Explore topic-wise InterviewSolutions in .

Correct answer is (c) is, make

The correct answer is Leaves.

• During the water cycle, leaves are involved in the process of evaporation.
• Plants absorb water from the soil. The water moves from the roots through the stems to the leaves.
• Once the water reaches the leaves, some of it evaporates from the leaves, adding to the amount of water vapor in the air.
• This process of evaporation through plant leaves is called transpiration. 

• Evaporation is the process by which water changes from a liquid to a gas or vapor.
• Evaporation is the primary pathway that water moves from the liquid state back into the water cycle as atmospheric water vapor. 
• A very small amount of water vapor enters the atmosphere through sublimation, the process by which water changes from a solid (ice or snow) to a gas, by passing the liquid phase. 

Correct answer is zero

\(\lim\limits_{x\to 0}\frac{a^x+ax+a^x-x-2a^x}{x^2}\)

 = \(\lim\limits_{x\to 0}\frac{2a^x+ax+x-x-2a^x}{x^2}\) 

= \(\lim\limits_{x\to 0}\frac{x(a-1)}{x^2}\) = \(\lim\limits_{x\to 0}\frac{a-1}{x}=\frac{a-1}0\) 

 = ∞ (not defined)

4x + 6y = 92        .....(1)

6x + 10 y = 148    ......(2)

⇒ 12x + 18y = 276   .....(3)   (On multiplying equation (1) by 3)

12x + 20y = 296      ......(4)   (On multiplying equation (2) by 2)

Equation (4) - Equation (3), we get

2y = 20

⇒ y = 10

∴ From (1), we get

4x + 60 = 92

⇒ \(x = \frac {92 - 60}{ 4} = \frac {32}{4} = 8\)

(i) The fare from Rajiv Chowk to Karol Bagh  is x = Rs 8

(ii) The fare from Rajiv chowk to Haus Khas is y = Rs 10

Given quadratic equation is

3x² - 2\(\sqrt2\) - 2\(\sqrt3\) = 0

By comparing with ax² + bx + c = 0 we get

a = 3, b = -2\(\sqrt2\) and c = -2\(\sqrt3\)

\(\therefore\) Discriminant = b² - 4ac

 = (-2\(\sqrt2\))² - 4 x 3 x - 2\(\sqrt3\)

 = 8 + 24\(\sqrt3\) 

= 8(1 + 3\(\sqrt3\)) > 0

We have, f(x) = x² ‒ √2x – 12

Now, put f(x) = 0

x² ‒ √2x – 12 = 0

x² - 3√2x + 2√2x - 12 = 0

x(x ‒ 3√2) + 2√2(x – 3√2)=0

(x ‒ 3√2) + (x + 2√2) = 0

Thus, x= 3√2 , -2√2

Correct answer is (D) 1/25

Correct option is 4. 48

Given: 12gM₂O₃ contains 8g of metal.

To find: We have find the atomic mass of the metal.

Solution:

Atomic mass of oxygen = 16

Here three oxygen atoms are present so, total atomic mass = 16 x 3 = 48

12gM₂O₃ contains 8g of metal.

The amount of oxygen is = (12 - 8) = 4g

Now, 4g of oxygen has total atomic mass 48g

8g of metal has total atomic mass = \(\frac{8\times 48}4\) = 96

Atomic mass of the metal can be given as-

= \(\frac{96}{2}\) 

= 48

Atomic mass of metal is 48.

Volume of the sphere = \(\Large \frac{4}{3} \pi r^{3}\ cm^{3} \)

Surface area of sphere = \(\Large 4 \pi r^{2}\ cm^{2}\)

The volume of the sphere and the surface area of the sphere are numerically equal (given), 

\(\Large \frac{4}{3} \pi r^{3}\ cm^{3} \) = \(\Large 4 \pi r^{2}\ cm^{2}\)

r = 3. 

Hence, diameter = 2r = 6cm. 

We have l = rθ, the radius and angle are inversely proportional. Therefore;

\(\frac{\theta_1}{\theta_2} = \frac{r_2}{r_1}\)

\(\Rightarrow\)\(\cfrac{\frac{\pi}{180} \times 65}{\frac{\pi}{180} \times 110}\) = \(\frac{r_2}{r_1}\)

\(\Rightarrow\) \(\frac{13}{22} = \frac{r_2}{r_1}\)

\(\Rightarrow\) r₁ : r₂ = 22: 13

E. is none of the above

C. 22^◦ above the horizontal

Concept:

Order: The order of a differential equation is the order of the highest derivative appearing in it.

Degree: The degree of a differential equation is the power of the highest derivative occurring in it, after the Equation has been expressed in a form free from radicals as far as the derivatives are concerned.

Calculation:

Given:

\(\rm y = x \frac{dy}{dx}+\left(\frac{dy}{dx}\right)^{-2} \\ \rm y = x\frac{dy}{dx}+\frac{1}{(\frac{dy}{dx})^2} \\ y(\frac{dy}{dx} )^2= x(\frac{dy}{dx})^3 + 1\)

For the given differential equation the highest order derivative is 1.

Now, the power of the highest order derivative is 3.

We know that the degree of a differential equation is the power of the highest derivative.

Hence, the degree of the differential equation is 3.

Concept:

Order: The order of a differential equation is the order of the highest derivative appearing in it.

Degree: The degree of a differential equation is the power of the highest derivative occurring in it, after the Equation has been expressed in a form free from radicals as far as the derivatives are concerned.

Calculation:

\(\dfrac{d^2y}{dx^2}+3\left(\dfrac{dy}{dx}\right)^2 =x^2 \log \left(\dfrac{d^2y}{dx^2}\right)\)

For the given differential equation the highest order derivative is 2.

The given differential equation is not a polynomial equation because it involved a logarithmic term in its derivatives hence its degree is not defined.

y = x³ = f (x)

f (2) = 8 and f (-2) = – 8

f’ (x) = 3x²

f’ (x) = [f (2) – f (-2)] / [2 – (-2)]

= [8 – (-8)] / [4] = 3x²

x = ± 2 / √3

Correct option is (B) α = –3

\(f'(x) = e^{(4x^3 - 12x^2 - 180x + 31)}(12(x^2 - 2x + 7)(x + 3)(x - 5)+ 2(x - 1))\)

for x ∈ [–3,0] 

⇒ f’(x) < 0

f(x) is decreasing function on [–3,0] 

The absolute maximum value of the function f(x) is at x = –3 

⇒ α = –3

Concept:

Simpson’s \(\frac{1}{3}\) Rule: It is numerical integration method which follows the three-point Newton quadrature rule and is given by,

\(\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = {\rm{\;}}\frac{{\rm{h}}}{3}\left[ {\left( {{{\rm{y}}_0} + {{\rm{y}}_{\rm{n}}}} \right) + 4\left( {{{\rm{y}}_1} + {{\rm{y}}_3} + \ldots } \right) + 2\left( {{{\rm{y}}_2} + {{\rm{y}}_4} + \ldots } \right)} \right]{\rm{\;}}\)

Where \({{\rm{x}}_0}{\rm{\;to\;}}{{\rm{x}}_{\rm{n}}}\) are equally spaced n number (must be even number) of integration points, \({{\rm{y}}_0}{\rm{\;to\;}}{{\rm{y}}_{\rm{n}}}\) are functional value at those points and h is the interval length or step length.

Calculation:

Given, a = 0, b = 4, step length, h = 1 and \({\rm{f}}\left( {\rm{x}} \right) = {\rm{\;}}\left( {{{\rm{x}}^4} + 10} \right)\)

Hence, the integration points and functional values at those points are listed in the following table:

∴ Integration by Simpson’s 1/3 Rule = \({\rm{I'}} = \frac{{\rm{h}}}{3}\left[ {\left( {{{\rm{y}}_0} + {{\rm{y}}_4}} \right) + 4\left( {{{\rm{y}}_1} + {{\rm{y}}_3}} \right) + 2\left( {{{\rm{y}}_2}} \right)} \right]\)

\(\therefore {\rm{I'}} = \frac{1}{3}\left[ {\left( {10 + 266} \right) + 4\left( {11 + 91} \right) + 2\left( {26} \right)} \right] = \frac{{736}}{3} = 245.33\)

∴ The Exact integration, \({\rm{I}} = \mathop \smallint \limits_0^4 \left( {{{\rm{x}}^4} + 10} \right){\rm{dx}} = \left[ {\frac{{{{\rm{x}}^5}}}{5} + 10{\rm{x}}} \right]_0^4 = 244.8\)

∴ The Error = \({\rm{\xi }} = {\rm{I'}} - {\rm{I}} = 245.33 - 244.8 = 0.53\)

Trick:

The absolute error by Simpson’s 1/3 Rule is given by,

\({\rm{\xi }} = \frac{{{{\rm{h}}^4}}}{{180}}\left( {{\rm{b}} - {\rm{a}}} \right){\rm{max}}\left[ {{{\rm{f}}^4}\left( {\rm{x}} \right)} \right]\)

where, x is some number between a and b and \({{\rm{f}}^4}\left( {\rm{x}} \right)\) is the fourth derivative of the function.

\({\rm{Given}},{\rm{\;\;f}}\left( {\rm{x}} \right) = {\rm{\;}}\left( {{{\rm{x}}^4} + 10} \right){\rm{\;\;}}\therefore {\rm{\;}}{{\rm{f}}^4}\left( {\rm{x}} \right) = 24{\rm{\;\;}}\therefore \max \left[ {{{\rm{f}}^4}\left( {\rm{x}} \right)} \right] = 24{\rm{\;\;}}\)

\(\therefore {\rm{\xi }} = \frac{{{1^4}}}{{180}}\left( {4 - 0} \right) \times 24 = 0.53\)

Concept:

Probability Density Function (PDF): It is the probability function which is represented for the density of a continuous random variable lying between a certain range of values. If x is the continuous random variable with density function f(x), then for a valid probability density function the area between the density curve and horizontal X-axis must be equal to 1. It means

\(\mathop \smallint \limits_{ - \infty }^\infty {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = 1\) 

Calculation:

\(\begin{array}{*{20}{c}}{{\rm{Given}},{\rm{\;\;f}}\left( {\rm{x}} \right) = {\rm{\lambda }}\left( {{\rm{x}} - 1} \right)\left( {2 - {\rm{x}}} \right)}&{{\rm{for\;}}1 \le {\rm{x}} \le 2}\\{ = 0}&{{\rm{otherwise}}}\end{array}\) 

\(\therefore {\rm{\;}}\mathop \smallint \limits_{ - \infty }^\infty {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \mathop \smallint \limits_1^2 {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \mathop \smallint \limits_1^2 {\rm{\lambda }}\left( {{\rm{x}} - 1} \right)\left( {2 - {\rm{x}}} \right){\rm{dx}} = {\rm{\lambda }}\mathop \smallint \limits_1^2 \left( { - {{\rm{x}}^2} + 3{\rm{x}} - 2} \right){\rm{dx}}\) 

\(= {\rm{\lambda }} \times \left[ { - \frac{{{{\rm{x}}^3}}}{3} + \frac{{3{{\rm{x}}^2}}}{2} - 2{\rm{x}}} \right]_1^2 = {\rm{\lambda }} \times \left[ { - \frac{{{2^3} - {1^3}}}{3} + \frac{{3\left( {{2^2} - {1^2}} \right)}}{2} - 2\left( {2 - 1} \right)} \right] = {\rm{\lambda }} \times \frac{1}{6}\) 

Now, to be a valid probability density function \(\mathop \smallint \limits_{ - \infty }^\infty {\rm{f}}\left( {\rm{x}} \right){\rm{dx\;\;must\;be\;equal\;to\;}}1.\)

Hence, \({\rm{\lambda }} \times \frac{1}{6} = 1{\rm{\;\;}}\therefore {\rm{\;\lambda }} = 6\)

Trisomy-21 is seen in Down syndrome. Trisomy 21 is a genetic disorder caused by the presence of all or part of a Klinefelter syndrome. A Trisomy 21 affected baby will have one extra chromosome at 21st position out of 23. This dramatically impacts the development of children and can cause serious illnesses like epilepsy, hearing loss, etc. This trisomy is found in about 1 infant in every 600-700 live births. 

Down syndrome  

• It is one of the well-known and best-studied examples of chromosome-caused abnormalities in human beings.
• This condition, which is an autosomal trisomy, was first described by Langdon Down in 1866 and was originally called mongoloid idiocy or mongolism.
• Down chose this name because of the prominence of the epicanthic fold in the eyelid a phenotype resembling members of the mono gold race. 

• Turner syndrome: This genetic disorder generally affects the females where the person has a missing/partially missing X chromosome, ( a healthy girl child has 2'X' chromosome.)
• Klinefelter syndrome: This genetic disorder generally affects the male child. A healthy male child has one 'X' and one 'Y' chromosome but in the case of Klinefelter syndrome, the child is born with the extra X chromosome. This affects the growth of the child as a normal adult male. 

Hence, we can conclude that Trisomy-21 is cause of Down syndrome.

प्रश्नानुवाद → 'विमुक्तः' इस पद का विलोमपद क्या है?

स्पष्टीकरण → 'विमुक्तः' इसपद का विलोमपद है → आबद्धः।

'विमुक्तः' अर्थात् मुक्त कराया गया, आजाद अथवा स्वतंत्र कराया गया। 'आबद्धः' अर्थात् बंधनकारक​।

विमुक्तः के अन्य विलोमपद → संयुत​, पाशित​, समुन्नद्ध, बन्धनीय​, अवनद्ध, अमुक्त​।

विमुक्तः के पर्यायीवाचक पद → विमोचित, निर्मुक्त​, विनिर्गत।

अतः स्पष्ट होता है कि, 'विमुक्तः' इसपद का विलोमपद है → आबद्धः।

अन्य विकल्पों का हिंदी अनुवाद →

• युद्ध → लडाई, संग्राम​।
• उद्विग्नम् → व्याकुल​, चिंतित​।
• पक्षतः → पक्ष से। पक्ष का अर्थ है → चंद्रमास के दो बराबर भागों में से प्रत्येक भाग जो पंद्रह दिनों का माना जाता है अथवा किसी विषय या वस्तु के दो या दो से अधिक परस्पर विरोधी तत्व, सिद्धांत या भाग।

Concept:

The formula for comfort/centrifugal criteria is,

\(e + f = \frac{{{V^2}}}{{125R}}\)

Where,

e = superelevation

but for taxiway no superelevation is required, hence e = 0

f = coefficient of friction between tyres and pavement surface

V = turning speed of an airplane

R = turning radius of taxiway

Calculation:

Given,

e = 0, f = 0.13, V = 40 km/hr

\(e + f = \frac{{{V^2}}}{{125R}}\)

\(0 + 0.13 = \frac{{{{40}^2}}}{{125 \times R}}\)

Explanation:

Effective gradient: As per ICAO Maximum effective gradient 

(i) For A, B and C type = Max 1%

(ii) For D and E type = Max 2%

Transverse gradient: Provided for drainge purpose. As per ICAO

(i) For A, B type = Max. value 1.5% and Min value 0.5%

(ii) For C, D and E type = Max. value 2% and min value 0.5%

Rate of change longitudinal gradient: AS per ICAO

(i) For A, B and C type = Max. 0.10% per 30 m for vertical curve

(ii) For E and D type = 0.4% per 30 m for vertical curve. 

Concept:

Correction for basic Runway length (l)

i) Correction for elevation

ICAO recommends that basic runway length should be increased at the rate of 7% per 300 m rise in elevation above mean sea level.

∴ Correction for elevation \( = \frac{7}{{100}} \times runway\;length \times \frac{{Airport\;elevation}}{{300}}\)

ii) Temperature correction

Airport reference Temperature (ART)

\({\rm{T}} = {{\rm{T}}_{\rm{a}}} + \frac{{{{\rm{T}}_{\rm{m}}} - {{\rm{T}}_{\rm{a}}}}}{3}\)

Tm → monthly mean of maximum daily temp of hottest month

Ta → monthly mean of the average daily temp of the hottest month

Rise in temperature = ART - SAT  

Where,

ART is Atmospheric reference temperature

SAT Standard atmospheric Temperature

As per ICAO, Basic runway length after correction for elevation should be further increased at the rate of 1% for every 1° C rise of the airport reference temperature

 correction for temperature \(= Correctedlength \times \frac{1}{{100}} \times Rise\;in\;temperature\)

Note: As per ICAO, the cumulative connection for elevation and temperature together should not be ≯ 35%.

Cumulative (%) correction \(= \frac{{{{\rm{l}}_2} - {{\rm{l}}_1}}}{{\rm{l}}} \times 100 \not > 35\% \)

If exceeds modify, l2 = 1.35 × l

iii) Gradient correction

For every 1% effective gradient, runway length will be increased by 20%

\({\rm{Final\;length\;}}\left( {{{\rm{l}}_3}} \right) = {{\rm{l}}_2} + \left( {20 \times \frac{{{{\rm{G}}_{{\rm{effective}}}}}}{{100}}{\rm{\% }}} \right){{\rm{l}}_2}\)

\({{\rm{G}}_{{\rm{effective}}}}\left( {\rm{\% }} \right) = \frac{{{\rm{R}}{{\rm{L}}_{{\rm{highest\;point}}}} - {\rm{R}}{{\rm{L}}_{{\rm{lower\;point}}}}}}{{{\rm{Runway\;unit\;}}\left( {{{\rm{l}}_2}} \right)}}\).

The correct answer is Trusts.

Infrastructure Investment Trust (InvITs)

• An Infrastructure Investment Trust (InvITs) is a Collective Investment Scheme similar to a mutual fund, which enables direct investment of money from individual and institutional investors in infrastructure projects to earn a small portion of the income as a return. 
• The InvIT is designed as a tiered structure with the Sponsor setting up the InvIT which in turn invests into the eligible infrastructure projects either directly or via special purpose vehicles (SPVs).
• The InvITs are regulated by the SEBI (Infrastructure Investment Trusts) Regulations, 2014. 

​Structure of InvITs

• Structured like mutual funds, they have a trustee, sponsor(s), investment manager and project manager.
  • Trustee (certified by Sebi) has the responsibility of inspecting the performance of an InvIT.
  • Sponsor(s) are promoters of the company that set up the InvIT.
  • The investment manager is entrusted with the task of supervising the assets and investments of the InvIT.
  • The project manager is responsible for the execution of the project.

Features and Benefits

• InvITs enable investors to buy a small portion of the units being sold by the fund depending upon their risk appetite. 
• Given that such trusts comprise largely of completed and operational projects with positive cash flow, the risks are somewhat contained.
• Unitholders also benefit from favourable tax norms, including exemption on dividend income and no capital gains tax if units are held for more than three years.

Explanation:

Basic Runway length:

It is the length of runway under the following assumed conditions at the aircraft

(i) Airport altitude at sea level.

(ii) Temperature at the airport is standard (15° C)

(iii) Runway is leveled in the longitudinal direction.

(iv) No wind is blowing on runway.

(v) Aircraft is loaded to its full loading capacity.

(vi) There is no wind blowing enroute to the destination.

(vii) Enroute temperature is standard.

Correction for Elevation, Temperature and Gradient:

(i) Correction for Elevation: Basic runway length is increased at the rate of 7% per 300 m rise in elevation above the mean sea level.

(ii) Correction for Temperature: 

Airport reference temperature = \({T_a} + \frac{{{T_m} - {T_a}}}{3}\)

Where, Ta = Monthly mean of average daily temperature

Tm = Monthly mean of the maximum daily temmperature for the same month of the year

Total correction for elevation plus temperature ⇒ 35% of basic runway length

(c) Correction for Gradient:

(i) Steeper gradient results in greater consumption of energy and as such longer length of runway is required to attain the desired ground speed.

(ii) After having been corrected for elevation and temperature should be further increased at the rate of 20% for every 1% of effective gradient.

Explanation:

Magnitude of earthquake:

(i) The time, location, and magnitude of an earthquake can be determined from the data recorded by a seismometer. Seismometers record the vibrations from earthquakes that travel through the Earth. Each seismometer records the shaking of the ground directly beneath it.

(ii) Sensitive instruments, which greatly magnify these ground motions, can detect strong earthquakes from sources anywhere in the world. Modern systems precisely amplify and record ground motion (typically at periods of between 0.1 and 100 seconds) as a function of time.

(iii) The Richter magnitude of an earthquake is determined from the logarithm of the “amplitude” of waves recorded by seismographs. Adjustments are included for the variation in the distance between the various seismographs and the epicenter of the earthquakes.

(iii) Magnitude is expressed in whole numbers and decimal fractions. For example, a magnitude 5.3 is a moderate earthquake, and a 6.3 is a strong earthquake. Because of the logarithmic basis of the scale, each whole-number increase in magnitude represents a tenfold increase in measured amplitude as measured on a seismogram.

(c) Resistor

(c) IDSS

The transconductance increases when the drain current approaches 'IDSS'.

(d) Mhos or Siemens

   The correct answer is 18 October 2010.

• The National Green Tribunal was formed on 18 October 2010.
  • The National Green Tribunal Act, 2010 is an Act of the Parliament of India which enables the creation of a special tribunal to handle the expeditious disposal of the cases pertaining to environmental issues.
  • It draws inspiration from India's constitutional provision of Article 21 Protection of life and personal liberty, which assures the citizens of India the right to a healthy environment.
  • NGT is proposed to be set up at five places of sittings and will follow circuit procedure for making itself more accessible.
  • New Delhi is the Principal Place of Sitting of the Tribunal and Bhopal, Pune, Kolkata, and Chennai shall be the other place of sitting of the Tribunal.
• Hence, Option 1 is correct.

• NGT Structure :
  • The Principal Bench of the NGT is in New Delhi.
  •  Each Bench has a specified geographical jurisdiction in a region. 
  • The Chairperson of the NGT is a retired Judge of the Supreme Court, head quartered in New Delhi.
  • On 18 October 2010, Justice Lokeshwar Singh Panta became its first Chairman.
  • Retired justice Adarsh Kumar Goel is the incumbent chairman.
  • Other Judicial members are retired Judges of High Courts.
  • Each bench of the NGT will comprise at least one Judicial Member and one Expert Member.
  • Expert members should have a professional qualification and a minimum of 15 years experience in the field of environment/forest conservation and related subjects.
• Green Bench :
  • A green bench is a judicial bench that hears and adjudicates disputes relating to the preservation of forests and the protection of the environment. 
  • The word green bench was coined by the Supreme Court in the 'Madras Tanneries' case, on August 28, 1996.
• Justice Adarsh Kumar Goel is a Chairperson of National Green Tribunal (NGT).
• Under the NGT Act 2010, the tribunal is to have a full-time chairperson and up to 20 judicial and expert members each.
• According to the Act, the tribunal is not supposed to have less than 10 of either.
• It currently has four judges, including the chair, and as many experts.

(C) Statement I is true, but Statement II is false 

We know that,

du = Tds - Pdv------(i)

dH = du + pdv + vdp-----(ii)

and dG = dH - Tds - sdT------(iii)

Putting the value of dv in equation (ii), we got---

dH = Tds - pdv + pdv + vdp

dH = Tds + vdp------(iv)

Putting the value of dH from equation (iv) to equation (iii), we got

dG = Tds + vdp - Tds - sdT

dG = vdp - sdT

at constant temperature dT = 0

\(\therefore\) dG = vdp, at constant temperature.