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BITSAT Mock Test - 3 - JEE MCQ


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30 Questions MCQ Test - BITSAT Mock Test - 3

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

A particle of mass m is attached to a spring (of spring constant k) and has a natural angular frequency . An external force F(t) proportional to is applied to the oscillator. The time displacement of the oscillator will be proportional to

Detailed Solution for BITSAT Mock Test - 3 - Question 1

BITSAT Mock Test - 3 - Question 2

A uniform but time varying magnetic field B(t) exists in a circular region of radius a and is directed into the plane of the paper as shown in the figure. The magnitude of the induced electric field at point P at a distance r from the centre of the circular region

Detailed Solution for BITSAT Mock Test - 3 - Question 2


The magnitude of the induced electric field at point P at a distance r from the centre of the circular region decreases as .

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BITSAT Mock Test - 3 - Question 3

The displacement y of a particle executing periodical motion is given by y = 4 cos2 (t/2) sin (1000t). This expression may be considered to be a result of the superposition of how many independent harmonic motions?

Detailed Solution for BITSAT Mock Test - 3 - Question 3

Hence above expression involves three different waves the displacement given is a result of superposition of three different waves.
BITSAT Mock Test - 3 - Question 4

Which of the following statements is correct about the circuit shown in the figure below, where 1 Ω and 0.5 Ω are the internal resistances of the 6 V and 12 V batteries, respectively?

Detailed Solution for BITSAT Mock Test - 3 - Question 4

Total resistance = 4 + 1 + 0.5 + 0.5 = 6 Ω
Net voltage in the circuit is 18 V. Current I = 18/6 = 3 A in the anti-clockwise direction
Current is being forced into the 6 V battery in the opposite direction.
Hence, V6 = E + Ir = 6 + 3 1 = 9 V
Hence, the correct choice is (3).

BITSAT Mock Test - 3 - Question 5

The motion of an object follows the differential equation , where 'a' and 'b' are constants. The time instant at which the velocity remains constant is

Detailed Solution for BITSAT Mock Test - 3 - Question 5


Solving for v in terms of t, we get
.
Applying the initial conditions at t = 0 and v = 0 (as the object is dropped), we get .
Thus, the expression for v is equal to , where the velocity is constant and acceleration is zero.
Hence, v =
Substituting for the value of v and solving for 't', we get the value of t as infinity.

BITSAT Mock Test - 3 - Question 6

A circular coil of radius R carries an electric current. The magnetic field due to the coil at a point on the axis of the coil, located at a distance r from the centre of the coil such that r >> R, varies as

Detailed Solution for BITSAT Mock Test - 3 - Question 6

Magnetic field at a distance 'r' along the axis of a circular wire having radius R is given by,

For r >> R,

or,

BITSAT Mock Test - 3 - Question 7
The speed of sound in a gas is v and the root mean square speed of gas molecules is vrms. If the ratio of the specific heats of the gas is 1.5, then the ratio v/vrms is
Detailed Solution for BITSAT Mock Test - 3 - Question 7
BITSAT Mock Test - 3 - Question 8
The halide having the highest melting point is
Detailed Solution for BITSAT Mock Test - 3 - Question 8
NaF has the highest melting point among all these halides. F is a highly electronegative element. Thus, the molecules are highly polar and have strong electrostatic force (ionic solid). Thus, large amount of heat is required to pull the molecules apart.
Moreover, as the size of the halide ion increases, the covalent character in the halides increases due to higher polarisability of anion.
BITSAT Mock Test - 3 - Question 9
If = 224 cm2gequiv-1, = 38.5 cm2gequiv-1 and = 203 cm2gequiv-1, then what is the value of ?
Detailed Solution for BITSAT Mock Test - 3 - Question 9
According to Kohlrausch's law,

From Kohlrausch's law,

Or, = 224 ohm-1 cm2gequiv-1 …(i)

Or, = 203 ohm-1 cm2gequiv-1 ... (ii)

Or, = 38.5 ohm-1 cm2gequiv-1 ...(iii)
Adding (i) and (ii), and subtracting (iii), we get

= 224 + 203 - 38.5
= 427 - 38.5
= 388.5 cm2gequiv-1
BITSAT Mock Test - 3 - Question 10
In an experiment, addition of 4.0 mL of 0.005 M BaCl2 to 16.0 mL of arsenious sulphide solution just causes complete coagulation in 2 hours. The flocculating value of the effective ion is
Detailed Solution for BITSAT Mock Test - 3 - Question 10
As2S3 solution is a negatively charged solution formed by the preferential adsorption of S2- ions on the passage of excessive H2S.
Therefore, the cation of the electrolyte, i.e. Ba+2 is the active ion.
Flocculating value = Minimum millimole of the effective ion per litre of solution
= (millimoles of Ba+2/total volume in L) =
BITSAT Mock Test - 3 - Question 11

The solubility of a particular compound depends upon the value of its lattice enthalpy and hydration enthalpy. The lattice and hydration enthalpies of four compounds are given below.

The pair of compounds that is soluble in water is

Detailed Solution for BITSAT Mock Test - 3 - Question 11

A compound is soluble in water when its hydration enthalpy is greater than its lattice enthalpy.

BITSAT Mock Test - 3 - Question 12

Potassium sulphite is used in preserving squashes and other mildly acidic foods because

Detailed Solution for BITSAT Mock Test - 3 - Question 12

Potassium metabisulphite and potassium sulphite are used as a source of sulphur dioxide, which reacts with H2O to form sulphurous acid.
Sulphurous acid inhibits the growth of bacteria, fungi, yeast and moulds.
K2S2O5 → K2SO3 + SO2
SO2 + H2O → H2SO3

BITSAT Mock Test - 3 - Question 13

Directions: Consider the following statements.
I. Bond length in is 0.02 greater than in N2.
II. Bond length in NO+ is 0.09 less than in NO.
III. has a shorter bond length than O2.
Which of the following statements are true?

Detailed Solution for BITSAT Mock Test - 3 - Question 13

In case of nitrogen,
N2 → N2+ + e-
An electron is lost from bonding MO and the bond order decreases from 3 to 2.5. As a result, bond dissociation energy decreases and the bond length increases.
The bond order of the nitrogen-oxygen bond in NO+ is three (triple bond), whereas in NO, the one additional electron is located in an antibonding MO. Therefore, the bond order in NO is 2.5. Thus, bond length in NO+ is less than in NO because bond length is inversely proportional to bond order.
Bond orders of and O2 are 1 and 2, respectively. Therefore, O2 bond must be shorter as bond length is inversely proportional to bond order.

BITSAT Mock Test - 3 - Question 14

A compound possesses 8% sulphur by mass. The least molecular mass of the compound is

Detailed Solution for BITSAT Mock Test - 3 - Question 14

8% sulphur by mass means 8 g of sulphur is present in 100 g of compound.
Therefore, 32 g of sulphur (1 mole atom) will be present in x 32 = 400 g of compound.
[∵ Compound must have at least one atom of sulphur]
Minimum molecular mass = 400 g

BITSAT Mock Test - 3 - Question 15
Which of the following statements account(s) for the acidic nature of acetylene?
Detailed Solution for BITSAT Mock Test - 3 - Question 15
The acidic nature of acetylene is due to the greater sigma electron density of C-H bond of acetylene towards carbon, which is sp hybridised and has 50% of s-character.
Hence, the electron pair of C-H bond gets displaced towards the carbon atom; and as a result, abstraction of H+ ion becomes easier.
BITSAT Mock Test - 3 - Question 16

What is the position of D from the bottom end?

Detailed Solution for BITSAT Mock Test - 3 - Question 16

G lives at the right end of the horizontal block. A lives at the top of the vertical block. D is at the point where the horizontal and vertical blocks meet.

F is third from the left end in the horizontal row. B is third from the bottom in the vertical column.

As the residing place for C and E is not defined, there are two possibilities:

D's position is 1st from the bottom end.

BITSAT Mock Test - 3 - Question 17

Directions: Select one figure from the 'Answer Figures' which will continue the same series as given in the 'Problem Figures'.

Detailed Solution for BITSAT Mock Test - 3 - Question 17

From Problem Figure (1) to (2), the figure on the bottom left position is eliminated and a new figure is formed on top of the bottom right figure. From figure (2) to (3), figure shifts to half-side in clockwise direction and one new figure is added on top of the right figure. The same method is repeated in the subsequent figures.

BITSAT Mock Test - 3 - Question 18

Directions: The following question consists of two sets of words. Each word in the two sets is related to the other in the same way. Find out the relationship in the first set of two words and then select the alternative that best fits the blank in the second set of words.
Etiquette : Social :: Protocol : ____

Detailed Solution for BITSAT Mock Test - 3 - Question 18

Etiquette' means 'the customs or rules governing behaviour regarded as correct or acceptable in social or official life'.
'Protocol' means 'the official procedure or system of rules governing affairs of state or diplomatic occasions'.

BITSAT Mock Test - 3 - Question 19

Directions: There is a certain relationship between the two terms to the left of the sign (: :). The same relationship exists between the two terms to its right, out of which one is missing. Find the missing term from the given alternatives.
ADE : FGJ : : KNO : ?

Detailed Solution for BITSAT Mock Test - 3 - Question 19

The first and third letters of the first term are each moved five steps forward, while the second letter is moved three steps forward to obtain the corresponding letters of the second term.
K + 5 = P
N + 3 = Q
O + 5 = T
Hence, option (2) is correct.

BITSAT Mock Test - 3 - Question 20

Directions: In this question, a set of Problem figures is given followed by a set of Answer figures (a, b, c and d). Select one figure from the Answer figures which will continue the same pattern as given in the Problem figures.

Detailed Solution for BITSAT Mock Test - 3 - Question 20

The line with dot at the end alternately rotates 90o and 135o in the clockwise direction. Hence, the correct option is (4).

BITSAT Mock Test - 3 - Question 21

Directions: In the following question, four statements are provided. These statements form a coherent paragraph when properly arranged. Select the alternative representing the proper and logical sequencing of these statements.
A. In effect, Proton cars are versions of Mitsubishi models and incorporate parts imported from Japan.
B. This is an alliance between the Mitsubishi Motor Corp. and Hicom, a Malaysian entity making the Proton brand of cars.
C. The most interesting of the joint ventures is ironically one that claims to be Malaysia's indigenous national car producer.
D. The fortunes of Proton have been decidedly mixed but the sometimes fraught joint venture demonstrated how on East Asian nation could build a car-making industry from scratch (hitherto there had been only auto assembly in Malaysia) and do so at great speed.

Detailed Solution for BITSAT Mock Test - 3 - Question 21

Out of statements D, A and C, only C makes the beginning as it talks about the joint venture which is explained in all the rest statements. As B starts with "this" and gives the names of two companies, it follows C. Now, we are left with A and D; as A explains proton cars, it follows B. This makes (4) as the best possible answer.

BITSAT Mock Test - 3 - Question 22

Directions: In the following question, the first and the last sentences of a paragraph are numbered 1 and 6, respectively. The remaining sentences are named P, Q, R and S. These four sentences are not given in their proper order. Rearrange them in the proper sequence to form a meaningful paragraph. Then, find the correct answer out of the given alternatives.
1. The highway bypass would have disastrous effects on the area's homeowners.
P. Finally, the new road would cause residential properties to depreciate.
Q. What is more, homeowners would have to deal with the increased noise and pollution.
R. This would increase vehicles in the neighbourhood.
S. The new road would cut directly through the middle of the subdivision.
6. This means that families who chose to move away would have to sell their homes for far less than their current value.

Detailed Solution for BITSAT Mock Test - 3 - Question 22

This passage is about the effects of highway bypass. 'The new road' in S refers to the 'highway bypass' in sentence 1. So, S is placed first in the sequence. 'This' in R refers to the 'new road' in S - new road would lead to more vehicles. Therefore, R follows S and we get the first pair SR.
The word 'finally' in P gives a clue that it should be the last in the sequence. That means Q precedes P and we get the pair QP. Moreover, 'increase vehicles' in R leads to 'increased noise and pollution' in Q.
Hence, we get the right sequence as SRQP. Option (1) is the correct answer.

BITSAT Mock Test - 3 - Question 23
Solving the differential equation + (sec x)y = tan x, we get
Detailed Solution for BITSAT Mock Test - 3 - Question 23
BITSAT Mock Test - 3 - Question 24

If g(x) = xf(x), where f(x) = x sin , then at x = 0,

Detailed Solution for BITSAT Mock Test - 3 - Question 24

g(x) = xf(x), where f(x) = x sin
at x = 0
g(0) = 0 f(0)
= 0 (0) = 0
g'(0) =
=
=
also g'(x) = xf'(x) + f(x)
g'(x) = [xf' (x) + f(x)]
= 0 + f(0) = 0 + 0 = g'(0)
g'(x) = g'(0)
Hence g' is continuous and g(x) is differentiable.

BITSAT Mock Test - 3 - Question 25
When the length of the shadow of a pole is equal to the height of the pole, then the angle of elevation of the source of light is
Detailed Solution for BITSAT Mock Test - 3 - Question 25
BITSAT Mock Test - 3 - Question 26
If , then r equals
Detailed Solution for BITSAT Mock Test - 3 - Question 26
BITSAT Mock Test - 3 - Question 27
If a, b, c are in AP as well as in GP, then
Detailed Solution for BITSAT Mock Test - 3 - Question 27
BITSAT Mock Test - 3 - Question 28
If 6 girls and 5 boys sit in a row, then the probability that no two boys sit together is
Detailed Solution for BITSAT Mock Test - 3 - Question 28
BITSAT Mock Test - 3 - Question 29
Detailed Solution for BITSAT Mock Test - 3 - Question 29
BITSAT Mock Test - 3 - Question 30

The area under the curve y = sin 2x + cos 2x between x = 0 and x = π/4 is

Detailed Solution for BITSAT Mock Test - 3 - Question 30

Required area
=
=
=
= 1 sq. unit

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