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Henery talks Aurora that the change from romantic boyhood to cynical maturity takes fifteen years. But when the change rapidly happens within fifteen minutes, the pace is very fast. The emotions are highly strerred and the person suffers unbearable pains. Such pains are called growing pains.

हैनरी औरोरा को बताता है कि लड़कपन की रोमांचकारी अवस्था से परिपक्व सनकपनं में परिवर्तन 15 वर्ष लेता है। परन्तु जब यह परिवर्तन तेजी से 15 मिनट में ही होता है तो रफ्तार बहुत तेज हो जाती है। भावनाओं में बड़ी हलचल मच जाती है और उस व्यक्ति को बहुत दर्द होते हैं। इन्हीं दर्दो को बढ़ता हुआ दर्द कहा जाता है।

Her husband was late for the dinner because his sister Geogina had urgently called him  to her house.

उसके पति को रात्री के भोजन के लिये देर हो गई थी क्योंकि उसकी बहन जार्जिना ने जरूरी काम के लिये उसको अपने घर बुला लिया था।

Poet regrets saying that he could not travel on both the roads and spend his whole life as a traveller He knows that one must have a destination in life and to this end, one must make choices. The word ‘sorry’ shows poet’s regret. The Poet took the road less travelled.

It depicts the feeling of regret. After making a choice, he was plagued with the idea of not making the other choice. He regrets not adopting the other cause, the other road, the other choice. Even when we do well often making a choice we keep thinking of the games and success which could have been ours if we had taken the other choice.

The poet faces a dilemma that every man faces in his life, i.e., making a right decision. One day during the walk the poet reached bifurcation in the road. Out of the two roads, he had to take only one. He decides to take the second road which was less frequented. Hence, the road is a symbolic of the choice. It is just not possible to make more than one choice and to take both the roads. So the poet decides to take the road less travelled.

Undisputedly the best-loved American poet, Robert Frost was born in San Fransisco on 26th March 1874. He studied as a farmer for eleven years at Derry, inNew Hemisphere. He migrated to England in 1912 where he met Rupert Brooke, Ezra Pound, Edward Thomas and many other distinguished poets. It was in England that his first volume of poetry, ‘A boy’s will’ was published. He returned to the United States in 1915.

The American editors of north of Boston’ established him as a poet. He was awarded the Pulitzer Prize for poetry four times, the medal of the American Academy of Art and letters in 1938, and the poetry society of America in 1941. His works include ‘A boy’s will, ‘North of Boston’, ‘Mountain interval’ ‘New hemisphere’, West-running Brook, Complete poems etc. He died on 29 Jan. 1963.

There were two roads feeing the poet and he had to make a choice He stood at the fork for a long time trying to decide. He looked down one road as far as he could see till a point where it bent and he could use no further. Then he decided to take another road.

The poem is written in response to the indecisiveness of the poet. On a deeper level it is about the choices one makes in life and the dilemma one faces in making these choices.

The poet, travelling alone one day, reaches a place where the road divides into two. He had to decide about which road to choose because both seen equally good. He stands there for a long time trying to decide. Both the roads have not been travelled on that day because both are covered with leaves After contemplating for a long time, he decides to take the road not trodden by anyone’s footsteps.

After choosing the road that appears to be less travelled. He decides to leave the other road for another day even though he knows that once a choice has been made one travels farther and farther along the chosen road until it is impossible to come back.

The poet says that after ages he will be telling his story of how he chose the road less travelled and that had made all the difference in his life. He feels sorrow because he will have regrets that he did not choose the other road. It is human nature to constantly wonder whether the other path would have given more fame, happiness and money. What the poet is saying is that it is difficult to take a decision and then, after making a choice, the alternative seems to be as good, if not better than the chosen one.

(a) Friedrich Hayek

Planning Commission was set up in the year 1950.

(c) Conjugation

Correct Answer is: (c) Sexual

Firstly, we will make groups of three digits starting from the rightmost digit of the number as bar1 bar(331).

There are 2 groups, 1 and 331, in it.

Considering 331,

The digit at its unit place is 1. We know that if the digit 1 is at the end of a perfect cube number, then its cube root will have its unit place digit as 1 only. Therefore, the unit place digit of the required cube root can be taken as 1. Taking the other group i.e., 1,

The cube of 1 exactly matches with the number of the second group. Therefore, the tens digit of our cube root will be taken as the unit place of the smaller number whose cube is near to the number of the second group i.e., 1 itself. 1 will be taken as tens place of the cube root of 1331.

Hence, 3√1331 = 11

The cube root of 4913 has to be calculated.

We will make groups of three digits starting from the rightmost digit of 4913, as bar4 bar(913). The groups are 4 and 913.

The number 913 ends with 3. We know that if the digit 3 is at the end of a perfect cube number, then its cube root will have its unit place digit as 7 only. Therefore, the unit place digit of the required cube root is taken as 7. Taking the other group i.e., 4,

We know that, 1³ = 1 and 2³ = 8

Also,1 < 4 < 8

Therefore, 1 will be taken at the tens place of the required cube root.

Thus,√4913 = 17

The cube root of 12167 has to be calculated.

We will make groups of three digits starting from the rightmost digit of the number 12167, as bar(12) bar(167).The groups are 12 and 167.

Considering the group 167,

167 ends with 7. We know that if the digit 7 is at the end of a perfect cube number, then its cube root will have its unit place digit as 3 only. Therefore, the unit place digit of the required cube root can be taken as 3. Taking the other group i.e., 12,

We know that,2³ = 8 and 33 = 27

Also, 8 < 12 < 27

2 is smaller between 2 and 3. Therefore, 2 will be taken at the tens place of the required cube root..

Thus, √312167 = 23

The cube root of 32768 has to be calculated.

We will make groups of three digits starting from the rightmost digit of the number 32768, as bar(32) bar(768).

Considering the group 768,

768 ends with 8. We know that if the digit 8 is at the end of a perfect cube number, then its cube root will have its unit place digit as 2 only. Therefore, the unit place digit of the required cube root will be taken as 2. Taking the other group i.e., 32,

We know that, 3³ = 27 and 4³ = 64

Also, 27 < 32 < 64

3 is smaller between 3 and 4. Therefore, 3 will be taken at the tens place of the required cube root.

Thus, 3√32768 = 32

-1331 = 3√-1331

= 3√-(11 x 11 x 11)

= -11

-512 = 3√-512 

= 3√-(8 x 8 x 8) 

= -8

64 x 27 = 3√64 x 27

= 3√(4 x 4 x 4) x (3 x 3 x 3)

= 4 x 3 = 12

27/64 = 3√27/64 = 3√((3 x 3 x 3)/(6 x 6 x 6))

= 3/4

-216 = 3√-216

= 3√-(6 x 6 x 6) 

= -6

We know that:

The minute hand of a watch describes an angle of 4 right angles i.e. 360° in one hour. the time taken by minute hand to trace an angle of 1° = \(\frac { 1 }{ 360^o }\) hours the time taken by minute hand to trace 120° angle

= \(\frac { 1 }{ 360^o }\) x 120

= \(\frac { 1 }{ 3 }\) hours

= \(\frac { 1 }{ 3 }\) x 60 minutes

= 20 minutes.

∵ 180° = π radians

∴ 1° = π/180° radians

(i) 45 = π/180° x 45° radians

= π/4 radians

(ii) 120° = π/180° x 120° radians

= 2π/3 radians

(iii) 135° = π/180° x 135° radians

= 3π/4 radians

(iv) 540° = π/180° x 540° radians

= 3π radians

We know that:

The time taken by the minute hand of a clock in tracing 4 right angles is 1 hour.

So, we convert 30 right angles in terms of the multiple of 4 right angles

i.e. 30 right angles

= 7 x (4 right angles) + 2 right angles

= 7 x 1 hr + \(\frac { 1 }{ 2 }\) hr = 7\(\frac { 1 }{ 2 }\) hours

= 7 hours 30 minutes

Hence, time = 7 : 30 p.m.

Answer is (B) 135°

Correct answer is (C) Ti-plasmid

Correct Answer is: (C) Ti-plasmid 

The most common cause is the Ti plasmid (tumour-inducing plasmid), which is carried by soil bacteria of the Agrobacterium group.

Emulsions are colloids in which both the dispersed phase and dispersion medium are liquids. Emulsions are of two types-oil-in-water type and water-in-oil type.

(i) Oil-in-water type emulsion:- In this type of emulsions the dispersed phase is oil and the dispersion medium is water. Examples are mild in which fats are dispersed in water, vanishing cream is an another example.

(ii) Water-in-oil type emulsion:- In this type of emulsions water in dispersed in oil which acts as dispersion medium.

Examples are butter, cold-liver oil, cold cream etc.

Correct Answer is: (a) 6

These ligands occupy two positions around the central metal atom. There are 3 oxalate ligands, so it will occupy = (2 × 3) = 6 positions around the central metal atom. Hence, the coordination number of chromium in given complex is 6

The process of purifying the crude metals is called refining of metals. Low melting metals like tin and lead are refined by liquation since the melting points of these metals are lower than those of refining. The impure metals are melted on a sloping hearth of a furnace. Pure metals melt earlier than impurities and flow on the loping surface leaving behind the unmelted impurities.

One molal solution is solution in which one mole of solute is present in 1000g of solvent.

Due to smaller +I effect of the alkyl group in a aldehydes as compared to large +I effect of two alkyl groups in ketones, the magnitude of positive charge on the carbonyl carbon is more in aldehydes that ketones. As a result nucleophilic addition reaction occur more readily in aldehydes that in ketones.

As the number and size of the alkyl groups around C = O group increases, there is steric hindrance which makes the attack of the nucleophile on the carbonyl carbon more difficult.

Antibiotics are chemical substances produced wholly or partially by chemical synthesis which in low concentration either will or inhibit the growth or micro-organisms by inter vening the metabolic process. Penicillin was the first antibiotic discovered.

Diamminedichloroplatinum (IV) chloride.

The basic theme of organisation in the periodic table is to classify the elements.

Correct option is C) a, c

Correct option is A) Mg(OH)₂

Correct option is B) Milk of Magnesium

Correct option is B) Bases
 

Correct option is A) Calcium hydroxide

Correct option is A) Glass

Correct option is C) Sodium Hydroxide

Correct option is B) Methyl orange

Correct option is A) Red

Correct option is A) Yellow

The clement with atomic number 19 is in 4th period and first group of the periodic table. 

Reason : 

1. Electronic configuration : 1s² 2s² 2p⁶ 3s² 3p⁶4s or [Ar]4s¹ 

2. The differentiating electron enters into 4th shell. Hence it belongs to 4th period. 

3. The differentiating electron is in ‘s’ orbital. So it belongs to ‘s’ block. 

4. The outermost orbital has only one electron. Hence it belongs to first group.

Correct option is  C) d_xy

An atom of an element ‘X’ having atomic number ‘Z’ and atomic mass number ‘A’ is represented as : \(^A_ZX\) 

22.4 L of the gas at STP will weigh 1.97 × 22.4 = 44.1 g ie., molecular mass = 44.1.

Hence vapour density V = mass/volume = 44.1/2 = 22.05

• Atomic number is defined as the number of protons present in the nucleus of an atom of a particular element.
• Atomic number is represented by Z.
• An atom is electrically neutral. Hence, the number of protons equals to the number of electrons. In other words, the atomic number of an atom is equal to the number of electrons.

∴ Atomic Number (Z) = Number of protons = Number of electrons.

• The sum of the total number of protons and neutrons present in the nucleus of an atom is called the atomic mass number of that atom.
• Atomic mass number is represented by A
• Mass number (A) = Number of protons (Z) + Number of neutrons (N) = Total number of nucleons 

∴ A = Z + N OR N = A – Z

22.4 L of the gas at STP will weigh

= 1.97 x 22.4 = 44.1g

i.e. molecular mass = 44.1