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VITEEE PCME Mock Test - 5 - JEE MCQ


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30 Questions MCQ Test - VITEEE PCME Mock Test - 5

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VITEEE PCME Mock Test - 5 - Question 1

A function f(x) is differentiable at x = c(c R). Let g(x) = |f(x)|, f(c) = 0 then

Detailed Solution for VITEEE PCME Mock Test - 5 - Question 1
g(x) = |f(x|
g'(c+) =


= f'(c)
for g (x) to be differentiable at x = c.
f'(c) must be 0. Else it is non-differentiable.
VITEEE PCME Mock Test - 5 - Question 2

If ABCD is a rhombus whose diagonals cut at the origin O, then + + + equals

Detailed Solution for VITEEE PCME Mock Test - 5 - Question 2

Since the diagonals of a rhombus bisect each other,
and
So,

VITEEE PCME Mock Test - 5 - Question 3

The coefficient of x in the expansion of (x² + cx)⁵ is:

Detailed Solution for VITEEE PCME Mock Test - 5 - Question 3

The general term of given expression (x² + cx)⁵ can be written as, 

Then for the coefficient of x,

∴10−3r=1
⇒ 3r = 9
⇒ r = 3 
The coefficient of 

VITEEE PCME Mock Test - 5 - Question 4

The area of the region bounded by the curves y = ex and y = e(-x) from x = 0 to x = 1 is:

Detailed Solution for VITEEE PCME Mock Test - 5 - Question 4

So, 

VITEEE PCME Mock Test - 5 - Question 5

Let A = {1, 2, 3}, B = {1, 3, 5}. A relation R: A → B is defined by R = {(1, 3), (1, 5), (2, 1)}. Then R⁻¹ is defined by:

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

Let A = {1, 2, 3}, B = {1, 3, 5}. A relation R: A → B is defined by R = {(1, 3), (1, 5), (2, 1)}.

Then, R⁻¹ (the inverse of R) is defined by:

R⁻¹ = {(3, 1), (5, 1), (1, 2)}

So, the correct answer is:

C. {(1, 2), (5, 1), (3, 1)}

VITEEE PCME Mock Test - 5 - Question 6

Let θ ∈ [-2π,2π] and 2 cos2θ + 3 sin θ = 0 then sum of all solutions is

Detailed Solution for VITEEE PCME Mock Test - 5 - Question 6


Hence sum = 2π

VITEEE PCME Mock Test - 5 - Question 7

Power gain for N-P-N transistor is 106, input resistance 100Ω and output resistance 10000Ω. Find current gain.

Detailed Solution for VITEEE PCME Mock Test - 5 - Question 7

Power gain = (Current gain)2

VITEEE PCME Mock Test - 5 - Question 8

At what distance from his face, a person should place concave mirror of focal length 0.4 m so that magnification is 5 times for a virtual image?

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

For virtual image, magnification m = 5 . Focal length f = −0.4 m = −40cm.

u + 40 = 8 ⇒u = −32cm

Distance: ∣u∣ = 32 cm

VITEEE PCME Mock Test - 5 - Question 9

A weight W rests on a rough horizontal plane. If the angle of friction be θ, the least horizontal force that will move the body along the plane will be

Detailed Solution for VITEEE PCME Mock Test - 5 - Question 9

Given, a block of weight W rests on a rough horizontal plane, with the angle of friction θ,
Let N be the normal force and f be the limiting friction acting on the block, then the angle of friction is defined as,

tanθ = f/N = μ; where μ is the coefficient of friction.
So, the minimum force required to move the block should be equal to the frictional force applied on the block, i.e.,
Fmin = μN = μW = W tanθ.

VITEEE PCME Mock Test - 5 - Question 10

The equivalent resistance between the points A and B of an infinite network of resistances, each of 1 Ω, connected as shown is

Detailed Solution for VITEEE PCME Mock Test - 5 - Question 10

Let x be the equivalent resistance of the entire network between A and B. Hence, we have

RAB = 1 + the resistance of the parallel combination of 1 Ω and x Ω


⇒ x + x2 = 1 + x + x
⇒ x2 − x − 1 = 0
⇒ 

VITEEE PCME Mock Test - 5 - Question 11

A force F is applied to a block of mass √3 kg resting on a horizontal surface with a coefficient of friction 1/√3. The maximum value of force F so that the block does not move is to be determined. (Take g = 10 m/s²).

Detailed Solution for VITEEE PCME Mock Test - 5 - Question 11

From acting on block are shown in adjoining figure
As the block does not move, hence
Fcos60° = f = μN = μ(Mg + Fsin60°)


On simplification, we get F = 20N

VITEEE PCME Mock Test - 5 - Question 12

A force F is required to break a wire of length l and radius r. What force is required to break a wire, of the same material, having twice the length and six times the radius?

Detailed Solution for VITEEE PCME Mock Test - 5 - Question 12

Breaking force does not depend upon length.

Breaking force = breaking stress × area of cross-section

For a given material, breaking stress is constant.

∴ F₂ / F₁ = A₂ / A₁ = π(6r)² / πr² = 36

Or F₂ = 36F₁ = 36F

VITEEE PCME Mock Test - 5 - Question 13

The figure shows a circuit diagram of a Wheatstone bridge to measure the resistance G of the galvanometer. The relation P/Q = R/G will be satisfied only when

Detailed Solution for VITEEE PCME Mock Test - 5 - Question 13

As the given condition is of balanced Wheatstone bridge, current can not flow through bridge arm whether switch is open or close. So, current in galvanometer will remain the same (zero) and would be independent of the position of switch.

In a balanced Wheatstone bridge:

Galvanometer current is zero.

VITEEE PCME Mock Test - 5 - Question 14

A conduction rod of 1 m length and 1 kg mass is suspended by two vertical wires through its ends. An external magnetic field of 2 T is applied normal to the rod. Now the current to be passed through the rod so as to make the tension in the wires zero is 
[Take g = 10 ms−2]

Detailed Solution for VITEEE PCME Mock Test - 5 - Question 14


Weight of the rod has to be balanced by the magnetic force such that tension in the strings become zero.
Let a current i is allowed to flow in the rod, such that,
magnetic force on the rod = Bil
where B is the magnetic field and ll is the length of the rod.
Now, weight of the rod = mg
Thus for no tension in the wire,
Bil = mg


= 5 A

VITEEE PCME Mock Test - 5 - Question 15

The temperature of an iron block is 140°F. Its temperature on the Celsius scale is

Detailed Solution for VITEEE PCME Mock Test - 5 - Question 15

The Celsius and Fahrenheit scales are related by the following equation:

C/5 = (F - 32)/9

where C is the temperature in degrees Celsius and F is in Fahrenheit.

Given: F = 140°F

Substituting the value:

(140 - 32)/9 = C/5

(140 - 32) = 108

108/9 = 12

C = 12 × 5 = 60°C

So, the temperature on the Celsius scale is 60°C.

VITEEE PCME Mock Test - 5 - Question 16

The half-life period and the mean-life period of a radioactive element are denoted respectively by Th and Tm. Then

Detailed Solution for VITEEE PCME Mock Test - 5 - Question 16

Half-life of a radioactive element is given by,
Th = 0.693 / λ
Where λ is called decay constant.
Mean-life of a radioactive element is given by Tm = 1/λ
Clearly, Th < Tm

VITEEE PCME Mock Test - 5 - Question 17

Calculate the emf of the cell in which the following reaction takes place.
Ni (s) + 2Ag+ (0.002 M) → Ni2+ (0.160 M) + 2Ag (s)
(Given:
= 1.05 V)

Detailed Solution for VITEEE PCME Mock Test - 5 - Question 17

According to Nernst equation,

VITEEE PCME Mock Test - 5 - Question 18

Which of the following reagents is/are used in the Hinsberg test of amines?

Detailed Solution for VITEEE PCME Mock Test - 5 - Question 18

The Hinsberg reaction is an excellent test for distinguishing primary, secondary and tertiary amines. In this test, the amine is shaken well with Hinsberg reagent in the presence of an aqueous alkali (either KOH or NaOH). A reagent containing an aqueous sodium hydroxide solution and benzenesulphonyl chloride is added to a substrate to form sulphonamides.
RNH2 (Primary amine) + C6H5SO2Cl (Hinsberg reagent) → R-NH-SO2-C6H5 R-N-Na+-SO2C6H5 (Soluble)
R2NH (Secondary amine) + C6H5SO2Cl (Hinsberg reagent) → R2NSO2C6H5 (precipitate) No reaction (Insoluble)
R3N (Tertiary amine) + C6H5SO2Cl (Hinsberg reagent) → No reaction

VITEEE PCME Mock Test - 5 - Question 19
The stability of +1 oxidation state increases in the sequence:
Detailed Solution for VITEEE PCME Mock Test - 5 - Question 19
Group 13 elements exhibit +3 and +1 oxidation states. Stability of the lower oxidation state increases on moving down the group due to inert pair effect. So, the order is: Al < Ga < In < Tl.
VITEEE PCME Mock Test - 5 - Question 20

Acidic hydrolysis of diethylmalonate, followed by heating, will yield which of the following carboxylic acids?

Detailed Solution for VITEEE PCME Mock Test - 5 - Question 20

Step 1: Acidic Hydrolysis of Diethylmalonate
Diethylmalonate has the structure:

 
When diethylmalonate undergoes acidic hydrolysis, the ester groups (–COOEt) are broken down by the acid, resulting in the formation of malonic acid:


Step 2: Heating the Malonic Acid
After hydrolysis, heating malonic acid leads to its decarboxylation (removal of a carbon dioxide molecule), which results in the formation of acetic acid (ethanoic acid):


Final Answer: After acidic hydrolysis followed by heating, the carboxylic acid formed is ethanoic acid (also known as acetic acid).

Thus, the correct answer is: D: Ethanoic acid

VITEEE PCME Mock Test - 5 - Question 21

Which of the following statements is/are true?
P. N3- has larger ionic radius than O2-.
Q. Cr has higher atomic radius than Co.
R. Technetium has only two stable isotopes.

Detailed Solution for VITEEE PCME Mock Test - 5 - Question 21

Both N3- and O2- are isoelectronic with each other, means they have the exact same electronic configuration. But O2- has one more proton than N3-; hence, the ionic radius of N3- is larger than O2-. Hence, statement P is correct.
In general, moving left to right across a period, atomic radius usually decreases, but in this case, both Cr (125 pm) and Co (125 pm) have exceptionally equal atomic radius. Thus, statement Q is false.
Technetium is the chemical element with atomic number 43 and symbol Tc. It is the lowest atomic number element without any stable isotope; every form of it is radioactive. Nearly all technetium is produced synthetically and only minute amounts are found in nature. Naturally occurring technetium occurs as a spontaneous fission product in uranium ore or by neutron capture in molybdenum ores. Thus, statement R is false.

VITEEE PCME Mock Test - 5 - Question 22

The most acidic α -hydrogen is in the following compound:

Detailed Solution for VITEEE PCME Mock Test - 5 - Question 22





After losing H  ions anion should be more stable and stability of anion is proportional to −M effect.
-M effect of 
Order of stability:

Thus order of acidity: A > B > C > D

VITEEE PCME Mock Test - 5 - Question 23

Zr and Hf have almost equal atomic and ionic radii because

Detailed Solution for VITEEE PCME Mock Test - 5 - Question 23

The size of lanthanide ions decreases as we move through the lanthanides from lanthanum to lutetium. This steady decrease in the ionic radii of M+3 cations in the lanthanide series is called lanthanide contraction. Due to the lanthanide contraction, the atomic and ionic radius of the next and the before, on the lanthanide elements with the same group, are about the same, i.e., the atomic radius of Zr(group-4, period-5) and Hf (group-4, period-6) are almost the same.

VITEEE PCME Mock Test - 5 - Question 24

According to Bohr's theory the energy required to remove electron from n = 2 of Li+2 ion is (given that the ground state ionization energy of hydrogen atom is 13.6 eV)

Detailed Solution for VITEEE PCME Mock Test - 5 - Question 24

Energy of electron in 2nd orbit of Li+2

Energy required = 0 − (−30.6) = 30.6 eV

VITEEE PCME Mock Test - 5 - Question 25

Activation energy of a chemical reaction can be determined by

Detailed Solution for VITEEE PCME Mock Test - 5 - Question 25

We know that the activation energy of a chemical reaction
is given by the formula, 

The formula is obtained from the Arrhenius equation, where k1 is the rate constant at the temperature T1, k2 is the rate constant at the temperature T2 and Ea is the activation energy.
Therefore, the activation energy of a chemical reaction is determined by evaluating the rate constant at two different temperatures.

VITEEE PCME Mock Test - 5 - Question 26

Arrange in the increasing order of oxidation state of nitrogen for following nitrogen oxides.
N2O, NO2, NO, N2O3

Detailed Solution for VITEEE PCME Mock Test - 5 - Question 26

VITEEE PCME Mock Test - 5 - Question 27

Read the following sentences carefully and choose the most appropriate phrase to join them.
I. Studies on nonlinear dependence on applied voltage can be expanded to more complex quantum transport experiments.
II. Experiments on artificial atoms are called quantum dots.
i. Ever since…
ii. Even though…
iii. For instance…

Detailed Solution for VITEEE PCME Mock Test - 5 - Question 27

Here, Statement II is a simple continuation of Statement I. It further extends sentence I, and can be connected by using ‘for instance’. The correct formation of the sentence will be:
Studies on nonlinear dependence on applied voltage can be expanded to more complex quantum transport experiments, for instance, those on artificial atoms are called quantum dots.

VITEEE PCME Mock Test - 5 - Question 28

Each of the following questions has two sentences A and B.
Mark (A) if you think sentence A has an error.
Mark (B) if you think sentence B has an error.
Mark (C) if you think both sentences A and B have errors.
Mark (D) you think neither sentence has an error.

A. Because he wanted it done right, he always did it himself.
B. One criteria that is invariably used is your score in the written test.

Detailed Solution for VITEEE PCME Mock Test - 5 - Question 28

In the first sentence, an adverb is required, which qualifies the verb. Replace ‘right’ with ‘rightly’.
Sentence B has an error (incorrect use of "criteria" instead of "criterion").
Hence, option C is the correct answer.

VITEEE PCME Mock Test - 5 - Question 29

The following sentence has a word or group of words missing. Four or five alternative words are given. You have to find out which one them would make the sentence grammatically correct and meaningful.
I left home "_____" a walk in the garden.

Detailed Solution for VITEEE PCME Mock Test - 5 - Question 29

The correct answer is 'for'
A Preposition is a word or set of words that indicates location (in,near,beside) or some other relationship between a noun or pronoun or other parts of a sentence. The preposition 'for' usually tell us about the use of something, a reason or purpose.
So, in the given sentence, the subject left home for a purpose (for a walk). Hence, the correct answer is 'for'.

VITEEE PCME Mock Test - 5 - Question 30
In a certain code language, 'come and go' is coded as 758, 'go home soon' is coded as 839, and 'good and bad' is coded as 521. Which of the following is code for 'come'?
Detailed Solution for VITEEE PCME Mock Test - 5 - Question 30
758 is code for 'come and go'.......(i)
839 is code for 'go home soon'......(ii)
521 is code for 'good and bad'.......(iii)
In statements (i) and (ii), 'go' is the common word and '8' is the common code. Thus, the code for 'go' is '8'.
In statements (i) and (iii), 'and' is the common word and '5' is the common code. Thus, the code for 'and' is '5'.
From statement (i), the code for 'come' is '7'.
Hence, answer option 4 is correct.
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