Bihar Public Service Commission Exam, 2017

Assistant Engineer (Preliminary)

GENERAL PAPER
Solved Paper
Time Allowed : 3 Hours ] [ Exam Date : 16 Sep., 2018

1. Light-emitting diode is an example of Lateral stress– It is the stress produced in lateral

(a) photonic devices direction. Generally Lateral direction is defined in
(b) mechanical devices comparison to the applied load.
(c) optoelectronic devices Thermal stress– It is defined as any change in the
thermal relation between temperature regulator and its
(d) sensing devices
environment that it is not compensated by thermo-
Ans. (c) : Light-emitting diode (LED) is an regulatory effectors reactions.
optoelectronic device which generates light via
4. National Science Day is celebrated on
electroluminescence. It contains a p-n junction,
(a) 26th December
through which an electric current is sent. The current
(b) 26th January
generates electrons and holes, which release their
(c) 28th February
energy portions as photons when they recombine.
(d) 5th September
Example of optoelectronic device are :
Ans. (c) : National Science day is celebrated on Feb,
(i) Light emitting diode
28 every year to honour the discovery of 'Raman
(ii) Laser diode
effect'. Dr. C.V. Raman was the first among those
(iii) Super luminescent device Indian's who got first "Bharat Ratna Award".
2. The premature ignition of fuel is called 26 January 1950, we adopted our constitution fully in
(a) engine knock action that's why this day is observed every year as
(b) autoignition "Republic day of India"
(c) detonation 5th Sep., we observe "National teacher's day on birth
(d) All of the above date of" Dr. Sarvapalli Radha Krishnan, the first Vice-
President of India. 26 Dec. Jayanti of shaheed Uddham
Ans. (c) : Detonation is a rapid and violent singh, who killed General Dyre in London (1940)
combustion. Detonation involves a supersonic responsible for Jallianwala Bagh massacre in 1919.
exothermic front accelerating through a medium that
5. Which of the following is not a part of
eventually drives a shock front propagating directly in
venturimeter
front of it.
(a) Diverging part
Autoignition– It is defined as the self - ignition of the
(b) Converging part
vapors emitted by a liquid heated above its ignition
(c) Working fluid
temperatures and that when escaping into the
(d) Throat
atmosphere enter into their explosive range.
Engine knock– It is an internal-combustion engine, Ans. (c) : Venturimeter is a device that is used to
measure the rate of flow of fluid through a pipe.
sharp sounds caused premature combustion of part of
A working fluid is a gas or liquid that primarily
the compressed air fuel mixture in the cylinder.
transfers force, motion or mechanical energy.
3. Skin stress is also called as Diverging Part, Converging Part, Throat are part of
(a) shear stress (b) bending stress Venturimeter.
(c) lateral stress (d) temperature stress 6. What is the principle of the 'Johansson
Ans. (b) : Bending stress– It is a normal stress that is Mikrokator'
induced at a point in a body subjected to loads that (a) Button spinning on a loop of string
cause into bend. (b) Principle of interference
Shearing stress– It is defined a type of stress that acts (c) Optical magnification
coplanar with cross section of material. (d) Principle of transformer

BPSC AE (Pre), 16 Sep., 2018 7 YCT

 Ans. (a) : Johansson Mikrokator works on the simple 11. When air passes through silica gel
principle of button spinning on a loop of string. Since (a) it absorbs water vapour molecules
H. Abrahamson developed this Instrument, this simple (b) latent heat of condensation is released
basic principle was also known as the Abrahamson (c) DBT of air increases
movement. When there is a vertical movement other
upward or downwind ward in the plunger is (d) All of the above
transmitted through the elbow of the bell crank lever. Ans. (a) : The water in the air actually absorbs
7. At 00C, silicon behaves as a/an between the tiny passages as the air passes through
(a) conductor (b) insulator them. The water molecules became trapped so that air
(c) semiconductor (d) superconductor is dried out as it passes through the filter. This process
is reversible. It is the silica gel desiccant is heated to
Ans. (b) : A semi-conductor acts like an ideal insulator
at zero (00C) temperature that is at 273 Kelvin. It is 180 0f it will release the trapped water.
because the free electrons in the valence band of semi- 12. Which of the following screw threads is
conductors will not carry enough thermal energy to stronger than other threads?
overcome the forbidden energy gap at absolute zero. (a) Square threads
8. Temperature stress is a function of (b) Trapezoidal threads
(a) coefficient of linear expansion (c) Buttress threads
(b) change in temperature (d) V threads
(c) modulus of elasticity
Ans. (c) The buttress thread form is designed to handle
(d) All of the above
extreme high axial thrust in one direction. The thread
Ans. (d) : Thermal stress is mechanical stress created shares the low friction properties of a square thread
by any change in temperature of a material. These
format roughly twice the shear strength.
stresses can lead to fracturing or plastic deformation
depending on the other variables of heating, which 13. In Physics, the Nobel Prize, 2014 was awarded
includes material types and constraints. for the discovery of
9. Who has served as the 11th President of India? (a) gravitational waves
(a) Shri Pranab Mukherjee (b) blue light LED
(b) Shri K. R. Narayanan (c) neutrino oscillations
(c) Shri A. P. J. Abdul Kalam (d) MRI
(d) Smt. Pratibha Patil Ans. (b) : The Royal Swedish Academy of sciences
Ans. (c) : Dr. A.P.J. Abdul Kalam knownas "missile announced that three Researchers will share this year's
man", was an Indian scientist and politician who Noble Prize in Physics for their Invention of blue light
played a leading role in the development of India's emitting diodes (LEDs). The prize committee
missiles and Nuclear weapons programme. He served emphasised that Isamu Akasaki, Hiroshi Amano and
India from 2002 to 2007 as President of India He was
shuji Nakamura at Santa Barbara launched a revolution
11th President of India.
in energy efficient lighting.
Pranab Mukharjee born in 1935 in west Bengal, served
in different Ministries and later on became 13th The researchers work has already made it into many
President of India from 2012 to 2017. He was Indian Common Electronic devices blue-LEDs can be found
National Congress leader. He was honoured Bharat in most touch screen devices. White LEDs usually use
Ratna in 2019. a blue LED to excite a phosphor to emit white light,
K.R. Narayan was a diplomat, academic and politician. and can be found in the camera flashes of most modern
He served India as the l0th President and 9th Vice smart phones.
President of India. 14. In India, 15th September is celebrated as
Pratibha Devi Singh Patil served as President of India
(a) Engineer's Day
from 2007 to 2012, she was the first woman President
of India. She was India's 12th President. (b) Scientist's Day
(c) Labour's Day
10. NASA was established in the year
(a) 19 (b) 1950 (d) Women's Day
(c) 1958 (d) 1985 Ans. (a) : Every year, the country celebrates 15th Sep.
Ans. (c) : National Aeronautics and Space as 'Engineers Day' to appreciate the contribution of M.
Administration is an American space agency as well Visveswarya, the Bharat Ratna Awardee. Visveswarya
as aeronautics and space research works for the born in Karnataka.
Civilian Space Programme, established in 1958 its 4th March world Engineers day is observed across
Headquarter is in Washington D.C. world.

BPSC AE (Pre), 16 Sep., 2018 8 YCT

 1th May is observed as International Labour day in Ans. (b) : Dr. Bhim Rao Ramji Ambedkar born on
several countries to commemorate the labour April 14 1891 in Madhya Pradesh's Mhow region. He
movement which brought (and bring) Various was not only, social reformer but above it fought
Reforms in labour laws and made various against atrocities against "Dalits" He is also known as
humanitarian change in labour working condition. "Father of Indian constitution" He became India's first
International women's day (march 8) is a global day law minister although he was not a congress man but
celebrating the social, Economic, cultural and political was in ministry of Pt. Jawahar Lal Nehru. He got
achievements of women. education from London school of Economics and
15. Ministry of Science and Technology was political science he died on 6 December 1956.
formed in the year Maulana Abul Kalam Azad the first education
(a) 1950 (b) 1971 minister of India. Later it became MHRD in 1985.
(c) 1985 (d) 1992 Chandra Sekhar singh, India's first textile minister
Ans. (b) : The Ministry of Science and Technology from 30 March 1985 to 15 November 1985.
which formed in 1971 is an Indian government India's first foreign minister was Pt. Jawahar Lal
ministry charged with formulation and Administration Nehru.
of the rules and regulations and laws relating to
science and technology in India. Its head quarter is in 18. Graphene is a
New Delhi. (a) one-dimensional material
The Union Ministry of Human Resource development (b) two-dimensional material
has been officially renamed as the ministry of (c) three-dimensional material
education in 2020, dedicated for the development and (d) All of the above
form of education and shaping future through overall Ans. (b) : Graphene is the name for an atom thick
development of its young population. Formation of honey comb sheet of carbon atoms. It is building block
HRD ministry was in 1985.
for other Graphitic material. It is two dimensional
The Ministry of Home Affairs Discharges multifarious materials. Each atom in a graphene sheet is connected
responsibilities, the important among them being-
to its three neighbours by a bond and contributes one
internal security, border infiltration stoppage etc.
electron to a conduction band that extends over the
formed in 1947
whole sheet.
16. Who among the following scientists has made
his contribution in the establishment of ISRO? 19. Bihar Diwas (Bihar Day) is observed every
(a) A. P. J. Abdul Kalam year on
(b) C. V. Raman (a) 25th March (b) 22nd March
(c) Vikram Sarabhai (c) 1st April (d) 1st March
(d) Aryabhatta Ans. (b) : Bihar Diwas or Bihar day is observed on 22
Ans. (c) : Indian Space Research also known as ISRO March every year. It marks the formation of the state
Indian space research wing that was created in 1969, of Bihar when the state was carved out from the
is a national space agency of India its head quarter is in Bengal Presidency of the British in the year 1912.
Bengaluru. It is celebrated in other countries as well as by the
ISRO built India's first satellite Aryabhatta which was Bihari Diaspora. Bihar Diwas was started and
launched by the Soviet Union on 19 April 1975. it was celebrated on large scale by Bihar government in the
named after the mathematician Aryabhatta. tenure of Nitish Kumar 1 April- Odisha day 25 March-
SAARC establishment of secretariat on 1 March 1987
Vikram Ambalal Sarabhai born in 1919 in
in Kathmandu Nepal.
Ahmadabad was on Indian physicist and industrialist
who initiated space Research and helped develop 20. Sardar Sarovar Dam is located on
Nuclear power in India. Dr. A.P.J. Abdul Kalam (a) Ganga river (b) Narmada river
known as "missile man of India" became 11th president (c) Sutlej river (d) Godavai river
from 2002 to 2007. Ans. (b) : Sardar Sarovar Dam (SSD) is build on the
Dr. C.V. Raman got noble prize because of its Indian Narmada River is located in the village of
"Raman effect" in Physics in 1930. kevadia in the state of Gujarat it is one of the largest
17. Dr. B. R. Ambedkar was independent India's and most controversial inter state, multipurpose river
first valley infrastructure project in the country.
(a) Textile Minister Narmada River–The Narmada river also called the
(b) Law Minister reva and previously also known as Narbada it is also
(c) HRD Minister known as life line of MP, for its huge contribution to
(d) Foreign Minister the state of MP. .

BPSC AE (Pre), 16 Sep., 2018 9 YCT

 Ganga– is a transnational river, it is national river of 24. Rateau turbine belongs to the category of
India it is said to be cultural glory of India. In (a) pressure-compounded turbine
Bangladesh it is Known as "Padma". It originates (b) reaction turbine
from Gangotri glacier of Himalaya and merge into (c) velocity-compounded turbine
Bay of Bangal. Its total length is around 2500 km. It is
(d) radial flow turbine
considered pious in Hinduism.
Sutlej river– longest of the 5 tributaries of the Indus Ans. (a) Rateau Turbine pressure compounding
river that give the Punjab meaning "Five rivers" Curtis Turbine velocity compounding.
(heeBÛeveoer) its name. It originates from Mount Kailash in 25. Gradually varied flow is
Tibet. Godavari river, river of central and south eastern (a) steady uniform
India is sacred to Hindus. Its total length is about (b) non-steady non-uniform
1465km. It is second longest River after Ganga in
(c) non-steady uniform
India.
(d) steady non-uniform
21. The First Bharat Ratna Award was given in
the year Ans. (d) If the rate of varied of depth with respect to
(a) 1951 (b) 1952 distance is small and rapidly varied flows if the rate of
(c) 1953 (d) 1954 variation is large.
Ans. (d) : Bharat Ratna is the highest civilian honor 26. The temperature of normal human body is
given for the service towards advancement for Art, (a) 38.60C (b) 370C
0
Literature and Science for the recognition of public (c) 37.6 C (d) 380C
service of the highest order. It is not mandatory that Ans. (b) : Normal body temperature varies depending
Bharat Ratna would be awarded every year. The on many factors, including a person's age, sex and
provision for Bharat Ratna was introduced in 1954.
activity levels. The Normal body temperature for an
The first ever Indian received this award was the
adult is around 98.60F(370C) but every person body
famous scientist, C.V. Raman. The two Non-Indians
Khan Abdul Gaffar khan (1990) and (1990) Nelson temperature is slightly different and may consistently
Mandela also received this award. The first ever be little higher or lower. A Normal body temperature
sports player 'Sachin Tendulkar' received this award in for children aged 3-10 ranges from 95.9–99.50F when
2017 taken orally. Children lend to have similar body
temperatures to adults.
22. Raxaul Airport is located in the State of
(a) Goa (b) Maharashtra 27. Who was the founder of Aligarh Muslim
(c) Bihar (d) Uttarakhand University?
Ans. (c) : Raxaul Airport is located at west of the town (a) Sir Syed Ahmad Khan
of Raxaul in Bihar state, India. The town is on the (b) Mohammad Ali Jinnah
border with Nepal. It was established after China-India (c) Abul Kalam Azad
war in 1962. (d) Ram Mohan Roy
Important Notes:- Ans. (a) : Aligarh Muslim University is an Indian
Goa– Goa Dabolim Airport. leading university [central]
Maharastra– Chattrapati Shivaji Mahraj International It was established in 1875 as Anglo-oriental
Airport, Shirdi International Airport, Osmanabad Mohmadden collage by Sir Syed Ahmad khan for
Airport. encouraging Muslim youth to learn English language
Uttarakhand– Jolly Grant Airport (Dehradun and promote scientific fervor among them. He wanted
Airport), Pant Nagar Airport a blend of education (Oriental+ English). That same
23. Which of the following weldings is used for college in 1920 became Aligarh Muslim University Sir
welding vertical section in one pass? Syed Ahmad Khan, a Muslim educator, jurist and
(a) Electroslag welding author and founder of Anglo-Oriental Mohamaddan
(b) Atomic hydrogen welding Collage in 1875 in Aligarh.
(c) Laser-beam welding Abul Kalam Azad– original name Abdul Kalam
(d) Electrogas welding Ghulam Muhiyuddin, also called Maulana Abul Kalam
Ans. (a) Electro slag welding is a welding process, in Azad or Maulana Azad (born Nov.11 1888 Mecca
which the heat is generate by an electric current Saudi Arabia), Islamic theologian who was one of the
passing between the consumable electrode (filler leaders of the Indian independence movement.
metal) and the work piece through a molten slag Mohammad Ali Jinnah, founder of Pakistan state
covering the weld surface. "responsible for partition"

BPSC AE (Pre), 16 Sep., 2018 10 YCT

28. Mr. Jagadish Chandra Bose is a famous 31. An instrument, that is used for the detection of
scientist for the invention of earthquake, is
(a) Bose-Einstein statistics (a) barometer (b) lactometer
(b) crescograph (c) seismograph (d) holograph
(c) X-rays Ans. (c) : A seismograph or seismometer is an
(d) scattering of light instrument used to detect and record earthquakes.
Ans. (b) : Mr. Jagadish Chandra Bose was an Indian Generally it consists of a mass attached to fixed base.
physicst, botanist and a pioneer in radio-science. He Earthquake is any sudden shaking of the ground
conducted experiments to prove plants feel heat, cold, caused by the passage of seismic waves through
light, noise, happiness and pain. His instrument the Earth's rocks.
"crescograph" can measure plant growth. Note:-
X-rays are form of electromagnetic similar to visible Barometer- Measurement instrument of air pressure
light, however, X-rays have higher energy and can Lactometer-Instrument for measuring the density of
pass through most object, including the body. X-rays milk.
was discovered by German Scientist Rontgen Holograph- A manuscript or other document written
Radiation. completely by the hand a person above whose name it
When light passes from one medium to any other appears.
medium let' say Air, a glass of water then a part of the 32. The Head Office of the Central Pollution
light is absorbed by particles of the medium preceded Control Board (CPCB) is located in
by its subsequent Radiation in a particular direction. (a) Mumbai (b) Kolkata
(c) Patna (d) None of the above
29. Albert Einstein was awarded the Nobel Prize
for Ans. (d) : Central Pollution Control Board which
(a) theory of relativity comes under ministry of environment, forest and
climate change. It serves a field formation and also
(b) quantum optics
provides technical services to the ministry of
(c) photoelectric effect environment and forests under the provisions of the
(d) Bose-Einstein theory environment (protection) Act, 1986.
Ans. (c) : Albert Einstein born March 14. 1879 in It coordinates the activities of the state pollution
Ulm, Germany was one of the most well known and control Boards by providing technical assistance and
influential physicist of the 20th century. On November guidance and also resolves disputes among them. It
9, 1922 he was named the winner of the 1921 Noble was formed in 1974 and head quarter is in New Delhi.
Prize in physics "for his services to theoretical Physics 33. RDX is a chemical compound. How is it used?
and especially for his disarray of the law of the photo (a) As a composition
electric effect". (b) As a reactor
30. Rana Pratap Sagar Dam is situated on (c) As an explosive
(a) Chambal river (d) As a nuclear weapon
(b) Yamuna river Ans. (c) : RDX abbreviation of research department
(c) Narmada river explosive or Royal demolition explosive discovered by
(d) Brahmaputra river Gorge Friedrich Henning of Germany and patented in
1898 but not used until world war II. It is relatively
Ans. (a) : The Rana Pratap Sagar Dam is a gravity safe and expensive to manufacture, RDX was
masonry Dam of 53.8 meters height built on the produced on a large scale in The United States by a
Chambal river at Rawatbhata in Rajasthan. It is part of secret process. The name RDX was coined by British.
Integrated scheme of a cascade development of the
The Germans Called it hexogen and the Italians called
river involving 4 projects starting with the Gandhi
it T4.
Sagar Dam in the upstream reach in MP.
The direct benefit from the Dam is hydropower 34. The planet Neptune was discovered by
generation of 172 MW and irrigation to Rajasthan. (a) Galle (b) Galileo
Madhya Pradesh especially. (c) Kepler (d) Newton
Yamuna river India's one of the primary river, flows Ans. (a) : Neptune is the only giant planet that is not
in Northern India. The river originates from Yamunotri visible without a telescope. Having a apparent
Glacier. Narmada river, important river of central magnitude of 7.8 it is approximately one fifth as bright
India. Brahmaputra river flows from India to as the faintest stars visible to the Unaided eye.
Bangladesh. In Bangladesh, it is also known as Galileo is credited as the first person to view the planet
"Jamuna river". with a telescope in 1609.

BPSC AE (Pre), 16 Sep., 2018 11 YCT

35. Resistance of which of the following is CERN- The European organisation for Nuclear
unaffected by temperature? Research. It is the world's largest Particle physics
(a) Manganin (b) Constantan laboratory.
(c) Nichrome (d) All of the above 39. Name an acid which is secreted in the stomach.
Ans. (d) : All of the above. (a) Sulphuric acid (b) Hydrochloric acid
(c) Carbonic acid (d) Nitric acid
Nichrome is a Nickle-chromium alloy with Non-
Magnetic properties Ans. (b) : Hydrochloric acid is the main component of
gastric Juice and is secreted by the parietal cells of the
Manganin an alloy of copper, manganese and nickle
gastric mucosa in the fund us and corpus. In healthy
used chiefly in electrical devices. adults. Intro gastric PH ranges between 1.5 and 3.5.
Constantan is a copper nickle alloy used in electrical
40. The term 'CTBT' is related to
work for its high resistance.
(a) nuclear weapons
36. Which of the following is the first calculating (b) taxes
device? (c) space research
(a) Abacus (b) Calculator (d) railway goods
(c) Turing machine (d) Pascaline
Ans. (a) : The comprehensive Test Ban treaty
Ans. (a) : Abacus is simple tool or a hardware used for prohibits 'any nuclear weapon test explosion or any
performing rapid arithmetic calculation. Calculation other Nuclear explosion' Anywhere in the world. The
based on Abacus was invented at ancient times and treaty was opened for signature on 24 Sep 1996 and
still in use today. has been signed by 184 nations and ratified by 168.
The first solid state electronic calculator was created in The treaty cannot be entered into force until it is
the Early 1960. Pascaline also called 'Arithmetic ratified by 44 specific nations, eight of which have yet
machine' the first calculator or Adding machine to be to do so China, India, Pakistan, North Korea.
produced in any quantity and actually used. Israel, Iran and the US.
37. Name the polymer used in making bulletproof 41. India's first mobile court was inaugurated in
glass. (a) Maharashtra (b) Haryana
(a) Melamine (b) Bakelite (c) Uttar Pradesh (d) Rajasthan
(c) Lexan (d) Vinyl rubber Ans. (b) : First ever mobile court was launched at
Ans. (c) : Lexan is the most popular polymer which is Punhana Haryana in India, an integrated system of
used for manufacturing different things due to its Judiciary we find, one Supreme court and 25 high
strong and Rigid Nature. They are hard and are court are currently functioning at higher level of
observably transparent. This is the reason why a Judiciary.
laminated Poly carbonate (Lexan) is used to make 42. In which year, railway finances were separated
bulletproof glass. from the general finances of the Central
Melamine is a nitrogen based compound used by Government?
many manufacturers to create a number of products (a) 1920 (b) 1972
especially plastic dishware. (c) 1923 (d) 1924
Bakelite is the commercial name for phenol Ans. (d) : Rail budget was separated from general
formaldehyde resin. It is usually brown/amber but can budget in 1924. The measure was taken on the
recommendations submitted by a panel led by British
be made in a variety of bright colors.
Economist William Acworth in 1920-21. In 2016
38. Where was the World Wide Web created and railway budget and general budget was merged.
in which year? 43. Logarithm tables were invented by
(a) CERN, 1989 (a) J. J. Thomson (b) John Napier
(b) Photonics 21, 1989 (c) Paul Ehrlich (d) A. G. Bell
(c) CLUSTER, 1995 Ans. (b) : In Mathematics the logarithm table is used
(d) Gikll, 1993 to find the value of the logarithmic function. The
Ans. (a) : World Wide Web, the leading information simplest way to find the value of the given logarithmic
retrieval service of the Internet (the world wide function is by using the log table
computer network). Tim burners Lee, a British Originator of logarithms is john Napier
Scientist, invented the www in 1989 while working at A.G.Bell- Graphophone, Telephone
CERN. Paul Ehrlich-Chemotherapy, Immunology.

BPSC AE (Pre), 16 Sep., 2018 12 YCT

44. What is India's per capita emission of 48. Rainbow Revolution is related to which sector
greenhouse gases (GHG)? of the economy?
(a) 0.8 tonne of CO2 (a) Small-scale industries
(b) 1.0 tonne of CO2 (b) Information technology services
(c) 1.2 tonne of CO2 (c) Overall development of agriculture sector
(d) 1.5 tonne of CO2 (d) Mining sector
Ans. (d) : Greenhouse gases:- any gas that has the Ans. (c) : Rainbow Revolution– is an integral
property of absorbing infrared radiation (net heat development programme of agriculture, horticulture,
Energy) emitted from earth's surface and radiating it forestry, sugarcane, fishery, poultry and animal
back to earth's surface thus contributing to the green husbandry, yellow revolution etc.
house effect. 49. Who among the following was the first
Carbon dioxide, methane and water vapour are the economist to hold the Office of Secretary,
most important green house gases. Department of Economic Affairs in the Union
Per capita emission of GHG. India is 1.5 tonne of CO2 Finance Ministry?
Paris agreement 2015 related to cut emission of green (a) Dr. I. G. Patel
house gas by double upto 2024 (b) Dr. Manmohan Singh
45. A new study provided the first evidence that (c) Rakesh Mohan
fatter people may be more affected by exposure (d) Dr. M. S. Ahluwalia
to Ans. (a) : I.G. Patel was an Indian Economist and a
(a) sunlight career civil servant who served as 14th governor of
(b) X-rays RBI. He was the first economist to hold the office of
(c) y-rays secretary, department of economic affairs in the union
(d) ozone finance ministry.
Ans. (b) : Obese people may be exposed to more Rakesh Mohan– Former Deputy governor of RBI.
radiation when they undergo a CT scans or X-rays Dr. M.S. Ahluwalia– Former deputy chairman of
compared with normal weight people, a new study former Planning Commission of India.
shows. Note– Now the work of planning commission has been
The study conducted by using computer models assigned by NITI Aayog (National Institution for
suggests that the organs of obese individuals receive Transforming India) since 1 January 2015.
up to 62% more radiation energy during a CT scan. 50. Who is the author of Soul and Structure of
46. Which of the following units is used for Governance in India?
measuring the speed of processor? (a) V. K. Duggal (b) Jairam Ramesh
(a) MPIS (b) MISP (c) Dr. I. G. Patel (d) Jagmohan
(c) MIPS (d) MSIP Ans. (d) : Jagmohan Malhotra, served as L. Governor
Ans. (c) : MIPS-million instruction per second is an of Delhi and Goa. wrote several books–
approximate measure of a computer's raw processing 1. Island of truth, 2. My frozen turbulence in
power. Kashmir 3. Soul and structure of governance in
MISP- Malware Information Sharing Platform India etc.
V.K. Duggal– Vinod Kumar Duggle was the home
47. Nerves from the eyes and ears are connected to
secretary from 2005 to 2007 in the government of
the
Manmohan singh.
(a) cerebrum
I.G.Patel– Former civil servant and 14th governor of
(b) cerebellum RBI.
(c) medulla oblongata Jairam Ramesh– Congress leader and former
(d) spinal cord agriculture minister in Manmohan Singh government.
Ans. (a) : Cerebrum- is the uppermost part of the 51. A Cantilever beam of rectangular cross-section
brain. It contains two hemisphere split by a central is subjected to a point load at its free end. If
fissure. It connects nerves from the eyes and ears. width and depth of the beam section are
Medulla Oblongata- simply medulla is a long stem doubled, then the deflection at free end of the
like structure which up the tower part of the brainstem. beam will be reduced to
Cerebellum-"Little brain" is a major structure of the (a) 6.25% (b) 15%
hind brain that is located near the brain stem. (c) 25.5% (d) 29%

BPSC AE (Pre), 16 Sep., 2018 13 YCT

 Ans. (a) M σ E
Ans. (b) : From Bending equation = =
I Y R

deflection MY
Bending stress σ =
WL3 I
∆= Where it is distance from neutral axis to where
3EI
bending stress to be found.
I = moment of Inertia
BD3 i.e σ × y
Rectangular section I =
12 Bending stress is symmetrical I section beams will be
WL 3
WL 3 maximum at the top & bottom of the beams section.
Case (i) ∆ = =
3EI BD3 54. A circular shaft is subjected to a twisting
3E ×
12 moment Mt and bending moment M. The ratio
Case (ii) B & D double of maximum stress developed due to bending
2B × ( 2D3 )
moment and that due to twisting moment is
I= equal to
12
2M 2M
(a) (b)
BD3 3M t Mt
I = 16
12 M M
(c) (d)
Beam will reduce x% Mt 2M t
WL3 WL3 Ans. (b) For circular shaft, bending stress due to
× x% =
3E × BD3 BD 3
bending moment (M)
3E × 16
12 32M
σb =
100 πd 3
x= = 6.25
16 M → Bending moment
σ=
52. The ratio of maximum shear stress to average Z → Section modulus
shear stress in a beam of rectangular cross- Shear stress due to Twisting moment(Mt)
section is
(a) 3.0 (b) 2.5 T 16T
τ= =
(c) 2.0 (d) 1.5 ZP πd 3
Ans. (d) 16M t
=
Beam maxm shear stress πd 3
ratio
avg. shear stress σb 32M 2M
= =
Rectangular 1.5 τ πd 3 Mt
Circular 4/3 = 1.34 16M t
Triangular 1.5 πd 3
53. In a symmetrical I-section beam, the bending 55. If a simply-supported beam of span L carries a
stress will be maximum at moment force at its mid-span, then the shear
(a) the neutral axis force diagram will be
(b) the top and bottom of the beam section (a) triangular
1 (b) rectangular
(c) tha depth from top and bottom of section
4 (c) parabolic
(d) the junction of flange and web (d) cubic parabolic

BPSC AE (Pre), 16 Sep., 2018 14 YCT

 Ans. (b) : Simply supported Beam span L moment at 59. A linear helical spring with spring constant K
mid-span. is cut into two equal halves. The spring
constants of the individual halves will be
(a) K / 2 (b) K / 2
(c) 2K (d) 2K
Ans. (d) Given,

SFD
Rectangular

We know that,
m+n
k1 = ×k
n
m+n
k2 = ×k (m = n)
m
k1 = k2 = 2k
56. According to maximum shear stress criterion,
60. In a body, loaded under plane stress
yielding in material occurs when
conditions, the number of independent stress
(a) maximum shear stress = 2 yield stress components in order to completely specify the
(b) maximum shear stress = 0.5 yield stress state of stress at a point is
(c) maximum shear stress = √2 yield stress (a) 1 (b) 3
2 (c) 4 (d) 6
(d) maximum shear stress = yield stress
3 Ans. (b) : Plane stress– There is no normal & shear
Ans. (b) : Maximum shear stress theory, yielding in stresses on the two planes perpendicular to the z-
material occurs when direction. This system is known as plane stress.
stresses, σz = 0, τxz = 0 & τyz = 0
1
maximum shear stress ( τ max ) ≤ × yield stress(σ y ) σx, σy & τxy may have non zero values so the no. of
2 independent stress is 3.
57. A frictionless pin joint transmits a 61. A shaft of 60 mm diameter is subjected to
(a) force which passes through the pin torsion has a shear strain of 0.0006. The rate of
(b) torque about the pin twist will be equal to
(c) moment about the pin (a) 0.00002 (b) 0.00025
(d) All of the above (c) 0.0036 (d) 0.00001
Ans. (a) : Pin joints transmit forces not the moment. A Ans. (a) : Given, d = 60mm, r = 30mm
frictionless pin joint transmits a force which posses Shear strain (γ) = 0.0006
through the pin. θ
Rate of twist =
58. A bar held between two rigid supports will be l
subjected to tensile stress if it is T Gθ τ
(a) heated Torsion equation = =
J l r
(b) cooled
Gθ τ
(c) heated or cooled =
(d) heated beyond the melting point l r
Ans. (b) : When both ends support will be rigid when shear stress ( τ )
G=
temperature ↑ then the material try to expand but due shear strain(γ )
to rigid support they will not expand. So material
θ τ shear strain(γ )
under compression. = =
If temperature decreases than the material will try to l Gr r
contract but due to rigid support not possible so it will θ 0.0006
= = 0.00002
be subjected to tensile stress. l 30

BPSC AE (Pre), 16 Sep., 2018 15 YCT

62. The equivalent spring constant for a bar of 65. A small plastic boat loaded with nuts and bolts
length L, cross-sectional area A and modulus of is floating in a bathtub. If the cargo is dumped
elasticity E is subjected to an axial force P is into water, allowing the boat to float empty, the
(a) AE / L water level in the tub will
(a) rise (b) fall
(b) L / AE
(c) remain same (d) None of the above
(c) PL / A
Ans. (b) : Given,
(d) P2L / 2AE
Ans. (a) :
Equivalent spring constant (K) = Spring stiffness
Force P
Spring stiffness = =
Unit deformation ∆
PL P AE
Deformation (∆) = ⇒ =
AE ∆ L
P P AE
Equivalent spring constant (K) = = =
∆ PL L Let, mass of boat = mboat, Mass of bolt = mbolt, weight
AE of fluid displaced = Wfd
63. The no-slip boundary condition applied in a According to given figure,
fluid In case (I)
(a) is a consequence of laminar behaviour of According to principle of floating,
fluid Wboat + Wbolt = FB = Wfd (in floating
(b) because the fluid is treated as continuous condition)
(c) because the fluid is incompressible mboat g + mbolt g = ρf g Vfd
(d) because the fluid is viscous m boat + m bolt
(mboat + mbolt) = ρf Vfd ⇒ Vfd =
Ans. (d) : When real fluid pass a solid. The velocity of ρf
the fluid on the surface of the object will be same as In case (II)
that of the surface. If the surface is at rest the fluid will Wboat = (FB)boat (in floating condition)
have zero velocity because of no slip condition at the mbolt g = ρf g(Vfd)bolt ⇒ mbolt = ρf (Vfd)bolt
boundary. Wbolt > (FB)boat
No slip boundary condition applied in a fluid because mbolt g > ρf g(Vfd)bolt
fluid is viscous. mbolt > ρf (Vfd)bolt
64. When a liquid rotates at constant angular (mboat + mbolt) > ρf (Vfd)bolt + ρf(Vfd)boat
velocity about a vertical axis as a rigid body, ρf Vfd > ρf[(Vfd)bolt + (Vfd)boat]
the pressure intensity Vfd > [(Vfd)bolt + (Vfd)boat]
(a) decreases as the square of radial distance So we can say if the cargo is dumped into water,
(b) increases linearly as radial distance allowing the boat to float empty then displaced volume
(c) varies inversely as the elevation along any of fluid in case (I) more than case (II) so the water
vertical line level in the bathtub will fall.
(d) varies as square of radial distance 66. Nusselt number is the ratio of
(a) temperature gradient of wall to that across the
Ans. (d) : Vertical surface cross-section of the surface
entire pipe
is the curve represented
(b) temperature difference to the temperature
r 2 ω2 gradient at the wall
Z=
2g (c) heat flux at the wall to that across the entire
This is equation of parabola so liquid surface is pipe
paraboloidal form. (d) None of the above
P = ρgz Ans. (c) : Nusselt No. is the ratio of heat flux at the
wall to that across the entire pipe.
P ω2 R 2 (or)
=Z=
ρg 2g It is the ratio of heat convected to the conducted.
P ∝ R 2 ω2 ; P = f(R, ω, ρ) Q hL
Nusselt no. = (Nu) = Conv. =
Pressure in tensing ∝ square of radial distance. Q Cond. k f

BPSC AE (Pre), 16 Sep., 2018 16 YCT

67. The momentum correction factor for laminar Ans. (a) : Effect of pressure gradient on boundary
flow through a circular pipe is dp
layer = pressure gradient
(a) 1.67 (b) 3.0 dx
(c) 0.85 (d) 1.33 dp
< 0 → flow not separate the entire boundary
Ans. (d) : momentum correction factor (β) dx
2
layer moves forward.
( momentum/sec.) act = 1  U 
( momentum / sec ) avg A ∫  U avg 
=   dA dp
> 0 → flow has separated the growth of
dx
for laminar flow → 1.33 boundary layer.
for turbulent flow → 1.2 70. Turbulent boundary layer thickness is
proportional to
Kinetic correction factor (α)
(a) 1 / x (b) x1/5
3
1  U  2/5
(c) x (d) x4/5
= ∫  dA
A  U avg  Ans. (d) : For turbulent boundary layer
δ 0.371
for laminar flow → 2 =
x ( R ex )1/ 5
for turbulent flow → 1.05
0.371× x 0.371× x
δ= 1/ 5
= 1/ 5
68. The velocity distribution for laminar flow  ρvx   ρv 
 µ  ×x
1/ 5
between two parallel plates  µ 
   
(a) is constant over the whole cross-section
0.371× x 4/5
(b) is zero at the boundary and increases linearly = 1/ 5
towards the centre line  ρv 
 
(c) varies linearly across the section with a  µ 
maximum at the centre line
δ ∝ x4/5
(d) varies parabolically across the section with a
maximum at the centre line 71. The value of fraction factor for smooth pipes
Ans. (d) : The velocity of flow b/w two parallel plates for Reynolds' number equal to 106 is
approximately
1  ∂P 
 −  ( By − y )
u= 2 (a) 0.0001 (b) 0.001
2µ  ∂x  (c) 0.01 (d) 0.1
B = distance b/w the plates Ans. (*) : Given, Re = 106
y = distance of point from one of the plate We know that,
µ = dynamic viscosity of flowing fluid Friction factor (f) =
64
∂P Re
− = pressure gradient along the flow.
∂x 64
∴ f= = 0.6037
→ From the expression that the velocity distribution 106
b/w two plates is parabolic with max value at the 16 16
Coefficient of friction = = = 0.1509
centre and minimum at boundary/plate surface. Re 106
69. The growth of boundary layer is supported 72. The time constant of an R-C circuit is one
when (p is the pressure and x is the distance second. Then in one second the capacitor is
from the leading edge) charged to
∂p (a) about 66%
(a) is positive (b) about 98%
∂x
∂p (c) 100%
(b) is zero (d) None of the above
∂x
∂p Ans. (a) : Given time constant τ = RC
(c) is negative
∂x Voltage across capacitor during charging can be given
(d) None of the above by– Vc (t) = V(1 – e–t/τ)

BPSC AE (Pre), 16 Sep., 2018 17 YCT

 Vc(t) = V(1 – e–t/RC) ∵ τ = RC = 1sec 76. In the Laplace transform
At t = 1 sec F(s) = (s+2) / s(2s + 1)
Vc = V(1 – e–1/1) the function f(t) as t → ∞ and t → 0
Vc = V(1 – e–1) respectively are
Vc = V(1 – 0.367) (a) 2, 0 (b) 0, 0.5
Vc = 0.6333 V (c) 2, 0.5 (d) 0.5, 2
0.633V Ans. (c) : Given function–
%Vc = × 100
V s+2
F(s) =
Vc = 63.33% s(2s + 1)
≅ 66% s+2 A B
= = +
73. A linear circuit must obey  1 s  1
2s  s +  s + 
(a) superposition theorem  2  2
(b) superposition theorem and Thevenin's By partial fraction–
theorem
s+2 2
(c) superposition, Thevenin's theorem and A= = =4
1 1
Norton's theorem s+
2 s =0 2
(d) superposition and Norton's theorem
1
Ans. (c) : A linear circuit must obey superposition − +2
s+2 2
theorem. Thevenin's theorem and Norton's theorem. B= = = −3
s s=− 1 1
Because these are applicable only for linear, bilateral 2 −
active circuit. 2
(s + 2) 4 3
74. In a parallel R-L-C circuit, the values of R, L F(s) = = −
and C are 40 ohms, 2 henries and 1/2 farad  1  2s  1
2s  s +  2 s + 
respectively. The quality factor Q of the circuit  2  2
will be 2 3
F(s) = −
(a) 1/20 (b) 20 s  1
(c) 40 (d) 80 2 s + 
 2
1 Taking inverse laplace transform–
Ans. (b) : Given, R = 40 Ω, L = 2H, C = F
2 2 3 
L−1F(s) = L−1  −
C s  1
Quality factor (Q) = R  2  s +  
L   2
1/ 2 3
= 40 f (t) = 2 − e− t / 2
2 2
1 At t → ∞
= 40 ×
2 3
f (∞) = 2 − e −∞
Q = 20 2
=2–0 {Q e–∞ = 0}
75. A negative resistance is an element
f(∞) = 2
(a) that can act only as a source of active power
At t → 0
(b) that can act as a source of both active as well
as reactive power 3
f(0) = 2 − e −0 / 2
(c) That can act only as a source of reactive 2
power 3
= 2 − ×1 {Q e–0 = 0}
(d) that will store energy 2
Ans. (c) : A negative resistance is an element that can = 2 – 1.5
act only a source of reactive power because positive f (0) = 0.5
resistance always consumed active power and
temperature coefficient of a mated is positive. A 77. Increasing the value of the coupling capacitor
negative resistance means it delivered the power to Cc in a common-emitter amplifier affects its
source. Therefore it act as reactive power source. (a) mid-band voltage gain
(b) fL (lower cut-off frequency)
BPSC AE (Pre), 16 Sep., 2018 18 YCT
 (c) fH (higher cut-off frequency) (b) by applying voltage across the stator with the
(d) fL and fH both help of autotransformer
Ans. (b) : Increasing the value of the coupling (c) by changing the number of poles in the stator
capacitor Cc in a common-emitter affects its lower cut- winding
off frequency. Because, (d) None of the above
Ans. (d) : A single phase induction motor is not self
starting inherently due to pulsating torque. It is made
self starting by using an auxiliary winding. There are
different method to start the 1 – φ motor–
(1) Split phase method
(2) Capacitor start
(3) Permanent split capacitor
(4) Capacitor start capacitor run
78. The base width of a junction transistor is
82. Transformer core is made of lamination to
chosen by design to be small so that
reduce
(a) the electric field becomes large
(a) eddy-current loss only
(b) the concentration gradient of injected carriers
is small (b) hysteresis loss only
(c) the recombination of injected minority (c) both hysteresis and eddy-current loss
carriers is reduced (d) None of the above
(d) the majority carriers easily reach the collector Ans. (a) : Transformer core is made of lamination to
Ans. (d) : The base width of a junction transistor is reduce eddy-current loss. To reduce hysteresis loss we
chosen by design to be small so that the majority use silicon steel core.
carriers easily reach the collector. In a transistor 83. When a two-winding transformer is connected
emitter have highly doped and collector is doped less as an auto-transformer, its efficiency (full-load)
than emitter and base is lightly doped. (a) remains the same
+
79. To increase the switching speed of a p n diode (b) increases
(a) the n region width should be made larger (c) decreases
(b) the n region width should be made smaller (d) rises to 100%
(c) the p region's bulk resistance should be larger Ans. (b) : When a two-winding transformer is
(d) None of the above is true connected as an auto-transformer, its efficiency (full-
Ans. (b) : To increase the switching speed of a p n + load) increases. An auto transformer has higher
diode the n region width should be made smaller efficiency than two winding transformer. This is
because if region width made large the time constant is because of less ohmic loss and core loss due to
+
increases then switching speed of a p n diode decrease. reduction of transformer material.
80. Threshold voltage of a MOSFET can be 84. Which of the following motors runs at constant
reduced by speed at all loads?
(a) increasing the oxide thickness (a) Synchronous motor
(b) reducing the dielectric constant of oxide (b) Induction motor
(c) increasing the oxide thickness, and increasing (c) DC shunt motor
the oxide dielectric constant (d) DC series motor
(d) reducing the oxide thickness and increasing Ans. (a) : Synchronous motor always runs at
the oxide dielectric constant synchronous speed.
Ans. (d) : Threshold voltage of a MOSFET can be 120f
reduced by reducing the oxide thickness and increasing NS =
P
the oxide dielectric constant. Threshold voltage is the
Where,
voltage applied between gate and source of a MOSFET
that is needed to turn the device on for linear and Ns = Synchronous speed
saturation region of operation. f = frequency
81. A single-phase induction motor starts P = No. of Poles.
(a) due to the development of rotating field for 85. Four-point starter is used for
single-phase a.c. supply (a) synchronous motor
(b) induction motor of large capacity
BPSC AE (Pre), 16 Sep., 2018 19 YCT
 (c) DC shunt motor with wide range of speed Q = C∆T = 1 × 300 = 300 Joule
(d) DC series motor with heavy load For finite body
Ans. (c) : Four-point starter is used for the armature of T 
a DC Shunt motor or compound wound DC motor ∆S = C ln  2 
 T1 
against the initially high starting current of the DC
∆S = 1×ln 
motor. 300 

 600 
86. The electromechanical energy conversion is
(∆S)body = ln   = –0.693 J/K
a/an 1
(a) irreversible process (b) reversible process 2
(c) isothermal process (d) None of the above For reservoir
Ans. (b) : The electromechanical energy conversion is Q 300
a reversible process. (∆S) Resb = = =1 J /K
T 300
87. The synchronous speed of a 3-phase induction (∆S) universe = ( ∆S )finitebody + ( ∆S )reservoir
motor having 12 poles and running on 50 Hz
supply is = 1– 0.693
(a) 1200 r.p.m. (b) 1000 r.p.m. = 0.3068 J/K
(c) 800 r.p.m. (d) 500 r.p.m. 90. A frictionless piston slowly compresses a gas in
Ans. (d) : Given, an adiabatic cylinder. The entropy change will
P = 12 poles be
f = 50 Hz (a) greater than zero
120f (b) less than zero
Ns = (c) equal to zero
P
120 × 50 (d) None of the above
Ns = = 500 r.p.m.
12 Ans. (c) When a frictionless piston slowly compressed
a gas in an adiabatic cylinder it approaches reversible
88. A liquid has surface tension σ. The minimum
adiabatic process which also known as isentropic
work required to divide a spherical drop of this
liquid of radius t into 8 equal-sized spherical process.
drops is So, for isentropic process
(a) πt2σ (b) 2πt2σ Change in Entropy (∆S) = 0
(c) 4πt2σ (d) 8πt2σ 91. A heat engine operates between 500 K and 300
Ans. (c) Given, K. The minimum heat absorption from the
source for every kilo joule of work is
4 3 4
πt = 8 × πr 3 (a) 1.5 kJ (b) 1.7 kJ
3 3
(c) 2.5 kJ (d) 3 kJ
t = 2r
Ans. (c) Given, T1 = 500 K, T2 = 300 K
Work done (W) For maximum efficiency of heat engine must be
surface tension (σ) =
Change in surface area (∆A) reversible so,
∆A = 8×4πr2 – 4πt2 For ηmax, Qs is minimum & Wnet is maximum
= 32πr2 – 4π×4×r2 We know that,
= 16πr2 = 4πt2 T
ηth = 1 − L
W TH
σ= ⇒ W = 4πt 2 σ
4πt 2 300
= 1− = 0.4
89. A metal block of heat capacity 1 J/K is cooled 500
from 600 K to 300 K by placing it in a large W
heat reservoir at 300 K. The entropy change of η = net = 0.4
Qs
the universe in this process is
(a) – 0.693 J/K (b) 1 J/K Wnet 1
( QS )min = = = 2.5 kJ
(c) –1.693 J/K (d) 0.307 J/K ( ηth ) 0.4
Ans. (d) Given, C = 1 J/k, T1 = 600 K, T2 = 300 K
92. A refrigerator maintains a temperature of 270
∆T = 600–300 = 300K K in a room at 300 K. If heat is removed from

BPSC AE (Pre), 16 Sep., 2018 20 YCT

 the interior at a rate of 900 J sec–1 and the (2) Boundary phenomenon
refrigerator operates at 50% of its maximum (3) Not a property
thermal efficiency, the power requirement is (4) Inexact differential
(a) 100 W (b) 150 W So heat and work transfer are path function.
(c) 200 W (d) 250 W 2

Ans. (*) Given, T2 = 270 K, T1 = 300 K

Note- ∫1
δQ = Q1− 2 = 1Q 2 ≠ Q 2 − Q1
2

(COP)R =
T2 T −T
& η= 1 2 ∫1
δW = W1−2 = 1W2 ≠ W2 − W1
T1 − T2 T1
95. Amorphous glass is expected to have zero value
T −T W of entropy at 0 K. The statement is
∴ 50% = 1 2 = net
T1 Q (a) true
(b) false
1
⇒ ηmax = (c) true if it is in the powder form
1 + (COP) R
(d) None of the above
Ans. (b) :
.
Nerst simon states that the entropy of a pure crystalline
substance is zero at absolute zero temperature (0 K).
So, for amorphous or non-crystalline substance it is
impossible to expected to have zero value of entropy at
If refrigerator operates on 50% of ηmax
0 K.
1 1
∴ = Amorphous glass is a solid in which there is no long-
2 1 + (COP)R range order of the positions of the atoms.
(COP)R = 1 All the bonds are not equally strong. These solids do
Desired effect (Q 2 ) not have a precise welding point.
=1 ex.– Rubbers, glass, plastic, cement & paraffin.
Work input (Win )
Win = Q2 = 900 J/sec = 900 kW
Win = 900 kW

93. Liquid water at 1 atmosphere and 00C freezes

to ice, transferring heat to the surroundings,
also at 00C. In this process.
(a) the entropy of the water decreases, but that of
the universe increases
(b) the entropy of water decreases, but that of the
universe remains constant
(c) the entropy of the water as well as that of the • at ok value of entropy not zero.
universe increase According to third law of thermodynamics–
(d) the entropy of the water increases, but that of It is impossible to achieve absolute 0 temperature in a
the universe decreases finite no. of process.
Ans. (b) : j 96. The efficiency of a reversible engine is
94. Q-W is a maximum and depends only on the
(a) path function temperature of the source and the sink. The
(b) state function statement is
(c) path as well as state function (a) correct
(b) wrong
(d) None of the above
(c) uncertain
(where q is specific heat transfer and w is specific
(d) correct if it is irreversible process
work done)
Ans. (a) : A reversible engine which has maximum
Ans. (a) : Heat (Q) and work (W) both are–
possible efficiency i.e. Carnot engine
(1) Path function
BPSC AE (Pre), 16 Sep., 2018 21 YCT
 T1 − T2 Labour etc.
ηmax =
T1
Efficiency of reversible engine is only the function of
temperature.
∴ ηmax = f (T1 ,T2 )

97. Heat and work are examples of

(a) thermodynamic properties
(b) states of thermodynamic systems
(c) mode of energy transfer Direct cost of an activity decreases with increase in
(d) None of the above duration.
Ans. (c) : Heat and work are two different ways of 101. Mean, median and the mode for the set of
from transferring energy between the system and values – 10, 9, 8, 10, 12, 9, 9, 10, 11, 14 and 8
surrounding. are
Heat is the transfer of thermal energy. (a) 10, 8, 14 (b) 10, 9, 9
Work is the transfer of mechanical energy. (c) 9, 10, 8 (d) 11, 9, 8
98. For an ideal gas, compressibility factor should Ans. (*) :
be 102. In case of PERT, if most pessimistic, optimistic
(a) 0 (b) 1 and most likely time are 10, 2 and 8 days
(c) –1 (d) close to 10 respectively, then the expected duration and
Ans. (b) : Compressibility Factor (z)– This factor variance are
represents the deviation of real gases from the (a) 8 and 4/3 (b) 20/3 and 16/9
behaviour of ideal gas. (c) 7.33 and 16/9 (d) 7.67 and 20/3
Mathematically, Ans. (c) : We know, expected duration
PV t o + 4t m + t p
Z= te =
RT 6
1 < Z < 1} for real gases Given, Optimistic time (to) = 2
Z = 1} for ideal gases most likely time (tm) = 8
99. The method which follows deterministic pessimistic time (tp) = 10
approach is tp − to
(a) CPM Standard deviation (σ) =
6
(b) PERT t o + 4t m + t p 2 + 4 × 8 + 10 44
(c) both PERT and CPM te = = =
6 6 6
(d) None of the above
= 7.33
Ans. (a) : CPM (Critical Path Method) 2
 t p − t o   10 − 2 
2
(a) Activity oriented Variance (V) = σ = 
2
 = 
(b) Deterministic approach  6   6 
(c) Normal distribution approach 4 4
8 × 8 16
(d) Suited to repetitive type of work. = =
6× 6 9
100. Direct cost of an activity 3 3

(a) increases with increase in duration 103. In case of cash-flow monitoring, it is

(b) decreases with increase in duration recommended to draw
(c) remains same (a) histogram
(d) Nothing can be said (b) cumulative diagram
Ans. (b) : Direct cost– Direct cost of the project are (c) bar chart
those expenses which are directly chargeable and can (d) homograph
be identified by activities. Ans. (b) :
ex.– Cost of Material 104. The total cost of a building is ` 3,00,000. The
Machine depreciated cost of the building after 30 years,

BPSC AE (Pre), 16 Sep., 2018 22 YCT

 if the life span is 90 years and scrap value is (a) 35 kg/cm2 (b) 50 kg/cm2
2
` 30,000, will be (by declining balance method) (c) 15 kg/cm (d) 20 kg/cm2
(a) ` 2,10,000 (b) ` 1,39,504 Ans. (a)
(c) ` 1,75,254 (d) ` 2,50,000 Crushing Strength of Brick
Ans. (b) : Declining Balance Method / Constant
Brick class Crushing Strength
Percentage Method
M/D 1st class > 105 kg/cm2
S
Book value after m years = C   (S ≠ 0) 2nd class > 70 kg/cm2
C
3rd class > 50 kg/cm2
C = original cost
Common Building Bricks > 35 kg/cm2
S = scrap value
30 109. The proportion of cement mortar used for 1
 30, 000  90 and 2 storeyed structure is
Book value after 30 years = 3,00,000  
 3, 00, 000  (a) 1 : 2 (b) 1 : 3
= 3,00000 × (0.1)1/3 (c) 1 : 6 (d) 1 : 1 : 2
= 3,00,000 × 0.464 Ans. (*) :
= 139247 110. The ingredient which imparts hardness and
105. An owner has installed an air conditioner at colour to cement is
the cost of ` 18,000. If the life of the conditioner (a) alkali (b) alumina
is 18 years, the coefficient of sinking fund (rate (c) magnesia (d) sulphur
of interest is 5%) is
(a) 0.055 (b) 0.0355 Ans. (c)
(c) 640 (d) 1.20 Constituents of Portland Cement
Ans. (b) Function Composition
(%)
r
coefficient of sinking fund (SC) = Lime Controls strength and 60–65
(1 + r) n −1
soundness
r = rate of interest 25% Silica gives strength 17–25
n = 18 years
Alumina responsible for quick 3–8
0.05 setting
=
(1 + 0.05 )
18
−1 Iron Oxide gives colour and helps 0.5–6
0.05 0.05 in fusion of different
= = ingredients
(1.05 ) 2.40 − 1
18
−1
Magnesia imports colour and 0.5–4
0.05 hardness
= = 0.035%
1.40 111. The compressive strength of the brick should
106. The average life of Class I timber is not be less than
(a) 60 months (b) 90 months (a) 3.5 MPa (b) 5 MPa
(c) 120 months (d) 150 months (c) 15 MPa (d) 20 MPa
Ans. (*) Class I–Natural durable heart wood timber Ans. (a) : Minimum compressive strength of brick is
having average life of 120 months or over. 35 kg/cm2 or 3.5 N/mm2 or 3.5 MPa for superior brick
Class II–Natural durable heart wood timber having it is very from 70 kg/cm2 to 140 kg/cm2.
average life of 60 months or over but less than 120
months. 112. Which one of the following is responsible for
red colour of brick?
Class III–Timber having average life less than 60
months. (a) Iron oxide (b) Magnesia
(c) Silica (d) Alumina
107. A good stone should have water absorption less
than Ans. (a) : Iron oxide– (<7%) improves
(a) 0.4 (b) 0.6 impermeability and durability.
(c) 0.8 (d) 0.9 → Lower fusion point of the clays especially if
Ans. (*) present as ferrous oxide.
→ Gives strength and hardness
108. The minimum crushing strength of brick
should be → Provides red colour.

BPSC AE (Pre), 16 Sep., 2018 23 YCT

 Magnesia (<1%)→ imparts yellow tint to the bricks (a) Light in weight
and decreases shrinkage. (b) Better soundproofing qualities
Silica (50–60%)– prevent cracking and shrinking of (c) Poor fire resistance
raw bricks. (d) Better thermal insulation
– Imports durability and uniform shape to bricks. Ans. (c) : Ribbed or Hollow tiled floor advantages–
Alumina (20–30%)– absorbs water and imports (a) Light in weight
plasticity to the earth. (b) Better thermal insulation
113. Enamel paint is prepared by adding (c) Better sound proofing qualities
(a) white lead or zinc (d) Better fire resistance
(e) Electrical, plumbing and other service can be
(b) alumina and zinc
conveniently installed through it without
(c) magnesia and alumina
affecting its appearance.
(d) white lead and alumina
118. Plywood is identified by
Ans. (a) : Enamel paints– is a mixture of base &
(a) thickness (b) volume
varnish.
(c) area (d) weight
– prepared by adding white lead or zinc.
Ans. (a) : Plywood– plywood is made by gluing
Babe– White lead, Red lead, lithophone, Aluminium
together thin sheets of wood.
powder.
Plywood thickness is the key specification of a
– These are acid resistant and water proof. plywood.
– Not affected by alkalis and gases. The thickness of an individual sheet is 0.2 – 3.2mm.
– Mainly used for wood works.
119. It is required to produce a small-scale map of
114. Pigments are added to an area in a magnetic zone by directly plotting
(a) give colour to paint and checking the work in the field itself. Which
(b) reduce the cost of the paint one of the following surveys will be most
(c) hold the ingredients of the paint appropriate for this purpose?
(d) make the paint thinner (a) Chain (b) Theodolite
(c) Plane table (d) Compass
Ans. (a) : Pigments– Pigments are added to give
colour to paint. Ans. (c) : Plane table– Instrument for survey by
graphical method in which field observation and
– Hide the surface irregularities.
plotting done simultaneously.
115. The base material of distemper is In this topographic details are mapped in full view
(a) iron oxide (b) lithopone hence no change of missing any details.
(c) chalk (d) lime use where greater accuracy is not required.
Ans. (c) : Distempers– It is a mixture of white chalk for small-medium scale map
(Base) & water (solvent) This graphical method of producing topographical
Base – white chalk maps – cartographic surveying.
Solvent – water 120. The technique of plotting all the accessible
116. In industrial building, hard wearing surface stations with a single setup of plane table is
can be achieved by called
(a) terrazzo flooring (a) radiation (b) intersection
(b) granolithic flooring (c) resection (d) traversing
(c) mosaic flooring Ans. (a) : Radiation– In this method instrument is
setup at station and rays are drawn to various stations
(d) tiled flooring
which area be plotted.
Ans. (b) : Granolithic Flooring– A mixture of river Suited– When survey area is small and all stations
sand and cement laid on wet or dry concrete subfloor. are clearly visible & accessible from instrument
– Suitable for heavy duty industrial floors, domestic station.
entrances, verandah & church floors. Intersection (Graphical triangulation)– used to
Advantages– It is cheap to construct hard wearing. plot/locate inaccessible point.
117. Which one of the following is not true with Traversing– When narrow strip of terrain surveyed
respect to ribbed tiled floors? (survey of road, railways)

BPSC AE (Pre), 16 Sep., 2018 24 YCT

 Resection– In this method orientation is used when 0.7%. The rate of change of grade is 0.05 per
the plane tables occupies a position not yet plotted chain. The length of the vertical curve is
on drawing sheet. (a) 30 chains (b) 40 chains
121. A 30 m chain is found to be 0.1 m short (c) 50 chains (d) 60 chains
throughout the measurement. If the distance Ans. (a)
measured is recorded as 300 m, then the actual g1 = + 0.80%
distance will be
g2 = – 0.70%
(a) 300.1 m (b) 301.0 m
Change in gradient N = g1–g2
(c) 299 m (d) 310.0 m
= 0.80 – (–0.70)
Ans. (c) : True length of the chain L = 30m
= 1.50%
correct length of the chain L1 = 30–0.1 = 29.9m
Since the rate of change of grade is 0.05% per chain.
measured length of the survey line l1 = 300m
∴ The length of vertical curve
L'
True or actual length = Measured length × 1
L Ls = × 1.5 = 30 chain
0.05
10
29.9
= 300 × 125. An angle measuring instrument reading up to
30 one-sixth of a degree on the main scale is
= 299m equipped with a vernier having 19 main scale
122. Offsets are divisions divided into 20 parts. The correct
(a) lateral measurements made with respect to least count for the instrument is
main survey lines (a) 60 seconds (b) 30 seconds
(b) perpendicular erected from chain lines (c) 20 seconds (d) 10 seconds
(c) taken to avoid unnecessary walking between Ans. (b) : Least count of a scale is
stations
x−y
(d) measurements which are not made at right LC = MS
x
angles to the chain line
Ans. (b) : Offsets– Lateral measurement from chain given x = 20
line. y = 19
(may be perpendicular or oblique) MS = least value measured by main scale
123. If fore-bearing of a line is S49052'E, then the 1
MS = degree
back bearing will be 6
0 0
(a) S 49 52' E (b) S 52 49' E 20 − 19 1 1
LC = × = degree
(c) N 49008' E (d) N 49052' W 20 6 120
Ans. (d) : Fore & Back Bearing 10 = 60 min = 60 × 60 = 3600 sec.
Fore Bearing (FB) = magnetic bearing of follower 1
30

view Lc = × 360 0 sec

12 0
Back Bearing (BB) = magnetic bearing of leader view
In W.C.B. System (F.B–B.B = 1800) = 30 sec.
In Q.B System (F.B = NOE ⇒ B.B = SOW 126. For a simple circular curve, which one of the
i.e. N & S interchange following gives the correct relation between the
E & W interchange radius R and degree of curve D, for 20 m are
length?
(a) R = 5729.6/D (b) R = 1718.9/D
(c) R = 1145.9/D (d) R = 572.9/D
Ans. (c) : For 20m
1145.9 1146
R= ≈
D D
Given,
For 30m
F.B = S490 52 ' E
↓ ↓ interchange 1720
R=
B.B = N490 52 ' W D
124. In a parabolic vertical curve, the rising grade 127. The radius of curvature of an ideal transition
g1 = +0.8% and the falling gradient g2 = – curve should be
BPSC AE (Pre), 16 Sep., 2018 25 YCT
 (a) inversely proportional to its length when the lighthouse whose height is 100m is
(b) directly proportional to its length visible just above the horizon from the ship?
(c) proportional to speed of vehicle (a) 30.68 km (b) 36.50 km
(d) proportional to super-elevation (c) 38.54 km (d) 40.54 km
Ans. (a) : Curve should be inversely proportional to its Ans. (c) : Distance of visible horizon
length.
d=3.85 h
πRθ
l= h = height of light
1800
Given,
l = length of curve h = 100m
R = radius of curve d = 3.85 100
1 = 3.85 × 10
= radius of curvature
R
= 38.54 km
1 πθ 1
= × 131. To find the RL of a roof slab of a building, staff
R 180 l
readings were taken from a particular setup of
1 the levelling instrument. The readings were
radius or curvature α
l 1.050 m with staff on the benchmark and 2.300
128. If the difference of height between two points is m with staff below the roof slab and held
1m and the slope distance between them is inverted. Taking the RL of the BM as
100m, then the accuracy of slope correction 135.150m, the RL of the roof slab will be
determination could be 1 in 100000 provided (a) 129.800 (b) 131.900
the heights are measured with an accuracy of (c) 134.400 (d) 138.500
(a) ± 0.1 cm (b) ± 0.5 cm Ans. (d)
(c) ± 1.0 cm (d) ± 5.0 cm R.L of roof slab = R.L of BM + staff reading
Ans. (a) : with staff om BM + (interveed
129. A and B are two traverse stations free from staff reading)
local attraction errors. If the true bearing of a
line AB is 890 and the magnetic declination at R.L of Bench mark (BM ) = 135.15m
point A is 10 west, then the magnetic bearing of Staff reading on B.M = 1.050m
line BA would be inverted staff reading = 2.300m
0 0
(a) 88 (b) 90 R.L of roof slab = 135.15+1.050+300
(c) 2680 (d) 2700 = 138.500m
Ans. (d) 132. For the scale of plotting 1 in 400, the
Magnetic Bearing permissible error in centring of plane table is
M.B = T.B + declination of AB about
(a) 0.5 m (b) 0.3 m
= 890 + 10
= 900 (c) 0.1 m (d) 0.01
MB of BA = 90 + 180 = 2700 Ans. (c) :
133. Ceylon Ghat Tracer is used to measure
(a) slope (b) reduced levels
(c) distances (d) depth of sea
Ans. (a)
measure by
slope – Ceylon Ghat-tracer
depth of sea – Fathometer
134. BOD test is standardized at
Declination = + (if west) (a) 10 0C and 10 days
= – (east)
(b) 20 0C and 5 days
130. Which one of the following gives the correct (c) 37 0C and 3 days
distance between the lighthouse and a ship, (d) 50 0C and 2 days
BPSC AE (Pre), 16 Sep., 2018 26 YCT
 Ans. (b) : The BOD can be determined by directing a (b) high nitrate content in water
known volume of a sample of waste water with a (c) high chloride content in water
known volume of aerated pure water and then (d) high iron content in water
calculating the DO of this diluted sample. The diluted Ans. (b) : The presence of too much of nitrate in water
sample is then incubated for 5-days at 200C. may adversely affect the health of infants, causing a
Standard 5-day BOD at 200C disease technically called "Mathemog lobineming"
( BoD )5 = ( BOD )4 (1 − 10k D ×5
) commonly called Blue baby disease.
139. The major constituent which causes alkalinity
135. Absolutely soft waters are required for in water is
(a) drinking
(a) dissolved O2
(b) boilers
(b) dissolved NH3
(c) washing with synthetic detergent soap
(c) dissolved CO2
(d) prevention of corrosion in pipe
(d) All of the above
Ans. (b) : Soft waters are required for Boilers.
Ans. (b) : Alkalinity– Alkalinity is defined as quantity
A water softner will reduce the risk of hard scaling
of ions in water that will react to neutralize hydrogen
inside the Boiler, improve the flow of water, ensure
ions (H+). Alkalinity is a measure of the ability of
maximum levels of efficiency and help your Boiler
avoid failures or any kind of damage that could reduce water to neutralize acids.
its lifespan. Most of common constituents of alkalinity are
→ it is important to supply your boiler with soft water CO32 , HCO3− , OH − & dissolved NH3.
at all times to avoid damage and costly repairs. 140. Sedimentation process is based on which of the
136. Permanent hardness of water is because of following physical laws?
(a) CaHCO3 (b) NaHCO3 (a) Newton's third law
(c) MgHCO3 (d) CaSO4 (b) Conservation of mass
Ans. (d) : Permanent Hardness– sulphate, chlorides (c) Stokes' law
and nitrates of Ca and Mg cause permanent hardness. (d) Conservation of energys
It is also called non Carbonate Hardness.
Ans. (c) : Sedimentation is a natural process by which
CaSO4, MgSO4, Cacl2 etc.
solids with higher density than the fluid in which they
Temporary Hardness or Carbonate Hardness– the
are suspended, settle under the action of gravity.
presence of carbonate and carbonate of calcium &
efficiency of ideal sedimentation tank
magnesium.
Vs
137. Zeolite process is used η= × 100
Vo
(a) for disinfection of water
Vs – settling velocity of discrete particle can be
(b) for colour removal from water
calculated by using stoke's law.
(c) for water softening
(d) for turbidity removal ( γs − γ w ) d2
Vs =
Ans. (c) : Softening of water–The reduction or 18µ
removal of hardness from water. 141. Fine sand is used as media in case of
(a) Removal of Temporary Hardness– Sedimentation (a) slow sand filter
by forming insoluble precipitations.
(b) rapid sand filter
(i) Boiling
(c) pressure filter
(ii) Addition of lime
(d) All of the above
(b) Removal of Permanent Hardness–
Ans. (a) : Slow sand filter has less D10 value so it
(i) lime-soda process
require finer sand.
(ii) zeolite or base exchange
Fine sand will help in better cleaning of water.
(c) Zeolite process is remove of permanent hardness
for water softening. Effective size D10 for slow sand filter is 0.2–0.3mm.
138. Blue baby disease results with 142. Sullage is
(a) high fluoride content in water (a) waste water from baths

BPSC AE (Pre), 16 Sep., 2018 27 YCT

 (b) drainage from road (b) ammonia
(c) industrial liquid waste (c) Both (a) and (b)
(d) All of the above (d) None of the above
Ans. (a) : Sullage– Combination of kitchen & Ans. (a) : Sludge digestion involves the treatment of
washroom waste is known as sullage. it does not highly concentrated organic wastes in the absence of
contain human excreta it does not create bad smell as oxygen by anaerobic bacteria. The organic matter
organic master content is less or in negligible amount. present in the primary and secondary sludge is
143. The end product of decomposed organic matter biologically converted to stable and products like.
is Methane (H4) & CO2
(a) CO2 (b) H2S So,
(c) NO3 (d) NH3 Methane is the gas mainly evolves in a anaerobic
Ans. (a) : Organic matter can be decomposed by sludge digestion.
anaerobic decomposition. 148. Self-purification of water body is mainly due to
For anaerobic respiration NO3− ,Sa12 , CO 2− or fumarate (a) dissolved O2
can serve as terminal electron acceptors, depending on (b) dissolved NO3
the bacterium studied. The end result of the respiratory (c) Both (a) and (b)
process is the complete oxidation of the organic
(d) None of the above
substrate molecule and the end products formed are
primarily CO2&H2O. Ans. (a) : The self-purification process of water bodies
or processes involving biological, chemical and
144. Grit is
physical processes working simultaneously on
(a) inert matter of specific gravity > 2.65
biological pollutants oxidicing them and increasing the
(b) organic matter of specific gravity 1
amount of dissolved oxygen.
(c) organic and inert matter combined
149. When bleaching powder is added to water, its
(d) colloidal matter of heavy specific gravity
pH value
Ans. (a) : Grit is the process to remove sand, silt from
(a) increases
water.
(b) decreases
Grit is often found in the head works of waste water
(c) remains unaffected
treatment plants.
(d) depends on characteristics of water
In Grit inert matter of specific gravity > 2.65
Ans. (a) : Bleaching powder reacts with water to form
145. Activated sludge process is a biological process Calcium Hydroxide or Ca(OH)2 also called slaked lime
involving & Hypochlorous acid (Hull) slaked lime in basic
(a) aerobic + anaerobic bacteria nature (pH = 12.8) concentrated Bleaching powder has
(b) aerobic bacteria + protozoa + algae a pH of 13 which makes its highly alkaline.
(c) anaerobic bacteria + fungi 150. If the total hardness of water is greater than its
(d) facultative bacteria + algae alkalinity, the carbonate hardness will be equal
Ans. (b) : The activated sludge process is the to
biological process by which non-settleable substances (a) total alkalinity
occurring in dissolved and colloidal forms are (b) total hardness
converted into settleable sludge which is removed (c) total hardness – total alkalinity
from the liquid carrier. (d) non-carbonate hardness
146. Sludge digestion is Ans. (a) : Carbonate Hardness =
(a) disposal of sludge  total or alkalin 
min or less  
(b) dilution of sludge Hardness 
(c) stabilization of sludge
(d) removal of sludge from waste CH = minimum of {TH, alkalinity}
Ans. (c) : The sludge withdrawn from the given, TH > alkalinity
sedimentation basins contains lot of putrescible matter Carbonate Hardness = alkalining
and it is stabilized by decomposing the organic master Non Carbonate Hardness (NCH)
under controlled anaerobic conditions. The process of
stabilization is called the sludge digestion. NCH = TH-CH
147. Anaerobic sludge digestion mainly yields
(a) methane
BPSC AE (Pre), 16 Sep., 2018 28 YCT

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