West Bengal Group D 2017 Official paper (Previous Year Paper)

The correct answer is option 3 i.e. Mahasweta Debi.

• Aranyer Adhikar (Rights over the Forest) is a Bangla novel by Mahasweta Debi, first published 1977 for which she received Sahitya Akademi Award in 1979.
• It narrates the life and struggle of tribal freedom fighter Birsa Munda.

 

The correct answer is option 2 i.e. 5 hours 30 minutes.

• Greenwich Mean Time or GMT
• The Prime Meridian or the 0° longitude passes through Greenwich near London and the local time calculated at the British Royal Observatory is considered as the accurate time for the Prime Meridian, and is called Greenwich Mean Time or GMT.
• Since the earth rotates from west to east, the places east of Greenwich are ahead of GMT and those to the west are behind it and all the places located on a particular meridian or a particular longitude have the same local time.
• Indian Standard Time or IST
• In order to maintain uniformity of time as far as possible within the territorial limits of a country, the time at the central meridian of the country is taken as the Standard Meridian and its local time is taken as the standard time for the whole country.
• The Standard Meridian is selected in a manner that it is divisible by 15° or 7° 30’ so that the difference between its standard time and the Greenwich Mean Time may be expressed as multiples of an hour or half an hour.
• The Indian Standard Time (IST) is calculated from 82°30’E meridian passing through Mirzapur.
• Therefore, IST is plus 5 hours 30 minutes from the GMT ((82°30’ x 4) (60 minutes=5 hours 30 minutes).
• Similarly, all countries of the world choose the standard meridian within their territory to determine the time within their administrative boundaries.

The correct answer is option 1 i.e. Lord Cornwallis.

• In pre-colonial India, the governments, under the Mughals and other native states, were autocratic in nature, and lacked a separate or formal police system. However, there have been watch guards since time immemorial protecting villages at night.
• Later, under the Mughal rule there were the faujdars who helped in maintaining law and order, amils who were basically revenue collectors but had to contend with rebels, if any and the kotwal was responsible for maintenance of law and order in the cities.
• Even during the dual rule in Bengal, Bihar and Orissa between 1765 and 1772 the zamindars were expected to maintain the staff including thanedars for law and order duties and for maintaining peace, as well as dealing with crime and criminals but very often, the zamidars neglected their duties and were even said to have colluded with dacoits and shared their loot.
• In 1770, the institution of the faujdar and amils were abolished.
• However, in 1774, Warren Hastings restored the institution of faujdars and asked the zamindars to assist them in suppression of dacoits, violence and disorder.
• In 1775, faujdar thanas were established in the major towns of large districts and were assisted by several smaller police stations.
• In 1791 Lord Cornwallis (Governor-General, 1786-1793) organised a regular police force to maintain law and order by going back to and modernising the old Indian system of thanas (circles) in a district under a daroga (an Indian) and a superintendent of police (SP) at the head of a district & relieved the zamindars of their police duties.
• Lord Bentinck (Governor-General, 1828-1835) abolished the office of the SP.
• The collector/magistrate was now to head the police force in his jurisdiction and the commissioner in each division was to act as the SP.
• This arrangement resulted in a badly organised police force, putting a heavy burden on the collector/magistrate.
• Presidency towns were the first to have the duties of collector/magistrate separated.

Calculation:

x(3 - 2/x) = 3/x

⇒ 3x - 2 = 3/x

⇒ 3x - 3/x = 2

⇒ 3(x - 1/x) = 2

⇒ x - 1/x = 2/3

Cubing both sides:

(a – b)³ = a³ - b³ - 3ab(a – b)

(x - 1/x)³ = 8/27

⇒ x³ - 1/x³ - 3(x – 1/x) = 8/27

⇒ x³ - 1/x³ - 3(2/3) = 8/27

⇒ x³ - 1/x³ = (8/27) + 2 = 62/27 or (2 \frac{8}{27})

Short Trick:

If x - 1/x = 2/3 = A (Let)

then x³ - 1/x³ = A³ + 3A

(2/3)³ + 3(2/3) = 62/27 or (2 \frac{8}{27})

The correct answer is option 2 i.e. Warren Hastings.

• Lord Robert Clive had come to Madras (now Chennai) from England in 1743 at the age of 18 and he was appointed Governor of Bengal in 1764.
• The Regulating Act of 1773 designated the Governor of Bengal as the ‘Governor-General of Bengal’. The first such Governor-General was Lord Warren Hastings from 1773 to 1785.
• The Charter Act of 1833 made the Governor-General of Bengal as the Governor-General of India and vested in him all civil and military powers. Lord William Bentinck was the first Governor-General of India from 1828-1835.
• The Government of India Act of 1858 changed the designation of the Governor-General of India to that of Viceroy of India. Lord Canning, thus, became the first Viceroy of India from 1858-1862.

The correct answer is option 3 i.e. China.

• Pulp and paper is a key industry in the world, and paper products are among the major export commodities in many countries.
• North America, Asia, and Europe account for most of the world’s paper and pulp.
• With around 25% share China has the world’s largest paper, and pulp production after its production exceeded that of the United States which had held top position for many years.
• Other leading paper and pulp producers in the world are Germany, Japan, Finland, and Canada.

Formula Used:

1/(n(n + 1)) = ((n + 1) - n)/(n(n + 1)) = (1/n) - (1/n + 1

Calculation:

Above equation can be written as:

1/(1 × 2) + 1/(2 × 3) + 1/(3 × 4) + 1/(4 × 5) + 1/(5 × 6) + 1/(6 × 7) + 1/(7 × 8) + 1/(8 × 9) + 1/(9 × 10)

⇒ (2 - 1)/(1 × 2) + (3 - 2)/(2 × 3) + (4 - 3)/(3 × 4) + (5 - 4)/(4 × 5) + (6 - 5)/(5 × 6) + (7 - 6)/(6 × 7) + (8 - 7)/(7 × 8) + (9 - 8)/(8 × 9) + (10 - 9)/(9 × 10)

In both the above steps, value is same as expression given in the question but we have modified it just to get the general expression:

1/(n(n + 1)) = ((n + 1) - n)/(n(n + 1)) = (1/n) - (1/n + 1)

∴ Step 2 can be written as:

1/1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 + 1/4 – 1/5 + 1/5 - 1/6 + 1/6 – 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 – 1/10

⇒ 1/1 - 1/10 = 9/10  (∵ All the middle term got cancelled out except for the first and last)

The correct answer is option 2 i.e. 1828 A.D.

• Raja Rammohan Roy (1772-1833), often called the the father of Indian Renaissance and the maker of Modern India, was a man of versatile genius who believed in the modern scientific approach and principles of human dignity and social equality and put his faith in monotheism.
• In 1814, he set up the Atmiya Sabha (or Society of Friends) in Calcutta to propagate the monotheistic ideals of the Vedanta and to campaign against idolatry, caste rigidities, meaningless rituals and other social ills.
• He stood for a creative and intellectual process of selecting the best from different cultures, over which, again, he faced orthodox reaction.
• Raja Rammohan Roy founded the Brahmo Sabha in August 1828, later renamed as Brahmo Samaj through which he wanted to institutionalise his ideas and mission.

Given:

30% of A is added to 40% of B

Calculation:

According to the question,

30% of A + 40% of B = 80% of B 

⇒ 30% of A = 80% of B - 40% of B 

⇒ 30% of A = 40% of B

⇒ A/B= 4/3 

∴ Required percentage = (B/A)× 100 = (3/4) × 100 = 75%

Given:

Passing marks = 33%

 got 125 marks and failed by 40 marks

Calculation:

Let the maximum marks be x

He got 125 marks and failed by 40 marks, It means that if he would have got 40 more marks he would have passed.

According to question:

125 + 40 = 33% of x = (33/100) × x

⇒ 165 × 100 = 33x

⇒ x = (165 × 100)/33 = 5 × 100 = 500

∴ Maximum marks = x = 500

Short trick:

If 33% marks = 125 + 40, then maximum marks = 100%

⇒ 100% = (165/33) × 100 = 500

∴ Maximum marks = 100% = 500

The correct answer is option 3 i.e. Large grassland.

Grasslands are known by different names in different regions.

Given:

Distance between Raja and Gopal = 48 km

Speed of Raja and Gopal = 7 km/hr and 5 km/hr respectively

Formula used:

When the two objects travel in the same direction their relative speed = (a – b) km/hr

When the two objects travel in the opposite direction their relative speed = (a + b) km/hr

Were a and b are speeds of two objects in km/hr and (a > b)

Time = distance/speed

Calculation:

As they meet after some time, means they are traveling towards each other and their speed will be added

∴ Time after they meet = 48/(7 + 5) = 48/12 = 4 hours

Calculation:

If the denominators are the same, then the fraction with the greater numerator is the greater fraction. The fraction with the lesser numerator is the lesser fraction.

Reverse the above fraction:

219/2 = 109.5       ---- (1)

343/3 = 114.3       ---- (2)

437/4 = 109.25     ---- (3)

687/6 = 114.5       ---- (4)

Since we have reversed the fraction, the lowest value will be the highest fraction:

437/4 = 109.25 has the lowest value after reversing and hence will have higher fraction:

∴ 4/437 is the greatest fraction.

(x = \frac{2n^2+3n}{5n^2},)

⇒ x = n(2n + 3)/5n²

⇒ x = (2n + 3)/5n

We can see that in the denominator, n is increasing by 5 times whereas, in the numerator, n is increasing by 2 times only.

∴ x will decrease, since denominator is greater than numerator

The correct answer is option 3 i.e. The Bengal Gazette.

• The Bengal Gazette also known as Calcutta General Advertiser started its publication in Calcutta, the centre of colonial India during that time was India’s first newspaper in 1780.
• It was an English language weekly that was started by James Augustus Hicky, an Irishman who acted as the paper’s writer, editor and publisher.

 

The correct answer is option 3 i.e. Bibhutibhushan Bandyopadhyay.

Calculation:

Total questions = 48

Let the correct question be x,

⇒ Positive marks = x × 1

According to question:

x + wrong answered question = 48

Wrong answered question = 48 - x,

⇒ Marks deducted = 0.5(48 - x)

According to the question

Marks obtained = 21,

⇒ 21 = x × 1 - 0.5(48 - x)

⇒ 21 = x - 24 + 0.5x

x + 0.5x = 45

⇒ 1.5x = 45

⇒ x = 30

∴ Number of questions answered wrongly = 48 - 30 = 18

Short Trick:

By hit and trial method

Let 30 question were correct, So marks obtained in correct answer = 30 × 1 = 30

So left 18 question must be wrong and marks deducted = 18 × 0.5 = 9

∴ Marks obtained = 30 - 9 = 21. which satisfies the question.

∴ 18 questions are wrongly answered

Given:

Ratio of 2 numbers is 3 : 4 

Each number is increased by 2

New ratio of numbers = 7 : 9 

Calculation:

Let the numbers be 3x and 4x

According to the question,

(3x + 2)/(4x + 2) = 7/9

⇒ 9(3x + 2) = 7(4x + 2)

⇒ 27x + 18 = 28x + 14

⇒ x = 4

∴ The numbers are 3 × 4 = 12 and 4 × 4 = 16

Concept:

a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc − ca)

Calculation: 

Using the above formula:

a³ + b³ + c³ = (a + b + c)(a² + b² + c² – ab – bc − ca) + 3abc

put x - y = a, y - z = b, z - x = c

⇒ (x - y)³ + (y - z)³ + (z - x)³ = (x - y + y - z + z - x)[(x – y)2 + ( y   - z)2 + (z - x)2 – (x – y) - (y – z)  - (z – x)]  + 3(x - y)(y - z)(z - x)

∴ 3(x - y)(y - z)(z - x) (∵ (x - y + y - z + z - x) = 0 and it cancels out the product )

The correct answer is option 4 i.e. Nuclear fusion reaction.

Explanation:

• Nuclear fusion reactions are the source of energy in the Sun and other stars.
• It takes considerable energy to force the nuclei to fuse. The conditions needed for this process are extreme – millions of degrees of temperature and millions of pascals of pressure.
• Nuclear fusion
• Fusion means joining lighter nuclei to make a heavier nucleus, most commonly hydrogen or hydrogen isotopes to create helium, such as

2H¹ + 2H¹ → ₂He⁴ (+ ₀n¹)

• It releases a tremendous amount of energy, according to the Einstein equation, as the mass of the product is a little less than the sum of the masses of the original individual nuclei.

Hence, solar energy is due to Nuclear fusion reaction.

Additional Information 

• Chemical reactions
• The transformation of chemical substances (known as reactants) into a new chemical substance (known as products) by making and breaking of chemical bonds between different atoms of reactants sometimes in presence of a catalyst (which changes the rate of reaction) is known as Chemical Reaction.
• Combustion Reaction 
• A chemical reaction between substances, usually including oxygen and usually accompanied by the generation of heat and light in the form of flame. Combustion reactions must involve O₂ as one reactant.
• Many combustion reactions occur with a hydrocarbon, a compound made up solely of carbon and hydrogen. The products of the combustion of hydrocarbons are carbon dioxide and water.
• Nuclear fission
• In a process called nuclear fission, the nucleus of a heavy atom (such as uranium, plutonium or thorium), when bombarded with low-energy neutrons, can be split apart into lighter nuclei.
• When this is done, a tremendous amount of energy is released if the mass of the original nucleus is just a little more than the sum of the masses of the individual products.
• The fission of an atom of uranium, for example, produces 10 million times the energy produced by the combustion of an atom of carbon from coal.
• In a nuclear reactor designed for electric power generation, such nuclear ‘fuel’ can be part of a self-sustaining fission chain reaction that releases energy at a controlled rate.
• The released energy can be used to produce steam and further generate electricity.

The correct answer is 1793 A.D.

Key Points

• Lord Charles Cornwallis was the Governor-General of Bengal when the Permanent Settlement was introduced in 1793.
• By the terms of the settlement, the rajas and taluqdars were recognised as zamindars and were asked to collect rent from the peasants and pay revenue to the Company.
• The amount to be paid was fixed permanently, that is, it was not to be increased ever in future.
• It was felt that this would ensure a regular flow of revenue into the Company’s coffers and at the same time encourage the zamindars to invest in improving the land.
• Since the revenue demand of the state would not be increased, the zamindar would benefit from increased production from the land.

The correct answer is option 2 i.e. AIDS.

Concept:

Angle opposite to the greater side is greater.

Calculation:

PQ = 4 - 0.270 = 3.730 cm

QR = 2 + 1.732 = 3.732 cm

PR = 1.35 + 2.381 = 3.731 cm

∴ Side QR is largest hence ∠P or ∠QPR has the maximum measure

The correct answer is Zn.

Key Points

• Only a few elements have been found to be absolutely essential for plant growth and metabolism.
• These elements are further divided into two broad categories based on their quantitative requirements: (i) Macronutrients, and (ii) Micronutrients.
• Macronutrients are generally present in plant tissues in large amounts.
• The macronutrients include carbon (C), hydrogen (H), oxygen (O), nitrogen (N), phosphorous (P), sulphur (S), potassium (K), calcium (Ca) and magnesium (Mg).
• Of these, carbon, hydrogen and oxygen are mainly obtained from CO₂ and H₂O, while the others are absorbed from the soil as mineral nutrition.
• Micronutrients or trace elements are needed in very small amounts.
• These include iron, manganese (Mn), copper (Cu), molybdenum (Mo), zinc (Zn), boron (B), chlorine (Cl) and nickel (Ni).

Additional Information

• In addition to the 17 essential elements named above, there are some beneficial elements such as sodium (Na), silicon (Si), cobalt (Co) and selenium (Se). They are required by higher plants

Given:

HCF of 3 numbers = 6

LCM = 630

Two numbers are 18 and 42

Concept:

HCF is the highest common factor of some given numbers. When we divide the numbers by their HCF, the quotients will be prime to each other.

LCM = HCF × Product of the quotients after dividing the numbers by the HCF

Calculation:

Let, the numbers be 6x, 6y and 6z where x, y and z are prime to each other

Also let, two numbers are 6x and 6y 

⇒ 6x = 18

⇒ x = 3

and 6y = 42

⇒ y = 7

LCM of the numbers = HCF × yxz

⇒ 630 = 6 × 3 × 7 × z

⇒ z = 5

The 3rd number is 6z = 6 × 5 = 30

∴ The third number will be 30.

The formula LCM × HCF = Product of the numbers is only applicable in the case of two numbers.

When it comes to three or more numbers it will not work.

For example:

HCF of 2, 4 and 6 is 2 and LCM = 12

Now LCM × HCF = 12 × 2 = 24

But, the product of the numbers is 2 × 4 × 6 = 48

The correct answer is Radon.

1. Radioactivity is a nuclear phenomenon in which an unstable nucleus undergoes decay and is referred to as radioactive decay.
2. Three types of radioactive decay occur in nature : (i) α-decay (ii) β-decay (iii) γ-decay
3. Noble Gases
4. There are six elements in Group 18 of the Periodic Table - Helium (He), Neon (Ne), Argon (Ar), Krypton (Kr) , Xenon (Xe) and Radon (Rn) which were earlier termed as inert gases or rare gases.
5. Both of these terms are now no longer used as they are not so rare and not completely inactive. They do show some chemical reactivity in the form of some xenon and krypton compounds and hence are termed as noble gases.
6. Occurrence
7. All the noble gases except Radon occur in the atmosphere. Xenon and radon are the rarest elements of the group.
8. Radon is a radioactive element. Radon (²²²Rn) is a decay product of Radium (²²⁶Ra) and is obtained by pumping off the gas from radium chloride solutions.
9. They are the minor constituent of the air, the atmospheric abundance being ~ 1% by volume of which argon is the major constituent. Ne, Ar, Xe and Kr are obtained by fractional distillation of liquid air.
10. Helium and sometimes neon is found in minerals of radioactive origin e.g., pitchblende, monazite, cleveite. The main commercial source of helium is natural gas to an extent of 2-7%.
11. They have closed valence shell electronic configuration due to which they have the least tendency to react.
12. The best-characterised compounds are those of xenon with fluorine and oxygen only under certain conditions and only with highly electronegative fluorine and oxygen.
13. Uses: These gases have several uses. Argon is used to provide an inert atmosphere, helium is used in filling balloons for meteorological observations, neon is used in discharge tubes and fluorescent bulbs.

The correct answer is Nucleus.

The correct answer is option 4 i.e. Black.

Key Points

The correct answer is Dinabandhu Mitra.

• Nil Darpan (The Indigo Planting Mirror) is a Bangla play by Dinabandhu Mitra in 1858–59 and published in Dhaka in 1860, under a pseudonym.
• It depicted the Indigo Revolt of February–March 1859 in Bengal, when farmers refused to sow indigo in their fields to protest against exploitative farming under the British Raj.
• It was also essential to the development of theatre in Bengal where the first play ever commercially staged was "Nildarpan".
• Michael Madhusudan Dutt translated the play into English immediately after it was published.
• Bankim Chandra Chattopadhyay compared Nil Darpan to Uncle Tom's Cabin for its role in arousing people's awareness of the evils of indigo plantations.

Additional Information

The correct answer is Potato.

Modifications of Root

1. For storage of food

• Roots are modified in some plants for storing reserve food materials.
• These modified roots usually are swollen and assume different forms such as spindle-shaped, e.g., radish; top-shaped, e.g., beet, turnip; cone-like, e.g., carrot; indefinite shape, e.g., sweet potatoes.
• Dahlia, Asparagus, Portulaca are some other examples of plants with modified roots for food storage.

2. Nodulated roots

• The roots of pea and other leguminous plants have numerous swollen nodules on fine branches of roots.
• These nodules are formed due to the symbiotic association of Rhizobium (bacterium) that live inside the root cortical cells of the roots.
• They fix nitrogen and an active nodule is pink in colour.

3. For mechanical support

• Roots are modified to provide mechanical support as seen in a banyan tree which has roots growing vertically/obliquely downwards (prop roots); sugarcane/maize in which roots arise from the nodes in the cluster at the base of the stem (stilt roots) and betel/black pepper in which nodes and internodes bear roots which help in climbing.

4. For gaseous exchange

• Pneumatophores or breathing roots are found in plants growing in mangroves or swamps with saline water for exchange of gases.
• They are erect peg-like structures with numerous pores through which air circulates e.g., Rhizophora mangle.
• Modifications of Stem

1. For storage of food

• Stems get modified into underground structures for storage of food as seen in potato (tuber), ginger (rhizome), garlic (bulb), yam (corm).
• Presence of an eye (node) in potato, distinct nodes with internodes and scaly leaves in ginger/yam, a cluster of roots at the base of the reduced stem in garlic/ onion, all indicate that these underground plant parts are modified stem.

2. For vegetative propagation

• Plants besides reproducing sexually also propagate through vegetative parts.
• For this purpose, stems may be modified into a runner (Cyanodon dactylon, Oxalis). Runners are slender prostrate branches arising from axillary buds;
• stolon (e.g., mint, strawberry) which is a slender lateral branch arising from the base of the stem grows upward and then down to develop new daughter plants;
• offset having a single long horizontal internode growing up to some distance and producing a tuft of leaves above and a cluster of roots below at the apex (Eichornia, Pistia) and
• sucker, which arises from the underground part of the stem, grow obliquely and gives rise to a new shoot. (Chrysanthemum, Banana, Pineapple).

3. For protection

• Some modified stem provides protection as thorns which are hard, pointed structures each representing a branch that arises from the axil of a leaf.
• Thorns are found in plants like Duranta, Pomegranate, Acacia, Ber, Prosopis, Bougainvillea, Citrus, etc.

4. For support

• Tendrils are modifications of the stem to provide support to plants, e.g., Vitis, passionflower, Bignonia etc.

5. For photosynthesis

• Stems are also modified into Phylloclade, to facilitate photosynthesis.
• Phylloclades are flattened/cylindrical stem or branches of unlimited growth (Cactus).

Concept:

Each angle of n-sided polygon = ((n - 2) × 180)/n

Number of diagonals of n - sided polygon = n(n - 3)/2

Calculation:

Each interior angle = 150° 

150° = (n - 2) × 180)/n

⇒ 6n - 12 = 5n

n = 12 = Total side of the polygon

∴ Number of diagonals of n - sided polygon = n(n - 3)/2 = 108/2

∴ Number of diagonals of n - sided polygon = 54

Given:

Sum of ages of Mallika and Uttam is 40 years

Five years ago, the ratio of their ages = 7: 8

Calculation:

Let the ages of Mallika and Uttam 5 years ago be  7x and 8x

Present age ratio = (7x + 5): (8x + 5)

According to the question,

Sum of ages of Mallika and Uttam is 40 years  

⇒ 7x + 5 + 8x + 5 = 40

⇒ 15x = 30 or x = 2 years

⇒ Age of Mallika 5 years ago = 7x = 14

∴ Present age of Mallika = 14 + 5 = 19 years.

Short Trick:

Given that 5 years ago, Age of Mallika was 7x

∴ The present age of Mallika will be that option from which if we subtract 5, it will be multiple of 7 i.e. (19 - 5) = 14 which is multiple of 7.

Hence, 19 years is the correct answer.

Concept: 

(a + b)(a − b) = a² – b²

Calculation:

In the product 9547 × 9545 we find that the difference between 9547 & 9545 is 2.

So after 9545 and before 9547, there is one integer 9546.

So we can write 9547 = 9546 + 1 and 9545 = 9546 – 1.

So, 9547 × 9545 = (9546 + 1)(9546 – 1) = 9546² – 1².

∴ It is clearly found that if 1 is added to the product of 9547×9545 the result will be a perfect square.

∴ If we add 1 to 9547 × 9545, sum will become perfect square

Given:

Along the stream means downstream, Downstream speed = 8/1 = 8 km/hour

Against the stream means Upstream, Upstream speed = 2/1 = 2 km/hour

Formula Used:

Let the speed of boat be x km/hour and speed of stream be y km/hour

Downstream speed = (x + y)

Upstream speed =  (x - y)

Calculation:

According to question:

 (x + y) = 8       ---- (1)

 (x - y) = 2       ---- (2)

Solving above equation we get:

2x = 10 or x = 5 km/hour 

⇒ y = 8 - 5 = 3 km/hour  

∴ Speed of stream is 3 km/hr

<br>

Shortcut Trick

If the speed of the Man/Boat/Ship along the stream is u km/hr and the speed against the current is v km/hr, then

Speed of stream = ((u - v)\over2) km/hr

Speed of Man/Boat/Ship = ((u + v)\over2) km/hr

Speed of stream = ((u - v)\over2) km/hr

⇒ ((8 - 2)\over2) = 3 km/hr

∴ Speed of stream is 3 km/hr

The correct answer is option 2 i.e. transportation of worn-out sediments and deposition.

Igneous Rocks

• As igneous rocks form out of magma and lava from the interior of the earth, they are known as primary rocks.
• The igneous rocks (Ignis – in Latin means ‘Fire’) are formed when magma cools and solidifies.
• When magma in its upward movement cools and turns into solid form it is called igneous rock.
• The process of cooling and solidification can happen in the earth’s crust or on the surface of the earth.
• Igneous rocks are classified based on texture which depends upon size and arrangement of grains or other physical conditions of the materials.
• If molten material is cooled slowly at great depths, mineral grains may be very large; sudden cooling (at the surface) results in small and smooth grains; intermediate conditions of cooling would result in intermediate sizes of grains making up igneous rocks.
• Granite, gabbro, pegmatite, basalt, volcanic breccia and tuff are some of the examples of igneous rocks.

Sedimentary Rocks

• The word ‘sedimentary’ is derived from the Latin word sedimentum, which means settling.
• Rocks (igneous, sedimentary and metamorphic) of the earth’s surface are exposed to denudational agents, and are broken up into various sizes of fragments.
• Such fragments are transported by different exogenous agencies and deposited.
• These deposits through compaction turn into rocks. This process is called lithification.
• In many sedimentary rocks, the layers of deposits retain their characteristics even after lithification. Hence, we see a number of layers of varying thickness in sedimentary rocks like sandstone, shale etc.
• Depending upon the mode of formation, sedimentary rocks are classified into three major groups:

1. mechanically formed — sandstone, conglomerate, limestone, shale, loess etc. are examples;
2. organically formed — geyserite, chalk, limestone, coal etc. are some examples;
3. chemically formed — chert, limestone, halite, potash etc. are some examples.

Calculation:

Considering the given equation,

25x² - 20x + t²

The equation can be written as 

(5x)² - 2 × (5x) × 2 + t²

⇒ (5x)² - 2 (5x) × 2 + 2² + t² - 2²

⇒ (5x – 2²)² + t² - 2

If t² = 2, then above equation will be (5x – 2²)² which is perfect square:

∴ For t = ± 2

The correct answer is option 3 i.e. Typhoon.

Key Points

• Thunderstorms
• Thunderstorms develop in hot, humid tropical areas like India very frequently.
• The rising temperatures produce strong upward rising winds. These winds carry water droplets upwards, where they freeze, and fall down again.
• The swift movement of the falling water droplets along with the rising air create lightning and sound. It is this event that we call a thunderstorm.
• Cyclones
• Before cloud formation, water takes up heat from the atmosphere to change into vapour. When water vapour changes back to liquid form as raindrops, this heat is released to the atmosphere.
• The heat released to the atmosphere warms the air around. The air tends to rise and causes a drop in pressure. More air rushes to the centre of the storm.
• This cycle is repeated. The chain of events ends with the formation of a very low-pressure system with very high-speed winds revolving around it.
• It is this weather condition that we call a cyclone. Factors like wind speed, wind direction, temperature and humidity contribute to the development of cyclones.
• The centre of a cyclone is a calm area. It is called the eye of the storm.
• A large cyclone is a violently rotating mass of air in the atmosphere, 10 to 15 km high.
• The diameter of the eye varies from 10 to 30 km. It is a region free of clouds and has light winds. Around this calm and clear eye, there is a cloud region of about 150 km in size. In this region there are high-speed winds (150–250 km/h) and thick clouds with heavy rain.
• Away from this region the wind speed gradually decreases. The formation of a cyclone is a very complex process.
• Tornadoes 
• A tornado is a dark funnel shaped cloud that reaches from the sky to the ground.
• Most of the tornadoes are weak. A violent tornado can travel at speeds of about 300 km/h.
• Tornadoes may form within cyclones.

A cyclone is known by different names in different parts of the world

The correct answer is Water pollution.

• The process in which nutrient-enriched water bodies support a dense plant population, which kills animal life by depriving it of oxygen and results in subsequent loss of biodiversity is known as Eutrophication.
• The organic chemicals are another group of substances that cause water pollution.

Key Points

• Petroleum products pollute many sources of water e.g., major oil spills in oceans; other organic substances with serious impacts are the pesticides that drift down from sprays or runoff from lands.
• Various industrial chemicals like polychlorinated biphenyls, (PCBs) which are used as cleansing solvent, detergents and fertilisers add to the list of water pollutants and are suspected to be carcinogenic.
• Nowadays most of the detergents available are biodegradable but the bacteria responsible for degrading biodegradable detergent feed on it and grow rapidly.
• While growing, they may use up all the oxygen dissolved in water and the lack of oxygen kills all other forms of aquatic life such as fish and plants.
• Fertilisers contain phosphates as additives and the addition of phosphates in water enhances algae growth.
• Such profuse growth of algae covers the water surface and reduces the oxygen concentration in water. This leads to anaerobic conditions, commonly with an accumulation of obnoxious decay and animal death.
• Thus, bloom-infested water inhibits the growth of other living organisms in the water body.

The correct answer is option 1 i.e. 20 Hz - 20,000 Hz.

• The audible sound for human beings extends from the frequency range of about 20 Hz to 20000 Hz (one Hz = one cycle/s).
• Children under the age of five and some animals, such as dogs can hear up to 25 kHz (1 kHz = 1000 Hz). The police use high frequency whistles which dogs can hear but humans cannot.
• Infrasonic sound or Infrasound
• Sounds of frequencies below 20 Hz are called infrasonic sound or infrasound.
• If we could hear infrasound we would hear the vibrations of a pendulum just as we hear the vibrations of the wings of a bee.
• Rhinoceroses communicate using infrasound of frequency as low as 5 Hz. Whales and elephants produce sound in the infrasound range. It is observed that some animals get disturbed before earthquakes. Earthquakes produce low-frequency infrasound before the main shock waves begin which possibly alert the animals.
• Ultrasonic sound or Ultrasound
• Frequencies higher than 20 kHz are called ultrasonic sound or ultrasound.
• Ultrasound is produced by dolphins, bats and porpoises. Moths of certain families have very sensitive hearing equipment. These moths can hear the high frequency squeaks of the bat and know when a bat is flying nearby, and are able to escape capture. Rats also play games by producing ultrasound.
• The ultrasound equipment, familiar to us for investigating and tracking many medical problems, works at frequencies higher than 20,000 Hz.
• Hearing Aid 
• As people grow older their ears become less sensitive to higher frequencies and people with hearing loss may need a hearing aid.
• A hearing aid is an electronic, battery operated device that receives sound through a microphone which converts the sound waves to electrical signals.
• These electrical signals are amplified by an amplifier and the amplified electrical signals are given to a speaker of the hearing aid.
• The speaker converts the amplified electrical signal to sound and sends to the ear for clear hearing.

Given:

Length of train = 300 m

Speed of walking man = 3 km/hour 

Train passed the man in time = 33 seconds

Formula used:

Speed = Distance/Time

Concept:

Length of man is neglected

Calculation:

Let the speed of train be x km/hr

Length of train = 300 meters

Relative speed in the same direction = (x - 3) km/hr

According to the question,

33 = 300/[(x - 3) × 5/18]

11x - 33 = 360

⇒ x = 393/11 km/hr = (35\frac{8}{11}) km/hr

The correct answer is A region of calm climate.

• The sun shines almost vertically on the equator throughout the year ans as a result the air gets warm and rises over the equatorial region and produce equatorial low pressure.
• This belt extends from equator to 10º N and 10º S latitudes.
• Due to excessive heating horizontal movement of air is absent here and only convectional currents are there.
• Therefore this belt is called doldrums (the zone of calm) due to virtual absence of surface winds.
• These are the regions of convergence because the winds flowing from sub tropical high pressure belts converge here. This belt is also known as Inter Tropical Convergence Zone (ITCZ).

The correct answer is Specific Gravity.

Given:

x/y + y/x = 1

Formula Used:

(a³ + b³) = (a + b)(a² + b² - ab)

Calculation:

x/y + y/x = 1

(x² + y²)/xy = 1

⇒ x² + y² = xy

⇒ x² + y² - xy = 0

Multiply by (x + y) both the side we get, 

⇒ (x + y)(x² - xy + y²) = 0(x + y) 

⇒ x³ + y³ = 0

∴ x³ + y³ = 0

Calculation:

Using Assumption:

Let (x + a)/(b + c) = -1 , (x + b)/(a + c) = -1 , (x + c)/(a + b) = -1

(x + a)/(b + c) + (x + b)/(a + c) + (x + c)/(a + b) = -1 + (-1) + (-1) = - 3

If (x + a)/(b + c) = - 1, then x + a = - b - c or x = - (a + b + c)

Similarly, (x + b)/(a + c) = -1

(x + b) = -a - c or x = - (a + b + c)

Similarly, (x + c)/(a + b) = - 1

(x + c) = -a - b or x = - (a + b + c)

We assumed (x + a)/(b + c) = -1 , (x + b)/(a + c) = -1 , (x + c)/(a + b) = -1 and we got each value x equal in all three conditions.

∴ Assumption is correct and x = -(a + b + c)

Given:

Ratio of male and female members of a club = 5 ∶ 6

After 22 female left, New ratio = 6 ∶ 5

Calculation:

Let the number of male be 5x and female be 6x

Number of female left = 22

According to question,

5x/(6x - 22) = 6/5

⇒ 36x = 25x + 132

⇒ 11x = 132 or x = 12

Male members = 5x = 60

∴ The number of male members is 60.

 

Inital ratio of Male to female = (5 ∶ 6) × 6 = 30 ∶ 36

After 22 female left, New ratio Male to female = (6 ∶ 5) × 5 = 30 ∶ 25

Only females left, value of male must remain same so, New ratio = 30 ∶ 25

36 - 25 = 11 unit = number of females left = 22 or 1 unit = 22/11 = 2

Number of male = 30 unit = 30 × 2 = 60

∴ The number of male members is 60.

• The correct answer is option 3 i.e. Total internal reflection of light.

Explanation:

• On hot summer days, the air near the ground becomes hotter than the air at higher levels.
• The refractive index of air increases with its density. Hotter air is less dense and has a smaller refractive index than cooler air. If the air currents are small, that is, the air is still, the optical density at different layers of air increases with height.
• As a result, light from a tall object such as a tree passes through a medium whose refractive index decreases towards the ground.
• Thus, a ray of light from such an object successively bends away from the normal and undergoes total internal reflection, if the angle of incidence for the air near the ground exceeds the critical angle.

• To a distant observer, the light appears to be coming from somewhere below the ground and naturally assumes that light is being reflected from the ground, say, by a pool of water near the tall object such inverted images of distant tall objects cause an optical illusion to the observer. This phenomenon is called a mirage.
• This type of mirage is especially common in hot deserts.
• Some of you might have noticed that while moving in a bus or a car during a hot summer day, a distant patch of road, especially on a highway, appears to be wet. But, you do not find any evidence of wetness when you reach that spot. This is also due to mirage.

Confusion Points

• Students are usually confused about whether mirage is the phenomenon of refraction or total internal reflection.
• It is a case of total internal reflection.

Calculation:

(x – 1)² + (y – 4)² = 0

(x – 1)² = - (y – 4)² = 0 

Put x – 1 = 0 and y – 4 = 0 

x = 1 and y = 4

As only above value can make the equation zero 

∴ x³ + y³ = 1 + 64 = 65

The correct answer is Germanium.

• The seed of the development of modern solid-state semiconductor electronics goes back to the 1930s when it was realized that some solid-state semiconductors and their junctions offer the possibility of controlling the number and the direction of flow of charge carriers through them.
• Simple excitations like light, heat, or small applied voltage can change the number of mobile charges in a semiconductor.
• Note that the supply and flow of charge carriers in the semiconductor devices are within the solid itself, while in the earlier vacuum tubes/valves, the mobile electrons were obtained from a heated cathode and they were made to flow in an evacuated space or vacuum.
• No external heating or large evacuated space is required by the semiconductor devices.
• They are small in size, consume low power, operate at low voltages, and have a long life and high reliability.
• Even the Cathode Ray Tubes (CRT) used in television and computer monitors which work on the principle of vacuum tubes are being replaced by Liquid Crystal Display (LCD) monitors with supporting solid-state electronics.
• Much before the full implications of the semiconductor devices were formally understood, a naturally occurring crystal of galena (lead sulphide, PBS) with a metal point contact attached to it was used as a detector of radio waves.
• On the basis of the relative values of electrical conductivity (σ) or resistivity ( ρ = 1/σ ), the solids are broadly classified as:

1. Metals: They possess very low resistivity (or high conductivity). ρ ~ 10^–2 – 10^–8 Ωm; σ ~ 10² – 10⁸ Sm^–1
2. Semiconductors: They have resistivity or conductivity intermediate to metals and insulators. ρ ~ 10^–5 – 10⁶ Ωm; σ ~ 10⁵ – 10^–6 Sm^–1
3. Insulators: They have high resistivity (or low conductivity). ρ ~ 10¹¹ – 10¹⁹ Ωm; σ ~ 10^–11 – 10^–19 Sm^–1

• The values of ρ and σ have given above are indicative of the magnitude and could well go outside the ranges as well.
• Relative values of the resistivity are not the only criteria for distinguishing metals, insulators and semiconductors from each other. There are some other differences too.
• Semiconductors could be:

1. Elemental semiconductors: Si and Ge
2. Compound semiconductors: Examples are:

• Inorganic: CdS, GaAs, CdSe, InP, etc.
• Organic: anthracene, doped phthalocyanines, etc.
• Organic polymers: polypyrrole, polyaniline, polythiophene, etc.
• Most of the currently available semiconductor devices are based on elemental semiconductors Silicon (Si) or Germanium (Ge) and compound inorganic semiconductors.
• Some semiconductor devices are junction diodes (a 2-electrode device) and bipolar junction transistor (a 3-electrode device).
• However, after 1990, a few semiconductor devices using organic semiconductors and semiconducting polymers have been developed signaling the birth of a futuristic technology of polymer-electronics and molecular-electronics.

Calculation:

Diagonal of square = √2a

If Diagonal increased by 10% then sides also increased by 10%

a' = a + 0.1a = 1.1a

(a')² = (1.1a)² = 1.21a²

Change in area = 1.21a² - a² = 0.21a²

∴ Change in area = 21%

The correct answer is option 2 i.e. Granite.

The correct answer is option 2 i.e. 1897 A.D.

Key Points

• Ramakrishna Paramahamsa
• The teachings of Ramakrishna Paramahamsa (1836- 1886), a poor priest at the Kali temple in Dakshineshwar, on the outskirts of Calcutta (who was known in childhood as Gadadhar Chattopadhyay) found many followers.
• Ramakrishna experienced spiritual trances (ecstasy) from a very early age and is considered to have attained the highest spiritual experience available to Hindus.
• He did not write books, but his conversations with people formed the basis of what were considered his teachings; spoke simply, in the form of parables and metaphors, drawn from the observation of ordinary life and nature. But what he said was of universal relevance.
• Two objectives of the Ramakrishna movement were—

1. To bring into existence a band of monks dedicated to a life of renunciation and practical spirituality, from among whom teachers and workers would be sent out to spread the universal message of Vedanta as illustrated in the life of Ramakrishna, and
2. In conjunction with lay disciples to carry on preaching, philanthropic and charitable works, looking upon all men, women and children, irrespective of caste, creed or colour, as veritable manifestations of the Divine.

• Paramahamsa himself laid the foundations of the Ramakrishna Math with his young monastic disciples as a nucleus to fulfil the first objective.
• The second objective was taken up by Swami Vivekananda after Ramakrishna’s death when he founded the Ramakrishna Mission in 1897.
• The headquarters of the Ramakrishna Math and Mission are at Belur near Calcutta. The two are twin organisations, though legally and financially separate.

Additional Information

• Swami Vivekananda
• Narendranath Datta (1863-1902), who later came to be known as Swami Vivekananda spread Ramakrishna’s message and tried to reconcile it to the needs of contemporary Indian society and emerged as the preacher of neo-Hinduism.
• Certain spiritual experiences of Ramakrishna, the teachings of the Upanishads and the Gita and the examples of the Buddha and Jesus are the basis of Vivekananda’s message to the world about human values.
• He subscribed to the Vedanta which he considered a fully rational system with a superior approach. His mission was to bridge the gulf between paramartha (service) and vyavahara (behaviour), and between spirituality and day-to-day life.
• At the Parliament of Religions held at Chicago in 1893, Swami Vivekananda made a great impression on people by his learned interpretations. The keynote of his opening address was the need for a healthy balance between spiritualism and materialism.
• Envisaging a new culture for the whole world, he called for a blend of the materialism of the West and the spiritualism of the East into a new harmony to produce happiness for mankind.
• Vivekananda gave several lectures on Vedanta in the USA and in London before returning to India in 1897.
• In India he delivered a series of lectures, the focus of which were to infuse into the new generation a sense of pride in India’s past, a new faith in India’s culture, and a rare sense of confidence in India’s future; to bring about a unification of Hinduism by pointing out the common foundation of its sects; to make the educated people see the misery of the downtrodden and work for their uplift by the application of practical Vedanta principles.
• His emphasis was not only on personal salvation, but also on social good and reform.
• In 1897 he founded the Ramakrishna Mission.
• Vivekananda was a great humanist and used the Ramakrishna Mission for humanitarian relief and social work.
• The Mission stands for religious and social reform.
• Vivekananda advocated the doctrine of service—the service of all beings.
• The service of jiva (living objects) is the worship of Siva.

The correct answer is Ishwar Chandra Vidyasagar.

Barnaparichay or Barna Parichay (1855) is a Bengali primer (a small book for teaching children to read) written by social reformer Ishwar Chandra Vidyasagar and considered as "the most influential primer of Bengal".

The correct answer is Earthworm.

• In earthworm, blood glands produce blood cells and haemoglobin which is dissolved in blood plasma and gives the blood-red colour.
• Animals such as sponges and hydra do not posses any circulatory system.
• The water in which they live brings food and oxygen as it enters their bodies and carries away waste materials and carbon dioxide as it moves out. Thus, these animals do not need a circulatory fluid like the blood.
• The blood of snail is blue in colour due to the presence of haemocyanin which contains copper.

Shortcut Trick

Ratio of interior angles ∠A : ∠B = 1 : (2/3) = 3 : 2.

Using Exterior Angle Theorem: ∠A + ∠B = ∠ACD.

Total units = 3 + 2 = 5 units.

5 units = 120° ⇒ 1 unit = 24°.

∠BAC = 3 units = 3 × 24° = 72°.

∴ The correct answer is 72°.

Alternate Method

Given: Exterior angle ∠ACD = 120° and ∠B = (2/3) ∠A.

Formula Used: Exterior angle of a triangle is equal to the sum of the two opposite interior angles.

Calculations:

⇒ Let ∠BAC = x.

⇒ Then, ∠ABC = (2/3)x.

⇒ According to the theorem, ∠BAC + ∠ABC = ∠ACD.

⇒ x + (2/3)x = 120°

⇒ (5/3)x = 120°

⇒ 5x = 120° × 3

⇒ 5x = 360°

⇒ x = 360° ÷ 5

⇒ x = 72°

∴ The correct answer is 72°.

Additional Information

Exterior Angle Theorem

The measure of an exterior angle of a triangle is equal to the sum of its two remote interior angles (∠Ext = ∠Int1 + ∠Int2).

Angle Sum Property

The sum of all interior angles of a triangle is always 180° (∠A + ∠B + ∠C = 180°).

Linear Pair

The sum of an interior angle and its adjacent exterior angle is 180° because they form a linear pair on a straight line.

The correct answer is option 4 i.e. along the lee ward slope.

• When the saturated air mass comes across a mountain, it is forced to ascend and as it rises, it expands; the temperature falls, and the moisture is condensed.
• The chief characteristic of this sort of rain is that the windward slopes receive greater rainfall.
• After giving rain on the windward side, when these winds reach the other slope, they descend, and their temperature rises.
• Then their capacity to take in moisture increases and hence, these leeward slopes remain rainless and dry.
• The area situated on the leeward side, which gets less rainfall is known as the rain-shadow area.
• It is also known as the relief rain.

The correct answer is option 2 i.e. Bass Channel.

• Tasmania is an island state of Australia located 240 km to the south of the Australian mainland and is separated by Bass Channel.
• George Bass left Sydney in an open boat in December 1797 to explore the southeast coast of Australia and reached Western Port in Victoria.
• On his return to Sydney, he reported his conviction that a strait existed, and was instructed to return for conclusive proof.
• Bass and Flinders sailed from Sydney on the Norfolk on 7 October 1798 and returned three months later having successfully circumnavigated Tasmania.
• The Cook Strait is a strait separating the North and South islands of New Zealand, extending northwest to southeast from the Tasman Sea to the south Pacific Ocean.
• The Bering Strait is a strait of the Pacific, which separates Russia and the United States slightly south of the Arctic Circle.
• The Luzon Strait is a strait extending for more than 320 km between the islands of Taiwan (north) and Luzon, Philippines (south) that connects the South China Sea (west) with the Philippine Sea (east) and is on a main shipping route.

Calculation:

2∠A = 3∠B = 6∠C

⇒ LCM(2, 3, 6) = 6

Divide by 6

⇒ 2∠A/6 = 3∠B/6 = 6∠C/6

(1/3)∠A = (1/2)∠B = ∠C

∠A = 3∠C, ∠B = 2∠C

Now, ∠A + ∠B + ∠C = 180° 

6∠C = 180° or ∠C = 30° , Now since ∠A = 3∠C, ∠B = 2∠C

∴ ∠C = 30° = smallest angle

Calculation:

In ΔABC, Construct AD median

In right-angled triangle ΔADB,

∠ADB = 90° 

Using Pythagoras, Hypotenuse AB will always be greater than Perpendicular AD and base BD

∴ AB > AD is always true.

Calculation:

Ratio of its internal to external angle is 7: 2

Interior angle = 7x and Exterior angle = 2x.

2x + 7x = 180° (Interior angle + Exterior angle = 180°)

⇒ 9x = 180

⇒ x = 180/9 = 20

⇒ Interior angle = 140°

⇒ Exterior angle = 40°

Number of sides = 360/exterior angle

∴ Number of sides = 360°/40° = 9

The correct answer is option 2 i.e. Nerve cell.

• In multicellular organisms, cell division is the means of tissue growth and maintenance.
• The growth and division of different cell populations are regulated in different ways, but the basic mechanisms are similar throughout multicellular organisms.
• Most tissues of the body grow by increasing their cell number, but this growth is highly regulated to maintain a balance between different tissues.
• In adults most cell division is involved in tissue renewal rather than growth, many types of cells undergoing continuous replacement.
• Skin cells, for example, are constantly being sloughed off and replaced; in this case, the mature differentiated cells do not divide, but their population is renewed by division of immature stem cells.
• In certain other cells, such as those of the liver, mature cells remain capable of division to allow growth or regeneration after injury.
• In contrast to these patterns, other types of cells either cannot divide or are prevented from dividing by certain molecules produced by nearby cells
• As a result, in the adult organism, some tissues have a greatly reduced capacity to renew damaged or diseased cells.
• Examples of such tissues include heart muscle, nerve cells of the central nervous system, and lens cells in mammals.
• Maintenance and repair of these cells is limited to replacing intracellular components rather than replacing entire cells.

Given:

(a/1 – a) + (b/1 – b) + (c/1 – c) = 1

Calculation:

Add 1 to 1st term on LHS, add 1 to second term and again 1 to last term. Balance this by adding 3 to RHS.

[1 + a/(1 – a)] + [1 + b/(1 – b)] + [1 + c/ (1 – c)] = 1 + 3

⇒ (a + 1 – a)/(1 – a) + (b + 1 – b)/(1 – b) + (c + 1 – c)/(1 – c) = 1 + 3

⇒ 1/(1 – a) +  (1/1 – b) + (1/ 1 – c) = 4

∴ The value of (1/1 – a) + (1/1 – b) + (1/ 1 – c) is 4.

The correct answer is Rabindranath Tagore.

Chitrangada is a dance-drama by Rabindranath Tagore published in 1892, based on the story of Chitrāngadā, the mythological princess of the Kingdom of Manipur and one of the wives of Arjuna according to the Mahabharata.

Given:

(a² + b²)³ = (a³ + b³)²

Formula Used:

(a + b)³ = a³ + b³ + 3a²b + 3ab²

(a + b)² = a² + b² + 2ab

Calculation:

a⁶ + b⁶ + 3a⁴ b² + 3a²b⁴ = a⁶ + b⁶ +2a³b³

Eliminating a⁶ + b⁶ from both sides:

3a²b²(a² +b²) = 2a³b³

⇒ 3a²b²(a² +b²) = 2ab(a²b²)

⇒ 3(a² +b²) = 2ab

⇒ a²/ab + b²/ab = 2/3

⇒ a/b + b/a = 2/3

∴ Required value is 2/3

The correct answer is Less.

• Static and sliding friction
• The impending motion of the body is opposed by static friction.
• When the external force exceeds the maximum limit of static friction, the body begins to move.
• Kinetic friction
• Once the body is in motion, it is subject to sliding or kinetic friction which opposes relative motion between the two surfaces in contact.
• Kinetic friction is usually less than the maximum value of static friction.

The correct answer is option 2 i.e. Machiavelli.

The correct answer is option 1 i.e. the Mediterranean Region.

Key Points

• Most of the east and northeast margins of the continents are covered by temperate evergreen and deciduous trees.
• The west and south-west margins of the continents are different. They have Mediterranean vegetation.
• It is mostly found in the areas around the Mediterranean sea in Europe, Africa and Asia, hence the name.
• This kind of vegetation is also found outside the actual Mediterranean region in California in the USA, south-west Africa, southwestern South America and Southwest Australia.
• These regions are marked for hot dry summers and mild rainy winters.
• Citrus fruits such as oranges, figs, olives and grapes are commonly cultivated here because people have removed the natural vegetation in order to cultivate what they want to.
• Mediterranean trees adapt themselves to dry summers with the help of their thick barks and wax-coated leaves which help them reduce transpiration.
• Mediterranean regions are known as ‘Orchards of the world’ for their fruit cultivation.
• There isn’t much wildlife here.

The correct answer is option 1 i.e. Babar.

• Babur successfully applied the Rumi (Ottoman) method of warfare in The First Battle of Panipat in 1526.
• As the Afghans advanced to attack the right flank; Babur ordered his reserve forces to move.
• The Afghans, greater in number, were unable to move forward nor backward. They were attacked from both sides.
• This created total confusion among the Afghan forces, Babur took full advantage of the situation and his right and left wings soon attacked the Afghan forces from the rear side.
• This was followed with the opening up of fireshots and this completely paralysed the Afghan army.
• Afghan casualties reported by Babur were approximately 20,000 including the Sultan Ibrahim Lodi.
• In the battle it was not Babur's artillery but his 'superb tactics' and the 'mounted archers' played the decisive role, a fact which Babur himself acknowledged.

Concept:

In an Isosceles triangle, Any two sides are always equal:

Distance formula: AB² = (x2 - x1)² + (y2 - y1)²

Calculation:

Points are A(1, 0), B(5, 0) and C(3,3) 

AB² = (5 - 1)² + (0 - 0)² = 16 unit

BC² = (3 - 5)² + (3 - 0)² = 13 unit

AC² = (3 - 1)² + (3 - 0)² = 13 unit

∴ 2 sides, AC and BC are equal and hence points are vertices of an isosceles triangle

Concept:

H.C.F. = H.C.F.(of numerator)/L.C.M.(of denominator)

Calculation:

Convert decimals into fractions:

9/10 , 9/100 , 9/1000

H.C.F. = H.C.F.(9, 9, 9)/L.C.M.(10, 100, 1000)

⇒ H.C.F. = 9/1000 = 0.009

The correct answer is option 4 i.e. H₂SO₄.

A vitriol (from the Latin word vitriolum, meaning "glassy") is a compound containing a sulphate as the crystals of several metallic sulphates resemble pieces of coloured glass.

Calculation:

Let unit digit be x

Ten's digit = 3x

Sum of the number = [10(3x) + 1(x)]

Sum of the number and the unit digit = 32

[10(3x) + x] + x = 32

⇒ 30x + x + x = 32

⇒ x = 1

Putting the value of x in  unit digit and Ten's digit 

Ten's digit = 3

Unit digit = 1

∴ Number is 3 × 10 + 1 = 31

The correct answer is option 4 i.e. 1817 AD.

• Rammohan Roy did much to disseminate the benefits of modern education to his countrymen.
• He supported David Hare’s efforts to found the Hindu College in 1817 in Calcutta (Kolkata).
• Roy’s English school taught mechanics and Voltaire’s philosophy.

Shortcut Trick

For any straight line in the intercept form x/a + y/b = 1, the area of the triangle enclosed by the line and the coordinate axes is given by the formula:

Area = 1/2 × |a × b|

Given equation: x + y = 2

Divide by 2 to convert to intercept form:

⇒ x/2 + y/2 = 1

Here, the x-intercept (a) = 2 and the y-intercept (b) = 2.

⇒ Area = 1/2 × |2 × 2| = 2 square units.

∴ The correct answer is 2 square unit.

Alternate Method

Given:

Equation of the straight line: x + y = 2

Formula Used:

Area of a right-angled triangle = 1/2 × Base × Height

To find the points where the line intersects the coordinate axes:

For the x-intercept, substitute y = 0 in the equation:

⇒ x + 0 = 2

⇒ x = 2

∴ The line intersects the x-axis at point A(2, 0).

For the y-intercept, substitute x = 0 in the equation:

⇒ 0 + y = 2

⇒ y = 2

∴ The line intersects the y-axis at point B(0, 2).

The vertices of the enclosed triangle are the origin O(0, 0), A(2, 0), and B(0, 2).

This forms a right-angled triangle OAB with the right angle at the origin O.

⇒ Base (OA) = 2 units

⇒ Height (OB) = 2 units

⇒ Area of ΔOAB = 1/2 × Base × Height

⇒ Area = 1/2 × 2 × 2

⇒ Area = 2 square units

∴ The correct answer is 2 square unit.

Additional Information

Intercept Form of a Line

The equation of a line with x-intercept 'a' and y-intercept 'b' is written as x/a + y/b = 1. The area of the triangle formed by this line and the axes is 1/2 × |a × b|.

General Line Equation Area

For a general line Ax + By + C = 0, the area of the triangle enclosed by the line and the coordinate axes is C² ÷ (2 × |A × B|).

Area of Triangle with Vertices

For a triangle with vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃), the area is given by 1/2 × |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|.

Calculation:

p/q = (2/3)³ × (3/2)³ 

⇒  p/q = ((2/3) × (3/2))³ 

⇒ p/q = (1)³ = 1

⇒ (p/q)^-10 = (1)^-10

⇒ (p/q)^-10 = 1

Calculation:

We are given Δ triangle ABC, and the side BC is extended to D so that CD = AC.

∠ACB = 72° and ∠BAD = 109°

Join A to D, We obtained a new triangle ACD.

In Δ ACD , AC = CD (Given)

∠ADC = ∠DAC = x (Angle opposite to equal side)

∠DAC + ∠ADC + ∠DCA = 180° (Angle sum property)

x + x + (180 - 72°) = 180°  (∵ ∠DCA + ∠ACB = 180°, ∠DCA = 180° - 72°)

⇒ 2x = 72° or x = 36° = ∠ADC = ∠DAC

∠BAD = 109° 

⇒ ∠BAC + ∠DAC = 109° 

⇒ ∠BAC + 36° = 109° or ∠BAC = 73° 

In Δ ABC 

73° + 72° + ∠ABC = 180° 

∴ ∠ABC = 35°

संस्कृत शब्द हिन्दी में आकर **‘**तद्भव ’ शब्द बनता है। अन्य विकल्प असंगत है। अतः सही विकल्प तद्भव  शब्द है।

Key Points

अन्य विकल्प

दिए गए विकल्पों में से ‘हार’ शब्द ‘आभूषण’ पर्यायवाची नहीं है। अन्य विकल्प आभूषण के पर्यायवाची हैं। अतः सही विकल्प हार है।

स्पष्टीकरण

हार के अन्य पर्यायवाची शब्द हैं – माला, लड़ी

'आभूषण' के अन्य पर्यायवाची शब्द - जेवर, गहना 

विशेष

जिस समास शब्द के पहले संख्या लगता है उसे द्विगु समास कहते हैं। अन्य विकल्प असंगत है। अतः सही विकल्प द्विगु समास है।

अन्य विकल्प

दिए गए विकल्पों में से ‘मनुष्यता’ भाववाचक संज्ञा है।अन्य विकल्प असंगत है। अतः सही विकल्प भाववाचक संज्ञा है।

अन्य विकल्प

दिए गए विकल्पों में से 'बिना विचार किया हुआ विश्वास' वाक्यांश के लिए शब्द अंधविश्वास है। अन्य विकल्प असंगत है। अतः सही विकल्प अंधविश्वास है।

अन्य विकल्प

• अविश्वास – विश्वास का न होना।
• सविश्वास तथा बेविश्वास निरर्थक शब्द हैं।

अन्य महत्वपूर्ण वाक्यांश के लिए एक शब्द

भाषा को शुद्ध-शुद्ध जानने के लिए व्याकरण की आवश्यकता है। अन्य विकल्प असंगत है। अतः सही विकल्प व्याकरण है।

अन्य विकल्प

दिए गए विकल्पों में से ‘इक’ प्रत्यय का प्रयोग कला में नहीं हो सकता है। अन्य विकल्प असंगत है। अतः सही विकल्प कला है।

Additional Information

 

Important Points

दिए गए विकल्पों में से 'कुछ लड़के वहाँ उपस्थित थे' में अनिश्चित संख्यावाचक विशेषण है। अनिश्चित संख्यावाचक विशेषण अर्थात वे विशेषण शब्द जो विशेष्य की निश्चित संख्या का बोध न कराते हों, वे अनिश्चित संख्यावाचक विशेषण कहलाते हैं। अन्य विकल्प असंगत है। अतः सही विकल्प अनिश्चित संख्यावाचक है।

अन्य विकल्प

दिए गए विकल्पों में से ‘घाव पर नमक छिड़कना’ मुहावरे का अर्थ दुखी को और दुखी करना है। अन्य विकल्प असंगत है। अतः सही विकल्प दुखी को और दुखी करना है।

अन्य महत्वपूर्ण मुहावरे

दिए गए विकल्पों में से 'राम-लक्ष्मण वन में गए’ वाक्य भूतकाल के सामान्य भूत का उदाहरण है। सामान्य भूत अर्थात क्रिया के जिस रूप से काम के सामान्य रूप से बीते समय में पूरा होने का बोध हो, उसे सामान्य भूतकाल कहते हैं। अन्य विकल्प असंगत है। अतः सही विकल्प सामान्य भूत है।

अन्य विकल्प