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SRMJEE Mock Test - 3 (Medical) - JEE MCQ


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30 Questions MCQ Test SRMJEEE Subject Wise & Full Length Mock Tests 2025 - SRMJEE Mock Test - 3 (Medical)

SRMJEE Mock Test - 3 (Medical) for JEE 2025 is part of SRMJEEE Subject Wise & Full Length Mock Tests 2025 preparation. The SRMJEE Mock Test - 3 (Medical) questions and answers have been prepared according to the JEE exam syllabus.The SRMJEE Mock Test - 3 (Medical) MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for SRMJEE Mock Test - 3 (Medical) below.
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SRMJEE Mock Test - 3 (Medical) - Question 1

If a screw gauge moves 1 mm in two rotations, then the pitch of the screw gauge is

Detailed Solution for SRMJEE Mock Test - 3 (Medical) - Question 1
Pitch = distance covered by screw head/number of rotations
= (1 mm)/2
= 0.5 mm
SRMJEE Mock Test - 3 (Medical) - Question 2

A hollow cylinder has a charge q coulomb within it. If ϕ is the electric flux in units of voltmeter associated with the curved surface B the flux linked with the plane surface A in units of voltmeter will be

Detailed Solution for SRMJEE Mock Test - 3 (Medical) - Question 2

The flux through the cylinder is

As ϕA = ϕC

SRMJEE Mock Test - 3 (Medical) - Question 3

A signal of frequency 20 kHz and peak voltage of 5 Volts is used to modulate a carrier wave of frequency 1.2 MHz and peak voltage 25 Volts. Choose the correct statement.

Detailed Solution for SRMJEE Mock Test - 3 (Medical) - Question 3

Modulation index defines the modulated signal's ratio of maximum to minimum voltage:

Modulation index = Vm / Vc = 5 / 25 = 0.2

Carrier Frequency fc = 1.2 × 10³ kHz = 1200 kHz

Lower side-band frequency, f₁ = 1200 - 20 = 1180 kHz

Upper side-band frequency, f₂ = 1200 + 20 = 1220 kHz

SRMJEE Mock Test - 3 (Medical) - Question 4

The upper half of an inclined plane with inclination ϕ is perfectly smooth, while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if coefficient of friction for the lower half is given by

Detailed Solution for SRMJEE Mock Test - 3 (Medical) - Question 4

For the smooth portion BC,

u = 0, s = l, a = g sinϕ

v = ?

From v² − u² = 2as,

v² − 0 = 2g sinϕ × l

For the rough portion CO:

u = v = √(2g sinϕ ⋅ l)

v = 0, a = g (sinϕ − μ cosϕ)

s = l

From v² − u² = 2as,

0 − 2gl sinϕ = 2g (sinϕ − μ cosϕ) l

−sinϕ = sinϕ − μ cosϕ

μ cosϕ = 2 sinϕ

μ = 2 tanϕ

SRMJEE Mock Test - 3 (Medical) - Question 5

 

A sphere of mass M and radius R2 has a concentric cavity of radius R1 as shown in the figure. The force F exerted by the sphere on a particle of mass mlocated at a distance r from the centre of sphere varies as (0 ≤ r ≤ ∞)

Detailed Solution for SRMJEE Mock Test - 3 (Medical) - Question 5

F = 0 when 0 ≤ r ≤ R₁
Because intensity is zero inside the cavity.

F increases when R₁ ≤ r ≤ R₂
As there will be more mass enclosed if the radius of the enclosure is increased.

F ∝ 1/r² when r > R₂
As all mass is enclosed, it will behave as a point mass.

SRMJEE Mock Test - 3 (Medical) - Question 6

The amount of energy released in the annihilation of an electron and positron is ____ (me = 0.00055 amu)

Detailed Solution for SRMJEE Mock Test - 3 (Medical) - Question 6

Electron and Positron is a matter and anti matter pair. The mass of an electron and of a positron is 0.00055 amu.
When matter and anti-matter annihilate their total mass gets converted into energy. Therefore, the energy released will be,
E = 2 × (0.00055) × 931
⇒ E = 1.02 MeV

SRMJEE Mock Test - 3 (Medical) - Question 7

At 300°C, the hole in a steel plate has a diameter of 0.99970 cm. A cylinder of diameter exactly 1 cm at 30°C is to be slid into the hole. To what temperature must the plate be heated? (Given: αₛₜₑₑₗ = 1.1 × 10⁻⁵ °C⁻¹)

Detailed Solution for SRMJEE Mock Test - 3 (Medical) - Question 7

Δl = 1 - 0.99970 = 0.00030 cm
Δl = αl Δt
Δt = Δl / (αl) = 0.0003 / (1.1 × 10⁻⁵ × 0.9997) = 27.3°C.
So, the plate temperature must be raised to 30°C + 27.3°C = 57.3°C.

SRMJEE Mock Test - 3 (Medical) - Question 8

The most suitable reagent for the conversion of 2-phenylpropanamide into 1-phenylethylamine is:

Detailed Solution for SRMJEE Mock Test - 3 (Medical) - Question 8

The most suitable reagent for the conversion of 2-phenylpropanamide into 1-phenylethylamine is Br2, NaOH.

SRMJEE Mock Test - 3 (Medical) - Question 9

For a first order chemical reaction,

Detailed Solution for SRMJEE Mock Test - 3 (Medical) - Question 9

Arrhenius equation for rate constant, k is k = Ae−Ea/RT where A is the pre-exponential factor.
Since, e−Ea/RT ​ is dimensionless, dimension of A = k.
For first order reaction, dimension of k = times−1 
∴ Also same for A.
As for the other options,
For a first order reaction, the product formation rate is directly proportional to reactant concentration, half-life = 0.693/k
or 69.3% of (1/k), and concentration of reactant changes with time by the equation which shows exponential decreases, not linear.

SRMJEE Mock Test - 3 (Medical) - Question 10
Which of the following forms a molecular solid?
Detailed Solution for SRMJEE Mock Test - 3 (Medical) - Question 10
A molecular solid is a solid composed of molecules held together by the Van der Waals forces. Because these forces are weaker than covalent or ionic bonds, molecular solids are soft and have relatively low melting temperatures. For example: SO2.
All the other are covalent (network) solids.
SRMJEE Mock Test - 3 (Medical) - Question 11
The major product in the following reaction is

Detailed Solution for SRMJEE Mock Test - 3 (Medical) - Question 11
SRMJEE Mock Test - 3 (Medical) - Question 12
An example of a vicinal dihalide is
Detailed Solution for SRMJEE Mock Test - 3 (Medical) - Question 12
Vicinal dihalides are compounds that have halogens at the adjacent carbons.
They can be prepared by a reaction between halogens and alkenes.
The simplest example of a vicinal dihalide is 1,2-dichloroethane (ethylene dichloride).

SRMJEE Mock Test - 3 (Medical) - Question 13

Which is non-reducing sugar

Detailed Solution for SRMJEE Mock Test - 3 (Medical) - Question 13

The carbohydrates or sugar where free aldehyde or ketonic group is absent (utilized in glycosidic bond formation) can not reduce the above reagents are called non-reducing sugar i.e., Sucrose, glycogen, starch.

SRMJEE Mock Test - 3 (Medical) - Question 14

Substances which can be stretched to couse large strain are called

Detailed Solution for SRMJEE Mock Test - 3 (Medical) - Question 14

Substances which can be stretched to couse large strain are called elastomers. e.g. tissue of aorta, rubber etc.

SRMJEE Mock Test - 3 (Medical) - Question 15

Which of the following is polycarbonate?

Detailed Solution for SRMJEE Mock Test - 3 (Medical) - Question 15

Lexan is a polymer of diethyl carbonate and bisphenol-A.
Lexan is a trademarked term for polycarbonate sheeting, one of the most widely used plastics in the world.
Lexan is not glass, but a polycarbonate resin thermoplastic. It is strong, transparent, temperature-resistant and easily formed, so is commonly used in place of glass.

SRMJEE Mock Test - 3 (Medical) - Question 16

The given figure shows development of an embryo that undergoes two phases A and B. Select the correct option regarding it.
886358

Detailed Solution for SRMJEE Mock Test - 3 (Medical) - Question 16

An embryo shows geometrical growth initially but later it passes into arithematic phase.

SRMJEE Mock Test - 3 (Medical) - Question 17

The given figure shows growth of two leaves over the period of one day. If, AG = absolute growth and RGR = relative growth rate, then select the correct option.

Detailed Solution for SRMJEE Mock Test - 3 (Medical) - Question 17

Absolute growth rate (AGR): Increase in total growth of two organs or organisms is measured and comparison of total growth per unit time is called absolute growth. Absolute growth rate is the total growth per unit time. Relative growth rate (RGR) : It is growth per unit time per unit initial growth.
RGR= Growth in given time period / Measurement at start of time period
AG for leaf A = 10cm2 = 5cm2 = 5cm2
RGR for leaf A = 5/5 ​× 100 = 100%
AG for leaf B = 55cm2 − 50cm2 = 5cm2
RGR for leaf B = 5/50 ​× 100 = 10%
Though the absolute growth is same for both the leaves (A and B), relative rate of growth is more in leaf A because of its initial small size.

SRMJEE Mock Test - 3 (Medical) - Question 18

Increase in girth (diameter).of plant as a result of the activities of lateral meristems is called _______________.

Detailed Solution for SRMJEE Mock Test - 3 (Medical) - Question 18

The Lateral meristem e.g. vascular cambium and  cork cambium(in dicotyledons and gymnosperms) are the meristems that cause increase in girth of the organs in which they are active, This is known as secondary growth of the plant.

SRMJEE Mock Test - 3 (Medical) - Question 19

Which of the following blood components play a major role in blood coagulation?

Detailed Solution for SRMJEE Mock Test - 3 (Medical) - Question 19
  • Platelets, also called thrombocytes, are cell fragments produced from megakaryocytes.
  • Platelets can release a variety of substances most of which are involved in the coagulation or clotting of blood.
SRMJEE Mock Test - 3 (Medical) - Question 20

In the systemic circulation, the blood vessel that carries blood from the intestine to the liver is named:

Detailed Solution for SRMJEE Mock Test - 3 (Medical) - Question 20

The hepatic portal vein carries blood from the intestine to the liver before it is delivered to the systemic circulation.

SRMJEE Mock Test - 3 (Medical) - Question 21

The first step in photosynthesis is

Detailed Solution for SRMJEE Mock Test - 3 (Medical) - Question 21

Photosynthesis takes place in the green leaves of plants and other green parts of plants like stem etc. The most active photosynthetic tissue in higher plants is the mesophyll of leaves. Mesophyll cells have many chloroplasts, which contain the specialized light-absorbing green pigments, the chlorophylls. When chlorophyll absorbs light, it gets excited and emits electrons. These chlorophylls are found in photosynthetic units called Photosystem I and Photosystem II. Each unit has a specific reaction centre which contains pigment molecules. These molecules absorb light of different wavelengths and emit electrons. Due to the photon of light, electrons of chlorophyll get excited. These electrons are picked up by an electron acceptor which passes them to an electron transport system of cytochromes. The excitement of electrons of chlorophyll b photon of light is the first step of photosynthesis.

SRMJEE Mock Test - 3 (Medical) - Question 22

 The ultimate gain of light reaction is :-

Detailed Solution for SRMJEE Mock Test - 3 (Medical) - Question 22

The energy of light captured by pigment molecules called chlorophylls, in chloroplasts is used to generate high- energy electrons with great reducing potential. These electrons are used to produce NADPH2 as well as ATP in a series of reactions called the light reactions because they require light.

SRMJEE Mock Test - 3 (Medical) - Question 23

Agrochemical based agriculture includes

Detailed Solution for SRMJEE Mock Test - 3 (Medical) - Question 23

Agrochemical based agriculture is used to increase the food production. It includes use of agrochemicals such as fertilisers and pesticides.

SRMJEE Mock Test - 3 (Medical) - Question 24

RNAi is used to produce resistance against the nematode____.

Detailed Solution for SRMJEE Mock Test - 3 (Medical) - Question 24

RNA interference (RNAi) is present in all eukaryotic organisms as a method of cellular defense. This method involves silencing of a specific mRNA due to complementary dsRNA molecules that binds and prevents translation of the mRNA.

SRMJEE Mock Test - 3 (Medical) - Question 25

ELISA is

Detailed Solution for SRMJEE Mock Test - 3 (Medical) - Question 25

ELISA is abbreviation of Enzyme linked immunosorbent assay. It is used to detected the disease at early stage.

SRMJEE Mock Test - 3 (Medical) - Question 26

Double helix model of DNA was proposed by proposed by

Detailed Solution for SRMJEE Mock Test - 3 (Medical) - Question 26

Double helix model of DNA was proposed by James Watson and Francis Crick in 1953 based on X-ray diffraction data produced by Maurice Wilkins and Rosalind Franklin.

SRMJEE Mock Test - 3 (Medical) - Question 27

What would happen if in a gene encoding a polypeptide of 50 amino acids, 25th codon (UAU) is mutated to UAA?

Detailed Solution for SRMJEE Mock Test - 3 (Medical) - Question 27

UAA is a nonsense codon. It signals for polypeptide chain termination. Hence, only 24 amino acids chain will be formed as the process of translation will be terminated at 25th codon.

SRMJEE Mock Test - 3 (Medical) - Question 28

The following are some major events in the early history of life
P. First heterotrophic prokaryotes
Q. First genes
R. First eukaryotes
S. First autotrophic prokaryotes
T. First animals
Which option below places these events in the correct order?

Detailed Solution for SRMJEE Mock Test - 3 (Medical) - Question 28

Organisms have evolved from simpler forms to complex form. Hence the order of the events are genes first, then heterotrophic prokaryotes, then autotrophic prokaryotes, then eukaryotes and then animals.

Topic in NCERT: Evolution of life forms - a theory

Line in NCERT: Error occcured while getting response from embedding

SRMJEE Mock Test - 3 (Medical) - Question 29

In the given figure, AB || CD and CD || MN.

Find the value of x + y + z.

Detailed Solution for SRMJEE Mock Test - 3 (Medical) - Question 29

AB || MN and AM is the transversal.
∠1 + (x + 65°) = 180° (Co-interior angles)
90° + x + 65° = 180°
Or x = 25°
Now, CD || MN and BM is the transversal.
y + 65° = 180° (Co-interior angles)
y = 115°
Now, AB || CD and BM is the transversal.
z = y (Corresponding angles)
z = 115°
Now, x + y + z = 25° + 115° + 115° = 255°

SRMJEE Mock Test - 3 (Medical) - Question 30

The pair of equations 15p – 45q = 24 and 6p – 18q = 48/5 has

Detailed Solution for SRMJEE Mock Test - 3 (Medical) - Question 30

Given equations: 15p – 45q = 24 and 6p – 18q = 48/5
We will compare these equations with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0.
15/6 = =
5/2 = 5/2 = 5/2
This implies

So, the given pair of equations has infinite solutions.

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