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.