The document contains 17 physics questions in a multiple choice format with single correct answers. Each question is followed by its corresponding solution. The questions cover topics in mechanics, thermodynamics, gravitation, and other core areas of physics.
Formulation and Evaluation of Liquid Shampoo.pdfRAHUL PAL
Shampoo is a preparation of a surfactant (i.e., surface active material) in a suitable from liquid, solid
or powder-which when used under the specified condition will remove surface grease, dirt, skin debris
from the hair shaft and scalp without adversely affecting the user.
Surfactant are the main components of shampoo preparation. Mainly anionic surfactant is used. A
surfactant consists of two-part are hydrophilic (water loving) while the other is hydrophobic in nature.
Shampoo for oily hair can been exactly the same detergent at the same concentration as shampoo for
dry hair.
Microbiological assay-Principles and methods of different microbiological assay.someshwar mankar
Principles and methods of different microbiological assay. Methods for standardization of
antibiotics, vitamins and amino acids. Assessment of a new antibiotic.
Disinfectant and antiseptic is used for kill the microbes or inhibit the growth of microbes and decreasing their numbers in such a low level that they become unable to impart any harmful effect.
This document discusses the four major types of adult tissues - epithelial, connective, muscle and nervous tissue. It provides details on the classification, structure and functions of each type of tissue. The key points are:
1) Tissues are classified based on structure, composition and function. The four major types are epithelial, connective, muscle and nervous tissue.
2) Epithelial tissues cover surfaces, line organs and form glands. They protect, absorb, secrete and transport. Connective tissues connect, support and protect other tissues. Muscle tissues contract to cause movement and pumping blood. Nervous tissues transmit electrical signals.
3) Each tissue contains different cell types and extracellular matrix to suit their functions.
This document contains 9 physics problems involving concepts like dimensional analysis, kinematics, forces, work, energy, and momentum.
Problem 1 asks students to determine the dimensional formula for surface tension using fundamental quantities of energy, velocity, and time. Problem 2 involves calculating the time taken for two ships moving towards each other to reach their shortest distance apart. Problem 3 asks students to determine the acceleration of a particle given its velocity varies with position.
The remaining problems involve calculating forces, work, energy, momentum, and velocities in various mechanical systems involving blocks, springs, and particles. Students must apply physics equations like Newton's laws, work-energy theorem, and conservation of momentum to arrive at the solutions.
1. The document contains 38 multiple choice questions related to physics concepts. The questions cover topics like dimensions, motion with constant power, work-energy theorem, forces, simple harmonic motion, capacitors, electromagnetism, and radiation.
2. The questions range from calculations involving kinematics, forces, energy, electromagnetism to conceptual questions about properties of waves, radiation, circuits and electromagnetism.
3. The multiple choice options provide quantitative or conceptual answers to problems formulated in the questions relating to various physics topics.
1. The document contains 38 multiple choice questions related to physics concepts. The questions cover topics like dimensions, motion with constant power, work-energy theorem, forces, simple harmonic motion, capacitors, electromagnetism, and radiation.
2. The questions range from calculations involving kinematics, forces, energy, electromagnetism to conceptual questions about properties of waves, radiation, circuits and more.
3. The document tests understanding of fundamental physics principles as well as ability to set up and solve quantitative problems across various domains of physics.
Formulation and Evaluation of Liquid Shampoo.pdfRAHUL PAL
Shampoo is a preparation of a surfactant (i.e., surface active material) in a suitable from liquid, solid
or powder-which when used under the specified condition will remove surface grease, dirt, skin debris
from the hair shaft and scalp without adversely affecting the user.
Surfactant are the main components of shampoo preparation. Mainly anionic surfactant is used. A
surfactant consists of two-part are hydrophilic (water loving) while the other is hydrophobic in nature.
Shampoo for oily hair can been exactly the same detergent at the same concentration as shampoo for
dry hair.
Microbiological assay-Principles and methods of different microbiological assay.someshwar mankar
Principles and methods of different microbiological assay. Methods for standardization of
antibiotics, vitamins and amino acids. Assessment of a new antibiotic.
Disinfectant and antiseptic is used for kill the microbes or inhibit the growth of microbes and decreasing their numbers in such a low level that they become unable to impart any harmful effect.
This document discusses the four major types of adult tissues - epithelial, connective, muscle and nervous tissue. It provides details on the classification, structure and functions of each type of tissue. The key points are:
1) Tissues are classified based on structure, composition and function. The four major types are epithelial, connective, muscle and nervous tissue.
2) Epithelial tissues cover surfaces, line organs and form glands. They protect, absorb, secrete and transport. Connective tissues connect, support and protect other tissues. Muscle tissues contract to cause movement and pumping blood. Nervous tissues transmit electrical signals.
3) Each tissue contains different cell types and extracellular matrix to suit their functions.
This document contains 9 physics problems involving concepts like dimensional analysis, kinematics, forces, work, energy, and momentum.
Problem 1 asks students to determine the dimensional formula for surface tension using fundamental quantities of energy, velocity, and time. Problem 2 involves calculating the time taken for two ships moving towards each other to reach their shortest distance apart. Problem 3 asks students to determine the acceleration of a particle given its velocity varies with position.
The remaining problems involve calculating forces, work, energy, momentum, and velocities in various mechanical systems involving blocks, springs, and particles. Students must apply physics equations like Newton's laws, work-energy theorem, and conservation of momentum to arrive at the solutions.
1. The document contains 38 multiple choice questions related to physics concepts. The questions cover topics like dimensions, motion with constant power, work-energy theorem, forces, simple harmonic motion, capacitors, electromagnetism, and radiation.
2. The questions range from calculations involving kinematics, forces, energy, electromagnetism to conceptual questions about properties of waves, radiation, circuits and electromagnetism.
3. The multiple choice options provide quantitative or conceptual answers to problems formulated in the questions relating to various physics topics.
1. The document contains 38 multiple choice questions related to physics concepts. The questions cover topics like dimensions, motion with constant power, work-energy theorem, forces, simple harmonic motion, capacitors, electromagnetism, and radiation.
2. The questions range from calculations involving kinematics, forces, energy, electromagnetism to conceptual questions about properties of waves, radiation, circuits and more.
3. The document tests understanding of fundamental physics principles as well as ability to set up and solve quantitative problems across various domains of physics.
AIPMST (Secondary) is Releasing Sample Question paper for study MBBS & BDS Aspirants.
AIPMST (Secondary) consists of four sections - Physics, Chemistry, Biology and English Proficiency & Logical Reasoning. There will be 50 questions in each section and each question carrying four marks. The questions will be of objective type for medical courses (MBBS, BDS). Candidates must use BLACK BALL POINT PEN only for filling up the OMR sheet. The mode of examination will be offline and the medium of examination will be English.
for more info:https://aipmstsecondary.co.in/
1. The document contains 38 multiple choice questions related to physics concepts. The questions cover topics like dimensions, kinematics, forces, energy, rotational motion, gravitation, fluids, thermodynamics, oscillations, waves, optics, electricity, and modern physics.
2. The questions test understanding of concepts like work, power, translational and rotational motion, gravitational potential, terminal velocity, moment of inertia, escape velocity, kinetic energy, heat radiation, electromagnetic waves, capacitance, magnetic field, electromagnetic induction, and blackbody radiation.
3. The multiple choice options provide quantitative answers involving calculations related to the various physics equations and concepts being tested in each question.
NEET 2016 Previous Year Question PaperStudMonkNEET
The document contains a 35 question physics exam from 2016 with multiple choice answers. It covers topics in mechanics, electricity, waves, thermodynamics and modern physics. The questions test conceptual understanding of topics like moment of inertia, electromagnetic waves, sound waves, capacitors, diffraction, gravitation, quantum mechanics and more.
Dear Students/Parents
We at 'Apex Institute' are committed to provide our students best quality education with ethics. Moving in this direction, we have decided that unlike other expensive and 5star facility type institutes who are huge investors and advertisers, we shall not invest huge amount of money in advertisements. It shall rather be invested on the betterment, enhancement of quality and resources at our center.
We are just looking forward to have 'word-of-mouth' publicity instead. Because, there is only a satisfied student and his/her parents can judge an institute's quality and it's faculty members coaching.
Those coaching institutes, who are investing highly on advertisements, are actually, wasting their money on it, in a sense. Rather, the money should be invested on highly experienced faculty members and on teaching gears.
We all at 'Apex' are taking this initiative to improve the quality of education along-with each student's development and growth.
Committed to excellence...
With best wishes.
S . Iqbal
( Motivator & Mentor)
This document contains a 50 question physics exam with multiple choice answers. The questions cover topics in physics including X-rays, electron motion in electric and magnetic fields, wavelength of electrons, projectile motion, optics, atomic structure, electricity, magnetism, and more. For each question there are typically 4 possible answers labeled a, b, c, or d.
This document contains 57 physics questions from an AIEEE past paper from 2004. Each question has 4 multiple choice answers. The questions cover topics in mechanics, properties of matter, heat and thermodynamics, waves and sound, electricity and magnetism, optics and modern physics.
1) This document contains a past physics exam with 33 multiple choice questions covering topics like kinematics, thermodynamics, optics, electricity and magnetism.
2) The questions test conceptual understanding of physics principles like SHM, collisions, thermal expansion, interference, electric potential, circuits and magnetic fields.
3) The format is single answer multiple choice with the correct answer indicated as a, b, c or d for each question.
1) The document describes the structure and scoring guidelines for a physics exam with two sections: Section 1 on physics concepts and Section 2 on single-digit answers. Section 1 contains 7 multiple choice questions with options to select one or more answers, while Section 2 contains 5 questions each with a single digit answer.
2) The exam also includes a Section 3 on matching questions containing two tables and 3 associated questions to test matching concepts to options.
3) Answers are provided for the sample questions, with the correct options indicated for Section 1 questions and single digit answers for Section 2 questions.
NEET Previous Year Question Paper | NEET 2016 Phase 2StudMonkNEET
This document contains the details of a 180 question exam for NEET (UG) Phase-II 2016, including important instructions, sample questions, and an answer key. The exam is 3 hours long and tests concepts in physics and biology. Correct answers receive 4 marks each, while incorrect answers deduct 1 mark. The maximum total score is 720 marks. The document provides examples of multiple choice questions on topics like mechanics, electricity and magnetism, waves, and thermodynamics.
1) Maxwell's equations are used to derive the wave equation for electromagnetic waves traveling in the x direction.
2) The electric and magnetic fields are shown to oscillate perpendicular to each other and the direction of propagation at the speed of light c.
3) The intensity of the electromagnetic wave, which determines its brightness or loudness, is directly proportional to the square of the amplitudes of the electric and magnetic fields.
This document contains a 50 question physics exam with multiple choice answers for each question covering topics like: sound, forces, electromagnetism, optics, thermodynamics, mechanics, and modern physics. The questions test understanding of concepts like speed of sound, types of forces, properties of inductors and capacitors, angles in geomagnetism, surface tension, photon properties, rigidity of liquids, radio wave propagation, relationships between charge, distance and force, SI units, nuclear reactions, induced charge, weightlessness, lens combinations, mercury drop potentials, gas properties, semiconductors, induced emf, gravitational acceleration on other planets, induced emf, gravitational acceleration and location, transistor circuits, diode applications, ion
This document contains information about JEE Advanced 2015 Paper 2 Code 3 for Physics. It includes 10 multiple choice questions testing concepts in physics. The questions cover topics such as optics, circuits, quantum mechanics, mechanics, thermodynamics and electromagnetism. For each question, students had to select the single correct answer ranging from 0 to 9. The document also provides two additional sections with more complex multi-concept questions, some requiring selecting one or more answers. The questions test a range of fundamental physics principles and problem solving abilities.
Jee advanced 2015 paper 1 code 1 final Pradeep Kumar
1. The document provides information about JEE Advanced 2015 paper 1, including 8 multiple choice questions in Section 1 and 10 multiple choice questions in Section 2.
2. Section 3 contains 2 matching questions matching concepts in Column I to statements in Column II.
3. The questions cover topics in physics including electromagnetism, quantum mechanics, thermodynamics, and nuclear physics.
The document contains 14 chemistry questions and their solutions. The questions cover topics like organic reactions, estimation of nitrogen content, precipitation of salts, uses of additives in soaps, properties of metal nitrates, bond orders, atomic radii trends, activation energy determination, isoelectronic and isostructural species, organic compound identification based on tests, and magnetic moments. The solutions provided explain the reasoning for the answers in 2-3 sentences each.
Aipmt 2015 question paper code e revisedPradeep Kumar
The document discusses the benefits of exercise for mental health. Regular physical activity can help reduce anxiety and depression and improve mood and cognitive functioning. Exercise causes chemical changes in the brain that may help boost feelings of calmness, happiness and focus.
Full jee mains 2015 online paper 10th april finalPradeep Kumar
The document provides solutions to 13 physics problems from a JEE Mains exam. Some key details:
- Problem 1 asks about the relationship between average force on container walls and temperature in an ideal gas, with the answer being directly proportional.
- Problem 2 asks why electrons diffuse from the n-region to the p-region in an unbiased n-p junction, with the answer being due to a higher electron concentration in the n-region.
- Problem 3 involves calculating the voltage reading of two batteries connected in parallel, with the answer being 13.1V.
- The remaining problems cover topics in physics including de Broglie wavelength, drift velocity, momentum, trajectories, moments of inertia
Tentative flow chart of councilling procedure 26.05Pradeep Kumar
1. The document outlines the tentative admission procedure for CSAB 2014, including 4 rounds of online seat allocation and options for candidates between July 11 and July 26, as well as spot round allocation on August 4.
2. It details the process for candidates to register, choose options for upgrading seats between rounds, pay initial fees, and report to institutions to complete admission if allocated a seat in any given round.
3. After the fourth round allocation on July 26, any candidates who have not secured admission can register for the spot round occurring August 4, with final admissions taking place between August 4-7 and classes starting August 8.
AIPMST (Secondary) is Releasing Sample Question paper for study MBBS & BDS Aspirants.
AIPMST (Secondary) consists of four sections - Physics, Chemistry, Biology and English Proficiency & Logical Reasoning. There will be 50 questions in each section and each question carrying four marks. The questions will be of objective type for medical courses (MBBS, BDS). Candidates must use BLACK BALL POINT PEN only for filling up the OMR sheet. The mode of examination will be offline and the medium of examination will be English.
for more info:https://aipmstsecondary.co.in/
1. The document contains 38 multiple choice questions related to physics concepts. The questions cover topics like dimensions, kinematics, forces, energy, rotational motion, gravitation, fluids, thermodynamics, oscillations, waves, optics, electricity, and modern physics.
2. The questions test understanding of concepts like work, power, translational and rotational motion, gravitational potential, terminal velocity, moment of inertia, escape velocity, kinetic energy, heat radiation, electromagnetic waves, capacitance, magnetic field, electromagnetic induction, and blackbody radiation.
3. The multiple choice options provide quantitative answers involving calculations related to the various physics equations and concepts being tested in each question.
NEET 2016 Previous Year Question PaperStudMonkNEET
The document contains a 35 question physics exam from 2016 with multiple choice answers. It covers topics in mechanics, electricity, waves, thermodynamics and modern physics. The questions test conceptual understanding of topics like moment of inertia, electromagnetic waves, sound waves, capacitors, diffraction, gravitation, quantum mechanics and more.
Dear Students/Parents
We at 'Apex Institute' are committed to provide our students best quality education with ethics. Moving in this direction, we have decided that unlike other expensive and 5star facility type institutes who are huge investors and advertisers, we shall not invest huge amount of money in advertisements. It shall rather be invested on the betterment, enhancement of quality and resources at our center.
We are just looking forward to have 'word-of-mouth' publicity instead. Because, there is only a satisfied student and his/her parents can judge an institute's quality and it's faculty members coaching.
Those coaching institutes, who are investing highly on advertisements, are actually, wasting their money on it, in a sense. Rather, the money should be invested on highly experienced faculty members and on teaching gears.
We all at 'Apex' are taking this initiative to improve the quality of education along-with each student's development and growth.
Committed to excellence...
With best wishes.
S . Iqbal
( Motivator & Mentor)
This document contains a 50 question physics exam with multiple choice answers. The questions cover topics in physics including X-rays, electron motion in electric and magnetic fields, wavelength of electrons, projectile motion, optics, atomic structure, electricity, magnetism, and more. For each question there are typically 4 possible answers labeled a, b, c, or d.
This document contains 57 physics questions from an AIEEE past paper from 2004. Each question has 4 multiple choice answers. The questions cover topics in mechanics, properties of matter, heat and thermodynamics, waves and sound, electricity and magnetism, optics and modern physics.
1) This document contains a past physics exam with 33 multiple choice questions covering topics like kinematics, thermodynamics, optics, electricity and magnetism.
2) The questions test conceptual understanding of physics principles like SHM, collisions, thermal expansion, interference, electric potential, circuits and magnetic fields.
3) The format is single answer multiple choice with the correct answer indicated as a, b, c or d for each question.
1) The document describes the structure and scoring guidelines for a physics exam with two sections: Section 1 on physics concepts and Section 2 on single-digit answers. Section 1 contains 7 multiple choice questions with options to select one or more answers, while Section 2 contains 5 questions each with a single digit answer.
2) The exam also includes a Section 3 on matching questions containing two tables and 3 associated questions to test matching concepts to options.
3) Answers are provided for the sample questions, with the correct options indicated for Section 1 questions and single digit answers for Section 2 questions.
NEET Previous Year Question Paper | NEET 2016 Phase 2StudMonkNEET
This document contains the details of a 180 question exam for NEET (UG) Phase-II 2016, including important instructions, sample questions, and an answer key. The exam is 3 hours long and tests concepts in physics and biology. Correct answers receive 4 marks each, while incorrect answers deduct 1 mark. The maximum total score is 720 marks. The document provides examples of multiple choice questions on topics like mechanics, electricity and magnetism, waves, and thermodynamics.
1) Maxwell's equations are used to derive the wave equation for electromagnetic waves traveling in the x direction.
2) The electric and magnetic fields are shown to oscillate perpendicular to each other and the direction of propagation at the speed of light c.
3) The intensity of the electromagnetic wave, which determines its brightness or loudness, is directly proportional to the square of the amplitudes of the electric and magnetic fields.
This document contains a 50 question physics exam with multiple choice answers for each question covering topics like: sound, forces, electromagnetism, optics, thermodynamics, mechanics, and modern physics. The questions test understanding of concepts like speed of sound, types of forces, properties of inductors and capacitors, angles in geomagnetism, surface tension, photon properties, rigidity of liquids, radio wave propagation, relationships between charge, distance and force, SI units, nuclear reactions, induced charge, weightlessness, lens combinations, mercury drop potentials, gas properties, semiconductors, induced emf, gravitational acceleration on other planets, induced emf, gravitational acceleration and location, transistor circuits, diode applications, ion
This document contains information about JEE Advanced 2015 Paper 2 Code 3 for Physics. It includes 10 multiple choice questions testing concepts in physics. The questions cover topics such as optics, circuits, quantum mechanics, mechanics, thermodynamics and electromagnetism. For each question, students had to select the single correct answer ranging from 0 to 9. The document also provides two additional sections with more complex multi-concept questions, some requiring selecting one or more answers. The questions test a range of fundamental physics principles and problem solving abilities.
Jee advanced 2015 paper 1 code 1 final Pradeep Kumar
1. The document provides information about JEE Advanced 2015 paper 1, including 8 multiple choice questions in Section 1 and 10 multiple choice questions in Section 2.
2. Section 3 contains 2 matching questions matching concepts in Column I to statements in Column II.
3. The questions cover topics in physics including electromagnetism, quantum mechanics, thermodynamics, and nuclear physics.
The document contains 14 chemistry questions and their solutions. The questions cover topics like organic reactions, estimation of nitrogen content, precipitation of salts, uses of additives in soaps, properties of metal nitrates, bond orders, atomic radii trends, activation energy determination, isoelectronic and isostructural species, organic compound identification based on tests, and magnetic moments. The solutions provided explain the reasoning for the answers in 2-3 sentences each.
Aipmt 2015 question paper code e revisedPradeep Kumar
The document discusses the benefits of exercise for mental health. Regular physical activity can help reduce anxiety and depression and improve mood and cognitive functioning. Exercise causes chemical changes in the brain that may help boost feelings of calmness, happiness and focus.
Full jee mains 2015 online paper 10th april finalPradeep Kumar
The document provides solutions to 13 physics problems from a JEE Mains exam. Some key details:
- Problem 1 asks about the relationship between average force on container walls and temperature in an ideal gas, with the answer being directly proportional.
- Problem 2 asks why electrons diffuse from the n-region to the p-region in an unbiased n-p junction, with the answer being due to a higher electron concentration in the n-region.
- Problem 3 involves calculating the voltage reading of two batteries connected in parallel, with the answer being 13.1V.
- The remaining problems cover topics in physics including de Broglie wavelength, drift velocity, momentum, trajectories, moments of inertia
Tentative flow chart of councilling procedure 26.05Pradeep Kumar
1. The document outlines the tentative admission procedure for CSAB 2014, including 4 rounds of online seat allocation and options for candidates between July 11 and July 26, as well as spot round allocation on August 4.
2. It details the process for candidates to register, choose options for upgrading seats between rounds, pay initial fees, and report to institutions to complete admission if allocated a seat in any given round.
3. After the fourth round allocation on July 26, any candidates who have not secured admission can register for the spot round occurring August 4, with final admissions taking place between August 4-7 and classes starting August 8.
Temple of Asclepius in Thrace. Excavation resultsKrassimira Luka
The temple and the sanctuary around were dedicated to Asklepios Zmidrenus. This name has been known since 1875 when an inscription dedicated to him was discovered in Rome. The inscription is dated in 227 AD and was left by soldiers originating from the city of Philippopolis (modern Plovdiv).
Gender and Mental Health - Counselling and Family Therapy Applications and In...PsychoTech Services
A proprietary approach developed by bringing together the best of learning theories from Psychology, design principles from the world of visualization, and pedagogical methods from over a decade of training experience, that enables you to: Learn better, faster!
Main Java[All of the Base Concepts}.docxadhitya5119
This is part 1 of my Java Learning Journey. This Contains Custom methods, classes, constructors, packages, multithreading , try- catch block, finally block and more.
Philippine Edukasyong Pantahanan at Pangkabuhayan (EPP) CurriculumMJDuyan
(𝐓𝐋𝐄 𝟏𝟎𝟎) (𝐋𝐞𝐬𝐬𝐨𝐧 𝟏)-𝐏𝐫𝐞𝐥𝐢𝐦𝐬
𝐃𝐢𝐬𝐜𝐮𝐬𝐬 𝐭𝐡𝐞 𝐄𝐏𝐏 𝐂𝐮𝐫𝐫𝐢𝐜𝐮𝐥𝐮𝐦 𝐢𝐧 𝐭𝐡𝐞 𝐏𝐡𝐢𝐥𝐢𝐩𝐩𝐢𝐧𝐞𝐬:
- Understand the goals and objectives of the Edukasyong Pantahanan at Pangkabuhayan (EPP) curriculum, recognizing its importance in fostering practical life skills and values among students. Students will also be able to identify the key components and subjects covered, such as agriculture, home economics, industrial arts, and information and communication technology.
𝐄𝐱𝐩𝐥𝐚𝐢𝐧 𝐭𝐡𝐞 𝐍𝐚𝐭𝐮𝐫𝐞 𝐚𝐧𝐝 𝐒𝐜𝐨𝐩𝐞 𝐨𝐟 𝐚𝐧 𝐄𝐧𝐭𝐫𝐞𝐩𝐫𝐞𝐧𝐞𝐮𝐫:
-Define entrepreneurship, distinguishing it from general business activities by emphasizing its focus on innovation, risk-taking, and value creation. Students will describe the characteristics and traits of successful entrepreneurs, including their roles and responsibilities, and discuss the broader economic and social impacts of entrepreneurial activities on both local and global scales.
How to Make a Field Mandatory in Odoo 17Celine George
In Odoo, making a field required can be done through both Python code and XML views. When you set the required attribute to True in Python code, it makes the field required across all views where it's used. Conversely, when you set the required attribute in XML views, it makes the field required only in the context of that particular view.
हिंदी वर्णमाला पीपीटी, hindi alphabet PPT presentation, hindi varnamala PPT, Hindi Varnamala pdf, हिंदी स्वर, हिंदी व्यंजन, sikhiye hindi varnmala, dr. mulla adam ali, hindi language and literature, hindi alphabet with drawing, hindi alphabet pdf, hindi varnamala for childrens, hindi language, hindi varnamala practice for kids, https://www.drmullaadamali.com
LAND USE LAND COVER AND NDVI OF MIRZAPUR DISTRICT, UPRAHUL
This Dissertation explores the particular circumstances of Mirzapur, a region located in the
core of India. Mirzapur, with its varied terrains and abundant biodiversity, offers an optimal
environment for investigating the changes in vegetation cover dynamics. Our study utilizes
advanced technologies such as GIS (Geographic Information Systems) and Remote sensing to
analyze the transformations that have taken place over the course of a decade.
The complex relationship between human activities and the environment has been the focus
of extensive research and worry. As the global community grapples with swift urbanization,
population expansion, and economic progress, the effects on natural ecosystems are becoming
more evident. A crucial element of this impact is the alteration of vegetation cover, which plays a
significant role in maintaining the ecological equilibrium of our planet.Land serves as the foundation for all human activities and provides the necessary materials for
these activities. As the most crucial natural resource, its utilization by humans results in different
'Land uses,' which are determined by both human activities and the physical characteristics of the
land.
The utilization of land is impacted by human needs and environmental factors. In countries
like India, rapid population growth and the emphasis on extensive resource exploitation can lead
to significant land degradation, adversely affecting the region's land cover.
Therefore, human intervention has significantly influenced land use patterns over many
centuries, evolving its structure over time and space. In the present era, these changes have
accelerated due to factors such as agriculture and urbanization. Information regarding land use and
cover is essential for various planning and management tasks related to the Earth's surface,
providing crucial environmental data for scientific, resource management, policy purposes, and
diverse human activities.
Accurate understanding of land use and cover is imperative for the development planning
of any area. Consequently, a wide range of professionals, including earth system scientists, land
and water managers, and urban planners, are interested in obtaining data on land use and cover
changes, conversion trends, and other related patterns. The spatial dimensions of land use and
cover support policymakers and scientists in making well-informed decisions, as alterations in
these patterns indicate shifts in economic and social conditions. Monitoring such changes with the
help of Advanced technologies like Remote Sensing and Geographic Information Systems is
crucial for coordinated efforts across different administrative levels. Advanced technologies like
Remote Sensing and Geographic Information Systems
9
Changes in vegetation cover refer to variations in the distribution, composition, and overall
structure of plant communities across different temporal and spatial scales. These changes can
occur natural.
LAND USE LAND COVER AND NDVI OF MIRZAPUR DISTRICT, UP
Aipmt 2015 answer key & solutions
1.
𝐏𝐡𝐲𝐬𝐢𝐜𝐬
Single
correct
answer
type:
1. If
energy
(E),
velocity
(V)
and
time
(T)
are
chosen
as
the
fundamental
quantities,
the
dimensional
formula
of
surface
tension
will
be:
(1) E V!!
T!!
(2) E V!!
T!!
(3) E V!!
T!!
(4) E!!
V!!
T!!
Solution:
(3)
2. A
ship
A
is
moving
Westwards
with
a
speed
of
10 kmh!!
and
a
ship
B
100
km
south
of
A,
is
moving
Northwards
with
a
speed
of
10 km h!!
.
The
time
after
which
the
distance
between
them
becomes
shortest,
is:
(1) 0 h
2.
(2) 5 h
(3) 5 2 h
(4) 10 2 h
Solution:
(2)
3. A
particle
of
unit
mass
undergoes
one-‐dimensional
motion
such
that
its
velocity
varies
according
to
v x = βx!!"
,
where
β
and
n
are
constants
and
x
is
the
position
of
the
particle.
The
acceleration
of
the
particle
as
a
function
x,
is
given
by:
3.
(1) −2nβ!
x!!"!!
(2) −2nβ!
x!!"!!
(3) −2β!
x!!"!!
(4) −2nβ!
e!!"!!
Solution:
(2)
4. Three
blocks
A,
B
and
C,
of
masses
4
kg,
2kg
and
1
kg
respectively,
are
in
contact
on
a
frictionless
surface,
as
shown.
If
a
force
of
14
N
is
applied
on
the
4
kg
block,
then
the
contact
force
between
A
and
B
is:
(1) 2N
(2) 6N
(3) 8N
(4) 18N
Solution:
(2)
4.
5. A
block
A
of
mass
m!
rests
on
a
horizontal
table.
A
light
string
connected
to
it
passes
over
a
frictionless
pulley
at
the
edge
of
table
and
from
its
other
end
another
block
B
of
mass
m!
is
suspended.
The
coefficient
of
kinetic
friction
between
the
block
and
the
table
is
µμ!.
When
the
block
A
is
sliding
on
the
table,
the
tension
in
string
is:
(1)
!!!!!!! !
!!!!!
(2)
!!!!!!! !
!!!!!
(3)
!!!! !!!! !
!!!!!
(4)
!!!! !!!! !
!!!!!
Solution:
(3)
5.
6. Two
similar
springs
P
and
Q
have
spring
constants
K!
and
K!,
such
that
K! > K!.
They
are
stretched,
first
by
the
same
amount
(case
a),
then
by
the
same
force
(case
b).
the
work
done
by
the
springs
W!
and
W!
are
related
as,
in
case
(a)
and
case
(b),
respectively:
(1) W! = W!; W! > W!
(2) W! = W!; W! = W!
(3) W! > W!; W! > W!
(4) W! < W!; W! < W!
Solution:
(3)
6.
7. A
block
of
mass
10
kg,
moving
in
x
direction
with
a
constant
speed
of
10 ms!!
,
is
subjected
to
a
retarding
force
F = 0.1 x
J/m
during
its
travel
from
x
=
20
m
to
30m.
Its
final
KE
will
be:
(1) 475
J
(2) 450
J
(3) 275
J
(4) 250
J
Solution:
(1)
7.
8. A
particle
of
mass
m
is
driven
by
a
machine
that
delivers
a
constant
power
k
watts.
If
the
particle
starts
from
rest
the
force
on
the
particle
at
time
t
is:
(1)
!"
!
t!
!
!
(2) mk t!
!
!
(3) 2mk t!
!
!
(4)
!
!
mk t!
!
!
Solution:
(1)
8.
9. Two particles of masses m!, m! move with initial velocities u! and u! . On collision, one of the
particles get excited to higher level, after absorbing energy ε If final velocities of particles be v! and
v! then we must have:
(1) m!
!
u! + m!
!
u! − ε = m!
!
v! + m!
!
v!
(2)
!
!
m!u!
!
+
!
!
m!u!
!
=
!
!
m!v!
!
+
!
!
m!v!
!
− ε
(3)
!
!
m!u!
!
+
!
!
m!u!
!
− ε =
!
!
m!v!
!
+
!
!
m!v!
!
(4)
!
!
m!
!
u!
!
+
!
!
m!
!
u!
!
+ ε =
!
!
m!
!
v!
!
+
!
!
m!
!
v!
!
Solution: (3)
10. A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a
horizontal position. The knives are at a distance d from each other. The centre of mass of the rod is
at distance x from A. the normal reaction on A is:
(1)
!"
!
(2)
!"
!
(3)
! !!!
!
(4)
! !!!
!
Solution: (4)
9.
11. A mass m moves in a circle on a smooth horizontal plane with velocity v! at a radius R!. The mass is
attached to a string which passes through a smooth hole in plane as shown.
The tension in the string is increased gradually and finally m moves in a circle of radius
!!
!
. The
final value of the kinetic energy is:
(1) mv!
!
(2)
!
!
mv!
!
(3) 2 mv!
!
(4)
!
!
mv!
!
Solution: (3)
10.
12. Three identical spherical shells, cach of mass m and radius r are placed as shown in figure. Consider
an axis XX!
which is touching to tow shells and passing through diameter of third shell. Moment of
inertia of the system consisting of these three spherical shell about XX!
' axis is :
(1)
!!
!
𝑚𝑟!
(2) 3𝑚𝑟!
(3)
!"
!
𝑚𝑟!
(4) 4 𝑚𝑟!
Solution: (4)
13. Kepler's third law states that square of period of revolution (T) of a planet around the sun, is
proportional to third power of average distance r between sun and planet i.e., T!
= Kr!
here K is constant
If the masses of sun and planet are M and m respectively then as per Newton's law of gravitation
force of attraction between them is
F =
!"#
!! , here G is gravitational constant
The relation between G and K is described as:
11.
(1) GK = 4π!
(2) GMK = 4π!
(3) K = G
(4) K =
!
!
Solution: (2)
14. Two spherical bodies of mass M and 5 M and radii R and 2R released in free space with initial
separation between their centres equal to 12 R. If they attract each other due to gravitational force
only, then the distance covered by the smaller body before collision is:
(1) 2.5 R
(2) 4.5 R
(3) 7.5 R
(4) 7.5 R
Solution: (3)
12.
𝑀𝑥 = 5 𝑀 (9𝑅 − 𝑥)
15. On observing light from three different stars P,Q and R, it was found that intensity of violet colour is
maximum in the spectrum of P, the intensity of green colour is maximum in the spectrum of R and
the intensity of red colour is maximum in the spectrum of Q. If T!, T! and T! are the respective
absolute temperatures of P, Q and R, then it can be concluded from the above observations that :
(1) T! > T! > T!
(2) T! > T! > T!
(3) T! < T! < T!
(4) T! < T! < T!
Solution: (2)
16. The approximate depth of an ocean is 2700 m. The compressibility of water is 45.4×10!!!
Pa!!
and
density of water is 10! !"
!!. What fractional compression of water will be obtained at the bottom of the
ocean?
(1) 0.8×10!!
(2) 1.0×10!!
(3) 1.2×10!!
(4) 1.4×10!!
13.
Solution: (3)
17. The two ends of a metal rod are maintained at temperatures 100!
C and 110!
C. The rate of heat flow
in the rod is found to be 4.0 J/s. If the ends are maintained at temperatures 200!
C and 210!
C, the
rate of heat flow will be:
(1) 44.0 J/s
(2) 16.8 J/s
(3) 8.0 J/s
(4) 4.0 J/s
Solution: (4)
In both cases the temperature difference between the ends of the rool is 10!
𝐶
∴ rate of heat flow is also 4J/s in the second case.
18. A wind with speed 40 m/s blows parallel to the roof of a house. The area of the roof is 250 m!
.
Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind
on the roof and the direction of the direction of the force will be : P!"# = 1.2
!"
!!
(1) 4.8×10!
N, downwards
(2) 4.8×10!
N, upwards
(3) 2.4×10!
N, upwards
(4) 2.4×10!
N, downwards
Solution: (3)
14.
=
1
2
×1.2×1600
19. Figure below shows two paths that may be taken by a gas to go from a state A to a state C.
In process AB, 400 J of heat is added to the system and in process BC, 100 J of heat is added to
the system. The heat absorbed by the system in the process AC will be:
(1) 380 J
(2) 500 J
(3) 460 J
(4) 300 J
Solution: (3)
15.
20. A Carnot engine, having an efficiency of η =
!
!"
as heat engine, is used as refrigerator. If the work
done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature
is:
(1) 100 J
(2) 99 J
(3) 90 J
(4) 1 J
Solution: (3)
16.
21. One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in
the figure,
17.
The change in internal energy of the gas during the transition is:
(1) 20 kJ
(2) −20 kJ
(3) 20 J
(4) −12 kJ
Solution: (2)
22. The ratio of the specific heats
!!
!!
= γ in terms of degrees of freedom (n) is given by:
(1) 1 +
!
!
(2) 1 +
!
!
(3) 1 +
!
!
(4) 1 +
!
!
Solution: (3)
18.
23. When two displacements represented by y! = a sin (ωt) and y! = b cos ωt are superimposed the
motion is:
(1) Not a simple harmonic
(2) Simple harmonic with amplitude
!
!
(3) Simple harmonic with amplitude a! + b!
(4) Simple harmonic with amplitude
!!!
!
Solution: (3)
24. A particle is executing SHM along a straight line. Its velocities at distances x! and x! from the mean
position are V! and V! respectively. Its time period is:
(1) 2π
!!
!!!!
!
!!
!!!!
!
(2) 2π
!!
!!!!
!
!!
!!!!
!
(3) 2π
!!
!!!!
!
!!
!!!!
!
(4) 2π
!!
!!!!
!
!!
!!!!
!
19.
Solution: (2)
25. The fundamental frequency of a closed organ pipe of length 20 cm is equal to the second overtone
of an organ pipe open at both the ends. The length of organ pipe at both the ends is:
(1) 80 cm
(2) 100 cm
(3) 120 cm
(4) 140 cm
Solution: (3)
20.
26. A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected
from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now
inserted in it. Which of the following is incorrect?
(1) The potential difference between the plates decreases K times
(2) The energy stored in the capacitor decreases K times
(3) The change in energy stored is
!
!
CV! !
!
− 1
(4) The charge on the capacitor is not conserved
Solution: (4)
27. The electric field in a certain region is acting radially outward and is given by E = Ar. A charge
contained in a sphere of radius ‘a’ centred at the origin of the filed, will be given by:
(1) 4πϵ!Aa!
21.
(2) A ϵ! a!
(3) 4πϵ! Aa!
(4) ϵ! Aa!
Solution: (3)
28. A
potentiometer
wire
has
length
4
m
and
resistance
8Ω.
The
resistance
that
must
be
connected
in
series
with
the
wire
and
an
accumulator
of
e.m.f.
2V,
so
as
to
get
a
potential
gradient
1
mV
per
cm
on
the
wire
is
:
(1) 32 Ω
(2) 40 Ω
(3) 44 Ω
(4) 48 Ω
Solution:
(1)
22.
29. A,
B
and
C
are
voltmeters
of
resistance
R,
1.5
R
and
3R
respectively
as
shown
in
the
figure.
When
some
potential
difference
is
applied
between
X
and
Y,
the
voltmeter
readings
are
V!, V!
and
V!
respectively.
Then:
(1) V! = V! = V!
(2) V! ≠ V! = V!
(3) V! = V! ≠ V!
(4) V! ≠ V! ≠ V!
Solution:
(1)
30. Across
a
metallic
conductor
of
non-‐uniform
cross
section
a
constant
potential
difference
is
applied.
The
quantity
which
remains
constant
along
the
conductor
is:
(1) Current
density
(2) Current
23.
(3) Drift
velocity
(4) Electric
field
Solution:
(2)
The
area
cross
section
of
conductor
is
non-‐uniform
so
current
density
will
be
different
but
the
numbers
of
flow
of
election
will
be
same
so
current
will
be
constant
31. A
wire
carrying
current
I
has
the
shape
as
shown
in
adjoining
figure.
Linear
parts
of
the
wire
are
very
long
and
parallel
to
X-‐axis
while
semicircular
portion
of
radius
R
is
lying
in
Y-‐Z
plane.
Magnetic
field
at
point
O
is
:
(1) B =
!!
!"
!
!
πı + 2k
(2) B = −
!!
!"
!
!
πı − 2k
(3) B = −
!!
!"
!
!
πı + 2k
(4) B =
!!
!"
!
!
πı − 2k
Solution:
(3)
24.
32. An
electron
moving
in
a
circular
orbit
of
radius
r
makes
n
rotations
per
second.
The
magnetic
field
produced
at
the
centre
has
magnitude
:
(1)
!!!"
!!"
(2) Zero
(3)
!!!!!
!
(4)
!!!"
!!
Solution:
(4)
25.
33. A
conducting
square
frame
of
side
'a'
and
a
long
straight
wire
carrying
current
I
are
located
in
the
same
plane
as
shown
in
the
figure.
The
frame
moves
to
the
right
with
a
constant
velocity
'V'.
The
emf
induced
in
the
frame
will
be
proportional
to
:
(1)
!
!!
(2)
!
!"!! !
(3)
!
!"!! !
(4)
!
!"!! !"!!
Solution:
(4)
26.
34. A
resistance
'R'
draws
power
'P'
when
connected
to
an
AC
source.
If
an
inductance
is
now
placed
in
series
with
the
resistance,
such
that
the
impedance
of
the
circuit
becomes
'Z',
the
power
drawn
will
be:
(1) P
!
!
!
(2) P
!
!
(3) P
!
!
(4) P
Solution:
(1)
27.
35. A
radiation
of
energy
‘E’
falls
normally
on
a
perfectly
reflecting
surface.
The
momentum
transferred
to
the
surface
is
(C
=
Velocity
of
light)
:-‐
(1)
!
!
(2)
!"
!
(3)
!"
!!
(4)
!
!!
Solution:
(2)
36. Two
identical
thin
plano-‐convex
glass
lenses
(refractive
index
1.5)
each
having
radius
of
curvature
of
20
cm
are
placed
with
their
convex
surfaces
in
contact
at
the
centre.
The
intervening
space
is
filled
with
oil
of
refractive
index
1.7.
The
focal
length
of
the
combination
is
:-‐
(1) −20 cm
(2) −25 cm
(3) −50 cm
(4) 50 cm
Solution:
(3)
28.
37. For
a
parallel
beam
of
monochromatic
light
of
wavelength
‘λ’
,
diffraction
is
produced
by
a
single
slit
whose
width
'a'
is
of
the
order
of
the
wavelength
of
the
light.
If
'D'
is
the
distance
of
the
screen
from
the
slit,
the
width
of
the
central
maxima
will
be
:
(1)
!"!
!
(2)
!!
!
(3)
!"
!
(4)
!"#
!
Solution:
(1)
29.
38. In
a
double
slit
experiment,
the
two
slits
are
1
mm
apart
and
the
screen
is
placed
1
m
away.
A
monochromatic
light
of
wavelength
500
nm
is
used.
What
will
be
the
width
of
each
slit
for
obtaining
ten
maxima
of
double
slit
within
the
central
maxima
of
single
slit
pattern?
(1) 0.2
mm
(2) 0.1
mm
(3) 0.5
mm
(4) 0.02
mm
Solution:
(1)
30.
39. The
refracting
angle
of
a
prism
is
A,
and
refractive
index
of
the
material
of
the
prism
is
cot
!
!
.
The
angle
of
minimum
deviation
is
:-‐
(1) 180!
− 3A
(2) 180!
− 2A
(3) 90!
− A
(4) 180!
+ 2A
Solution:
(2)
31.
40. A
certain
metallic
surface
is
illuminated
with
monochromatic
light
of
wavelength,
λ.
The
stopping
potential
for
photo-‐electric
current
for
this
light
is
3V!.
If
the
same
surface
is
illuminated
with
light
of
wavelength2λ,
the
stopping
potential
is
V!.
The
threshold
wavelength
for
this
surface
for
photoelectric
effect
is:
(1) 6λ
(2) 4λ
(3)
!
!
(4)
!
!
Solution:
(2)
32.
41. Which
of
the
following
figures
represent
the
variation
of
particle
momentum
and
the
associated
de-‐
Broglie
wavelength?
(1)
(2)
(3)
(4)
Solution:
(2)
33.
42. Consider
3!"
orbit
of
He!
(helium),
using
non-‐relativistic
approach,
the
speed
of
electron
in
this
orbit
will
be
[given
K = 9×10!
constant,
Z = 2
and
h
(Planck’s
Constant)
=
6.6×10!!"
Js]
(1) 2.92×10!
m s
(2) 1.46×10!
m s
(3) 0.73×10!
m s
(4) 23.0×10!
m s
Solution:
(2)
43. If
radius
of
the
Al!"
!"
nucleus
is
taken
to
be
R!"
then
the
radius
of
Te!"
!"#
nucleus
is
nearly:
(A)
!"
!"
!
!
R!"
(B)
!
!
R!"
(C)
!
!
R!"
(D)
!"
!"
!
!
R!"
Solution:
(2)
34.
44. If
in
a
p-‐n
junction,
a
square
input
signal
of
10
V
is
applied
as
shown,
then
the
output
across
R!
will
be:
(1)
(2)
(3)
35.
(4)
Solution:
(4)
45. Which
logic
gate
is
represented
by
the
following
combination
of
logic
gates?
(1) OR
(2) NAND
(3) AND
(4) NOR
Solution:
(3)
36.
Chemisty
46. Which
of
the
following
species
contains
equal
number
of
σ
–
and
π
–
bonds?
(1) HCO!
!
(2) XeO!
(3) CN !
(4) CH! CN !
Solution:
(2)
XeO! ⇒
hybridization
of
central
atom
H =
!
!
8 = 4 (sp!
)
There
are
4σ
bonds
&
4π
bonds
as
central
Xe
atom
joined
to
O
atoms
at
corners
of
regular
tetrahedron
by
double
bonds.
47. The
species
Ar, K!
and
Ca!!
contain
the
same
number
of
electrons.
In
which
order
do
their
radii
increase?
(1) Ar < K!
< Cr!!
(2) Ca!!
< Ar < K!
(3) Ca!!
< K!
< Ar
(4) K!
< Ar < Ca!!
Solution:
(3)
For
isoelectronic
species
as
!
!
ratio
increases
ionic
size
decreases.
E
(total
no
of
electrons)
=
18
For
Ar
!
!
=
!"
!"
K! !
!
=
!"
!"
Ca!! !
!
=
!"
!"
i.e.
Ca!!
< K!
< Ar
48. The
function
of
“Sodium
pump”
is
a
biological
process
operating
in
each
and
every
cell
of
all
animals.
Which
of
the
following
biologically
important
ions
is
also
a
constituent
of
this
pump?
(1) Ca!!
(2) Mg!!
(3) K!
(4) Fe!!
Solution:
(3)
Memory
based
37.
K!
ions
is
constituent
of
this
pump.
49. “Metals
are
usually
not
found
as
nitrates
in
their
ores”.
Out
of
the
following
two
(a
and
b)
reasons
which
is
or
are
true
for
the
above
observation?
a
–
Metal
nitrates
are
highly
unstable.
b
–
Metal
nitrates
are
highly
soluble
in
water.
(1) a
and
b
are
true
(2) a
and
b
are
false
(3) a
is
false
but
b
is
true
(4) a
is
true
but
b
is
false
Solution:
(3)
Metal
nitrates
are
not
unstable
metal
nitrates
are
highly
soluble.
50. Solubility
of
the
alkaline
earth’s
metal
sulphates
in
water
decreases
in
the
sequence
:
(1) Mg
>
Ca
>
Sr
>
Ba
(2) Ca
>
Sr
>
Ba
>
Mg
(3) Sr
>
Ca
>
Mg
>
Ba
(4) Ba
>
Mg
>
Sr
>
Ca
Solution:
(1)
Lattice
energies
of
alkaline
earth
sylphates
are
almost
constant
but
hydration
energy
∝
!
!"#$ !" !"#$%&
from
Mg!!
→ Ca!!
→ Sr!!
→ Ba!!
,
cationic
size
increases,
hydration
energy
decreases
i.e.
MgSO!
is
soluble
&
BaSO!
is
a
precipitate.
MgSO! > CaSO! > SrSO! > BaSO!
51. Because
of
lanthanoid
contraction,
which
of
the
following
pairs
of
elements
have
nearly
same
atomic
radii?
(Number
in
the
parenthesis
are
atomic
numbers).
(1) Ti
(22)
and
Zr
(40)
(2) Zr
(40)
and
Nb
(41)
(3) Zr
(40)
and
Hf
(72)
(4) Zr
(40)
and
Ta
(73)
Solution:
(3)
!"Zr
is
[Kr]!" 5s!
4d!
!"Hf
is
[Xe]!" 6s!
4f!"
5d!
38.
Both
lie
in
period
IV(B)
Lanthanide
contraction
&
additional
shell
introduction
cancell
size
effects
&
both
metals
have
same
radii.
52. Which
of
the
following
processes
does
not
involve
oxidation
of
iron?
(1) Rusting
of
iron
sheets
(2) Decolourization
of
blue
CuSO!
solution
by
iron
(3) Formation
of
Fe CO !
from
Fe
(4) Liberation
of
H!
from
steam
by
iron
at
high
temperature
Solution:
(3)
Rusting
of
iron
Fe → Fe!!
(oxidation)
Cu!!
+ Fe → Fe!!
+ Cu oxidation
Fe + H!SO! → FeSO! + H! ↑ g (oxidation)
Fe CO ! ⇒ oxidation state of Fe is zero
53. Which
of
the
following
pairs
of
iron
are
isoelectronic
and
isostructural?
(1) CO!
!!
, SO!
!!
(2) CIO!
!
, CO!
!!
(3) SO!
!!
, NO!
!
(4) CIO!
!
, SO!
!!
Solution:
(4)
Isoelectronic
(same
no
of
electrons)
CIO!
!
, SO!
!!
⇒ 17 + 24 + 1 = 42
and
16 + 24 + 2 = 42
Isostructural
(same
type
of
hybridization)
CIO!
!
⇒ H =
1
2
7 + 1 = 4 sp!
SO!
!!
⇒ H =
1
2
6 + 2 = 4 (sp!
)
39.
54. Which
of
the
following
options
represents
the
correct
bond
order?
(1) O!
!
> O! > O!
!
(2) O!
!
< O! < O!
!
(3) O!
!
> O! < O!
!
(4) O!
!
< O! > O!
!
Solution:
(2)
B. O =
!! !!!
!
B.O
for
O!
!
= 1.5 ⇒
!" ! !
!
= 1.5
B.O
for
O! = 2.0 ⇒
!" – !
!
= 2
B.O
for
O!
!
= 2.5 ⇒
!" ! !
!
= 2.5
As
per
M.O.T
electronic
configurations
O!
!
⇒ σ1s!
, σ1∗
s!
, σ2s!
, σ2∗
s!
, σ2pz!
, π2px!
, π2py!
, π2∗
px!
, π2∗
py!
O! ⇒ σ1s!
, σ1∗
s!
, σ2s!
, σ2∗
s!
, σ2pz!
, π2px!
, π2py!
, π2∗
px!
, π2∗
py!
O!
!
⇒ σ1s!
, σ1∗
s!
, σ2s!
, σ2∗
s!
, σ2pz!
, π2px!
, π2py!
, π2∗
px!
, π2∗
py!
O!
!
< O! < O!
!
55. Nitrogen
dioxide
and
sulphur
dioxide
have
some
properties
in
common.
Which
property
is
shown
by
one
of
these
compounds,
but
not
by
the
other?
(1) Forms
‘acid-‐rain’
40.
(2) Is
a
reducing
agent
(3) Is
soluble
in
water
(4) Is
used
as
a
food-‐preservative
Solution:
(4)
SO!
is
used
as
food
preservative
NO!
is
not
used
as
food
preservative
56. Maximum
bond
angle
at
nitrogen
is
present
in
which
of
the
following?
(1) NO!
(2) NO!
!
(3) NO!
!
(4) NO!
!
Solution:
(3)
NO!
!
ion
H =
!
!
5 − 1 = 2 sp hybridisation
Linear
geometry
Bond
angle
180!
57. Magnetic
moment
2.84
B.M.
Is
given
by:
(At.
nos,
Ni
=
28,
Ti
=
22,
Cr
=
24,
Co
=
27)
(1) Ni!!
(2) Ti!!
(3) Cr!!
(4) Co!!
Solution:
(1)
41.
58. Cobalt
(III)
chloride
forms
several
octahedral
complexes
with
ammonia.
Which
of
the
following
will
not
give
test
for
chloride
ions
with
silver
nitrate
at
25!
C?
(1) CoCl! ∙ 3NH!
(2) CoCl! ∙ 4NH!
(3) CoCl! ∙ 5NH!
(4) CoCl! ∙ 6NH!
Solution:
(1)
In
CoCl! ∙ 3NH!
the
complex
can
be
written
as
[CoCl! NH! !]
With
Co!!
oxidation
state
3,
Cl!
ions
&
3,
NH!
molecules
be
with
in
the
co-‐ordination
sphere
Co___Cl
bonds
in
co-‐ordination
sphere
not
ionisable.
59. Which
of
these
statements
about
Co CN !
!!
is
true?
(1) Co CN !
!!
has
no
unpaired
electrons
and
will
be
in
a
low-‐spin
configuration.
(2) Co CN !
!!
has
four
unpaired
electrons
and
will
be
in
a
low-‐spin
configuration.
(3) Co CN !
!!
has
four
unpaired
electrons
and
will
be
in
a
high-‐spin
configuration.
(4) Co CN !
!!
has
no
unpaired
electrons
and
will
be
in
a
high-‐spin
configuration.
Solution:
(1)
Due
to
strong
Ligand
field
of
CN!
ions
pairing
of
electrons
takes
place
in
inner
3d
orbital
it
is
a
low
spin
complex
with
no
unpaired
electrons.
42.
60. The
activation
energy
of
a
reaction
can
be
determined
from
the
slope
of
which
of
the
following
graphs?
(1) In K vs. T
(2)
!" !
!
vs. T
(3) In K vs.
!
!
(4)
!
!" !
vs.
!
!
Solution:
(3)
61. Which
one
is
not
equal
to
zero
for
an
ideal
solution?
(1) ∆H!"#$%&'
(2) ∆S!"#$%&'
(3) ∆V!"#$%&'
(4) ∆P = P!"#$%&$' − P!"#$%&
Solution: (2) ΔS!"#(per mole) = −Σx! log x!
x! is mole fraction of i!"
component in solution ΔS!"# mole is +ve and not zero for an ideal solution.
43.
62. A
mixture
of
gases
contains
H!
and
O!
gases
in
the
ratio
of
1
:
4
(w/w).
What
is
the
molar
ratio
of
the
two
gases
in
the
mixture?
(1) 1
:
4
(2) 4
:
1
(3) 16
:
1
(4) 2
:
1
Solution: (2) Let the masses of H! and O! be x g and 4x g
∴ Molar ratio =
n!!
N!!
=
x
2
4x
32
=
4
1
63. A
given
metal
crystallizes
out
with
a
cubic
structure
having
edge
length
if
361
pm.
If
there
are
four
metal
atoms
in
one
unit
cell,
what
is
the
radius
of
one
atom?
(1) 40
pm
(2) 127
pm
(3) 80
pm
(4) 108
pm
Solution: (2) 4 atoms/unit cell is FCC
i.e. 4r = 2 ∙ a
r =
2 ∙ a
4
= 1.414×
361
4
= 127 pm
64. When
initial
concentration
of
a
reactant
is
doubled
in
a
reaction,
its
half-‐life
period
is
not
affected.
The
order
of
the
reaction
is:
(1) Zero
(2) First
(3) Second
(4) More
than
zero
but
less
than
first
Solution: (2)
44.
t! ! ∝
1
a!
!!!
t! ! is independent of initial concentration i.e.
t! ! ∝
!
!!
!!! (is constant)
i.e n = 1 (order of reaction)
65. If
the
value
of
an
equilibrium
constant
for
a
particular
reaction
is
1.6×10!"
,
then
at
equilibrium
the
system
will
contain:
(1) All
reactants
(2) Mostly
reactants
(3) Mostly
products
(4) Similar
amounts
of
reactants
and
products
Solution: (3) For any reaction of equilibrium
K =
!"#$%&' !
!"#$%#&% ! is K is 1.6×10!"
(very high)
Then equilibrium mixture shall mostly contain products.
66. A
device
that
converts
energy
of
combustion
of
fuels
like
hydrogen
and
methane,
directly
into
electrical
energy
is
known
as:
(1) Fuel
cell
(2) Electrolytic
cell
(3) Dynamo
(4) Ni-‐Cd
cell
Solution:
(1)
Fuel
cell
Burns
duels
like
H!(g)
&
CH!(g)
&
converts
chemical
energy
into
electrical
energy.
67. The
boiling
point
of
0.2 mol kg!!
solution
of
X
in
water
is
greater
than
equimolal
solution
of
Y
in
water.
Which
one
of
the
following
statements
is
true
in
this
case?
(1) X
is
undergoing
dissociation
in
water
45.
(2) Molecular
mass
of
X
is
greater
than
the
molecular
mass
of
Y.
(3) Molecular
mass
of
X
is
less
than
the
molecular
mass
of
Y.
(4) Y
is
undergoing
dissociation
in
water
while
X
undergoes
no
change
Solution:
(1)
∆T! = i K!. m
K!
&
m
are
constant
for
solutions
of
X
and
y.
∆T!
is
greater
for
solution
of
X
imples
i(van’t
Hoff
factor
for
X
>
1)
i.e
X
undergoes
dissociation.
68. Which
one
of
the
following
electrolytes
has
the
same
value
of
van’s
Hoff’s
factor
(i)
as
that
of
Al! SO! !
(if
all
are
100%
ionized)?
(1) K!SO!
(2) K! Fe CN !
(3) Al NO! !
(4) K! Fe CN !
Solution:
(4)
69. The
number
of
d-‐electrons
in
Fe!!
(Z = 26)
is
not
equal
to
the
number
of
electrons
in
which
one
of
the
following?
(1) s-‐electrons
in
Mg
(Z
=
12)
(2) p-‐electrons
in
Cl
(Z
=
17)
(3) d-‐electrons
in
Fe
(Z
=
26)
(4) p-‐electrons
in
Ne
(Z
=
10)
Solution:
(2)
!"Fe!!
⇒ !" Ar 3d!
(d electrons = 6)
!"Cl ⇒ 1s!
2s!
2p!
3s!
3p!
(p − electrons = 11)
46.
70. The
correct
bond
order
in
the
following
species
is:
(1) O!
!!
< O!
!
< O!
!
(2) O!
!!
< O!
!
< O!
!
(3) O!
!
< O!
!
< O!
!!
(4) O!
!
< O!
!
< O!
!!
Solution: (4) Bond order =
!!!!!
!
as per M.O.T
O!
!
⇒ B. O =
10 − 7
2
= 1.5
O!
!
⇒ B. O =
10 − 5
2
= 2.5
O!
!!
⇒ B. O =
10 − 4
2
= 3.0
O!
!
< O!
!
< O!
!!
71. The angular momentum of electron ‘d’ orbital is equal to:
(1) 6 ℏ
(2) 2 ℏ
(3) 2 3 ℏ
(4) 0 ℏ
Solution: (1) For d-‐electrons l = 2
Angular orbital momentum = l(l + 1)
!
!"
= 6
h
2π
= 6 ℏ
72. The K!" of Ag!CrO!, AgCl, AgBr and AgI are respectively, 1.1×10!!"
, 1.8×10!!"
, 5.0×
10!!"
, 8.3×10!!"
. Which one of the following salts will precipitate last if AgNO! solution is
added to the solution containing equal moles of NaCl, NaI and Na!CrO! ?
(1) AgI
(2) AgCl
(3) AgBr
(4) Ag!CrO!
Solution: (4) Let the concentration of each
47.
Cl!
= Br!
= I!
= CrO!
!!
= x M
Then for precipitation concentration of [Ag!
] in case of Ag!CrO! will be
Ag!
=
K!"(Ag!CrO!)
x
=
1.1×10!!"
x
i.e. maximum and therefore Ag!CrO! salt will precipitate out last.
73. Which property of colloidal solution is independent of charge on the colloidal particles?
(1) Coagulation
(2) Electrophoresis
(3) Electro-‐osmosis
(4) Tyndall effect
Solution: (4) Tyndall effect is related to scattering of light by colloidal particles and not dependent
on charge.
74. Which of the following statements is correct for a reversible process in a state of
equilibrium?
(1) ΔG = −2.30 RT log K
(2) ΔG = 2.30 RT log K
(3) ΔG!
= −2.30 RT log K
(4) ΔG!
= 2.30 RT log K
Solution: (3) For a reversible process oat equilibrium
ΔG = 0 = ΔG!
+ RT ln K Q = K
i.e. ΔG!
= −2.303 RT log K
75. Bithional is generally added to the soaps as an additive to function as a/an:
(1) Softener
(2) Dryer
(3) Buffering agent
(4) Antiseptic
Solution: (4) Bithional is added as antiseptic to soaps.
48.
76. The electrolytic reduction of nitrobenzene in strongly acidic medium produces:
(1) p-‐aminophenol
(2) Azoxybenzene
(3) Azobenzene
(4) Aniline
Solution: (1)
77. In Duma’s method for estimation of nitrogen, 0.25 g of an organic compound gave 40 mL of
nitrogen collected at 300 K temperature and 725 mm pressure. If the aqueous tension at 300
K is 25 mm, the percentage of nitrogen in the compound is:
(1) 17.36
(2) 18.20
(3) 16.76
(4) 15.76
Solution: (3) n!!
=
!"
!"
=
725 − 25 40×10!!
760×0.082×300
=
28
760×24.6
=
28×28
760×24.6
g of N!
% N! =
28×28
760×24.6
×
1
0.25
×100
= 16.77%
78. In which of the following compounds, the C − Cl bond ionization shall give most stable
carbonium ion?
49.
(1)
(2)
(3)
(4)
Solution: (2)
Carbocation is most stable as 3!
carbocation.
79. The reaction
is called:
50.
(1) Williamson synthesis
(2) Williamson continuous etherification process
(3) Etard reaction
(4) Gatterman-‐Koch reaction
Solution:(1) It is called Williamson’s synthesis for ether formation.
80. The reaction of C!H!CH = CHCH! with HBr produces:
(1)
(2)
(3) C!H!CH!CH!CH!Br
(4)
Solution: (1)
81. A single compound of the structure
51.
is obtainable from ozonolysis of which of the following cyclic compounds?
(1)
(2)
(3)
(4)
Solution: (1)
52.
82. Treatment of cyclopentanone with methyl lithium gives which of the following
species?
(1) Cyclopentanonyl anion
(2) Cyclopentanonyl cation
(3) Cyclopentanonyl radical
(4) Cyclopentanonyl biradical
Solution: (1)
83. Consider the following compounds
Hyperconjugation occurs in:
(1) I only
(2) II only
(3) III only
(4) I and III
Solution: (C) There is no α − H in the structure I and II.
84. Which of the following is the most correct electron displacement for a nucleophile reaction to
take place?
(1)
53.
(2)
(3)
(4)
Solution: (3)
85. The enolic form of ethyl acetoacetate as below has:
(1) 18 sigma bonds and 2 pi-‐bonds
(2) 16 sigma bonds and 1 pi-‐bonds
(3) 9 sigma bonds and 2 pi-‐bonds
(4) 9 sigma bonds and 1 pi-‐bonds
Solution: (1) 18 σ and 2π bonds in both keto and enol form of ethyl acetoacetate.
54.
86. Given,
Which of the given compounds can exhibit tautomerism?
(1) I and II
(2) I and III
(3) II and III
(4) I, II and III
Solution: (4) All compounds show tautomerism.
In 3!"
ketone
55.
87. Given,
The enthalpy of hydrogenation of these compounds will be in the order as:
(1) I > II > III
(2) III > II > I
(3) II > III > I
(4) II > I > III
Solution: (2) Enthalpy of hydrogenation is inversely proportional to stability of alkene.
III > II > I
88. Biodegradable polymer which can be produced from glycine and aminocaproic acid is:
(1) Nylon-‐2-‐nylon-‐6
(2) PHBV
(3) Buna-‐N
(4) Nylon-‐6,6
Solution: (1) It is an alternating polyamide copolymer of glycine (H!N − CH! − COOH) and amino
caproic acid [H!N CH! !COOH].
89. The total number of π-‐bond electrons in the following structure is:
(1) 4
(2) 8
56.
(3) 12
(4) 16
Solution: (2) π bond electrons =
No. of π bonds×2 = 4×2 = 8
90. An organic compound ‘X’ having molecular formula C!H!"O yields phenyl hydrazine and
gives negative response to the iodoform test and Tollen’s test. It produces n-‐pentane on
reduction. ‘X’ could be:
(1) Pentanal
(2) 2-‐pentanone
(3) 3-‐pentanone
(4) N-‐amyl alcohol
Solution: (3) For the iodoform test there must be a terminal −CH! group, similarly for the Tollen’s
test −CHO group is required. Hence, as per question the structure would be
Biology
91. Which
one
of
the
following
matches
is
correct?
(1) Phytophthora
-‐
Aseptate
mycelium
-‐
Basidiomycetes
(2) Alternaria
–
Sexual
reproduction
absent
–
Deuteromycetes
(3) Mucor
–
Reproduction
by
Conjugation
–
Ascomycetes
(4) Agaricus
–
Parasitic
fungus
–
Basidiomycetes
Solution:
(2)
As
Alternaria
belongs
to
Class
Deuteromycetes
and
have
asexual
reproduction
by
conidia,
thus
sexual
reproduction
absent.
92. Read
the
following
five
statements
(A
to
E)
and
select
the
option
with
all
correct
statements:
(A) Mosses
and
Lichens
are
the
first
organisms
to
colonise
a
bare
rock.
(B) Selaginella
is
a
homosporous
pteridophyte.
(C) Coralloid
roots
in
Cycas
have
VAM.
(D) Main
plant
body
in
bryophytes
is
gametophytic,
whereas
in
pteridophytes
it
is
sporophytic.
57.
(E) In
gymnosperms,
male
and
female
gametophytes
are
present
with
in
sporangia
located
on
sporophytes.
(1) (A),
(C)
and
(D)
(2) (B),
(C)
and
(D)
(3) (A),
(D)
and
(E)
(4) (B),
(C)
and
(E)
Solution:
(3)
Mosses
are
found
in
moist,
shady
places
,moist
rocks,
damp
walls
or
on
tree
trunks
as
well
as
lichens
grow
on
rock
surfaces
&
named
by
Theophrastus
are
the
first
organisms
to
colonise
bare
rock.
Main
plant
body
in
bryophytes
is
gametophyte
because
gametophyte
is
dominant
over
sporophyte
as
gametophyte
is
branched,
haploid,
long
lived
and
independent
whereas
sporophyte
is
diploid
short
lived
and
dependent
upon
gametophyte.
Main
plant
body
in
pteridophyte
is
sporophyte.
93. In
which
of
the
following
gametophyte
is
not
independent
free
living?
(1) Funaria
(2) Marchantia
(3) Pteris
(4) Pinus
Solution:
(4)
In
plants
(gymnosperms
and
angiosperms),
gametophytes
is
dependent
and
develop
into
multicellular
organisms
while
still
enclosed
within
the
sporangium
while
in
bryophytes
(mosses,
liverworts,
and
hornworts),
the
gametophyte
free
living
(photosynthetic)
eg.
Funaria,Marchantia
In
ferns,
gametophytes
are
photosynthetic
free
living
organism
called
a
prothallus
(Pteris).
94. Which
one
of
the
following
statements
is
wrong?
(1) Algin
and
carrageen
are
products
of
algae
(2) Agar-‐agar
is
obtained
from
Gelidium
and
Gracilaria
(3) Chlorella
and
Spirulina
and
used
a
space
food
(4) Mannitol
is
stored
food
in
Rhodophyceae
Solution:
(4)
Mannitol
is
the
stored
food
in
Phaeophyceae
(brown
algae)
In
RHODOPHYCEAE
stored
food
is
floridean
starch.
95. The
guts
of
cow
and
buffalo
possess:
(1) Fucus
spp.
(2) Chlorella
spp.
(3) Methanogens
(4) Cyanobacteria
58.
Solution:
(3)
The
Bovine
rumen
is
a
niche
that
has
a
narrow
pH
range
and
is
kept
fairly
stabilized,
so
that
the
bacteria
in
the
rumen
as
well
as
the
intake
sacs
for
food
and
water
are
not
too
disturbed
The
organisms
responsible
for
producing
methane
through
bovine
rumen
are
termed
methanogens
and
do
so
in
order
to
reduce
the
amount
of
carbon
in
the
rumen
system
for
fermentation.
There
are
two
major
divisions
within
the
methanogens
found
in
the
bovine
rumen.
The
Methanobrevibacter
ruminatium
and
the
Methanosphaera
stadtmanae.
96. Male
gametes
are
flagellated
in:
(1) Polysiphonia
(2) Anabaena
(3) Ectocurpus
(4) Spirogyra
Solution:
(3)
Ectocarpus
gametes
and
spores
are
characterized
by
two
flagella
.Male
and
female
gametes
are
morphologically
identical
in
Ectocarpus
but
differ
with
respect
to
their
physiology
and
their
behavior.
Female
gametes
settle
sooner
and
produce
a
pheromone
whilst
male
gametes
swim
for
longer
and
are
attracted
to
the
pheromone
produced
by
the
female.
97. Vascular
bundles
in
monocotyledons
are
considered
closed
because:
(1) A
bundle
sheath
surrounds
each
bundle
(2) Cambium
is
absent
(3) There
are
no
vessels
with
perforations
(4) Xylem
is
surrounded
all
around
by
phloem
Solution:
(2)
The
vascular
bundles
of
monocotyledonous
plants
do
not
contain
a
layer
of
meristematic
tissue
(cambium)
as
the
dicots
do.
Thus
no
new
cells
can
be
formed
inside
the
vascular
bundles
of
monocots
and
their
vascular
bundles
are
termed
closed
whereas
those
of
dicot
plants
are
open.
98.
is
the
floral
formula
of:
(1) Allium
(2) Sesbania
(3) Petunia
(4) Brassica
Solution:
(3)
59.
Given
floral
formula
is
of
Solanaceae
family.
Petunia
is
a
plant
of
Solanaceae
family.
99. A
major
characteristic
of
the
monocot
root
is
the
presence
of:
(1) Open
vascular
bundles
(2) Scattered
vascular
bundles
(3) Vasculature
without
cambium
(4) Cambium
sandwiched
between
phloem
and
xylem
along
the
radius
Solution:
(3)
Vascular
cambia
are
found
in
dicots
and
gymnosperms
but
not
in
monocots
because
radial
vascular
bundle
is
present
in
monocot
root
in
which
cambium
is
not
present.
100. Keel
is
the
characteristic
feature
of
flower
of:
(1) Tulip
(2) Indigogera
(3) Aloe
(4) Tomato
Solution:
(2)
As
Indigofera
belongs
to
Family
Fabaceae.
This
family
was
earlier
called
Papilionoideae,
a
sub-‐family
of
family
Leguminosae.
Keel
is
a
characteristic
of
Family
Fabaecae
(enclosing
stamens
and
pistil)
101. Perigynous
flowers
are
found
in:
(1) Guava
(2) Cucumber
(3) China
rose
(4) Rose
Solution:
(4)
Perigynous
flower
means
thalamus
is
either
disc/cup/flasked
shaped.
Thalmus
of
rose
is
cup
shaped,
ovary
lies
in
the
centre
of
thalamus,
all
the
whorls
arise
from
periphery
and
remain
at
the
same
level.
Flowers
of
guava
and
cucumber
are
epigynous(i.e.
gynoecium
is
completely
inserted
within
thalamus
60.
102. Leaves
become
modified
into
spines
in:
(1) Opuntia
(2) Pea
(3) Onion
(4) Slik
Cotton
Solution:
(1)
Opuntia
leaves
get
modified
into
spikes
to
protect
them
from
grazing
animals,
also
reduce
area
of
transpiration.
Pea
leaves
are
modified
into
tendrils.
Onion
leaves
become
fleshy
since
food
is
stored
in
it.
Silk
cotton
get
modified
from
unifoliate
leaves
to
multifoliate
leaves.
103. The
structures
that
are
formed
by
stacking
of
organized
flattened
membranous
sacs
in
the
chloroplasts
are:
(1) Cristae
(2) Grana
(3) Stroma
lamellae
(4) Stroma
Solution:
(2)
Cristae
found
in
mitochondria.
Stroma
lamellae,
they
are
thylakoids
that
cross
the
stroma
of
a
chloroplast,
interconnecting
the
grana.
Stroma
is
the
matrix
of
a
chloroplast.
104. The
chromosomes
in
which
centromere
is
situated
close
to
one
end
are:
(1) Metacentric
(2) Acrocentric
(3) Telocentric
(4) Sub-‐metacentric
Solution:
(2)
Metacentric
chromosomes
has
centromere
in
the
middle
position.
Telocentric
chromosomes
has
centromere
at
the
terminal
position.
61.
Submetacentric
chromosomes
has
centromere
near
the
centre
but
not
in
the
middle.
105. Select
the
correct
matching
in
the
following
pairs:
(1) Smooth
ER
-‐
Oxidation
of
phospholipids
(2) Smooth
ER
-‐
Synthesis
of
lipids
(3) Rough
ER
–
Synthesis
of
glucogen
(4) Rough
ER
–
Oxidation
of
fatty
acids
Solution:
(2)
Synthesis
of
lipids
is
the
main
function
of
smooth
ER,
besides
this
smooth
ER
also
engaged
in
synthesis
of
glycogen
and
steroids.
Rough
ER
is
responsible
for
protein
synthesis.
Oxidation
of
fatty
acids
takes
place
in
microbodies(Glyoxysomes).
106. True
nucleus
is
absent
in:
(1) Anabaena
(2) Mucor
(3) Vaucheria
(4) Volvox
Solution:
(1)
Anabaena
cells
do
not
have
any
organelles
so
they
do
not
have
a
nucleus.
This
is
because
they
can
perform
photosynthesis
in
their
own
cell
membrane.
107. Which
one
of
the
following
is
not
an
inclusion
body
found
in
prokaryotes?
(1) Phosphate
Granule
(2) Cyanophycean
Granule
(3) Glycogen
Granule
(4) Polysome
Solution:
(4)
Polysomes
are
the
cell
organelle
found
in
cytoplasm
in
a
free
floating
form.
108. Transpiration
and
root
pressure
cause
water
to
rise
in
plants
by:
62.
(1) Pulling
it
upward
(2) Pulling
and
pushing
it,
respectively
(3) Pushing
it
upward
(4) Pushing
and
pulling
it,
respectively
Solution:
(2)
In
transpiration
water
rises
in
plant
due
to
strong
cohesive
force
of
transpiration
pull
in
which
water
is
pulled
(absorbed)
from
soil
to
roots,
roots
to
stem,
stem
to
xylem
of
leaf
and
lastly
xylem
to
vein
of
leaf
which
is
then
evaporated.
Whereas,
root
pressure
causes
passive
absorption
of
water
resulting
in
rise
of
water
in
plants
i.e.
it
creates
tension
in
xylem
elements
which
is
transmitted
downwards
upto
the
roothair,
as
a
result
roots
are
subjected
to
tension
and
suction
is
set
up
in
xylem;
hence
water
is
pulled
inside
the
roots.
109. Minerals
known
to
be
required
in
large
amounts
for
plant
growth
include:
(1) Phosphorus,
potassium,
sulphur,
calcium
(2) Calcium,
magnesium,
manganese,
copper
(3) Potassium,
phosphorus,
selenium,
boron
(4) Magnesium,
sulphur,
iron,
zinc
Solution:
(1)
Major/macroelements/macronutrients/meganutrients
are
required
in
large
amounts
which
include:-‐
C,
H,
O,
N,
P,
K,
Ca,
S,
Mg
and
Fe.
While
minor/micronutrients/trace
elements
are
required
in
very
small
amounts,
these
incude:-‐
Cu,
Zn,
Mn,
B,
Mo
and
Cl.
Therefore
correct
option
is
(1)
which
includes
only
major
nutrients/megaelements.
110. What
causes
a
green
plant
exposed
to
the
light
on
only
one
side,
to
bend
toward
the
source
of
light
as
it
grows?
(1) Green
plants
need
light
to
perform
photosynthesis.
(2) Green
plants
seek
light
because
they
are
phototropic.
(3) Light
stimulates
plant
cells
on
the
lighted
side
to
grow
faster.
(4) Auxin
accumulates
on
the
shaded
side,
stimulating
greater
cell
elongation
there.
63.
Solution:
(4)
Auxins
accumulates
on
the
shaded
side,
stimulating
grater
cell
elongation
there.Since
auxins
move
from
morphological
apex
to
morphological
base
therefore
gets
accumulated
on
the
shaded
side.
Thus
plants
exposed
to
the
light
on
only
one
side
bends
towards
the
source
of
light.
It
is
also
due
to
the
tropic
movement
of
auxins
which
causes
phototropism
and
geotropism
in
plants.
111. In
a
ring
girdled
plant:
(1) The
shoot
dies
first
(2) The
root
dies
first
(3) The
shoot
and
root
die
together
(4) Neither
root
nor
shoot
will
die
Solution:
(2)
In
a
ring
girdled
plant
the
root
dies
first.
Like
all
vascular
plants,
trees
use
two
vascular
tissues
for
transportation
of
water
and
nutrients:
the
xylem
(also
known
as
the
wood),
and
the
phloem
(the
innermost
layer
of
the
bark).
Girdling
results
in
the
removal
of
the
phloem,
and
death
occurs
from
the
inability
of
the
leaves
to
transport
sugars
(primarily
sucrose)
to
the
roots.
112. Typical
growth
curve
in
plants
is:
(1) Sigmoid
(2) Linear
(3) Stari-‐steps
shaped
(4) Parabolic
Solution:
(1)
Growth
pattern
of
cell,
organisms
is
uniform
under
favorable
conditions
&
a
typical
growth
curve
in
plants
is
sigmoidal.
Thus
following
phases
of
growth
are
recognized.
(i)
Lag
phase:
In
lag
period
the
growth
is
slow.
It
represents
formative
or
cell
division
phase.
(ii)
Log
phase:
Also
called
as
exponential
phase.
During
this
phase
growth
is
maximum
&
most
rapid.
It
represents
cell
elongation
phase.
(iii)
Steady
or
stationary
phase:
It
represents
cell
maturation
phase.
Time
taken
in
growth
phases
(mainly
log
phase)
is
called
as
grand
period
of
growth.
64.
113. Which
one
given
the
most
valid
and
recent
explanation
for
stomatal
movements?
(1) Transpiration
(2) Potassium
influx
and
efflux
(3) Starch
hydrolysis
(4) Guard
cell
photosynthesis
Solution:
(2)
The
most
recent
&
valid
explanation
of
stomatal
movements
is
given
by
potassium
influx
&
efflux/potassium
ion
pump
theory
which
is
also
as
the
modern
theory.
The
theory
was
given
by
Levitt
in
1974.
This
theory
states
that
the
accumulation
K+
ions
leads
to
opening
of
stomata
during
the
day
while
the
reverse
situation
prevails
during
dark
when
the
stomata
is
closed.
114. The
hilum
is
a
scar
on
the:
65.
(1) Seed,
where
funicle
was
attached
(2) Fruit,
where
it
was
attached
to
pedicel
(3) Fruit,
where
style
was
present
(4) Seed,
where
micropyle
was
present
Solution:
(1)
-‐
Hilum
is
a
scar
on
the
surface
of
a
seed
marking
its
point
of
attachment
to
the
seed
stalk
(funicle)
115. Which
one
of
the
following
may
require
pollinators,
but
is
genetically
similar
to
autogamy?
(1) Geitonogamy
(2) Xenogamy
(3) Apogamy
(4) Cleistogamy
Solution:
(1)
Geitonogamy
is
functionally
cross-‐pollination
involving
a
pollinating
agent,
genetically
it
is
similar
to
autogamy
since
the
pollen
grains
come
from
the
same
plant.
116. Which
one
of
the
following
statements
is
not
true?
(1) Pollen
grains
are
rich
in
nutrients,
and
they
are
used
in
the
form
of
tablets
and
syrups
(2) Pollen
grains
of
some
plants
cause
severe
allergies
and
bronchial
afflictions
in
some
people
(3) The
flowers
pollinated
by
files
and
bats
secrete
foul
odour
to
attract
them
(4) Honey
is
made
by
bees
by
digesting
pollen
collected
from
flowers
Solution:
(4)
Honey
is
made
by
bees
by
digesting
the
nectar
collected
from
flowers.
117. Transmission
tissue
is
characteristic
feature
of:
(1) Hollow
style
(2) Solid
style
(3) Dry
stigma
(4) Wet
stigma
Solution:
(2)
Transmission
tissue
is
characteristic
feature
of
solid
style.
They
are
located
in
centre
of
style
and
cytoplasm
of
these
cells
are
rich
in
organelles.
It
is
essential
for
pollen
tube
growth,
66.
because
of
the
nutrients
and
guidance.
It
also
regulates
GSI
(Gametophytic
self
–
Incompatibility)
in
style.
118. In
ginger
vegetative
propagation
occurs
through:
(1) Rhizome
(2) Offsets
(3) Bulbils
(4) Runners
Solution:
(1)
In
Ginger
vegetative
propagation
occurs
through
rhizome.
Rhizome
which
is
a
modified
subterranean
stem
of
the
plant
that
is
usually
found
underground.
The
rhizome
retains
the
ability
to
allow
new
shoots
to
grow
upwards.
The
plant
uses
rhizome
to
store
starch
&
protein.
If
Rhizome
is
separated
into
pieces
each
piece
may
be
able
to
give
rise
to
a
new
plant.
This
is
called
as
vegetative
propagation.
119. Which
of
the
following
are
the
important
floral
rewards
to
the
animal
pollinators?
(1) Colour
and
large
size
of
flower
(2) Nectar
and
pollen
grains
(3) Floral
fragrance
and
calcium
crystals
(4) Protein
pellicle
and
stigmatic
exudates
Solution:
(2)
The
important
floral
rewards
to
the
animal
pollinators
are
nectar
&
pollen
grain.
Plants
attract
pollinators
to
their
flowers
by
advertising
their
floral
rewards
i.e
nectar
&
pollen
grain.
They
take
the
advantage
of
the
fact
that
the
animals
can
see,
smell
&
taste
by
evolving
different
different
flower
sizes,
shapes,
colors
and
scents
to
selectively
attract
pollinators.
120. How
many
pairs
of
contrasting
characters
in
pea
plants
were
studied
by
Mendel
in
his
experiments?
(1) Five
(2) Six
67.
(3) Eight
(4) Seven
Solution:
(4)
Mendel
had
studied
7
pairs
of
contrasting
characters
in
pea
plants
in
his
experiments
which
are
stem
height,
seed
colour,
seed
shape,
pod
colour,
pod
shape,
flower
position
&
flower
colour.
121. Which
is
the
most
common
mechanism
of
genetic
variation
in
the
population
of
sexually-‐
reproducing
organism?
(1) Transducting
(2) Chromosomal
aberrations
(3) Genetic
Drift
(4) Recombination
Solution:
(4)
The
most
common
mechanism
of
genetic
variation
is
recombination.
In
sexually
reproducing
organism
during
gametogenesis,
the
homologus
chromosomse
exchanges
genetic
matrial
by
process
of
crossing
over.This
produces
new
combination.It
is
responsible
for
variation.
122. A
technique
of
micropropagation
is
(1) Somatic
hybridization
(2) Somatic
Embryogenesis
(3) Protoplast
fusion
(4) Embryo
rescue
Solution:
(2)
Micropropagation
or
PTC
(Plant
tissue
culture)
is
a
techinique
of
producing
thousands
of
plantlets
from
explant
in
aseptic
environment,
it
can
be
performed
by
callus
culture
or
somatic
embryogenesis.
123. The
movement
of
a
gene
from
one
linkage
group
to
another
is
called:
(1) Inversion
(2) Duplication
(3) Translocation
(4) Crossing
over
68.
Solution:
(3)
A set of genes at different loci on the same chromosome tend to act as a single pair of genes in
meiosis instead of undergoing independent assortment. Chromosome translocation is a chromosome
abnormality caused by rearrangement of parts between non-homologous chromosomes. Hence in
translocation there is movement of gene from one linkage group to another and lead to change the position
of gene.
124. Multiple
alleles
are
present;
(1) On
different
chromosomes
(2) At
different
loci
on
the
same
chromosome
(3) At
the
same
locus
of
the
chromosomes
(4) On
non-‐sister
chromatids
Solution:
(3)
Multiple
alleles
is
a
type
of
non-‐Mendelian
inheritance
pattern
that
involves
more
than
just
the
typical
two
alleles
that
usually
code
for
a
certain
characteristic
in
a
species.
125. Which
body
of
the
Government
of
india
regulates
GM
research
and
safety
of
introducing
GM
organisms
for
public
services/
(1) Bio-‐safety
committee
(2) Indian
Council
of
Agricultural
Research
(3) Genetic
Engineering
Approval
Committee
(4) Research
Committee
on
Genetic
Manipulation
Solution:
(3)
126. In
Bt
cotton,
the
Bt
toxin
present
in
plant
tissue
as
pro-‐toxin
is
converted
into
active
toxin
due
to:
(1) Alkaline
pH
of
the
insect
gut
(2) Acidic
pH
of
the
insect
gut
(3) Action
of
gut
micro-‐organisms
(4) Presence
of
conversion
factors
in
insect
gut
Solution:
(1)
Bacillus
thurigensis
which
contain
cry
protein
in
inactive
form
known
as
pro-‐toxin.
when
this
is
ingested
by
insect
it
get
activated
by
alkaline
ph
of
gut,
which
solubilise
the
protein
crystal.
This
activated
toxin
bind
to
the
epithelial
midgut
cell
result
into
lysis
of
epithelial
cell
and
eventually
lead
to
the
death
of
insect.
69.
127. The
crops
engineered
for
glyphosate
are
resistant/tolerant
to;
(1) Fungi
(2) Bacteria
(3) Insects
(4) Herbicides
Solution:
(4)
Glyphosate
(N-‐(phosphonomethyl)
glycine)
is
a
broad-‐spectrum
systemic
herbicide
used
to
kill
weeds.
Herbicide
tolerant
crops
are
designed
to
tolerate
specific
broad-‐spectrum
herbicides,
which
kill
the
surrounding
weeds,
but
leave
the
cultivated
crop
intact.
128. DNA
is
not
present
in:
(1) Chloroplast
(2) Ribosomes
(3) Nuclues
(4) Mitochondria
Solution:
(2)
The
ribosome
is
a
cellular
machine
which
is
highly
complex
and
is
made
up
of
dozens
of
distinct
protein
as
well
as
a
few
specialized
RNA
molecules
known
as
ribosomal
RNA
(rRNA)
and
does
not
contain
DNA
129. Which
of
the
following
enhances
or
induces
fusion
of
protoplasts?
(1) Sodium
chloride
and
potassium
chloride
(2) Polyethylene
glycol
and
sodium
nitrate
(3) IAA
and
kinetin
(4) IAA
and
Gibberellins
Solution:
(2)
The high molecular weight polymer (1000-6000) of PEG acts as a molecular bridges
connecting the protoplasts. Calcium ions linked the negatively charged PEG and membrane
surface. When PEG elute, the surface potential are disturbed, leading to intramembrane contact
and subsequent fusion due to the strong affinity of PEG for water which may cause local
dehydration of the membrane and increase fluidity, thus inducing fusion.
70.
130. The
UN
Conference
of
Parties
on
climate
change
in
the
year
2011
was
held
in:
(1) Poland
(2) South
Africa
(3) Peru
(4) Qatar
Solution:
(2)
UN
conference
of
parties
on
climate
change
(2011)
was
held
at
Durban
South
Africa.
131. Vertical
distribution
of
different
species
occupying
different
levels
in
a
biotic
community
is
known
as:
(1) Divergence
(2) Stratification
(3) Zonation
(4) Pyramid
Solution:
(2)
By
the
definition
of
stratification,
it
is
the
distribution
of
a
community
in
different
levels
by
various
socioeconomic
means.
132. In
which
of
the
following
both
pairs
have
correct
combination?
(1) In
situ
conservation:
National
Park
Ex
situ
conservation:
Botanical
Garden
(2) In
situ
conservation:
Cryopreservation
Ex
situ
conservation:
Wildlife
Sanctuary
(3) In
situ
conservation:
National
Park
Ex
situ
conservation:
National
Park
(4) In
situ
conservation:
Tissue
culture
Ex
situ
conservation:
Sacred
groves
Solution:
(1)
In
situ
conservation
is
the
conservation
of
resources
in
its
natural
populations.
Ex
situ
conservation
is
the
conservation
of
resources
outside
its
habitat
maybe
wild
area
or
within
human
care.
Best
example
is
of
botanical
gardens.
133. Secondary
Succession
takes
place
on/in:
(1) Bare
rock
(2) Degraded
forest
(3) Newly
created
pond
(4) Newly
cooled
lava
71.
Solution:
(2)
Secondary
succession
usually
occurs
on
pre-‐existing
soil.
This
is
mainly
triggered
due
to
various
mechanisms
that
cause
forest
degradation
etc,.
134. The
mass
of
living
material
at
a
trophic
level
at
a
particular
time
is
called:
(1) Gross
primary
productivity
(2) Standing
state
(3) Net
primary
productivity
(4) Standing
crop
Solution:
(2)
Biomass
is
the
mass
of
the
living
material
present
at
a
given
time
at
a
particular
trophic
level,
also
called
as
standing
crop.
135. In
an
ecosystem
the
rate
of
production
of
organic
matter
during
photosynthesis
is
termed
as:
(1) Net
primary
productivity
(2) Gross
primary
productivity
(3) Secondary
productivity
(4) Net
productivity
Solution:
(2)
Gross
primary
productivity
is
amount
of
organic
matter
produced
at
given
length
of
time
during
photosynthesis.
136. Which
of
the
following
characteristics
is
mainly
responsible
for
diversification
of
insects
on
land?
(1) Segmentation
(2) Bilateral
symmetry
(3) Exoskeleton
(4) Eyes
Solution:
(3)
Exoskeleton
plays
a
role
in
defense
from
the
prey
and
is
also
an
important
characteristic
in
diversification
of
many
species.
Eg:
Insects
having
chitin
as
the
form
of
exoskeleton.
137. Which
of
the
following
endoparasites
of
humans
does
show
viviparity?
(1) Ancylostoma
duodenale
(2) Enterobius
vermicularis
(3) Trichinella
spiralis
(4) Ascaris
lumbricoides
72.
Solutions:
(3)
Viviparity
means
development
of
an
embryo
inside
the
body
of
mother
rather
than
laying
eggs.
Eg:
Trichinella
spiralis
is
an
endoparasite
of
human
body
that
shows
viviparity
whereas
the
remaining
endoparasites
lay
eggs
(oviparity).
138. Which
of
the
following
represents
the
correct
combination
without
any
exception?
(1)
Characteristics
Class
Mammary
gland;
hair
on
body;
pinnae;
two
pairs
of
limbs
Mammalia
(2)
Characteristics
Class
Mouth
ventral;
gills
without
operculum;
skin
with
placoid
scales;
persistent
notochord
Chondrichthyes
(3)
Characteristics
Class
Sucking
and
circular
mouth;
jaws
absent,
integument
without
scales;
paired
appendages
Cyclostomata
(4)
Characteristics
Class
Body
covered
with
feathers;
skin
moist
and
glandular;
fore
–
limbs
form
wings;
lungs
with
air
sacs
Aves
Solution:
(2)
Mammalia
:
pinnae
absent
in
whales.
Cyclostomata
:
paired
appendages
absent
in
Agnatha.
Aves
:
Almost
all
the
aves
show
presence
of
nonglandular
skin.
139. Which
of
the
following
animals
is
not
viviparous?
(1) Flying
Fox
(Bat)
(2) Elephant
(3) Platypus
(4) Whale
73.
Solution:
(3)
Platypus
is
an
oviparous
mammal
i.e.
it
lays
eggs
whereas
the
remaining
are
viviparous.
140. Erythropoiesis
starts
in:
(1) Kidney
(2) Liver
(3) Spleen
(4) Red
bone
marrow
Solution:
(4)
Erythropoiesis
starts
in
liver
and
spleen
in
a
foetus
whereas
in
adults
it
starts
in
the
red
bone
marrow.
As
we
cannot
opt
for
two
options
(spleen
and
liver)
simultaneously
which
are
for
foetus,
therefore
the
answer
is
red
bone
marrow
141. The
terga,
sterna
and
pleura
of
cockroach
body
are
joined
by
:
(1) Cementing
glue
(2) Muscular
tissue
(3) Arthrodial
membrane
(4) Cartilage
Solution:
(3)
Tough flexible cuticle between the sclerotized parts (skeletal elements) that allows
relative movement is called as arthrodial membrane. The terga, sterna and pleura are chitinous
plates which covers cockroach body. These three are linked together by thin arthrodial
membrane.
142. Nuclear
envelope
is
a
derivative
of:
(1) Smooth
endoplasmic
reticulum
(2) Membrane
of
Golgi
complex
(3) Microtubules
(4) Rough
endoplasmic
reticulum
Solution:
(4)
Rough endoplasmic reticulum form the nuclear envelop during karyokinesis. The
nuclear membrane (outer) is contiguous with the endoplasmic reticulum.
143. Cytochromes
are
found
in:
74.
(1) Matrix
of
mitochondria
(2) Outer
wall
of
mitochondria
(3) Cristae
of
mitochondria
(4) Lysosomes
Solution:
(3)
Cytochromes are the Iron containing electron acceptors, which are present on inner
mitochondrial membrane, called cristae, helpful in ETS.
144. Which
one
of
the
following
statements
is
incorrect?
(1) A
competitive
inhibitor
reacts
reversibly
with
the
enzyme
to
form
an
enzyme-‐inhibitor
complex.
(2) In
competitive
inhibition,
the
inhibitor
molecule
is
not
chemically
changed
by
the
enzyme.
(3) The
competitive
inhibitor
does
not
affect
the
rate
of
breakdown
of
the
enzyme-‐substrate
complex.
(4) The
presence
of
the
competitive
inhibitor
decreases
the
Km
of
the
enzyme
for
the
substrate.
Solution:
(4)
Competitive inhibition is a form of enzyme inhibition where binding of the inhibitor
to the active site on the enzyme prevents binding of the substrate and vice versa. Presence of
competitive inhibitor, increase the Km constant of enzyme (and not decrease) while does not
affects the Vmax as the competitive inhibitor binds at active site, so decrease the affinity of
enzyme for its substrate, so Km constant increase. If substrate concentration is increased,
inhibition over comes and attains normal Vmax.
145. Select
the
correct
option:
I
II
75.
(a)
Synapsis
aligns
homologous
chromosomes
(i)
Anaphase
–
II
(b)
Synthesis
of
RNA
and
protein
(ii)
Zygotene
(c)
Action
of
enzyme
recombinase
(iii)
𝐺!
–
phase
(d)
Centromeres
do
not
separate
but
chromatids
move
towards
opposite
poles
(iv)
Anaphase
–
I
(v)
Pachytene
(1)
(𝑎)
(𝑖𝑖)
(𝑏)
(𝑖)
(𝑐)
(𝑖𝑖𝑖)
(𝑑)
(𝑖𝑣)
(2)
(𝑎)
(𝑖𝑖)
(𝑏)
(𝑖𝑖𝑖)
(𝑐)
(𝑣)
(𝑑)
(𝑖𝑣)
(3)
(𝑎)
(𝑖)
(𝑏)
(𝑖𝑖)
(𝑐)
(𝑣)
(𝑑)
(𝑖𝑣)
(4)
(𝑎)
(𝑖𝑖)
(𝑏)
(𝑖𝑖𝑖)
(𝑐)
(𝑖𝑣)
(𝑑)
(𝑣)
Solution:
(2)
Synapsis of homologous chromosome – Zygotene
Synthesis of RNA & protein – 𝐺! phase
Action of enzyme recombinase – Pachytene
Centromeres do not separate but chromatid move towards opposite pole – Anaphase-I
146. A
somatic
cell
that
has
just
completed
the
S
phase
of
its
cell
cycle,
as
compared
to
gamete
of
the
same
species,
has:
(1) Twice
the
number
of
chromosomes
and
twice
the
amount
of
DNA
(2) Same
number
of
chromosomes
but
twice
the
amount
of
DNA
(3) Twice
the
number
of
chromosomes
and
four
times
the
amount
of
DNA
(4) Four
times
the
number
of
chromosomes
and
twice
the
amount
of
DNA
76.
Solution:
(3)
S-phase (synthesis phase) is the part of the cell cycle in which DNA is replicated, but
number of chromosome does not change.
If a cell is diploid then after S phase DNA content in cell will be 4 C and Chromosome in cell
will be 2 N.
While in gamete,
DNA content in gamete = C
Chromosome in gamete = N
Hence, Number of chromosome in somatic cell will be twice than gamete while DNA content
will be four times.
147. Which
of
the
following
statements
is
not
correct?
(1) Brunner’s
glands
are
present
in
the
submucosa
of
stomach
and
secrete
pepsinogen
(2) Goblet
cells
are
present
in
the
mucosa
of
intestine
and
secrete
mucus
(3) Oxyntic
cells
are
present
in
the
mucosa
of
stomach
and
secrete
HCL
(4) Acini
are
present
in
the
pancreas
and
secrete
carboxypeptidase
Solution:
(1)
Brunner's glands (or duodenal glands) are compound tubular submucosal glands
found in that portion of the duodenum which is above the hepatopancreatic sphincter (Sphincter
of Oddi). The main function of these glands is to produce a mucus-rich alkaline secretion
(containing bicarbonate), which is the non-enzymatic part of intestinal juice.
148. Gastric
juice
of
infants
contains:
(1) Maltase,
pepsinogen,
rennin
(2) Nuclease,
pepsinogen,
lipase
(3) Pepsinogen,
lipase,
rennin
(4) Amylase,
rennin,
pepsinogen
Solution:
(3)
Gastric acid is a digestive fluid, formed in the stomach. In infants, it primarily has
the ability to digest milk protein by enzyme rennin, along with small amounts of pepsinogen &
lipase.
149. When
you
hold
your
breath,
which
of
the
following
gas
changes
in
blood
would
first
lead
to
the
urge
to
breathe?
(1) Falling
𝑂!
concentration
(2) Rising
𝐶 𝑂!
concentration
(3) Falling
𝐶 𝑂!
concentration
(4) Rising
𝐶 𝑂!
and
falling
𝑂!
concentration
77.
Solution:
(2)
Rise in 𝐶𝑂! concentration stimulates chemoreceptors present in aorta and carotid
artery which stimulates respiratory centre. Respiratory centre is not directly sensitive to oxygen
concentration & hence desire to breath is induced by rise in 𝐶𝑂! concentration of blood.
150. Blood
pressure
in
the
mammalian
aorta
is
maximum
during:
(1) Systole
of
the
left
atrium
(2) Diastole
of
the
right
ventricle
(3) Systole
of
the
left
ventricle
(4) Diastole
of
the
right
atrium
Solution:
(3)
Left ventricular systole drives blood through the aortic valve (AoV) to the body and
organs excluding the lungs. Hence B.P. in Aorta will be maximum when left ventricle pumps the
stroke volume into its lumen during its systole.
151. Which
one
of
the
following
is
correct?
(1) Plasma
=
Blood
–
Lymphocytes
(2) Serum
=
Blood
+
Fibrinogen
(3) Lymph
=
Plasma
+
RBC
+
WBC
(4) Blood
=
Plasma
+
RBC
+
WBC
+
Platelets
Solution:
(4)
Blood is a liquid connective tissue which contains plasma, RBCs, WBCs & Platelet.
In animals it delivers necessary substances such as nutrients and oxygen to the cells and
transports metabolic waste products away from those same cells.
152. Removal
of
proximal
convoluted
tubule
from
the
nephron
will
result
in:
(1) More
diluted
urine
(2) More
concentrated
urine
(3) No
change
in
quality
and
quantity
of
urine
(4) No
urine
formation
Solution:
(1)
The question is wrongly framed in concept. Hence no appropriate answer which is
absolutely correct, can be found.
Still the least incorrect answer can be (1) because maximum reabsorption of filterate (70%)
occurs from P.C.T.
Hence removal of PCT will increase the urine volume.
153. Sliding
filament
theory
can
be
best
explained
as:
(1) When
myofilaments
slide
pass
each
other
Actin
filaments
shorten
while
Myosin
filament
do
not
shorten
(2) Actin
and
Myosin
filaments
shorten
and
slide
pass
each
other
78.
(3) Actin
and
Myosin
filaments
do
not
shorten
but
rather
slide
pass
each
other
(4) When
myofilaments
slide
pass
each
other,
Myosin
filaments
shorten
while
Actin
filaments
do
not
shorten
Solution:
(3)
In the sliding filament model, the thick and thin filaments past each other, shortening
the sarcomere, i.e., actin & myosin filaments do not shorten, rather actin filaments slide over
myosin filaments.
154. Glenoid
cavity
articulates:
(1) Clavicle
with
acromion
(2) Scapula
with
acromion
(3) Clavicle
with
scapula
(4) Humerus
with
scapula
Solution:
(4)
The glenoid cavity (or glenoid fossa of scapula) is a part of the shoulder, located on
the lateral angle of the scapula. It is directed laterally and forward and articulates with the head
of the humerus.
155. Which
of
the
following
regions
of
the
brain
is
incorrectly
paired
with
its
function?
(1) Medulla
oblongata
–
homeostatic
control
(2) Cerebellum
–
language
comprehension
(3) Corpus
callosum
–
communication
between
the
left
and
right
cerebral
cortices
(4) Cerebrum
–
calculation
and
contemplation
Solution:
(2)
The cerebellum is involved in the coordination of voluntary motor movement,
balance and equilibrium and not language comprehension. It is located just above the brain stem
and toward the back of the brain.
156. A
gymnast
is
able
to
balance
his
body
upside
down
even
in
the
total
darkness
because
of:
(1) Cochlea
(2) Vestibular
apparatus
(3) Tectorial
membrane
(4) Organ
of
corti
Solution:
(2)
The vestibular system is the sensory system that provides the sense of balance and
spatial orientation for the purpose of coordinating movement with balance.
79.
157. A
chemical
signal
that
has
both
endocrine
and
neural
roles
is:
(1) Melatonin
(2) Calcitonin
(3) Epinephrine
(4) Cortisol
Solution:
(3)
Epinephrine (also known as adrenaline) is a hormone and a neurotransmitter.
158.
Which
of
the
following
does
not
favour
the
formation
of
large
quantities
of
dilute
urine?
(1) Alcohol
(2) Caffeine
(3) Renin
(4) Atrial
–
natriuretic
factor
Solution:
(3)
Renin does not favor for the formation of large quantities of dilute urine as it
activates RAAS (Renin angiotensin activating system) so it causes reabsorption of sodium which
leads to formation of concentrated urine.
159. Capacitation
refers
to
changes
in
the:
(1) Sperm
before
fertilization
(2) Ovum
before
fertilization
(3) Ovum
after
fertilization
(4) Sperm
after
fertilization
Solution:
(1)
Sperm capacitation refers to the physiological changes spermatozoa must undergo in
order to have the ability to penetrate and fertilize an egg. The changes take place via the sperm
cell membrane in which it may be that receptors are made available through the removal of a
glycoprotein layer. The area of the acrosomal cap is also so altered thereby that the acrosome
reaction becomes possible.
160. Which
of
these
is
not
an
important
component
of
initiation
of
parturition
in
humans?
(1) Increase
in
estrogen
and
progesterone
ratio
(2) Synthesis
of
prostaglandins
(3) Release
of
oxytocin
(4) Release
of
prolactin
Solution:
(4)
Prolactin has no role in parturition, it helps in lactation process, (lactogenic
hormone), development of mammary glands (mammotropin) & maintenance of corpus luteum
(leutotropin).
161. Which
of
the
following
viruses
is
not
transferred
through
semen
of
an
infected
male?
(1) Hepatitis
B
virus
(2) Human
immunodeficiency
virus
(3) Chikungunya
virus
80.
(4) Ebola
virus
Solution:
(3)
Chikungunya is a viral disease transmitted to humans by infected mosquitoes. It
causes fever and severe joint pain.
162. Which
of
the
following
cells
during
gametogenesis
is
normally
diploid?
(1) Primary
polar
body
(2) Spermatid
(3) Spermatogonia
(4) Secondary
polar
body
Solution:
(3)
Spermatogonia or the sperm mother cell is diploid while all other cells are formed in
later steps of spermatogenesis during mitotic cell division.
163. Hysterectomy
is
surgical
removal
of
(1) Uterus
(2) Prostate
gland
(3) Vas-‐deference
(4) Mammary
glands
Solution:
(1)
Hystero is the term used for uterus. Hysterectomy is the surgical removal of the
uterus. It may also involve removal of the cervix, ovaries, fallopian tubes and other surrounding
structures.
164. Which
of
the
following
is
not
a
sexually
transmitted
disease?
(1) Syphilis
(2) Acquired
Immuno
Deficiency
Syndrome
(AIDS)
(3) Trichomobiasis
(4) Encephalitis
Solution:
(4)
Encephalitis is an acute inflammation (swelling up) of the brain resulting either from
a viral infection or when the body's own immune system mistakenly attacks brain tissue. The
most common cause is a viral infection.
165. An
abnormal
human
baby
with
‘XXX’
sex
chromosomes
was
born
due
to:
(1) Formation
of
abnormal
sperms
in
the
father
(2) Formation
of
abnormal
ova
in
the
mother
(3) Fusion
of
two
ova
and
one
sperm
(4) Fusion
of
two
sperms
and
ovum
81.
Solution:
(2)
Formation of abnormal ova i.e., 22 + XX in the mother will lead to birth of baby
with ‘XXX’ genotype, due to chromosomal non disjunction in ova. Nondisjunction is the failure
of the chromosomes to disjoin and move to opposite poles.
166. Alleles
are
(1) Different
phenotype
(2) True
breeding
homozygotes
(3) Different
molecular
forms
of
a
gene
(4) Heterozygotes
Solution:
(3)
An allele or an allelomorph is one of a number of alternative forms of the same gene
or same genetic locus.
167. A
man
with
blood
group
‘A’
marries
a
woman
with
blood
group
‘B’.
What
are
all
the
possible
blood
groups
of
their
offsprings
:
(1) A
and
B
only
(2) A,
B,
and
AB
only
(3) A,
B,
AB
and
O
(4) O
only
Solution:
(3)
168. Gene
regulation
governing
lactose
operon
of
E.coli
that
involves
the
lac
I
gene
product
is:
(1) Positive
and
inducible
because
it
can
be
induced
by
lactose
(2) Negative
and
inducible
because
repressor
protein
prevents
transcription
(3) Protein
and
repressible
because
repressor
protein
prevents
transcription
(4) Feedback
inhibition
because
excess
of
𝛽-‐
galactosidase
can
switch
off
transcription
Solution:
(2)
In negative regulation, a repressor molecule binds to the operator of an operon and
terminates transcription. The lac operon is a negatively controlled inducible operon, where the
inducer molecule is allolactose.
82.
169. In
sea
urchin
DNA,
which
is
double
stranded,
17%
of
the
bases
were
shown
to
be
cytosine.
The
percentages
of
the
other
three
bases
expected
to
be
present
in
this
DNA
are
(1) G
34%,
A
24.5%,
T
24.5%
(2) G
17%,
A
16.5%,
T
32.5%
(3) G
17%,
A
33%,
T
33%
(4) G
8.5%,
A
50%,
T
24.5%
Solution:
(3)
According to Chargaff’s rule,
A = T, G = C
C = 17% So G = 17%
A+G / T+C =1
A + G + C + T = 100
A + 17 + 17 + T = 100
A + T = 100 – 34
So A = 33%, T = 33% = 66
170. Which
of
the
following
had
the
smallest
brain
capacity?
(1) Homo
erectus
(2) Homo
sapiens
(3) Homo
neanderthalensis
(4) Homo
habilis
Solution:
(4)
Homo habilis (Handy man) is considered to be the first human which evolved from
Australopithecus its cranial capacity was smallest (650 – 800cc) among humans.
171. A
population
will
not
exist
in
Hardy-‐Weinberg
equilibrium
if
:
(1) Individuals
mate
selectively
(2) There
are
no
mutation
(3) There
is
no
migration
(4) The
population
is
large
Solution:
(1)