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Uses of KMnO₄: KMnO₄ is used,

i. in volumetric analysis for the estimation of ferrous salts, oxalates, iodides and hydrogen peroxide. 

ii. as a strong oxidising agent in the laboratory as well as an effective oxidising agent in organic synthesis. Alkaline potassium permanganate is used for testing unsaturation in organic compounds and is known as Baeyer’s reagent. 

iii. as a disinfectant and germicide. A very dilute solution of permanganate is used for washing wounds and gargling for mouth sore. It is also used for purifying water of stinking wells. 

iv. for bleaching of wool, cotton, silk and other textile fibres because of its strong oxidizing power and also for decolourisation of oils.

a) Fe – Haber’s process forthe manufacture of NH Ni – Hydrogenation of Oil for the manufacture of vanaspati ghee

b) Because of 

• variable oxidation state of metals 
• ability to form complexes 

c) The outer electronic conguration remains almost same and hence they show horizontal similarity

1. The variable valency is due to the following reason. The energy difference between inner ‘d’ level and outers level is so small. Therefore electrons from ‘d’ level as well as ‘s’ level can take part in reaction.

2. Sc to Cr – Atomic radius decreases due to increase in nuclear charge. Cr to Cu – Due to a balance between the increased nuclear charge and increased screening effect atomic radii become almost constant.

Cu to Zn – Repulsive interactions between the paired electrons in d-orbitals become very dominant towards the end of the period and cause the expansion of electron cloud and thus result in increased atomic size.

3. Lanthanoid, Contraction.

1. The variable oxidation states of transition elements is due to the participation of ns and (n-1) d electrons in bonding,

2. E.g. : The outer configuration of Mn is 3d⁵,4s² it exhibits all the oxidation states from +2 to +7.

i) This is because each time a new electron enters a ‘d’ orbital the nuclear change increases by unity. The shielding effect of a ‘d’ electron is not that effective. Hence, the net electrostatic attraction between the nuclear charge and the outermost electron increases and the ionic radius decreases.

ii) The effect of addition of new shells in 4d series overtakes the effect of increase in nuclear charge.
Thus, electrostatic attraction between nucleus and valence electron decreases and hence atomic size increases.

iii) This phenomenon is due to the intervention of the 4f – orbitals which must be filled before the 5d series of elements begins. The filling of 4f before 5d – orbitals results in a regular decrease in atomic radii called Lanthanoid Contraction due to imperfect shielding of intervening 4f – orbital electrons which compensate for the expected increase in atomic size on moving down the group, with increasing atomic number. The net result of Lanthanoid Contraction is that the second and the third d series exhibit similar radii.

Due to poor shielding by 5f electrons in the Actinoids than that of the 4f electron in the Lanthanoids.

(a) Pauling scale

1. ionic 

2. 6th (sixth) period 

3. Atomic number 

4. 0.99 Å 

5. A⁺ 

6. Dimitri Mendeleev 

7. decreases 

8. Lanthanides, Actinides 

9. bauxite 

10. hydrated ferric oxide

B) 2nd period and 17th group

Cu shows variable valency.

(a) Due to smaller size and high poalrising power of Li+ ion ,LiCl is more covalent than KCl. 

(b) Small cation Li+ ion highly polarizes large anion I^-, LiI is more covalent and hence has lower melting point. 

(c) Due to pseudo inert gas configuration, Cu⁺ is more polarizing than Na⁺ and hence CuCl is more Covalent NaCl. 

Symbol of Na is Na.

1. Mercury 

2. Iron 

3. Copper 

4. Lead

Atomicity refers to the total number of atoms present in one molecule of an element. 

Example: Argon is a noble gas and exists in a free state. One molecule of argon comprises only one atom and therefore the atomicity of argon is one.

Hg – Mercury 

Pb – Lead 

Au – Gold 

Ag – Silver 

Sn – Tin

Application of chromatography -

• In the separation of colours from a dyes.
• In the separation of pigments from natural colours.
• In the separation of drugs from blood for pathological tests.

Basic principle : 

Coloured components of a mixture can be separated by using an Absorbent on which they are absorbed at different rates.

The correct answer is

(c) f_v < f_r

Numbers of girls in a class = 20

Numbers of boys in a class = 15

Total numbers of students in a class = 20 + 15 = 35

(a) Ratio of number of girls to number of boys

= 20/15 = 4/3 = 4 : 3

(b) Ratio of numbers of girls to total number of students

= 20/35 = 4/7 = 4 : 7

(a) Present age of father = 42 years

Present age of son = 14 years

Required ratio = 42/14 = 3/1 = 3 : 1

(b) Two years ago, the age of the son was 12 years and the age of the father was 42 – 2 = 40 years

Required ratio = 40/12 = (4 x 10)/(4 x 3) = 10/3

= 10 : 3

(c) After 10 years, the age of the father and son will be 52 years and 24 years respectively

Required ratio  = 52/24 = (4 x 13)/(4 x 6) =  13/6 = 13 : 6

(d) 12 years ago, the father was 30 years old, At that time age of son = 14 – 12 = 2 years

Required ratio = 30/2 = 15/1 = 15 : 1

(a) 16 : 24 :: 20 : 30

16/24 = 2/3, 20/30 = 2/3

Therefore 16 : 18 = 28 : 30

Hence, True

(b) 21 : 6 :: 35 : 10

21/6 = 7/2, 35/10 = 7/2

Therefore 21 : 6 = 34 : 10

Hence, True

(c) 12 : 18 :: 28 : 12

12/18 = 2/3, 28/12 = 7/3

Therefore 21 : 18 ≠ 28 : 12

Hence, True

(d) 8 : 9 :: 24 : 27

As, 24/27 = (3 x 8)/(3 x 9) = 8/9

Therefore, True

(e) 5 : 239 :: 3 : 4

As, 5.2/3.9 = 4/3

Therefore 5.2 : 3.9 ≠ 3 : 4

Therefore 0.9 : 0.36 = 10 : 4

Hence, True

(f) 0.9 : 0.36 :: 10.4

0.9/0.36 = 90/36 = 10/4

Therefore 0.9 : 0.36 = 10.4

Hence, True

(A) b/(b + g)

From the question,

Number of boys in the class = b

Number of girls in the class = g

Total number of students in the class = b + g

Therefore, The ratio of the number of boys to the total number of students in the class

= b/(b + g)

(a) 15, 45, 40, 120

15/45 = 1/3; 40/120 = 1/3

Therefore, 15 : 45 = 40 : 120

Hence, these are in proportion

(b) 33, 121, 9, 96

24/28 = 6/7, 36/48 = 3/4

Therefore 33 : 121 ≠ 9 : 96

Hence, these are not in proportion

(c) 24, 28, 36, 48

33/121 = 3/11, 9/96 = 3/32

Therefore, 24 : 28 ≠ 36 : 48

Hence, these are not in proportion

(d) 32, 48, 70, 210

32/48 = 2/3, 70/210 = 1/3

Therefore 32 : 48 ≠ 70 : 210

Hence, these are not in proportion

(e) 4, 6, 8, 12

4/6 = 2/3, 8/12 = 2/3

Therefore 4 : 6 = 8 : 12

Hence, these are in proportion

(f) 33, 44, 75, 100

33/44 =3/4, 75/100 = 3/4

Therefore, 33 : 44 = 75 : 100

Hence, these are in proportion

(D) 40 : 25

Consider the given ratios, 2 : 3, 5 : 8, 75 : 121 and 40 : 25.

Simplified form of 2: 3 = 2/3 = 0.67

Simplified form of 5 : 8 = 5/8 = 0.625

Simplified form of 75: 121 = 75/121 = 0.61

Simplified form of 40: 25 = 40/25 = 1.6

Therefore, the greatest ratio among the given ratios is 40 : 25

(C) 27

From the question it is given that,

On a shelf, books with green cover and that with brown cover are in the ratio 2:3

There are 18 books with green cover

So, let us assume the common factor of 2 and 3 be y.

Then, 2x = 18

X = 18/2

Divide both numerator and denominator by 2.

X = 9

Therefore, the number of books with brown cover is = 3x = 3 × 9

= 27

The correct answer is (D) 40 : 25

Let the cost of English textbook = Rs. x 

Twice of it = 2x Rs. 10 less to above = 2x – 10

Then cost of Mathematics textbook y = 2x – 10 is the required linear equation.

The correct answer is (C) 27

(D) 22

From the question it is given that, the ratio of red marbles to blue marbles is 7:4.

Now, let us assume the common factor of 7 and 4 be x.

So, the total number of marbles in the box = 7x + 4x = 11x

Hence number of marbles in the box is a multiple of 11.

Therefore, 11 × 2 = 22

The correct answer is (D) 22

(C) 3 : 80

From the question it is given that,

Total number of pages in the Mathematics textbook for Class VI = 320 pages

The chapter ‘symmetry’ runs from page 261 to page 272

Number of pages contains symmetry chapter = 12

So, the ratio of the number of pages of this chapter to the total number of pages of the book is,

= 12/320

Divide both numerator and denominator by 12.

= 6/160

Again, divide both numerator and denominator by 2.

= 3/80

Therefore, the ratio of the number of pages of this chapter to the total number of pages of the book is 3: 80.

The correct answer is (A) 1:8 

The correct answer is (C) 3:80

The downward force of the elephant’s weight would be applied over a much smaller area if it were balancing on a thumbtack, so the pressure would be greater.

 juicy fruit forests are found in Mediterranen forest 

khejadi sacrifice is related to Forest consevation.

Correct option is  A) 0°

1. (b) 

2. (a) 

3. (d) 

4. (c).

Anaconda is found in south America

The trees, small plants, creepers, vines, bushes, grass etc. which grow in a particular geographical environment are called vegetation.

Grasslands are of two types : 

Hot Zone/Tropical Grasslands : 

• Savanna in Africa 
• Compose grassland in Brazil 
• Llianos grassland in Venezuela. 

Cold Zone/Temperate Grasslands : 

• Steppes grasslands, Eurasia 
• Prairies grasslands, North America
• Veles grassland in South Africa
• Downs grasslands in Australia.

The main types of forests are namely : 

1. Evergreen forest: 

• Hot Zone/Tropical evergreen forest 
• Cold Zone/Temperate evergreen forest

2. Deciduous forest

• Hot Zone/Tropical monsoon forest 
• Cold Zone/Temperate deciduous forest

5. Desert forest 

• Hot Zcme/Tropical desert forest 
• Cold desert forest 

6. Grasslands 

• Hot Zone/Tropical grasslands 
• Cold Zone/Temperate grasslands. 

Hot Zone/Tropical Evergreen Forest:

The unregulated and irregular cutting of forests has resulted in increase in levels of carbon dioxide in atmosphere, scanty rainfall and ecological imbalances which have consequently resulted in reduction of biodiversity, damage to vegetation and threat to survival of wild animals, damage to cottage industry, increase in natural hazards, change in weather and climate and global warming.

1. Khejdli Movement : This movement depicts the love of people of western Rajasthan for nature. The residents of western Rajasthan have immense love for forest and wild animals’ therefore hunting of wild animals is prohibited. The village contractors of Khejdli village in Jodhpur district of Rajasthan were cuttings trees for their personal benefits. In the year 1730, under the supervision of Amrita Devi Bishnoi, 363 men and women embraced trees for their protection and lost their lives. A park has been established as a tribute to their sacrifice. 

2. Chipko Movement : This movement depicts the love of people of Uttarakhand for nature. In the year 1972 women of Uttarakhand embraced trees and showed their discontent for cutting of trees. Hence, this movement is since then called Chipko Movement. This movement was initiated under the supervision of “Gaura Devi” and later on environmentalist “Sundar Lai Bahuguna” also joined the movement. Chipko movement opposed cutting trees, and supported planting trees in the sloping land of Himalayan region so that 60% of Himalayan region got forested.

Appiko Movement was started in Karnataka.

Power = mgh/T = ((100 x 9.8 x 10)/20)W = 49.W

The angular velocity is given by

ω₁ = \(\frac {2πn1}{t}\)rads^-1

Here n₁ = 240,

t = 1 min = 60 s.

∴ Initial angular velocity

= \(\frac{2π \times 240}{60}\)= 8π rads^-1

Initial linear velocity

v₁ = rω₁;

Here, r = 0.2 m

∴ v₁ = 0.20 × 8π

= 5.02 ms^-1

∴ Final angular velocity

ω₂ = \(\frac{2π \times n2}{t}\)

= \(\frac{2π \times330}{60}\) = 11π rads^-1

∴ Final linear velocity v₂ = rω₂

= 0.2 × 11π = 6.91 ms^-1

From the equation of motion, we have,

ω₂ = ω₁ + at

∴ a = \(\frac{w_2-w_1}{t}\)

Here, ω₂ = 11π,

ω₁ = 8 π,

∴ α =\(\frac{w_2-w_1}{t}\)

= \(\frac {11π-8π}{10}\)

=\(\frac {3π}{10}\) = 0.3π rads^-2

Linear acceleration is

a = rα = 0.2 × 0.3π

= 0.188 ms^-2.

The turning effect of a force is τ = r × F. when arm of the wrench is long, r is larger. So, smaller force is required to produce the same turning effect.