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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 2024 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 2024 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
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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 1

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 2
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The point charges and 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 2

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

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BITSAT Mock Test - 6 - Question 3
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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 for BITSAT Mock Test - 6 - Question 3

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 4
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Two long, parallel, horizontal rails, distance d apart and each having a resistance per unit length are joined at one end by a resistance R. A perfectly conducting rod MN of mass m is free to slide along the rails without friction (see the figure). There is a uniform magnetic field of induction B normal to the plane of the paper and directed into the paper. A variable force F is applied to the rod MN such that as the rod moves, constant current flows through R.

The magnitude of the emf induced in the loop is
Detailed Solution for BITSAT Mock Test - 6 - Question 4

Let the distance from R to MN be x. Then the area of the loop between MN and R is xd and the magnetic flux linked with the loop is Bxd.
As the rod moves, the emf induced in the loop is given by
|e| = (Bxd) = Bd = Bvd
where v = velocity of MN.

BITSAT Mock Test - 6 - Question 5
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When one of the slits in Young's experiment is covered with a transparent sheet of thickness 3.6 10-3 cm, the central fringe shifts to a position originally occupied by the 30th bright fringe. If = 6000 , then the refractive index of the sheet is
Detailed Solution for BITSAT Mock Test - 6 - Question 5
BITSAT Mock Test - 6 - Question 6
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The product of the given reaction is
Detailed Solution for BITSAT Mock Test - 6 - Question 6

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

BITSAT Mock Test - 6 - Question 7
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Which of the following is aromatic?
Detailed Solution for BITSAT Mock Test - 6 - Question 7

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

BITSAT Mock Test - 6 - Question 8
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The equation for Freundlich adsorption isotherm is
Detailed Solution for BITSAT Mock Test - 6 - Question 8

Freundlich adsorption isotherm
= kp1/n
Here, = Amount of gas adsorbed by adsorbent per gram
k and n = Constant
p = Pressure
Other forms are x = mkp1/n
= kp-n
All equations represent Freundlich adsorption isotherm.

BITSAT Mock Test - 6 - Question 9
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The compound used for gravimetric estimation of copper (II) is
Detailed Solution for BITSAT Mock Test - 6 - Question 9
The compound used for gravimetric estimation of Cu (II) is Cu2 (SCN)2. Gravimetric analysis is concerned with the transformation of element or radical to be determined into a pure stable compound, which can be readily converted into a form suitable for weighing.
BITSAT Mock Test - 6 - Question 10
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For the estimation of nitrogen, 1.4 g of an organic compound was digested by Kjeldahl method and the evolved ammonia was absorbed in 60 mL of sulphuric acid. The unreacted acid required 20 ml of sodium hydroxide for complete neutralisation. The percentage of nitrogen in the compound is
Detailed Solution for BITSAT Mock Test - 6 - Question 10
BITSAT Mock Test - 6 - Question 11
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Which of the following statements is false for the reaction H2 + Br2 2HBr?
The rate law is = k[H2][Br2]½ .
Detailed Solution for BITSAT Mock Test - 6 - Question 11

The given reaction H2 + Br2 2HBr is a bi-molecular reaction, hence the molecularity is 2.
The order can be determined as
Order =
If concentrate of Br2 is increased 4 times,
R2 = k[H2][4Br2]1/2
R2 = 2k[H2][Br2]1/2
R2 = 2R1
i.e., rate increases 2 times.
Hence, all the statements are correct.

BITSAT Mock Test - 6 - Question 12
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When a 5 g piece of charcoal is added to a 0.75 M CH3COOH, the concentration drops to 0.73 M. The extent of adsorption is
Detailed Solution for BITSAT Mock Test - 6 - Question 12
Number of moles of CH3COOH adsorbed = 0.75 - 0.73 = 0.02
Mass of CH3COOH adsorbed = 0.02 × 60 = 1.2 g
x/m = 1.2/5 = 0.24
BITSAT Mock Test - 6 - Question 13
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The compound [Co(H2O)6]2+, having three unpaired electrons, shows the experimental magnetic moment of 4.40 BM, while the theoretical calculated value is 3.87 BM. This is because of
Detailed Solution for BITSAT Mock Test - 6 - Question 13

From Quantum Mechanics viewpoint, the magnetic moment is dependent on both spin and orbital angular momentum contributions. The spin-only formula used is given by

And this can be modified to include the orbital angular momentum as follows.

Therefore, the value of the magnetic moment can differ because of the contribution of the total orbital motion of the electron to the magnetic moment. Thus, this is the correct answer.

BITSAT Mock Test - 6 - Question 14
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Directions: In the given question, the first and the last parts of the paragraph are numbered 1 and 6. The rest of the paragraph is split into four parts and named P, Q, R and S. These four parts are not given in their proper order. Find out the correct combination.
1. Our pleasures should be healthy so that they can impart a sense of well-being.
P. This applies very much to the passion for sports.
Q. Some people become slaves to an enthusiasm and regard it as their real and only pleasure in life.
R. It is quite possible that indulging in this passion is doing them great harm.
S. Modern sports have become so exaggerated that they can damage and sometimes destroy one's health.
6. An enthusiasm for violent sports may well dig an early grave for the participant.
Detailed Solution for BITSAT Mock Test - 6 - Question 14

The correct choice is (1). Q follows 1 as 'regard it as ... pleasure in life' and 'our pleasures should be healthy' relate to each other. P follows Q as 'an enthusiasm' in Q relates with 'passion for sports' in sentence P. R follows next as the 'this passion' in R refers to 'passion for sports' in P. S follows next as 'destroy one's health' refers to 'great harm' in R, and it is also a reason for why people could die early as mentioned in 6.

BITSAT Mock Test - 6 - Question 15
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Directions: In the following question, choose the word which best fills the blank from the four options given.
Leisure should be utilised not in _______ away the time saved, but in engaging in some pleasurable activity.
Detailed Solution for BITSAT Mock Test - 6 - Question 15

Leisure means free time and most of the people like spending it doing nothing. The idea conveyed by the sentence is that instead of idling or wasting the leisure time, one should engage in some sort of pleasurable activity.

BITSAT Mock Test - 6 - Question 16
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When she learned that she could not attend the university in Warsaw, Marie felt
Detailed Solution for BITSAT Mock Test - 6 - Question 16

Disgruntled' means 'disappointed; displeased; unhappy'.
The following sentence, 'She became disgruntled, however, when she learned that the university in Warsaw was closed to women' justifies that option (2) is the correct answer.

BITSAT Mock Test - 6 - Question 17
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Name the person who is a politician and magician only.
Detailed Solution for BITSAT Mock Test - 6 - Question 17


It is clear from the above table that Suresh is a politician and magician only.

BITSAT Mock Test - 6 - Question 18
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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. Even in the depths of Japan's late-1990s recession, the country was still home to industrial world leaders and to some of the most respected brands in the global market.
B. They may well have done so as a consequence of government bullying to ensure that resources were poured into research and development but the results speak for themselves.
C. The Japanese economic lift-off did not come when companies were busy trying to make cheaper cars and cheaper electronic goods but when they started making products that were better than anything else in the market.
D. Japanese companies and the larger Korean companies are much more like Western organisations in terms of the resources they devote to business development.
Detailed Solution for BITSAT Mock Test - 6 - Question 18

The context is about Japanese companies; sentence D begins the sequence. 'They' in B refers to 'companies' in D; the word 'resources' also connects these two sentences. The idea of 'development' in sentence B is further emphasised in sentence A. Both A and C are in past tense; C follows A and becomes the additional feature of Japanese companies. Sequence DBAC is correct.

BITSAT Mock Test - 6 - Question 19
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Directions: This problem consists of a Question Figure followed by four Answer Figures (a), (b), (c) and (d). A part of the Question Figure is missing. Choose the Answer Figure which will complete the pattern in Question Figure.
Detailed Solution for BITSAT Mock Test - 6 - Question 19

BITSAT Mock Test - 6 - Question 20
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Directions: Find the odd one out.
7, 8, 18, 57, 228, 1165, 6996
Detailed Solution for BITSAT Mock Test - 6 - Question 20

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 21
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Solve the equation 2 sin2 x + 3 cos x = 0.
Detailed Solution for BITSAT Mock Test - 6 - Question 21
2 sin2 x + 3 cos x = 0
⇒ 2(1 - cos2 x) + 3 cos x = 0
⇒ 2 - 2 cos2 x + 3 cos x = 0
⇒ 2 cos2 x - 3 cos x - 2 = 0
⇒ 2 cos2 x - 4 cos x + cos x - 2 = 0
⇒ 2 cos x(cos x - 2) + 1(cos x - 2) = 0
⇒ (cos x - 2)(2 cos x + 1) = 0
Either cos x - 2 = 0 or 2 cos x + 1 = 0
But cos x - 2 = 0
i.e. cos x = 2, which is not possible.
Now, from 2 cos x + 1 = 0, we get:
cos x = -1/2
⇒ cos x = cos ()

Therefore, general solution of the equation is:
BITSAT Mock Test - 6 - Question 22
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The radius of a cylinder is increasing at the rate of 3 m/s and its altitude is decreasing at the rate of 4 m/s. What is the rate of change of volume when radius is 4 m and altitude is 6 m?
Detailed Solution for BITSAT Mock Test - 6 - Question 22

Let h and r be the height and radius of cylinder.
Given that, = 3 m/s, = -4 m/s
Also, V = πr2h
On differentiating with respect to t, we get

At r = 4 m and h = 6 m
∴ = 80π cu m/s

BITSAT Mock Test - 6 - Question 23
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If y = f(x), passing through (1, 2), satisfies the differential equation y(1 + xy) dx - xdy = 0, then
Detailed Solution for BITSAT Mock Test - 6 - Question 23
Since y = f(x) and the given differential equation is
y(1 + xy) dx - xdy = 0

Or,
.
Integrating, we get
, which passes through .

Then, the curve is

Or,
.
BITSAT Mock Test - 6 - Question 24
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The matrix is a
Detailed Solution for BITSAT Mock Test - 6 - Question 24

Let A =
Now A' =
[where A' is the transpose of A calculated by interchanging the rows and columns of A]
Since A = A′, so matrix A is symmetric.

BITSAT Mock Test - 6 - Question 25
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The value of k such that f(x) = sin x − cos x − kx + b decreases for all real values and is given by
Detailed Solution for BITSAT Mock Test - 6 - Question 25
BITSAT Mock Test - 6 - Question 26
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The probability that at least one of the events A and B occurs is 0.6. If A and B occur simultaneously with probability 0.2, then P () + P () is equal to
Detailed Solution for BITSAT Mock Test - 6 - Question 26
BITSAT Mock Test - 6 - Question 27
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All the letters in the word ''AGAIN'' are permuted in every possible way and the words so formed (with or without meaning) are written in alphabetical order. Then, the 50th word will be
Detailed Solution for BITSAT Mock Test - 6 - Question 27
Starting with letter A and arranging the other 4 letters, we can form 4! = 24 words.
These are the first 24 words.
Then, starting with G and arranging A, A, I and N in different ways, we can form words.
Now, the 37th word will start with I. There will be 12 words starting with I.
So, all of these make up the first 48 words.
The 49th word will be NAAGI, followed by NAAIG at 50th.
BITSAT Mock Test - 6 - Question 28
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If then which of the following holds true?
Detailed Solution for BITSAT Mock Test - 6 - Question 28

By LMVT: for some
as
BITSAT Mock Test - 6 - Question 29
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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 29

BITSAT Mock Test - 6 - Question 30
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Consider the lines given by L1: x + 3y - 5 = 0, L2: 3x - ky - 1 = 0 and L3: 5x + 2y - 12 = 0. If one of L1, L2 and L3 is parallel to at least one of the other two, then
Detailed Solution for BITSAT Mock Test - 6 - Question 30

According to the question, L1 and L3 are not parallel.
L2 is parallelto L1 if k = -9
L2 is parallel to L3 if k = .
So the required values of k are given by (k + 9) (k + ) = 0
or 5k2 + 51k + 54 = 0

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