All Courses
Get the App Signup / Login
Sign in Open App
JEE Exam  >  JEE Tests  >  BITSAT Mock Tests Series & Past Year Papers 2025  >  BITSAT Mock Test - 7 - JEE MCQ

BITSAT Mock Test - 7 - JEE MCQ


Test Description

30 Questions MCQ Test BITSAT Mock Tests Series & Past Year Papers 2025 - BITSAT Mock Test - 7

BITSAT Mock Test - 7 for JEE 2024 is part of BITSAT Mock Tests Series & Past Year Papers 2025 preparation. The BITSAT Mock Test - 7 questions and answers have been prepared according to the JEE exam syllabus.The BITSAT Mock Test - 7 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for BITSAT Mock Test - 7 below.
Solutions of BITSAT Mock Test - 7 questions in English are available as part of our BITSAT Mock Tests Series & Past Year Papers 2025 for JEE & BITSAT Mock Test - 7 solutions in Hindi for BITSAT Mock Tests Series & Past Year Papers 2025 course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt BITSAT Mock Test - 7 | 130 questions in 180 minutes | Mock test for JEE preparation | Free important questions MCQ to study BITSAT Mock Tests Series & Past Year Papers 2025 for JEE Exam | Download free PDF with solutions
BITSAT Mock Test - 7 - Question 1
Save

The refractive index of material of a prism and liquid are 1.56 and 1.32, respectively. What will be the value of θ for the following refraction?

Detailed Solution for BITSAT Mock Test - 7 - Question 1

BITSAT Mock Test - 7 - Question 2
Save

A charge Q is placed at the mouth of a conical flask. The flux of the electric field through the flask is

Detailed Solution for BITSAT Mock Test - 7 - Question 2

Charge associated with the conical flask is , Hence, flux through the conical flask is .

1 Crore+ students have signed up on EduRev. Have you? Download the App
BITSAT Mock Test - 7 - Question 3
Save

A thin plano-convex lens, acts like a concave mirror of focal length 0.2 m, when silvered from its plane surface. The refractive index of the material of the lens is 1.5. The radius of curvature, of the convex surface, of the lens will be

Detailed Solution for BITSAT Mock Test - 7 - Question 3

After silvering the plane surface, plano convex lens behaves as a concave mirror of focal length

But F = 0.2 m
∴ flens = 2F = 2 × 0.2 = 0.4 m
Now from lens maker's formula

= (1.5 - 1) ×
⇒ R1 = 0.5 × 0.4 = 0.2 m

BITSAT Mock Test - 7 - Question 4
Save

Inside a uniform sphere of mass M (M is mass of completer sphere) and radius R, a cavity of radius R/3 is made in the sphere as shown.

Which of the following statements is true?
Detailed Solution for BITSAT Mock Test - 7 - Question 4

Gravitational field inside the cavity is , where is mass density and is separation between centre of sphere and centre of cavity.
Applying law of conservation of energy, we get
= 0
Solving, Vesc =

BITSAT Mock Test - 7 - Question 5
Save

A right-angled prism is to be made by selecting a proper material and angles A and B (B ≤ A), as shown in the figure. It is desired that a ray of light incident on face AB emerges parallel to the incident direction after two internal reflections. What should be the minimum refractive index (n) for this to be possible?
Detailed Solution for BITSAT Mock Test - 7 - Question 5

Referring to Fig. the path of the ray is PQRS suffering internal reflections at Q and R.
It is clear from the figure that angles α and β should be greater than the critical angle given by
sin ic =

Also angle A = α and angle B = β. Since A ≥ B; β ≤ α, the minimum value of n is given by ≤ sin β
∴ nmin =
So the correct choice is (2).

BITSAT Mock Test - 7 - Question 6
Save
In the shown figure, the mass m sticks to the spring just after it strikes it. Then the minimum value of h so that lower mass bounces off the ground during its rebound, is :
Detailed Solution for BITSAT Mock Test - 7 - Question 6
BITSAT Mock Test - 7 - Question 7
Save
In a resonance tube experiment, the first and the second resonance with a given tuning fork were observed at 16.7 cm and 51.7 cm, respectively. The wavelength as deduced from the data is:
Detailed Solution for BITSAT Mock Test - 7 - Question 7
BITSAT Mock Test - 7 - Question 8
Save

A body of mass 6 kg is acted upon by a force which causes a displacement in it by x = metres, where t is time in seconds. The work done by the force in 2 seconds is
Detailed Solution for BITSAT Mock Test - 7 - Question 8


BITSAT Mock Test - 7 - Question 9
Save

If a photon of wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected with a velocity of 1.5 × 107 ms-1, then what is the energy with which it is bound to the nucleus?
Detailed Solution for BITSAT Mock Test - 7 - Question 9

Energy of the photon,
E = =
= 1.32 × 10-15 J
Energy of the ejected electron,
E' = mv2
=
= 1.024 × 10-16
Total energy of the photon = Binding energy of the electron + Energy of the ejected electron
1.32 x 10-15 = Binding energy + 1.024 x 10-16
∴ Binding energy = (1.32 x 10-15) - (1.024 x 10-16)
= 1.2176 x 10-15 J
=
= 7.6 x 103 eV

BITSAT Mock Test - 7 - Question 10
Save

Which of the following oxides is expected to exhibit paramagnetic behaviour?
Detailed Solution for BITSAT Mock Test - 7 - Question 10

The structure of ClO2 is represented as follows:

Due to the presence of three electron bonds, ClO2 has odd number of electrons. Hence, ClO2 is paramagnetic in nature.

BITSAT Mock Test - 7 - Question 11
Save
Select the incorrect statement.
Detailed Solution for BITSAT Mock Test - 7 - Question 11
Physical as well as chemical both the adsorptions are favoured by high pressure. However, the point of difference is that decrease in pressure causes desorption in case of physical adsorption, but not in case of chemical adsorption.
BITSAT Mock Test - 7 - Question 12
Save

The rate of the reaction between two reactants A and B decreases by a factor of 4, if the concentration of reactant B is doubled. The order of this reaction with respect to reactant B is
Detailed Solution for BITSAT Mock Test - 7 - Question 12


The rate decreases by a factor of 4, if the concentration of reactant B is doubled.




Hence, the order of reaction with respect to reactant B is –2.

BITSAT Mock Test - 7 - Question 13
Save

Which among the following molecules have the same number of lone pairs on Xe?
(i)
(ii)
(iii)

Detailed Solution for BITSAT Mock Test - 7 - Question 13


All the compounds have one lone pair of electrons on the xenon atom.

BITSAT Mock Test - 7 - Question 14
Save

One mole of magnesium in the vapour state absorbs 1200 kJ mol-1 energy. If the first and second ionisation energies of Mg are 750 kJ mol-1 and 1450 kJ mol-1, respectively, then the final composition of the mixture is
Detailed Solution for BITSAT Mock Test - 7 - Question 14

One mole of Mg in the vapour state absorbs 1200 kJ/mol of energy.
Energy absorbed in the ionisation of 1 mole of Mg to Mg+(g) = 750 kJ
Energy left unused = 1200 - 750 = 450 kJ
Percentage of Mg+ (g) converted into Mg2+ (g)
=
Hence, the percentage of Mg+(g) = 100 - 31 = 69%

BITSAT Mock Test - 7 - Question 15
Save

An electrochemical cell is represented as follows.
Pt (H2, 1 atm) | 0.1 M HCl || 0.1 M acetic acid | (H2, 1 atm) Pt
The emf of this cell will not be zero because the
Detailed Solution for BITSAT Mock Test - 7 - Question 15

The emf of this cell will not be zero because the pH values of 0.1 M HCl and 0.1 M acetic acid are not the same.

BITSAT Mock Test - 7 - Question 16
Save
x moles of potassium dichromate oxidise 1 mole of ferrous oxalate in the acidic medium. Here, x is
Detailed Solution for BITSAT Mock Test - 7 - Question 16
The reaction of oxidation of ferrous oxalate by potassium dichromate in an acidic medium is written as
2FeC2O4 + + 14H+ 2Fe3+ + 2Cr3+ + 4CO2 + 7H2O
2 moles of FeC2O4 are oxidised by 1 mole of .
1 mole of FeC2O4 will be oxidised by 0.5 mole of .
BITSAT Mock Test - 7 - Question 17
Save

Directions: In this question, a figure (X) is given followed by four alternative figures (a), (b), (c) and (d) such that figure (X) is embedded in one of them. Choose the alternative figure in which figure (X) is embedded.
Detailed Solution for BITSAT Mock Test - 7 - Question 17

BITSAT Mock Test - 7 - Question 18
Save

Directions: In the question given below, there are four question figures marked (A), (B), (C) and (D), followed by four other answer figures marked (1), (2), (3) and (4). Select a figure from amongst the answer figures which will continue the same series as established by the four question figures.
Detailed Solution for BITSAT Mock Test - 7 - Question 18

The arrow is rotating anticlockwise in pattern 45°, 90°, 135°, 180°….Figure (D) needs to rotate by 180° to get the next figure.

BITSAT Mock Test - 7 - Question 19
Save

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 - 7 - Question 19

First figure resembles the figure in fourth step, second figure resembles the figure in fifth step, third figure resembles the figure in (c) and each time a figure reappears, it rotates through 90o anti-clockwise.

BITSAT Mock Test - 7 - Question 20
Save

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 - 7 - Question 20

In going from Problem Figure 1 to Problem Figure 2, the double element, which is at the bottom right, goes to the centre and becomes a single bigger element. The element, which was on the top left, gets doubled and is enclosed by the previous. Going by this logic, answer figure (b) is correct.

BITSAT Mock Test - 7 - Question 21
Save

Directions: The second figure of the Problem figures bears a certain relationship to the first figure. Similarly, one of the figures in Answer Figures bears the same relationship to the third figure. You have to select the figure from the set of Answer figures which would replace the sign of question mark (?).
Detailed Solution for BITSAT Mock Test - 7 - Question 21

The main figure rotates 45o clockwise and two leaves are added to the main figure in a specific order.

BITSAT Mock Test - 7 - Question 22
Save

Directions: In the following problem, four figures are given. Three of them are similar in some way, but one figure is dissimilar. Find out which figure does not belong to the group.
Detailed Solution for BITSAT Mock Test - 7 - Question 22

(1), (2) and (3) are the different orientations of the same arrow, whereas (4) is a different arrow.

BITSAT Mock Test - 7 - Question 23
Save

Directions: In the following problem, four figures are given. Three of them are similar in some way, but one figure is dissimilar. Find out which figure does not belong to the group.
Detailed Solution for BITSAT Mock Test - 7 - Question 23

Except (4), all the other options include one right angle.

BITSAT Mock Test - 7 - Question 24
Save

Directions: In the following question, there is some relationship between the two terms on the left of (: :) and the same relationship exists between the two terms on its right. Find the missing term.
6 : 222 : : 7 : ?
Detailed Solution for BITSAT Mock Test - 7 - Question 24


Option (4) is the correct answer.

BITSAT Mock Test - 7 - Question 25
Save
The degree of the differential equation
Detailed Solution for BITSAT Mock Test - 7 - Question 25
BITSAT Mock Test - 7 - Question 26
Save

Universal set U = (x : x5 - 6x4 + 11x3 - 6x2 = 0)
A = (x : x2 - 5x + 6 = 0)
B = (x : x2 - 3x + 2 = 0)
(A ∩ B)' is equal to
Detailed Solution for BITSAT Mock Test - 7 - Question 26

BITSAT Mock Test - 7 - Question 27
Save
The distance between the pair of parallel lines represented by the expression x2 + 2xy + y2 - 8ax - 8ay - 9a2 = 0 is
Detailed Solution for BITSAT Mock Test - 7 - Question 27
The given equation can be written as (x + y)2 - 8a(x + y) - 9a2 = 0
Now, put t = x + y, the equation becomes t2 - 8at - 9a2 = 0
i.e., t2 - 9at + at - 9a2 = 0
(t - 9a)(t + a) = 0
Hence, the equations are x + y - 9a = 0 and x + y + a = 0
The given pair of straight lines are x + y - 9a = 0 and x + y + a = 0
Hence, the distance between them is .
BITSAT Mock Test - 7 - Question 28
Save

is equal to
Detailed Solution for BITSAT Mock Test - 7 - Question 28

Given series can be rewritten as

Now,
= tan-1 (r + 1) - tan-1(r)

= tan-1 (n + 1) - tan-1 (1)

BITSAT Mock Test - 7 - Question 29
Save
The sum of the series terms is
Detailed Solution for BITSAT Mock Test - 7 - Question 29
BITSAT Mock Test - 7 - Question 30
Save
,
then is equal to
Detailed Solution for BITSAT Mock Test - 7 - Question 30
View more questions
2 videos|17 docs|85 tests
Join Course
Information about BITSAT Mock Test - 7 Page
In this test you can find the Exam questions for BITSAT Mock Test - 7 solved & explained in the simplest way possible. Besides giving Questions and answers for BITSAT Mock Test - 7, EduRev gives you an ample number of Online tests for practice

Top Courses for JEE

Download as PDF

Top Courses for JEE

Important Questions for BITSAT Mock Test - 7

Find all the important questions for BITSAT Mock Test - 7 at EduRev.Get fully prepared for BITSAT Mock Test - 7 with EduRev's comprehensive question bank and test resources. Our platform offers a diverse range of question papers covering various topics within the BITSAT Mock Test - 7 syllabus. Whether you need to review specific subjects or assess your overall readiness, EduRev has you covered. The questions are designed to challenge you and help you gain confidence in tackling the actual exam. Maximize your chances of success by utilizing EduRev's extensive collection of BITSAT Mock Test - 7 resources.

BITSAT Mock Test - 7 MCQs with Answers

Prepare for the BITSAT Mock Test - 7 within the JEE exam with comprehensive MCQs and answers at EduRev. Our platform offers a wide range of practice papers, question papers, and mock tests to familiarize you with the exam pattern and syllabus. Access the best books, study materials, and notes curated by toppers to enhance your preparation. Stay updated with the exam date and receive expert preparation tips and paper analysis. Visit EduRev's official website today and access a wealth of videos and coaching resources to excel in your exam.

Online Tests for BITSAT Mock Test - 7 BITSAT Mock Tests Series & Past Year Papers 2025

Practice with a wide array of question papers that follow the exam pattern and syllabus. Our platform offers a user-friendly interface, allowing you to track your progress and identify areas for improvement. Access detailed solutions and explanations for each test to enhance your understanding of concepts. With EduRev's Online Tests, you can build confidence, boost your performance, and ace BITSAT Mock Test - 7 with ease. Join thousands of successful students who have benefited from our trusted online resources.
© EduRev
Education Revolution
Follow Us
Signup to see your scores go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup