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Let the radius of base be r m. Then,

2πr = 22

⇒ 2 x 22/7 x r = 22

r = 7/2

h = 3 m

Total Surface Area

= 2 πrh + πr²

= 2 x 22/7 x 7/2 x 3 + 2/7 x 7/2 x 7/2

= 66 + 38.5

= 104.5 m²

For cylinder A

Radius of base (r) = 14/2 cm. = 7 cm.

Height (h) = 7 cm.

∴ Volume = πr²h

= 22/7 . (7)² . (7)

= 1078 cm³

For cylinder B

Radius of Base (R) = 7/2 cm.

Height (H) = 14 cm.

∴ Volume = πR²H

= 22/4 x 7/2 x 7/2 x 14

= 539 cm³

Hence, volume of cylinder A is 1078 – 539 = 539 cm³ is greater than cylinder B.

Total surface area of cube is 600 m².

(i) Surface area of cubical wooden log

= 6 a²

= 6 x (8)²

= 6 x 64

= 384 cm²

(ii) Surface area of brick

= 2 (lb + bh + hl)

= 2(9 x 5 + 5 x 3 + 3 x 9)

= 2 (45 + 15 + 27)

= 2 (87)

= 174 cm²

(iii) Surface area of box

= 2 (lb + bh + hl)

= 2 (150 x 40 + 40 x 70 + 70 x 150)

= 2 (6000 + 2800 + 10500)

= 2 (19300)

= 38600 cm²

Given,

Radius of cylinder (r) = 7 cm

Total Surface Area = 968 sq. cm.

Height (h) = ?

Total surface area of cylinder = 2πr(h + r)

By putting values

968 = 2 x 22/7 x 7(h + 7)

968 = 44(h + 7)

968/44 = h + 7

22 = h + 7

h = 22 – 7

h = 15 cm.

∴ Height of cylinder = 15 cm.

Side of a cube is √6 cm.

(i) 1.5 cm

Here, a = 1.5 cm

∴ Volume of cube = a³

= (1.5)³

= 1.5 x 1.5 x 1.5

= 3.375 cm³

(ii) 4 cm

Here, a = 4 cm

∴ Volume of cube = a³

= (4)³

= 4 x 4 x 4

= 64 cm³

The volume of cuboid is 140 x 3 m³.

(i) Here, l = 8 cm.

b = 3 cm.

h = 2 cm.

∴ Volume of cuboid

= l x b x h

= 8 x 3 x 2

= 48 cm³

(ii) Here, l = 1 cm.

b = 1 cm.

h = 7 cm.

∴ Volume of cuboid

= l x b x h

= 1 x 1 x 7

= 7 cm³

Surface area of cube of side 3 cm.

= 6 a²

= 6(3)²

= 54 cm²

Surface area of 5 cubes

= 54 x 5

= 270 cm²

Here, l = 10 cm.

b = 4 cm.

h = 4 cm.

∴Lateral surface area of cuboid

= 2 (l + b) h

= 2 (10 + 4) 4

= 112 cm²

Position I

Length = 3x
Breadth = x
Height = x

Position II

Length = x
Breadth = x
Height = 3x

The total surface area is 10 cm².

Total surface area of 5 cubes each of 3 side = 270 cm²

If 5 cubes are joined one after one, then it becomes a cuboid whose length (l)

= 3 cm. + 3 cm. + 3 cm. + 3 cm. + 3 cm.

= 15 cm.

Breadth (b) = 3 cm.

Height (h) = 3 cm.

∴ Surface Area= 2 (lb + bh + hl).

= 2 (15 x 3 + 3 x 3 + 3 x 15)

= 2 (45 + 9 + 45)

= 198 cm²

∴ Hence, surface area will be reduced and the difference = 270 cm² – 198 cm² = 72 cm²

Shivangi medical bill for 15 days in Rs.51

• The parabolic shape of dish reflects the signals to the dish’s focal point.
• Mounted on brackets at the dish’s focal point is a device called a feed horn. 
• This feed horn is a wave guide that gathers the signals at or near the focal point and conducts them to a LNB (Low-Noise Block down converter). 
• The LNB converts these electromagnetic waves into electrical signals and shifts to the receiver i.e. T.V. set. 
• This is all possible only with the help of parabolic dish antennas (concave shape). 
• We know, the T.V. plays an important role in our daily life. So the inventions of the utilities of concave shapes are more helpful for us.

Votes receive is 14810

Seven monts bill is Rs.1260

The given equations are as follows: 

23x + 29y = 98 ………….(i) 

29x + 23y = 110 …………(ii) 

On adding (i) and (ii), we get: 

52x + 52y = 208 

⇒ x + y = 4 …………(iii) 

On subtracting (i) from (ii), we get: 

6x – 6y = 12 

⇒ x – y = 2 …………(iv) 

On adding (iii) and (iv), we get: 

2x = 6 

⇒ x = 3 

On substituting x = 3 in (iii), we get: 

3 + y = 4 

⇒ y = 4 – 3 = 1 

Hence, the required solution is x = 3 and y = 1.

Given pair of linear equations is

29x – 23y = 110 …(i)

And 23x – 29y = 98 …(ii)

On multiplying Eq. (i) by 23 and Eq. (ii) by 29 to make the coefficients of x equal, we get the equation as

667x – 529y = 2530 …(iii)

667x – 841y = 2842 …(iv)

On subtracting Eq. (iii) from Eq. (iv), we get

⇒ 667x – 841y – 667x + 529y = 2842 – 2530

⇒ – 312y = 312

⇒ y = – 1

On putting y = 2 in Eq. (ii), we get

⇒ 29x – 23( – 1) = 110 ⇒ 29x + 23 = 110

⇒ 29x = 110 – 23

⇒ 29x = 87

⇒ x = 3

Hence, x = 3 and y = – 1 , which is the required solution.

We have,

f(x) = 9x³ - 3x² + x - 5 and g(x) = x = \(-\frac{2}{3}\)

Therefore, by remainder theorem when f (x) is divided by g (x) = x - \(\frac{2}{3},\) the remainder is equal to f \((\frac{2}{3})\)

Now, f(x) = 9x³ - 3x² + x - 5

f \((\frac{2}{3})=9(\frac{2}{3})^{3}-3(\frac{2}{3})^{2}+\frac{2}{3}-5\)

\(=\frac{8}{3}-\frac{4}{3}+\frac{2}{3}-5\)

= 2 – 5 = - 3

Hence, the required remainder is - 3.

f (x) = x² + mx + 14

Since, (x + 2) is a factor of f (x), so

f (-2) =0

(-2)² + m (-2) + 14 = 0

18 - 2m = 0

m = 9

Let f (x) = xⁿ +1 

Since, x + 1 is a factor of f (x), so f (-1) = 9 

Thus, 

n is an odd integer.

The last Mughal emperor spent the last years of his life in a jail in Rangoon; along with his wife.

Within hundred years of assuming the Diwani of Bengal, the British became the unquestioned rulers over a large part of the Indian subcontinent. They rarely had to use force to take control over a territory. They were doing profitable trade which was their main objective. They also succeeded in winning the support of most of the local rajas and chieftains. These are some of the possible reasons for the confidence of the British rulers about their position in India before May 1857.

The decision by Bahadur Shah Zafar to support the rebellion had dramatically changed the entire situation. Most of the smaller rulers and chieftains had been ruling over their territories on behalf of the Mughal ruler. They hoped that if the Mughal ruler could once again resume power, they would also be able to rule their own territories once again. People were emboldened by an alternative possibility.

In 1850, a new law allowed an Indian who had converted to Christianity to inherit property of his ancestors. This law made it easier to convert to Christianity.

The British made various changes in their policies after the revolt of 1857. Some of them are as follows:

• The British Crown took the direct responsibility of governance in India.
• All the ruling chiefs and kings were assured that their territory would never be annexed in future. They could pass on their kingdom to their heirs; including the adopted sons. However, they were to accept the suzerainty of the British Crown.
• A decision was taken to increase the proportion of Europeans in the army. More people were to be recruited from among the Gurkhas, Sikhs and Pathans.
• It was decided to respect the local customs and traditions.

• The first structure to successfully use the dome was called the _____________.
  
  Answer: Tomb of Balban was the first structure with a dome to be built in India.
  Note: This question is not clear.
• The two architects who designed New Delhi and Shahjahanabad were _____________ and _____________.
  
  Answer: Edward Lutyens and Herbert Baker
• The British saw overcrowded spaces as _____________.
  
  Answer: Unhygienic
• In 1888 an extension scheme called the _____________ was devised.
  
  Answer: Lahore Gate Improvement Scheme

Answer: (1) True, (2) False, (3) False, (4) True, (5) False

After the 1857 revolt, the British understood the symbolic importance of Delhi for the Indians. Hence, they choose to hold a grand Durbar in Delhi although it was not the capital.

Answer: The British

The famed water canal system of the Old Delhi deteriorated. Growth in population led to further overcrowding. Many havelis were subdivided and sold. The street front of the havelis were turned into shops and warehouses. The drainage system also collapsed.

The process by which a city declines is called de-urbanisation.

Partition resulted in large scale migration of people from both sides of the new border. Most of the Muslims left for Pakistan and Hindus and Sikhs came from Pakistan. The occupations of people changed. A predominantly urban culture which was based on Urdu had changed with the arrival of new people. Many settlement colonies developed for the refugees.

William Jones had a deep respect for ancient cultures, both Indian and Western. He thought it important to discover the sacred texts in order to understand India. He was of the view that a new study of these texts could form the basis of future development in India. He felt that this would not only help the British learn from the Indian culture but would also help Indians rediscover their own heritage.

Mahatma Gandhi gave more emphasis on skill development rather than an ability to read and write. He felt that a person should learn the necessary skills so that he could earn a livelihood. Hence, he wanted children to be taught handicrafts.

They thought that education should be useful and practical. They felt that Indians should be made familiar with the advances being made by the West. They thought that the Indians were uncivilized and needed to be civilized. This could be done only through English education.

Answer: (1) True, (2) True, (3) False, (4) False

Answer:(a) → 2, (b)→ 4, (c) → 1, (d) → 5, (e) → 3

Two vectors are said to be equal, if they have (i) the same magnitude and (ii) the same direction.

When the two vectors are unequal in magnitude and are in the same direction.

Magnitude of them is same, but direction is not same.

The vectors are acting in the same directions.

K₃[Fe(CN)₆] → Pottassium Hexa-cyanoferate

Order of reaction:- The sum of the power of concentration terms of reactants which involved in rate equation of the reaction is called order of reaction. Let us suppose that rate of reaction is given by

Rate = K.C₁^n₁ x C₂^n₂ x C₃^n₃ where n₁ n₂ and n₃ power of the concentration terms C₁, C₂, C₃ respectively.

∴ order of reaction = (n₁ + n₂ + n₃)

(i) A → Product, Rate = KC¹_A

∴ order = 1

(ii) A + B → Product, Rate = KC¹_A x C¹_B

∴ order of reaction = 1 + 1 = 2

(iii) CH₃COOCH₃ + H₂O + H+ → CH₂COOH + CH₃OH

Rate = K.CCH₃COOCH₃ x CH₂O

(∵ CH₂O = constant ∴ order of reaction = 1)

Mineral: Compounds which occur Naturally are called minerals.

Point defect:- Schottky defect, non-schottky defect and impurities are found in a crystal, which is called point defect.

The defect has been found to be present in silver halides such as AgCl, AgBr, AgI etc. because the size of Ag+ ion is quite small. It is rather easy for the ion to leave its position and occupy a new position in the crystal lattice.