Previous Year Paper

UPPSC Polytechnic Lecturer Civil 23 Dec 2021 Official Paper II (Previous Year Paper)

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General Studies (25 questions)

Question 1

The "Chandra Prabha Wildlife Sanctuary" a major wildlife sanctuary of Uttar Pradesh is situated in which of the following?

  1. AOption (a)

    Pilibhit District

  2. BOption (b)

    Chandauli District

  3. COption (c)

    Unnao District

  4. DOption (d)

    Maharajganj District

View answer & explanation

Correct answer

Answer: (b)B.

Chandauli District

Why this answer is correct

The Correct Answer is Chandauli District.

Key Points

  • The Chandra Prabha Wildlife Sanctuary is situated in the Chandauli district of Uttar Pradesh state in central India.
  • It is well endowed with beautiful picnic spots, dense forests, and scenic waterfalls like Rajdari and Devdari that attract tourists every year to its vicinity.
  • Chandra Prabha Wildlife Sanctuary is situated about 70 kilometers from the historic city of Varanasi.
  • The area was made a hunting preserve for the rulers of Benares in the second half of the 18th century.
  • The wildlife sanctuary was established in May 1957. Asiatic lions were introduced at Chandra Prabha in 1958.

Additional Information

  • Importance of Wildlife Sanctuaries: There are a number of reasons for establishing wildlife sanctuaries. Some of the reasons are listed below:
  • The wildlife sanctuaries are established to protect endangered species.
  • It is quite difficult to always relocate the animals from their natural habitat, therefore, protecting them in their natural environment is advantageous.
  • The endangered species are specially monitored in the wildlife sanctuaries. If they reproduce and grow in number while under protection, few specimens can be kept for breeding in the conservation parks for their survival.
  • Biologist activities and researches are permitted in the wildlife sanctuaries so that they can learn about the animals living there.
  • A few sanctuaries take in injured and abandoned animals and rehabilitate them to health before releasing them in the forest.
  • Wildlife sanctuaries preserve endangered species and protect them from humans and predators.

Question 2

How many lok sabha members are there in the Committee on Privileges?

  1. AOption (a)

    12

  2. BOption (b)

    15

  3. COption (c)

    20

  4. DOption (d)

    22

View answer & explanation

Correct answer

Answer: (b)B.

15

Why this answer is correct

The correct answer is 15.

Key Points

S. No.CommitteeMembersObjective
1Rules committeeIn Lok sabha 15 members including the speaker as its chairman. In Rajya sabha 16 members including the chairman as its ex officio chairman.Consider the matters of procedure and conduct of business in the House and recommends necessary amendments or additions to the rules of the house.
2Library committeeIt consists of 9 members ( six from Lok sabha and three from Rajya sabha)Considers all matters related to the library of the parliament
3General-purpose committeeIn each house, it consists of the presiding officer, deputy speaker/ vice-chairman, the panel of chairpersons in both houses, chairpersons of all departmental committees, leaders of the recognized party.It considers and advises on matters concerning affairs of the house which do not fall under any committee
4Committee on privilegesIt consists of 15 members from Lok Sabha and 10 from Rajya sabhaExamines the cases of breach of privileges of the House

Question 3

With reference to Nobel Peace Prize announced in October 2021, which of the following statement(s) is/are correct?

  1. Maria Ressa and Dmitry Muratov have jointly won this prize.
  2. Both the winners of this year's award are journalists.

Select the correct answer from the codes given below:

  1. AOption (a)

    Only 1

  2. BOption (b)

    Only 2

  3. COption (c)

    Both 1 and 2

  4. DOption (d)

    Neither 1 nor 2

View answer & explanation

Correct answer

Answer: (c)C.

Both 1 and 2

Why this answer is correct

The correct answer is Only 1.

Key Points

  • Journalists Maria Ressa, of the Philippines, and Dmitry Muratov, of Russia, have won the 2021 Nobel Peace Prize.
  • They have been honored for their efforts to safeguard freedom of expression, which is a precondition for democracy and lasting peace.
  • Maria Angelita Ressa is a Filipino-American journalist and author, the co-founder and CEO of Rappler, and the first Filipino recipient of the Nobel Peace Prize.
  • Dmitry Andreyevich Muratov is a Russian journalist, television presenter, and editor-in-chief of the Russian newspaper Novaya Gazeta.

Additional Information 

  • Nobel prize in chemistry was won by Benjamin List and David MacMillan in 2021.
  • Nobel prize in Medicine was won by David Julius and Ardem Patapoutian in 2021.

Question 4

Which of the following Articles empowers the President to grant pardon?

  1. AOption (a)

    Article 72

  2. BOption (b)

    Article 74

  3. COption (c)

    Article 78

  4. DOption (d)

    Article 80

View answer & explanation

Correct answer

Answer: (a)A.

Article 72

Why this answer is correct

The correct answer is Article 72.

Key Points

  • Pardoning Power of the President:
  • Constitution’s Article 72 empowers the President to grant pardons to persons tried and convicted of any offence where:
  • The Sentence is for an offence against a Union Law;
  • A sentence is by a court-martial (military court); and
  • It is a sentence of death.
  • This pardoning power is independent of the Judiciary;
  • It is executive power.
  • But, the President does not sit as a court of appeal, while exercising this power.
  • We may point out that, it is a power which the Sovereign exercises against its judicial mandate.
  • Hence option 4 is NOT correct.

 Important Points

  • The pardoning power of the President includes the following:
  • Pardon: Completely frees the convict from all sentences, punishments, and disqualifications.
  • Commutation: Substitution of punishment into a lighter form.
  • For example, a death sentence may be substituted for rigorous imprisonment.
  • Remission: Implies reducing the period of a sentence without changing its character.
  • For example, a sentence of rigorous imprisonment for two years may be remitted to rigorous imprisonment for one year.
  • Respite: It means awarding a lesser sentence in place of one originally awarded due to some special fact, such as the pregnancy of a woman offender or the physical disability of a convict.
  • Reprieve: It denotes a stay of the execution of a sentence (especially that of death) for a temporary period to enable the convict to have time to seek pardon or commutation.

Question 5

The Tropic of Cancer passes through which of the following States?

  1. Tripura
  2. Mizoram
  3. Manipur
  4. West Bengal

Select the correct answer from the codes given below:

  1. AOption (a)

    1, 2 and 3

  2. BOption (b)

    2, 3 and 4

  3. COption (c)

    1, 2 and 4

  4. DOption (d)

    1, 3 and 4

View answer & explanation

Correct answer

Answer: (c)C.

1, 2 and 4

Why this answer is correct

The correct answer is 1, 2, and 4.

 Key Points

  • The latitude approximately 23°27′ N of the terrestrial equator is called tropic of cancer.
  • There are 17 countries(Including Taiwan) through which the tropic of cancer passes.
  • The tropic of cancer is the only latitude pass through India.
  • The tropic of cancer passes through 8 Indian states.

Important Points

  • The 8 Indian states through which tropic of cancer passes through are:
  1. Gujarat.
  2. Rajasthan.
  3. Madhya Pradesh.
  4. Chhatisgarh.
  5. Jharkhand.
  6. West Bengal.
  7. Tripura.
  8. Mizoram.
  • Mahi River is the only river in India that passes the tropic of cancer twice.
  • The city nearest to the tropic of cancer in India is Udaipur city in Tripura.

Question 6

Match List-I with List-II and select the correct answer using the codes given below the lists.

List-I
(Vedas)
List-II
(Brahmans)
A.Rigveda1.Gopath Brahman
B.Yajurveda2.Kaushitaki Brahman
C.Samveda3.Satpath Brahman
D.Atharvaveda4.Panchvish Brahman
  1. AOption (a)

    A - 2, B - 3, C - 4, D - 1

  2. BOption (b)

    A - 3, B - 4, C - 2, D - 1

  3. COption (c)

    A - 2, B - 1, C - 3, D - 4

  4. DOption (d)

    A - 1, B - 2, C - 4, D - 3

View answer & explanation

Correct answer

Answer: (a)A.

A - 2, B - 3, C - 4, D - 1

Why this answer is correct

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

Key Points

List - IList - II
RigvedaKaushitaki Brahman
SamavedaPanchvisha
AtharvavedaGopatha
YajurvedaShatpatha

Additional Information 

  • Rigveda - The oldest Veda is the Rigveda. It has 1028 hymns called ‘Suktas’ and is a collection of 10 books called ‘Mandalas.’
  • Samaveda - Known as the Veda of melodies and chants, Samaveda dates back to 1200-800 BCE. This Veda is related to public worship.
  • Yajurveda - Stands to mean ‘Worship Knowledge’, Yajurveda dates back to 1100-800 BCE; corresponding with Samaveda. It compiles ritual-offering mantras/chants. These chants were offered by the priest alongside a person who used to perform a ritual (in most cases yajna fire.)
  • Atharvaveda - Stands to mean a tatpurusha compound of Atharvan, an ancient sage, and knowledge (atharvan+knowledge), it dates back to 1000-800 BCE.

Question 7

Which Part of the Indian Constitution contains the Union-State Relations?

  1. AOption (a)

    Part - V

  2. BOption (b)

    Part - VII

  3. COption (c)

    Part - VIII

  4. DOption (d)

    Part - XI

View answer & explanation

Correct answer

Answer: (d)D.

Part - XI

Why this answer is correct

The correct answer is Part XI.

Key Points

  • India is a  "Union of States" according to the Constitution of India.
  • Center-state relation is mentioned in Part XI of the constitution.
  • It divides into the legislative, executive, and financial powers between the centre and the states.
  • The center-state relations are divided into three parts - 
  • Legislative Relations (Article 245-255)
  • Administrative Relations (Article 256-263)
  • Financial Relations (Article 268-293)
  • Part XIII - consists of Articles on trade and Commerce inside the territory of India.
  • Articles 301 - 305 on Freedom of Trade and Commerce.
  • Article 306 - Repealed - Replaced by the Constitution (Seventh Amendment) Act, 1956.

Additional Information 

  • Originally the constitution contained 395 articles divided into 22 parts and 8 schedules.
  • At present, there are 448 articles in 25 parts, 12 schedules.​

 

Sl No.PartHeadingArticles
1Part IUnion and its Territory1 to 4
2Part IICitizenship5 to 11
3Part IIIFundamental Rights12 to 35
4Part IVDirective Principles of State Policy36 to 51
5Part IVAFundamental Duties51A
6Part VThe Union52 to 151
7Part VIThe States152 to 237
8Part VII (Repealed)The State in Part B of the Schedule238
9Part VIIIThe Union Territories239 to 242
10Part IXThe Panchayats243 to 243O
11Part IXAThe Municipalities243P to 243ZG
12Part IXBThe Co-operative Societies243ZH to 243ZT
13Part XThe scheduled and Tribal Areas244 to 244A
14Part XIRelations between the Union and the States245 to 263
15Part XIIFinance, Property, Contracts, and Suits264 to 300A
16Part XIIITrade and Commerce within the territory of India301 to 307
17Part XIVServices Under the Union, the States308 to 323
18Part XIVATribunals323A to 323B
19Part XVElections324 to 329A
20Part XVISpecial Provisions Relating to certain Classes330 to 342
21Part XVIILanguages343 to 351
22Part XVIIIEmergency Provisions352 to 360
23Part XIXMiscellaneous361 to 367
24Part XXAmendment of the Constitution368
25Part XXITemporary, Transitional, and Special Provisions369 to 392
26Part XXIIShort title, date of commencement, etc.393 to 395

Question 8

Which of the following country was elected to the United Nations Human Rights Council for 2022-24 period its election held in October 2021?

  1. AOption (a)

    Honduras

  2. BOption (b)

    France

  3. COption (c)

    Canada

  4. DOption (d)

    Japan

View answer & explanation

Correct answer

Answer: (a)A.

Honduras

Why this answer is correct

The correct answer is Honduras.

Key PointsUN Human Rights Council Elections for 2022-2024 and the Responsibility to Protect

  • On 14 October 2021, the UN General Assembly elected Argentina, Benin, Cameroon, Eritrea, Finland, The Gambia, Honduras, India, Kazakhstan, Lithuania, Luxembourg, Malaysia, Montenegro, Paraguay, Qatar, Somalia, United Arab Emirates, and the United States of America to the Human Rights Council (HRC) for the 2022-2024 term. With the elections of Argentina, Finland, Lithuania, Luxembourg, Montenegro, Paraguay, Qatar, and the United States of America, 19 of the 47 Council members in 2022 will be “Friends of the Responsibility to Protect” – having appointed an R2P Focal Point and/or joined the Group of Friends of R2P in New York and Geneva.
  • The Human Rights Council and its mechanisms and procedures – including the Universal Periodic Review (UPR), investigative mechanisms, special procedures and treaty bodies, as well as the technical assistance provided by the Office of the High Commissioner for Human Rights (OHCHR) – all play an essential role in providing early warning of the risk factors that can lead to crimes against humanity, ethnic cleansing, war crimes and genocide, and provide recommendations to prevent their recurrence. 

Additional Information

The Global Centre for the Responsibility to Protect encourages all HRC members to:

  • Improve the link between human rights and peace and security:
  • Ask the UN Secretary-General to bring to the attention of the Security Council relevant decisions and reports by the HRC;
  • Encourage the UN General Assembly to ensure information collected by HRC-mandated special procedures and international investigative mechanisms, is relayed to the Security Council in a timely manner;
  • Encourage the Security Council to request regular briefings by OHCHR and relevant HRC mechanisms and procedures;
  • Promote continued dialogue on human rights and the prevention of mass atrocities.
  • Make better use of the UPR to detect early warning signs of potential mass atrocity crimes:
  • In preparation for your national report, consider what action your government has taken to uphold its primary responsibility to protect all populations in your territory;
  • Use the UPR to ask relevant questions regarding the ratification and implementation of core human rights treaties, as well as regarding risk factors related to systematic violations and abuses of human rights.
  • Ensure the HRC responds in a timely and effective manner to atrocity situations:
  • Actively support the adoption of resolutions addressing serious violations of human rights;
  • Support HRC-mandated investigative mechanisms;
  • Mandate that special procedures and investigative mechanisms conduct country-specific atrocity risk assessments, including by utilizing the UN Framework of Analysis for Atrocity Crimes;
  • Where an imminent risk of mass atrocity crimes is detected, hold a special session in order to help mobilize appropriate international diplomatic action.
  • Make use of the Irish Principles, which layout independent and objective considerations to guide decisions on whether and when the HRC should respond to a country-specific situation.

Question 9

Who was the advisor of Constituent Assembly of India?

  1. AOption (a)

    Dr. B. R .Ambedkar

  2. BOption (b)

    Dr. Rajendra Prasad

  3. COption (c)

    Sh. B. N. Rao

  4. DOption (d)

    Dr. K. M. Munshi

View answer & explanation

Correct answer

Answer: (c)C.

Sh. B. N. Rao

Why this answer is correct

The correct answer is B. N. Rao.

Key Points

The drafting of the document called the constitution was done by an assembly of elected representatives called the Constituent Assembly.

  • Elections to the Constituent Assembly were held in July 1946.
  • Its first meeting was held in December 1946.

Important Points

Sir Benegal Narsing Rau, CIE, (26 February 1887 - 30 November 1953) was an Indian civil servant, jurist, diplomat and statesman known for his key role in drafting the Constitution of India.

  • He was the Constitutional Advisor to Constituent Assembly.
  • He was also India's representative to the United Nations Security Council from 1950 to 1952.
  • One of the foremost Indian jurists of his time, Rau helped draft the constitutions of Burma in 1947 and India in 1950.
  • As India's representative on the United Nations Security Council (1950–52), he was serving as president of the council when it recommended armed assistance to South Korea (June 1950).
  • Later he was a member of the Korean War post-Armistice United Nations Command Military Armistice Commission (UNCMAC).

Thus, we can say that B. N. Rao was the constitutional advisor to the Constituent Assembly of India.

Additional Information

  • Kavalam Madhava Panikkar (3 June 1895 - 10 December 1963), popularly known as Sardar K. M. Panikkar, was an Indian statesman and diplomat. He was also a professor, newspaper editor, historian and novelist.
  • K.T Shah was elected to the Constituent Assembly from Bihar. He was an active member of the Assembly who made immense contributions during the debates. He was a socialist and intervened on issues ranging from freedom of speech, presidential elections, citizenship and separation of powers.

Question 10

Which one of the following Revolutionaries was NOT associated with the "Kakori Train Incident"?

  1. AOption (a)

    Rajendra Lahiri

  2. BOption (b)

    Ashfakullah Khan

  3. COption (c)

    Khudiram Bose

  4. DOption (d)

    Roshan Singh

View answer & explanation

Correct answer

Answer: (c)C.

Khudiram Bose

Why this answer is correct

The correct answer is Khudiram Bose.

​ Important Points

  • Khudiram Bose was hanged for his attempt to assassinate a British judge, Magistrate Kingsford.
  • Bose was executed on August 11, 1908, after being held guilty for the famous Muzaffarpur conspiracy where he, made an attempt on the life of Douglas Kingsford, a British judge in colonial India.
  • Prafulla and Khudiram Bose tried to assassinate the District Judge, Mr. Kingsford by throwing bombs at the carriage in which Kingsford was supposed to travel but he was not in the carriage and two British women were killed.

 Key Points

  • Khudiram Bose was born on 3 December 1889, He was an Indian revolutionary from the Bengal Presidency who opposed British rule of India.
  • At the age of 15, Bose joined the Anushilan Samiti, an early 20th-century organization that propounded revolutionary activities in Bengal.
  • At the time of his hanging, Khudiram was 18 years old making him one of the 2nd youngest revolutionaries in India.
  • On July 13, 1908, Bose was finally sentenced to death.

Additional Information

  • Douglas Kingsford colonial British judicial officer, former
  • Chief Presidency Magistrate of  Calcutta was a Kingsford was the Chief Magistrate of the Presidency Court of Alipore and had overseen the trials of Bhupendranath Dutta and other editors of Jugantar, sentencing them to rigorous imprisonment.
  • During  Kingsford time Bengal was a hotbed of revolutionaries and freedom fighters, and freedom movements were more visible in Bengal than in other regions.

Important Points about Kakori

  • Kakori Conspiracy is also known as Kakori Kand, Kakori Conspiracy Case, or The Kakori Train Robbery.
  • It was an armed robbery that took place on August 9, 1925, on a train in central Uttar Pradesh.
  • The robbery occurred at the town of Kakori, about 16 km from Lucknow which was where the train was headed.
  • The objectives of the Kakori Kand/Conspiracy were to:
  • Gain funds for the HRA by taking money from the British Administration through force.
  • Create a positive image of the HRA among Indians by attacking a high-profile British government target with minimum collateral damage.
  • The robbery was planned by Ram Prasad Bismil and Ashfaqullah Khan.
  • It was executed by Bismil, Khan, Chandrashekhar Azad, Rajendra Lahiri, Shachindra Bakshi, Keshab Chakravarty, Murari Lal Khanna(Gupta), Banwari Lal, Mukundi Lal Gupta, and Manmathnath Gupta.
  • Though no passengers were targeted by the revolutionaries, one passenger named Ahmed Ali was killed in the crossfire between the guards and revolutionaries.
  • This made it a manslaughter case.
  • The final verdict of the Kakori Kand was pronounced in July 1927. Around 15 people were let off by the court for lack of evidence. Death sentences were awarded to:
  • Ram Prasad Bismil
  • Ashfaqullah Khan
  • Thakur Roshan Singh
  • Rajendra Lahiri

Question 11

When was "Ek Bharat Shreshtha Bharat" announced?

  1. AOption (a)

    2014

  2. BOption (b)

    2015

  3. COption (c)

    2017

  4. DOption (d)

    2018

View answer & explanation

Correct answer

Answer: (b)B.

2015

Why this answer is correct

The correct answer is 2015.

Important Points

  • Ek Bharat Shreshtha Bharat’s campaign was started in October 2015 A. D.
  • The initiative Ek Bharat Shreshtha Bharat was announced by the Hon'ble Prime Minister on the occasion of the 140th birth anniversary of Sardar Vallabhbhai Patel.
  • Ek Bharat Shreshtha Bharat (EBSB) has been one of the major flagship programs of the Government to bring in cultural integration of the country as a part of the vision to build a New India by 2022.

Key Points

  • Ek Bharat Shreshtha Bharat program aims to enhance interaction & promote mutual understanding between people of different states/UTs through the concept of state/UT pairing.
  • The aim of online Ek Bharat Shresth Bharat Saptah was to foster the spirit of National Integration and solidarity by overcoming cultural, religious, geographical, environmental, and linguistic barriers from all state NCC cadets.
  • Ek Bharat Shrestha Bharat of NCC is an annual camp for national integration conducted between paired states.

Additional Information 

  • M/o Human Resource Development (HRD) is the nodal ministry for the EBSB program.
  • The Minister of Education, formerly the Minister of Human Resources Development (1985-2020), is the head of the Ministry of Education and one of India's cabinet ministers.
  • The current minister is Dharmendra Pradhan and its Headquater is located in New Delhi.

Question 12

In the reign of which Chola King first trade delegation was sent to China?

  1. AOption (a)

    Rajendra I

  2. BOption (b)

    Raja Raja

  3. COption (c)

    Vikram Chola

  4. DOption (d)

    Raja Mahendra

View answer & explanation

Correct answer

Answer: (b)B.

Raja Raja

Why this answer is correct

The correct answer is Raja Raja.

 Key Points

  • A delegation sent to China by the great Chola emperor Rajaraja over trade relations 1,000 years ago, a Shiva temple in China, and the use of Chinese umbrellas centuries ago shed light on ancient links between Tamil Nadu and the dragon land.
  • During the reign of the Song dynasty (960-1279), Raja Raja Chola the Great sent a delegation of emissaries to China by 1014 following a representation from the Chinese rulers, seeking trade ties.

Additional Information

  • Rajaraja Chola(985-1014AD)
  • The original name, Arumolivarman, who crowned himself as Rajaraja I in 985 A D the next thirty years of his rule formed the formative period of Chola imperialism.
  • He was the successor of Parantaka II,
  • The Chola kingdom grew under him into an extensive and well-knit empire, efficiently organized and administered and possessing a powerful standing army and navy.
  • Rajaraja began his conquests by attacking the confederation between the rulers of the Pandya and Kerala kingdoms and of Ceylon.
  • Polonnaruva became the capital of Chola province in North Ceylon after the defeat of Mahinda V, the Ceylonese king.
  • He also annexed the Maldives. Elsewhere, several parts of modern Mysore were conquered and annexed which intensified their rivalry with the Chalukyas.
  • Rajaraja built the magnificent Shiva temple of Brihadeshwara or Rajaraja temple at Thanjavur which was com­pleted in 1010.
  • It is considered a remarkable piece of architecture in the South Indian style.
  • Rajaraja I also encouraged Sri Mara Vijayottungavarman, the Sailendra ruler of Sri Vijaya to build a Buddhist Vihara at Negapatam.
  • This vihara was called ‘Chudamani Vihara’ after the father of Sri Mara.

Question 13

The Citizenship (Amendment) Bill 2019 was passed by the Parliament on which date?

  1. AOption (a)

    11 December 2019

  2. BOption (b)

    10 December 2019

  3. COption (c)

    9 December 2019

  4. DOption (d)

    8 December 2019

View answer & explanation

Correct answer

Answer: (a)A.

11 December 2019

Why this answer is correct

The correct answer is 11 December 2019.

Key Points 

  • The Citizenship Amendment Bill, 2019 has been passed by the Parliament, with the Rajya Sabha approving it on 11 December 2019.
  • The bill amends the Citizenship Act, 1955 to make illegal migrants of six communities from Afghanistan, Bangladesh, and Pakistan eligible for Indian citizenship.
  • The communities are:
  • Hindus, Sikhs, Buddhists, Jains, Parsis, and Christians.

Question 14

Which one of the following pairs is NOT correctly matched?

Wildlife SanctuaryLocation
a.Mahavir Swami SanctuaryLalitpur District
b.Lakh Bahosi SanctuaryKannauj District
c.Kishanpur Wildlife SanctuaryBalrampur District
d.Kaimoor SanctuaryMirzapur and Sonbhadra District
  1. AOption (a)

    a

  2. BOption (b)

    b

  3. COption (c)

    c

  4. DOption (d)

    d

View answer & explanation

Correct answer

Answer: (c)C.

c

Why this answer is correct

The correct answer is c.

Key Points

Wildlife SanctuaryLocation
a.Mahavir Swami SanctuaryLalitpur District
b.Lakh Bahosi SanctuaryKannauj District
c.Kishanpur Wildlife SanctuaryLakhimpur Kheri District.
d.Kaimoor SanctuaryMirzapur and Sonbhadra District

Additional Information

  • Mahavir Swami Wildlife Sanctuary:-
  • Established in the year 1977.
  • Total area: 05 sq km
  • It is located in the Lalitpur district.
  • This sanctuary is home to Wolves, Wild cats, Hyenas, Wild Dogs, Mongoose, etc.
  • Kaimur Wildlife Sanctuary:-
  • ​Established in the year 1982.
  • Total area: 501 sq km.
  • It is located in the Mirzapur district.
  • It is famous for Black Bucks.
  • Kishanpur Wildlife Sanctuary:
  • It is in the Lakhimpur Kheri district not in Lucknow.
  • Kishanpur Wildlife Sanctuary was founded in 1972.
  • Katarniaghat Wildlife Sanctuary is in Bahraich.
  • Katarniaghat Wildlife Sanctuary was established in 1975.
  • Lakh Bahosi Sanctuary is in Kannauj.
  • Lakh Bahosi Sanctuary was founded in 1972.
  • ​Lakh Bahosi Sanctuary.
  • The largest bird Sanctuary of Uttar Pradesh is Lakh Bahosi Sanctuary.
  • Lakh Bahosi Sanctuary is situated in the Kannauj district of UP
  • Lakh Bahosi Sanctuary covers 80 square km.
  • Lakh Bahosi Sanctuary is home to various migratory birds.
  • Jackal, blue bull, mongoose, fishing cat, and monkeys are also found in Lakh Bahosi Sanctuary.

Important Points

  • There is a total of 11 Wildlife sanctuaries in Uttar Pradesh.
  • First Wildlife Sanctuary: Chandraprabha
  • Set up in 1957 in Chandauli.
  • Largest Wildlife Sanctuary: Hastinapur
  • Total area: 2073 sq km.
  • Smallest Wildlife Sanctuary: Mahavir Swami
  • Total area: 05 sq km.

Question 15

With reference to the inauguration of the International Airport at Kushinagar in October 2021, which of the following statement(s) is/are correct?

  1. The first inaugural flight to land was of Sri Lankan Airlines.
  2. Kushinagar became the Fourth International Airport of Uttar Pradesh.

Select the correct answer from the codes given below:

  1. AOption (a)

    Only 1

  2. BOption (b)

    Only 2

  3. COption (c)

    Both 1 and 2

  4. DOption (d)

    Neither 1 nor 2

View answer & explanation

Correct answer

Answer: (a)A.

Only 1

Why this answer is correct

The correct answer is Only 1.

Key Points

  • Prime Minister Narendra Modi inaugurated the Kushinagar International Airport on 20 October 2021.
  • The inaugural flight will land at the airport from Colombo, Sri Lanka carrying 125 dignitaries & Buddhist Monks.
  • Inauguration of the Kushinagar Airport will facilitate the pilgrims from across different parts of the world by providing seamless connectivity to various Buddhist sites in the region.
  • Tourism inflow is expected to rise by up to 20% with the inauguration of the flight.
  • The Kushinagar International Airport will not only put the pilgrimage site on International Aviation Map but also give a boost to the economic development of the region.
  • It will have a multiplier effect on the hospitality industry by promoting hotel business, tourism agencies, restaurants, etc.
  • It will create employment for local people by opening an immense amount of opportunities in feeder transport services, local guide jobs, etc.
  • The development of the airport at Kushinagar will help in developing Kushinagar as one of the four principal places of Buddhist Pilgrimage.
  • The airport will serve a population of more than two crores since the airport has a hinterland of around10-15 Districts and will be a great support for the large migrant population of Eastern Uttar Pradesh and Western / Northern Part of Bihar.
  • This will also boost the opportunities for the export of horticultural products like bananas, strawberries, and mushrooms.
  • The Kushinagar Airport is spread across 3600sqm. It is built at an estimated cost of ₹260 crores.
  • The new terminal is equipped to handle 300 passengers during peak hours.
  • At present, two international airports - Chaudhary Charan Singh International Airport at Lucknow and Lal Bahadur Shastri International Airport at Varanasi -- are operational in the state while another one is coming up at Jewar in Gautam Budh Nagar.
  • Now Kushinagar became the Third International Airport of Uttar Pradesh.

Question 16

Which one of the following city was selected as 100th Smart City in India?

  1. AOption (a)

    Srinagar

  2. BOption (b)

    Shivamogga

  3. COption (c)

    Shillong

  4. DOption (d)

    Satna

View answer & explanation

Correct answer

Answer: (c)C.

Shillong

Why this answer is correct

The correct answer is Shillong.

Key Points

  • Shillong, the capital city of Meghalaya, has been selected as the 100th Smart City of India by the Ministry of Housing and Urban Affairs after evaluating the proposal submitted by it.
  • With the selection of Shillong, the total proposed investment in the finally selected 100 cities under the Smart Cities Mission would be Rs.2,05,018 crores.
  • Till now, 99 smart cities have been selected in four rounds of competition, and with this announcement; the selection of 100 cities has been completed under the Smart Cities Mission.
  • Previously, 20 cities were selected in January 2016, 13 cities in May 2016, 27 cities in September 2016, 30 cities in June 2017, and nine cities in January this year.
  • The Smart Cities Mission was launched on June 25, 2015, with the objective of strengthening urban infrastructure through the application of smart solutions and giving a decent quality of life to citizens.

Question 17

To whom the credit is given to organize the labour movements in India?

  1. AOption (a)

    Bal Gangadhar Tilak

  2. BOption (b)

    Narain Medhji Lokhande

  3. COption (c)

    N. M. Joshi

  4. DOption (d)

    Swami Sahjanand Saraswati

View answer & explanation

Correct answer

Answer: (b)B.

Narain Medhji Lokhande

Why this answer is correct

The correct answer is  Narayan Meghaji Lokhande.

Key Points

  • Narayan Meghaji Lokhande was the pioneer in organizing the labor movement in British India.
  • Lokhande is acclaimed as the Father of the Trade Union Movement in India.
  • The first labor association in India,  'Bombay Mill Hands Association' was started by him.
  • The Government of India issued a postage stamp with his photograph in 2005.

Question 18

Which one of the following is NOT a Andaman and Nicobar groups of Islands?

  1. AOption (a)

    Rutland Island

  2. BOption (b)

    Narcondum Island

  3. COption (c)

    Bompoka Island

  4. DOption (d)

    Amin Divi Island

View answer & explanation

Correct answer

Answer: (d)D.

Amin Divi Island

Why this answer is correct

The correct answer is Amin Divi Island.

 Key Points

Amindivi Islands, Lakshadweep

  • Amindivi Islands build up the northern part of Lakshadweep and consists of several small islands that are popular tourist attractions around the place.
  • These islands are namely – Chetlat, Kadmat, Bitra, Amini, and Kiltan.
  • Amindivi Islands are one of the three major island groups in Lakshadweep.
  • Lakshadweep is an exotic archipelago, sun-kissed amidst the gigantic Arabian Sea boasting soft white sands, and crystal blue water accompanied with some of the best seafood delights.
  • The main attraction of Amindivi Islands is hidden in its very name, the name comes from the word ‘Amini’ which roughly translates to a heavenly place
  • Kadmat Island offers a bountiful of water sports options like scuba diving, swimming.

Nearby Attractions of Amindivi Islands

  • Amindivi is a group of islands with each island being slightly more unique than the other. So, most nearby attractions include natural beauty at its best with friendly natives, eager to help, and the overall vibe of complete relaxation.
  • The nearby attractions in Amindivi are:
  • Dolphin Dive Centre, Kavaratti
  • Minicoy Dive Centre
  • Kadmat Scuba Diving Centre
  • Kiltan
  • Chetlat
  • Bitra

Additional Information

  • Rutland Island is an island of the Andaman Islands. It belongs to the South Andaman administrative district, part of the Indian union territory of Andaman and Nicobar Islands. the island lies 20 km south from Port Blair.
  • Narcondam, India's easternmost island, is a small volcanic island located in the northern Andaman Sea. The island's peak rises to 710 m above mean sea level, and it is formed of andesite.
  • Bompoka island lies about 1,8 NM east from the southern end of nearby Teressa Island in the Nicobar group of islands. The hills on the island rise up to 193 m. The island has dense forest and some portions have grass growth.

Question 19

In which year was the National Development Council formed?

  1. AOption (a)

    1947

  2. BOption (b)

    1952

  3. COption (c)

    1965

  4. DOption (d)

    1966

View answer & explanation

Correct answer

Answer: (b)B.

1952

Why this answer is correct

The correct answer is 1952.

Key Points

  • National Development Council (NDC) is the apex body that gives approval to five-year plans in India.
  • It was established on 6th August 1952.
  • The Prime minister is serving as the chairman of the National Development Council.
  • The headquarters of the National Development Council (NDC) is in New Delhi.
  • The first five-year plan is the only five-year plan adopted without the consent of the National Development Council.
  • The first meeting of the NDC was chaired by Jawaharlal Nehru in November 1952.

Question 20

Who among the following became the first Indian women wrestler to win silver medal in World Wrestling Championship in October 2021?

  1. AOption (a)

    Vinesh Phogat

  2. BOption (b)

    Gita Phogat

  3. COption (c)

    Anshu Malik

  4. DOption (d)

    Jyoti Tripathi

View answer & explanation

Correct answer

Answer: (c)C.

Anshu Malik

Why this answer is correct

The correct answer is Anshu Malik.

Key Points

  • At the 2021 World Wrestling Championships, Indian wrestler Anshu Malik created history as she became the first Indian women finalist at the World Championship.
  • And also the first female player from India to claim a silver medal.
  • The 19-year old Anshu settled for a silver medal after losing to 2016 Olympic champion Helen Lousie Maroulis of the United States in a 57kg freestyle bout.

Additional Information

  • To date, Indian women wrestlers Alka Tomar (2006), Geeta Phogat (2012), Babita Phogat (2012), Pooja Dhanda (2018), and Vinesh Phogat (2019) have been able to win a bronze each.
  • Wrestler Sushil Kumar (2010) is the lone wrestler from India to win World Championship to date.
  • Besides this, Sarita Mor claimed her maiden medal by winning bronze in the 59kg category at the 2021 World Wrestling Championships being held at Oslo, Norway from October 02 to 10, 2021.

Question 21

Patron of Literature, Jahangir awarded the title of "Pandit Raj" to whom?

  1. AOption (a)

    Munishwar Das

  2. BOption (b)

    Bhagwathi Swami

  3. COption (c)

    Jagannath

  4. DOption (d)

    Raghunath

View answer & explanation

Correct answer

Answer: (c)C.

Jagannath

Why this answer is correct

The correct answer is Jagannath.

Key Points

  • Jagannātha, also known as Jagannātha Paṇḍita or Jagannātha Paṇḍitarāja, or Jagannatha Pandita Rayalu, was a famous poet and literary critic who lived in the 17th century.
  • As a poet, he is known for writing the Bhāminī-vilāsa. He was a Telugu Brahmin from the Khandrika family and a junior contemporary of Emperor Akbar.
  • Pandit Jagannath was a famous poet and literary critic in the courts of the Mughal emperors Jehangir and Shah Jahan.
  • As a literary theorist or rhetorician, he is renowned for his Rasagagadhara, a work on poetic theory. He was granted the title of Paanditaraja by the Mughal emperor, at whose court he received patronage.
  • Sanskrit scholars of Jahangir’s time, Jaganath, who later became  Jahangir’s court pandit and was called ‘Pandit Raj’ was patronized by Asaf Khan 1 and wrote ‘Asaf Vijaya’ a eulogy of Asaf Khan.

Additional Information

Question 22

Consider the following events and arrange these in chronological order.

  1. Simon Commission
  2. Cabinet Mission Plan
  3. First Meeting of Indian Constitution Assembly
  4. Quit India Movement

Select the correct answer from the codes given below:

  1. AOption (a)

    1, 4, 2, 3

  2. BOption (b)

    1, 2, 3, 4

  3. COption (c)

    2, 1, 3, 4

  4. DOption (d)

    2, 1, 4, 3

View answer & explanation

Correct answer

Answer: (a)A.

1, 4, 2, 3

Why this answer is correct

The correct answer is 

Key Points

  • Simon Commission:​

  • In 1927, the British government appointed Simon Commission to look into the working of the Government of India Act,1919.

  • All its seven members were Englishmen.

  • On 3 February 1928 when the Commission reached Bombay, a general hartal was observed all over the country.

  • And they were greeted with black flags and the cries of 'Simon Go Back'.

  • The Simon Commission Report became the basis for enacting the Government of India Act of 1935.

  • Quit India Movement

  • On 8 August 1942 at the All-India Congress Committee session in Bombay, Mohandas Karamchand Gandhi launched the 'Quit India' movement.

  • The next day, Gandhi, Nehru, and many other leaders of the Indian National Congress were arrested by the British Government.

  • The Quit India Movement, also known as the August Movement, was a movement launched at the Bombay session of the All-India Congress Committee by Mahatma Gandhi during World War II, demanding an end to British rule in India.

  • The slogan 'Quit India' was given by Yusuf Meherally.

  • The viceroy during the Quit India movement was Lord Linlithgow.

  • During the 'Quit India' movement, Mahatma Gandhi gave the slogan 'Do or Die'​

  • The cabinet mission plan was formed in the year 1946.

  • The main purpose of this plan was to transfer the power from the British government to the Indian government.

  • The three British cabinet members were - Pethick Lawrence, Stafford Cripps, and A.V Alexander.

  • The elections were held for the first time for the constituent assembly.

  • The members of the constituent assembly were elected by the provincial assemblies by a single, transferable-vote system of proportional representation.

  • First Meeting of Indian Constitution Assembly​

  • The first meeting of the Constituent Assembly of India was held on 9 December 1946 at the Constitutional Hall, New Delhi.

  • The Indian National Congress won 208 seats, the Muslim League won 73 seats.

  • The Constitution came into force on 26 January 1950 which became the Republic Day of India.

  • It was passed by the Constituent Assembly on 26 November 1949.

  • In the first meeting, the Assembly adopted an 'Objectives Resolution' which later became the Preamble to the Constitution.

  • The constitution-making body was elected by the Provincial Legislative Assembly consisting of 389 members, 93 from princely states, and 296 from British India.

  • Dr. Satchidanand Sinha was the first president of the Constituent Assembly.

Question 23

Match List-I with List-II and select the correct answer from the codes given below.

List-I
(Minerals)
List-II
(Mines)
A.Copper1.Bailadila
B.Iron ore2.Balaghat
C.Manganese3.Taloja
D.Mica4.Nellor
  1. AOption (a)

    A - 3, B - 2, C - 4, D - 1

  2. BOption (b)

    A - 2, B - 1, C - 4, D - 3

  3. COption (c)

    A - 4, B - 1, C - 2, D - 3

  4. DOption (d)

    A - 3, B - 1, C - 2, D - 4

View answer & explanation

Correct answer

Answer: (d)D.

A - 3, B - 1, C - 2, D - 4

Why this answer is correct

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

Confusion Points

  • Although the Balaghat mine produces 52% of India's Copper production.
  • However, Manganese ore fines are generated at Balaghat and Ukwa mines of MOIL.

Key Points

List-I
(Minerals)
List-II
(Mines)
A.CopperTaloja
B.Iron oreBailadila
C.ManganeseBalaghat
D.MicaNellor

Additional Information

  • Balaghat is the largest source of manganese in Madhya Pradesh.
  • Madhya Pradesh is India's foremost source of manganese ore with a reserve of around 23.64 million tonnes.
  • It accounts for near about 14.13% of the national reserve.
  • The country's largest Open Cast Copper Mine is at Malajkhand in the Balaghat district of the state.
  • The "Bharveli manganese mine" in Balaghat district is the largest underground mine operating in the Asian sub-continent.
  • Rajasthan has the highest reserve of copper in India followed by Madhya Pradesh.
  • Copper plants in India:
  • Ghatsila (Jharkhand).
  • Korba (Chattisgarh).
  • Taloja (Maharastra).
  • Chilpi, Malanjkhand (Madhya Pradesh).
  • Bailadila-14 mine is the first large-scale open cast mechanized iron ore mine in India for which DPR was prepared by NMDC.
  • The mine has the distinction of having a unique down-the-hill conveyor system passing through a tunnel to transport iron ore from the crushing plant to the processing plant.
  • The iron ore mined at Bailadila is mostly haematite.
  • Nellore is a city located on the banks of the Penna River.
  • It is in the Nellore district of the Indian state of Andhra Pradesh.
  • ​Nellore district produces most of the crude mica in India.

Question 24

Consider the following events and arrange them into chronological order.

I. Construction of Dhai Din Ka Jhopra

II. Construction of Quwat-ul-Islam Mosque

III. Construction of Siri Fort

IV. Construction of Sultan Garhi

Select the correct answer from the codes given below:

  1. AOption (a)

    II, I, III and IV

  2. BOption (b)

    II, I, IV and III

  3. COption (c)

    I, II, IV and III

  4. DOption (d)

    I, II, III and IV

View answer & explanation

Correct answer

Answer: (c)C.

I, II, IV and III

Why this answer is correct

The correct answer is 

Key Points

  • Dhai din ka Jhopra is situated in Ajmer.
  • Dhai din ka Jhopra was constructed by Qutb-Ud-Din-Aibak in 1192 CE.
  • ​Groundbreaking: 1192 CE
  • Completed: 1199 CE

​

  • The  Quwwat-Ul-Islam Mosque was built along with the Qutub Minar in 1193 AD and finished in 1197 AD by Sultan Qutb-ud-din Aibak of the Slave dynasty as a Jami Masjid.
  • Quwat-ul-Islam mosque in the Qutub Minar Premises was converted from a Jain temple.

​

​​

  • Sultan Ghari was the first Islamic Mausoleum built in 1231 AD for Prince Nasiru'd-Din Mahmud, eldest son of Iltumish, in the "funerary landscape of Delhi" in the Nangal Dewat Forest, Near Nangal Dewat Vasant Kunj.

  • Iltumish was the third Sultan of the Slave Dynasty who ruled in Delhi from 1210 to 1236 AD.

  • Siri Fort was built by Allauddin Khilji to defend the city of Delhi from the Mongol attack.

  • Siri Fort is one of the many forts of Delhi, which were built during the time of the Delhi Sultanate (AD 1191-1526) in the medieval age.

  • It is situated between Hauz Khas and Mehrauli.

  • The fort was said to be a part of the city of Siri.

  • Built: 1303

  • Demolished: 1540 - 1545

​

Question 25

When did World Health Organisation declare COVID-19 a pandemic?

  1. AOption (a)

    11 March 2020

  2. BOption (b)

    13 January 2020

  3. COption (c)

    22 February 2020

  4. DOption (d)

    21 January 2020

View answer & explanation

Correct answer

Answer: (a)A.

11 March 2020

Why this answer is correct

The correct answer is 11 March 2020.

Key Points

  • Pandemic 
  • A pandemic is an outbreak of infectious disease that occurs over a wide geographical area and that is of high prevalence, generally affecting a significant proportion of the world’s population, usually over the course of several months.
  • Pandemics arise from epidemics, which are outbreaks of disease confined to one part of the world, such as a single country.
  • Pandemics, especially those involving influenza, sometimes occur in waves, so that a post-pandemic phase, marked by decreased disease activity, may be followed by another period of high disease prevalence.
  • In March 2020, an ongoing outbreak of a novel coronavirus known as severe acute respiratory syndrome coronavirus-2 (SARS-CoV2) was declared a pandemic by WHO officials on 11 March 2020.
  • Infection with SARS-CoV2 produced an illness known as coronavirus disease 2019 (COVID-19).
  • The illness was characterized primarily by fever, cough, and shortness of breath.

Technical (100 questions)

Question 26

Which one of the following processes takes place in trickling filter ?

  1. AOption (a)

    Sedimentation

  2. BOption (b)

    Coagulation and flocculation

  3. COption (c)

    Disinfection

  4. DOption (d)

    Bio-filtration

View answer & explanation

Correct answer

Answer: (d)D.

Bio-filtration

Why this answer is correct

Concept:

Trickling Filters:

  • A trickling filter is an attached growth process i.e. process in which microorganisms responsible for treatment are attached to inert packing material.
  • The packing material used in attached growth processes includes rock, gravel, slag, sand, redwood, and a wide range of plastic and other synthetic materials.

Figure: Trickling Filter

The process of trickling filter is as follows:

  • The wastewater in the trickling filter is distributed over the top area of a vessel containing non-submerged packing material.
  • Air circulation in the void space, by either natural draft or blowers, provides oxygen for the microorganisms growing as an attached biofilm.
  • During operation, the organic material present in the wastewater is metabolized by the biomass attached to the medium. The biological slime grows in thickness as the organic matter abstracted from the flowing wastewater is synthesized into new cellular material.
  • The thickness of the aerobic layer is limited by the depth of penetration of oxygen into the microbial layer.
  • The micro-organisms near the medium face enter the endogenous phase as the substrate is metabolized before it can reach the micro-organisms near the medium face as a result of increased thickness of the slime layer and lose their ability to cling to the media surface.
  • The liquid then washes the slime off the medium and a new slime layer starts to grow. This phenomenon of losing the slime layer is called sloughing.
  • The sloughed-off film and treated wastewater are collected by an under drainage which also allows circulation of air through the filter.
  • The collected liquid is passed to a settling tank used for solid-liquid separation.

Question 27

When two roads with two lane, two way traffic cross, at an uncontrolled intersection, the total no. of potential major conflict points would be

  1. AOption (a)

    32

  2. BOption (b)

    24

  3. COption (c)

    16

  4. DOption (d)

    4

View answer & explanation

Correct answer

Answer: (c)C.

16

Why this answer is correct

Explanation:

Intersection:

  • Intersection is the area where two or more roads join or cross
  • At traffic intersection change in direction of movements may occur.
  • Due to movement of traffic at intersection various types of conflicts occur like crossing conflict, merging conflict and diverging conflict.
  • The movement of one vehicle can conflict with the movements of other vehicles in the same stream, the cross-streams, the opposing stream, and pedestrians in crosswalks. Consider a typical four-legged intersection as shown in figure.
  • The number of conflicts for competing through movements are 4, while competing right turn and through movements are 8.
  • The conflicts between right turn traffics are 4, and between left turn and merging traffic is 4.
  • The conflicts created by pedestrians will be 8 taking into account all the four approaches.
  • Diverging traffic also produces about 4 conflicts.
  • Therefore, a typical four-legged intersection has about 32 different types of conflicts of which 24 are vehicular conflict out of which 16 are major and 8 are minor conlfict and 8 are Pedestrian conflict.

Important Points

Number of lanesNumber of potential conflicts
Road ARoad BBoth roads Two-wayA-One way B-two wayBoth roads one-way
2224116
2324118
24321710
33241311
44442518

Question 28

The maximum depth allowed for an aerobic stabilization tank is

  1. AOption (a)

    0.9 m

  2. BOption (b)

    0.7 m

  3. COption (c)

    0.5 m

  4. DOption (d)

    0.3 m

View answer & explanation

Correct answer

Answer: (c)C.

0.5 m

Why this answer is correct

Concept:

Stabilization Ponds:

  • The stabilization ponds are open flow through basins specifically designed and constructed to treat sewage and biodegradable industrial wastes.
  • They provide long detention periods extending from a few to several days.
  • Pond systems, in which oxygen is provided through mechanical aeration rather than algal photosynthesis are called aerated lagoons.
  • Lightly loaded ponds used as a tertiary step in waste treatment for polishing of secondary effluents and removal of bacteria are called maturation ponds.

Classification of Stabilization Ponds is as follows:

1. Aerobic ponds:

  • It is a shallow pond with a depth of less or equal to 0.5 m and BOD loading of 40-120 kg/ha/d so as to maximize penetration of light throughout the liquid depth.
  • Such ponds develop intense algal growth.

2. Anaerobic ponds:

  • It is used as pretreatment of high-strength wastes with a BOD load of 400-3000 kg/ha.d.
  • Such ponds are constructed with a depth of 2.5-5m as light penetration is unimportant.

3. Facultative ponds:

  • Facultative pond functions aerobically at the surface while anaerobic conditions prevail at the bottom.
  • They are often about 1 to 2 m in depth.
  • The aerobic layer acts as a good check against odor evolution from the pond.

Question 29

Lucknow Plan of Indian Roads Congress is related to the following

  1. AOption (a)

    1st 20 year road plan

  2. BOption (b)

    2nd 20 year road plan

  3. COption (c)

    3rd 20 year road plan

  4. DOption (d)

    3rd five year plan

View answer & explanation

Correct answer

Answer: (c)C.

3rd 20 year road plan

Why this answer is correct

Explanation:

Three road plans developed for the planning and construction of the road network are:

1st 20-year road plan2nd 20-year road plan3rd 20-year road plan
Nagpur PlanBombay PlanLucknow Plan
1943 - 19631961 - 19811981 - 2001
Total road length - 5,32,700 kmTotal road length - 10,57,330 km
Density - 16 km road length / 100 km2 area.Density - 32 km road length / 100 km2 area.Density - 82 km road length / 100 km2 area.
Pattern - Star, and grid-Square and block pattern
-1600 km Expressways have been considered.2000 km Expressways have been considered.
Development allowance - 15 %Development allowance - 5 %-
Roads are divided into 5 categories: NH, SH, MDR, ODR, VR.Every town with a population above 2000 in plains and 1000 in semi-hilly areas and 500 in hilly areas should be connected by a metalled road.Roads are divided into 3 categories: Primary, secondary, and Tertiary roads system.

Note: Lucknow road plan does not give the length of NH and SH. There is a formula given to calculate the length of different road plans.

Question 30

The perpendicular offset from a tangent to the junction of transition curve and circular curve is equal to:

(Where L = Length of transition curve, R = Radius of the circular curve)

  1. AOption (a)

    L/6R

  2. BOption (b)

    L/24R

  3. COption (c)

    L2/(6R)L^2/(6 R)

  4. DOption (d)

    L2/(24R)L^2/(24 R)

View answer & explanation

Correct answer

Answer: (c)C.

L2/(6R)L^2/(6 R)

Why this answer is correct

Concept:

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

 S=L224R{\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: -

offset;(y)=ℓ36RL{\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

y=l36RL=L36RL=L26R{\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

Question 31

The radial offset at a distance x from the beginning of circular curve of radius R is given by

  1. AOption (a)

    R2−x2−R\sqrt{R^2-x^2} - R

  2. BOption (b)

    R−R2−x2R-\sqrt{R^2-x^2}

  3. COption (c)

    R−R2+x2R-\sqrt{R^2+x^2}

  4. DOption (d)

    R2+x2−R\sqrt{R^2+x^2}-R

View answer & explanation

Correct answer

Answer: (d)D.

R2+x2−R\sqrt{R^2+x^2}-R

Why this answer is correct

Concept:

The methods of setting out curves:

The methods of setting out curves may be divided into two classes according to the instrument employed for setting out of curves as follows:

  1. Linear or chain and tape method:- Linear methods are those in which the curve is set out with chain and tape only. It contains sub-methods as follows:
  • By offsets or ordinates from the long chord
  • By successive bisections of arcs
  • By offsets from the tangents
  • By offsets from chord produced
  1. Angular or instrumental method:- Instrumental methods are those in which theodolite with or without a chain is employed to set out a curve. It contains sub-methods as follows:
  • Rankine’s method of tangential angle
  • Two theodolite method​

By offsets from the tangents: In this method, the offsets are set out either radially or perpendicular to the tangents. It contains sub-methods as follows:

  • By Radial offsets
  • By offsets perpendicular to tangents

Explanation:

By Radial offsets:

The radial offset at a distance X from the point of commencement of curve of radius R

Let, T1 be the first tangent point and R is the radius of the curve

EE1 = Ox= the radial offset at E at a distance of X from T1 along with the tangent AB.

Now in the ∆OT1E,

OT1 = R  and  T1E = X

OE = OE1+ EE1 =R + Ox

Now, OE2 = (OT1) 2 + (T1E)2

(R + Ox)2 = R2 + x 2 

OX=R2+X2−RO_X = \sqrt {R^2 + X^2} - R

The radial offset at a distance X from the point of commencement of curve of radius R is given by OX=R2+X2−RO_X = \sqrt {R^2 + X^2} - R

Question 32

Pick the correct statement for evaporation from water surface.

  1. AOption (a)

    increases with humidity

  2. BOption (b)

    decreases with wind speed

  3. COption (c)

    is proportional to the deficit of vapour pressure

  4. DOption (d)

    increases if there is Salinity or pollution

View answer & explanation

Correct answer

Answer: (c)C.

is proportional to the deficit of vapour pressure

Why this answer is correct

Concept:

Evaporation:

  • The phenomenon of change of a liquid into vapours at any temperature below its boiling point is called evaporation.

The factors that affect the process of evaporation are as follows:

1. Surface area:

  • Evaporation is a surface phenomenon. If the surface area is increased, the rate of evaporation increases. For example, while putting clothes for drying up we spread them out.

2. Temperature:

  • With the increase in temperature, more particles get enough kinetic energy to go into the vapour state, hence evaporation increases.

3. Humidity:

  • Humidity is the amount of water vapour present in the air. The air around us cannot hold more than a definite amount of water vapour at a given temperature.
  • If the amount of water in the air is already high, the rate of evaporation decreases.

3. Wind speed:

  • With the increase in wind speed, the particles of water vapour move away from the wind, decreasing the amount of water vapour in the surroundings and its increases evaporation.

4. Salinity or pollution:

  • Increasing water salinity reduces evaporation since the dissolved salt ions lower the free energy of the water molecules, i.e., reduce the water activity, and hence reduce the saturation vapour pressure above the saline water at a given water temperature.

Question 33

The discharge per unit drawdown at a well is known as

  1. AOption (a)

    Specific yield

  2. BOption (b)

    Specific storage

  3. COption (c)

    Safe yield

  4. DOption (d)

    Specific capacity

View answer & explanation

Correct answer

Answer: (d)D.

Specific capacity

Why this answer is correct

Explanation:

Specific Capacity:

  • The specific capacity of a well is defined as the rate of flow through the well per unit drawdown.
  • The specific capacity is not the same for all the unit drawdown, but it is computed only for the initial first-meter fall of the height of water in the well.

QSpecific;Capacity=Discharge;through;the;well;(Q)Unit;Drawdown;(ST) {Q_{Specific;Capacity}} = \frac{{Discharge;through;the;well;\left( Q \right)}}{{Unit;Drawdown;\left( {{S_T}} \right)}}

Factors affecting the specific capacity of the well are:

a) Diameter of well: On increasing the diameter of the well, its specific capacity increases as the yield of well increases.

b) Aquifer thickness: The higher the thickness of the aquifer, the higher will be the specific capacity.

c) Duration of pumping and pumping rate: It decreases with the increase in pumping rate and its duration.

Specific Retention (Sr):

Specific retention of a rock or soil is the ratio of the volume of water a rock can retain against gravity to the total volume of the rock.

(SR)=Volume;of;water;retain;against;gravity;(VRG)Volume;of;soil×100\rm\left( {{S_R}} \right) = \frac{{{Volume;of;water;retain;against;gravity;(V_{RG})}}}{Volume;of;soil} \times 100

Question 34

Flow of air can be considered to be incompressible within 1% error if the mach number of flow is less than _____.

  1. AOption (a)

    0.4

  2. BOption (b)

    0.6

  3. COption (c)

    0.1

  4. DOption (d)

    0.3

View answer & explanation

Correct answer

Answer: (d)D.

0.3

Why this answer is correct

Explanation:

Mach number (M or Ma)

  • Dimensionless quantity in fluid dynamics
  • The ratio of the flow velocity of the object to the local speed of sound.
  • If the Mach number is less than 0.3, flow can be treated as incompressible​

Important Points

  • The magnitude of the compressibility effect can be judged with the flow velocity.
  • For air, when the flow velocity is 100 m/s or less, the air is treated as an incompressible fluid.
  • If the velocity is greater than 100 m/s, the air is treated as **compressible fluid.**​
Mach NumberType of flow
M < 1Subsonic
M = 1Transonic
M > 1Supersonic
M > 5Hypersonic

Mistake PointsStudents get confused between supersonic and hypersonic if M > 1.

M > 1 - Supersonic 

M > 5 - Hypersonic​

Question 35

Distance travelled by a moving vehicle during the perception and brake reaction time is known as

  1. AOption (a)

    Sight distance

  2. BOption (b)

    Lag distance

  3. COption (c)

    Stopping distance

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (b)B.

Lag distance

Why this answer is correct

Explanation:

Sight distance:

It is the length of the road visible ahead to the driver at any distance

Three sight distance situations are considered

1) Stopping or non-passing sight distance or absolute minimum sight distance

2) Overtaking or passing sight distance

  1. Sight distance at an uncontrolled intersection

Lag distance:

During the total reaction time or PIEV time the vehicle may be assumed to proceed forward with a uniform speed at which the vehicle has been moving and this speed may be taken as the design speed. If 'V' is the design speed in m/sec and 't' is the total reaction time of the driver in seconds, then the lag distance will be (V × t) m.

If the design speed V is in kmph then

Lag distance = (0.278 × V × t) m

Question 36

A rectangular channel section is most economical when the depth of flow and bottom width is in the ratio of ____________.

  1. AOption (a)

    2 : 1

  2. BOption (b)

    1 : 1

  3. COption (c)

    1 : 2

  4. DOption (d)

    1 : 4

View answer & explanation

Correct answer

Answer: (c)C.

1 : 2

Why this answer is correct

Concept:

Most economical section is the one whose wetted perimeter is minimum for the given value of discharge.

For minimum wetted perimeter,

d = b/2

∴ d/b = 1 : 2

Important Points

For most economical Channel
S.No.ShapeHydraulic Radius
1Rectangular Channelym=y02{y_m} = \frac{y_0}{2}
2Trapezoidal channelym=y02{y_m} = \frac{y_0}{2}
3Triangular Channelym=y022{y_m} = \frac{y_0}{{2\sqrt 2 }}
4Circular channelym=0.29d{y_m} = 0.29d

Question 37

Minimum per capita water supply in Indian towns and cities as per National Building Code of India should be

  1. AOption (a)

    70 lpcd

  2. BOption (b)

    270 lpcd

  3. COption (c)

    210 lpcd

  4. DOption (d)

    135 lpcd

View answer & explanation

Correct answer

Answer: (d)D.

135 lpcd

Why this answer is correct

Explanation:

 As per NBC 2016, BIS

The following rates per capita per day may be considered the minimum for domestic and non-domestic needs:

  1. For communities with a population of up to 20,000 and without a flushing system : 

a) water supply through standpost: 40 lpcd (Min) 

b) water supply through house service connection:  70 to 100 lpcd.

2. For communities with a population of 20,000 to 100,000 together with a full flushing system: 100 to 135 lpcd. 3. For communities with a population above 100000  together with a full flushing system: 150 to 200 lpcd.

From the above given options answer should be 135lpcd.

Question 38

The angle between true meridian and the magnetic meridian at the time of observation is known as

  1. AOption (a)

    Orientation angle

  2. BOption (b)

    Magnetic declination

  3. COption (c)

    Magnetic bearing

  4. DOption (d)

    Angle of dip

View answer & explanation

Correct answer

Answer: (b)B.

Magnetic declination

Why this answer is correct

Explanation:

Declination: 

The horizontal angle between the True meridian and magnetic meridian is called declination.

Local attraction: 

Local attraction is the phenomenon by which the magnetic needle is constantly prevented to point towards the magnetic north at a place. This is because that these magnetic compass is influenced by other magnetic objects at that locality such as wires carrying electric current, rails, steel and iron structures, steel tapes etc.

Magnetic Bearing: 

The horizontal angle which a line makes with the magnetic meridian measured from Magnetic North line is called magnetic bearing. It varies with time. Magnetic meridian of a line can be measured in the field by using prismatic compass.

Azimuth: 

Azimuth of a line is its horizontal angle measured clockwise from geographic or true meridian.

Question 39

Which of the following is a cumulative elemental toxin?

  1. AOption (a)

    Cadmium

  2. BOption (b)

    Sodium

  3. COption (c)

    Barium

  4. DOption (d)

    Bromine

View answer & explanation

Correct answer

Answer: (a)A.

Cadmium

Why this answer is correct

Concept:

Toxicity and health effects of some heavy metals:

  • Heavy metal toxicity has proven to be a major threat and there are several health risks associated with it.
  • The toxic effects of these metals, even though they do not have any biological role, remain present in some of the other forms harmful for the human body and its proper functioning.
  • They sometimes act as a pseudo-element of the body while at certain times they may even interfere with metabolic processes.
  • Few metals, such as aluminum, can be removed through elimination activities, while some metals get accumulated in the body and food chain, exhibiting a chronic nature.
  • The most commonly found heavy metals in wastewater include arsenic, cadmium, chromium, copper, lead, nickel, and zinc, all of which cause risks for human health and the environment.
  • Heavy metals are significant environmental pollutants and their toxicity is a problem of increasing significance for ecological, evolutionary, nutritional, and environmental reasons.
  • Various public health measures have been undertaken to control, prevent and treat metal toxicity occurring at various levels, such as occupational exposure, accidents, and environmental factors.

Question 40

The water balance equation for a catchment area in terms of rainfall (P), run off (R), evaporation (E) and storage (S) is given as

  1. AOption (a)

    R = P - E ± Δ S

  2. BOption (b)

    R = P + E ± Δ S

  3. COption (c)

    R = E - P ± Δ S

  4. DOption (d)

    R = E + R ± Δ S

View answer & explanation

Correct answer

Answer: (a)A.

R = P - E ± Δ S

Why this answer is correct

Explanation:

Water - Budget Equation:

The Water - Budget equation is based on conservation of mass in a closed system and states that any water entering a system, must be transferred into either evaporation, surface runoff or stored in the ground.

A general water balance equation is

Mass inflow - Mass outflow = Change in storage

or

P=R+E±ΔSP = R + E\pm ΔS

where

P is precipitation (gain)

E is evapotranspiration (loss)

R is streamflow (loss)

ΔS is the change in storage.

Question 41

The construction of an optical square is based on the principle of

  1. AOption (a)

    Reflection

  2. BOption (b)

    Refraction

  3. COption (c)

    Double refraction

  4. DOption (d)

    Double reflection

View answer & explanation

Correct answer

Answer: (d)D.

Double reflection

Why this answer is correct

Concept:

An optical square is used for setting out right angles. It consists of a small circular metal box of diameter 5 cm and depth 1.25 cm. It has a metal cover that slides around the box to cover the slits.

Principle of optical square:

  • According to the principle of reflecting surfaces, the angle between the first incident ray and the last reflected ray is twice the angle between the mirrors. In this case, the angle between the mirrors is fixed at 45˚. So, the angle between the horizon sight and index sight will be 90˚.

Construction of Optical Square:

  • A horizon glass H is fixed at the bottom of the metal box. The lower half of the glass is unsilvered and the upper half is silvered.
  • An index glass “I” is also fixed at the bottom of the box which is completely silvered.
  • The angle between the index glass and horizon glass is maintained at 45˚.

The horizon glass is placed at an angle of 120˚ with the horizon sight. The index glass is placed at an angle of 105˚ with the index sight.

Question 42

Lacey's scour depth for a stream carrying a discharge of 3 cumecs/m width having a silt factor 1.2 is

  1. AOption (a)

    2.10

  2. BOption (b)

    2.64

  3. COption (c)

    2.89

  4. DOption (d)

    3.10

View answer & explanation

Correct answer

Answer: (b)B.

2.64

Why this answer is correct

Concept: 

Lacey's scour depth is given by:

R;=;1.35(q2f)13;R; = ;1.35{\left( {\frac{{{q^2}}}{f}} \right)^{\frac{1}{3}}};

Where, q = Discharge per meter width, f = silt factor

Calculations:

Given, q = 3 cumec, f = 1.2

R;=;1.35(321.2)13;=;2.64;mR; = ;1.35{\left( {\frac{{{3^2}}}{1.2}} \right)^{\frac{1}{3}}};=;2.64;m

Important Points 

Various formulas deduced by lacey is given below:

1) silt factor ⇒ f=1.76dmm\rm{f = 1.76\sqrt {{d_{mm}}}}

  1. velocity of flow ⇒V=[Qf2140]16V = {\left[ {\frac{{Q{f^2}}}{{140}}} \right]^{\frac{1}{6}}}

3) hydraulic mean depth ⇒R=5V22fR = \frac{{5{V^2}}}{{2f}}

  1. wetted perimeter ⇒ P=4.75QP = 4.75\sqrt Q

Question 43

The wave height (m) generated on the surface of a reservoir, having a fetch length F = 30 km, due to wind blowing on the surface of the reservoir at a velocity of 30 km/h is

  1. AOption (a)

    1.75 m

  2. BOption (b)

    0.96 m

  3. COption (c)

    0.52 m

  4. DOption (d)

    1.2 m

View answer & explanation

Correct answer

Answer: (a)A.

1.75 m

Why this answer is correct

Concept:

Forces due to wave pressure on a gravity dam:

  • Due to the generation of waves on the surface by blowing winds.

Wave height(hw) is given by

hw=0.032VF+0.763−0.271F3/4{h_w} = 0.032\sqrt {VF} + 0.763 - 0.271{F^{3/4}}    (For F < 32 km)

hw=0.032VF{h_w} = 0.032\sqrt {VF}     (For F > 32 km)

where, 

V = Wind velocity (km/hr)

F = Fetch or straight length of water (km) 

Calculation:

Given data

F  = 30 km

V = 30 km/hr

For F < 32 km, Wave height(hw)

hw=0.032VF+0.763−0.271F3/4{h_w} = 0.032\sqrt {VF} + 0.763 - 0.271{F^{3/4}}​​

hw=0.03230×30+0.763−0.271(30)3/4{h_w} = 0.032\sqrt {30 \times 30} + 0.763 - 0.271{(30)^{3/4}}

hw = 1.75 m

Question 44

The runway length is increased on every rise of 300 m from MSL for designing of airport by

  1. AOption (a)

    4 percent

  2. BOption (b)

    6 percent

  3. COption (c)

    7 percent

  4. DOption (d)

    10 percent

View answer & explanation

Correct answer

Answer: (c)C.

7 percent

Why this answer is correct

Explanation:

Correction for basic Runway length (I)

1) 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.

2) Temperature correction

Airport reference Temperature (ART)

Ta+13(Tm−Ta)T_a + {1\over3}(T_m-T_a)

Tm = Monthly mean of the maximum daily temperature of the hottest month.

Ta = Monthly mean of the average daily temperature of the hottest month

Rise in temperature = ART - SAT

Where,

SAT is Standard Atmospheric Temperature

NOTE: Due to the elevation, there is a decrease in standard temperature at the rate of -6.50 c per 1000 m rise.

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

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

If exceeds modify, I2 = 1.35 x I

3) Gradient correction

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

Gradient=Elevation differenceOriginal length of runwayGradient = {Elevation \space difference\over Original \space length \space of \space runway}

Question 45

Mercury is used in the barometer because:

  1. AOption (a)

    it is a perfect fluid

  2. BOption (b)

    its volume changes with temperature

  3. COption (c)

    it is a liquid metal

  4. DOption (d)

    it gives less height of column for high pressure

View answer & explanation

Correct answer

Answer: (d)D.

it gives less height of column for high pressure

Why this answer is correct

Mercury is used in the barometer because it is a high-density fluid which gives less height of column for high pressures. A barometer using water, for instance, would need to be 13.6 times taller than a mercury barometer to obtain the same pressure difference. This is because mercury is 13.6 times denser than water.

Question 46

If g1 and g2 are the two gradients r is the rate of change of grade in % per chain, the length of the vertical curve will be

  1. AOption (a)

    (g1 + g2)/r2

  2. BOption (b)

    (g1 - g2)/r-0.5​

  3. COption (c)

    (g1 - g2)/r​

  4. DOption (d)

    (g1 + g2) 0.5/r3

View answer & explanation

Correct answer

Answer: (c)C.

(g1 - g2)/r​

Why this answer is correct

Explanation:

  • Vertical curves are used in highway and street vertical alignment to provide a gradual change between two adjacent grade lines.
  • The gradient is the rate of rising or falling along the length of the road with respect to the horizontal.
  • The length of the vertical curve can be obtained by dividing the algebraic difference of the two grades by the rate of change of grade.

Length of curve(L)=Total change of gradeRate of change of gradeLength \ of \ curve(L) = \frac{{Total \ change \ of \ grade}}{{{Rate \ of\ change \ of \ grade}}}

Length of curve(L) = (g2-g1)**/**r

Where g1, g2 = Grade in percentage

r= rate of change of grade in percentage

Question 47

Which of the following tests is used to make the horizontal axis perpendicular to the vertical axis ?

  1. AOption (a)

    Azimuth test

  2. BOption (b)

    Diaphragm test

  3. COption (c)

    Spire test

  4. DOption (d)

    Vertical arc test

View answer & explanation

Correct answer

Answer: (c)C.

Spire test

Why this answer is correct

Concept:

The adjustments of the theodolite  are as follows:

Azimuth test:

  • The line of sight should be perpendicular to the horizontal axis.

​​Spire test:

  • The horizontal axis should be perpendicular to the vertical axis.

​Vertical circle index test:

  • The axis of the bubble tube attached to the telescope should be parallel to the line of sight. This is done so that the vertical circle reads zero when the telescope is horizontal.
  • This test is also known as the bubble tube adjustment test.

Vertical arc test:

  • When the line of sight is perpendicular to the vertical axis, the vertical circle should read zero.
  • This test is also known as the vertical circle test.

Cross-Hair ring test:

  • The vertical cross-hair should lie in a plane perpendicular to the horizontal axis, i.e. it is to be made truly vertical.

Plate level test:

  • The vertical axis of the instrument should be perpendicular to the axis of each plate bubble tube.

Question 48

Stability of an excavated portion of a road on a hill can be provided by construction of

  1. AOption (a)

    Parapet wall

  2. BOption (b)

    Retaining wall

  3. COption (c)

    Breast wall

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (c)C.

Breast wall

Why this answer is correct

Explanation:

Breast wall:

  • A breast wall is built to sustain the face of a natural bank of earth
  • A breast wall is usually constructed on the hillside of the roadway.

Retaining wall: 

  • Retaining walls are relatively rigid walls used for supporting soil laterally so that they can be retained at different levels on the two sides.

Parapet wall: 

  • A wall around the edge of a roof, balcony, terrace, or stairway is designed either to prevent those behind it from falling over or to shelter them from attack from the outside.

Difference between Breast wall and Retaining wall:

Retaining wallBreast wall
Retaining wall used to retain earth pressure against an artificially cut edge or excavation edge in soil.
A breast wall on the other hand is used to sustain earth against a natural slope such as the hillside of a mountain road.

Question 49

Infiltration capacity is

  1. AOption (a)

    A constant factor

  2. BOption (b)

    Changes with time

  3. COption (c)

    Changes with location

  4. DOption (d)

    Changes with both time and location

View answer & explanation

Correct answer

Answer: (d)D.

Changes with both time and location

Why this answer is correct

Explanation :

Infiltration:

Infiltration is the flow of water through the soil surface into a porous medium under gravity action and pressure effects.

The infiltration capacity is the maximum rate at which water can be absorbed by a given soil per unit area under given conditions.

Question 50

A light house of 120 m height is just visible above the horizon from a ship. The correct distance (m) between the ship and the light house considering combined correction for curvature and retraction is 

  1. AOption (a)

    39.098 m

  2. BOption (b)

    42.226 m

  3. COption (c)

    39098 m

  4. DOption (d)

    42226 m

View answer & explanation

Correct answer

Answer: (d)D.

42226 m

Why this answer is correct

Concept:

Combined correction to take into the effect of error due to collimation and refraction is given by:

Cd = 0.0673 × d2

Distance to visible horizon (H): = 0.0673 × D2 or D = 3.85 × √H

Where,

H is the height of the tower/ lighthouse in meters and D is the distance between the object and lighthouse in km.

Calculation:

Given: H = 120 m

120 = 0.0673 × D2

D=1200.0673=42.226;kms=42226;meters{\rm{D}} = \sqrt {\frac{{120}}{{0.0673}}} = 42.226{\rm{;kms}} = 42226{\rm{;meters}}

Important Point:

Combined correction is the combined effect of curvature correction and refraction correction.

Curvature correction (Cc) = - 0.07849 × D2, Refraction correction (CR) = 1/7 × Cc

If the observer has a height of H1 (m) and the object has a height of H2 (m) then, the distance between horizons will be

D = 3.85(√H1+√H2)

Where D is in km.

Question 51

A lysimeter is used to measure

  1. AOption (a)

    infiltration

  2. BOption (b)

    evaporation

  3. COption (c)

    evapotranspiration

  4. DOption (d)

    radiation

View answer & explanation

Correct answer

Answer: (c)C.

evapotranspiration

Why this answer is correct

Measurement of evapotranspiration is done in two ways either by using lysimeter or by the use of field plots. Lysimeter studies are time consuming and expansive.

Important Points

Instruments used in the measurement

InstrumentUse
PsychrometerRelative humidity
HygrometerHumidity
TensiometerCapillary potential
AnemometerWind velocity
Ombrometer/pluviometerRainfall depth
LysimeterEvapotranspiration
AtmometerEvaporation
PhytometerTranspiration
PermeameterHydraulic conductivity
Rainfall simulatorInfiltration capacity

Confusion point:

The pitot tube is used to measure the velocity of fluid whereas an anemometer is used to measure gas and air velocity

Question 52

In brick lining, the layer, which makes the canal water tight is

  1. AOption (a)

    Lower layer of bricks

  2. BOption (b)

    Sandwiched layer of cement mortar

  3. COption (c)

    Upper layer of the bricks

  4. DOption (d)

    Deposited soil

View answer & explanation

Correct answer

Answer: (b)B.

Sandwiched layer of cement mortar

Why this answer is correct

Explanation:

Canal Lining:

  • Canals are lined to reduce seepage losses, and to the chances of waterlogging in a given area.
  • Various types of lining options are available, and the selection of a particular type is related to cost effectiveness and ready availability.
  • Lined channels are non-erodible, and hence their design theories are different from those of unlined canals.

Brick Lining:

  • To make the lining successful the lining may be constructed of two layers of bricks laid flat in mortar. The first layer is laid on a 12 mm layer of 1: 6 cement mortar. This 12 mm thick layer of cement mortar is spread on the properly compacted and wetted subgrade.
  • Then on the top of first brick layer, a 12 mm layer of 1: 3 cement mortar is given. It is allowed to cure for two days.
  • Finally second layer of bricks in mortar is laid on the top. Thus a 12 mm layer of rich cement mortar (1: 3) is sandwiched between two brick layers and this layer is actually responsible to make the lining practically watertight.

Important Points Types of canal lining: Canals are lined by any of the following materials:

  • Concrete (in-situ construction).
  • Precast concrete (tiles).
  • Lime concrete.
  • Shotcrete or guniting.
  • Brick tiles.
  • Asphalt- (a) Buried asphalt membrane (b) Asphaltic concrete.
  • Stone blocks, concrete blocks or undressed stones.
  • Earth materials- (a) Compacted earth (b) Soil cement.
  • Prefabricated light-weight membranes.
  • Bentonite - soil and clay membranes.
  • PVC films.

Question 53

If super elevation is NOT provided on a horizontal curve of a highway in which portion of the road, potholes will be developed ?

  1. AOption (a)

    Outer edge of road

  2. BOption (b)

    Inner edge of road

  3. COption (c)

    Center of road

  4. DOption (d)

    Shoulder of road

View answer & explanation

Correct answer

Answer: (a)A.

Outer edge of road

Why this answer is correct

Concept:

Superelevation:

  • It is the transverse slope to counteract the centrifugal force and to reduce the tendency of the vehicle to overturn or skid.
  • It is the rise of the outer edge of the pavement with respect to the inner edge on a horizontal curve thus providing a transverse slope throughout the length of the horizontal curve.
  • It is also known as cant or banking.

Pothole:

  • The appearance of bowl-shaped holes, usually after rain.
  • This indicated progressive disintegrations of bituminous premix carpet surfacing by loss of aggregates.
  • When a vehicle negotiates a vertical curve on a highway without super elevation the pressure on the tires on the outer edge is more than on the inner edge. Thus the chance of potholes is more on the outer edge of the road.

Question 54

​If the base period of a 6 hr. unit hydrograph of a basin is 84 hr. then, the base period of a 12 hr. unit hydrograph of the same basin will be

  1. AOption (a)

    90 hr.

  2. BOption (b)

    168 hr.

  3. COption (c)

    72 hr.

  4. DOption (d)

    84 hr.

View answer & explanation

Correct answer

Answer: (a)A.

90 hr.

Why this answer is correct

Explanation:

Unit hydrograph: 

A unit hydrograph is a direct runoff hydrograph resulting from one unit (one cm) of constant intensity uniform rainfall occurring over the entire watershed.

Base period of 12 hr unit hydrograph = Base period of 6 hr unit hydrograph + 6

Base period of 12 hr unit hydrograph = 84 + 6 = 90h

Question 55

The ratio of actual evapotranspiration to potential evapotranspiration is in the range of

  1. AOption (a)

    0.0 to 0.4

  2. BOption (b)

    0.6 to 0.9

  3. COption (c)

    0.0 to 1.0

  4. DOption (d)

    1.0 to 2.0

View answer & explanation

Correct answer

Answer: (c)C.

0.0 to 1.0

Why this answer is correct

Explanation:

Actual evapotranspiration: Actual Evapotranspiration (ETa) is the quantity of water that is removed from a surface due to the processes of evaporation and transpiration.

Potential evapotranspiration: Potential evapotranspiration is the demand or maximum amount of water that would be evapotranspirate  if enough water were available.

The ratio of actual evapotranspiration to potential evapotranspiration is in the range of 0.0 to 1.0

For soil moisture at field capacities, AET = PET

For soil moisture at permanent willing point, AET ≈ 0 

Question 56

For a 8° curve track diverging from a main curve of 5° in an opposite direction in the layout of a B.G. yard, how much cant is to be provided for the branch track for maximum speed of 45 km/hr. on the main line ? Permitted cant deficiency for the main line is 7.6 cm, value of gauge G = 1.676m, cant on diverging track

  1. AOption (a)
    • 0.168 cm
  2. BOption (b)

    7.432 cm

  3. COption (c)

    0.168 cm

  4. DOption (d)

    7.768 cm

View answer & explanation

Correct answer

Answer: (a)A.
  • 0.168 cm

Why this answer is correct

Concept:

Superelevation/Cant:

  • It is the difference between the height of the outer rail and the inner rail on a curve. The inner rail is taken as the reference rail and with respect to the inner rail, the outer rail is raised.
  • The objective of providing a superelevation to neutralize the effect of centrifugal force, achieve equal distribution of wheel loads, improve the comfort level of passengers, etc.

Superelevation or cant 

e=GV2127Re = {GV^2 \over 127R}

Where, R = radius of rail in meters, V = Speed of train in km/hr, G = gauge of track

Cant deficiency:

  • The actual cant provided on a railway track is based on the average speed of trains.
  • However, in the case of high-speed trains, the cant requirement will be more than the actual value of the cant provided so the train will be forced to move on a lower value of cant than the maximum cant is required for the high-speed train. This deficiency in the cant for high-speed trains is called a cant deficiency.

Calculation:

Given data:

G = 1.676 m, V = 45 km/hr,

R = 1720 / D where D = degree of curve

R = 1720 / 5 

So, the cant for the main line is 

e=GV2127R=1.676×452×5127×1720e = {GV^2 \over 127R} = {1.676 × 45^2 \times 5\over 127× 1720 }  = 0.0778 m = 7.78 cm

The permitted cant deficiency for the mainline = 7.6 cm

Hence, cant for mainline = 7.78 - 7.60 = 0.18 cm

The cant for the branch track on the main line is - 0.18 cm, which is closest to option B ie - 0.168 cm

Additional Information Radius or degree of a curve:

  • A curve is defined either by its radius or by its degree. The degree of a curve is the angle subtended at its center by a 100 ft or 30.5 m cord.
  • The value of the degree of the curve can be determined as below:

Circumference of a circle = 2 × π × R 

The angle subtended at the center by a circle with this circumference = 360° 

The angle subtended at the center by a circle by a 30 m chord or degree of curve = 360×302×π×R360 \times 30 \over {2\times\pi\times R } = 1750R1750 \over R (Where R is in meters).

In general, we use a 30 m chord, but for exact calculations, a 30.5 m chord is used in railways, and accordingly, in place of 1720 we have to use 1750 in the formula.

But, the given options were set according to 1720, so we have provided a solution accordingly.

Question 57

Water Year in India starts from the first day of

  1. AOption (a)

    January

  2. BOption (b)

    June

  3. COption (c)

    April

  4. DOption (d)

    September

View answer & explanation

Correct answer

Answer: (b)B.

June

Why this answer is correct

Concept:

Water Year:

  • Generally, 1 October to 30 September in the Northern Hemisphere, 1 July to 30 June in the Southern Hemisphere; the annual cycle that is associated with the natural progression of the hydrologic seasons.
  • It commences with the start of the season of soil moisture recharge, includes the season of maximum runoff (or season of maximum groundwater recharge), if any, and concludes with the completion of the season of maximum evapotranspiration (or season of maximum soil moisture utilization).
  • Each country has fixed a definite 12-month period to begin and in a relatively dry season as the water year.
  • In India, the water year starts from June 1st and ends on May 31st of the following calendar year.

 Important Points

  • The World Water Day is observed every year on 22nd March to mark the importance of freshwater.

Question 58

A plot of land 60m x 20m is measured by a steel tape. If the standard error of length and width measurement is taken as ± 1 cm, then the standard error of the area of the plot would be

  1. AOption (a)

    ± 0.1414 m2

  2. BOption (b)

    ± 0.566 m2

  3. COption (c)

    ± 0.632 m2

  4. DOption (d)

    ± 0.8484 m2

View answer & explanation

Correct answer

Answer: (c)C.

± 0.632 m2

Why this answer is correct

Explanation:

Concept: Standard error in computation of area is given by:

  eA=[(eldAdl)2+(ebdAdb)2]1/2e_{A}={[(e_{l}\frac{\mathrm{d}A }{\mathrm{d} l})^{2}}+(e_{b}\frac{\mathrm{d}A }{\mathrm{d} b})^{2}]^{1/2}    

where, el and eb = standard error of length and width measurement,

l and b = length and width of plot respectively, A = Area of plot = l × b

∴     dAdl=b,,,and,,dAdb=l\frac{\mathrm{d} A}{\mathrm{d} l} = b ,,, and ,, \frac{\mathrm{d} A}{\mathrm{d} b} = l\textbf{}

Solution: 

given , l = 60 m , b = 20 m , el = eb = ±\pm 1 cm = ±\pm0.01 m

∴ Standard error in computation of area 

eA=[(eldAdl)2+(ebdAdb)2]1/2=[(0.01×20)2+(0.01×60)2]1/2=±0.632,,m2e_{A}={[(e_{l}\frac{\mathrm{d}A }{\mathrm{d} l})^{2}}+(e_{b}\frac{\mathrm{d}A }{\mathrm{d} b})^{2}] ^ {1/2}={[(0.01\times20)^{2}}+(0.01\times60)^{2}]^{1/2}= \pm0.632,,m^2

Question 59

A sewer is laid from a manhole A to a manhole B, 250m apart along a downward grade of 1 in 125. If the R.L. of the invert at A is 205.75m and the length of boning rod is 3 m, then R.L. of the sight rail at B is

  1. AOption (a)

    202.75 m

  2. BOption (b)

    206.75 m

  3. COption (c)

    208.75 m

  4. DOption (d)

    211.75 m

View answer & explanation

Correct answer

Answer: (b)B.

206.75 m

Why this answer is correct

Explanation:

Distance between sewers = 250 m

Slope = 1:125 

Difference of height between 2 sewers = 250/125 = 2 m = ΔH 

RL of manhole B = RL of invert at A - Difference of height between 2 manholes + Height of boning rod

RLB = RLA - ΔH + HB

RLB = 205.75 - 2 + 3 = 206.75 m

Question 60

If EC is specific energy at critical depth YC in a triangular channel, the value of ECYC\frac{E_C}{Y_C} is

  1. AOption (a)

    3.33

  2. BOption (b)

    2.5

  3. COption (c)

    1.5

  4. DOption (d)

    1.25

View answer & explanation

Correct answer

Answer: (d)D.

1.25

Why this answer is correct

Explanation:

The total head or energy head (H), at any location in an open channel, can be defined as

  • H=z+y+V22gH = z + y + \frac{{{V^2}}}{{2g}}

By assuming kinetic energy factor (α =1) and pressure distribution is hydrostatic.

Using channel bottom as datum (z = 0), The total head or energy head above the channel bottom is called specific energy.

  • E=y+V22gE = y + \frac{{{V^2}}}{{2g}}

For critical flow in triangular channel

Ec=yc+Vc22g{E_c} = {y_c} + \frac{{V_c^2}}{{2g}} {eqn (i)}

∵ We know that, Froude number for critical flow, Fr = 1

and, Fr=VcgD=1{F_r} = \frac{{{V_c}}}{{\sqrt {gD} }} = 1 {eqn(ii)}

Where, D = Hydraulic depth, D = Area of section/Top width of section

Hydraulic depth for triangular section, D = yc/2 

From eqn (ii), Vcgyc2=1\frac{{{V_c}}}{{\sqrt {g\frac{{{y_c}}}{2}} }} = 1 ⇒ 2Vc2gyc=1\frac{{2V_c^2}}{{g{y_c}}} = 1

Vc2g=yc2\frac{{V_c^2}}{g} = \frac{{{y_c}}}{2}

from (i), Ec=yc+yc4=54yc{E_c} = {y_c} + \frac{{{y_c}}}{4} = \frac{5}{4}{y_c}

Ec = 1.25 Yc

so, EcYc=1.25{E_c\over Y_c}=1.25

Question 61

The concept of boundary layer was first introduced by

  1. AOption (a)

    Newton

  2. BOption (b)

    Prandtl

  3. COption (c)

    Reynold

  4. DOption (d)

    Kutter

View answer & explanation

Correct answer

Answer: (b)B.

Prandtl

Why this answer is correct

Explanation:

(i) Boundary layer concept was first introduced by L. Prandtl in 1904.

(ii) In ideal fluid viscous force is neglected but in actual viscous stresses are prominent within the boundary layer.

(iii) When fluid flows past a solid body the fluid particles adhere to the boundary and the condition of no-slip occurs.

(iv) No-slip conditions: It means the velocity of the fluid at body surface will be same as the velocity of body.

(v) Within the boundary layer viscous stress decreases from maximum to negligible value in which in turn causes an increase in velocity from zero at body surface to upstream velocity.

(vi)Outside the boundary layer effect of shear stress is very low and velocity gradient is negligible.

Question 62

Aeration in water is done for the removal of

  1. AOption (a)

    Colour

  2. BOption (b)

    Turbidity

  3. COption (c)

    Hardness

  4. DOption (d)

    Bad odour

View answer & explanation

Correct answer

Answer: (d)D.

Bad odour

Why this answer is correct

Concept:

Aeration:

  • Aeration brings water and air in close contact in order to remove dissolved gases (such as carbon dioxide) and oxidizes dissolved metals such as iron, hydrogen sulfide, and volatile organic chemicals (VOCs).
  • It also removes bad odor from the water.
  • Aeration is often the first major process at the treatment plant.
  • The efficiency of aeration depends on the amount of surface contact between air and water, which is controlled primarily by the size of the water drop or air bubble.

The chemicals removed or oxidized by aeration are as follows:

  • Volatile organic chemicals, such as benzene (found in gasoline), trichloroethylene, dichloroethylene, and perchloroethylene (used in dry-cleaning or industrial processes)
  • Ammonia
  • Chlorine
  • Carbon dioxide
  • Hydrogen sulfide
  • Methane
  • Iron and Manganese

Question 63

An aquifer confined at the bottom, but NOT at the top is called

  1. AOption (a)

    Semiconfined aquifer

  2. BOption (b)

    Unconfined aquifer

  3. COption (c)

    Confined aquifer

  4. DOption (d)

    Perched aquifer

View answer & explanation

Correct answer

Answer: (b)B.

Unconfined aquifer

Why this answer is correct

Explanation:

Unconfined aquifer: 

  • An unconfined aquifers is one in which the water table forms the upper surface of zone of saturation.
  • The term unconfined signifies the absence of any geological layer confining the zone of saturation (above the water table).
  • The unconfined aquifer is in direct contact with atmosphere through the zone of aeration. At the water table, the hydraulic pressure head or ground water pressure head is equal to atmospheric pressure (i.e. zero).
  • This aquifer is also known as water table aquifer or phreatic aquifer.

Confined aquifer:

  • The confined aquifer a geological formation qualifying as an aquifer is overlained by a impermeable layer or an aquiclude.
  • The confining layers show very low intrinsic permeability (k).Here the overlying layer acts as layer confining the underlying aquifer.
  • The water in the confined aquifer is not in direct contact with the

atmosphere. This is because of the presence of confining layer overlying the confined aquifer.

Semiconfined aquifer:

  • An aquifer partially confined by soil layers of low permeability through which recharge and discharge can still occur.
  • Semi-confined aquifers can usually be located between aquifers and the ground surface with the water seeping into via the internal layers to move freely through the main aquifers.

Perched aquifer:

  • A special case of unconfined aquifer is known as perched aquifer. A perched aquifer is formed when the infiltrated rain water is intercepted within the zone of Aeration by an impermeable layer and a local zone of saturation is formed.
  • The upper surface of such local zone of saturation is known as perched water table. The perched aquifer occurs at higher elevation than the regional water table.

Question 64

The relationship between the length I and radius r of an ideal transition curve is stated as:

  1. AOption (a)

    l  ∝ r

  2. BOption (b)

    l ∝ r2

  3. COption (c)

    l ∝ 1r\frac{1}{r}

  4. DOption (d)

    l ∝ 1r2\frac{1}{r^{2}}

View answer & explanation

Correct answer

Answer: (c)C.

l ∝ 1r\frac{1}{r}

Why this answer is correct

Explanation:

A transition curve is provided at the junction of the straight and curved portion of a railway track to serve the following purposes:

i) to provide the superelevation in a gradual manner

ii) to avoid the sudden jerking of the passengers

iii) to reduce the radius of curvature gradually

The requirement of an ideal transition curve:

i) The length of the transition curve is inversely proportional to the radius of curvature ( l∝1r;l \propto \frac{1}{r};).

ii) The curve should be tangential to its junction points.

iii) The rate of change of superelevation provided should be equal to the rate of change of curvature so that full superelevation can be provided within the length of the transition curve.

iv) The rate of change of centrifugal acceleration should be consistent, this implies the radius of the transition curve should be consistently decreased from infinity at the tangent point to the radius R of the circular curve.

Question 65

Infiltration capacity of the ground does NOT depend on

  1. AOption (a)

    Gravity

  2. BOption (b)

    Capillarity

  3. COption (c)

    Water table

  4. DOption (d)

    Temperature

View answer & explanation

Correct answer

Answer: (a)A.

Gravity

Why this answer is correct

Concept:

Infiltration:

It is the flow of water into the ground through the soil surfaces.

Infiltration capacity:

It is defined by Horton as the maximum rate at which rain can be absorbed by the soil in the given condition.

Horton’s equations:

Horton expressed the decay of infiltration capacity with time as an exponential decay

ft = fc + (f0 - fc) e-kt for 0 < t < tc

where,

ft = Infiltration capacity at any time t from the start of rainfall

f0 = Initial infiltration capacity at t = 0

fc = Final infiltration capacity occurring at t = tc

k = Horton’s decay constant which depends upon soil characteristic and vegetation cover

The main factors that influence the infiltration are as follows:

1. The soil type i.e. texture, structure, hydrodynamic characteristics:

  • The soil characteristics influence capillary forces and adsorption.

2. The soil coverage and vegetation:

  • They increase the time of water penetration in soil.

3. The topography and morphology of slopes: 

  • The steep slope has low infiltration capacity as water flow rapidly due to gravity.

4. Rain intensity: 

  • High rainfall implies more rate of infiltration.

5. Soil humidity or moisture:  

  • Dry soil has more infiltration as compared to wet soil.

6. Soil compaction:

  • High compaction reduces infiltration capacity.

7. Soil thickness:

  • Higher the thickness soil can hold more water.

8. Surface retention or runoff: 

  • More surface run-off means less water is available to infiltrate into the soil.

9. Temperature:

  • If the saturated soil temperature is very low, the soil mass becomes impermeable so that the infiltration rate will be below.

Hence, The infiltration capacity of the ground does NOT depend on gravity.

Question 66

The holding capacity of the water quantity by geological strata depends on the property of soil

  1. AOption (a)

    Permeability

  2. BOption (b)

    Porosity

  3. COption (c)

    Transmissibility

  4. DOption (d)

    Yield

View answer & explanation

Correct answer

Answer: (d)D.

Yield

Why this answer is correct

Concept:

Permeability:

  • The ability of soil to allow the flow of water through it is called the permeability of the soil.
  • It is a very important factor for the structures which are in contact with water.
  • The flow of water in soil takes place through void spaces, which are interconnected.

Porosity:

  • The pore space of soil is defined as the portion of the soil volume occupied by air and water.
  • It refers to the percentage of soil volume occupied by pore spaces.

Transmissibility:

  • Transmissibility (or transmissivity) is a property closely related to hydraulic conductivity that describes the capacity of a specific water-bearing unit of a given thickness, such as an aquifer, to transmit water.
  • Transmissibility is most simply defined as the effective hydraulic conductivity of an aquifer or other water-bearing unit multiplied by the thickness of that unit.

Specific retention:

  • It is that volume fraction of water that is held back by adhesion and capillary forces when an aquifer is drained.

Specific yield:

  • It is the amount of water that is actually available for groundwater pumping when sediments or rocks are drained due to the lowering of the water table.

Question 67

What is the permissible velocity of water in cement concrete lined canal as per IS 10430-2000?

  1. AOption (a)

    2.7 m/s

  2. BOption (b)

    1.8 m/s

  3. COption (c)

    3.1 m/s

  4. DOption (d)

    1.5 m/s

View answer & explanation

Correct answer

Answer: (a)A.

2.7 m/s

Why this answer is correct

As per IS 10430 : 2000, the maximum permissible velocity in different types of canals is given below in tabulated form: The velocity is limited to avoid scouring in canals.

Type of CanalMaximum Permissible Velocity
Stone Pitched Lining1.5 m/s
Burnt Clay tile or Brick Lining1.8 m/s
Cement Concrete Lining2.7 m/s

Question 68

In a 12 cm dia pipe, the minor losses are equal to 5V22g5 \frac{V^2}{2g} the equivalent length of pipe to account for the minor losses for f = 0.04 is

  1. AOption (a)

    5 m

  2. BOption (b)

    15 m

  3. COption (c)

    10 m

  4. DOption (d)

    20 m

View answer & explanation

Correct answer

Answer: (b)B.

15 m

Why this answer is correct

Concept:

Types of losses in pipe:

Minor losses: 

  • Whenever there is a change in the cross-section, minor losses occur.
  • For e.g. sudden expansion, sudden contraction, or bend in the pipes.

Major losses: 

  • Whenever the losses in the pipes are because of friction they are considered as major losses because there is a significant loss of energy because of friction.

According to Darcy’s Weisbach equation, A major loss (hL) is the head loss due to friction

hL=fLV22gD{h_L} = \frac{{fL{V^2}}}{{2gD}}

where f = friction factor, L = length of pipe, V = velocity of flow, D = diameter of the pipe

Calculation:

Given data

D = 12 cm

f = 0.04

Minor losses =  5V22g5 \frac{V^2}{2g}

Loss due to friction(hL)

hL=fLV22gD{h_L} = \frac{{fL{V^2}}}{{2gD}}

To account for the minor losses

fLV22gD=5V22g \frac{{fL{V^2}}}{{2gD}} = 5 \frac{V^2}{2g}

fLD=5{fL \over D} = 5

L=5×Df=5×0.120.04L = {5 \times D \over f} = {5 \times 0.12 \over 0.04} = 15 m

Question 69

What is the theoretical oxygen demand (mg/l) of a glucose solution of concentration 500 mg/l?

  1. AOption (a)

    250.33

  2. BOption (b)

    380.65

  3. COption (c)

    533.33

  4. DOption (d)

    650.21

View answer & explanation

Correct answer

Answer: (c)C.

533.33

Why this answer is correct

\(\begin{matrix} {{\text{C}}{6}}{{\text{H}}{12}}{{\text{O}}{6}} \ \left( 180\text{ }!!~!!\text{ part} \right)\left( \text{Glucose} \right) \ \end{matrix}+\begin{matrix} 6{{\text{O}}{2}} \ 6\times 32\text{ }!!~!!\text{ parts}\left( \text{Oxygen} \right) \ \end{matrix}\to 6\text{C}{{\text{O}}{2}}+6{{\text{H}}{2}}\text{O}\)

180 parts of C6H12O6 gives 192 parts of O2

∴ 1 part of C6H12O6 gives 192/180 parts of O2

⇒ 500 mg/l of C6H12O6 = 192/180 × 500 mg/l of O2

Theoretical oxygen demand = 533.33 mg/l

∴ The theoretical oxygen demand of a glucose solution of concentration 500 mg/l is 533.33 mg/l

Question 70

The method of design of flexible pavement as recommended by IRC is:

  1. AOption (a)

    Group index method

  2. BOption (b)

    Westergard method

  3. COption (c)

    CBR method

  4. DOption (d)

    Benkelman beam method

View answer & explanation

Correct answer

Answer: (c)C.

CBR method

Why this answer is correct

Concept:

CBR (California Bearing Ratio) Method: (IRC 37-1984)

  • The method of design of flexible pavement as recommended by IRC is the CBR method.
  • The CBR method was developed originally by the California State Highway Department.
  • The advantage of the CBR method is that it can be used to find the pavement and that of individual courses in addition thickness of the CBR values of the materials of the courses are also known.

The following methods are used for the design of flexible pavements:-

  1. Empirical methods
  2. Semi-empirical methods
  3. Theoretical methods

Empirical methods:

They are based on the strength properties of the soil subgrade.

  1. The Group index method,
  2. CBR method and
  3. Mcleod method

Semi-empirical or Semi theoretical method:

These methods are based upon the stress-strain function.

  1. Triaxial test method

Theoretical Methods:

These methods totally depend upon theoretical analysis and mathematical computations.

  1. Burmeister method

​.

Additional Information

Group-Index Method:

This method was devised by the U.S. Highway Engineers (Highway Research Board) in1945. The group index of a soil is an arbitrary index assigned to different soil types based on their percentage of fines, liquid limit and plasticity index and is defined by:

GI = 0.2a + 0.005ac + 0.01bd

where

a = (P - 35) < 40

b = (P -15) < 40

c= (WL - 40) < 20

d = (1p - 10) < 20

where

P = Percentage fines passing from 0.074 mm sieve.

WL = Liquid Limit

Ip = Plasticity Index

Gmax= 20 and Gmin = 0

Key Points

  • Westergard method is used for rigid pavement

Question 71

Confined aquifer is NOT associated with

  1. AOption (a)

    Phreatic zone

  2. BOption (b)

    Peizometric surface

  3. COption (c)

    Free water table

  4. DOption (d)

    Free flowing well

View answer & explanation

Correct answer

Answer: (a)A.

Phreatic zone

Why this answer is correct

Explanation:

Confined aquifer:

  • When some amount of water is present in between two impermeable layers, it is known as a confined aquifer.
  • It is also known as the Artesian aquifer.
  • The pressure in this aquifer is greater than hydrostatic pressure

.

  • Groundwater in a confined aquifer is under pressure and will rise up inside a borehole drilled into the aquifer. The level to which the water rises is called the piezometric surface.
  • Hence, a confined aquifer is NOT associated with the phreatic zone because it is associated with an unconfined aquifer.

Question 72

The sleeper density of a broad gauge track is (M + 7) in metric units, where M is length of rail in metres. The number of sleepers for 1.024 km length of track is:

  1. AOption (a)

    1700

  2. BOption (b)

    1500

  3. COption (c)

    1600

  4. DOption (d)

    1900

View answer & explanation

Correct answer

Answer: (c)C.

1600

Why this answer is correct

On Indian railways standard length of BG rail = 13 m 

So, total no. of rail required in 1.024 km or 1024 m

= 1024/13 = 78.78 = 80 (in round figure)

Sleeper density = M + 7

M = length of rail

So, sleeper density = 13 + 7 = 20

No. of sleepers = 20 × 80 = 1600

Question 73

The magnetic bearing of line AB is 132° 45'. If the magnetic declination is 10° 20'W, the true bearing of the line AB will be

  1. AOption (a)

    122° 25'

  2. BOption (b)

    143° 5'

  3. COption (c)

    121° 25'

  4. DOption (d)

    122° 30'

View answer & explanation

Correct answer

Answer: (a)A.

122° 25'

Why this answer is correct

Concept:

Magnetic declination: At any place horizontal angle between true meridian and magnetic meridian is called magnetic declination.

For Eastern Declination:

α = β + θe or T.B = M.B + θe

Here, α = True bearing (T.B)

β = Magnetic bearing (M.B), and

θE = Eastern declination

For Western Declination:

α = β – θw or T.B = M.B – θw

Here, θw = western declination.

Calculation:

Magnetic bearing of a line AB = 132°45’

Declination = 10°20’ West

∴ True bearing = 132°45’ - 10°20’ = 122°25' = S57°35’E

Question 74

Which of the following rivers in alluvial flood plains flows in two or more channels around alluvial islands?

  1. AOption (a)

    Braided river

  2. BOption (b)

    Flashy river

  3. COption (c)

    Virgin river

  4. DOption (d)

    Tidal river

View answer & explanation

Correct answer

Answer: (a)A.

Braided river

Why this answer is correct

Explanation:

Braided river: 

  • It is formed when a stream contains more sediment than it can readily transport.
  • It is the stream that divides into branching and intertwining subchannels separated by islands or sandbars.
  • It is also characteristic of mature stage of river development where erosion and deposition both are going on.

Flashy river:

  • In this type of rivers the flood occurs and recedes very suddenly.
  • The flood may rise and recede in a short period of one or two days.

Virgin river:

  • Those types of rivers that may completely dry before they join another river or ocean.

Tidal river:

  • A tidal river is a river whose flow and level are influenced by tides. A section of a larger river affected by the tides is a tidal reach.
  • Generally, tidal rivers are short rivers with relatively low discharge rates but high overall discharge, which generally implies a shallow river with a large coastal mouth.

Question 75

The number of spectral bands in the enhanced thematic mapper sensor on the remote sensing satellite LANDSAT-7 is

  1. AOption (a)

    64

  2. BOption (b)

    10

  3. COption (c)

    8

  4. DOption (d)

    24

View answer & explanation

Correct answer

Answer: (c)C.

8

Why this answer is correct

Explanation:

  • The Enhanced Thematic Mapper Plus (ETM+) is the main instrument on board Landsat-7 and has been operational since 1999.
  • Landsat 7 Enhanced Thematic Mapper Plus (ETM+) images consist of eight spectral bands with a spatial resolution of 30 meters for Bands 1 to 7. The resolution for Band 8 (panchromatic) is 15 meters.
  • All bands can collect one of two gain settings (high or low) for increased radiometric sensitivity and dynamic range, while Band 6 collects both high and low gain for all scenes.
  • Approximate scene size is 170 km north-south by 183 km east-west (106 mi by 114 mi).

Question 76

The fixing of a plane table from three known points is good if

  1. AOption (a)

    The middle station is nearest

  2. BOption (b)

    The middle station is farther than the other two stations

  3. COption (c)

    Either of the extreme stations is the nearest

  4. DOption (d)

    The middle station is close to the great circle

View answer & explanation

Correct answer

Answer: (a)A.

The middle station is nearest

Why this answer is correct

Explanation:

Fixing of plane table by three known points:

Fix of plane table: 

  • The accuracy with which a plane table station can be located through three-point problem is known as its fix.
  • The accuracy of the fix depends on the relative positions of the plotted points and that of the location of the plane table station.
  • The degree of accuracy of the solution of three point problem is its strength.
  • For high accuracy, fix is termed as strong.
  • For low accuracy, the fix is termed as poor.

​The strength of fix is good if:

  1. Location of station is chosen within the great triangle formed by joining the three well defined object.
  2. The middle object is nearer to the position of the plane table than the other two objects
  3. Of the two interior angles subtended by the three objects at the plane table stations, one is small and other is large.
  4. However, the object subtending small angles should be widely separated from each other.

​The strength of fix is Poor if:

  1. The location of the plane table is on (or) near the circumference of the great circle.
  2. Both the interior angles, subtended by well defined objects, at the plane table stations, are small.

Question 77

Which of the following equations represents the d/s profile of "ogee" spillway with vertical u/s face? (x, y) are the coordinates of the point on the d/s profile with origin at the crest of the spillway and Hd is the design head.

  1. AOption (a)

    YHd=−0.5(XHd)1.85\frac{Y}{H_d}=-0.5\left(\frac{X}{H_d}\right)^{1.85}

  2. BOption (b)

    YHd=−0.5(XHd)11.85\frac{Y}{H_d}=-0.5\left(\frac{X}{H_d}\right)^{\frac{1}{1.85}}

  3. COption (c)

    YHd=−2.0(XHd)1.85\frac{Y}{H_d}=-2.0\left(\frac{X}{H_d}\right)^{1.85}

  4. DOption (d)

    YHd=−2.0(XHd)11.85\frac{Y}{H_d}=-2.0\left(\frac{X}{H_d}\right)^{\frac{1}{1.85}}

View answer & explanation

Correct answer

Answer: (a)A.

YHd=−0.5(XHd)1.85\frac{Y}{H_d}=-0.5\left(\frac{X}{H_d}\right)^{1.85}

Why this answer is correct

Explanation:

  • The downstream profile can be represented by the equation:
  • xn=K.Hdn−1.yx^n=K.H_{d} ^{n-1}. y
  • where (x, y) are the co-ordinates of the points on the crest profile with the origin at the highest point C of the crest, called the apex.

          Hd = design head including the velocity head.

          K and n are constants depending upon the slope of the upstream face.

          For vertical upstream face of the spillway k =2.0 and n = 1.85

  • Thus, for a spillway having a vertical u/s face, the downstream crest is given by the equation:  x1.85=2.Hd0.85.yx^{1.85}=2.H_{d} ^{0.85}. y

             yHd=−0.5(xHd)1.85\frac{y}{H_{d}} = -0.5 (\frac{x}{H_{d}})^{1.85}

Question 78

A road bend deflecting by 80° is to be designed. The maximum speed is to be 2 km/min, maximum centrifugal ratio 0.25 and rate of change of radial acceleration of 300 mm/sec3. The circular arc is combined with two cubic spirals. The radius of the circular arc will be about

  1. AOption (a)

    272.42 m

  2. BOption (b)

    905 m

  3. COption (c)

    360.15 m

  4. DOption (d)

    453 m

View answer & explanation

Correct answer

Answer: (d)D.

453 m

Why this answer is correct

Explanation:

Concept: Centrifugal ratio =v2Rg= \frac{v^2}{Rg}  ⇒Radius of circular arc, R=v2centrifugal,,ratio×gR= \frac{v^2}{centrifugal,,ratio\times g}

Solution: given, Deflection angle = 80° , maximum centrifugal ratio = 0.25

                 Maximum speed of vehicle = 2 km/min = 120 km/hour 

                 Rate of change of radial acceleration = 300 mm/sec3

Centrifugal ratio =v2Rg=v2127R= \frac{v^2}{Rg} =\frac{v^2}{127R} , where v is in km/hour

∴ Radius of circular arc,  R=v2127×centrifugal,,ratio=1202127×0.25=,,453.54,,mR= \frac{v^2}{127\times centrifugal,,ratio }=\frac{120^2}{127\times 0.25}=,, 453.54 ,, m

Hence, The radius of the circular arc will be about 453 m.

Question 79

A fluid is one which can be defined as a substance that

  1. AOption (a)

    has same shear stress at all points

  2. BOption (b)

    can deform indefinitely under the action of smallest shear force

  3. COption (c)

    has the small shear stress in all directions

  4. DOption (d)

    is practically incompressible

View answer & explanation

Correct answer

Answer: (b)B.

can deform indefinitely under the action of smallest shear force

Why this answer is correct

Explanation:

Fluid:

  • Substance in liquid or gaseous phase is called fluid.
  • They are capable of deforming continuously under the action of shear stress.
  • In case of fluid, stress is proportional to strain rate.
  • Fluids flow under the action of shear (tangential) forces. That is, they do not resist the shear stresses as solids do.​

Important Points

​Types of fluid: 1. Ideal Fluid   2. Real Fluid

  • Idea fluids do not have viscosity, surface tension and are incompressible.
  • Fluids which are not ideal are called real fluids.
  • As ideal fluids do not exist practically, so all are real fluids.

Question 80

Match List - I with List - II and select the correct answer using codes given below in the Lists.

List - IList - II
A.Hardness1.Reverse Osmosis
B.Brackish water from sea2.Chlorination
C.Residual MPN from3.Zeolite Treatment
D.Turbidity4.Coagulation and flocculation
5.Flocculation, filtration and coagulation
  1. AOption (a)

    (A) - (1), (B) - (2), (C) - (4), (D) - (5)

  2. BOption (b)

    (A) - (3), (B) - (4), (C) - (2), (D) - (5)

  3. COption (c)

    (A) - (2), (B) - (1), (C) - (3), (D) - (5)

  4. DOption (d)

    (A) - (3), (B) - (1), (C) - (2), (D) - (5)

View answer & explanation

Correct answer

Answer: (d)D.

(A) - (3), (B) - (1), (C) - (2), (D) - (5)

Why this answer is correct

Explanation:

The hardness of water: 

  • The term hardness was originally applied to waters that were hard to wash in, referring to the soap-wasting properties of hard water.
  • Dissolved calcium and magnesium salts are primarily responsible for most scaling in pipes and water heaters and cause numerous problems in laundry, kitchen, and bath.

General guidelines for the classification of waters are:

  • 0 to 60 mg/L (milligrams per liter) as calcium carbonate is classified as soft
  • 61 to 120 mg/L as moderately hard
  • 121 to 180 mg/L as hard; and more than 180 mg/L as very hard.
  • Acceptable limit - 200 mg/l and Cause of rejection - 600 mg/l

Method of removal of hardness

  • Chemical Process of Boiling Hard Water.
  • Adding Slaked Lime (Clark's Process)
  • Adding Washing Soda.
  • Calgon Process.
  • Ion Exchange Process(Zeolite treatment)
  • Using Ion Exchange Resins.

Turbidity: 

  • Turbidity is the measure of the relative clarity of a liquid.
  • Acceptable limit - 1 NTU, Cause of rejection - 5 NTU

​

Type of water impurityMethod of treatment
HardnessZeolite treatment
Brackish water from the seaReserve osmosis
Residual MPN from filtersChlorination
TurbidityCoagulation, Flocculation & Filtration

Question 81

To generate 10,000 kW under a head of 81m while working at a speed of 500 rpm, the turbine of choice would be

  1. AOption (a)

    Pelton

  2. BOption (b)

    Bulb

  3. COption (c)

    Kaplan

  4. DOption (d)

    Francis

View answer & explanation

Correct answer

Answer: (d)D.

Francis

Why this answer is correct

Explanation:

Classification of turbine based on specific speed as follows:

FlowEnergyHeadSpecific speedExample
TangentialImpulseHigh head (300 m and above)Low (0 – 60 RPM)Pelton Wheel turbine
RadialReactionMedium (30 m to 300 m)Medium (60 – 300) RPMFrancis turbine
AxialReactionLow (less than 30 m)High
(300 – 600) RPMKaplan turbine
(600 – 1000) RPM

 

Specific speed of turbine: 

The specific speed of a turbine may be defined as the speed of an imaginary turbine, identical to the given turbine which will develop a unit power under a unit head, and is given by:

 Ns=NPH54N_{s}=\frac{N\sqrt{P}}{H^{\frac{5}{4}}}

Where, N = Speed of the runner in r.p.m. , H =Head of water, and P = Power produced.

Solution:

given, Power developed (P) = 10000 kW , H = 81 m, N = 500 rpm

∴ Specific speed of Turbine

 Ns=NPH54=500×1000081(54)=205.76N_{s}=\frac{N\sqrt{P}}{H^{\frac{5}{4}}} = \frac{500\times\sqrt{10000}}{81^{(\frac{5}{4})}}=205.76

 

Here specific speed of turbine is in the range of 60 to 300, hence Francis turbine should be choose.

Important Points

 Specific speed of Pump : Ns=NQH34N_{s}=\frac{N\sqrt{Q}}{H^{\frac{3}{4}}}

        Where, Ns = specific speed, N = pump shaft rotational speed (rpm),

                     Q= flow rate (m3/s), h = Head of water (m)

Question 82

Top of rails of a track are placed at an inward slope of

  1. AOption (a)

    1 in 30

  2. BOption (b)

    1 in 20

  3. COption (c)

    1 in 15

  4. DOption (d)

    1 in 10

View answer & explanation

Correct answer

Answer: (b)B.

1 in 20

Why this answer is correct

Explanation:

  • Maximum load of the wheel acts near the inner edge of the rail. This causes wear and tear of rails.
  • To reduce wear on rails, Top of rails of a track are placed at an inward slope of 1 in 20.

     

 

Additional Information

 Wear on Rail:

  • Wear is one of the prominent defects of rails.
  • When the axle loads are abnormally heavy and the train moves with very fast speed then the concentrated stresses exceed the elastic limit resulting in metal flow.
  • On the gap or joint the ends are battered and at the curves the occurrence of skidding, slipping and striking of wheel flanges with rails results in wear and tear of rails.
  • On the basis of position of wear, wear on rail classified as:
  1. ​Wear on the top or head of rail.
  2. Wear at the ends of rails.
  3. **Wear on the sides of the head.**​

Question 83

A triangular direct run off hydrograph due to a storm has a time base of 90 hour. The peak flow of 60 m3/s occurs at 20 hours from the start of the storm. The area of catchment is 300 km2. The rainfall excess of the storm (cm) is

  1. AOption (a)

    5.40

  2. BOption (b)

    6.48

  3. COption (c)

    2.00

  4. DOption (d)

    3.24

View answer & explanation

Correct answer

Answer: (d)D.

3.24

Why this answer is correct

Explanation:

 

Concept: 

Area of Direct Runoff Hydrograph = Area of catchment × Rainfall excess

Solution:

Given, Peak flow = 60 m3/sec, Area of catchment (Acatchment) = 300 km2

Area of Direct Runoff Hydrograph = Area of catchment × Rainfall excess

12×60×90×3600=300×106×h\frac{1}{2}\times 60\times 90\times 3600=300\times 10^{6}\times h ⇒ h=60×90×36002×300×106=0.0324,,m=3.24,,cmh= \frac{60\times90\times3600}{2\times300\times10^{6}}= 0.0324,,m=3.24,, cm

∴ Rainfall excess of the storm is 3.24 cm.

Question 84

Shock waves will NOT occur in the

  1. AOption (a)

    Convergent section of a nozzle

  2. BOption (b)

    Divergent section of a nozzle

  3. COption (c)

    Diffuser throat of a supersonic wind tunnel

  4. DOption (d)

    In front of blunt nosed bodies

View answer & explanation

Correct answer

Answer: (a)A.

Convergent section of a nozzle

Why this answer is correct

Explanation:

Shock wave:

  • A shock wave can be considered as a discontinuity in the properties of the flow field.
  • The fluid crossing a shock wave, normal to the flow path, will experience a sudden increase in pressure, temperature, and density, accompanied by a sudden decrease in speed, from a supersonic to a subsonic range.
  • In case of convergent divergent nozzle shock weave form in the divergent section of nozzle.
  • If the shock wave is perpendicular to the flow direction it is called normal shock wave.

Question 85

In an anaerobic stabilization basin which of the following is NOT generated?

  1. AOption (a)

    Phosphine

  2. BOption (b)

    Carbondioxide

  3. COption (c)

    Nitrate

  4. DOption (d)

    Methane

View answer & explanation

Correct answer

Answer: (a)A.

Phosphine

Why this answer is correct

Explanation:

Anaerobic Stabilization Basin: 

  • Anaerobic Stabilization basins or ponds are larger in depth (2.5 - 4m) and are suitable for industrial wastes from refineries, dairies, slaughter houses, poultry etc.
  • These are devoid of oxygen and employ anaerobic methane producing bacteria to stabilize the wastewater.
  • They are used as a pretreatment stage for BOD, COD, and SS removal, for both domestic and industrial wastewaters.
  • Anaerobic microorganisms in the absence of dissolved oxygen convert organic materials into stable products such as carbon dioxide and methane.
  • The degradation process involves two separate but interrelated phases: acid formation and methane production.
  • The following gases generated during stabilization of waste in Anaerobic stabilization basin: Methane, Carbon-di-oxide, Nitrate, Ammonia etc.

Question 86

Based on Fuller's maximum density criterion, for 4 mm maximum size of soil particles, what is the percentage of particles between 4 mm and 2 mm, by weight?

  1. AOption (a)

    80

  2. BOption (b)

    50

  3. COption (c)

    30

  4. DOption (d)

    20

View answer & explanation

Correct answer

Answer: (c)C.

30

Why this answer is correct

Explanation:

Fuller's Maximum Density Criteria:

  • This method is economical to design for a minimum amount of cement-water paste in a concrete mix.
  • Fuller assumes that the greater the amount of solid particles that can be packed in a given volume of concrete, the higher will be its strength.
  • According to Fuller's maximum density criteria, Percentage of total material smaller than size d is given as:

 P=100(dD)0.5P= 100(\frac{d}{D})^{0.5}

Where P = % passing sieve size d , D= Maximum size of particle

Solution:

Given, maximum size of particle = 4 mm 

∴ Percentage particles between 4 mm and 2 mm, by weight can be calculated by: P4mm - P2mm  =100×(44)0.5−100×(24)0.5=100−70.71=29.28≈30= 100 \times ( \frac{4}{4})^{0.5}-100 \times (\frac{2}{4})^{0.5}= 100-70.71 = 29.28 \approx 30​

Question 87

Traps are used in house-hold drainage system

  1. AOption (a)

    To trap the solid wastes

  2. BOption (b)

    To restrict the flow of water

  3. COption (c)

    To provide a partial vacuum

  4. DOption (d)

    To prevent entry of foul gases

View answer & explanation

Correct answer

Answer: (d)D.

To prevent entry of foul gases

Why this answer is correct

Explanation:

Traps:

  • A trap is a device which is used to prevent sewer gases from entering the buildings.
  • The traps are located below or within a plumbing fixture and retains small amount of water.
  • The retaining water creates a water seal which stops foul gases going back to the building from drain pipes. Therefore all plumbing fixtures such as sinks, washbasins, bathtubs and toilets etc. are equipped with traps.

   

 Additional Information

Various types of traps:

  1. Gully Trap
  2. P- Trap
  3. S - Trap
  4. Floor Trap or Nahani Trap
  5. Intercepting Trap

Question 88

The population of a town as per census records were : 2 lakhs, 2.1 lakhs and 2.3 lakhs for the years 1981, 1991 and 2001 respectively. The projected population of the town in the year 2011, using arithmetic increase method, will be

  1. AOption (a)

    2.5 lakhs

  2. BOption (b)

    2.55 lakhs

  3. COption (c)

    2.40 lakhs

  4. DOption (d)

    2.45 lakhs

View answer & explanation

Correct answer

Answer: (d)D.

2.45 lakhs

Why this answer is correct

Explanation:

Concept:

As per arithmetic increase method of population forecasting, population of town after n decades is:  Pn=P0+nx‾P_{n}= P_{0}+n \overline x

where, Pn = Population of town after n decades

            P0 = Last known Population of town  

            n =  number of decades between P0 and Pn

            x‾\overline x = Rate of population growth 

Solution:

YearPopulation (P)Increase in population (x)
19812 lakhs-
19912.1 lakhs0.1 lakhs
20012.3 lakhs0.2 lakhs

Rate of population growth,  x‾=x1+x22=0.1+.022=0.15\overline x = \frac{x_{1}+x_{2}}{2} = \frac{0.1+.02}{2}= 0.15 , 

Here n = 1 (i.e. 10 years)

P2011=P2001+1×x‾P_{2011}=P_{2001}+ 1 \times \overline x  ⇒ P2011=2.3+1×0.15=2.45 LakhsP_{2011}=2.3+ 1 \times0.15 = 2.45 \ Lakhs

Question 89

As per IRC: 37-2012 in order to control sub grade rutting in flexible pavements the parameter to be considered is

  1. AOption (a)

    Horz. tensile strain at the bottom of bituminous layer

  2. BOption (b)

    Vertical compressive strain on top of subgrade

  3. COption (c)

    Vertical compressive stress on top of granular layer

  4. DOption (d)

    Vertical deflection at the surface of the pavement

View answer & explanation

Correct answer

Answer: (b)B.

Vertical compressive strain on top of subgrade

Why this answer is correct

Explanation:

As per ,IRC: 37-2012, "TENTATIVE GUIDELINES FOR THE DESIGN OF FLEXIBLE PAVEMENT".

  • Rutting is the permanent deformation in pavement usually occurring longitudinally along the wheel path. The rutting may partly be caused by deformation in the subgrade and other non-bituminous layers which would reflect to the overlying layers to take a deformed shape.
  • Vertical subgrade strain, ϵv\epsilon_{v} , is an index of bearing capacity of the subgrade and it needs to be kept below a certain value for a given traffic to limit rutting in the subgrade.
  • The limiting value of rutting is recommended as 20 mm in 20 per cent of the length for design traffic up to 30 msa and 10 per cent of the length for the design traffic beyond.

Important Points

IRC 37: 2012, recommendation to control fatigue cracking;

  • Tensile strain, ϵt\epsilon_{t} , at the bottom of the bituminous layer is considered as critical parameter for pavement design to limit Fatigue cracking in bituminous layer.
  • The limiting value of Fatigue cracking is recommended as cracking in 20 per cent area has been considered for traffic up to 30 msa and 10 per cent for traffic beyond that.

Question 90

The axle load including weight of wheels and axle, provided that rail has not lost more than 5% of its original section, for a 44.5 kg/m rail section for broad gauge should be

  1. AOption (a)

    17.5 tonnes

  2. BOption (b)

    23.0 tonnes

  3. COption (c)

    44.5 tonnes

  4. DOption (d)

    89.0 tonnes

View answer & explanation

Correct answer

Answer: (b)B.

23.0 tonnes

Why this answer is correct

Explanation:

Concept: The ratio of weight of the rail and the locomotive axle load:

      Weight,,of,,the,,rail,,in,,tonnesLocomotive,,axle,,load,,in,,tonnes=1510\frac{Weight ,,of ,,the ,,rail ,,in,, tonnes}{Locomotive ,,axle,, load,, in,, tonnes } = \frac{1}{510}

Solution: Given, Weight of rail = 44.5 kg/m = 44.5 × 10-3 tonnes/m

        Weight,,of,,the,,rail,,in,,tonnesLocomotive,,axle,,load,,in,,tonnes=1510\frac{Weight ,,of ,,the ,,rail ,,in,, tonnes}{Locomotive ,,axle,, load,, in,, tonnes } = \frac{1}{510}

Locomotive,,axle,,load,,in,,tonnes=Weight,,of,,the,,rail,,in,,tonnes×510Locomotive ,,axle,, load,, in,, tonnes =Weight ,,of ,,the ,,rail ,,in ,, tonnes × 510

∴ Locomotive axle load = 44.5× 10-3 × 510 = 22.695 ≈ 23.0 tonnes.

Question 91

The reading of differential manometer of a venturimeter, placed at 45° to the horizontal is 11 cm. If the venturimeter turned in horizontal position, the manometer reading will be

  1. AOption (a)

    Zero

  2. BOption (b)

    112\frac{11}{\sqrt2} cm

  3. COption (c)

    11 cm

  4. DOption (d)

    112\sqrt2 cm

View answer & explanation

Correct answer

Answer: (c)C.

11 cm

Why this answer is correct

Explanation: 

Venturimeter: Venturimeter is a type of flowmeter that works on the principle of Bernoulli's Equation.

In case of inclined venturimeter with differential manometer, the piezometric head difference (h) can be obtain by :

  h=(G2G1−1)xh= (\frac{G_{2}}{G_{1}}-1)x , if G2 > G1  and  h=(1−G2G1)xh= (1-\frac{G_{2}}{G_{1}})x , if G2 < G1 

Where G2 = Specific gravity of manometric fluid; 

           G1 = Specific gravity of flowing fluid; 

            x = Reading on manometer.

Manometer reading does not depends upon inclination of venturimeter. Hence, if the reading of differential manometer of a venturimeter, placed at 45° to the horizontal is 11 cm. If the venturimeter turned in horizontal position, the manometer reading will be 11 cm.

Important Points

  • Venturimeter is used to find discharge from a large diameter pipe.
  • Reduction in area leads to increase in velocity and decrease in pressure, this decrease in pressure is noted and used in Bernoulli's equation to calculate discharge.
  • Actual discharge through venturimeter is calculated by:

  Qactual=cda1a22gha12−a22Q_{actual}= c_{d}\frac{a_{1}a_{2}\sqrt{2gh}}{\sqrt {a_{1}^{2}-a_{2}^{2}}}  

Where a1 and a2 = cross section areas at section 1 and section 2 , cd = Discharge coefficeint (≈ 0.98) , h = Piezometric head difference​

Question 92

First watering of cropped area is also known as

  1. AOption (a)

    Duty

  2. BOption (b)

    Delta

  3. COption (c)

    Kor depth

  4. DOption (d)

    Kor watering 

View answer & explanation

Correct answer

Answer: (d)D.

Kor watering 

Why this answer is correct

Explanation:

Kor Watering:

  • The first watering which is given to a crop when the crop is few cm high is called kor watering.
  • It is usually the maximum single watering followed by the watering at usual interval as required by the crop.
  • The optimum depth of core watering for :Wheat = 13.5 cm, Rice =19 cm.

Duty: 

  • It is defined as area in hectares irrigated for given crop for its full growth by supply of 1m3/sec of water continuously during the entire base period of the crop.

​Delta: 

  • The depth of water required by the crop for its full growth is termed as Delta.
  • Average approximate value of delta of different crop:

​​           Sugar can : 120 cm

           Rice : 120 cm

           Fruits : 60 cm

           Vegetables: 45 cm 

           Wheat :  40 cm 

           Barley: 30 cm 

           Peas: 15 cm

Question 93

The Indian Road Congress (IRC) was founded in the year

  1. AOption (a)

    1929

  2. BOption (b)

    1934

  3. COption (c)

    1939

  4. DOption (d)

    1943

View answer & explanation

Correct answer

Answer: (b)B.

1934

Why this answer is correct

Explanation:

Indian Road Congress (IRC): 

The Indian Roads Congress (IRC) is the premier technical body of Highway Engineers in the country.

It was founded and constituted with Head Quarters at New Delhi by Jayakaar Committee in 1934

It publishes various codes standardizing the specifications.

The IRC provides a national forum for sharing of knowledge and pooling of experience on the entire range of subjects dealing with the construction and maintenance of roads and bridges, including technology, equipment, research, planning, finance, taxation, organization and all connected policy issues.

Important Points

1927 → Jayakar committee (Indian road development committee)

1929 → Central Road Fund (CRF)

1943 → Nagpur Road plan

1952 → Central Road Research Institute (CRRI)

Question 94

The difference between the observed value and its most probable value of a quantity is called

  1. AOption (a)

    Conditional error

  2. BOption (b)

    True error

  3. COption (c)

    Total error

  4. DOption (d)

    Residual error

View answer & explanation

Correct answer

Answer: (d)D.

Residual error

Why this answer is correct

Concept:

True Error = Actual Value-Observed Value 

Weighted Observations=Observations taken with proportion to any quantity like an area

Conditional Error= Error occurring due to certain conditions prevailing. Like temperature error.

Residual Error = Most Probable Value- Observed Value

Additional Information 

Permissible error:

• It is defined as a maximum allowable limit up to that a measurement may vary from the true value.

Residual error:

These are the values that cannot be avoided during measurement it can be counter balance by suitable correction measures.

• The difference between the most probable value of a quantity and its observed value is residual error.

Important Points:

Systematic errors are caused by the surveying equipment, observation methods, and certain environmental factors.

Random (or accidental) errors are not directly related to the conditions or circumstances of the observation.

For a single measurement or a series of measurements, it is the error remaining after all possible systematic errors and blunders have been eliminated.

Natural errors are caused by environmental conditions or significant changes in environmental conditions.

Instrument errors are caused by imperfectly constructed, adjusted, or calibrated surveying equipment.

Human errors are caused by physical limitations and inconsistent setup and observation habits of the surveyor.

Question 95

The use of unit hydrographs for estimating floods is generally limited to catchments of size less than

  1. AOption (a)

    5000 km2

  2. BOption (b)

    500 km2

  3. COption (c)

    106 km2

  4. DOption (d)

    5000 ha

View answer & explanation

Correct answer

Answer: (a)A.

5000 km2

Why this answer is correct

Explanation:

Limitation of unit hydrographs:

  • The unit hydrograph for estimating floods is generally limited to the catchment of size less than 5000 km2.
  • Unit hydrograph assumes uniform distribution of rainfall over the catchment.
  • Also the intensity is assumed constant for the duration of rainfall excess.
  • Very large sizes of catchment can give different direct runoff hydrograph (DRH) under otherwise identical situations.

Question 96

A culvert is designed for a peak flow θp on the basis of rational formula. If a storm of the same intensity as used in the design and twice the duration occurs, then the resulting peak discharge will be

  1. AOption (a)

    θp

  2. BOption (b)

    θp2\frac{θ_p}{2}

  3. COption (c)

    2\sqrt2 θp

  4. DOption (d)

    2 θp

View answer & explanation

Correct answer

Answer: (a)A.

θp

Why this answer is correct

Explanation:

Rational method:

According to the Rational method, the stormwater or runoff is defined by

Q=K×A×I×RQ=K\times A\times I\times R

where Q is expressed in liter per second,

'K' i a constant,

'A' is the catchment area in hectares,' I' is the impermeability factor

and 'R' is the intensity of rainfall in mm per hour.'

 

There is no term for duration of rainfall in the rational formula. So peak discharge​ will not change.

Question 97

A parabolic vertical curve is setout connecting a + 0.7% grade to a - 0.6% grade. The chainage and R. L. of the point of intersection are 1000 m and 250 m respectively. The admissible rate of change of grade is 0.05% per 20 m. The chainage of the tangent points will be

  1. AOption (a)

    600m and 1400m

  2. BOption (b)

    650m and 1350m

  3. COption (c)

    700m and 1300m

  4. DOption (d)

    740m and 1260m

View answer & explanation

Correct answer

Answer: (d)D.

740m and 1260m

Why this answer is correct

Concept:

Vertical Curves: Vertical curves are used in highway and street vertical alignment to provide a gradual change between two adjacent grade lines.

Length of a vertical curve= (Total grade/Rate of change of grade per chain length) * Length of chain

Vertical Curves are of two types:

  1. Summit Curve and

  2. Valley Curve

Solution :

Length of the vertical curve,

= 0.7−(−0.6)0.0520\frac{0.7 -\left ( -0.6 \right )}{\frac{0.05}{20}} = 520 m

Length of the curve on either side of the apex,

= 520/2 = 260m

Chainage of first tangent point = 1000 - 260 = 740 m

Chainage of second tangent point = 1000 + 260 = 1260 m

Question 98

A highway designed for 80 km/hr. speed has a horizontal curve section with radius 250m. If the design lateral friction is assumed to develop fully, the required superelevation is

  1. AOption (a)

    0.09 m

  2. BOption (b)

    0.07 m

  3. COption (c)

    0.05 m

  4. DOption (d)

    0.02 m

View answer & explanation

Correct answer

Answer: (c)C.

0.05 m

Why this answer is correct

Concept:

Superelevation(e): Superelevation inroads is basically provided on the horizontally curved portion of the roads in which the outer edge of the road pavement is raised with respect to the inner edge, thus providing a transverse slope throughout the length of the horizontal curve of the road.

   e+f=v2127Re+f=\frac{v^{2}}{127R}

where,

f = lateral friction = 0.15, V = Designed speed,

R = Radius of curve

Calculation:  Given,

Design Speed (V) = 80 kmph

Radius of curve (R) = 250 m,

Lateral friction is assumed to develop fully i.e. f = 0.15

   e+f=v2127R;e+0.15=802127×250=0.052e+f=\frac{v^{2}}{127R}; e+0.15 = \frac{80^{2}}{127\times 250}=0.052

 the required superelevation is 0.052

Question 99

Distance between channelizing Islands of a roadway is known as

  1. AOption (a)

    Sight distance

  2. BOption (b)

    Overtaking distance

  3. COption (c)

    Weaving distance

  4. DOption (d)

    Safe stopping distance

View answer & explanation

Correct answer

Answer: (c)C.

Weaving distance

Why this answer is correct

Explanation :

Weaving Length :

  • ​Distance between channelizing Islands of a roadway.
  • Weaving width at the rotary is in between 6 and 18 metres. The ratio of average width of the carriage way at entry and exit to the weaving width is in the range of 0.4 to 1.
  • The maximum traffic which a rotary can efficiently handle is 3000 vehicle/hour and a minimum of 500 vehicle/hour.
  • Design speed is 40 km/hour for rural areas and 30 km/hr for urban areas.

Question 100

A horizontal angle θ is measured by four observers multiple times and the reported values are as follows:

SurveyorAngle θNo. of obs.
136°30'4
236°00'3
335°30'8
436°30'4

The most probable value of the angle θ is approximately:

  1. AOption (a)

    35°

  2. BOption (b)

    36°

  3. COption (c)

    37°

  4. DOption (d)

    35°30'

View answer & explanation

Correct answer

Answer: (b)B.

36°

Why this answer is correct

Explanation

Most Probable Value;

  • It is the value of the quantity which has more chances of being true than any other value.
  • Most probable value is thus very close to (but NOT equal to) the true value of an observed quantity.
  • The most probable value of an angle is equal to its weighted arithmetic mean.

​Solution :

                  M.P.V.=36∘30′×4+36∘00′×3+35∘30′×8+36∘30′×44+3+8+4=36∘00M.P.V.=\frac{36^{\circ}30'\times 4+36^{\circ}00'\times 3+35^{\circ}30'\times 8+36^{\circ}30'\times 4}{4+3+8+4}=36^{\circ}00

Question 101

As per IRC recommendation, the maximum limit of super elevation for mixed traffic in plain terrain is

  1. AOption (a)

    1 in 15

  2. BOption (b)

    1 in 12.5

  3. COption (c)

    1 in 10

  4. DOption (d)

    equal to camber

View answer & explanation

Correct answer

Answer: (a)A.

1 in 15

Why this answer is correct

Explanation:

Super elevation- When a roadway (or railway) is laid on curved path, then its outer edge is always made higher than inner edge, to keep the vehicle in equilibrium while in motion. 

The amount by which outer edge is raised, is known as cant or super elevation. In case of roadways, the process of providing super elevation is known as banking of the road. 

As per IRC , the maximum limit of super elevation for mixed traffic in plain terrain is 1 in 15. 

Confusion PointsThe general practice to define super elevation in roadways, is to mention the angle of inclination of the road surface.

In case of railways, the general practice to define super elevation is to mention the difference of levels between the two rails.

Question 102

For a turbulent flow the Reynold's number should be more than

  1. AOption (a)

    4000

  2. BOption (b)

    2700

  3. COption (c)

    2000

  4. DOption (d)

    1200

View answer & explanation

Correct answer

Answer: (a)A.

4000

Why this answer is correct

Explanation:

  • Streamline flow: In fluids the flow in which the fluids flow in parallel layers such that there is no intermixing of the layers and also, the velocity of each fluid particle at a given point passing by remains constant with time is known as Streamline Flow.
  • Reynold's number: The ratio of fluid's inertial force and its viscous force is called Reynolds number.

Mathematically Reynolds number is given by:

R=ρvDμR = \frac{\rho v D}{\mu}

where R is Reynolds number, ρ is the fluid density,v is the velocity, D is the diameter of the pipe, μ is the viscosity of the fluid.

  • For Reynold's number greater than 4000, the flow is turbulent flow. Therefore option is correct.

Additional Information 

  • For laminar flow, the value of Reynold's number should be less than 2000.
  • For Reynold's number between 2000 to 4000, the flow is transient flow.

Question 103

An aerial photograph of a terrain having an average elevation of 1400m is taken at a scale of 1 : 7500. The focal length of camera is 15 cm. The flight altitude above MSL is

  1. AOption (a)

    1225 m

  2. BOption (b)

    2525 m

  3. COption (c)

    3025 m

  4. DOption (d)

    3535 m

View answer & explanation

Correct answer

Answer: (b)B.

2525 m

Why this answer is correct

Detailed Solution:

Average elevation (h) =1400 m,

focal length (f) = 15 cm,

scale (S) = 1:7500

Let the altitude of the flight above mean sea level be 'H'

Now

S=fH−h;17500=0.15H−1400S=\frac{f}{H-h} ;\frac{1}{7500}=\frac{0.15}{H-1400}

 H=2525 m

.. The altitude of the flight above mean sea level is 2525 m

Question 104

Hydraulic jump occurs when there is a break in grade from a

  1. AOption (a)

    Steep to steeper slope

  2. BOption (b)

    Mild to milder slope

  3. COption (c)

    Steep to mild slope

  4. DOption (d)

    Mild to steep slope

View answer & explanation

Correct answer

Answer: (c)C.

Steep to mild slope

Why this answer is correct

Explanation:

Near the break in grade, there is a transition from a supercritical flow to a sub-critical flow. This transition is called a hydraulic jump.

  • When the flow conditions changes form super critical (F, > 1) to subcritical (F, < 1), this results to an abrupt rise of water accompanied by turbulent rollers is called as hydraulic Jump or standing wave.

                       

The hydraulic Jump occurs when the grade changes from steep to mild.

Question 105

The classification categories of roads network was finalized in the Nagpur Plan of Indian Roads Congress as such

  1. AOption (a)

    NH, SH, MDR and VR

  2. BOption (b)

    NH, SH, MDR only

  3. COption (c)

    NH, MDR, ODR and VR

  4. DOption (d)

    NH, SH, MDR, ODR and VR

View answer & explanation

Correct answer

Answer: (d)D.

NH, SH, MDR, ODR and VR

Why this answer is correct

Concept:

The first 20-year road plan is called the Nagpur Road plan (1943 - 1963)

Feature:

  • This plan was a major attempt in planning for road development in a scientific manner.
  • The road length of 5,32,700 km with a density of 16 km of road length per 100 km area would be available by 1963.

All roads were classified into 5 categories

i) National Highway

ii) State Highway

iii) Major District Roads

iv) Other district roads

v) Village roads

They recommended the construction of star and grid pattern of road throughout the country.

 Additional Information

Types of Road Plan India :

(1)   Nagpur road plan (1943 – 63) or first 20 years road plans.

(2)   Bombay road plans (1961 – 1981) or second 20 years road plans.

(3)   Lucknow Road plans (1981- 2001) or third 20 years road plans.

Nagpur Road plan (1943 – 63) or first 20-year Road plan →

→ Duration = 1943 – 63

→ Road density = 16 km/100 km²

→ Road pattern = star and grid pattern

→ Development allowance = 15%

Bombay road plan (1961- 81) or second 20-year road plan →

→ Duration = 1961 -81

→ Road density = 32 km/100 km²

→ Development allowance = 5%

→ Length of express ways = 1600 km

Lucknow road plan (1981 – 2001) or third 20-year road plan→

→ Duration = 1981- 2001

→ Road density = 82 km/100 km²

→ Road pattern = Square grid

→ Length of express ways = 2000 km.

 Road classification

(i)     Primary system or 1° system = Expressway, National highway

(ii)   Secondary system or 2° system = State highway, major district road

(iii)  Tertiary system or 3° system = Other district roads, village road.

Question 106

In an external focussing tacheometer the fixed hairs interval is 5 mm, the focal length of the objective lens is 25 cm and the distance of the vertical axis of the instrument from the optical centre of the objective lens is 15 cm. The constants of the tacheometer will be

  1. AOption (a)

    30, 0.15

  2. BOption (b)

    30, 0.40

  3. COption (c)

    50, 0.25

  4. DOption (d)

    50, 0.40

View answer & explanation

Correct answer

Answer: (d)D.

50, 0.40

Why this answer is correct

Solution:

For a tacheometer multiplying constant,

K = fi\frac{f}{i}      K=25/0.5=50

Additive constant,

C=f + d = 0.25 + 0.15 = 0.40

So  The constants of the tacheometer will be 50 and 0.40.

Question 107

Creep is the

  1. AOption (a)

    Lateral movement of rail 

  2. BOption (b)

    Longitudinal movement of rail

  3. COption (c)

    Vertical movement of rail 

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (b)B.

Longitudinal movement of rail

Why this answer is correct

Explanation

Creep of rail:

Creep is defined as the longitudinal movement of the rail with respect to the sleepers. Rails tend to gradually move in the direction of dominant traffic. 

Theories for the development of creep

  1. Wave motion theory:

According to this theory the rails have a wavy formation, as the wheels move forward, the depression also moves with them and the previously depressed portion springs back to the original one.

  1. Percussion theory:

According to this theory, creep is developed due to the impact of wheels at the rail end ahead of the joint. The continuous movement of several wheels passing over the joint pushes the facing or landing rail forward, thereby causing creep

Major causes for creep in rail are as follows:

a) Forces developed when the train is starting or stopping by application of brakes.

While starting, the wheels push the rail backwards and hence the direction of creep is in the backward direction.

When brakes are applied then the wheels of the vehicles push the rails in the forward direction and hence the creep is in the forward direction.

b) Due to wave motions developed when the wheels of the vehicle hit the crest.

c) Unequal expansion/contraction due to the different temperature gradients along the track.

Question 108

Free surface width of open channel section is known as

  1. AOption (a)

    Wetted perimeter

  2. BOption (b)

    Hydraulic depth

  3. COption (c)

    Hydraulic radius

  4. DOption (d)

    Top width of section

View answer & explanation

Correct answer

Answer: (d)D.

Top width of section

Why this answer is correct

Free surface width of open channel section is known as Top width of section.

 

 

Depth, y

Area, A 

Wetted perimeter, P

Top width, T

Hydraulic Radius, R = Area / Wetted perimeter

Hydraulic Depth, D = Area / Top width

Question 109

Two straight lines deflect through an angle of 60°, the radius of a curve joining the two straight lines is 600 m. The length of the long chord and mid ordinate (in m) of the curve respectively are

  1. AOption (a)

    80.4, 600.0

  2. BOption (b)

    600.0, 80.4

  3. COption (c)

    600.0, 39.89

  4. DOption (d)

    49.89, 300.00

View answer & explanation

Correct answer

Answer: (b)B.

600.0, 80.4

Why this answer is correct

Concept:

For the given curve:

 

Tangent length (T)=Rtan⁡;Δ;2\left( {\rm{T}} \right) = {\rm{R}}\tan \frac{{{\rm{;}}{\rm{Δ }}{\rm{;}}}}{2}

Length of curve (l)=;π;R;Δ;180\left( {\rm{l}} \right) = \frac{{{\rm{;}}\pi {\rm{;R;}}{\rm{Δ }}{\rm{;}}}}{{180}} 

Long chord (L)=2Rsin⁡;Δ;2\left( {\rm{L}} \right) = 2{\rm{R}}\sin \frac{{{\rm{;}}{\rm{Δ }}{\rm{;}}}}{2} 

External distance (E)=R(sec⁡;Δ;2;;;−1)\left( {\rm{E}} \right) = {\rm{R}}\left( {\sec \frac{{{\rm{;}}{\rm{Δ }}{\rm{;}}}}{2}{\rm{;}}{\rm{;}}{\rm{;}} - 1} \right)  

Mid-ordinate (M)=R(1−cos⁡;Δ;2)\left( {\rm{M}} \right) = {\rm{R}}\left( {1 - \cos \frac{{{\rm{;}}{\rm{Δ }}{\rm{;}}}}{2}{\rm}} \right) 

Calculation:

Given:

∴ Deflection angle, Δ = 60° 

Length of long chord = 2 × 600 × sin (60°/2) = 600 m.

Length of the mid ordinate (m) = 600 (1 – cos(60°/2)) = 80.4 m

Question 110

If the depth of an irrigation canal is below the level of a natural stream, the type of cross drainage work is

  1. AOption (a)

    Aqueduct

  2. BOption (b)

    Superpassage

  3. COption (c)

    Sluice gate

  4. DOption (d)

    Level crossing

View answer & explanation

Correct answer

Answer: (b)B.

Superpassage

Why this answer is correct

Explanation:

Aqueduct:

When the HFL (Highest Flood Level) of the drain is sufficiently below the bottom of the canal such that the drainage water flows freely under gravity, the structure is known as Aqueduct.

Sluice gate: 

It is used to control the flow in a horizontal channel of unit width.

Super passage:

A super passage is just like a bridge in which the natural drain is carried over the canal. A super passage is constructed where the bed of the drain is well above the canal F.S.L (Full Supply Level).

Level Crossing:

The level crossing is an arrangement provided to regulate the flow of water through the drainage and the canal when they cross each other approximately at the same bed level.

Question 111

Traffic volume is measured for traffic study in terms of

  1. AOption (a)

    Vehicle per lane

  2. BOption (b)

    Vehicles per day

  3. COption (c)

    Vehicles per kilometre

  4. DOption (d)

    Vehicles per kilometre / hour

View answer & explanation

Correct answer

Answer: (b)B.

Vehicles per day

Why this answer is correct

Explanation

Traffic Volume is defined as the procedure to determine the volume of traffic or no of vehicles moving on the roads at a particular section during a particular time period. This time period can be in ‘minutes’, ‘hours’ or  ‘days’ etc.

Traffic volume study is very important because:

  1. It is generally accepted as a true measure of the relative importance of roads and in deciding the priority for improvement and expansion.
  2. It is used in planning, traffic operation and control of existing facilities and also for planning the new facilities.
  3. It is used in the analysis of traffic patterns and trends.
  4. Useful in structural design of pavement.
  5. Pedestrian traffic volume study is used for planning side walk, cross walks, subway and pedestrian signals.

Question 112

The difference between the sum of the angles of a plane triangle and that of a spherical triangle is 1 second, when the triangle of the earth surface has an area of about

  1. AOption (a)

    105 sq.km

  2. BOption (b)

    139 sq.km

  3. COption (c)

    159 sq.km

  4. DOption (d)

    195 sq.km

View answer & explanation

Correct answer

Answer: (d)D.

195 sq.km

Why this answer is correct

Practical distance is measured considering the plane triangle while actual distance is measured using a spheroidal triangle.

When we deal with only a small portion of earth's surface, for the length of an arc of 18.5 km the error introduced is 0.0152 m greater than the subtended chord and the difference between the sum of the angles of spherical triangle and that of plane triangle is only one second at the earth's mean surface for an area of 195 square km.

∴ The difference between the sum of interior angles of a plane triangle and spherical triangle for the area of triangle 195 square kilometre on the Earth’s surface is one second.

Question 113

Isolated storm is represented on a hydrograph as

  1. AOption (a)

    Single peak

  2. BOption (b)

    Complex peak

  3. COption (c)

    Multiple peak

  4. DOption (d)

    Without any peak

View answer & explanation

Correct answer

Answer: (a)A.

Single peak

Why this answer is correct

Hydrograph:

  • A hydrograph is the graphical representation of the discharge flowing in a river at the given location, with the passage of time.
  • A hydrograph is capable of representing discharge fluctuation in the river at a given site over a given time period.
  • Hydrograph gives us the peak flow, which is important for the design of hydraulic structure at that site.
  • The hydrograph result, due to an isolated storm generally shows the singly peaked skew distribution of discharge and is known as storm hydrograph, flood hydrograph, or simply hydrograph.
  • The hydrograph is the response of a given catchment to the rainfall input on the same catchment. Hydrograph consists of surface runoff interflow and baseflow.

Question 114

When any water enters in the earth strata, it is called

  1. AOption (a)

    Run off

  2. BOption (b)

    Precipitation

  3. COption (c)

    Water table

  4. DOption (d)

    Infiltration

View answer & explanation

Correct answer

Answer: (d)D.

Infiltration

Why this answer is correct

Explanation:

Infiltration: 

  • Infiltration is defined also as the movement of water into the soil matrix, starting from the soil surface.
  • Water that gets deposited on the earth's surface due to rainfall, tends to be held there by surface forces.
  • However, if the surface is permeable, water may move into the soil under the forces of gravity and capillarity, and the process is called infiltration.
  • Infiltration rate is the rate of flow at which water penetrates the ground surface at any time under the combined forces of gravity, viscosity, and capillarity.

​Runoff: 

  • Runoff is the quantity of water discharged in surface streams.
  • Types of runoff: 1. Surface runoff  2. Subsurface Runoff  (a) Prompt interflow (b) Delayed flow
  • The flow in where it travels all the time over the surface as overland flow and through the channels as open channel flow and reaches the catchment outlet is called surface runoff.

Precipitation:

  • It represents all form of water that reaches the earth surface from the atmosphere.
  • For Precipitation to form, the conditions required are:
  • Presence of moisture in the atmosphere.
  • Weather condition must be optimum for condensation to takes place.
  • The product of condensation must reach the earth surface.

​Water table:

  • When rain falls, a part of the rainwater seeps through the soil and gets collected on the non-porous rocks.
  • This collected water is known as underground water and the level of groundwater is known as the water table.​

Question 115

In an adjusted level, when the bubble is centred, the axis of the bubble tube becomes parallel to

  1. AOption (a)

    Horizontal axis

  2. BOption (b)

    Line of collimation

  3. COption (c)

    Axis of telescope

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (b)B.

Line of collimation

Why this answer is correct

Concept:

Line of sight:

  • The sighting or pointing line of a telescope, defined by the optical center of the objective and the intersection of crosshairs.

Line of collimation:

  • It is an imaginary line passing through optical centre of the objective glass and its continuation.

Axis of Telescope:

  • The axis is an imaginary line passing through the optical centre of the object glass and the optical centre of the eye-peace.

Axis of the Bubble Tube:

  • It is an imaginary line tangential to the longitudinal curve of the bubble tube at its middle point.

When the bubble of the level tube of a level remains central, the line of collimation or line of sight is horizontal, and the axis of the bubble tube becomes parallel to the Line of sight.

Question 116

A canal was designed for a full supply discharge of 60 cumecs. The average supply discharge in the canal is 40 cumecs. The capacity factor of the canal is

  1. AOption (a)

    6.0

  2. BOption (b)

    0.67

  3. COption (c)

    0.45

  4. DOption (d)

    4.00

View answer & explanation

Correct answer

Answer: (b)B.

0.67

Why this answer is correct

Concept:

Capacity Factor:

a) The capacity factor of a canal is the duty based on the discharge at the canal headworks.

b) The capacity factor is defined as the ratio of mean supply discharge in a canal during a period to its design full capacity.

Capacity Factor =( Average supply discharge/Full supply discharge)

Calculation: Given Data:

Full supply discharge = 60 cumec

Average supply discharge = 40 cumec

 Capacity Factor of the canal = 40/60

Capacity Factor of the canal = 0.666

Capacity Factor of the canal = 0.67

Additional Information

Time Factor: It is the ratio of the number of days the canal actually runs during a watering period to the total number of days of the watering period.

Outlet Discharge Factor: The outlet discharge factor is the duty of water of a crop at the outlet of the canal.

Question 117

A pumped storage plant is

  1. AOption (a)

    High head plant

  2. BOption (b)

    Run off river plant

  3. COption (c)

    Peak load plant

  4. DOption (d)

    Base load plant

View answer & explanation

Correct answer

Answer: (c)C.

Peak load plant

Why this answer is correct

Explanation:

Pumped-storage plants are peaking plants, designed to be used primarily (but not exclusively) during the peak periods – the hottest parts of the summer and the coldest parts of the winter.

A pumped-storage plant works much like a conventional hydroelectric station, except the same water can be used repeatedly. Water power uses no fuel in the generation of electricity, making for meager operating costs.

 

Upper reservoir

  • When power from the plant is needed, water stored in an upper reservoir is released into an underground tunnel.

Intake tunnel

  • The water rushes down the intake tunnel.

Turbines

  • The force of the water drives huge turbines, which are underground at the base of a dam. The spinning turbines are connected to large generators, which produce electricity.

Discharge tunnel

  • The water then flows through a discharge tunnel into a lower reservoir.

Recharging

  • When the demand for electricity is low, the turbines spin backward and pump the water back up into the upper reservoir to make it available to generate electricity when it's needed.

Question 118

Which of the following types of resolution corresponds to the smallest difference of radiant energy detected by a sensor?

  1. AOption (a)

    Spectral resolution

  2. BOption (b)

    Radiometric resolution

  3. COption (c)

    Temporal resolution

  4. DOption (d)

    Spatial resolution

View answer & explanation

Correct answer

Answer: (b)B.

Radiometric resolution

Why this answer is correct

1. Radiometric Resolution:-

  • The radiometric resolution of an imaging system describes its ability to discriminate very slight differences in energy.
  • The finer the radiometric resolution of a sensor, the more sensitive it is to detect small differences in reflected or emitted energy.
  • It refers to the sensitivity of sensor to incoming radiance.

2 Spatial Resolution:-

  • Measurement of the minimum distance between two objects that will allow them to differentiate from one another in an image.
  • Spatial resolution of the sensor refers to the size of the smallest possible feature that can be detected.
  • Spatial resolution is a measure of smallest object that can be resolved by the sensor, or the size of area on the ground represented by each pixel. Pixel size is used to determine resolution quality.
  1. Spectral Resolution:-
  • Spectral resolution describes the ability of a sensor to define fine wavelength intervals. The finer the spectral resolution, the narrower the wavelength range for a particular channel or band.
  • Spectral resolution refers to the specific wavelength intervals in electromagnetic spectrum that a sensor can record.
  • Narrow bands have higher spectral resolution.

4. Temporal Resolution:-

  • Each satellite has its own unique revisit schedule for obtaining image of a particular area.
  • The frequency at which the sensor revisits an area is known as temporal resolution.
  • For example, if a satellite images the same area every 10 days, then its temporal resolution would be 10 days.
  • Temporal resolution is an important factor to consider in change detection studies i.e. crop growth etc.
  • Hence, temporal resolution is the amount of time taken by sensor to return to a previously recorded location.

           Examples – Landsat, GOES and NOAA

Question 119

Generally the rail section is designed by assuming that it can bear a load equal to 'n' times of its own weight per meter length, the value of 'n' is

  1. AOption (a)

    100

  2. BOption (b)

    350

  3. COption (c)

    460

  4. DOption (d)

    560

View answer & explanation

Correct answer

Answer: (d)D.

560

Why this answer is correct

Explanation:

A rail is designated by its weight per unit length. The various important factors to be considered in deciding the weight of the rail to be used are as follows:

  1. Speed of the train;
  2. The gauge of the track;
  3. The axle load and nature of traffic;
  4. Type of rails; i.e. whether D.H or B.H or F.F;
  5. Spacing of sleepers (Sleeper density);
  6. Maximum permissible wear on top of rails (5% of the weight of rail is allowed);

It is evident that the heavier the rail, the bigger the section, and the higher would be its load-carrying capacity.

A general rule is adopted to specify a certain constant value of the ratio between the weight of the rail and the locomotive axle load. In India, this ratio is 510.

Weight,,of,,the,,rail,,in,,tonnesLocomotive,,axle,,load,,in,,tonnes=1510\frac{Weight ,, of ,,the, ,rail,, in ,,tonnes}{Locomotive , ,axle , ,load , ,in , ,tonnes}=\frac{1}{510}\textbf{}

∴ The nearest answer is 560.

Question 120

The coefficient of velocity (Cv) for an orifice is

  1. AOption (a)

    4x2yH\sqrt{\frac{4x^2}{yH}}

  2. BOption (b)

    2x4yH\frac{2x}{\sqrt{4yH}}

  3. COption (c)

    x24yH\sqrt{\frac{x^2}{4yH}}

  4. DOption (d)

    2x4yH\sqrt{\frac{2x}{4yH}}

View answer & explanation

Correct answer

Answer: (c)C.

x24yH\sqrt{\frac{x^2}{4yH}}

Why this answer is correct

Explanation:

Coefficient of velocity:

It is defined as the ratio of the actual velocity of the jet at vena contracta V to the theoretical velocity Vth.

Experimentally it is determine by using the following relation

Cv=x24yHC_{v}=\sqrt{\frac{x^{2}}{4yH}}

where x= Horizontal distance

y= vertical distance

h= constant head water

Generally the value of Cv varies from 0.95 to 0.99 for different orifices depending upon their size, shape, head etc.

for sharp edged orifices the value of Cv is 0.98.

Question 121

Over irrigation results in

  1. AOption (a)

    Water logging

  2. BOption (b)

    Rapid growth

  3. COption (c)

    Fertility

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (a)A.

Water logging

Why this answer is correct

Waterlogging is simply the saturation of soil with water, either temporarily or permanently.

The soil ought to absorb the water on the surface and regain its natural state.

The soil is either unable to do that, or human activity makes it difficult for groundwater to go anywhere.

It leads to a condition referred to as waterlogging, where the land is soaked up and water just rises up.

The soil might become waterlogged and fail to dry up.

Additional Information

Soil salinization:

Soil salinization involves a combination of processes like evaporation, salt precipitation and dissolution, salt transport, and ion exchange etc.

The salt-affected soils contain excessive concentrations of either soluble salts or exchangeable sodium or both to inadequate leaching of base-forming cations.

The major soil salinity effect on plant growth is tampering with water absorption.

Question 122

A mild sloped channel is followed by a steep sloped channel. The profiles of gradually varied flow in the channel are

  1. AOption (a)

    M3, S2

  2. BOption (b)

    M3, S3

  3. COption (c)

    M2, S1

  4. DOption (d)

    M2, S2

View answer & explanation

Correct answer

Answer: (d)D.

M2, S2

Why this answer is correct

Concept:

From dydx=S0−Sf1−Q2TgA3=[1−(yny)1031−(ycy)3]S0\frac{dy}{dx}=\frac{S_{0}-S_{f}}{1-\frac{Q^{2}T}{gA^{3}}}=\left [ \frac{1-\left ( \frac{y_{n}}{y} \right )^{\frac{10}{3}}}{1-\left ( \frac{y_{c}}{y} \right )^{3}} \right ] S_{0}

Using Mannings equation for wide rectangular channel.

(i) As  y→y0 ;  dy/dx → 0 ;i.e, the water surface approaches normal depth line asymptotically

(ii) As  y→yc ;  dy/dx → ∞ ; i.e, the water surface meets the critical depth line vertically in region 2 and 3

(iii) As  y→∞ ; dy/dx → 0 ; i.e, the water surface meets a very large depth as horizontal asymptote in region 1.

(iv) As  y→0 ;  dy/dx → ∞ ;i.e, means that the surface

  • M1, M2, S2 and S3 meet y0 lines asymptotically
  • M1, and S1 curve tend to horizontal as y → ∞
  • M2, M3 and S2 meet yc line normally
  • M3 and S3 meet channel bed normally.

So When A mild-sloped channel is followed by a steep-sloped channel

We Find the M2, S2 Profiles.

Question 123

Reciprocal levelling eliminates

  1. AOption (a)

    Collimation error

  2. BOption (b)

    Collimation, curvature and refraction error

  3. COption (c)

    Curvature and refraction error

  4. DOption (d)

    Curvature and collimation error

View answer & explanation

Correct answer

Answer: (b)B.

Collimation, curvature and refraction error

Why this answer is correct

Explanation

Reciprocal levelling

Reciprocal levelling is adopted to accurately determine the level difference between two points which are separated by obstacles like a river, ponds, lakes, etc.

It eliminates the following errors:

  • error in instrument adjustments i.e error due to collimation
  • the combined effect of Earth's curvature and the refraction of the atmosphere
  • variation in the average refraction

Important Points 

  • Reciprocal levelling eliminates the error due to collimation and error due to curvature of Earth completely, but as the refraction depends upon the atmosphere which may change every minute.

Question 124

Residual chlorine can be determined by

  1. AOption (a)

    Orthotolidine test

  2. BOption (b)

    Iodometric test

  3. COption (c)

    Mohr's argentometric method

  4. DOption (d)

    Winkler's method

View answer & explanation

Correct answer

Answer: (a)A.

Orthotolidine test

Why this answer is correct

Explanation:

Orthotolidine Test

  • The Orthotolidine test is used for the determination of Residual Chlorine.
  • Total, free and combined chlorine in water can be determined with speed and accuracy by this test.
  • The reagent is analytical grade Orthotolidine, dissolved in 10% solution of HCl.
  • When this reagent is added to the water containing chlorine, it turns yellow.
  • The intensity of yellow color is proportional to the concentration of chlorine
  • This yellow color is produced by both free and combined residuals.
  • Add 0.1 ml of reagent to I ml of water, Yellow color is produced and matched against charts within 10 sec, this gives the amount of free chlorine.
  • The yellow color appearing after a few minutes, gives the amount of Total chlorine.

Additional Information Iodo-Metric Test (Starch Iodide Test)

  • It is used when the presence of nitrite N2, manganic compound makes orthotolidene test unsuitable.
  • 10ml potassium iodide is added to 5ml starch in One litre water thus producing blue colour complex solution which is titrated with N/100 Sodium Thio Sulphate (Na2S2O3) to determine the residual chlorine.

Winkler's Test

Oxygen in water is determined by Winkler's test

  1. If Dissolved oxygen (DO) is present

Water sample + MnSO4 + NaOH + KI => Mn(OH)2 (white ppt)

  1. If Dissolved oxygen (DO) is absent

Water sample + MnSO4 + NaOH + KI => MnO2 (red ppt)

Mohr's argentometric method

Chlorides are estimated by Mohr's argentometric test in which raw water is titrated with standard AgNO3 solution using K2CrO4 as indicator.

Question 125

Using ferrous alum for coagulation in water treatment changes pH

  1. AOption (a)

    Lowers

  2. BOption (b)

    Increases

  3. COption (c)

    No variation

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (a)A.

Lowers

Why this answer is correct

Concept:

When alum is added in water, it reacts with alkalinity present in water and leads to formation of sticky gelatinous precipitate of Aluminium hydroxide which attracts fine suspended impurities in water over its surface and gets easily settled in the following sedimentation process.

The following typical chemical reaction takes place:

From above chemical reaction following may be concluded:

  1. The carbon dioxide is formed which further reacts with and leads to formation of carbonic acid and hence, reduces the pH of water.
  2. The bicarbonate hardness or temporary hardness (i.e. Ca (HCO3)2) is removed by alum, however, non-carbonate hardness (CaSO4) is introduced by alum.

125 questions in the full paper