UPPSC AE Civil 13 Dec 2020 Official Paper - II (Previous Year Paper)

The correct answer is option 4.

Key Points

• Gond is a tribe of Madhya Pradesh.
• Gond tribe is one of the largest Adivasi communities.
• The Gond called themselves Koi or Koiture.
• They inhabited the dense forests of the Vindhyas, Satpura and Mandla in the Narmada region of the Amarkantak range for centuries.
• Tharus is a tribe of Uttarakhand.
• They are live in Khatima and Sitarganj tehsil of UdhamSingh Nagar district (Uttarakhand).
• Todas is a tribe of Tamil Nadu.
• Todas is a tribal group in the higher altitudes of the Nilgiris.
• Santhal is a tribe of Jharkhand.
• They are also called Manjhi.
• The third-largest tribes of India are the Santhal Tribe.
• This tribe habitation is mainly in the states of West Bengal, Bihar, Odisha, Jharkhand and Assam.

Additional Information

Important tribes of :

The correct answer is September.

Key Points

• Rashtriya Poshan Maah was celebrated in September 2020.
• It was celebrated to promote nutrition among Women and Children.
• Poshan Maah is celebrated every year to encourage Jan Bhagidaari in order to ensure health and nutrition for everyone.
• Every year Poshan Maah is celebrated all over the country under POSHAN Abhiyaan, launched in 2018.
• The importance of Nutrition in our lives was emphasized by Prime Minister Shri Narendra Modi in his popular programme “Mann ki Baat” on 30^th August 2020.

The correct answer is (A) - (4), (B) - (3), (C) - (2), (D) - (1).

• The highest Peak of Tamil Nadu is Doda Betta.
• The highest peak of Tamil Nadu is Doddabetta and is a part of the Nilgiri Hills at the junction of the Western Ghats as well as the Eastern Ghats.
• ​The highest peak of Rajasthan is Guru Shikhar.
• ​Guru Shikhar, a peak in the Arbuda Mountains of Rajasthan, is the highest point of the Aravalli Range. The peak is named Guru Shikhar or “peak of the guru” after Dattatreya, an incarnation of Vishnu.
• ​The highest Peak of Nagaland is Saramati.
• ​Standing at an altitude of 3841 m, Saramati is the highest peak in the state of Nagaland and occupies a place of pride in the heart of its residents.
• ​The highest peak of Madhya Pradesh is Dhoopgarh.
• The Dhupgarh Peak (1,350 metres), near Pachmarhi in south-central Madhya Pradesh, is the state's highest point.

The correct answer is Postal Service Schemes.

• The Department of Posts launched a scheme called Five Star Villages.
• It is to ensure universal coverage of flagship postal schemes in rural areas of the country.
• The scheme is being launched on a pilot basis in Maharashtra.
• It bridges the gaps for generating public awareness and reach of postal products and services, mainly in interior villages.
• Under the Five Star Village scheme, all postal products and services are going to be made available and marketed, and publicized at the village level.
• The schemes covered under the Five Star scheme include:
• ​Savings Bank accounts, Recurrent Deposit Accounts, NSC / KVP certificates.
• Sukanya Samridhi Accounts/ PPF Accounts.
• Funded Post Office Savings Account linked India Post Payments Bank Accounts.
• Postal Life Insurance Policy/Rural Postal Life Insurance Policy.
• Pradhan Mantri Suraksha Bima Yojana Account / Pradhan Mantri Jeevan Jyoti Bima Yojana Account.

The correct answer is (A) - (4), (B) - (3), (C) - (1), (D) - (2).

Key Points

 In the given table, Biosphere reserves are given.  These are biosphere reserve of states:

• Nokrek is a biosphere reserve in Meghalaya.

• The Nokrek Biosphere Reserve is located in the northeast of India on the Tura Range, which forms part of the Meghalaya Plateau.

• ​Agasthyamalai is a biosphere reserve in Kerala.

• ​The Agasthyamala Biosphere Reserve (ABR) has got its name into UNESCO’s World Network of Biosphere Reserves (BR).

• It was established in 2001 and straddles the border of Kollam and Thiruvananthapuram Districts in Kerala and Tirunelveli and Kanyakumari Districts in Tamil Nadu, South India at the southern end of the Western Ghats.

• Nandadevi is a biosphere reserve in Uttarakhand.

• Nanda Devi Biosphere Reserve, located in the Himalayan Mountains in the northern part of the country.

• It includes core areas of the Nanda Devi and Valley of Flowers National Parks, which are one World Heritage site.

• ​​​Dehang Debang is a biosphere reserve in Arunachal Pradesh.

• Dihang-Dibang or Dehang-Debang is a biosphere reserve constituted in 1998.

• It is in the Indian state of Arunachal Pradesh.

• Biosphere reserves are areas of terrestrial and coastal ecosystems promoting solutions to reconcile the conservation of biodiversity with its sustainable use.

• They are internationally recognized, nominated by national governments, and remain under the sovereign jurisdiction of the states where they are located.

• They often include one or more National Parks and/or preserves, along with buffer zones that are open to some economic uses. Protection is granted not only to the flora and fauna of the protected region, but also to the human communities who inhabit these regions, and their ways of life.

The correct answer is Three.

Key Points

• Three teachers from Uttar Pradesh were selected for the 'National Award' on teacher day.
• The three from UP are — Mohammed Ishrat Ali, head teacher in a primary school, Rajwana in Mainpuri; Snehil Pandey, head teacher in a primary school Soharamau, Nawabganj in Unnao; and Vikas Kumar, principal, SD Inter College Meerapur, in Muzaffarnagar
• Union Education Ministry inferred the National Award to Teachers as a way of recognizing their achievements in the field of education.
• The awards were conferred by President Ram Nath Kovind on the occasion of Teacher's Day on September 5, 2020.
• 47 teachers have been selected by the government for 'National Awards to Teachers 2020.

Additional Information

• Teacher’s Day is celebrated across India today to honour Dr. Sarvepalli Radhakrishnan.
• It is observed annually on September 5, as Dr. Radhakrishnan was born on September 5, 1888.
• Dr. Sarvepalli Radhakrishnan was India's first Vice President and second President.
• Teachers' Day was first celebrated in 1962, the year he became the second President of India.
• Since then, the day has been celebrated every year. Dr. Radhakrishnan highly believed that teachers should be the best minds in the country.

The correct answer is Environmental Impact Assessment Method.

• 'Leopold Matrix’ is associated with Environmental Impact Assessment Method.
• It is a procedure for the evaluation of the effect or impact of a proposed development on the environment.​

Important Points

• Environmental Impact Assessment (EIA) is a process of evaluating the likely environmental impacts of a proposed project or development, taking into account inter-related socio-economic, cultural and human-health impacts, both beneficial and adverse.
• It is statutorily backed by the Environment Protection Act, 1986 which contains various provisions on EIA methodology and process.
• UNEP defines Environmental Impact Assessment (EIA) as a tool used to identify the environmental, social and economic impacts of a project prior to decision-making.
• It aims to predict environmental impacts at an early stage in project planning and design, find ways and means to reduce adverse impacts, shape projects to suit the local environment and present the predictions and options to decision-makers.

The correct answer is Lucknow​.

Key Points

• As announced by the Ministry of Housing and Urban Affairs, Lucknow is the cleanest city in Uttar Pradesh.
• Swachh Survekshan 2020 report was released by the Union Housing and Urban Affairs Ministry.
• Swachh Survekshan 2020 covered 4,242 cities, 62 cantonment boards and 92 Ganga towns.
• This survey was carried out in 28 days.

Additional Information

Swachh Survekshan 2020

• Indore was ranked the cleanest city in the overall category this year followed by Surat and Navi Mumbai.
• Chhattisgarh is the cleanest state in the category of states with more than 100 urban local bodies.
• Ahmedabad is India’s cleanest Megacity.
• New Delhi is the cleanest capital city.
• Chhattisgarh’s Ambikapur is the cleanest smallest city.
• Cleanest Cantonment: Jalandhar Cantt, Punjab.

The correct answer is Excise tax.

Key Points

• The state gets a share of the revenue from the Excise tax.
• Excise tax is a form of indirect tax that is levied by the Central Government of India for the production, sale, or license of certain goods.
• It is also collected by state governments for alcohol and narcotics.
• GST has now subsumed a number of indirect taxes including excise tax.
• This means excise tax, does not exist in India except on a few items such as **liquor and petroleum. **​

Additional Information

The correct answer is Dominic Thiem.

Key Points

• US Open Men's single title- Dominic Thiem (from Austria).
• US Open Women's single- Naomi Osaka (from Japan).
• The 2020 US Open was the 140^th edition of tennis US open and the second Grand Slam event of the year.
• It was held on outdoor hard courts at the USTA Billie Jean King National Tennis Center in Queens, New York.
• It was held without spectators due to the COVID 19 pandemic.

The correct answer is The speaker of the Lok Sabha.

Key Points

• The joint sitting of the Indian Parliament for transacting a legislative business is presided over by The speaker of the Lok sabha.
• The joint sitting is called by the President.
• In his absence of the speaker of the Lok sabha, the sitting is presided over by the Deputy Chairman of the Rajya Sabha.
• The quorum to constitute a joint sitting: 1/10^th of the total number of members of the House.
• A joint sitting of both Houses of Parliament of the Indian Constitution given in Article 108.

Additional Information

Speaker of Lok Sabha

• The speaker of Lok Sabha is a member who elected from amongst the members of the house. He chairs the house and no proceedings in the house take place in his absence.
• 1^st Lok Sabha Speaker – Ganesh Vasudev Mavalankar
• 17^th Lok Sabha elections (2019 General Elections), Om Birla is the speaker of Lok Sabha.

The correct answer is Esipattan-Saranath.

Key Points

• The anicient name of Kannauj is Mahoday.
• Kannauj is one of the most ancient place in India having a rich archeological and cultural heritage.
• The ancient name of this place is Kanyakubja or Mahodya (as per Balmiki Ramayana, Mahabharat and Puran) later name kanyakubja was changed as Kannauj the present name of the District.
• ​The ancient name of Mandsaur is Dashpur.
• ​Mandsaur was known as Dashpur in ancient times.
• It is also said that the name originated from Mandodri, wife of Ravana. And the ancient name Dashpur was named after her husband Dashanand, the other name of Ravana.
• The ancient name of Sarnath is Isipatana.
• ​Sarnath has been variously known as Mrigadava.
• Just 10 km north-east of Varanasi, near the confluence of the Ganga and Varuna Rivers in Uttar Pradesh, was Initially known as Isipatana.
• The ancient name of Talkad is Banvasi.
• Banavasi is an ancient town in Uttara Kannada in the state of Karnataka.

The correct answer is Article 253.

• Article 253 of the Indian Constitution empowers Parliament to make laws for implementing international agreements.
• Article 253- empowers Parliament to make any law for the whole or any part of the territory of India for implementing treaties, international agreements, and conventions.
• It enables the Government of India to implement all international obligations and commitments.

Additional Information

• Article 249- empowers the parliament to legislate with respect to a matter in the State List in the national interest.
• Article 250- According to this parliament is empowered to make laws on any item that is included in the State List for the whole or part of India, while a Proclamation of emergency is in operation.
• Article 252- It is about the Power of Parliament to legislate for two or more States by consent and adoption of such legislation by any other State.

The correct answer is (A) - (3), (B) - (1), (C) - (4), (D) - (2)​.

Key Points

• Kiratarjuniyam- It is a Sanskrit epic written in the 6^th century by Bharavi.
• It is his only recognized work.
• It is an epic poem with eighteen cantos in which Lord Shiva tests the power of Arjuna, one of the Pandavas in the Mahabharata.
• Dashakumar Charitam-  It is authored by renowned Sanskrit poet Dandi.
• It relates the adventures of ten princes in their pursuit of love and royal power.
• Buddha Charitam-  It is an epic poem in the Sanskrit maha kavya style on the life of Gautam Budha by Ashvagosha, composed in the early second century.
• Vikramorvashiya-  It is a five-act Sanskrit play by ancient Indian poet Kalidasa who flourished in the 4^th Century CE, on the Vedic love story of king Pururavas and an Apsara, a celestial nymph named Urvashi.

The correct answer is the President of India.

Key Points

• The appointment of acting Chief Justice is to be made by the President under Article 126 of the Constitution.
• Vacancy in the office of the Chief Justice must be filled whatever the period of vacancy.
• Article 126 – Appointment of acting Chief justice.
• Justice D Y Chandrachud is the current and 50^th CJI.

Additional Information

Role of Chief Justice of India

• The Chief Justice is responsible for the allocation of work to other judges.
• The Chief Justice administers the oath of office to the President In the absence of the President and the Vice-President, the Chief Justice sits as the Acting President of India
• The Chief Justice is the ex-officio Chancellor to most autonomous law schools in India.

Eligiblity

• The individual must be a citizen of India.
• The person must not exceed 65 years of age.
• In order to be eligible to be Chief Justice of India, the person must serve as a judge at High Court for at least five years or the person should be an advocate in the High court or the Supreme court for at least 10 years or a distinguished jurist.

The correct answer is Chirand​.

Key Points

• Chirand is the early farming site located on the northern bank of the river Ganga together with two other rivers Sarayu and Son.
• Chirand is an archaeological site in the Saran district of Bihar.
• It is the first known site of the Neolithic age.
• It has a large pre-historic mound (a rounded mass projecting above a surface) containing various artifacts.
• The mound is known for its continuous archaeological record from the Neolithic age (about 2500–1345 BC) to the reign of the Pala Dynasty, which ruled this place during the pre-medieval period.
• This is probably one of the youngest date of early farming known from the Gangetic basin.

Additional Information

• T. Narsipur is a place located in the Mysore district in the Indian state of Karnataka.
• This is the place in South India where Kumbhamela is held every three years.
• The word 'Narasipura' is the name of the town, which is derived from the famous Gunja Narasimhaswamy temple that is located on the right bank of the Kabini river.

The correct answer is the Constitution.

Key Points

• The term 'Office of profit' has not been directly defined by the Constitution.
• But The Constitution of India, apart from enlisting certain qualifications for Members of Parliament and Legislative Assembly, also enumerates certain disqualifications. One of these disqualifications is holding an “office of profit”.
• Under the provisions of Article 102 (1) and Article 191 (1) of the Constitution, an MP or an MLA (or an MLC) is barred from holding any office of profit under the central or state government.
• The articles clarify that “a person shall not be deemed to hold an office of profit under the government of India or the government of any state by reason only that he is a minister”.
• A person will be disqualified if he holds any office of profit under the central or state government, other than an office declared not to disqualify its holder by a law passed by Parliament or state legislature.

 Principles for including ‘office of profit’ as a criterion for disqualification are:

• Makers of the Constitution wanted that legislators should not feel obligated to the Executive in any way, which could influence them while discharging legislative functions.
• In other words, an MP or MLA should be free to carry out her duties without any kind of governmental pressure. The intent is that there should be no conflict between the duties and interests of an elected member.
• The office of profit law simply seeks to enforce a basic feature of the Constitution- the principle of separation of power between the legislature and the executive.

The correct answer is Rahil. 

• A large tank near Mahoba, temples at Ajaygarh and Mahoba and the city of Rajavsini were built by Chandella King Rahil.
• The town Rahilya which is three km southwest of Mohaba was named after Rahil.
• The Chandela or Chandel is a Rajput clan in Central India.
• An area of the Chandelas which was otherwise called Chandela Dynasty led a great part of the Bundelkhand district of central India for a long period of time between the 10^th and the thirteenth century AD.
• Chandel Dynasty is known in Indian history for Maharaja Rao Vidyadhara, who spurned the assaults of Mahmud of Ghazni.

The correct answer is Election Commission.

• While deciding any question relating to the disqualification of a Member of Parliament, the President shall obtain the opinion of the Election Commission.
• Election Commission of India is a permanent Constitutional Body.
• The Constitution provides for an independent election commission to ensure free and fair election to the Parliament, the State legislature,and the offices of President and Vice-President.
• Consists of Chief Election Commissioner +2 Election Commissioners.
• The Chief Election Commissioner is appointed by the President and the other Election Commissioners are appointed by the President after consultation with the Chief Election Commissioner, Article 324 also provides for the appointment of Regional Commissioners at the time of General Elections after consultation with the Election Commission.
• Election Commissioners are appointed for a term of 6 years.
• They are not eligible for re-appointment. Also, they cannot hold any office of profit after their retirement.
• In article 324 of the Indian constitution, it envisages the Election Commission of India (ECI).

Additional Information

The correct answer is Mohsin Fanni.

• The author of the ‘Dastane Mazahib’ which discusses the Din-i-Ilahi of Akbar, was Mohsin Fanni.
• Din-i Ilahi “the religion of God,” was a system of religious beliefs introduced by the Mughal emperor Akbar in 1582 CE.
• Akbar's idea was to combine Islam and Hinduism into one faith, but also to add aspects of Christianity, Zoroastrianism, and Jainism.
• Akbar took a deep personal interest in religious matters.
• He founded an academy, the Ibadat Khana, “the House of Worship,” in 1575, where representatives of all major faiths could meet to discuss questions of theology.

​

Akbar

• Akbar was born as Abu'l-Fath Jalal ud-din Muhammad Akbar on October 15, 1542.
• Akbar succeeded his father, Humayun, in 1556 when he was 13 years old.
• Akbar was given the nickname 'the Great' because of his many accomplishments.
• Akbar started a new religion that was based on the teachings of major religions like Hinduism, Christianity, Zoroastrianism, etc.
• This religious path was later referred to as the Din-i-illahi or Divine Monotheism.
• Historian Vinset Smith called Akbar's Din-i-ilahi as a monument of his folly not of wisdom.

The correct answer is Lentil.

• Lentil is not a Kharif Crop among the given options.

Key Points

• India has many growing seasons due to the prevalence of high temperatures over a long period. The different crop seasons are:​
• Kharif Crop​
• Crops are sown at the beginning of the southwest monsoon and harvested at the end of the South-West monsoon.
• Sown in June-July when rains first begin.
• Harvested in September-October.
• Requires a lot of water and hot weather to grow.
• Important crops: Jowar, Bajra, Soybean, Rice, Maize, Cotton, Jute, Groundnut, Sugarcane.
• Rabi crop
• Crops need a relatively cool climate during the period of growth but a warm climate during the germination of their seed and maturation.
• Sown in October-November.
• Harvested in April-May.
• Important Crops: Wheat, Barley, Gram, Linseed, Mustard, Masur, Lentil and Potatoes, Tobacco.

The correct answer is Right to mortgage and gift.

• Cultivators during the Mughal period enjoyed the Right to mortgage and gift their land.
• The Mughal dynasty was founded by Babur.
• He was a Chagatai Turkic prince and was descended from the Turkic conqueror Timur (Tamerlane) on his father’s side and Genghis Khan, on his mother’s side.
• The Mughal period was between 1526-1857.
• Mughal Empire controlled almost the entire Indian subcontinent.
• It stretched from the edges of the Indus river basin, northern Afghanistan and Kashmir in the north west to the now-Assam and Bangladesh highlands in the east, and down to the uplands of Deccan plateau in the south.

The correct answer is 40% protein and 20% oil.

Key Points

• Soybean seeds contain 40% of crude protein and about 20% of fat.
• The soybean is one of the most economical and valuable agricultural commodities because of its unique chemical composition.
• Since time immemorial, Soyabean has served as meat, milk, cheese, bread as well as oil for people across the globe.
• Soyabean is India’s one of the fastest-growing crops and a significant foreign exchange earner.

The correct answer is KC Niyogi.

Key Points

• KC Niyogi was appointed the Minister of 'Ministry of Rehabilitation' set up on 6 September 1947.
• Kshitish Chandra Neogy (1888–1970), also known as KC Neogi, was an Indian politician from West Bengal.
• He was a member of the Constituent Assembly of India, a member of the first Cabinet of independent India, and the chairman of the first Finance Commission of India.
• Neogy was elected as a member of the Constituent Assembly of India and after independence became a member of the First Cabinet of Independent India under Jawaharlal Nehru as the Minister of Relief and Rehabilitation and later as Minister for Commerce.

The correct answer is 300%.

• Crop intensity is referred to as the number of crops grown in the same field in one agriculture year.
• The formula to calculate any crop intensity is:
• ​Crop intensity= Total cropped area / Total sown Area)X100
• ​In this case, the number of different crops is 3.
• So, here the rotation intensity of Maize – Mustard – Mung crop is 300%.

Concept:

As per Lacey’s theory-

Hydraulic radius (R) is given by,

  $R =\frac{5}{2}\frac{{{{\rm{V}}^2}}}{{\rm{f}}}$

 Where,

 V is the flow velocity

 f is silt factor

Calculation:

 Given,

 v = 1.5 m/s, f = 1

Hydraulic mean radius,

  $R =\frac{5}{2} \times \frac{{{{\rm{V}}^2}}}{{\rm{f}}} = \frac{5}{2} \times \frac{{{1.5^2}}}{1}$ 

 = 5.625 m

Important Points

Following relations are used in Lacey's Theory

The velocity of flow is given by V = ${\left( {\frac{{Q{f^2}}}{{140}}} \right)^{\frac{1}{6}}}$

Where,

f = silt factor = 1.76√d and d = average size of soil particles in mm.

Q is discharge , Area of the section is given by A = Q/V

Perimeter is given by, P = 4.75√Q

Bed slope(S), is given by S = $\frac{{{{\rm{F}}^{\frac{5}{3}}}}}{{3340{{\rm{Q}}^{\frac{1}{6}}}}}$

Concept:

Hydraulic jump: Whenever supercritical flow merges into a subcritical flow, a sudden jump is formed to reduce the energy of the water depth, is known as a hydraulic jump.

For a rectangular channel, energy loss is given as:

\({{Δ E}}=\frac{{{\left( {{\text{Y}}{2}}-{{\text{Y}}{1}} \right)}^{3}}}{4{{\text{Y}}{1}}{{\text{Y}}{2}}}\)

Here,  

Y2 = post jump depth, and Y1 = pre-jump depth

$\text{Efficiency of jump}=\frac{\text{Actual energy loss in jump}}{\text{Required energy loss in jump}}\times 100$

Calculation:

Head loss during a hydraulic jump

${\rm{Δ }}E = \frac{{{{\left( {{y_2} - {y_1}} \right)}^3}}}{{4{y_1}{y_2}}}$

where y1 = pre-jump depth and y2 = post-jump

∴ ΔE = 0.45 m

Explanation:

(i) The ratio of contact pressure to tyre pressure is defined as the rigidity factor.

Rigidity factor = Contact pressure (Pc)/Tyre pressure (P)

R = P_c/P

Where,

Contact pressure = Load on wheel/Contact area

Tyre pressure = Pressure inside the tyre

(ii) Rigidity factor shows degree of tension developed in the tyre. When tyre pressure is low, walls of the tyre are subjected to compression and when tyre pressure is high walls are subjected to tension.

(iii) Tyre pressure is effective in surface layer while contact pressure is effecting throughout the thickness of the pavement i.e. thickness of pavement is affected by contact pressure, not by tyre pressure but contact pressure depends on tyre pressure.

Concept:

Burmister’s Method:

Assumptions

• Materials in each layers are isotropic, homogeneous and elastic.
• Pavement forms a stiffer layer having higher value of E than that of subgrade.
• Layers are in constant contact.
• Surface layer is infinite in horizontal direction but finite in vertical direction.

This method gives the following design deflection values for pavements with different layers –

1. For flexible pavement –

$\Delta = 1.5\frac{{pa}}{{{E_s}}} \times {F_2}$

1. For rigid pavement –

$\Delta = 1.18\frac{{pa}}{{{E_s}}} \times {F_2}$

Where, p = contact pressure at road surface due to wheel load (kg/m2)

 Es = modulus of elasticity of subgrade (kg/cm2),

 a = radius of contact area(cm) and

 F2 = deflection factor which depends upon Es/EP  and h/a,  for single layer system F2 = 1.

Formula used:-

Design deflection for single-layered flexible pavement is given by,

$\Delta = 1.5\frac{{pa}}{{{E_s}}}$

Radius of contact area is given by,

$a = \sqrt {\left( {\frac{P}{p}} \right) \times \frac{1}{\pi }}$

Calculation:

Wheel Load, P = 40 kN

Modulus of elasticity for subgrade and pavement, ES = EP = 25 MPa

Tyre pressure, p = 0.60 MPa

 Radius of contact (a) 

$a = \sqrt {\left( {\frac{P}{p}} \right) \times \frac{1}{\pi }} \Rightarrow a = \sqrt {\frac{{40 \times {{10}^3}}}{{0.6 \times {{10}^6}}} \times \frac{1}{{3.14}}} = 0.14567 m$

Design deflection for flexible pavement

$\Delta ; = 1.5\frac{{pa}}{{{E_s}}}$

$\Rightarrow \Delta ; = 1.5 \times \frac{{0.6 \times 0.1457}}{{25}} = 0.005244 m; \approx 5.244mm;$

Concept:

Maximum theoretical capacity on the basis of space headway:

Theoretical capacity of a lane, $C = \frac{{1000V}}{S}$

where, C = Capacity of single lane (vehicle per hour)

V = Speed of vehicle (kmph)

S = Average center to center spacing of vehicles = SSD + L

SSD = Stopping sight distance

L = Average length of vehicle

Calculation:

Design speed, V = 70 km/hr = 19.44 m/sec

Total reaction time, t = 2 sec

The average length of the vehicle, l = 8 m

Theoretical capacity per hour, $C = \frac{{1000;V}}{S}$

S = average centre to centre spacing between vehicles

S $= ((0.278\ \times V\ \times t)\ +\ (\frac{V^2}{254\ \times f}))\ +\ 8$

S $= ((0.278\ \times 70\ \times 2)\ +\ (\frac{70^2}{254\ \times 0.36}))\ +\ 8$

S = 100.5 m

$\therefore C = \frac{{1000×70}}{{100.5}} = 696.5;veh/hr$

So, the Nearest correct option is b.

Concept:

Stopping Sight Distance:

(i) Stopping sight distance is the minimum distance over which the driver travelling at design speed can apply brakes and bring the vehicle to stop position safely without collision with any other obstruction.

(ii) It is also known as minimum sight distance or non passing sight distance. Stopping sight distance should be provided throughout the length of all roads.

(iii) Stopping sight distance composed of two components:

(a) Lag distance: It is the distance travelled by the vehicle in total reaction time if v is the design speed in m/s and t is the total reaction time in sec, then

lag distance = vt meters

t = total reaction time = 2.5 second (As per IRC)

(b) Braking distance: It is the distance travelled by vehicle after the application of brakes.

Braking distance, $L = \frac{{{v^2}}}{{2gf}}$

Where, v =  design speed of vehicle in m/s

f = Coefficient of friction

Therefore,

Stopping sight distance (SSD) = Lag distance + Braking distance

SSD = $vt + \frac{{{v^2}}}{{2gf}}$

Calculation:

Given, v = 100 kmph = 100/3.6 = 27.78 m/s , (1 kmph = 1/3.6 m/s)

f = 0.10

t = 2.5 second (As per IRC)

$SSD = vt + \frac{{{v^2}}}{{2gf}} = 27.78 \times 2.5 + \frac{{{{27.78}^2}}}{{2; \times ;9.81; \times; 0.10}}$

SSD = 462.78 m

So as per given option most appropriate answer is 464 m

Explanation:

The selection of a suitable site for an airport depends upon the class of airport under consideration. However, if such factors as required for the selection of the largest facility are considered the development of the airport by stages will be made easier and economical. The factors listed below are for the selection of a suitable site for a major airport installation:

(i) Regional plan

(ii) Airport use

(iii) Proximity to other airports

(iv) Ground accessibility

(v) Topography

(vi) Obstructions

(vii) Visibility

(viii) wind

(ix) Noise nuisance

(x) Grading, drainage and soil characteristics

(xi) Future development

(xii) Availability of utilities from town

(xiii) Economic consideration

(xiv) Runway orientation

Explanation:

Types of pavement structure:

Pavements are generally classified into two categories based on structural behavior:

(i) Flexible pavement

(ii) Rigid pavement

Flexible pavement:

(i) The pavements which have very less flexural strength are called flexible pavements.

(ii) This type of pavements transmit the load to the lower layer by grain to grain transfer.

(iii) A typical flexible pavement consists of four components as shown in the figure below

• Soil subgrade
• Sub-base course
• Base course
• Surface course

Rigid pavement:

(i) Rigid pavement are those which possess worthy flexural strength.

(ii) The rigid pavement transmits the wheel load stresses through a wider area below by the slab action.

Important point:

(i) IRC–37:2012 Design code for the design of flexible pavement.

(ii) Flexible pavement are commonly designed using empirical charts or equations.

(iii) IRC-58:2012  Design code for the design of rigid pavement.

(iv) Rigid pavements are designed using elastic theory

Concept:

Braking distance:

(i) It is the distance travelled by vehicle after the application of brakes.

(ii) It is obtained by equating the work done in stopping the vehicle and kinetic energy stored in the vehicle.

Let us consider, F is the maximum frictional force developed and L is the braking distance, then the work done against friction in stopping the vehicle will be F × L.

Where, frictional force, F = f × W

Where W is the total weight of vehicle and f is the coefficient of friction

So, work done in stopping the vehicle = f × WL

While the kinetic energy stored in vehicle = $\frac{1}{2}m{v^2} = \frac{1}{2}\frac{{W{v^2}}}{g}$  (∵ W = mg)

Now by using the law of conservation of energy,

Work done in stopping the vehicle = Kinetic energy stored in the vehicle

$fWL = \frac{1}{2}\frac{{W{v^2}}}{g}$

$L = \frac{{{v^2}}}{{2gf}}$

Braking distance, $L = \frac{{{v^2}}}{{2gf}}$

If the braking efficiency is η,

Braking distance, $L = \frac{{{v^2}}}{{2gf\eta }}$

Calculation:

Given,

v = 50 kmph = 50/3.6 = 13.89 m/s  (∵ 1 kmph = 1/3.6 m/s)

L = 18 m

f = 0.75

∵ we know that, braking distance, $L = \frac{{{v^2}}}{{2gf\eta }}$

$18 = \frac{{{{13.89}^2}}}{{2; \times ; 9.81; \times ;0.75; \times ;\eta }}$

η = 0.7284 = 72.84 %

Concept

Sleeper density is denoted by (n + x)

These vary from (n+3) to (n+6)

Where ‘n’ is the length of one rail

Generally length of one rail (n) = 13 m (Need to remembered)

Calculation

Given,

 Length of rail = 650 m

 For BG track, length of rail = 13 m

 Number of rails $= \frac{{650}}{{13}} = 50$ 

 Sleeper Density = n + 5

 Number of sleepers per rail length = 13 + 5 = 18

 So number of sleepers needed = Number of rails × Sleeper density

 = 50× 18 = 900 

 ∴ Number of sleepers need = 900

Explanation:

Camber:

• Camber is the slope provided in the transverse direction of the road to drain off the rainwater from the road surface.
• Usually, a camber is provided in the straight roads by raising the center of the carriageway with respect to the edges forming the highest point at the center.
• The straight slope of the camber is easier to be cast and maintained in the case of concrete pavement.
• Parabolic shaped camber is difficult to construct and maintain. It has steeper edges which are inconvenient to use. However, it is kept in bituminous pavements for better drainage of water.

Explanation:

The various road patterns may be classified as follows:

(a) Rectangular or block pattern

(b) Radial or star and block pattern

(c) Radial or star and circular pattern

(d) Radial or star and grid pattern

(e) Hexagonal pattern

Important points:

(i) The rectangular pattern has been adopted in the city roads of Chandigarh

(ii) Radial and circular pattern is the road network of Cannaught Place in New Delhi

Concept:

Alkalinity is defined as ability of water to neutralize the acid.

Major constituents are CO32- and HCO3 –

and Minor constituent is OH-

[OH-] is negligible so we don’t consider in calculation  

Hardness of water is defined as the concentration of multivalent metallic cations which destroy the surfaced property of soap (Concentration of ions) which reacts with Soap and leads to the formation of a precipitate.

Ca 2+, Mg2+ are major constituents

Hardness is categorized into 2

Carbonate hardness and non-carbonate hardness

a) If TH > TA

Then, CH = TA

b) If TH ≤ TA

Then CH = TH

 As in the question it is mentioned if methyl alkalinity exceeds Total Hardness then all of the hardness is carbonate Hardness

Note:

The relative quantity of alkalinity species are pH-dependent. It is obtained using a titration curve.

The titration curve is plotted by taking measurement of pH at every stage.

• The conversion of CO32-  to HCO3-  is essentially complete at pH = 8.3, but  HCO3- formed also requires acid. Hence it is thought that half of CO32-  have been neutralized up to pH = 8.3.
• The neutralization of OH- is completed up to pH = 8.3. Hence $\left( {O{H^ - } + \frac{1}{2}CO_3^{2 - }} \right)$ alkalinity is completely neutralized up to pH = 8.3
• At pH = 4.5 all the bicarbonates are neutralized. Hence, the amount of acid required to titrate a sample of water to pH = 4.5 is equivalent to the total alkalinity of water.

P = Phenolphthalein alkalinity

M = Methyl orange alkalinity

• If P = M all alkalinity is caustic alkalinity.
• If P = M/2, all alkalinity is carbonate alkalinity.
• If P < M/2, predominant species are carbonate and bicarbonate.
• If P > M/2, the predominant species are carbonate and Hydroxide.
• If P = 0 total alkalinity is bicarbonate alkalinity

Explanation

Correct order is

Zone A – Zone of clear water

Zone B – Zone of Degradation

Zone C – Zone of Active decomposition

Zone D – Zone of Recovery

Zone E – Zone of recovery

From the curve, we can say in the zone of active decomposition there is a minimum level of dissolved oxygen.

ZONES OF POLLUTION: A polluted stream undergoing self-purification presents the following four distinct zones of pollution

(1) ZONES OF DEGRADATION: This usually occurs below the output sewer when discharging its contents into the stream. The zone is characterized by water becoming dark and turbid with the formation of sludge deposits on the bottom. Dissolved oxygen gets reduced to 40%.

(2) ZONES OF ACTIVE DECOMPOSITION: This is marked by heavy pollution. It is characterized by the absence of dissolved oxygen; water is grayish and darker with active anaerobic organic decomposition accompanying and with the evolution of methane (CH₄), Hydrogen sulphide (H₂S), Carbon dioxide (CO₂) and Nitrogen (N₂), bubbling to the surface with masses of sludge forming black scum. Fish life is practically absent, fungi and bacteria disappear. As the organic decomposition slackens, reaction sets in and D.O. again rises to its original level (i.e. 40%).

(3) ZONE OF RECOVERY: In this zone, the stream tries to recover its former appearance. Most of the organic matter has been settled as sludge, B.O.D. falls and the D.O. content rises above 40% microscopic aquatic life reappears. Water becomes clearer, fungi decrease and algae reappear. Mineralization is active and products such as nitrates, sulphates, and carbonates are formed.

(4) CLEAR WATER ZONE: In this, the natural stream condition is restored, the D.O. is higher than the BOD. Oxygen balance is attained and recovery is said to be complete. Water becomes attractive in appearance. Some pathogenic organisms may, however, be present.

Explanation:

(i) The function of the fine hair inside our nose is to keep foreign or unwanted particles away from entering the lungs through the nasal cavity.

(ii) The function of mucus inside our nose is it aids in moisturizing the inhaled air and prevents tissues such as the nasal and the airway epithelia from drying out.

(iii) Particles that are 10 microns or less are considered respirable and can settle deep into the lungs often causing adverse health effects. So Hairs of human nose can remove all the particles of size greater than 10 micron.

Explanation:

Detention time for different type of Treatment unit

Note: Septic tank has high detention time whereas Grit chamber has less detention time.

Explanation

Dead-end System:

• It is also called a tree system, it consists of one main supply line, from which originates (generally at right angles) a number of sub-main pipes; Each sub-main is then divided into several branch pipes, called laterals
• This type of distribution system is suitable for older towns which have developed in a haphazard manner, without properly planned road e.g. old city, irregularly grown city, etc; It is economical and simple

Gridiron System:

• It is also known as an interlaced system or reticulation system, the mains, sub-mains, and branches are all connected with each other
• There is no dead-end in the system as looping is provided and is most suitable for well-planned cities; There is equal pressure in all pipe and multiple flow path

Radial system:

• If a city or town is having a system of radial roads emerging from different centers, the pipeline can be best laid in a radial method by placing the distribution reservoir at these centers
• The water from the distribution reservoir is then supplied radially to distribution pipes and thus also known as zonal distribution

Ring System:

• It is also called a circular system; In this system, a closed ring, either circular or rectangular, of the main pipes, is formed around the area to be served; The distribution area is divided into rectangular or circular blocks, and the main water pipes are laid on the periphery of the blocks
• This system is very suitable for towns and cities having well-planned roads; It is economical and pressure at the ends are reasonably equal

The Railway Board has decided that on Indian Railways, transition curves will normally be laid in the shape of a cubic parabola.

• In this curve, both the curvature and the cant increase at a linear rate
• The cant of the transition curve from the straight to the curved track is so arranged that the inner rail continues to be at the same level while the outer rail is raised in the linear form throughout the length of the curve
• A straight-line ramp is provided for such transition curves

Explanation

A conference of the chief engineers of all the states and provinces was convened in 1943 by the government of India at Nagpur, at the initiative of the Indian Roads Congress to finalize the first road development plan for the country as a whole.

This is the landmark in the history of road development in India, as it was the first attempt to prepare a coordinated road development programme in a planned manner. At the end of this plan, the target road length aimed at was 16 km per 100 square km area of the country.

The first 20-year plan is also known as the Nagpur road plan. It classified the roads in India based on location and function into the following five categories:

1. National Highways
2. State Highways
3. Major District roads
4. Other District roads
5. Village Roads.

While National Transport Policy committee is appointed by the government of India in the year 1978 to prepare a national transport policy for the country for the next decade.

Flexible pavements: Flexible pavements are so named because the total pavement structure deflects, or flexes, under loading. Flexible pavement distributes the wheel load through a set of layers grain by grain.

Flexible pavement may be constructed in a number of layers and the top layer has to be of the best quality to sustain maximum compressive stress, in addition, to wear and tear. The lower layers will experience the lesser magnitude of stress and low-quality material can be used.

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%

\({\rm{Final;length;}}\left( {{{\rm{l}}_3}} \right) = {{\rm{l}}2} + \left( {20 \times \frac{{{{\rm{G}}{{\rm{effective}}}}}}{{100}}{\rm{% }}} \right){{\rm{l}}_2}\)

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

Concept:

Geometric incremental method:

This method is suitable for young and rapidly developing cities as they record logarithmic or exponential increases in population.

If population increase is given per decade:

Population after n decades (P_n) is given by

${{\rm{P}}_{\rm{n}}} = {P_o} \times {\left( {1 + \frac{r}{{100}}} \right)^n}$

If the population increase is given per year:

\({{\rm{P}}{\rm{t}}} = {{\rm{P}}{\rm{o}}} \times {e^{\left( {\frac{k}{{100}} \times t} \right)}}\)

where,

P_n → Population after ‘n’ number of decades

P_t → Population after ‘t’ number of years

P_o → Initial population considered

n → number of decades

t → number of years

r → geometric rate constant per decade

k → geometric rate constant per year

Calculation:

\({{\rm{P}}{\rm{t}}} = {{\rm{P}}{\rm{o}}} \times {e^{\left( {\frac{k}{{100}} \times t} \right)}}\)

Since given data is in increase per year:

Geometric rate constant per year (k) = $\sqrt {0.20 \times 0.4}$ = 0.2828 = 28.28%

${{\rm{P}}_{\rm{t}}} = 8400 \times {e^{\left( {0.2828 \times 1} \right)}} = 11145.93; \approx 11146$

Confusion Points

If the population increase is given per year: Exponential method is used

If population increase is given per decade: Geometric increase method is used

Concept

We know, pH is defined as the negative logarithm of [H]⁺ ion.

So, 

pH = -log10 [H+]

-log10 [H+] = pH

[H+] = 10 -pH

Calculation

 pH of water samples are 

 pH = 4 ⇒[H⁺] = 10 ^{- 4}

 pH = 7 ⇒ [H+] = 10^-7 

Comparing Both [H⁺]

\(\frac{{{{\left[ {{H^ + }} \right]}I}}}{{{{\left[ {{H^ + }} \right]}{II}}}} = \frac{{{{10}^{ -4}}}}{{{{10}^{ - 7}}}} = {10^{3}} = 1000\)

Hence, first sample is 1000 times acidic than second sample.

Explanation

Sludge volume index (SVI)

• It is the volume occupied in mL by one gm of solids in the mixed liquor after settling for 30 minutes.
• It indicates the physical state of sludge produced in a biological aeration system.
• It is used to decide the rate of recycling of sludge (Q) required to maintain the desired Mixed Liquor Suspended Solids (MLSS) and Food to Microorganism (F/M) ratio in the aeration tank to achieve the desired degree of purification. So by reducing the recycling ratio SVI can be controlled.

Calculation

The settled sludge volume Vob = 30 cm3/lit 

 = 30 mL/lit

The concentration of suspended solids in mixed liquor, Xob = 3 g/lit

\(\therefore {\rm{SVI}} = \frac{{{{\rm{V}}{{\rm{ob}}}}}}{{{{\rm{X}}{{\rm{ob}}}}}} = \frac{{30}}{3} = 10{\rm\space{mL}}/{\rm{gm}}\)

Explanation:

COD (Chemical oxygen demand):

It is the amount of oxygen required for the decomposition of both biodegradable & non-biodegradable organic matter present in the system.

BOD (Biochemical oxygen demand):

• It is the amount of oxygen required for the decomposition of biodegradable organic matter present in the system.
• BOD during 5 days at 20°c is taken as standard BOD and is approximately 68% of the ultimate BOD.

​ThOD (Theoretical oxygen demand):

If the type of organic matter and its concentration present in the system is known then the amount of oxygen required for its decomposition can be computed theoritacly and is termed as ThOD.

TOC (Total organic carbon):

TOC is just another way to express the concentration of organic matter present in the system in terms of its carbon content.

BOD to COD ratio:

• It signifies the quality of wastewater with respect to the type of organic matter present in it that is either biodegradable or non-biodegradable.
• If the value of this ratio high that means the biodegradable organic matter is high and vice-versa.

Explanation:

Sedimentation:

(i) The process of removal of suspended solids from the water is termed as sedimentation.

(ii) Sedimentation is a natural process by which solids with higher density than the fluid in which they are suspended, settles under the action of gravity.

(iii) Sedimentation tanks are designed to reduce the velocity of flow to reduce turbulence.

(iv) Efficiency of sedimentation is comparatively less if water consists of fine suspended particle, hence in such cases coagulation aided sedimentation is adopted.

(v) Coagulation aided sedimentation is performed in 3 stages:

• Coagulation (Fast mixing)
• Flocculation (slow mixing)
• Sedimentation

Coagulation:

(i) It is the process in which certain chemical terms as coagulants are added in the water so as the neutralise negative protective charge over the particle to increase their opportunity to come in contact with each other, thus grow in size and get easily settled in following sedimentation process.

(ii) To neutralise the charge over the suspended impurities minimum energy (threshold energy) is required which is provided by induced fast mixing in this process

Flocculation:

(i) Flocculation is the process in which neutralise particle is brought in intimate contact with each other to promote their agglomeration resulting in increase size of flocs, which can get easily settled in the following sedimentation process.

(ii) To increase the opportunity of the neutralise particle to come in contact with each other slow mixing is induced during flocculation.

Explanation:

coliform bacteria:

(i) Coliform bacteria are organisms that are present in the environment and in the feces of all warm-blooded animals and humans.

(ii) Coliform bacteria will not likely cause illness. However, their presence in drinking water indicates that disease-causing organisms (pathogens) could be in the water system. Most pathogens that can contaminate water supplies come from the feces of humans or animals.

(iii) Testing drinking water for all possible pathogens is complex, time-consuming, and expensive. It is relatively easy and inexpensive to test for coliform bacteria.

(iv) If coliform bacteria are found in a water sample, water system operators work to find the source of contamination and restore safe drinking water.

(v) There are three different groups of coliform bacteria; each has a different level of risk, Total coliform, fecal coliform, and E. coli

(vi) Total coliform, fecal coliform, and E. coli are all indicators of drinking water quality.

(vii) The total coliform group is a large collection of different kinds of bacteria. Fecal coliforms are types of total coliform that mostly exist in feces. E. coli is a sub-group of fecal coliform.

(Viii) When a water sample is sent to a lab, it is tested for total coliform. If total coliform is present then the Presumptive coliform test is confirmed, the sample will also be tested for fecal coliform and if fecal coliform is present then the Confirmed coliform test is complete and then again water sample is tested for E. coli, if E. coli is present then the Completed coliform test is done.

Explanation:

Relative stability:

The ratio of oxygen available in the wastewater (as dissolved oxygen) to the total oxygen required to satisfy its first state BOD is called relative stability.

$Relative;stability; = ;\frac{{BOD;removed}}{{Total;BOD}} \times 100$

Note:

As the DO level of the sewage is equal to the total oxygen required to decompose the organic matter present, total BOD will be removed.

∴ Relative stability = 100%

Explanation:

Trickling filter is a device that is used for the treatment of sewage. It is based on an aerobic attached growth system. Following factor affecting the rate of removal of organic matter in T.F.

(i) Temperature

(ii) Hydraulic loading rate

(iii) Organic loading rate

Following design parameters used in the trickling filter.

(i) **Hydraulic loading rate (m³/m²/day) (**Gives the area of the filter required)

(ii) Organic loading rate (kg/m³/day) ( Decide the volume of the filter)

(iii) Depth

Given:

d = 1.2 m

$s = \frac{1}{{1000}}$

Velocity when sewer runs full (V_full) = 0.9 m/s

${V_{full}} = \frac{1}{n};R_f^{\frac{2}{3}}\frac{{{S^1}}}{2}$

${V_{half}} = \frac{1}{n}R_h^{\frac{2}{3}};{S^{\frac{1}{2}}}$

${R_f} = \frac{{{A_f}}}{{{P_f}}} = \frac{{\left( {\frac{{\pi {D^2}}}{4}} \right)}}{{\pi D}} = \frac{D}{4}$

${R_h} = \frac{{{A_h}}}{{{P_h}}} = \frac{{\left( {\frac{{\pi {D^2}}}{8}} \right)}}{{\left( {\frac{{\pi D}}{2}} \right)}} = \frac{D}{4}$

$\frac{{{V_{half}}}}{{{V_{full}}}} = \frac{{\left( {\frac{1}{n}} \right){{\left( {\frac{D}{4}} \right)}^{\frac{2}{3}}}{{\left( S \right)}^{\frac{1}{2}}}}}{{\left( {\frac{1}{n}} \right){{\left( {\frac{D}{4}} \right)}^{\frac{2}{3}}}{{\left( S \right)}^{\frac{1}{2}}}}} = 1$

V_half = V_full = 0.9 m/s

Explanation:

Self-purification of running stream:

When the sewage effluent is disposed of in a river stream it undergoes the process of self-purification by the following natural mechanism;

Dilution & Dispersion:

In real terms, it is not the mechanism of self-purification it only reduces the potential hazard of the sewage effluent into the river stream.

Sedimentation:

As per the turbulence available in the river stream organic suspended particles get settled in it thereby helping the river water to get self purified.

Sunlight:

In the presence of sunlight, a photosynthesis reaction takes place during which O₂ is released which is utilized by the micro-organism for the decomposition of organic matter thereby helping the river water to get self purified.

Oxidation & Reduction:

Decomposition of the organic matter by the action of micro-organism over the surface of the river stream and by the anaerobic micro-organism on the bed of the river collectively helps the river to get self purified.

Concept:

Relief Displacement:

(i) Relief is the elevation of object above or below the datum. Relief displacement is the lateral

displacement (also called shift) in the position of image on the photograph of a ground object.

(ii) Relief displacement is given by;

$Relief;displacement;\left( d \right) = \frac{{rh}}{H}$

Where, h = Height of the object above datum

H = Flying height above datum

r = Radial distance of the image of the object top 

Calculation:

Given,

h = 50 m, H = 5000 m, r = 20 cm

Now,

$Relief;displacement;\left( d \right) = \frac{{20 \times 50}}{{5000}} = 0.2;cm$

Concept:

The distance through which the curve is shifted is known as shift (S) of the curve and is equal to

 ${\bf{S}} = \frac{{{{\bf{L}}^2}}}{{24{\bf{R}}}}$

where,

L is the length of each transition curve and R is the radius of the desired circular curve

The offset for the transition curve is given by: -

${\bf{offset}};\left( {\bf{y}} \right) = \frac{{{\ell ^3}}}{{6{\bf{RL}}}}$

Where,

ℓ = measured along the curve (m),

R = radius of the curve (m), and

L = length of the curve (m)

∴ For offset at the junction point, ℓ = L

${\bf{y}} = \frac{{{{\bf{l}}^3}}}{{6{\bf{RL}}}} = \frac{{{{\bf{L}}^3}}}{{6{\bf{RL}}}} = \frac{{{{\bf{L}}^2}}}{{6{\bf{R}}}}$         

Important Points

 The relation between Shift of curve S and offset of transition curve is:

 4 × S = Offset

Explanation:

(i) If the distance of the point or detail from the chain line is very large, then long offsets need to be taken.

(ii) Tie lines are the lines that are run to locate details in order to avoid long offsets.

So line CD represented in the figure is a tie line.

Important point:

Base line: It is the long survey line which is run through the middle of the area to be surveyed. The frame work of triangles is built on the base line.

Here AB is the base line

Check line:

(i) These lines are run to check the accuracy of the traverse consisting of a frame work of triangles

(ii) Check lines are measured in the field during the survey process of the land.

(iii) It is also known as Proof lines.

Explanation:

(i) In strike direction case, that is, when the tunnel is driven parallel to strike of the bed. The pressure distribution to the exposed layers is unsymmetrical along the periphery of the tunnel opening.

(ii) In case of parallel to the dip direction, the layers offer a uniformly distributed load on the excavation, even relatively weaker rocks might act as self-supporting in such cases.

Note:

Two expressions commonly used to describe the rock structure are ‘strike’ and ‘dip’. Strike is the horizontal direction of the structure on the rock surface.

Dip is the angle of the structure relative to the horizontal rock surface.

Explanation:

Tachometer:

(i) Tacheometer is nothing but a transit theodolite fitted with a stadia diaphragm.

(ii) The stadia diaphragm generally consists of two hairs, one above and one below and equidistant from a central horizontal hair.

Systems of Tacheometric measurements:

(a) Stadia system

(b) Tangential system

(c) Subtense bar system

(a) Stadia system:

(i) In the stadia system of tacheometry of, the tacheometer is set up station A and staff at station B.

(ii) The staff intercept between the upper and lower stadia is measured along with vertical angle θ made with horizontal.

(iii) The horizontal distance D between the instrument station A and staff station B and the difference of elevation between stations A and B are determined from the staff intercept (s) and the vertical angle.

(iv) Refraction correction may occur during measurement.

(b) Tangential system:

(i) Here in this system of tacheometry, the hairs are not at all required.

(ii) This method can even be used when telescope is not provided with a diaphragm.

(iii) The staff had two targets at a fixed distance (s) apart.

(iv) The vertical angles θ1 and θ2 are measured to the two targets.

(v) Now these vertical angles and the fixed distance on staff are used to determine the horizontal distance and difference of elevation.

(C) Subtense bar system:

(i) Here in this system of tacheometry, a bar of fixed length is used which is known as subtense bar as shown in fig.

(ii) The subtense bar has two targets at the ends at a fixed distance (s) apart.

(iii) The horizontal angle subtended between the instrument station A and the two targets on the subtense bar is measured.

(iv) In subtense bar system less chance of refraction error because the line of sight does not change the path

Concept:

Distance of visible Horizon:

Let there is a point C at an elevation h from the mean sea level. A is the point on the horizon as observed from C. The tangent from C meets the mean sea level at A. The distance AC is the distance of visible horizon.

h = 0.0673d²

d = 3.855 √h

Where,

d = Distance AC (in km)

h = height of the object (in m)

Calculation:

Given,

 h₁ = h₂ = 6.75 m

So total distance = d₁ + d₂

d₁ + d₂ = 3.855 √h₁ + 3.855 √h₂

d₁ + d₂ = 3.855 √6.75 + 3.855 √6.75 = 20.03 km

d₁ + d₂ ≈ 20 km

The correct answer is 'Passive Remote Sensing'

Key Points

• passive remote sensing:
• Passive remote sensing is a system that relies on the Sun as the natural source of electromagnetic energy.
• It records the energy that is naturally reflected or emitted by objects on the Earth’s surface.
• Sensors in passive remote sensing do not generate their own energy; instead, they depend on available solar radiation or thermal emission.
• This system is widely used for observing landforms, vegetation, water bodies, and atmospheric conditions.

Additional Information

• geographical information system:
• GIS is a computer-based system used for storing, analyzing, and displaying spatial data.
• It does not involve sensing or recording electromagnetic energy from objects.
• global positioning system:
• GPS is a satellite-based navigation system used to determine precise locations on Earth.
• It does not record reflected or emitted electromagnetic radiation from objects.
• active remote sensing:
• Active remote sensing systems emit their own electromagnetic energy toward the Earth’s surface.
• They record the energy reflected back, rather than relying on natural solar radiation.

Important Points

• Passive remote sensing depends on natural energy sources, mainly sunlight.
• It records both reflected solar radiation and naturally emitted thermal energy.
• Most optical satellite images are obtained using passive remote sensing techniques.

Explanation

We know,

Curvature correction is given by:

${C_c} = \frac{{{d^2}}}{{2R}}$

Where,

d = distance between points under consideration

R = Radius of earth = 6370 km

Refraction correction:

${C_R} = \frac{1}{7} \times \frac{{{d^2}}}{{2R}} = \frac{{{d^2}}}{{14R}}$

So,

The ratio of curvature correction to refraction correction is 1/(1/7) = 7

∴ Ratio is 7

Explanation

Given,

 Upgrade of 2.0 % followed by the downgrade of 2 % 

 N1 = 2 % , N2 = - 2%

  Rate of change of grade is 0.05 % per 20 m chain.

 Total change in grade (N) = N1 – N2

 N= 2 % - (- 2%) = 2% + 2%

 N = 4 %

${\rm{Total;Length}} = \frac{4}{{0.05}} \times 20 =1600;m$

Explanation:

Structural geology:

Structural geology is the study of factors such as origin, occurrence, classification, type, and effects of various secondary structures like folds, faults, joints, rock cleavage and are different from those primary structures such as bedding and vesicular structure, which develop in rocks at the time of their formation.

Folds:

(i) Folds are one of the most common geological structures found in rocks. When a set of horizontal layers are subjected to compressive forces, they bend either upward or downward. The bend noticed in rocks are called folds.

(ii) In terms of their nature too, folds may occur as single local bends or may occur repeatedly and intricately folded to the tectonic history of the region.

joints and Faults:

(i) Structurally, faults may be described as fractures along which relative displacement of adjacent blocks has taken place.

(ii) If such relative displacement does not take place on either side of the fracture plane, it is called a joint. Thus both joints and faults are fractures in rocks but with differences in the kind of displacement. Joints may be described as a set of aligned parallel cracks or openings in geological formations.

Explanation:

Velocity Distribution for Turbulent Flow in pipes.

Prandtl’s universal velocity distribution equation.

$v = {v_{max}} + 2.5{V^*}{\log _e}\left( {\frac{y}{R}} \right)$

Where,  ${V^*} = \sqrt {\frac{{{\tau _0}}}{\rho }} = Shear;or;friction;velocity.$

y = distance from the pipe wall

ρ = Density of fluid.

On observing the above equation we can say that velocity distribution in the turbulent boundary layer follows logarithmic law.

Important Points

The above equation is valid for both smooth and rough pipes.

Concept:

Flow over a Rectangular notch:

Discharge measured through rectangular notch:

 $Q = \frac{2}{3}{C_d}\sqrt {2g} L{H^{3/2}}$

Where, Cd = Coefficient of discharge

L = Length of crest or sill

H = Still water head

Effect on computed discharge over a weir or notch due to error in the measurement of head and length:

 $Q = \frac{2}{3}{C_d}\sqrt {2g} L{H^{3/2}}$

Or, $Q = KL{H^{3/2}}$

Where, $K = \frac{2}{3}{C_d}\sqrt {2g}$

Differentiating the above equation

dQ = KH^3/2dL + KL × (3/2) H^1/2 dH

 $\frac{{dQ}}{Q} = \frac{{dL}}{L} + \frac{3}{2}\frac{{dH}}{H};$

Calculation:

Give,

$\frac{{dH}}{H} = 2\%$

$\frac{{dL}}{L} = 3\%$

So, the Percentage error in discharge,

$\frac{{dQ}}{Q} = \frac{{dL}}{L} + \frac{3}{2}\frac{{dH}}{H}$

$\frac{{dQ}}{Q} = 3\% + \frac{3}{2} \times ;2\% = 6\%$

Explanation:

• The equivalent length method (L/D ratio) allows describing the pressure drop through the pipe due to friction loss corresponding to flow rate i.e. if in the two pipe of a system having same flow rate and same friction loss then this system of the pipe is said to be equivalent.
• An equivalent pipe is defined as the pipe of uniform diameter having loss of head and discharge equal to the loss of head and discharge of a compound pipe consisting of several pipes of different lengths and diameters.
• There are two combinations of equivalent pipe system:

1. Pipes connected in series
2. Pipes connected in parallel

For pipes connection in series:

• The discharge through all the pipes is the same and equivalent head loss is the sum of head loss in all the individual pipes.
• Qeq = Q1 = Q2 = … and heq = h1 + h2 + …..
• $\frac{{{l_{eq}}}}{{d_{eq}^5}} = \frac{{{l_1}}}{{d_1^5}} + \frac{{{l_2}}}{{d_2^5}} + \ldots$
• For pipes of the same diameter as all the pipes equivalent length of the pipe is given by the sum of the length of all the pipes.

For pipes connected in parallel:

• The head loss through all the pipes is the same and equivalent discharge is the sum of discharge in all the individual pipes.
• heq = h1 = h2 = …. and Qeq = Q1 + Q2 + …
• $\frac{{{l_{eq}}}}{{d_{eq}^5}} = \frac{{{l_1}}}{{d_1^5}} = \frac{{{l_2}}}{{d_2^5}}{\rm{;and;}}{\left( {\frac{{d_{eq}^5}}{{{l_{eq}}}}} \right)^{\frac{1}{2}}} = {\left( {\frac{{d_1^5}}{{{l_1}}}} \right)^{\frac{1}{2}}} + {\left( {\frac{{d_2^5}}{{{l_2}}}} \right)^{\frac{1}{2}}} + \ldots$
• For pipes of equal length and diameter deq = n2/5d