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BITSAT Mock Test - 2 Free Online Test 2026


Full Mock Test & Solutions: BITSAT Mock Test - 2 (130 Questions)

You can boost your JEE 2026 exam preparation with this BITSAT Mock Test - 2 (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: 180 minutes
  • - Total Questions: 130
  • - Analysis: Detailed Solutions & Performance Insights
  • - Sections covered: Physics, Chemistry, English Proficiency & Logical Reasoning, Engineering Entrance (Mathematics)

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BITSAT Mock Test - 2 - Question 1

If a source emitting waves of frequency f moves towards an observer with a velocity v/4 and the observer moves away from the source with a velocity v/6. The apparent frequency as heard by the observer will be
(v = Velocity of sound)

Detailed Solution: Question 1

Let f' be the frequency observed by the observer,
v0 be the velocity of the observer, vs be the velocity of the source,
v be the velocity of the sound and f be the actual frequency.

BITSAT Mock Test - 2 - Question 2

If a vector is perpendicular to the vector , then the value of α is

Detailed Solution: Question 2

When two vectors are perpendicular to each other, then their dot product is zero.

BITSAT Mock Test - 2 - Question 3

A cube is subjected to a uniform volume compression. If the side of the cube is decreased by 2%, then the bulk strain is

Detailed Solution: Question 3

BITSAT Mock Test - 2 - Question 4

The length of a wire of a potentiometer is 100 cm and the emf of its standard cell is E volt. It is employed to measure the emf of a battery whose internal resistance is 0.5 W. If the balance point is obtained at l = 30 cm from the positive end, then the emf of the battery is

Detailed Solution: Question 4

Potential gradient of the potentiometer wire is
V/cm
Emf of the cell is
, l = 30 cm

BITSAT Mock Test - 2 - Question 5

The expression up to correct significant figures is equal to

Detailed Solution: Question 5


In the expression, we have minimum three significant figures, hence answer should be up to three significant figures.
Rounding off the result 1.612979, we have

BITSAT Mock Test - 2 - Question 6

What is the potential difference between points C and D in the circuit shown in the figure?

Detailed Solution: Question 6

In DC capacitors behave as open circuit, so current will pass through the resistances.
VCD = VC - VD, VAD = VA - VD, VAC = VA - VC
VC - VD = (VA - VD) - (VA - VC)
VCD = VAD - VAC    

Current in the main circuit is
= 1.2 A
Thus, the potential difference across AB is:

Also, the potential difference across AD is:

Effective capacitance is:

Since the capacitors are in series, charge Q is the same and is given by:


Therefore, the potential difference across AC is:

Thus, the potential difference between points C and D is:

Hence, the correct choice is (1).

BITSAT Mock Test - 2 - Question 7

If the acceleration due to gravity is 'g' on the surface of Earth, then the value of acceleration due to gravity at a height of 32 km above the surface of Earth is (radius of Earth = 6400 km)

Detailed Solution: Question 7

The acceleration due to gravity at a height h above the surface of Earth is given by

For h = 32 km, we can use

BITSAT Mock Test - 2 - Question 8

The wavelength of a spectral line emitted by hydrogen atom in the Lyman series is cm. What is the value of n2?
(R = Rydberg constant)

Detailed Solution: Question 8

According to Reyberg's formula,

For Lyman series, n1 = 1

BITSAT Mock Test - 2 - Question 9

At 25°C, the dissociation constant of a base, BOH, is 1.0 × 10-12. The concentration of hydroxyl ions in 0.01 M aqueous solution of the base would be

Detailed Solution: Question 9

Base BOH is dissociated as follows.

So, dissociation constant of BOH base,

At equilibrium,
Kb =
Given, and [BOH] = 0.01 M
Thus,

BITSAT Mock Test - 2 - Question 10

Which compound acts as an oxidising as well as reducing agent

Detailed Solution: Question 10

The minimum and maximum oxidation number of are and respectively. Since the oxidation number of in is , therefore it can be either increased or decreased. Therefore behaves both as an oxidising as well as reducing agent.

BITSAT Mock Test - 2 - Question 11

The de-Broglie wavelength of the electron in the ground state of hydrogen atom is
(KE = 13.6 eV and 1 eV = 1.602 x 10-19 J)

Detailed Solution: Question 11

According to de-Broglie relation,

h = 6.62 x 10-34 kgm2s-1
KE = 13.6 x 1.602 x 10-19 J
KE =
v =
= 2.18824 x 106 m/s

= 0.3328 x 10-9 m = 0.3328 nm

BITSAT Mock Test - 2 - Question 12

Consider the ground state configuration of Cr atom (Z = 24). The total numbers of electrons with azimuthal quantum numbers l = 1 and l = 2, respectively, are

Detailed Solution: Question 12

E.C. of Cr (Z = 24) is

Thus, number of electrons with l = 1 is 12
and number of electrons with l = 2 is 5
Note: Chromium shows exceptional configuration, [Ar], 3d5, 4s1 so as to acquire a stable configuration of half-filled 3d orbitals.

BITSAT Mock Test - 2 - Question 13

Ferrous oxide forms a lattice structure in which the length of edge of the unit cell is 5 and the density of the oxide is 4.0 g cm-3. Then, the numbers of Fe2+ and O2- ions present in each unit cell will be

Detailed Solution: Question 13



a = 5 cm
V = (5 × 10-8)3 cm3
Molecular mass, M = 56 + 16 = 72 g mol-1

z = 4.18 4
Four molecules of ferrous oxide, FeO per unit cell, means that 4Fe2+ and 4O2- ions are present in each unit cell.
Hence, FeO forms a CCP lattice.

BITSAT Mock Test - 2 - Question 14

Which among the following species have square planar geometry around the central atom?
(i) XeF4
(ii) SF4
(iii) [NiCI4]2-
(iv) [PtCI4]2-

Detailed Solution: Question 14

XeF4 has square planner geometry having sp3d2 hybridisation in the central atom as shown below:

SF4 has distorted trigonal bipyramidal structure with sp3d hybridisation in the central atom.

[NiCI4]2- has sp3 hybridisation in the central atom with a tetrahedral structure.

[PtCI4]2-has dsp2 hybridisation in the central atom with square planar structure.

BITSAT Mock Test - 2 - Question 15

'A' and 'C' in the following conversion are

Detailed Solution: Question 15

BITSAT Mock Test - 2 - Question 16


'G' and 'H' are a mixture of

Detailed Solution: Question 16

Step I ('A' + 'B') - Nitration of toluene gives o- and p-nitrotoluene.
Step II ('C' + 'D') - Reduction of nitro-group to amine group gives o- and p-methyl aniline.
Step III ('E' + 'F') - Diazonation gives diazonium salt.
Step IV ('G' and 'H') - Sandmeyer reaction gives mixture of o- and p-bromobenzene.

BITSAT Mock Test - 2 - Question 17

Directions: In this question, the second figure of the problem figures bears a certain relationship to the first figure. Similarly, one of the figures in the answer figures bears the same relationship to the third problem figure. You have to select the figure from the set of answer figures which would replace the sign of question mark (?).

Detailed Solution: Question 17

Problem figure (2) is the mirror image of problem figure (1), but with inverse operations. Hence, to get the missing figure, the operations in problem figure (3) will become opposite, that is, '÷' will become '×' and vice versa.
Hence, option 4 is the correct answer.

BITSAT Mock Test - 2 - Question 18

Who among the following is the shortest?

Detailed Solution: Question 18

Uma is taller than Poonam and Sonika, but shorter than Qurashi and is the second tallest. Waseem plays chess with Sonika and is the fourth tallest.

Qurashi does not play golf. The shortest does not play hockey and the tallest does not play golf. Uma is taller than Poonam and Sonika, but shorter than Qurashi and is the second tallest. Varun is shorter than Sonika, but only taller than Poonam. Risha is taller than Tanvi who is shorter than sonika.

Varun does not play hockey or golf.

BITSAT Mock Test - 2 - Question 19

Directions: In the following question, there is a certain relationship between the two terms given 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.
ACH : BDI : : MOT : ?

Detailed Solution: Question 19

As,
A + 1 = B
C + 1 = D
H + 1 = I
Similarly,
M + 1 = N
O + 1 = P
T + 1 = U
Hence, NPU is the correct answer.

BITSAT Mock Test - 2 - Question 20

Directions: There is some 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 one from the given alternatives.
acE : bdF : : fhJ : ?

Detailed Solution: Question 20

Each letter of the first group is replaced by its successor in the English alphabet to get the second group. Hence, fhJ will become giK to replace the question mark (?).

BITSAT Mock Test - 2 - Question 21

Directions: Select the phrase/connector from the given three options which can be used to form a single sentence from the two sentences given below, implying the same meaning as expressed in the statement sentences.
Frequent elections impose a huge burden on a nation's human resources and also impede the development process. The president of the country has proposed the concept of 'one nation, one election'.
a. However the frequent elections...
b. In lieu of huge burden imposed...
c. As the development process is impeded...

Detailed Solution: Question 21

Here, the second sentence is the result of the first. Burden of frequent elections leads to the concept of 'one nation, one election'.
Use of 'however' and 'in lieu' in options a and b are not correct as they are used to create a contrasting meaning when we have to talk about two different sides. So only c is correct; As the development process is impeded and huge burden is imposed on a nation's human resources by frequent elections, the president of the country has proposed the concept of 'one nation, one election'.

BITSAT Mock Test - 2 - Question 22

One ticket is selected at random from 100 tickets numbered 00, 01, 02, ….., 98 and 99. If X and Y denote the sum and the product of the digits on the tickets, respectively, then P(X = 9|Y = 0) is equal to

Detailed Solution: Question 22

BITSAT Mock Test - 2 - Question 23

is equal to

Detailed Solution: Question 23

BITSAT Mock Test - 2 - Question 24

If ,and are non-coplanar unit vectors such that ( x ) = , and are non-parallel, then the angle between and is

Detailed Solution: Question 24

BITSAT Mock Test - 2 - Question 25

If z1 = 8 + 4i, z2 = 6 + 4i and arg , then z satisfies

Detailed Solution: Question 25

BITSAT Mock Test - 2 - Question 26

If , then is equal to

Detailed Solution: Question 26

BITSAT Mock Test - 2 - Question 27

If the coefficient of correlation between x and y is 0.28, covariance between x and y is 7.6 and variance of x is 3, then the SD in y-series is

Detailed Solution: Question 27

BITSAT Mock Test - 2 - Question 28

The domain of the real-valued function f(x) = is

Detailed Solution: Question 28

BITSAT Mock Test - 2 - Question 29

If the roots of the equation ax2 + bx + c = 0 are of the form and , then the value of (a + b + c)2 is

Detailed Solution: Question 29


BITSAT Mock Test - 2 - Question 30

The value of eccentricity for the hyperbola x2 - 2y2 - 2x + 8y - 1 = 0 is

Detailed Solution: Question 30

x2 - 2y2 - 2x + 8y - 1 = 0
or,
Comparing it with the general equation of hyperbola, we get
a2 = 6 and b2 = 3
Eccentricity for hyperbola is given by 'e' =
e =

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