mixed cranial nerve, located in medulla oblongata. This is sensory to posterior of throat (pharynx) and is associated with taste sensation from rest 1/3 part of tongue. This is also associated with gag reflexes.Vagus X: This is a sensory cranial nerve that leads from many visceral organs to medulla oblongata. It carries somatosensory information from organs of thoracic, abdominal cavity including heart and from that of gastrointestinal tract.Spinal Accessory Nerve XI: This motor nerve arises from medulla oblongata and leads to muscles of neck, back and larynx. It controls the head movement.Hypoglossal Nerve XII: This motor nerve arises from medulla oblongata and controls the muscles of tongue.algebra factoring
algebra factoring
Sec. 19-2 □ Thermometers and Temperature Scales1. Fahrenheit and Celsius At what temperature is the Fahrenheit scale reading equal to (a) twice that of the Celsius and (b) half that of the Celsius?2. Oyntyakon (a) In 1964, the temperature in the Siberian village of Oymyakon reached — 71°C. What temperature is this on the Fahrenheit scale? (b) The highest officially recorded temperature in the continental United States was 134°F in Death Valley, California. What is this temperature/on the Celsius scale?Sec. 19-4 □ Heating, Cooling, and Temperature3. Hot and Cold It is an everyday observation that hot and cold objects cool down or warm up to the temperature of their surroundings. If the temperature difference AT between an object and its surroundings (AT = Tobj - Tsur) is not too great, the rate of cooling or warming of the object is proportional, approximately, to this temperature difference; that is,where A is a constant. (The minus sign appears because AT decreases with time if AT is positive and increases if AT is negative.) This is known as Newton's law of cooling, (a) On what factors does A depend? What are its dimensions? (b) If at some instant f, = 0 the temperature difference is Ar1; show that it isat a later time t2.4. House Heater The heater of a house breaks down one day when the outside temperature is 7.0°C. As a result, the inside temperature drops from 22°C to 18°C in 1.0 h. The owner fixes the heater and adds insulation to the house. Now she finds that, on a similar day, the house takes twice as long to drop from 22°C to 18°C when the heater is not operating. What is the ratio of the new value of constant A in Newton's law of cooling (see Problem 3) to the previous value?Sec. 19-5 □ Thermal Energy Transfer to Solids and Liquids5. A Certain Substance A certain substance has a mass per mole of 50 g/mol. When 314 J of thermal energy is transferred to a 30.0 g sample, the sample's temperature rises from 25.0°C to 45.0°C. What are (a) the specific heat and (b) the molar specific heat of this substance? (c) How many moles are present?6. Diet Doctor A certain diet doctor encourages people to diet by drinking ice water. His theory is that the body must burn off enough fat to raise the temperature of the water from 0.00°C to the body temperature of 37.0°C. How many liters of ice water would have to be consumed to burn off 454 g (about 1 lb) of fat, assuming that this much fat burning requires 3500 Cal be transferred to the ice water? Why is it not advisable to follow this diet? (One liter = 103 cm3. The density of water is 1.00 g/cm3.)7. Minimum Energy Calculate the minimum amount of energy, in joules required to completely melt 130 g of silver initially at 15.0°C.8. How Much Unfrozen How much water remains unfrozen after 50.2 kJ of thermal energy is transferred from 260 g of liquid water initially at its freezing point?9. Energetic Athlete An energetic athlete can use up all the energy from a diet of 4000 food calories/day where a food calorie = 1000 cal. If he were to use up this energy at a steady rate, how would his rate of energy use compare with the power of a 100 W bulb? (The power of 100 W is the rate at which the bulb converts electrical energy to thermal energy and the energy of visible light.)10. Four Lightbulbs A room is lighted by four 100 W incandescent lightbulbs. (The power of 100 W is the rate at which a bulb converts electrical energy tff thermal energy and the energy of visible light.) Assuming that 90% of the energy is converted to thermal energy, how much thermal energy is transferred to the room in 1.00 h?11. Drilling a Hole A power of 0.400 hp is required for 2.00 min to drill a hole in a 1.60 lb copper block, (a) If the full power is the rate at which thermal energy is generated, how much is generated in Btu?_(b) What is the rise in temperature of the copper if the copper absorbs 75.0%-of this energy? (Use the energy conversion 1 ft- lb = 1.285 X lO^Btu.)12. How Much Butter How man^ grams of butter, which has a usable energy content of 6.0 Cal/g (= 6000 cal/g), would be equivalent to the change in gravitational potential energy of a 73.0 kg man who ascends from sea level to the top of Mt. Everest, at elevation 8.84 km? Assume that the average value of g is 9.80 m/s2.13. Immersion Heater A small electric immersion heater is used to heat 100 g of water for a cup of instant coffee. The heater is labeled "200 watts," so it converts electrical energy to thermal energy that is transferred to the water at this rate. Calculate the time required to bring the water from 23°C to 100°C ignoring any thermal energy that transfers out of the cup.14. T\ib of Water One way to keep the contents of a garage from becoming too cold on a night when a severe subfreezing temperature is forecast is to put a tub of water in the garage. If the mass of the water is 125 kg and its initial temperature is 20°C, (a) how much thermal energy must the water transfer to its surroundings in order to freeze completely and (b) what is the lowest possible temperature of the water and its surroundings until that happens?15. A Chef A chef, on finding his stove out of order, decides to boil the water for his wife's coffee by shaking it in a thermos flask. Suppose that he uses tap water at 15°C and that the water falls 30 cm each shake, the chef making 30 shakes each minute. Neglecting any transfer of thermal energy out of the flask, how long must he shake the flask for the water to reach 100°C?16. Copper Bowl A 150 g copper bowl contains 220 g of water, both at 20.0°C. A very hot 300 g copper cylinder is dropped into the water, causing the water to boil, with 5.00 g being converted to steam. The final temperature of the system is 100°C. Neglect energy transfers with the environment, (a) How much energy (in calories) is transferred to the water? (b) How much to the bowl? (d) What is the original temperature of the cylinder?17. Ethyl Alcohol Ethyl alcohol has a boiling point of 78°C, a freezing point of -114°C, a heat of vaporization of 879 kJ/kg, a heat of fusion of 109 kJ/kg, and a specific heat of 2.43 kJ/kg • C°. How much thermal energy must be transferred out of 0.510 kg of ethyl alcohol that is initially a gas at 78°C so that it becomes a solid at -114°C?18. Metric-Nonmetric Nonmetric version: How long does a 2.0 x 105 Btu/h water heater take to raise the temperature of 40 gal of water from 70°F to 100°F? Metric version: How long does a 59 kW water heater take to raise the temperature of 150 L of water from 21°C to 38°C?19. Buick A 1500 kg Buick moving at 90 km/h brakes to a stop, at a uniform rate and without skidding, over a distance of 80 m. At what average rate is mechanical energy transformed into thermal energy in the brake system?20. Solar Water Heater In a solar water heater, radiant energy from the Sun is transferred to water that circulates through tubes in a rooftop collector. The solar radiation enters the collector through a transparent cover and warms the water in the tubes; this water is pumped into a holding tank. Assume that the efficiency of the overall system is 20% (that is, 80% of the incident solar energy is lost from the system). What collector area is necessary to raise the temperature of 200 L of water in the tank from 20°C to 40°C in 1.0 h when the intensity of incident sunlight is 700 W/m2?21. Steam What mass of steam at 100°C must be mixed with 150 g of ice at its melting point, in a thermally insulated container, to produce liquid water at 50°C?22. Iced Tea A person makes a quantity of iced tea by mixing 500 g of hot tea (essentially water) with an equal mass of ice at its melting point. If the initial hot tea is at a temperature of (a) 90°C and (b) 70°C, what are the temperature and mass of the remaining ice when the tea and ice reach a common temperature? Neglect energy transfers with the environment.23. Ice Cubes (a) Two 50 g ice cubes are dropped into 200 g of water in a thermally insulated container. If the water is initially at 25°C, and the ice comes directly from a freezer at -15°C, what is the final temperature of the drink when the drink reaches thermal equilibrium? (b) What is the final temperature if only one ice cube is used?24. Thermos of Coffee An insulated Thermos contains 130 cm3 of hot coffee, at a temperature of 80.0°C. You put in a 12.0 g ice cube at its melting point to cool the coffee. By how many degrees has your coffee cooled once the ice has melted? Treat the coffee as though it were pure water and neglect energy transfers with the environment.Sec. 19-8 ■Some Special Cases of the First Law of Thermodynamics25. Gas Expands A sample of gas expands from 1.0 m3 to 4.0 m3 while its pressure decreases from 40 Pa to 10 Pa. How much work is done by the gas if its pressure changes with volume via each of the three paths shown in the P-V diagram in Fig. 19-30?26. Work Done Consider that 200 J of work is done on a system and 70.0 cal of thermal energy is transferred out of the system. In the sense of the first law of thermodynamics, what are the values (including algebraic signs) of (a) W, (b) Q, and (c) A£""?27. Closed Chamber Gas within a closed chamber undergoes the cycle shown in the P-V diagram of Fig. 19-31. Calculate the net thermal energy added to the system during one complete cycle.28. Thermodynamic A thermodynamic system is taken from an initial state A to another state B and back again to A, via state C, as shown by path ABCA in the P-V diagram of Fig. 19-31a. (a) Complete the table in Fig. 19-32b by filling in either + or - for the sign of each thermodynamic quantity associated with each step of the cycle, (b) Calculate the numerical value of the work done by the system for the complete cycle ABCA.29. From i to / When a system is taken from state i to state / along path iaf in Fig. 19-33. Q = 50 cal and W = 20 cal. Along path ibf, Q = 36 cal. (a) What is W along path ibfl (b) If W = -13 cal for the return path fi, what is Q for this path? (c) Take Efl = 10 cal. What is E'fl (d) If Ef' = 22 cal, what are the values of Q for path ib and path bp.30. Gas Within Gas within a chamber passes through the cycle shown in Fig. 19-34. Determine the thermal energy transferred by the system during process CA if the thermal energy added Qab>Sec. 19-9 □ More on Temperature Measurementalveolar gas equation
alveolar gas equation
The state of a given mass of gas is determined by three parameters namely, pressure, volume and temperature. When a gas is heated both its pressure and volume undergo a change. Thus a gas has two different coefficients of expansion.Volume coefficient" the coefficient of expansion of a gas at constant pressure.Pressure coefficient" coefficient of expansion of a gas at constant volume.The volume coefficient of a gas is defined as the ratio of increase in volume of a given mass of gas per degree rise of temperature to its volume at 0°C when pressure is kept constant. It is denoted by a. If VQ and V be the volumes of a given mass of gas at 0°C and t°C respectively and at the same pressure thenIf Vj and V2 be the volumes of a given mass of gas at C and t2 C respectively pressure remaining constant then using equation (2) it can be proved thatThe pressure coefficient of a gas is defined as the ratio of increase in pressure of a given mass of gas per degree rise of temperature to its pressure at 0°C when volume is kept constant. It is denoted by (3.force acts
force acts
Now let us move from forces' in solids to forces in liquids. If a solid is placed into a liquid the solid will push the liquid out of its path with the result that the liquid will push back on the solid. This is an example of Newton's Third law of motion which states 'action and reaction are equal and opposite'. The upward force exerted by liquids on objects placed in the liquid is known as the upthrust. Upthrust acts on a body in a liquid even if the body is resting on some support such as the bottom of a vessel, provided the space between the body and the support is not evacuated. A marble dropped into water in a beaker will sink to the bottom of the beaker as in Fig. 3.13(a). The forces acting on the marble are its weight W, the upthrust U of the water and the reaction R of the base of the beaker; W = R + U. A marble resting in a hole in the container will not have an upthrust acting on it. This time the downward forces on the marble are its weight W and the weight w of the liquid above it pressing the marble into the hole. The total reaction on the edges of the hole R2 = W + w, which is much greater than the reaction R in the beaker where R = W — U.hybridized
hybridized
3. Simple interaction (9:3:3:1 ratio)Bateson and Punnett gave classical example in fowl to show two genes influencing the same character. There are four types of combs recognised among fowls—pea, rose, walnut and single. These are caused by two allelomorphic pairs of genes symbolised by PPrr (pea), ppRR (rose). A cross between the two produces a new comb form in Fj generation called walnut (RrPp). The F[ generation on selfing produced F2 generation as follows—9 walnut, 3 pea, 3 rose and 1 single. Thus when dominant gene P is present alone (without other dominant gene R) the comb is pea type. When R is present alone, the comb is green or red or rose. But when both P and R are present there is an interaction and the comb becomes walnut. When no dominant gene is present, the comb is of double recessive, i.e., single type (Fig. 65.4).A Summary Chart of Common Types of Interaction of Genes Observeds.No.FactorsExamplesF2Phenotypic ratioP,Parent-1P2Parent-2FiIndividualF2 Population (Genotype and Phenotype)1.Simple interactionComb colour in fowl9:3:3:1RRpp Rose combrrPP Pea combR-P-Walnut combR-P-Walnut (9)R-pp Rose (3)rrP-Pea(3)rrpp Single (1)2.Epistasis(dominantepistasis)Body colour in dogsaaBB BlackAAbb WhiteA-B-White| A-B-| (9) WhiteA-MT~j (3) White 1aaB-(3) Blackaabb (1) BrownSeed coat colour in Cajanus cajan12:3:1BBrr Black seededbbRR Red seededB-R-Black seededir B-R-1 (9) Black L_____B-rr 1 (3) Black 1bbR-(3) Redbbrr (1) White3.Supplementary factors (Recessive epistasis)Flower colour in Antirrhinum9:3:4AAbb IvoryaaBB WhiteA-B-MagentaA-B- (9) MagentaA-bb (3) Ivory, aaB-1 (3) Whiteaabb (1) White jSkin colour in riceAAbb Black skinaaBB Albino skinA-B-Agauti colourA-B-(9) AgautiA-bb (3) BlackaaB1-" I (3) Albinoaabb j (1) Albino JAComplementary factors. IV V W V r 111colour of paddyAAbb GrayaaBB GrayA-B-RedA D ri u(9) Redi "/^bb1 (3) Gray1aaB-(3) Grayaabb ! (1) Gray JFlower colour in Lathyrus odoratus9 : 7CCpp White flowersccPP White flowersC-P-Red flowersC-P-(9) Redr c-pp1 (3) WhiteccP-(3) Whiteccpp J (1) White j5.Inhibitory factorsPigmentation in leaves of paddy13 : 3Ilpp Green pig.iiPP Purple pig.I-P-Green pig.I-P-(9) GreenI-pp (3) GreeniiP-(3) Purpleiipp (1) Green6.Duplicate factorsFruit shape of Capsella15 : 1AABB Triangularaabb Top shapedA-B-Triangular shapeA-B- (9) TriangularA-bb (3) TriangularaaB- (3) Triangularaabb (1) Top shapedAwned character in paddy grainsA1A1a2a2 Awneda j A2A2 AwnedA, - A2 AwnedA, - A2 (9) AwnedA j a2a2 (3) Awneda,a| A2-(3) Awneda^ aja2a2 (1) Awnless7.Polymorphism additive effectPericarp colour in wheat9:6:1Light redLight redR, - R2-Deep redR, - R2 (9) Deep redR, - r2r2 (3) Light redr,r,R2- (3) Light redrlrlr2r2 (1) WhiteC j C j . Deep redC|C|C2C2ColourlessC,-C2-Deep redC,-C2 Deep redC j c2c2 Light redLight redc1c,c2c2 White4. Complementary factors (9 : 7 ratio)Sometimes a trait is produced by the interaction of two or more genes situated on separate chromosomes which complement one another. In some varieties of sweet pea the following interaction of two pairs of genes has been noted, (Fig. 65.5) ccR-, C-rr and ccrr- white colour C-R-Red colour. It was observed by Bateson and Punnett in Lathyrus odoratus.Here gene C and R are complementory to each other and necessary for the production of colour in flower. When they are together (C - R -), red colour in petals develop.
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