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BITSAT Mock Test - 6 - JEE MCQ


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30 Questions MCQ Test - BITSAT Mock Test - 6

BITSAT Mock Test - 6 for JEE 2025 is part of JEE preparation. The BITSAT Mock Test - 6 questions and answers have been prepared according to the JEE exam syllabus.The BITSAT Mock Test - 6 MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for BITSAT Mock Test - 6 below.
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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 for BITSAT Mock Test - 6 - 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 for BITSAT Mock Test - 6 - 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 for BITSAT Mock Test - 6 - 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 for BITSAT Mock Test - 6 - 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 for BITSAT Mock Test - 6 - 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
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 for BITSAT Mock Test - 6 - Question 6
BITSAT Mock Test - 6 - Question 7
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 for BITSAT Mock Test - 6 - Question 7
BITSAT Mock Test - 6 - Question 8


The product of the given reaction is

Detailed Solution for BITSAT Mock Test - 6 - Question 8

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

BITSAT Mock Test - 6 - Question 9

Which of the following is aromatic?

Detailed Solution for BITSAT Mock Test - 6 - Question 9

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

BITSAT Mock Test - 6 - Question 10

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 for BITSAT Mock Test - 6 - Question 10

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 11

Two stereoisomers are given below. These are known as

Detailed Solution for BITSAT Mock Test - 6 - Question 11

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 12

The major product in the following reaction is

Detailed Solution for BITSAT Mock Test - 6 - Question 12

BITSAT Mock Test - 6 - Question 13

Identify the product S in the following sequence of reactions:

Detailed Solution for BITSAT Mock Test - 6 - Question 13

BITSAT Mock Test - 6 - Question 14

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 for BITSAT Mock Test - 6 - Question 14

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 15

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

Detailed Solution for BITSAT Mock Test - 6 - Question 15

BITSAT Mock Test - 6 - Question 16
Identify the correct statement for change of Gibbs energy for a system (Gsystem) at constant temperature and pressure.
Detailed Solution for BITSAT Mock Test - 6 - Question 16
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 17

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 for BITSAT Mock Test - 6 - Question 17


Hence, option 4 is the correct answer.

BITSAT Mock Test - 6 - Question 18

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 for BITSAT Mock Test - 6 - Question 18

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 19

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

Detailed Solution for BITSAT Mock Test - 6 - Question 19

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 20
Directions: In the given question, there are four different words out of which one is wrongly spelt. Find the wrongly spelt word.
Detailed Solution for BITSAT Mock Test - 6 - Question 20
Correctly spelled word: 'collusion.' It refers to 'a secret or illegal conspiracy in order to deceive others'.
BITSAT Mock Test - 6 - Question 21

Name the person who is a politician, an educationist and a magician.

Detailed Solution for BITSAT Mock Test - 6 - Question 21


Romesh is the only person who is a politician, an educationist and a magician.

BITSAT Mock Test - 6 - Question 22

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 for BITSAT Mock Test - 6 - Question 22

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

BITSAT Mock Test - 6 - Question 23

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 for BITSAT Mock Test - 6 - Question 23

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

BITSAT Mock Test - 6 - Question 24

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

Detailed Solution for BITSAT Mock Test - 6 - Question 24

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 25

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 for BITSAT Mock Test - 6 - Question 25

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 26

If = 0, then sin is equal to

Detailed Solution for BITSAT Mock Test - 6 - Question 26

BITSAT Mock Test - 6 - Question 27
The shortest distance from the point (1, 2, -1) to the surface of the sphere x2 + y2 + z2 = 24 is
Detailed Solution for BITSAT Mock Test - 6 - Question 27
BITSAT Mock Test - 6 - Question 28

is equal to

Detailed Solution for BITSAT Mock Test - 6 - Question 28

BITSAT Mock Test - 6 - Question 29

is equal to

Detailed Solution for BITSAT Mock Test - 6 - 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 for BITSAT Mock Test - 6 - Question 30

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