Explore topic-wise InterviewSolutions in Current Affairs.

1. There is no life zone without ozone. 

2. Avoid Plastics, use biodegradable materials. 

3. Save Water. 

4. Don’t pollute water resources like Pond, Lake, river, sea and oceans. 

5. Let’s make our world a greener place.

We are living in the polluted environment. If we do not take any action, we may suffer in the future. We are the main cause to pollute the environment, So we should correct it. Everyone should be aware of this, without the environment man cannot live. So purifying the environment is our responsibility. Planting of more trees (plants), keeping our surroundings clean, and minimize the use of non-bio- degradable substances like plastic, chemicals, glass etc., To save ourselves, we should save the environment. The poem gives this message.

The clock and I (poet) are the two characters in the poem

Now-a-days the number of industries are more. These more number of manufacturing industries expel the smoke, Airlines emitting the caustic gases from fuels, the nuclear weapons destruction cause more air pollution. Cutting of trees also added up the air pollution. Air pollution is increased day by day.

No, the clock doesn’t stop its exercise.

The pesticides and insecticides that are sprayed on the crops will mix with our food grains. Oceans are filled with crude oils and sea-creatures (life) are contaminated. Thus the terrestrial and seafood get contaminated.

The unwanted things like broken bottles, half burnt glass pieces, bundles of newspaper causes the environmental pollution. The Plastics, glass pieces are non-biodegradable substances, so it might cause soil pollution. Cutting of trees and making concrete buildings spoil the environment.

Our lives are at risk. It may cause the destruction of man.

The industrial waste fumes, Airline emission, fuel smoke, the nuclear weapons destruction, invisible rays etc., are polluting the environment.

Both the speaker and the clock can swing their arms. And they both exercise daily.

The poem ‘The Gymnastic Clock’ is written by M.C. Davies. It was a little cock and it was friends with him. It talks to him in a very clear and plain way. Each morning as the poet rises, the clock reminds him not to forget his exercises. The clock makes an example of itself and shows the poet how to exercise. It asks the poet to raise both his hands above his head and then lower them slowly. It cautions him not to get tired and stop exercising.

It boasts that it does such exercise all day. The poet feels humiliated and defining himself says to the clock right in its face that he wouldn’t boast like the clock and like it he can also swing his arms around too. So both the clock and the poet look at each other for a long time. Now both the clock and the poet rise up early in the morning and just exercise and exercise.

This poem emphasizes that man should always be active and industrious like a clock to stay healthy, wise and strong.

Poisons and insecticides are sprayed on our food.

Yes, the speaker or poet says the above line in the competitive mood.

He is boasting about the continuous exercise of his hands above and below.

you is referred to clock.

It does it in the morning, (all the time)

The clock raises its hands above and lowers it very slowly all the day.

Here ‘I’ is referred to clock.

The clock reminds the speaker not to forget his daily exercise.

As the root mean square speed of hydrogen is much less than that of nitrogen, it easily escapes from the earth’s atmosphere. 

In fact, the presence of nonreactive nitrogen instead of highly combustible hydrogen deters many disastrous consequences.

The correct answer is 1/2

(i) Bag I (ii) Bag II (iii) Bag III (iv) Bag I (v)Bag I (vi) Bag I (vii) Bag II (viii) Bag I (ix) Bag I (x) Bag I

Given,

ABCD is a rectangle with AB = 14 cm and BC = 7 cm

It’s seen that,

The area of shaded region = Area of rectangle ABCD + 2 x area of semi-circle with AD and BC as diameters – area of the semi-circle with DC as diameter

= 14 x 7 + 2 x 1/2 π(7/2)² – 1/2 π(7)²

= 98 + 22/7 x (7/2)² – 1/2 (22/7)(7)²

= 98 + 77/2 – 77

= 59.5 cm²

The nature of medium for salivary amylase to act on food component is alkaline.

The contractions of the stomach muscles squeeze and mix the food with the acids and juices of the stomach.

(b) 800

Let one number = x 

Then, the other number = 4000 – x

Given, 10% of x = 6\(\frac{2}{3}\)% of (4000-x)

\(\Rightarrow\) \(\frac{10}{100}\) x x = \(\frac{20}{3}\) x \(\frac{1}{100}\) x (4000-x)

\(\Rightarrow\) 10x = \(\frac{20}{3}\) x 4000 - \(\frac{20X}{3}\)

\(\Rightarrow\) 10x + \(\frac{20X}{3}\) = \(\frac{20}{3}\) x 4000

\(\Rightarrow\) \(\frac{50X}{3}\) = \(\frac{20}{3}\) x 4000

\(\Rightarrow\) x = \(\frac{20\times4000}{50}\) = 1600

The two numbers are 1600 and 2400.

\(\therefore\) Their difference = 2400 – 1600 = 800.

Let the required ratio be x : y. Then, \(\frac{15X}{100}\) + \(\frac{40y}{100}\) = \(\frac{30(X+y)}{100}\)

\(\therefore\) 15x + 40y = 30x + 30y

\(\Rightarrow\) 10y = 15x  \(\Rightarrow\) \(\frac{X}{y}\)= \(\frac{10}{15}\) = \(\frac{2}{3}\) \(\Rightarrow\) x : y = 2 : 3.

(c) 3 : 1

Let the two numbers be x and y.

Then, x²  - y² = 80% of (x² + y²)

\(\Rightarrow\) x² - y² = \(\frac{4}{5}\)(x² + y²) \(\Rightarrow\) x² - \(\frac{4}{5}\)x² = \(\frac{4}{5}\) y² + y²

\(\Rightarrow\)\(\frac{1}{5}\)x² = \(\frac{9}{5}\)y²  \(\Rightarrow\) \(\frac{X^2}{y^2}\) = \(\frac{9}{1}\) \(\Rightarrow\) \(\frac{X}{y}\)= \(\frac{3}{1}\) \(\Rightarrow\) x:y = 3:1

Since the series of data given is of inclusive type, first we shall convert the class intervals into exclusive type using a correction factor.

C.F – LL of a CI – UL of the previous CI

= \(\frac{15 - 14}{2}\) = 0.5

Where LL = Lower Limit

UL = Upper Limit

CI = Class Interval

CF = Cumulative frequency

Median class = Size of \([\frac{N}{2}]\) th item

= \([\frac{65}{2}]\)th item = 32.5 th item so, it lies between 24.5 = 29.5

Medain = L + \([\frac{\frac{N}{2}-Cf}{f}]\times i\) Where L – Lower limit

C.f = Cumulative frequency of class preceding the median class

L – Length of the median class

N = Total of frequency

f = frequency of the median class

Median class = 24.5 – 29.5

I = 24.5; i = 5; c.f = 30; N = 65; f = 20

Medain = 24.5 +\([\frac{\frac{65}{2}-30}{20}]\times 5\)

Medain = Rs. 25.125

Highest income of lowest 50% of workers

We know that, 

As per the properties of square, 

All the numbers that end with digits 2, 3, 7 or 8 are not a perfect square 

Hence, 

Considering the property, we get 

The number 5478 is not a perfect square 

As the last digit of the number is 8. 

Therefore, 

Option (C) is the correct option

Length = 825 cm = 3 × 5 × 5 × 11

Breadth = 675 cm = 3 × 3 × 3 × 5 × 5

Height = 450 cm = 2 × 3 × 3 × 5 × 5

Longest tape = HCF of 825, 675, and 450 = 3 × 5 × 5 = 75 cm

Therefore, the longest tape is 75 cm.

Power bipolar junction transistor.

(i) Active region.

(ii) Saturation region

(iii) Cut off.

Second breakdown phenomenon is that where thermal runaway quickly takes place when the junction temperature beyond safe limit and the device is destroyed.

Silicon controlled rectifier.

The other name for SCR is Thyristor.

We know that

∵ Alternate segment of ∠BCP = ∠BAC

∴ ∠BCP = ∠BAC

⇒ ∠BCP = 80°

There are two PN junctions in SCR.

S.P. of 6 articles = Re 1, Loss = 20 %

\(\therefore\) C.P. of 6 articles = \(\frac{1\times100}{(100-20)}\)= \(\frac{100}{80}\) = \(\frac{5}{4}\) = 1.25

If the gain is 20%, then S.P. of 6 articles = \(\frac{1.25\times120}{100}\) = Rs 1.50

\(\therefore\) For Re 1, 4 articles are sold.

Light activated silicon controlled rectifier.

Reverse conducting thyristor.

Answer is (A) Symmetric

Answer is (B) |A|

Total number of soldiers = 8160 

Number of soldiers left out = 60 

Number of soldiers arranged in rows to form a perfect square = 8160 – 60 

= 8100 

Hence, number of soldiers in each row = \(\sqrt{8100}\)

= \(\sqrt{9\times9\times10\times10}\)

= 90

 The total number of soldiers = 8160

Number of soldiers left out = 60

Number of soldiers arranged in rows to form a perfect square 

= 8160 – 60 

= 8100

Hence ,number of soldiers in each row 

= √8100

= √(9×9×10×10)

= 9×10

= 90

(1) Iron pyrite

The answer is (2) Al

The answer is (2) Sb₂S₃

चतुर्भुज के चारों अंतः कोणों का योग 360° होता है । 

माना चौथा कोण x

⇒ 75° + 95° + 110° + x = 360°

⇒ 280° + x = 360°

⇒ x = 360° – 280° = 80°

(1) nucleus 

(2) (2, 8, 7) 

(3) acidic radical 

(4) composite radical

(5) K₂SO₄ 

(6) Hg²⁺

(7) anion 

(8) 1 gross

(9) © 

(10) Antoine Lavoisier.