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UGEE SUPR Mock Test- 3 - JEE MCQ


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30 Questions MCQ Test UGEE Mock Test Series 2025 - UGEE SUPR Mock Test- 3

UGEE SUPR Mock Test- 3 for JEE 2024 is part of UGEE Mock Test Series 2025 preparation. The UGEE SUPR Mock Test- 3 questions and answers have been prepared according to the JEE exam syllabus.The UGEE SUPR Mock Test- 3 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for UGEE SUPR Mock Test- 3 below.
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UGEE SUPR Mock Test- 3 - Question 1

In amplitude modulation, the amplitude of the carrier wave is Ac, and that of the modulating signal is Am. In practice, the ratio of Am to Ac is kept less than or equal to one, to avoid

Detailed Solution for UGEE SUPR Mock Test- 3 - Question 1
Modulation index, μ= Am/Ac ≤ 1 is carried by the carrier wave, i.e., there is no signal distortion.
UGEE SUPR Mock Test- 3 - Question 2

The frequency of a sonometer wire is f but when the weights producing tension are completely immersed in water the frequency becomes f/2 and on immersing the weights in certain liquid the frequency becomes f/3. The specific gravity of the liquid is

Detailed Solution for UGEE SUPR Mock Test- 3 - Question 2

Let ρ the density of weight.

σ be the density of liquid.

and V be the volume of the weights.

From (1) and (2)

From (1) and (3)

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UGEE SUPR Mock Test- 3 - Question 3

A heat engine has an efficiency η. Temperatures of source and sink are each decreased by 100 K. Then, the efficiency of the engine.

Detailed Solution for UGEE SUPR Mock Test- 3 - Question 3

where T1 and T2 are the temperatures of a source and sink respectively.

When T1 and T2 both are decreased by 100 K each, (T1 - T2) stays constant. T1 decreases.

∴ η increases.

UGEE SUPR Mock Test- 3 - Question 4

Which of these is correct in regard to a magnet?

Detailed Solution for UGEE SUPR Mock Test- 3 - Question 4
Magnetic length is the distance between the two equal and opposite poles of the magnet.

Poles are not actually at the end of the magnet they are little inside from the end this makes the magnetic length of the magnet little less than the Geometric length.

Geometric length is the actual length of the magnet.

So when the relationship between these two measured it comes to be,

Magnetic length = 0.8× Geometric length

UGEE SUPR Mock Test- 3 - Question 5

The mean lives of radioactive substances are 1620 y and 405 y for α- emission and β- emission respectively. Find out the time during which three-fourth of a sample will decay if it is decaying both by α- emission and β- emission simultaneously.

Detailed Solution for UGEE SUPR Mock Test- 3 - Question 5
Let at some instant of time tt, number of atoms of the radioactive substance are N. It may decay either by α - emission on by β- emission. So, we can write,

If the effective decay constant is λ, then

Now,

UGEE SUPR Mock Test- 3 - Question 6

When the velocity of an electron increases, its de-Broglie wavelength

Detailed Solution for UGEE SUPR Mock Test- 3 - Question 6
The de Broglie wavelength is given by

So if the velocity of the electron increases, the de Broglie wavelength decreases.

UGEE SUPR Mock Test- 3 - Question 7

A charge q is located at the centre of a cube. The electric flux through any face is

Detailed Solution for UGEE SUPR Mock Test- 3 - Question 7
The total flux through the cube

∴ The electric flux, through any face

UGEE SUPR Mock Test- 3 - Question 8

Glass has refractive index μ with respect to air and the critical angle for a ray of light going from glass to air is an if a ray of light is incident from the air on the glass with the angle of incidence, a corresponding angle of refraction is

Detailed Solution for UGEE SUPR Mock Test- 3 - Question 8
Refractive index of glass with respect to air = μ Critical angle = θ

∴ μ = (1/sinθ) …(i)

When a ray of light incident from the air on the glass with the angle of incidence θ. If r be the angle of refraction, then by Snell’s law,

μ = sinθ/sinr

[fromEq. (i)]

⇒ sinr = 1/μ2r = sin-1(1/μ2)

UGEE SUPR Mock Test- 3 - Question 9

A molecule of water on the surface experiences a net

Detailed Solution for UGEE SUPR Mock Test- 3 - Question 9
A molecule of the surface experiences a nat force (Cohesive force) in a downward direction due to the net number of molecules pulling it downward (molecule C as shown in figure underneath). In contrast, a molecule well inside is equally pulled in all directions by molecules around at (molecule A as shown in the figure).

UGEE SUPR Mock Test- 3 - Question 10

The S I unit and dimensions of Stefan's constant σ in case of Stefan's law of radiation is

Detailed Solution for UGEE SUPR Mock Test- 3 - Question 10

According to Stefan's law, the energy emitted by a body per unit area per second is proportional to the fourth power of the absolute temperature.

E = σT4where E = energy emitted area/second,

T = absolute temperature in kelvin

and σ = Stefan's constant ⇒ σ = E/T4

Unit of Stefan's constant

⇒ J/m2s/K4 = J/m2sK4Dimensions of Stefan’s constant,

UGEE SUPR Mock Test- 3 - Question 11

V-I characteristics of LED is shown correctly by graph

Detailed Solution for UGEE SUPR Mock Test- 3 - Question 11
Since, an LED is forward biased diode, so its characteristics are those of forward bias, i,e„ Here, option (a) is characteristics of a diode in forward and reverse bias.

Option (c) characteristics of the solar cell.

Option (d) → reverse bias characteristics 1 drawn in the first quadrant can be photodiode or Zener diode.

UGEE SUPR Mock Test- 3 - Question 12

The refractive index of the crystal material is 1.68, and that of castor oil is 1.2. When a ray of light passes from oil to glass, its velocity will change by a factor

Detailed Solution for UGEE SUPR Mock Test- 3 - Question 12
We know, the refractive index of crystal w.r.t. Oil, where, μc = absolute refractive index of crystal w.r.t. Vacuum and μo = absolute refractive index of oil w. r. t, vacuumThe fractionalonal change in velocity, when light travel from oil to crystal = speed in crystal/speed in oil = 5/7.
UGEE SUPR Mock Test- 3 - Question 13

Six very long insulated (topper wires are bound together to form a cable. The currents carried by the wires are l1 = +10 A, l2 = - 13A , l3 = +10A, l4 = + 7A, l5 = 12A and l6 = 18A. The magnetic induction at a perpendicular distance of 10 cm from the cable is (μ0 = 4π x 10-7 Wb/A-m)

Detailed Solution for UGEE SUPR Mock Test- 3 - Question 13
Net current due to all wires, inet = i1 + i2 + i3 + i4 + i5 + i6

Inet = 10 - 13 + 10 + 7 - 12 + 18 = 20A

We know, magnetic field due to an infinitely long straight conductor at a perpendicular distance r from it is given by

Where, i = current in wire and r = perpendicular distance.

B = 40μT

UGEE SUPR Mock Test- 3 - Question 14

In case of dimension of electric field and electric dipole moment the power of mass is respectively,

Detailed Solution for UGEE SUPR Mock Test- 3 - Question 14
Electric field, E = F/q

Where, F = force and q = electric charge

Dimensions of E or [E] = [MLT-2]/[AT] = [MLT-0A-1]

Power of mass = 1

Electric dipole moment, m = l x A

Where, l = current and A = area

Dimensions of m = [M] = [A L2] = [M0L2T0A]

Hence, power of mass = 0

UGEE SUPR Mock Test- 3 - Question 15

For homogeneous isotropic material, which one of the following cannot be the value of Poisson’s ratio?

Detailed Solution for UGEE SUPR Mock Test- 3 - Question 15

The Poisson’s ratio for homogeneous isotropic material must be between - 10 to + 05 because of the requirement. For Young’s modulus, the shear modulus and bulk modulus have positive values. Hence, 0.8 cannot be the value of Poisson’s ratio for homogeneous isotropic material.

UGEE SUPR Mock Test- 3 - Question 16

A circular coil and a square coil is prepared from two identical metal wires and a current is passed through The ratio of magnetic dipole moment associated with circular coil to that of square coil is

Detailed Solution for UGEE SUPR Mock Test- 3 - Question 16
Let l be the length of metal wire. When wire is bent into a circular coil of radius r, then r = l/2π

∴ Area,

∴ Magnetic dipole moment associated with circular coil,

μc = il2/4π …(i)

When metal wire is bent into a square coil then side of square a = l/4

∴ Area, A = a2 = l2/16

∴ Magnetic dipole moment associated with square coil, μc = iA = i(l2/16)

Hnce, the ratio of magnetic dipole moment of circular coil and square coil is 4 : π.

UGEE SUPR Mock Test- 3 - Question 17

Figure show the circular coil carrying current l kept very close but not touching at a point A on a straight conductor carrying the same current I. The magnitude of magnetic induction at the centre of the circular coil will be

Detailed Solution for UGEE SUPR Mock Test- 3 - Question 17
Magnetic induction at the centre of circular current carrying loop,

Magnetic induction at the centre O due to straight wire,

Net magnetic induction at O,

UGEE SUPR Mock Test- 3 - Question 18

A depolarizer used in dry cell batteries is

Detailed Solution for UGEE SUPR Mock Test- 3 - Question 18
Depolarizer is a substance, which is used in primary cells to prevent a build-up of hydrogen gas bubble.

A battery depolarizer absorbs electrons during discharge of a cell, so it acts as an oxidising agent.

In dry cells, manganese dioxide act as a depolariser.

UGEE SUPR Mock Test- 3 - Question 19

Which of the following is the correct order of thermal stability of hydrides?

Detailed Solution for UGEE SUPR Mock Test- 3 - Question 19

Thermal stability is the stability of the molecule towards heat. Thermal stability of hydrides decreases down the group, as on going down the group, the size of the element E (side element with hydrogen) increases. So, the E−H bond length increases, hence- bond strength becomes weak and breaks easily to liberate hydrogen gas.

Thus, the order of thermal stability of hydrides is NH3 > PH3 > AsH3 > SbH3 > BiH3

UGEE SUPR Mock Test- 3 - Question 20

Which undergoes catalytic hydrogenation much faster?

Detailed Solution for UGEE SUPR Mock Test- 3 - Question 20

he reagent metal/Hydrogen is used in the catalytic hydrogenation. It is a syn addition to the pi-bond.

Hence, the reactivity depends on the steric hindrance of the groups present in the molecule.

The double bond present in

is less sterically hindered than

UGEE SUPR Mock Test- 3 - Question 21

The temporary hardness of water due to calcium bicarbonate and can be removed by adding

Detailed Solution for UGEE SUPR Mock Test- 3 - Question 21

Lime treatment removes bicarbonate hardness by forming insoluble CaCO3 as,

UGEE SUPR Mock Test- 3 - Question 22

KO2 is used in space and submarines because it,

Detailed Solution for UGEE SUPR Mock Test- 3 - Question 22
KO2 absorbs CO2 and increases O2 concentration, so it is used in space and submarines.
UGEE SUPR Mock Test- 3 - Question 23

Which of the following oxides can act both as an oxidising agent as well as reducing agent?

Detailed Solution for UGEE SUPR Mock Test- 3 - Question 23
SO2 oxides can act both as an oxidising agent as well as reducing agent. It behaves as a reducing agent in moist conditions. It converts iron (III) ions to iron (II) ions and decolourises acidified potassium permanganate (VII) solution.

2Fe3+ + SO2 + 2H2O → 2Fe2+ + SO2-4 + 4H+

When SO2 reacts with H2S, it produces sulphur and water. In this reaction, ‘SO2’ act as an oxidising agent.

UGEE SUPR Mock Test- 3 - Question 24

Which among the following is correct for electrolysis of brine solution?

Detailed Solution for UGEE SUPR Mock Test- 3 - Question 24
In electrolysis of brine solution, H2 gas is liberated at cathode. The solution contains four ions Na+, Cl+ , H+ and OH ions. There occurs a race amongst them for their discharge at their respective electrodes.

Following electrode reactions are possible:

In this electrolysis, H2 at cathode and Cl2, at anode are given off.

UGEE SUPR Mock Test- 3 - Question 25

Let f be a function satisfying f(xy) = f(x)/y for all positive real numbers. If f(500) = 3, then what is the value of f(600)?

Detailed Solution for UGEE SUPR Mock Test- 3 - Question 25

Given, f(xy) =

Also, f(500) = 3

⇒ f(100 × 5) = = 3

⇒ f(100) = 15

∴ f(600) = f(100 × 6)

=

= 2.5

UGEE SUPR Mock Test- 3 - Question 26

If z = , then

Detailed Solution for UGEE SUPR Mock Test- 3 - Question 26

Let

Then

∴ Im(z) = 0, Re(z) < />

UGEE SUPR Mock Test- 3 - Question 27

The sum of the cubes of three consecutive natural numbers is divisible by

Detailed Solution for UGEE SUPR Mock Test- 3 - Question 27

Let n, n + 1, n + 2 be three consecutive numbers.

Then, P(n) = n3 + (n + 1)3 + (n + 2)3

We prove by induction on n.

P(1) = 1 + 8 + 27 = 36, which is divisible by 9.

Let the result be true for k, i.e. P(k) = k3 + (k + 1)3 + (k + 2)3 is divisible by 9.

For k + 1, we have

P(k + 1) = (k + 1)3 + (k + 2)3 + (k + 3)3

= (k + 1)3 + (k + 2)3 + k3 + 9k2 + 27k + 27

= k3 + (k +1)3 + (k + 2)3 + 9(k2 + 3k + 3) = P(k) + 9(k2 + 3k + 3), which is divisible by 9.

Hence, P(n) is divisible by 9 for all natural numbers.

UGEE SUPR Mock Test- 3 - Question 28

What is the sum of the series 1.2.3 + 2.3.4 + 3.4.5 + ……. up to n terms?

Detailed Solution for UGEE SUPR Mock Test- 3 - Question 28

nth term = n(n + 1)(n + 2) = n(n2 + 3n + 2) = n3 + 3n2 + 2n. Let the sum be S, then

S =

S = [{n(n + 1)/(2)}2 + 3n(n + 1)(2n + 1)/6 + 2n(n + 1)/2]

S = n(n + 1)[n(n + 1)/4 + (2n + 1)/2 + (1)]

S = n(n + 1)[(n2 + n + 4n + 2 + 4)/4]

S = n(n + 1)(n + 2)(n + 3)/4

UGEE SUPR Mock Test- 3 - Question 29

Let f(x) = . If f(x) is continuous in , then is

Detailed Solution for UGEE SUPR Mock Test- 3 - Question 29
Since f is continuous, so

= -1/2

UGEE SUPR Mock Test- 3 - Question 30

The locus of the mid-points of the focal chords of the parabola is another parabola with which of the following equations?

Detailed Solution for UGEE SUPR Mock Test- 3 - Question 30

Let PQ be the focal chord of the parabola. Let the coordinates of P and Q be (at12, 2at1) and (at22, 2at2), respectively.

Then, t1t2 = -1

Let (h, k) be coordinates of the mid-point of PQ. Then,

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