Neeraj sir

1-Rajat Arora 2-iat classes

Features of the Harappan culture:

1. Urban Civilisation: The sight of the towns of Mohenjodaro and Harappa reflect the supremacy of the then architecture and town planning. There were wide roads ranging from 9 to 34 feet in width. The horizontal roads cut the vertical ones on right angles. The towns had brick houses, both big and small ones. There was provision of cross ventilation and sufficient lighting. The sight of drains and huge dustbins outside the houses revealed that there must have been some local institution to manage sanitation and such other activities smoothly.

2. Architecture: There were varieties of building that were unearthed:

(i) Buildings for the dwelling of the citizens.

(ii) Buildings for public purposes.

(iii) Public fire places.

3. The Great Bath: The Great Bath, discovered during excavations was 39 feet in length, 23 feet in width and 8 feet in depth.

Verandahs and rooms were constructed around it. There was a provision of filling the bath with water and emptying it. There was a well closely which might have been serving as the source of filling fresh water into the bath. This structure is an excellent symbol of the Indus Valley Civilisation.

4. Food: The people of Indus Valley ate barley, wheat, fruit and flesh also.

5. Clothes: The available needles and spinning and weaving tools and button show that the people knew spinning and weaving. Both the cotton and woollen clothes were worn. The people wore coloured clothes. There was a close similarity in the clothes of men and women-folk.

6. Ornaments: Both the men and the women wore ornaments. The chief ornaments used to be armlets, necklace, ear-rings, metal ring (bangle), belt etc.

7. Cosmetics: Hair used to be decorated in many forms. Brass combs, mirrors, piece of decoration made of ivory and a type of colour used to redden the lips like lipstick too have been found.

8. Amusements: The people used to utilise their time in hunting, playing chess, music, dance, drawing and painting, taming birds and catching fish. These formed the sources of their recreation.

9. Art and craft: Patterns of fine drawing are available on the seals of the times. The clay carts and toys too are fine specimen of the grand art.

10. Religion: There are various hypothesis regarding religion but we still do not know, there are indication that they could have worshipped nature, mother goddess, proto shiva etc.

Tournament is a large contest of many rounds among various teams. It is a competition held among various teams in a particular activity according to a fixed schedule where a winner is decided.
Number of teams participating are 27.
Number of matches = n – 1 = 27 – 1 = 26 matches.
Teams in upper half = n+1/2=27+1/2= 28/2
Teams in lower half = n−1/2=27−1/2= 26/2 = 13 teams
Number of teams participating are odd so number of byes will be = 32 – 27 = 5 byes.
In Knockout system the team which loses the match is eliminated and the winners continue to play. When we have less time usually we conduct single elimination tournament.
Date/time and court number, is specified for each match so that teams/players report for their matches well in advance.

The current flowing through the circuit is given as: I=E/(r+R), where r is the internal resistance of the cell and R is the external resistance. Maximum current can be drawn only when external resistance is zero. Hope It Helps You?

In a circuit the current flowing through the circuit is given as :
I=E/(r+R)
Where r is the internal resistance of the cell and R is the external resistance.
Maximum current can be only drawn when external resistance is zero.

Monopolist

Ans: Functional structure as the people are being grouped on the basis of functions

Explanation: Functional Structure: In functional structure activities are grouped and departments are created on the basis of specific functions to be performed. For example, all the jobs related to production are grouped under production department, sales departments etc.

Capital reciept

Dna ligase enzyme

DNA ligase

Hey Gaurav singh please yar tell me were do you live. Please.

The year 1947 was the year of one of the largest, most abrupt, unplanned and tragic transfer of population which human history has known.

1. There were killings and atrocities on both sides of the border. Cities such as Lahore, Amritsar and Kolkata (Calcutta) became divided into 'communal zones'. Muslims would avoid going into an area where mainly Hindus or Sikhs lived and vice versa.
2. Minorities on both sides of the border fled their home and often secured temporary shelter in 'refugee camps'. They travelled to the other side of the new border by all sorts of means, often by foot. Even during this journey, they were often attacked, killed or raped. Thousands of women were abducted on both sides of the border. Often women were killed by their own family members to preserve the 'family honour'. Many children were separated from their parents.
3. People who managed to cross the border did not have a place they could call home. For lakhs of these 'refugees', the country's freedom meant life in 'refugee camps' for months and sometimes for years.
4. Partition was not merely a division of properties, liabilities and assets, or a political division of the country and the administrative apparatus. Also divided were things such as tables, chairs, typewriters, paper-clips and books.
5. Even after large-scale migration of Muslims to the newly created Pakistan, the Muslim population in India accounted for 12% of the total population in 1951. The Partition had already created severe conflict between the two communities.

Above all, it was a violent separation of communities who had hitherto lived together as neighbours for centuries.

HIV is natural firstly it is found in a species of animal and one day some people eat them from that day it spread fastly..I hope u understand ?

Wrong

HIV has been seen to have evolved from or be closely related to Simian immunodeficiency virus ( SIV ) in West Central Africa. Some time in early 20th century HIV was discovered in the 1980s by the French scientist Luc Montagnier.Before the 1980s HIV was an unknown deadly disease.

When catalyst is added to the reaction then it lowers the activation energy of reactants and hence it increases the rate of reaction increases. Since on adding catalyst, enthalpy, entropy and temperature remain same therefore there is no effect on Gibbs free

Bhawana puri

Electric flux :expansion of electric field across area vector. SI unit Nm/C Magnetic flux :expansion of magnetic field across area vector. SI unit Tesla metre

Electric flux is no. of eletric field lines passing through unit area whereas magnetic flux is no. of magnetic field linez passing through unit area

Amino acid are activated in the presence of ATP and linked

Aminoacylation:It is a two stage process whereby amino acids are 'activated' in order to be incorporated into proteins; energy in the form of ATP is required and one of a family of 20 aminoacyl-tRNA synthetases, one specific for each amino acid, links the carboxyl group of the amino acid to either the 2′- or 3′-hydroxyl group of the ribose unit at the 3′-end of tRNA.

Amino acids are activated in the presence of ATP and linked to (cognate) t-RNA

Thanks

Henry law states whenever solid is dissolved in liquid or gas is dissolved in liquid its solubility is determined by pressure and temperature. it states that partial pressure is directly proportional to mole fraction. ie . p=khx where kh is Henry law constant. Raoult law states that when two volatile liquid are mixed their vapour pressure is directly proportional to mole fraction ie p1 = p*1 x

Abe dekh Ek strand 5' AAATTGC 3' hai To dusra 3' TTTAACG 5' ho jayega mtlb agar dono template ki tarah act karenge to ek RNA 3' UUUAACG 5' AUR DUSTE KA 5' AAAUUGC 3' ho jayega aur dono ek duste ke complimentary hain .. mtlb ya to RNA double strand nhi to ek protein dusra anti protein banayaga

a) If both the strands act as template, they would code for RNA molecule with different sequences, and in turn they code for different proteins. 

b) the two RNA molecules if produced simultaneously would be complementary to each other , hence would for a double stranded RNA.

The order from Berlin is called a thunderclap by Franz, because it was a complete shock for him. He had never thought that he would be deprived of the right of learning his native language. He realised he had so much more yet to learn and was going to lose the opportunity. When Franz reached the school, he saw a strange stillness in the class. Then the teacher, M. Hamel, announced that it was their last lesson in French and the new German teacher will take the charge on the following day. These words of M. Hamel came as thunderclap to Franz since the announcement was unexpected and sudden for him.

Intermediate goods refer to those goods which are used either for resale or for further production in the same year. They are not ready for use in the sense some value has to be added to the intermediate goods. They are still within the production boundary.

two examples of intermediate goods

(i) Milk used in dairy shop for resale; (ii) Coal used In factory for further production.

Furniture purchased by a school is a final product because it is purchased for investment. School buy furniture for long - term use and it is considered as an investment. School is the final user of the furniture and no value is to be added to the furniture. This will be deemed as investment expenditure because furniture is used by the school for several years and is of high value.

It contain 4 microspore

The faces of these children are pale and lifeless. They and their hair are like ‘rootless weeds’. The burden of fife makes them sit with their head ‘weighed down’. The stunted growth is depicted by ‘the paper-seeming bo/ and ‘the stunted unlucky heir of twisted bones’. Their weak bodies recite their fathers’ ‘gnarled disease’.

NOTICE                                                      National Public School 15 January 20XX Change in School Timings This is to inform all the students that there is a change in the school timings. From 20 January onwards, the school shall start from 9:00 a.m. instead of 8:00 a.m. The change is made due to the extreme cold weather and the fog in the morning. All the students are requested to adhere to the new timing.    Priyanka Jha (Principal)

 

1. Calculate the mass percentage of benzene () and carbon tetrachloride () if 22 g of benzene is dissolved in 122 g of carbon tetrachloride. Ans. Mass percentage of   = 15.28% Mass percentage of =  =84.72% Alternatively, Mass percentage of = (100 - 15.28) % = 84.72% 2. Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride. Ans. Let the total mass of the solution be 100 g and the mass of benzene be 30 g. ∴Mass of carbon tetrachloride = (100 - 30) g = 70 g Molar mass of benzene () = (612 + 6)  = 78  ∴Number of moles of  = 0.3846 mol Molar mass of carbon tetrachloride ( ) = 112 + 4355 = 154  ∴Number of moles of  = 0.4545 mol Thus, the mole fraction of is given as:  = 0.458 3. Calculate the molarity of each of the following solutions: (a) 30 g of Co(NO3)2 .6H2O in 4.3 L of solution (b) 30 mL of 0.5 M diluted to 500 mL. Ans. Molarity is given by:  (a) Molar mass of Co(NO3)2 = 59 + 2 (14 + 3 16) + 618 = 291  Therefore, Moles of Co(NO3)2 = 30/291 mol = 0.103 mol Therefore, molarity  = 0.023 M (b) Number of moles present in 1000 mL of 0.5 M  ∴Number of moles present in 30 mL of 0.5 M  = 0.015 mol Therefore, molarity = 0.03 M 4. Calculate the mass of urea () required in making 2.5 kg of 0.25 molal aqueous solution. Ans. Molar mass of urea  = 60  0.25 molar aqueous solution of urea means: 1000 g of water contains 0.25 mol =  of urea = 15 g of urea That is, (1000 + 15) g of solution contains 15 g of urea Therefore, 2.5 kg (2500 g) of solution contains = = 36.95 g = 37 g of urea (approximately) Hence, mass of urea required = 37 g 5. Calculate (a)molality (b)molarity and (c)mole fraction of KI if the density of 20% (mass/mass) aqueous KI is. Ans. (a) Molar mass of KI = 39 + 127 = 166  20% (mass/mass) aqueous solution of KI means 20 g of KI is present in 100 g of solution. That is, 20 g of KI is present in (100 - 20) g of water = 80 g of water Therefore, molality of the solution =  = 1.506 m = 1.51 m (approximately) (b) It is given that the density of the solution =  Therefore, Volume of 100 g solution =  = = 83.19 mL =  Therefore, molarity of the solution =  = 1.45 M (c) Moles of KI =  Moles of water =  Therefore, mole fraction of KI =  =  = 0.0263 6. , a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of  in water at STP is 0.195 m, calculate Henry's law constant. Ans. It is given that the solubility of in water at STP is 0.195 m, i.e., 0.195 mol of is dissolved in 1000 g of water. Moles of water =  = 55.56 mol ∴Mole fraction of , x =  =  = 0.0035 At STP, pressure (p) = 0.987 bar According to Henry's law: p= KHx  = 282 bar 7. Henry's law constant for  in water is 1.67 x 108 Pa at 298 K. Calculate the quantity of in 500 mL of soda water when packed under 2.5 atm pressure at 298 K. Ans. It is given that: KH = 1.67 x 108 Pa = 2.5 atm = 2.5 x 1.01325 x 105 = 2.533 x 105 Pa According to Henry's law: PCO2 = KHx x=  = 0.00152 We can write,  [Since, is negligible as compared to] In 500 mL of soda water, the volume of water = 500 mL [Neglecting the amount of soda present] We can write: 500 mL of water = 500 g of water = mol of water = 27.78 mol of water Now,  Hence, quantity of in 500 mL of soda water =  = 1.848 g 8. The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase. Ans. It is given that:  = 450 mm of Hg = 700 mm of Hg P total= 600 mm of Hg From Raoult's law, we have:  Therefore, total pressure,   600= (450-700)xA + 700 -100 = -250 xA xA = 100/250 = 0.4 Therefore,  = 1 - 0.4 = 0.6 Now,  =  = 180 mm of Hg  =  = 420 mm of Hg Now, in the vapour phase: Mole fraction of liquid A =   = 0.30 And, mole fraction of liquid B = 1 - 0.30 = 0.70 9. Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea () is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering. Ans. It is given that vapour pressure of water, of Hg Weight of water taken,  Weight of urea taken,  Molecular weight of water,  Molecular weight of urea,  Now, we have to calculate vapour pressure of water in the solution. We take vapour pressure as p1. Now, from Raoult's law, we have:    Hence, the vapour pressure of water in the given solution is 23.4 mm of Hg and its relative lowering is 0.0173. 10. Boiling point of water at 750 mm Hg is . How much sucrose is to be added to 500 g of water such that it boils at . Molal elevation constant for water is. Ans. Here, elevation of boiling point = (100 + 273) – (99.63 + 273) = 0.37 K Mass of water,  Molar mass of sucrose ( ),  =  Molal elevation constant, Kb =  We know that:  = 121.67 g (approximately) Hence, 121.67 g of sucrose is to be added. 11. Calculate the mass of ascorbic acid (Vitamin C,) to be dissolved in 75 g of acetic acid to lower its melting point by.. Ans. Mass of acetic acid,  Molar mass of ascorbic acid  =  Lowering of melting point,  We know that:   = 5.08 g (approx) Hence, 5.08 g of ascorbic acid is needed to be dissolved. 12. Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at. Ans. It is given that: Volume of water, V= 450 mL = 0.45 L Temperature, T = (37 + 273) K = 310 K Number of moles of the polymer,  We know that: Osmotic pressure,   = 30.98 Pa = 31 Pa (approximately)

Of which chapter and subject