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BITSAT Mock Test - 6 Free Online Test 2026


Full Mock Test & Solutions: BITSAT Mock Test - 6 (130 Questions)

You can boost your JEE 2026 exam preparation with this BITSAT 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: 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 - 6 - Question 1

A block of mass m slides down a wedge of mass M as shown. The whole system is at rest, when the height of the block is h above the ground. The wedge surface is smooth and gradually flattens. There is no friction between wedge and ground.

If there is no friction anywhere, the speed of the wedge, as the block leaves the wedge, is

Detailed Solution: Question 1

P1 = P2
o = mv1 - MV2
mV1 = MV2
k1 + v1 = k2 + v2
o + mgh =
mgh =
mgh =

=

BITSAT Mock Test - 6 - Question 2

In a Carnot engine, efficiency is 40% at hot reservoir temperature T. For efficiency to be 50%, what shall be the temperature of the hot reservoir?

Detailed Solution: Question 2

The efficiency of a heat engine is defined as the ratio of work done to the heat supplied, i.e.,

where T2 is temperature of sink,
and T1 is temperature of hot reservoir.

BITSAT Mock Test - 6 - Question 3

The magnetic field in a region is given by . A square loop of side d is placed with its edges along the x and y-axis. The loop is moved with a constant velocity . The emf induced in the loop is

Detailed Solution: Question 3

φi = ∫ from 0 to d of (B0 x / a) d * dx = (B0 d2) / (2a)

Final Flux (φf) after time dt:

x changes to x + v0 dt, so
φf = ∫ from 0 to d of (B0 (x + v0 dt) / a) d * dx
≈ (B0 d) / a * (x + v0 dt) (linear approximation, correct differential flux change)

EMF Calculation:

EMF = - dφ / dt

Flux (φ):

φ = (B0 d / a) ∫ x dx from 0 to d = (B0 d2) / (2a) at x = 0,
At x = v0 t, φ = (B0 d / a) ∫ from 0 to d of (x + v0 t) dx (incorrect approach)

Correct EMF:

EMF = B0 v0 d (rate of change of area swept)

BITSAT Mock Test - 6 - Question 4

A thin equi-convex lens is made of glass with refractive index 1.5, and its focal length is 0.2 m. If it acts as a concave lens of 0.5 m focal length when dipped in a liquid, the refractive index of the liquid is:

Detailed Solution: Question 4

The focal length of a convex lens of refractive index μg in air is
...(i)
Where R1 and R2 are the radius of curvatures of its first and second surface.
When lens immersed in a liquid of refractive index μl then refractive index of material of lens (glass) with respect to liquid is
... (ii)
∴ Focal length of lens in liquid is
...(iii)
Dividing (i) by (iii), we get

Putting f' = -0.5 m
fair = 0.2 m
= 1.5
= ?
=




∴ Refractive index of liquid =

BITSAT Mock Test - 6 - Question 5

The point charges 3μC and  4μC are placed at a separation of 7 m. The medium between them is of two type as shown in figure. What is the electric force acting between them?

Detailed Solution: Question 5

Force between 2 charges in a medium = F' =
∴ F' =
In the given question, total effective distance =
Therefore, force F =
=
= 1.73 × 10-4 N

BITSAT Mock Test - 6 - Question 6

Spheres A and B of equal radius and of masses 2 m and m, respectively, are moving towards each other and strike directly. The speeds of A and B before the collision are 3u and u, respectively. The collision is such that B experiences an impulse of 4mcu, where c is constant.
What is the coefficient of restitution?

Detailed Solution: Question 6

Magnitude of the Impulse received || = 4mcu

V1 and V2 are velocities after collision.
Magnitude of impulse is the same on both A and B.
For body A:
2mv1 - 6mu = -4mcu
2mv1 = 6mu - 4cum
V1 = 3u - 2cu()
For body B:
mv2 - (-mu) = 4mcu
mv2 = 4mcu - mu
v2 = 4cu - u() - speed of B
Coefficient of restitution,

BITSAT Mock Test - 6 - Question 7

The air of density ρ and moving with a velocity v strikes perpendicularly the inclined surface of area A and of a wedge kept on a horizontal surface. The mass of the wedge is m. Assuming the collisions to be perfectly inelastic, the minimum value of the coefficient of friction between the wedge and the ground, so that the wedge does not move is
(Assume mass of particles of air is negligible)

Detailed Solution: Question 7

As the collision is completely inelastic, so change in momentum is ρAv2.
Resolving it into components as shown, we have

BITSAT Mock Test - 6 - Question 8

A transverse sinusoidal wave of amplitude a, wavelength and frequency f is travelling on a stretched string. The maximum speed at any point on the string is v/10, where v is the speed of propagation of the wave. If a = 10-3 m and v = 10 ms-1, then and f are given by

Detailed Solution: Question 8

BITSAT Mock Test - 6 - Question 9

In a series circuit, L, C and R are connected with an alternating voltage source of frequency f. The current leads the voltage by 45o. The value of C is

Detailed Solution: Question 9

As the current is leading the voltage, then we have



BITSAT Mock Test - 6 - Question 10

A transformer has 200 windings in the primary and 400 windings in the secondary. The primary is connected to an AC supply of 110 V and a current of 10 A flows in it. The voltage across the secondary and the current in it respectively are

Detailed Solution: Question 10

BITSAT Mock Test - 6 - Question 11


The product of the given reaction is

Detailed Solution: Question 11

Oxymercuration-demercuration of alkenes yields a non-rearranged Markovnikov product.

BITSAT Mock Test - 6 - Question 12

Which of the following is aromatic?

Detailed Solution: Question 12

According to Huckle, an aromatic compound must be cyclic planar having (4n + 2)π electrons.

BITSAT Mock Test - 6 - Question 13

A hypothetical reaction X2 + Y2 → 2XY follows the following mechanism:
X2 →  X + X ... fast
X + Y2 →​​​​​​​ XY + Y … slow
X + Y →​​​​​​​ XY … fast
The order of the overall reaction is

Detailed Solution: Question 13

Rate = k[X][Y2]
But the rate law expression cannot be expressed in terms of a reaction intermediate.

Rate =
Thus, order of the reaction

BITSAT Mock Test - 6 - Question 14

Two stereoisomers are given below. These are known as

Detailed Solution: Question 14

Diastereomerism occurs when two or more stereoisomers of a compound have different configurations at one or more (but not all) of the equivalent stereocentres and are not mirror images of each other. Thus,

BITSAT Mock Test - 6 - Question 15

Identify the product S in the following sequence of reactions:

Detailed Solution: Question 15

BITSAT Mock Test - 6 - Question 16

Consider the reaction:
P + 2Q → R
When concentration of Q alone was tripled, the half-life did not change. When the concentration of P alone was made four times, the rate increased two times. The unit of rate constant for this reaction would be

Detailed Solution: Question 16

For the reaction: P + 2Q → R, when concentration of 'Q' is doubled, the half-life does not change, hence reaction is first-order with respect to 'Q'.
This is because for the first order reaction, rate constant is independent from initial concentration of the reactant.
When the concentration of 'P' is made four times, the rate of reaction becomes double. It means the reaction is half-order with respect to 'P'.
Hence, the rate law expression is:

Overall order = 1 + 1/2 = 3/2
Units of the rate constant = (mol L-1)(1 - n) s-1
So, unit of rate constant is L1/2 mol-1/2 s-1.

BITSAT Mock Test - 6 - Question 17

CH2=CH-CH2-CH2-NH2 Product
The major product is

Detailed Solution: Question 17

BITSAT Mock Test - 6 - Question 18

Identify the correct statement for change of Gibbs energy for a system (Gsystem) at constant temperature and pressure.

Detailed Solution: Question 18

In the alternatives, (3) is most confusing as when G > 0, the process may be spontaneous when it is coupled with a reaction whose G < 0 and total G is negative. So, the correct answer is (1).

BITSAT Mock Test - 6 - Question 19

A piece of paper is folded and punched cut as shown below in the Question figures. From the given Answer figures, indicate how it will appear when opened.

Detailed Solution: Question 19


Hence, option 4 is the correct answer.

BITSAT Mock Test - 6 - Question 20

Directions: A sentence has been given in Active/Passive Voice. Out of the four alternatives suggested, select the one which best expresses the same sentence in Passive/Active Voice.
The author wrote the book in one month.

Detailed Solution: Question 20

The context sentence is in active voice. A sentence is in passive form when the subject (person/place/thing or idea) receives the action.
The sentence is in simple past tense. We add 'was' in the passive form of sentence (was + written).
The doer of the action may or may not be mentioned at the end of the sentence with the preposition 'by' as it is understood that the book is written by an author only.
Option 4 has all the modifications in the sentence as per the rules of active/passive voice. Thus, it is the correct answer.

BITSAT Mock Test - 6 - Question 21

Directions: Out of the four alternatives, choose the correct antonym of the given word.
Fervent

Detailed Solution: Question 21

The antonym of 'fervent' is 'dispassionate'.
'Fervent' means 'passionate; sincere; emotional'.
'Dispassionate' means 'emotionless; impassive'.
Thus, option 2 is the correct answer.

BITSAT Mock Test - 6 - Question 22

Directions: In the given question, there are four different words out of which one is wrongly spelt. Find the wrongly spelt word.

Detailed Solution: Question 22

Correctly spelled word: 'collusion.' It refers to 'a secret or illegal conspiracy in order to deceive others'.

BITSAT Mock Test - 6 - Question 23

Directions: In the following question, four figures are given. Three of them are similar in some way but one figure is dissimilar. Point out the figure which does not belong to the group.

Detailed Solution: Question 23

Only option (4) has an 'x' instead of a '+'.

BITSAT Mock Test - 6 - Question 24

Directions: In the following question, which one set of letters when sequentially placed at the gaps in the given letter series shall complete it?
ab_c_c_a _ ab_a _cc

Detailed Solution: Question 24

The series is abcc | bcaa | cabb | abcc.
Hence, option 3 is correct.

BITSAT Mock Test - 6 - Question 25

Directions: Find the odd one out.
7, 8, 18, 57, 228, 1165, 6996

Detailed Solution: Question 25

7, 8, 18, 57, 228, 1165, 6996
228 is the odd one out.
7 × 1 + 1 = 8
8 × 2 + 2 = 18
18 × 3 + 3 = 57
57 × 4 + 4 = 232
232 × 5 + 5 = 1165
The correct term should be 232.

BITSAT Mock Test - 6 - Question 26

For all the real values of x, if f(x) + f(x + 8) = f(x + 4) + f(x + 12), g(x) = f(x) dx and g(1) = 2, then the value of g(12) is

Detailed Solution: Question 26

f(x) + f(x + 8) = f(x + 4) + f(x + 12)
Replacing x by x + 4, we get
f(x + 4) + f(x + 12) = f(x + 8) + f(x + 16)
Subtracting one from the other, we get
f(x) = f(x + 16)
g(x) = f(x) dx
g'(x) = f(x + 16) - f(x)
g'(x) = 0
g(x) = k
g(1) = 2
∴ k = 2
∴ g(12) = 2

BITSAT Mock Test - 6 - Question 27

If = 0, then sin is equal to

Detailed Solution: Question 27

BITSAT Mock Test - 6 - Question 28

The shortest distance from the point (1, 2, -1) to the surface of the sphere x2 + y2 + z2 = 24 is

Detailed Solution: Question 28

BITSAT Mock Test - 6 - Question 29

is equal to

Detailed Solution: Question 29

BITSAT Mock Test - 6 - Question 30

If and are the roots of the equation x2 − px + 36 = 0 and 2 + 2 = 9, then the value of p is

Detailed Solution: Question 30

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