UGEE SUPR Mock Test- 6 - JEE MCQ

# UGEE SUPR Mock Test- 6 - JEE MCQ

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## 50 Questions MCQ Test UGEE Mock Test Series 2024 - UGEE SUPR Mock Test- 6

UGEE SUPR Mock Test- 6 for JEE 2024 is part of UGEE Mock Test Series 2024 preparation. The UGEE SUPR Mock Test- 6 questions and answers have been prepared according to the JEE exam syllabus.The UGEE SUPR Mock Test- 6 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for UGEE SUPR Mock Test- 6 below.
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UGEE SUPR Mock Test- 6 - Question 1

### A solid sphere rolls down from the top of an inclined plane, 7m high, without slipping. Its Linear speed at the foot of plane is g = 10 m/s2)

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 1
Given, the height of the inclined plane, h = 7m

g = 10 m/s2

By conservation of energy, Potential energy lost by the solid sphere in rolling down the inclined plane = Kinetic energy gained by the sphere.

UGEE SUPR Mock Test- 6 - Question 2

### Which of the following is the dimensional formula for electric polarisation?

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 2
Bedric polarisation (P) is given by

P = ?o?E

where, E is the electric field, and x is electric susceptibility, and it is a dimensionless quantity, Dimensional formula of P = Dimensional formula of (ε0) x Dimensional formula of E

= [M-1 L-3 T4 I2] [ML T-3I-1] = [M0 L-2 T-1 I1]

UGEE SUPR Mock Test- 6 - Question 3

### A vector P has X and Y components of magnitude 2 units and 4 units respectively. A vector Q along the negative X-axis has magnitude 6 units. The vector (Q - P) will be

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 3
Recording to question, vectors P and Q can be written as

and

UGEE SUPR Mock Test- 6 - Question 4

When a large bubble rises from the bottom of a water lake to its surface, then its radius doubles, if the atmospheric pressure is equal to the pressure of height Hof a certain water column, then the depth of the lake will be

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 4
When a large bubble rises from the bottom of a water lake to its surface, then its radius becomes double.

i.e., r2 = 2r1

Since, volume V ∝ r3

∴ V2/V1 = (r2/r1)3 = (2r1/r1)3 = 8

⇒ V2 = 8V1 ….(i)

From Boyle's law,

p1V1 = p2V2

Where, p2 = atmospheric pressure.

p1V1 = p2 8V1

(from Eq. (i)J

p2 = p1/8

Where. p2 = atmospheric pressure and p1 = pressure at depth d.

∴ p1 = patmospheric + ρ gd

P1 = p2 + ρ gd

From Eqs. (ii) and (iii), we have

p2 = p2 + ?gd

7p2 = ρ gd

7.ρgH = ρ gd

7H = d ⇒ d = 7 H

UGEE SUPR Mock Test- 6 - Question 5

In the network shown cell E has internal resistance r and the galvanometer shows zero deflection. If the cell is replaced by a new cell of 2E and internal resistance 3r keeping ling everything else identical, then

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 5
In the given circuit diagram, the galvanometer shows zero deflection, hence it is a Wheatstone ba circuit.

Hence, VA = VB

On replacing a new cell of emf 2E and internal resistance 3r, no effect occurs on balanced condition of bridge, i.e., VA = VB.

Hence, the galvanometer wil shows zero deflection.

UGEE SUPR Mock Test- 6 - Question 6

[L2 M1 T-2] are the dimensions of

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 6
Dimension of torque is [L2 M1 T-2]

Torque = Force x Perpendicular distance

i.e., τ = F x r

∴ Dimensional formula of τ = [M L T-2] [L] = [ML2 T-2]

UGEE SUPR Mock Test- 6 - Question 7

If the radius of the circular path and frequency of revolution of a particle of mass m are doubled, then the change in its kinetic energy will be (Ei and Er are the Initial and final kinetic energies of the particle respectively.)

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 7
Initial kinetic energy of the body,

Ei = 2π2 mr21f21 ………..(i)

Where, f1 = frequency of revolution of the body.

When, r2 = 2r1 and f2 = and 2f1 , then

Change in kinetic energy,

ΔE = Er -Ei

= 16Ei - Ei = 15 Ei

UGEE SUPR Mock Test- 6 - Question 8

The rms speed of oxygen molecule in a gas is u, If the temperature is doubted and the molecules dissociates into two atoms, the rms speed will be

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 8
rms speed erf O2 molecule = u

i.e …(i)

where, T = temperature and M = molecular mass. When the temperature is doubled, then O2 dissociates into two atoms.

Hence, the atomic mass of the oxygen atom.

(M') = M/2

∴ rms velocity of the oxygen atom =

UGEE SUPR Mock Test- 6 - Question 9

Eight identical drops of water faking through the air with a uniform velocity of 10 cm/s combines to form a single drop of big size, then the terminal velocity of the big drop will be

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 9
Given, terminal velocity of each small drop v1 = 10cm/s

When 8 small drops are combined to form a bigger drop, then volume will be same.

If r1 be radius of small drop and r2 be the radius of biggordrop, then v2 = v1

∴ Terminal velocity of the drop

i.e., v ∝ r2

⇒ v1/v2 = 1/4

v2 = 4v1 x 10 = 40 cm/s

UGEE SUPR Mock Test- 6 - Question 10

A body of mass m is performing a UCM in a circle of radius r with speed v. The work done by the centripetal force in moving it through (2/3)rd of the circular path is

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 10
When a body perform uniform circular motion, then centripetal force acts always in perpendicular direction to its velocity, hence work done by centripetal force is always 2ero.

Hence, work done by the centripetal force is zero, when it through (2/3)rd of the circular path.

UGEE SUPR Mock Test- 6 - Question 11

The phenomenon of interference is based on

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 11
In interference phenomenon, energy is distributed. The intensity of the resultant wave is maximum at some points and mirk mum at another point; therefore, the phenomenon of interference is based on energy conservation.
UGEE SUPR Mock Test- 6 - Question 12

A magnetizing field of 5000 A/m produces a magnetic flux of 4 x 10-5 Wb in an iron rod of cross-sectional area 0.4 cm2. The permeability of the rod in Wb/A-m, is

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 12
Given, magnetising field, H = 5000A/m

Magnetic flux, ϕ = 4 x 10-5 Wb

Cross-sectional area, A = 0.4 cm2 = 4 x 10-5 m2

∴ Permeability of rod.

UGEE SUPR Mock Test- 6 - Question 13

A block of mass M is pulled along a smooth horizontal surface with a rope of mass m by force F. The acceleration of the block will be

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 13
When a block of mass m is pulled along the smooth horizontal surface with rope erf mass m by force F , then (M + m) will be the total mass of the system.

∴ F = (M + m)a

⇒ a = F/M+m

UGEE SUPR Mock Test- 6 - Question 14

In hydrogen emission spectrum , for any series, the principthe al quantum number is n.

Corresponding maximum wavelength λ is (R = Rydbergs constant)

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 14

UGEE SUPR Mock Test- 6 - Question 15

For formation of beats, two sound notes must have

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 15
In a sound note, various frequencies are mixed, hence beats can be produced by two sound notes only when they have nearly equal frequencies and equal amplitudes.
UGEE SUPR Mock Test- 6 - Question 16

The total energy of a simple harmonic oscillator is proportional to

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 16
Total energy of simple harmonic oscillator is given by

E = (1/2) mω2A2

Where, m = mass of body performing SHM.

ω = angular velocity

and A = amplitude.

E ∝ A2

UGEE SUPR Mock Test- 6 - Question 17

Which of the following is not the mineral of iron?

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 17
Corundum is not a mineral of iron. It is a crystalline form of aluminium oxide typically captaining traces of Fe, Ti, V and Cr. On the other hand, limonite, magnetite and haematite are minerals of iron. The name and a molecular formula of given minerals is given below.

UGEE SUPR Mock Test- 6 - Question 18

How many primary amines are possible for the molecular formula C3H9N?

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 18
Amine with molecular formula C3H9N have two primary structures. These are as follows

(a)

(b)

Other isomers are not primary amines.

UGEE SUPR Mock Test- 6 - Question 19

A polymer which becomes soft on heating and hard on cooling, belongs to class of

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 19
A polymer which becomes soft on heating and hard on cooling belongs to the class of thermoplastics. When these polymers are heated, the polymer becomes soft enough to be moulded into various shapes, e.g., polyethylene, polystyrene, etc.
UGEE SUPR Mock Test- 6 - Question 20

Which bond in a molecule of ethyl magnesium bromide is ionic in nature?

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 20
Mg — Br Bond in a molecule of ethyl magnesium bromide is ionic in nature because the difference in electronegativity is large. As a result, it possesses a high magnitude of lattice energy and hence maximum ionic character.
UGEE SUPR Mock Test- 6 - Question 21

IUPAC name of the complex Ba[CuCl4] is

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 21
IUPAC name of the complex Ba [CuCl4] is barium tetrachlorocuprate (II). While naming a complex, the cation is named first and ligands are named in an alphabetical order before the name of the central atom or ion.
UGEE SUPR Mock Test- 6 - Question 22

Which among the following polymer does not show cross linking in it?

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 22
Nylon - 6 polymer does not show cross linking in it. It is a linear polymer. It is obtained by heating caprolactam with water at a high temperature. The structural formula and name of given polymers are shown below

UGEE SUPR Mock Test- 6 - Question 23

Which of the following is not present in DNA?

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 23
Uracil is not present in DNA. It is present in RNA. DNA contains four bases, i.e . adenine (A), guanine (G), cytosine (C), and thymine (T). In RNA, thymine (T) is replaced by uracil (U).
UGEE SUPR Mock Test- 6 - Question 24

The correct order of boiling points of alkyl halides is

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 24
The correct order of boiling point of alkyl halides is Rl > RBr > RCl > RF. This is because with the increase in size and mass of the halogen atom, the magnitude of van der Waals' forces increases.
UGEE SUPR Mock Test- 6 - Question 25

The conversion of 2-methylpropan-1-ol to 2-methylpropan-2-ol is

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 25
The conversion of 2-methylpropan-1-ol to 2-methylpropan-2-ol is a rearrangement reaction. In these reactions, the carbon skeleton of a molocuto is rearranged to give a structural isomer of the original molecule. Here, the alcohol group of 2-methylpropan-1 -ol moves from one atom to another atom in the same molecule to give 2 -methyl propan -2 -ol.

UGEE SUPR Mock Test- 6 - Question 26

The effective atomic number of Iron (Z = 26) in [Fe{CN)6)-3 is

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 26
 Key Idea The total number of electrons which the central metal Ion appears to possess in the complex including those gained by it in bonding is called effective atomic number of central metal ions.

In the given complex ion [Fe(CN)6]3-, the Fe is in +3 oxidation state. As we know that atomic number of iron is 26. The number of electron in Fe3+ is 23. Each of the six cyanide molecules donates a pair of elections so that EAN becomes 23 + 2 x 6 ⇒ 35.

UGEE SUPR Mock Test- 6 - Question 27

Based on first Jaw of thermodynamics which of the following is correct

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 27
In the isobaric process, pressure is constant, and volume increases from V1 to V2 due to heat absorption. According to the first law of thermodynamics, ΔU = q + W

At constant pressure, qp = ΔU -W.

qp = ΔU - pΔV

qp = U2 - U1 - (-p(V2 - V1))

qp = (U2 + pV2) - (U1 + pV1)

qp = H2 - H1

qp = ΔH

As we know, ΔH = ΔU + pΔV

ΔH = ΔU + W

Putting the value of ΔH in (1)

qp = ΔU + W

Other corrected options are as follows:

• For adiabatic process ΔU = W

• For an isochoric process ΔU = qv

• For an isothermal process q = -W

UGEE SUPR Mock Test- 6 - Question 28

Hinsberg's reagent is

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 28
Hinsberg’s reagent is benzene sulphonyl chloride. Its molecular formula is C6H5SO2Cl. This reagent is used to distinguish primary (10), secondary (20), and tertiary (30) amines.
UGEE SUPR Mock Test- 6 - Question 29

The vitamin that belongs to the aromatic series is

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 29
The vitamin that belongs to the aromatic series is vitamin K . It has an aromatic structure. The structure of vitamin K is given below

UGEE SUPR Mock Test- 6 - Question 30

The SI unit of electrochemical equivalent is

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 30
(b) The electrochemical equivalent of a substance is the mass Liberated in electrolysis by 1 coulomb, it can be calculated by assuming Faraday's second law and using the experimental fad that the equivalent weight in grams of all substances is liberated by 96500 C.

Electrochemical equivalent = Equivalent weight/96500

gm per coulomb. Its unit is kgC-1.

UGEE SUPR Mock Test- 6 - Question 31

The molar conductivities at infinite dilution for sodium acetate, NaAc, HCI and NaCI are 91 S cm2 mol-1, 425.9 S cm2 mol-1 and 126.4 S cm2 mol-1 respectively. The molar conductivity of acetic acid at infinite dilution is

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 31
 Key idea Kohlrausch law of independent migration of ions can be used to calculate ∧*m for weak electrolytes such as acetic acid.

Given

Using Kohlrausch law, to find molar conductivity of acetic acid.

UGEE SUPR Mock Test- 6 - Question 32

A flavouring agent found in oil of whether green is

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 32
A flavouring agent found in the oil of wintergreen is methyl salicylate. It is an organic compound with the formula C6H4(OH) (CO2CH3). It is the methyl ester of salicylic acid. It is a colourless, viscous liquid. The structure of methyl salicylate is as follows

UGEE SUPR Mock Test- 6 - Question 33

Which of the following metal halides is more covalent?

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 33
 Key idea Fajans rules help in predicting the covalent character in a bond.

SbCl5 metal halide is more covalent than other given options. It can be explained on the basis of Fajan’s rule. The higher the charge on the cation, the ,,greater is its polarising power. In SbCt6, Antimony (Sb) is in +5 oxidation state, it has maximum polarising power. Hence, possess a the higher covalent character.

UGEE SUPR Mock Test- 6 - Question 34

Which of the following is not equal to w·(u × v)?

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 34
Key Idea: Use

(a × b)·c = (b × c)·a = (c × a)·b

∴ w·(u × r) = -v·(u × w)

UGEE SUPR Mock Test- 6 - Question 35

If the foot of the perpendicular drawn from the point (0,0,0) to the plane b (4, - 2 , - 5), then the equation of the plane is ..........

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 35
We have, foot of the perpendicular drawn from the point (0, 0, 0) to the plane is (4 , - 2, -5).

∴ Direction ratios of normal to the required plane is (0-4, 0 + 2, 0 + 5 i.e -4, 2, 5)

∴ Required aquation of plane passing through (4, - 2, 5) is a(x - x1) + b(y - y1) + c(z - z1) = 0

= (-4) (x - 4) + (-2) (y + 2) + (5)(z + 5) = 0

= -4x + 2y+ 5z + 16 + 4 + 25 = 0

⇒ 4x - 2y - 5z = 45

UGEE SUPR Mock Test- 6 - Question 36

If xy = ex-y , then dy/dx at x = 1 is ….

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 36
Given have. xy = ex-y

taking log on both sides, we get

y log x = (x - y) log e = (x - y)

when x = 1, then y (log 1) = (1 - y)

⇒ y = 1

On differentiating both sides,

when x = 1 the

UGEE SUPR Mock Test- 6 - Question 37

Which of the following statements is a contingency?

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 37
Key Idea: A statement that is neither a tautology nor a contradiction is a contingency.

Option (1) (p v q) v ~ q

= p v(q v ~ q) = p v T

= T, which is a tautology.

Option (2) (p v q) v ~

= (p v ~ p) v q = T v q

= T which is tautology.

Option (3)(p v q) ∧ ~ q

= (p ∧ ~ q) v (q ∧ ~ q) = (p ∧ ~ g) v F

= (p ∧ ~ q)

if p → T, q → T

Then p∧ ~ q → F

if p → T, q → F

Then p∧ ~ q → T

So, (p∧ ~ q) is a contingency.

Therefore statement (p v q)∧ ~ q is contingency.

UGEE SUPR Mock Test- 6 - Question 38

The intercept on the line y = x by the circle x2 + y2 - 2x = 0 is AB. The equation of the circle with AB as a diameter is ..........

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 38
We have equation of line y = x and equation of circle x2 + y2 - 2x = 0

Now. intersecting points of given line and circle, x2 + x2 - 2x = 0

⇒ 2x2 - 2x = 0

⇒ 2x(x - 1) = 0

⇒ x = 0, 1

when x = 0 then y = 0 and when x = It then y = 1

∴ Coordinates of endpoints of diameter AB are (0, 0) and (1, 1)

Required equation of the circle with diameter AB

(x - 0)(x - 1) + (y - 0) (y - 1) = 0

x2 - 2 + y2 - y = 0

⇒ x2 + y2 - x - y = 0

UGEE SUPR Mock Test- 6 - Question 39

= …….

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 39

put 1 - x = t ⇒ x = 1 - t

⇒ -dx = dt

⇒ dx = -dt

when x = 0, then t = 1 and

when x =1 , then t = 0

UGEE SUPR Mock Test- 6 - Question 40

then K = …...

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 40
We have

Let

put x = a sin2 θ

⇒ dx = a(2 sinθ cos θ)dθ

when, x = 0, θ = 0 and x = a , θ = π/2

∴ k = πa

UGEE SUPR Mock Test- 6 - Question 41

The solution of the differential equation dθ/dt = - k(θ - θ0) where k is constant, is …….

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 41
We have a differential equation

dθ/dt = -k(θ - θ0), where k is constant

⇒ (dθ/dt) + kθ = kθ0

Which is linear differential equation in the form of

(dy/dx) + Py = Q

IF = e∫kt = ekt

before the required solution,

(θ)(ekt) = ∫(ekt x kθ0)dt

⇒ θekt = ekt θ0 + a

⇒ θ = θ0 + ae-kt

UGEE SUPR Mock Test- 6 - Question 42

A particle moves so that x = 2 + 27t - t3. The Erection of motion reverses after moving a distance of ... units.

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 42
We have, x = 2 + 27t - t3

⇒ dx/dt = 27 - 3t2

Since, the direction will be reverse,

∴ dx/dt = 0

⇒ 27 - 3t2 = 0

⇒ t = 3

∴ Distance x = 2 + 27t - t3

= 2 + 27 x 3 - (3)3

= 2 + 81 - 27 = 56 units

UGEE SUPR Mock Test- 6 - Question 43

The value of sin 18º is….

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 43
Let, θ = 18º then, 2θ = 36º = 90º - 54° = 90º - 3θ

Now. sin2θ = sin(90º - 3θ)

⇒ 2sinθ cosθ = cos 3θ

⇒ 2 sinθ cosθ = 4cos3θ- 3cosθ

⇒ 2sinθ cosθ = cosθ(4cos2θ - 3)

⇒ 2sinθ = 4 cos2θ - 3(as cos θ ≠ 0)

⇒ 2sinθ = 4 - 4 sin2θ - 3

⇒ 4sin2 + θ + 2sinθ - 1 = 0

But as sinθ > 0 we have sinθ > 0 we have

UGEE SUPR Mock Test- 6 - Question 44

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 44

UGEE SUPR Mock Test- 6 - Question 45

If A= where i = √-1, then A(adj A) = …..

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 45

A =

Therefor, |A| =

= (1 + 2i)(1 - 2i) + i2

= 1 - (2i)2 + (-1) = 1 + 4 -1 = 4

∴ A(adj A) = |A| = 4i

UGEE SUPR Mock Test- 6 - Question 46

= ……….

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 46

On adding Eqs. (i) and (ii), we get

UGEE SUPR Mock Test- 6 - Question 47

The equation of the circle concentric with the circle x2 + y2 - 6x - 4y -12 = 0 and touching the Y-axis is ..........

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 47
Given equation of circle x2 + y2 - 6x + 4y -12 = 0 ...(i)

Centre of circle (i) te (3, 2)

Equation of circle concentric with the circle (i) and touching the Y- axis is

(x - 3)2 + (y - 2)2 = (3)2

⇒ x2 + 9 - 6x + y2 + 4 - 4y = 9

⇒ x2 + y2 - 6x - 4y + 4 = 0

UGEE SUPR Mock Test- 6 - Question 48

If 4 sin-1x + 6 cos-1 x = 3π then x

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 48
We have , 4 sin-1 x 6 cos-1x = 3π

⇒ 4 sin-1 x + 3π - 6 sin-1x = 3π

⇒ -2 sin-1 x = 0 ⇒ sin-1x = 0

⇒ x = sin(0) = 0

UGEE SUPR Mock Test- 6 - Question 49

In Δ ABC, with usual notations,

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 49
We have

= k sin(B + C)

= k sin(180º - A)

= k sin A = a

UGEE SUPR Mock Test- 6 - Question 50

The eccentricity of the hyperbola 25x2 - 9y2 = 225 is..

Detailed Solution for UGEE SUPR Mock Test- 6 - Question 50

We have equation of hyperbola

25x2 - 9y2 = 225

On comparing with we get

A2 = 9, b2 = 25

Therefore:

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