Explore topic-wise InterviewSolutions in Current Affairs.

The correct answer is Anthracite.

• Coal is a sedimentary rock usually occurring in rock strata in layers or veins called coal beds.
• Coal is a fossil fuel that forms when dead plant matter is converted into peat, which in turn is converted into lignite, bituminous, and anthracite.
• Anthracite
  • It is a high-grade coal
  • It has the highest heat content due to less amount of water present in it.
  • It is hard and dark black in colour.
  • Sedimentary Rock- Conglomerate, Shale, Limestone and Sandstone.
• Quartzite
  • It is a hard, non-foliated metamorphic rock which was originally pure quartz sandstone.
  • Sandstone is converted into quartzite through heating and pressure usually related to tectonic compression within orogenic belts.

• Granite
  • Granite is hard, coarse-grained rocks of crystalline structure.
  • It is a type of igneous rocks (plutonic rocks).
  • Granites can be predominantly white, pink, or grey in colour, depending on their mineralogy.

• 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 area 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:
    • mechanically formed — sandstone, conglomerate, limestone, shale, loess etc. are examples;
    • organically formed — geyserite, chalk, limestone, coal etc. are some examples;
    • chemically formed — chert, limestone, halite, potash etc. are some examples. 

The correct answer is Dost Mohammed Khan.

• Dost Mohammad Khan was one of the prominent rulers of Islamnagar in 1715.
• He was the founder of the modern city of Bhopal.
• The famous Rani Mahal and Chaman Mahal in Islamnagar are built by Dost Mohammad Khan.
• Chaman Mahal is a palace in Bhopal built-in 1715.
• Chaman Mahal is also known as Islamnagr fort.
• Islamnagar effectively became the capital of the Bhopal region in 1719.
• The Rani Mahal was constructed in 1720 as the residence for Dost Mohammad Khan's queens.

• Sultan Mohammad Khan is the brother of Dost Muhammad Khan(Emir of Afghanistan).
• Yar Mohammad Khan was a founder of the Awami League of Bangladesh.

(d) above 2.5 mm in 24 hours

Soil of Western Rajasthan have a high content or Calcium.

During respiration, about 97% of oxygen is transported by RBC’s by forming oxyhaemoglobin and the remaining 3% gets dissolved in the plasma.

 Since haemoglobin pigment has less affinity for CO₂, it is mainly transported in the dissolved form. Around 20 -25% of CO₂ is carried by haemoglobin as carbamino-haemoglobin, 7% is in a dissolved state in the plasma and the remaining is carried as bicarbonate. This deoxygenated blood gives CO₂ to lung alveoli.

Option : B. I, II & IV

• In India, the concept of constitutional ombudsman was first proposed by the then law minister Ashok Kumar Sen in parliament in the early 1960s.
• The term Lokpal and Lokayukta were coined by Dr. L. M. Singhvi.
• In 1966, the First Administrative Reforms Commission recommended the setting up of two independent authorities- at the central and state level, to look into complaints against public functionaries, including MPs.
• In 1968, Lokpal bill was passed in Lok Sabha but lapsed with the dissolution of Lok Sabha and since then it has lapsed in the Lok Sabha many times.
• Till 2011 eight attempts were made to pass the Bill, but all met with failure.
• "India Against Corruption movement" led by Anna Hazare put pressure on the United Progressive Alliance (UPA) government at the Centre and resulted in the passing of the Lokpal and Lokayuktas Bill, 2013, in both the Houses of Parliament.
• It received assent from President on 1 January 2014 and came into force on 16 January 2014.

•  As per the year Statistical Review of World Energy 2015, India is the 4th largest producer of coal in the world.

1. Peat coal- It contains less than 40% carbon content. It is the lowest grade of coal.
2. Lignite coal- It contains 65- 70% carbon. It is s low-grade brown coal, which is soft with high moisture content. The principal lignite reserves are in Neyveli in Tamil Nadu and are used for the generation of electricity.
3. Bituminous coal- It contains 60 to 80 % Carbon Content. It is the second-best quality of coal. It is the most popular coal in commercial use. Metallurgical coal is high-grade bituminous coal that has a special value for smelting iron in blast furnaces.
4. Anthracite coal- It contains more than 80% carbon content. It is the best quality of coal. In India, it is found only in Jammu and Kashmir.

The Correct Answer is Sundari.

• Littoral and Swamp Vegetation are called Mangrove Vegetation or Tidal Forests.
• They are found in the tidal deltas of Ganga, Mahanadi, Godavari and Krishna rivers.
• These vegetations are adapted to High water salinity and Floods at regular intervals.
• Sundari trees are found in the Sundarban Delta.
• Avicennia, Bruguiera, Ceriops, and Rhizophora are few tree species found in the Mangrove region.

• The ebony tree is found in tropical evergreen forests.
• Cypress is a common name for various coniferous trees found in colder regions.
• The Chinar is a deciduous tree.
• The chinar sheds its leaves once a year.

The cropping season in which the cultivation is started by the beginning of winter and harvested by the beginning of summer.

Option : (a).Pulses

Option : (c).Rubber

(b) Bering Strait is nearest to the International Date line, because the international Date line runs equidistant between the American continents, on its East and Asia, Australia, and Europe on its west.

(c) B is a cyclone in the northern hemisphere

The correct answer is Mahatma Gandhi.

• "I take holy vows in the name of God and say that I will continue to fight for the freedom of India and its thirty-eight crores inhabitants till my last breath."
• It was given by Mahatma Gandhi.

• Bhagat Singh was an Indian socialist revolutionary whose two acts of dramatic violence against the British in India and execution at age 23 made him a folk hero of the Indian independence movement.
• Subhas Chandra Bose was an Indian nationalist whose defiant patriotism made him a hero in India, but whose attempt during World War II to rid India of British rule with the help of Nazi Germany and Imperial Japan left a troubled legacy.
• Lajpat Rai was an Indian independence activist. He played a pivotal role in the Indian Independence movement. He was popularly known as Punjab Kesari.
  • He was one of the three Lal Bal Pal triumvirates.

Performance-enhancing substances (PESs) are used commonly by children and adolescents in attempts to improve athletic performance. More recent data reveal that these same substances often are used for appearance-related reasons as well. PESs include both legal over-the-counter dietary supplements and illicit pharmacologic agents. This report reviews the current epidemiology of PES use in the pediatric population, as well as information on those PESs in most common use.

The performance enhancing substances are as follows: The use of drugs to enhance performance is considered unfair and puts the health of athlete at high risk like 

(a) Mechanical aids :- It includes altitude training, aqua training, elastic cord, treadmills, vibration training, weight training etc. 

(b) Pharmacological aids :- It includes Anabolic steriod, beta blockers, caffeine, choline, sodium bicarbonate. These are all banned by IOC in sports. 

(c) Physiological aids :- It includes Herbal medicines, sports massage, sauna, Human Growth hormones. 

(d) Nutritional aids :- They are like Bicarbonate of soda, carbohydrate loading, creatine, sports drinks. 

(e) Psychological aids :- These includes mediation, motivation, centering, cheering, Relaxation. Most of these are valid and applicable in sports.

Given differential equation is

y'' - 4y' + 4y = \(\frac{e^{2x}}x\)

It's auxiliarly equation is

m² - 4m + 4 = 0

⇒ (m - 2)² = 0

⇒ m - 2 = 0

⇒ m = 2, 2

\(\therefore\) w(y₁, y₂) = \(\begin{vmatrix}y_1&y_2\\y'_1&y'_2\end{vmatrix}\) = \(\begin{vmatrix}e^{2x}&xe^{2x}\\2e^{2x}&(2x+1)e^{2x}\end{vmatrix}\)

 = (2x + 1 - 2x)e^4x = e^4x

_u1 = \(\int\cfrac{\begin{vmatrix}0&xe^{2x}\\ \frac{e^{2x}}x&(2x+1)e^{2x}\end{vmatrix}}{w(y_1y_2)}dx\)

\(=\int\frac{-e^{4x}}{e^{4x}}dx=-x\)

u₂ = \(\int\cfrac{\begin{vmatrix}e^{2x}&0\\ 2e^{2x}&\frac{e^{2x}}x\end{vmatrix}}{w(y_1y_2)}dx\) 

 = \(\int\frac{e^{4x}}{x.e^{4x}}dx\) = \(\int\frac1{x}dx=log x\)

\(\therefore\) P.I. = u₁y₁ + u₂y₂ = -xe^2x + xe^2xlog x

\(\therefore\) Complete solution is y = C.F. + P.I.

 = (C₁ + C₂x)e^2x - xe^2x + xe^2x log x

Given differential equation is \(x^2\frac {d^2y}{dx^2} + x\frac {dy}{dx} -y = \frac {x^3}{1+x^2}\)

Then \(x \frac {dy}{dx} = \frac {dy}{dz} = dy\), where D = \(\frac {y}{dz}\)

\(x^2 \frac {d^2y}{dx^2} \) = D (D-1) y.

Then given differential equation converts to 

D(D-1)y + Dy-y = \(\frac {e^{3z}}{1+e^{2z}}\)

= (D² - D + D-1)y = \(\frac {e^{3z}}{1+e^{2z}}\)

= (D² -1)y = \(\frac {e^{3z}}{1+e^{2z}}\)

It's auxiliarly equation is m²-1 = 0

= m = ± 1

∴ C.F = C₁ e^-z+ C₂ e^z

= \(\frac {C_1}{x} + C_2 x\) (∵e^z = x)

Let y₁ = \(\frac 1x\) & y₂ = x

∴ W(y₁, y₂) = \(\begin {vmatrix}y_1&y_2\\y_1^1&y_2^1\end {vmatrix} = \begin {vmatrix}\frac 1x&x\\\frac {-1}{x^2}&1\end {vmatrix}\)

= \(\frac1x + x . \frac {1}{x^2} = \frac 1x + \frac 1x = \frac 2x\)

∴ y₁ =\(\int\) \(\frac {\begin {vmatrix}0&x\\\frac {x^3}{1+x^2}&1\end{vmatrix}}{W(y_1,y_2)}dx\)

=  \(\int \frac {-\frac {x^4}{1+x^2}}{\frac 2x}dx\)

= \(\frac {-1}{2}\int \frac {x^5}{1+x^2}dx = \frac {-1}{2} \int (x^3-x+\frac {x}{1+x^2})dx\)

= \(\frac {-1}{2}(\frac {x^4}{4} - \frac {x^2}{2} + 12 log (1 +x^x2))\) & 

y₂ = \(\int \frac {\begin {vmatrix}\frac 1x&0\\\frac {-1}{x^2} & \frac {x^3}{1+x^2}\end {vmatrix}}{W (y_1,y_2)}\)

= \(\int \frac {\frac {x^2}{1+x^2}}{\frac 2x}dx = \frac 12\int\frac {x^3}{1+x^2}dx\)

= \(\frac 12 \int (x- \frac {x}{1+x^2})dx\)

 = \(\frac 12(\frac {x^2}{2} - \frac 12 log (1+x^2))\)

∴ P.I = y₁y₁ + y₂y₂

= \(\frac {-1}{2x} (\frac {x^4}{4} - \frac {x^2}{2} + \frac 12 log (1+x^2)) + \frac x2 (\frac {x^2}{2} - \frac 12 log (1+x^2))\)

= \(\frac {-x^3}{8} + \frac x4 - \frac {1}{4x}log (1+x^2) + \frac {x^3}{4} - \frac x4 log (1+x^2)\)

= \(\frac {x^3}{8} + \frac x4 - \frac x4log (1+x^2) \frac {-1}{4x}log (1+x^2)\)

∴ Complete solution is 

y = C.F + P.I

= \(\frac {C_1}{x} + C_2x + \frac {x^3}{8} + \frac x4 - \frac x4 log (1+x^2) - \frac {1}{4x} log (1+x^2)\)

Given differential equation is

y¹¹ - 6y¹ + 9y = 4e^x

It's auxiliarly equation is

m² - 6m + 9 = 0

⇒ (m - 3)² = 0

⇒ m = 3, 3

\(\therefore\) C.F. = (C₁ + C₂x)e^3x

Let y₁ = xe^3x, y₂ = e^3x

\(\therefore\) w(y₁,y₂) = \(\begin{vmatrix}xe^{3x}&e^{3x}\\e^{3x}(3x + 1)&3e^{3x}\end{vmatrix}\)

= e^6x(3x - 3x - 1) = -e^6x

^{\(\therefore\)} u₁ = \(\int\cfrac{\begin{vmatrix}0&e^{3x}\\4e^x&3e^{3x}\end{vmatrix}}{w(y_1y_2)}dx\) 

 = \(\int\frac{-4e^{4x}}{-e^{6x}}dx\) = 4\(\int e^{-2x}dx\)

 = \(\frac{4e^{-2x}}{-2}\) = -2e^-2x

= 2xe^-2x + e^-2x

 ⁼ (2x + 1)e^-2x

\(\therefore\) P.I. = u₁y₂ + u₂y₂

 = -2e^-2x x xe^3x + (2x + 1)e^x

 = e^x

\(\therefore\) Complete solution is y = C.F. + P.I.

 = (C₁ + C₂x)e^3x + e^x

We need to find the Greatest Common Factor (HCF).

Thus, (1471 – 1231) = 240

(1711 - 1471) = 240

(1711 – 1231) = 480 = 2 × 240

Therefore,

The greatest number is 240.

So, Numbers are –

1231 = (240 × 5) + 31

1471 = (240 × 6) + 31

1711 = (240 × 7) + 31

The remainder is 31.

According to question –

⇒ (a – b) = 240 – 31 = 209

GIVEN:

When four positive numbers a, b, c, and d are divided by 41, the remainders are 3, 4, 6, and 2, respectively.

CONCEPT:

Dividend = Divisor x Quotient + Remainder

CALCULATION:

a = 41w + 3, b = 41x + 4, c = 41y + 6, d = 41z + 2

Here, 41w, 41x, 41y and 41z is divisible by 41.

So, according to question –

(5a + 6b – 2c + 10d)

⇒ (15 + 24 – 12 + 20)

⇒ 59 – 12

⇒ 47

When (5a + 6b – 2c + 10d) is divided by 41.

⇒ Remainder (47/41) = 6

Correct answer is (d) they tethered a lamb to the outskirts of the village.

Correct answer is (b) the lion

Correct answer is zoo

Correct answer is Physical education

Correct answer is art museum

The Cartesian equation of line passing through two points A(3, 4, 1) and B(5, 1, 6) is \(\frac{x -3}{5-3} = \frac{y-4}{1-4} = \frac{z-1}{6-1} = \frac{y-4}{-3} = \frac{z - 1}{5}\). (By Cartesian equation of line passing through two points)

Since, the line passes through points A (3,4,1) and B (5,1,6) crosses the xy - plane.

Hence, z - coordinates of line is zero.

Therefore, \(\frac{x-3}{2} = \frac{y-4}{-3} = \frac{0-1}{5} = -\frac{1}{5}\)

⇒ \(\frac{x-3}{2} = -\frac{1}{5}\) and \(\frac{y-4}{-3} = -\frac{1}{5}\).

⇒ x = \(-\frac{2}{5} + 3\) and y = \(\frac{3}{5} + 4 = \frac{23}{4}\).

Therefore, the co-ordinates of the point is \((\frac{13}{5}, \frac{23}{5},0)\)

Hence, the co-ordinates of the point where the line through the points A(3,4,1)and B(5,1,6) crosses the xy - plane is \((\frac{13}{5}, \frac{23}{5},0)\).

Prime factors of 10010 = 2 × 5 × 7 × 11 × 13.

So, the number of prime factors is 5.

Concept

Value of 0.333… = 3/9

Value of 0.66… = 6/9

Explanation

Value of 0.4333….

Let x = 0.4333…     ----(1)

Multiply 0.4333… by 10

10x = 4.333….     ----(2)

Subtract (1) from (2)

⇒ 10x – x = 4.333… - 0.4333…

⇒ 9x = 3.9

⇒ x = 39/90      ----(A)

Value of 0.7666….

Let y = 0.7666….      ----(3)

Multiply 0.766… by 10

⇒ 10y = 7.6666      ----(4)

Subtract (3) from (4)

⇒ 10y – y = 7.666… - 0.7666

⇒ 9y = 6.9

⇒ y = 69/90     ----(B)

Difference of (A) and (B)

⇒ (69/90) – (39/90)

⇒ 30/90

⇒ 1/3

Here, we have to find the square of 0.02

Calculation:

(0.02)²

⇒ (2/100)²

⇒ 4/10000

⇒ 0.0004

Hence, the square of  0.02 is 0.0004.

Concept used:

Product of two odd natural numbers is always odd.

Difference and sum of two odd numbers is always even.

Calculation:

Statement 1:

Let x and y be two odd numbers

As, product of two odd numbers is always odd

⇒ x² is odd

⇒ y² is odd

|x² – y²| is even, because difference of two odd numbers is always even.

Hence, Statement 1 is not correct

Statement 2

We can express 30 as 2 × 3 × 5

⇒ 30 = 2 × (2 + 1) × (2 + 3)

For n = 2, 30 can be expressed as n × (n + 1) × (n + 3)

Hence, Statement 2 is correct

Statement 3

3/4 = (3 × 73/(4 × 73) = 219/292

Hence, Statement 3 is not correct

∴ Statement 1 and Statement 3 are not correct

Let’s assume a, b, c be the hundreds, tens and units digits of the desired number.

So, the desired number ‘abc’ = 100a + 10b + c

After reversing it, it becomes ‘cba’ = 100c + 10b + a

Now, ‘abc’ – ‘cba’ = (100a + 10b + c) - (100c + 10b + a)

⇒ 396 = 99(a – c)

⇒ (a – c) = 396/99 = 4

Thus, possible combination of (a, c) = {(9, 5), (8, 4), (7, 3), (6, 2), (5, 1), (4, 0)}

For every combination of (a, c) only one value of b is possible.

So in all, there are 6 such numbers possible.

correct option:

(B) 3x² sin(x³) 

correct option:

(B) 1/t

(A) Diagonal matrix

Correct option is: C. AB=BA

Option (D) is correct.

Given that point (2α, 3α) lies inside the ellipse \(\frac{x^2}{4} + \frac{y^2}{9} = 1.\)

Therefore, \(\frac{(2 \alpha)^2}{4} + \frac{(3 \alpha)^2}{9} < 1\)

\(\Rightarrow\) α² + α² < 1

\(\Rightarrow\) 2α² < 1

\(\Rightarrow\) α² < \(\frac{1}{2}\)

\(\Rightarrow\) α ∈ \(\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)\)

\(|2x - 1| = |x| + |x - 1|\)

holds for \(x\in [-\infty , 0] \cup [1, \infty)\) 

\(\therefore |2x - 1| = |x + (x - 1)|\)

\(\le |x| + |x - 1|\)

But equality holds for \(x\in(\infty, 0] \cup [1, \infty)\)

Alternatively:

Case I: \(x \in (-\infty, 0]\) 

Then |x| = -x

& \(|x - 1| = -(x - 1)\)

Also \(|2x - 1| = -(2x - 1)\)

\(\because -(2x - 1) = -2x + 1 = - x - x + 1 = -x -(x - 1)\)

⇒ \(|2x - 1| = |x | + |x - 1| \)

given equality holds.

Case II: \(x\in \left(0, \frac 12\right)\)

Then \(|x| = x\)

\(|x - 1| = -(x - 1) = 1 - x\)

\(|2x - 1| = -(2x - 1)\)

\(\because |2x -1| = -(2x - 1) = 1- 2x = 1 -x - x \)

\(= |x - 1| - |x|\)

given equality not hold

Case III: \(x\in\left(\frac 12 , 1\right)\) 

Then \(|x| = x\)

\(|x - 1| = -(x - 1) = 1 -x\)

\(|2x - 1| = 2x - 1\) 

\(\because |2x- 1| = 2x - 1 = x + x - 1 =|x| - (1 -x)\)

\(=|x| - |x - 1|\)

given inequality not hold.

Case IV: \(x\in[1, \infty)\)

Then \(|x| = x\) 

\(|x - 1| = x - 1\)

\(|2x| - 1 = 2x - 1\)

\(\because |2x - 1| = 2x -1 = x + x - 1 = |x | + |x - 1|\)

given equality holds.

Hence, given equality holds for \(x\in(-\infty, 0]\cup[1,\infty)\).