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A Real Time Operating system is a computing environment that react to input within a specific period of time.

A real-time operating system is a computing environment that reacts to input within a specific period of time. 

It manages the resources of the computer so that any particular operation is executed in the same amount of time every time it is executed. Windows CE and Lynx OS are examples of real-time operating systems.

1. b. Workshop on awareness about environmental concerns

2.d. The students of classes XI-XII are informed that------------

3. d. 2,3,4 and 5

4. b. No, it is optional

5. c. No, the details not available

Telling him that Gupta had not given back his money and he had only returned the paper.

HCF = 2
Product of  2 no:s = 120

LCM = x

We know that ,

HCF × LCM = Product of  2 no:s
2x = 120
x = 120/2
x = 60

LCM = x = 60

If you solve and make the two equations inthe form of ax²+bx+c, thrn you will see that the 2nd eqn. is double of 1st eqn.

So, 2r = p

So, 2r-p=2r-2r= 0

Hope It Helped You

(1.2)⁵ = (1 + 0.2)⁵

= (1 + \(\frac{2}{10}\))⁵

= (1 + \(\frac{1}{5}\))⁵

= ⁵C₀1⁵ + ⁵C₁1⁴ (\(\frac{1}{5}\)) + ⁵C₂1³( \(\frac{1}{5}\))² + ⁵C₃1² (\(\frac{1}{5}\))³ + ⁵C₄1 x (\(\frac{1}{5}\))⁴ + ⁵C₅ (\(\frac{1}{5}\))⁴ + ⁵C₅ (\(\frac{1}{5}\))⁵

= 1 + 5 x \(\frac{1}{5}\) + \(\frac{5\times 4}{2\times 1}\) x \(\frac{1}{25}\) + \(\frac{5\times 4}{2\times 1}\) x \(\frac{1}{125}\) + 5 x \(\frac{1}{5}\) x (0.2)³ + (0.2)⁵

= 1 + 1 + 2 x 0.2 + 2 x (0.2)² + (0.2)³ + (0.2)⁵

= 2 + 0.4 + 2 x 0.04 + 0.008 + 0.00032

= 2.4 + 0.08 + 0.00832

= 2.48832

Correct option is D) Twice a week

Correct option is (B) The particle is at rest t = 0

Given position vector is: x = 5 + 3t

Hence, at t = 0,x = 5 m i.e., particle is east of origin. Option A is correct.

Comparing with second equation of motion,

s = ut + 1/2 at²

u = 3 m/s

Option B is false

a = 0 Thus, option D is true.

Hence, particle's velocity is constant. Option C is true.

y=5x^4+3x^5+2x^2

wkt,

d/dx (x^n)=nx^n-1

dy/dx=5.4x^3+3.5x^4+2.2x

dy/dx=20x^3+15x^4+4x

Geeta, on her way to Gopalji’s haveli, saw from her car some village women. They were carrying their baskets of vegetables on their heads. Their faces were uncovered and their arms were full of colourful bangles. Their heavy skirts swayed as they shouted to sell their wares in their sweet rural dialect. Geeta envied their freedom and carefree life. This stood in contrast to the caged life at the haveli.

गोपाल जी की हवेली को जाते हुए अपनी कार से गीता जी ने कुछ ग्रामीण महिलाओं को देखा। वे अपने सिर पर सब्जियों की टोकरियां ले जा रही थीं। उनके चेहरे बिना ढके हुए थे और उनकी भुजायें रंग-बिरगी चूड़ियों से भरी हुई थीं। जब अपनी मधुर ग्रामीण आवाज में अपना सौदा बेचने के लिये जोर से बोलती थी तो उनके भारी लहंगे/घाघरे इधर उधर हिलते थे। गीता को उनकी स्वतन्त्रता और बेपरवाह जीवन से जलन हुई। यह हवेली की पिंजरे में बन्द जिन्दगी के बिल्कुल विपरीत था।

Last week, we celebrated English Week in our school. It was decided that only English would be used in school. The teachers and students had a wonderful time. Many activities were held during the week. There was an essay-writing competition on ‘The Advantages of English Language’. There were poetry recitation competitions, debates and elocution.

All the students and teachers enjoyed various games that were based on the English language. I won the Spelling-Bee competition. It was great fun. I honestly feel that English, as a language, should be taught in all schools. It is the international language of communication. Everyone should have the basic knowledge of English.

1. E .dA=qenc/3o

1. We must preserve our culture and traditions.

2. There is a ten-storeyed building in our town.

3. All the girls envied Mugdha for her beauty.

THERE ARE MANY TYPES OF MIRRORS WE USE IN OUR DAILY LIFE.some important are 

#.reflecting mirror which is used to see our faces

#.in car we use side mirror made up of concave

Heat is the transfer of energy from one body to another due to the difference in temperature between the two. A hotter object placed next to a cooler object will always transfer heat from itself into the cooler object, until both objects are of equal temperature. For example, when we place ice cubes in hot water, the heat from the hot water transfers to the ice cubes. This transfer of heat energy will continue until equilibrium is reached between the hot water and the ice.

Heat is a measure of the internal energy that has been absorbed or transferred from one body to another. It is not conserved; it can be either created or destroyed. There are two general ways that heating can occur: from a temperature difference, with energy moving from the region of higher temperature, and from an object gaining energy by way of an energy-form conversion.

The SI unit of heat is the joule. The metric unit of heat is called the calorie (cal), which is defined as the amount of heat required to raise the temperature of one gram of water from 14.5oC to 15.5oC. The English system’s measure of heating is called the British thermal unit (Btu), which is the amount of heat needed to increase the temperature of 1 pound of water 1 degree Fahrenheit.

Some examples of heat energy are: chemical energy from the foods is converted into heating our bodies; light from the sun is converted to heat as the sun’s rays warm the earth’s surface; energy from friction creates heat, like when we rub our hands; in light bulbs, electrical energy is converted into heat energy, etc.

• In the manufacture of rayon 
• In the rubber and silk industry. 
• Vinegar – Impart Sour taste for food item. 
• Used as preservative.

1. CH₃ – CH₂ – COO – CH₃ 

7. CH₃ – COO – CH₂ – CH₂ – CH₃ are esters 

1. CH₃ – CH₂ – COO – CH₃

CH₃ – CH₂ – COOH + OH – CH₃ + Conc.

H₂SO₄→ CH₃ – CH₂ – COO – CH₃ + H₂O 

7. CH₃ – COO – CH₂ – CH₂ – CH₃ 

CH₃ – COOH + OH – CH₂ – CH₂ – CH₃ + Conc.

H₂SO₄ → CH₃ – COO – CH₂ – CH₂ – CH₃ + H₂O

CH₃ – CH₃ – Ethane 

CH₂ = CH₂ – Ethene

In ethene, There is carbon – carbon double bond

a.  10 Mole of water = 10 × 18 = 180g, 10 × 6.022 × 10²³ Molecules 

5 mole of CaO = 280g, 5 × 6.022 × 10²³ Molecules 

2 Mole of H₂SO₄ = 196g, 2 × 6.022 × 10²³ Molecules 

2- Mole of AlO = 100g, 1/2 × 6.022 × 10²³

The correct option is (3) μmg.

Given:

Block of mass is 'm'

Coefficient of friction ' μ'

We have to apply a force that is just greater than static friction force on a body.

So the required force is F = μ.N

= μ mg.

a. When 1 mole of methane bum in air 1 mole of CO₂  gas is formed. 

When 20 moles methane burn 20 moles of CO₂  are formed. 

The molecular weight of 20 moles = 20 × 44 = 880 g

b. When 2 moles of ethane is burned 4 moles of CO₂ is formed. 

The number of moles of CO₂  when 1 mole of ethane is burned = 4/2 = 2 mole. 

The number of moles when 10 moles of ethane bums = 2 x 10 = 20 mole. 

Weight of 20 moles = 20 × 44 = 880 g

a. 1:3:2 

b. 2 Mole

c. No 

d. 10 Mole Nitrogen 30 Mole Hydrogen

\(\frac{69}{23}\) = 3 GAM,

2000 oxygen atoms.

2 x 16 = 32g

Answer is Ethane

Saturated Compounds

a. GMM= 18 g 

∴ Number of molecules

\(\frac{90}{18}\) x 6.022 x 10²³ = 5 x 6.022 x 10²³

b. ∴ Number of atoms = 3 × 5 × 6.022 × 10²³ 

= 15 x 6.022 x 10²³

c. Electrons in 1 H atom = 1 

Electrons in 1 O atom = 8 

Total electrons in H₂O molecule 10 

Total electrons in 90 g H₂O = 10 × 5 × 6.022 × 10²³ = 50 × 6.022 × 10²³

a. P – CH₂ = CH₂ 

Q – CH₃ – CH₃

R – CH₃ – CH₂Cl 

b. ii. Addition reaction 

iii. Substitution Reaction 

c. Chloroethane

Unsaturated propene

a. Substitution reaction 

b. CH = CH₂ 

c. HBr 

d. Addition reaction 

e. H₂O 

f. Combustion 

g. [CH₂- CH₂]_n

h. Polymerization

a. CH₄ + O₂ → CO₂ + 2H₂O 

2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O 

b. Carbon dioxide and Water

GAM of Helium = 4 g 

Number of mole atoms in 4g of Helium = 6.022 x 10²³ 

Number of atoms in 1g of Helium = \(\frac{1}{4}\) x 6.022 x 10²³

GAM of Carbon = 12 g 

∴ Number of atoms in 12g of Carbon  = 6.022 x 10²³ 

∴ Mass required for \(\frac{1}{4}\) x 6.022 x 10²³

Carbon atoms = 3g (12 x \(\frac{1}{4}\))

GAM of Oxygen = 16 g. 

∴ Number of atoms in 16 g of Oxygen = 6.022 × 10²³ 

∴ Mass required for ^{\(\frac{1}{4}\)} x 6.022 × 10²³ 

Oxygen atoms = 4g  (16 x \(\frac{1}{4}\))

6.022 x 10B carb 

2 x 6.022 x 10²³ oxygen atoms

a.  CH₂ = CH₂ + H₂ → CH₃ -CH₃(Addition reaction)

b. CH₃ - CH₃ + Cl₂ + sunlight⟶ CH₃ - CH₂ - Cl + HCl(Substitution Reaction)

CH₃ – CH₂Cl, CH₃ – CHCl₂ , 

CH₃ – CCl₃ , CH₂Cl – CCl₃, 

CHCl₃ – CCl₃ , CCl₃ – CCl₃

2 Oxygen atoms

a. C₆H₆ 

b. CH₃ – CH₂ – Cl 

c. C₂H₄ 

d. CH₃COOH

The mass of an atom is compared to the mass, of another atom and expressed as a number which shows how many times it is heavier than the other atom. The atomic mass of elements are expressed by considering 1/12 mass of an atom of carbon-12 as one unit.

Fill some more gas.

According to Boyles law, Pv = a constant 

Therefore, P₁V₁ = P₂V₂ 

Here, P₁ = 2atm V₁ = 20L P₂ = 3atm V₂ =? 

∴ 2 × 20 = 3 × V₂

Thus, v₂ = \(\frac{2\times20}{3}\) = 13.3L