Explore topic-wise InterviewSolutions in Current Affairs.

A greenwood turns to a yellow wood in j autumn. The stage in our life that can be compared to that season is middle age.

Correct option is A) Situation

Some basic concepts are to be understood for this. 

Those are :

1. site 

2. situation and 

3. the history of the place.

Correct option is A) Sedentary

Correct option is D) Immigrants

Correct option is A) Urban

C) City / Class 1 city

(A) Karnataka

India is the largest producer of oil seeds in the world.

B) Maharashtra

D) Andhra Pradesh

Gene therapy can completely cure this disease. 

The disease is due to ADA (Adenosine deaminase) deficiency. 

Lymphocytes from the blood of the patient are grown in a culture. A functional ADA cDNA is introduced into these lymphocytes, which are subsequently returned to the patient. The permanent cure is done by introducing ADA cDNA into cells at early embryonic stages.

(c) representative assemblies started insisting on a share in the decision-making process.

(a) representatives are ultimately responsible to the people.

These words tell us about Mr Lamb’s positive thoughts and behaviour. He is equally interested in man and Nature. He is satisfied with what life gives him. He looks at and loves everything.

ये शब्द हमें श्रीमान् लैम्ब के सकारात्मक विचारों और व्यवहार के विषय में बताते हैं। उन्हें प्रकृति और मानव में समान रूप से रुचि है। जीवन उन्हें जो कुछ भी देता है वह उससे सन्तुष्ट हैं। वह हर चीज को देखते हैं और उससे प्यार करते हैं।

House of Commons

Indian National Congress (it is a political party)

Answer s: (d) Aryabhatt

By saying so, Mr Lamb means that all people in the world are not indifferent to the sufferers. There are sensible people too who behave normally with physically impaired persons. He wants Derry to see such people’s company.

ऐसा कहने से, श्रीमान् लैम्ब का अभिप्राय है कि संसार में सभी लोग पीड़ितों के कष्ट के प्रति उदासीन नहीं होते हैं। समझदार लोग भी हैं जो शारीरिक रूप से अक्षम लोगों के प्रति सामान्य व्यवहार करते हैं। वह (श्रीमान् लैम्ब) चाहते हैं कि डैरी ऐसे लोगों का साथ ढूंढे।

White stone (marble) has been used in the construction of Amravati Stoopa.

Representative Democracy

1. dime 

2. Bharhut Stoopa 

3. Kushan 

4. Maharishi Valmiki 

5. Shipakiram (Tamil)

The Amravati Stoopa is situated in Amravati town situated on the banks of river Krishna in present day Andhra Pradesh.

The semi-circular dunes built of clay and bricks erected over the earthly remains of great saints are known as Stoopas.

The Wahan (Vehicle) of God Bhairwa is Dog.

Vegetation and wild life.

term “royal centre” is an appropriate description for the part of the city The for which it is used because the Royal center had more than 60 temples. Most of these temples were constructed by the ruler of Vijayanagara Empire to express their supremacy. The royal centre had 30 palaces. These were made of perishable material. 

A brief description of the building of Royal centre are as given below: 

(i) One of the most beautiful buildings in the royal centre is the Lotus Mahal. It was named by British travellers in the nineteenth century. While the name is certainly romantic, historians are not quite sure what the building was used for. One suggestion, found in a map drawn by Mackenzie, is that it may have been a council chamber, a place where the king met his advisers. 

(ii) Most temples were located in the sacred centre. One of the most spectacular of these is the Hazara Rama Temple. This was probably meant to be used only by the king and his family.

Stars twinkle on account of atmospheric refraction.

Light travels from rarer to denser medium. Rays always bend towards the normal.

There was a civil war in America in 1861.

कीमत अर्थात् वस्तु या सेवा के बदले में चुकाया जानेवाला मुद्राकीय इकाई का प्रमाण ।

मूल्य अर्थात् वस्तु या सेवा का अन्य वस्तु के प्रमाण में विनिमय मूल्य ।

(i) 81 = 3 × 3 × 3 × 3

Here, one 3 is left which is not in a triplet.

If we divide 81 by 3, then it will become a perfect cube.

Thus, 81 ÷ 3 = 27 = 3 × 3 × 3 is a perfect cube.

Hence, the smallest number by which 81 should be divided to make it a perfect cube is 3.

(ii) 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2 Here, one 2 is left which is not in a triplet.

If we divide 128 by 2, then it will become a perfect cube. Thus, 128 ÷ 2 = 64 = 2 × 2 × 2 × 2 × 2 × 2 is a perfect cube.

Hence, the smallest number by which 128 should be divided to make it a perfect cube is 2.

(iii) 135 = 3 × 3 × 3 × 5

Here, one 5 is left which is not in a triplet.

If we divide 135 by 5, then it will become a perfect cube.

Thus, 135 ÷ 5 = 27 = 3 × 3 × 3 is a perfect cube.

Hence, the smallest number by which 135 should be divided to make it a perfect cube is 5.

(i) 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3

Here, one 3 is left which is not in a triplet.

If we divide 192 by 3, then it will become a perfect cube. Thus, 192 ÷ 3 = 64 = 2 × 2 × 2 × 2 × 2 × 2 is a perfect cube.

Hence, the smallest number by which 192 should be divided to make it a perfect cube is 3.

(ii) 704 = 2 × 2 × 2 × 2 × 2 × 2 × 11

Here, one 11 is left which is not in a triplet.

If we divide 704 by 11, then it will become a perfect cube.

Thus, 704 ÷ 11 = 64 = 2 × 2 × 2 × 2 × 2 × 2 is a perfect cube.

Hence, the smallest number by which 704 should be divided to make it a perfect cube is 11

i. 2x + y = 5 …(i) 

3x – y = 5 …(ii)

 Adding equations (i) and (ii), 

2x + y = 5 

3x – y = 5 

5x = 10 

∴ x = 10/5 

∴ x = 2 

Substituting x = 2 in equation (i),

 2(2) + y = 5

 4 + y = 5 

∴ y = 5 – 4 = 1 

∴ (2, 1) is the solution of the given equations.

ii. x – 2y = -1 

∴x = 2y – 1 … .(i) 

∴ 2x – y = 7 ….(ii)

 Substituting x = 2y – 1 in equation (ii), 

2(2y – 1) – y = 7 

∴ 4y – 2 – y = 7 

∴ 3y = 7 + 2

∴ 3y = 9 

∴ y = 9/3 

∴ y = 3 

Substituting y = 3 in equation (i),

 x = 2y – 1 

∴ x = 2(3) – 1

 ∴ x = 6 – 1 = 5 

∴ (5, 3) is the solution of the given equations.

i. 2x – y = 5 

∴ -y = 5 – 2x 

∴ y = 2x – 5 …(i)

 3x + 2y = 11 ……(ii) 

Substituting y = 2x – 5 in equation (ii), 

3x + 2(2x – 5) = 11 

∴ 3x + 4x- 10= 11 

∴ 7x = 11 + 10 

∴ 7x = 21 

∴ x = \(\frac{21}{7}\)

∴ x = 3 

Substituting x = 3 in equation (i), 

y = 2x – 5 

∴ y = 2(3) – 5 

∴ y = 6 – 5 = 1 

∴(3,1) is the solution of the given equations.

ii. x – 2y = -2 

∴ x = 2y – 2 …(i)

 x + 2y = 10 …..(ii) 

Substituting x = 2y – 2 in equation (ii),

2y – 2 + 2y = 10 

∴ 4y = 10 + 2 

∴ 4y= 12 

∴ y = \(\frac{12}{4}\)

∴ y = 3 

Substituting y = 3 in equation (i),

 x = 2y – 2 

∴ x = 2(3) – 2 

∴ x = 6 – 2 = 4 

∴ (4, 3) is the solution of the given equations

i. x + y = 11 

∴ x = 11 – y …(i) 

2x – 3y = 7 …….(ii) 

Substituting x = 11 -y in equation (ii),

2(11 – y) – 3y = 7 

∴ 22 – 2y – 3y = 1 

∴ 22 – 5y = 7

∴ 22 – 7 = 5y 

∴ 15 = 5y 

∴ y = \(\frac{15}{5}\)

∴ y = 3 

Substituting y = 3 in equation (i),

x = 11 – y 

∴ x = 11 – 3 = 8 

∴ (8, 3) is the solution of the given equations.

ii. 2x + y = -2 …(i) 

3x – y = 7 …(ii) 

Adding equations (i) and (ii),

 2x + y = -2

+ 3x – y = l 

5x = 5 

∴ x = \(\frac{5}{5}\)

∴ x = 1 

Substituting x = 1 in equation (i),

 2x + y = -2 

∴ 2(1) +y = -2 

2 + y = -2 

∴ y = – 2 – 2 

∴ y = -4 

∴ (1, -4) is the solution of the given equations.

Correct answer is

(D) (mn + 7pq)

Hint :

Here, square root of the first term = mn 

Square root of the last term = 7pq 

∴ Required binomial = (mn + 7pq)²

No because there is air in the tumbler. Air need space to occupy so water cannot enter into the tumbler.

Matter needs space to occupy and has mass.

When zinc reacts with dilute solution of strong acid, it forms salt and hydrogen gas is evolved.
Zn + 2HCl ------> 2nCl₂ + H₂
When a burning splinter is brought near the mouth of the test tube, the gas burns with a pop sound.

Small displacement of sigma bond pair of electrons towards substituent.

The answer is (CH₃)₂CH-

Alkyl group (methyl group)

Small displacement of sigma bond pair of electrons away from substituent.

Shift of electron pair away from substituent in conjugate system.

Oxygen exists as diatomic molecule (O₂) whereas sulphur exists as polyatomic molecule (S₈). The van der Waals forces of attraction between O₂ molecules are relatively weak owing to their much smaller size. The large van der Waals attractive forces in the S₈ molecules are due to large molecular size. Therefore oxygen has low m.p. and b.p. as compared to sulphur.

• Oxygen has a smaller atomic size than sulphur. 
• It is more electronegative than sulphur. 
• It has a larger electron density. 
• Due to high electron density, oxygen does not accept the incoming electron easily and therefore has less electron gain enthalpy than sulphur.

(i) n = 4, l = 3

\(\therefore\) Sub shell is 4f

for l = 3, m_l , −3, −2, −1, 0, +1, +2, +3

Each orbital has maximum two electrons. 

So, maximum number of electron in 4f sub shell is 14.

(ii) n = 3, l = 1 and m_l = −1

According to this set of quantum numbers, the outermost subshell is 3p_x

\(\therefore\) Electronic configuration is 1s² , 2s²2p⁶ , 3s²3p_x1

Thus, maximum number of electrons is 13.

(iii) n = 4 

Maximum number of electrons 

= 2n² 

= 2(4)² 

= 32

• The atomic radii of group 18 elements is larger than the atomic radii of group 17 elements. 
• Down the group from He to Xe, atomic radii increases due to increase in the number of quantum shells. 
• The atomic radii increase in the order He < Ne < Ar < Kr < Xe