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VITEEE PCME Mock Test - 8 - JEE MCQ


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30 Questions MCQ Test VITEEE: Subject Wise and Full Length MOCK Tests - VITEEE PCME Mock Test - 8

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VITEEE PCME Mock Test - 8 - Question 1

If , cos2α = and sin2α = , find the value of .

Detailed Solution for VITEEE PCME Mock Test - 8 - Question 1

VITEEE PCME Mock Test - 8 - Question 2

A matrix is chosen at random from a set of all 2 x 2 matrices with elements 0 and 1 only. What is the probability that the value of the determinant of the matrix chosen is positive?

Detailed Solution for VITEEE PCME Mock Test - 8 - Question 2
Since each entry (element) of a 2 × 2 matrix, with elements 0 and 1 only, can be filled in 2 ways, the total number of 2 × 2 matrices = 24 = 16
There are three 2
× 2 matrices whose determinants are positive viz. .
Hence, the required probability is 3/16.
VITEEE PCME Mock Test - 8 - Question 3

If a plane passes through intersection of planes 2x - y - 4 = 0 and y + 2z - 4 = 0 and also passes through the point (1, 1, 0). Then the equation of plane is

Detailed Solution for VITEEE PCME Mock Test - 8 - Question 3

P₁ + λP₂ = 0
(2x - y - 4) + λ(y + 2z - 4) = 0
It passes through (1, 1, 0)
⇒ 1 + λ = 0
⇒ λ = -1
Equation of plane is x - y - z = 0

VITEEE PCME Mock Test - 8 - Question 4

is equal to:

Detailed Solution for VITEEE PCME Mock Test - 8 - Question 4

Let 

Similarly, 

VITEEE PCME Mock Test - 8 - Question 5

If , then find the domain and the range of f. Show that f is one-one. Also find the function  and its domain.

Detailed Solution for VITEEE PCME Mock Test - 8 - Question 5

domain = R - {0,-2)

Since, 
Also, 

Hence f is one - one

VITEEE PCME Mock Test - 8 - Question 6

Region formed by |x - y| 2 and |x + y| ≤ 2 is

Detailed Solution for VITEEE PCME Mock Test - 8 - Question 6




Side =

VITEEE PCME Mock Test - 8 - Question 7

Let numbers a1, a2, ............. a16 are in AP and a1 + a4 + a7 + a10 + a13 + a16 = 114 then a1 + a5 + a12 + a16 is equal to

Detailed Solution for VITEEE PCME Mock Test - 8 - Question 7

a1, a2, .......... a16 are in A.P.
a1 + a4 + a7 + a10 + a13 + a16 = 114
a1 + a16 = a4 + a13 = a7 + a10 = a5 + a12
3(a5 + a12) = 114
a5 + a12 = 38
(a1 + a5 + a12 + a16) = 2(a5 + a12) = 2 × 38 = 76

VITEEE PCME Mock Test - 8 - Question 8

The figure shows three circuits with identical batteries, inductors and resistances. Rank the circuits according to the currents through the battery just after the switch is closed, greatest first.

Detailed Solution for VITEEE PCME Mock Test - 8 - Question 8

In circuit (1), on closing the switch, the current in the inductor is zero due to self induction, i.e., i1 = 0.
In circuit (2), on closing the switch, the current in the inductor is zero due to self induction.

Therefore,

In circuit (3), on closing the switch, the current in the inductor is again zero due to the same reason.

Therefore,

Thus, it is obvious that,
i2 > i3 > i1 (= 0)

VITEEE PCME Mock Test - 8 - Question 9

The magnetic field at a point due to a current-element is directly proportional to

Detailed Solution for VITEEE PCME Mock Test - 8 - Question 9

We know that any moving charge will produce magnetic field. Thus, a current-carrying conductor will also produce a magnetic field which is given by Biot-Savart's law,

where,  is the proportionality constant,
r is the distance of the point where the magnetic field is to be determined,
i is the current flowing in the conductor,
dl is the length of the conductor in consideration and its direction is considered to be in direction of the flow of current.
Thus, from the above equation, we find that, B ∝ i.

VITEEE PCME Mock Test - 8 - Question 10

An inductance of 1 mH, a capacitor of 10 μF, and a resistance of 50 Ω are connected in series. The reactances of the inductor and capacitor are the same. The reactance of either of them will be:

Detailed Solution for VITEEE PCME Mock Test - 8 - Question 10

Given:
Inductance, L = 1 mH = 1 × 10⁻³ H
Capacitance, C = 10 μF = 10 × 10⁻⁶ F
Resistance, R = 50 Ω

Since the reactance of the inductor and capacitor are the same:
XL = XC
ωL = 1 / (ωC)

Squaring both sides:
ω² = 1 / (LC)

Substituting values:
ω = √(1 / (L × C))
= √(1 / (10⁻³ × 10 × 10⁻⁶))
= 10⁴ rad/s

Now, reactance of either component:
XL = ωL = (10⁴) × (10⁻³) = 10 Ω

Therefore, XC = 10 Ω.

VITEEE PCME Mock Test - 8 - Question 11

A point charge +q is placed at the center of a cube of side L. The electric flux emerging from the cube is:

Detailed Solution for VITEEE PCME Mock Test - 8 - Question 11

According to Gauss's law, the total electric flux Φ through a closed surface enclosing a charge q is given by:
Φ = q / ε₀

Since the charge +q is placed at the center of the cube, the entire charge is enclosed within the surface. Thus, the total electric flux emerging from the cube is:
Φ = q / ε₀
Correct answer: A) q / ε₀

VITEEE PCME Mock Test - 8 - Question 12

The figure shows the variation of photocurrent with anode potential for a photosensitive surface for three different radiations. Let Ia, Ib and Ic be the intensities and fa, fb and fc be the frequencies for the curves a, b and c respectively

Detailed Solution for VITEEE PCME Mock Test - 8 - Question 12

Saturation current is proportional to intensity while stopping potential increases with increase in frequency.
Hence, fa = fb while Ia < Ib

VITEEE PCME Mock Test - 8 - Question 13

Two identical metallic spheres A and B carry charges +Q+ and −2Q respectively. The force between them is F Newton, when they are separated by a distance dd in air. The spheres are allowed to touch each other and are moved back to their initial positions. The force between them now is -

Detailed Solution for VITEEE PCME Mock Test - 8 - Question 13

VITEEE PCME Mock Test - 8 - Question 14

In which of the following process, convection does not take place primarily? 

Detailed Solution for VITEEE PCME Mock Test - 8 - Question 14

The convection is process of heat transfer via medium but when glass surface is heated due to filament of bulb, it heated only due to radiation. It is because when glass bulb is manufactured, inside glass bulb, vacuum is maintained means there is no medium inside the bulb.

VITEEE PCME Mock Test - 8 - Question 15

If potential gradient on wire PQ is 0.01 V/m, then the reading of voltmeter is:

Detailed Solution for VITEEE PCME Mock Test - 8 - Question 15

Reading = Potential gradient × length
= 0.01 × 0.5
= 5 × 10-3 V
= 5 mV

VITEEE PCME Mock Test - 8 - Question 16

For 2A + B → C, find the rate law expression.

Detailed Solution for VITEEE PCME Mock Test - 8 - Question 16

Rate = K [A]x [B]y
r1 = k[A]1x [B]1y
r2 = k[A]2x [B]2y

8 = 212y
2y = 4
y = 2
R = k[A] [B]2

VITEEE PCME Mock Test - 8 - Question 17
Which of the following statements is not correct about the order of a reaction?
Detailed Solution for VITEEE PCME Mock Test - 8 - Question 17
The order of a reaction is not always equal to the sum of the stoichiometric coefficients of reactants in the balanced chemical equation. Only the elementary (single-step) reactions have reaction orders equal to the sum of the stoichiometric coefficients of each reactant in the balanced equation. But for the complex (multi-step) reactions, the order of the reaction is the sum of the stoichiometric coefficients of the reactants in the balanced equation of the slowest step in the mechanism, which is the rate determining step.
VITEEE PCME Mock Test - 8 - Question 18

A colourless aqueous solution, on adding water and on heating, gave a white precipitate. This precipitate, when reacted with NH4Cl and NH4OH in excess, resulted in dissolution of some of the precipitate and a gelatinous precipitate is obtained. What is the hydroxide formed in aqueous solution?

Detailed Solution for VITEEE PCME Mock Test - 8 - Question 18

Aluminium will form a gel-like precipitate when we add NH4Cl and NH4OH. Also, Al(OH)3 forms gelatinous mass.

VITEEE PCME Mock Test - 8 - Question 19

The number of isomers for the compound with molecular formula C2BrClFl is

Detailed Solution for VITEEE PCME Mock Test - 8 - Question 19

Molecular formula C2BrClFI has one degree of unsaturation.
It exists in three positional isomeric forms, each of which exists as two geometrical isomers (E-Z forms).
For example:

VITEEE PCME Mock Test - 8 - Question 20

Statement I: [Ti(H2O)6]4+  is coloured while[Sc(H2O)6]3+ is colourless.
Statement II: d − d transition is not possible in [Sc(H2O)6]3+.

Detailed Solution for VITEEE PCME Mock Test - 8 - Question 20


Both [Ti(H2O)6]4+ and [Sc(H2O)6]3+ are colourless due to the absence of free electrons in 3d− subshell. d-d transition is not possible because d-orbital is vacant, and all other electrons are completely paired.
so, statement I is false and statement II is correct.

VITEEE PCME Mock Test - 8 - Question 21

The change in the optical rotation of a freshly prepared solution of glucose is known as

Detailed Solution for VITEEE PCME Mock Test - 8 - Question 21

When there is a change in the optical rotation due to the change in the equilibrium between two anomers, then the corresponding stereocenters interconvert, and it is called as mutarotation.
When either of glucose's two forms is dissolved in the water, there is a change in the rotation till the equilibrium value of +52.5°. 

VITEEE PCME Mock Test - 8 - Question 22

The proper scientific name cellobiose is

Detailed Solution for VITEEE PCME Mock Test - 8 - Question 22

Cellobiose is the repeating disaccharide unit of cellulose having the glyocosidic linkage β (1→4); its full name is thus 4-O-β-D-glucopyranosyl D-α- glucopyranose

VITEEE PCME Mock Test - 8 - Question 23

Which of the following expressions correctly represents the equivalent conductance at infinite dilution of Al2(SO4)3? Given that ΛAl3+ and  are the equivalent conductances at infinite dilution of the respective ions?

Detailed Solution for VITEEE PCME Mock Test - 8 - Question 23

Apply Kohlrausch's law, which states that at an infinite dilution the equivalent conductivity of an electrolyte is equal to the sum of the equivalent conductances of anions and cations. So, we can calculate the equivalent conductance only for ions. Hence, the equivalent conductance of Al2(SO4)3 at infinite dilution is,

VITEEE PCME Mock Test - 8 - Question 24

Which element is used for making a transistor?

Detailed Solution for VITEEE PCME Mock Test - 8 - Question 24

Silicon is used for making a transistor because the introduction of a small amount of impurities like phosphorus, arsenic, gallium increases the conductivity of Silicon. This process is called as doping.

VITEEE PCME Mock Test - 8 - Question 25

An organic compound X on treatment with acidified K₂Cr₂O₇ gives a compound Y, which reacts with I₂ and sodium carbonate to form tri-iodomethane. The compound X is:

Detailed Solution for VITEEE PCME Mock Test - 8 - Question 25

VITEEE PCME Mock Test - 8 - Question 26

What is the ratio of Δv = vmax - vmin , for spectral lines corresponding to Lyman and Balmer series for hydrogen?

Detailed Solution for VITEEE PCME Mock Test - 8 - Question 26

VITEEE PCME Mock Test - 8 - Question 27


X is mixed with a mixture of phenol and aniline in acidic medium. The product obtained is:

Detailed Solution for VITEEE PCME Mock Test - 8 - Question 27

This is an example of azotisation reaction and the product is p-Amino azobenzene.
Hence, the correct answer is:

VITEEE PCME Mock Test - 8 - Question 28

Directions: In the following problem, a question is followed by three statements I, II and III. Read all the statements carefully and find out which of them are sufficient to answer the given question. Choose the correct alternative.

What is Meera's rank from the top in a class of 40 students?
Statements:
I. Meera is 3 ranks below Rajesh from the top.
II. Rajesh's rank from the bottom is 23.
III. Meera is 3 ranks below Rajesh from the bottom.

Detailed Solution for VITEEE PCME Mock Test - 8 - Question 28

From II, we get that Rajesh ranks 23rd from the bottom, i.e. 18th from the top, as it is a class of 40 students.
From I and II, we find that Meera is 3 ranks below 18th rank from the top, i.e. she ranks 21st from the top.
From II and III, we find that Meera is 3 ranks below 23rd rank from the bottom, i.e. she ranks 20th from the bottom. So, she ranks 21th from the top.
Therefore, either I and II or II and III are sufficient to answer the question.

VITEEE PCME Mock Test - 8 - Question 29

What least number must be added to 752 so that the sum is completely divisible by 59?

Detailed Solution for VITEEE PCME Mock Test - 8 - Question 29

The least number that must be added to 752 so that the sum is completely divisible by 59 will be 15 as when we divide 752 by 59, we get 44 as the remainder.
The required number will be obtained by subtracting the remainder from the divisor = 59 − 44 = 15

VITEEE PCME Mock Test - 8 - Question 30

Read the following table carefully and answer the following question.
Literacy Rate of India 1951−2011

In which year was the female literacy rate at its minimum?

Detailed Solution for VITEEE PCME Mock Test - 8 - Question 30

According to the given data:

  • The female literacy rate was at its lowest in 1951, at 8.86%.
  • From 1951 to 2011, the female literacy rate showed a consistent increasing trend, rising from 8.86% in 1951 to 65.46% in 2011.

Thus, the correct answer is 1951.

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