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UGEE SUPR Mock Test- 6 Free Online Test 2026


Full Mock Test & Solutions: UGEE SUPR Mock Test- 6 (40 Questions)

You can boost your JEE 2026 exam preparation with this UGEE SUPR Mock Test- 6 (available with detailed solutions).. This mock test has been designed with the analysis of important topics, recent trends of the exam, and previous year questions of the last 3-years. All the questions have been designed to mirror the official pattern of JEE 2026 exam, helping you build speed, accuracy as per the actual exam.

Mock Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 60 minutes
  • - Total Questions: 40
  • - Analysis: Detailed Solutions & Performance Insights

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UGEE SUPR Mock Test- 6 - Question 1

If a, b, c are in A.P., then the value of 

Detailed Solution: Question 1

Given a, b, c are in A.P.
∴ 2b = a + c…(i)

[Applying R2 → 2R2]


[using equation (i)]

[Applying R2 → R2 - (R1 + R3)]

UGEE SUPR Mock Test- 6 - Question 2

The equation y2 + 3 = 2(2x+y) represents a parabola with the vertex at

Detailed Solution: Question 2

The given equation can be rewritten as 
 which is a parabola
with its vertex  , axis along the line
y = 1, hence axis parallel tox-axis.
Its focus is 

UGEE SUPR Mock Test- 6 - Question 3

∫ (27e9x + e12x)1/3 dx is equal to

Detailed Solution: Question 3


UGEE SUPR Mock Test- 6 - Question 4

The conic represented by x = 2(cos t + sin t), y = 5(cos t − sin t) is

Detailed Solution: Question 4

From given equation
x/2 = cos t + sin t  ....(i) 
y/5 = cos t - sin t   ....(ii)
Eliminating t from (i) and (ii), we have
 which is an ellipse.

UGEE SUPR Mock Test- 6 - Question 5

The work done in ergs for the reversible expansion of one mole of an ideal gas from a volume of 10 litres to 20 litres at 25°C is

Detailed Solution: Question 5



UGEE SUPR Mock Test- 6 - Question 6

Arrange the following particles in increasing order of values of e/m ratio: electron (e), proton (p), neutron (n) and α-particle (α) -

Detailed Solution: Question 6

UGEE SUPR Mock Test- 6 - Question 7

For the reaction N2 + 3H2 → 2NH3 
if  mol L−1 s−1, then value of  would be

Detailed Solution: Question 7



UGEE SUPR Mock Test- 6 - Question 8

The value of Kc ​for the reaction:
A + 3B ⇌ 2C at 400C is 0.5. Calculate the value of Kp

Detailed Solution: Question 8

Kp = Kc [RT]Δn        Δn = 2 - 4 = -2
T = 673K, Kc = 0.5,
R = 0.082 L·atm·mol-1K-1
Kp = 0.5 × (0.082 × 673)-2
= 1.64 × 10-4 atm.

UGEE SUPR Mock Test- 6 - Question 9

The electrode potentials for
Cu²⁺(aq) + e⁻ → Cu⁺(aq)
and Cu⁺(aq) + e⁻ → Cu(s)
are +0.15 V and +0.50 V, respectively.
The value of E°Cu²⁺/Cu will be:

Detailed Solution: Question 9





UGEE SUPR Mock Test- 6 - Question 10

The wavelengths of two photons are 2000Å and 4000Å respectively. What is the ratio of their energies?

Detailed Solution: Question 10

UGEE SUPR Mock Test- 6 - Question 11

At low pressure, the van der Waals equation is reduced to

Detailed Solution: Question 11

UGEE SUPR Mock Test- 6 - Question 12

A direct current of 2A and an alternating current having a maximum value of 2A flow through two identical resistances. The ratio of heat produced in the two resistances will be

Detailed Solution: Question 12

Heat = I2 Rt.
DC: I = 2A, HDC = 4Rt.

UGEE SUPR Mock Test- 6 - Question 13

A particle of mass m is under a potential v(x) = (1/2)k1x2 for x > 0 and v(x) = k1x2 for x < 0. When disturbed a little from the position x = 0, it will

Detailed Solution: Question 13

T in general = 2π√(m/k); Half the cycle on one side has period π√(m/k) and the other half π√(m/2k)

UGEE SUPR Mock Test- 6 - Question 14

A far-sighted person can see clearly object beyond 100 cm distance. If he wants to see clearly an object at 40 cm distance, then the power of the lens he shall require is

Detailed Solution: Question 14

For a far-sighted person with a far point at 100 cm, the required lens power to focus an object at 40 cm is calculated using the lens formula. The parameters are as follows:

  • Object distance (u): 40 cm
  • Image distance (v): -100 cm (for a virtual image)

Using the lens formula:

1/f = 1/u - 1/v

Substituting the values:

1/f = 1/40 - 1/(-100) = 0.015

Therefore, the focal length (f) is:

f ≈ 66.67 cm

To convert this to lens power (P), we use the formula:

P = 1/f (in metres)

Converting f to metres:

f ≈ 0.6667 m

Thus, the lens power is:

P = 1/0.6667 ≈ +1.5 D

In conclusion, the correct lens is a convex lens with a power of +1.5 D.

UGEE SUPR Mock Test- 6 - Question 15

A hydrogen atom emits green light when it changes from n = 4 energy level to the n = 2 level. Which colour of light would the atom emit when it changes from n = 5 level to the n = 2 level?

Detailed Solution: Question 15





Hence, f₂ > f₁. Among the options, only violet is one which has frequency more than green. Hence, answer is violet.

UGEE SUPR Mock Test- 6 - Question 16

A 20 cm long capillary tube is dipped in water. The water rises up to 8 cm. If the entire arrangement is put in a freely falling elevator the length of water column in the capillary tube will be ______ cm.

Detailed Solution: Question 16

In a freely falling elevator, the effective gravity is zero. This scenario leads to some intriguing effects:

  • Effective Gravity: When in free fall, there is no gravitational pull acting on the objects inside the elevator.
  • Capillary Action: Without this gravitational influence, capillary action can occur more freely.
  • Water in the Tube: As a result, water can fill the entire length of the tube, reaching its full length of 20 cm.

UGEE SUPR Mock Test- 6 - Question 17

How many degree of freedom the gas molecules have if under STP the gas density ρ = 1.3 kg/m3 and the velocity of sound propagation in it is 330 m/sec?

Detailed Solution: Question 17


γ = 1.4 for diatomic gas molecules, so the degree of freedom is 5.

UGEE SUPR Mock Test- 6 - Question 18

The work function of a metallic surface is 5.01 eV. Photoelectrons are emitted when light of wavelength 2000 Å falls on it. What is the minimum potential difference required to stop the fastest photoelectrons? (h = 4.14 × 10-15 eV·s)

Detailed Solution: Question 18

First, calculate the energy of the incident photon using E = hc/λ.
Here, h = 4.14 × 10-15 eV·s, c = 3 × 108 m/s, λ = 2000 Å = 2000 × 10-10 m = 2 × 10-7 m.
So,
E = (4.14 × 10-15 × 3 × 108) / (2 × 10-7)
E = (12.42 × 10-7) / (2 × 10-7)
E = 6.21 eV

Maximum kinetic energy (K.E.max) = Energy of photon - Work function
K.E.max = 6.21 eV - 5.01 eV = 1.2 eV

The stopping potential (V0) is given by:
K.E.max = eV0, so V0 = 1.2 V

Therefore, the minimum potential difference required to stop the fastest photoelectrons is 1.2 V.

UGEE SUPR Mock Test- 6 - Question 19

A solid sphere of mass M and radius R is divided into two unequal parts. The first part has a mass of 7M/8 and is converted into a uniform disc of radius 2R. The second part is converted into a uniform solid sphere. Let I1 be the moment of inertia of the new sphere about its axis. The ratio I1/I2 is 140/x. Find the value of x.
Note: I1 is the inertia of the disk & I2 is the inertia of the new sphere about the diameter.

Detailed Solution: Question 19


UGEE SUPR Mock Test- 6 - Question 20

A block of mass 10 kg is kept on a rough inclined plane as shown in the figure. A force of 3 N is applied on the block. The coefficient of static friction between the plane and the block is 0.6. What should be the minimum value of force P (in newton), such that the block does not move downward? (Take g = 10 m/s2)

Detailed Solution: Question 20


UGEE SUPR Mock Test- 6 - Question 21

A plane electromagnetic wave of frequency 50 MHz travels in free space along the positive x-direction. At a particular point in space and time, . The corresponding magnetic field , at that point is . Find the value of x.

Detailed Solution: Question 21

As we know,


 [ ∵ EM wave travels along + (ve) x-direction]

UGEE SUPR Mock Test- 6 - Question 22

Two concentric coplanar circular loops made of wire, with resistance per unit length 10 Ω/m have diameters 0.2 m and 2 m. A time varying potential difference (4 + 2.5 t) volt is applied to the larger loop. Calculate the current (in A) in the smaller loop.

Detailed Solution: Question 22

The resistance of the loops, 

and 

Flux in the smaller loop, φ = B2A1

The induced current, 
After substituting the value and simplifying we get
i = 1.25 A.

UGEE SUPR Mock Test- 6 - Question 23

Derivative of  with respect to sin-1(3x - 4x3) is

Detailed Solution: Question 23


and v = sin-1(3 sin θ - 4 sin3 θ)

⇒ u = tan-1(tan θ) and v = sin-1(sin 3θ)
⇒ u = θ and v = 3θ
⇒ u = sin-1 x and v = 3 sin-1 x.
On differentiating both sides w.r.t. x, we get

UGEE SUPR Mock Test- 6 - Question 24


a11A21 + a12A22 + a13A23 is equal to
Note: axy represents element whereas Axy represents Cofactor matrix.

Detailed Solution: Question 24


Since, the sum of the product of elements other than the corresponding cofactor is zero.
∴ a11A21 + a12A22 + a13A23 = 0

UGEE SUPR Mock Test- 6 - Question 25

If Rolle's theorem for 
f(x) = ex (sin x − cos x) is verified on [ then the value of c is

Detailed Solution: Question 25

  1. Verify the conditions of Rolle’s Theorem:

    • f(x) = ex(sin x - cos x) is continuous and differentiable.

    • Check endpoints:

  2. Apply Rolle’s Theorem:

    • Since f(π/4) = f(5π/4) = 0, there exists c in (π/4, 5π/4) such that f'(c) = 0.

  3. Find f'(x):

    • f'(x) = d/dx (ex(sin x - cos x)) = ex(2 sin x)

  4. Set f'(c) = 0:

    • ec(2sin c) = 0 ⟹ sin c = 0

  5. Solve for c in (π/4, 5π/4):

    • c = π

UGEE SUPR Mock Test- 6 - Question 26

The joint equation of lines passing through the origin and trisecting the first quadrant is:

Detailed Solution: Question 26

  • The lines trisecting the first quadrant through the origin will have angles of 0°, 30°, 60°, and 90° with the x-axis.
  • The lines at 30° and 60° have slopes of tan(30°) = 1/√3 and tan(60°) = √3, respectively.
  • The joint equation of these lines, y = (1/√3)x and y = √3x, is obtained by multiplying the equations: (y - (1/√3)x)(y - √3x) = 0, which simplifies to x2 + √3xy - y2 = 0.

UGEE SUPR Mock Test- 6 - Question 27

If 2 tan-1(cos x) = tan-1(2 cosec x), then sin x + cos x is equal to:

Detailed Solution: Question 27

UGEE SUPR Mock Test- 6 - Question 28

Direction cosines of the line , z = -1 are

Detailed Solution: Question 28

Equation of given line is 


So, DR of given line are 

∴ DC of given line are 

UGEE SUPR Mock Test- 6 - Question 29

 is equal to 

Detailed Solution: Question 29


UGEE SUPR Mock Test- 6 - Question 30

The approximate value of f(x) = x3 + 5x2 - 7x + 9at x = 1.1 is

Detailed Solution: Question 30

Since, f(x) = x³ + 5x² - 7x + 9
After differentiating on both sides w.r.t.x, we get
f'(x) = 3x² + 10x - 7
As, f(x + Δx) = f(x) + Δx × f'(x)
= x³ + 5x² - 7x + 9 + Δx × (3x² + 10x - 7)
After putting x = 1 and Δx = 0.1, we get
f(1 + 0.1)
= 1³ + 5(1)² - 7(1) + 9 + 0.1 × (3 × 1² + 10 × 1 - 7)
So, f(1.1) = 1 + 5 - 7 + 9 + 0.1(3 + 10 - 7)
= 8 + 0.1(6) = 8.6

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