Explore Exams
Login Signup
Sign in Open App
JEE Exam  >  JEE Test  >  BITSAT Mock Tests Series & Past Year Papers 2026  >  BITSAT Mock Test - 9 - JEE MCQ

BITSAT Mock Test - 9 Free Online Test 2026


Full Mock Test & Solutions: BITSAT Mock Test - 9 (130 Questions)

You can boost your JEE 2026 exam preparation with this BITSAT Mock Test - 9 (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)

Sign up on EduRev for free and get access to these mock tests, get your All India Rank, and identify your weak areas to improve your marks & rank in the actual exam.

BITSAT Mock Test - 9 - Question 1
Save

A spherical capacitor consists of an inner sphere of diameter 6 cm and an outer sphere of diameter 10 cm. The space between the two concentric spheres is filled with a medium of dielectric constant 80. What is the capacitance of the capacitor?

Detailed Solution: Question 1

Capacity of the spherical capacitor having inner radius 'a' and outer radius 'b'
b = 5 cm, a = 3 cm


C = 667 × 10-12 F or C = 667 pF

BITSAT Mock Test - 9 - Question 2
Save

Two wires made of same material having lengths 'L' and '2L' and radii 'r and 2r', respectively, are clamped rigidly at one end and stretched by applying forces equal to 'F' and '2F', respectively. If extension in length of first wire is l, then the extension in second wire is

Detailed Solution: Question 2

Clear all your doubts with EduRev
BITSAT Mock Test - 9 - Question 3
Save

For the post office box arrangement to determine the value of unknown resistance, the unknown resistance should be connected between

Detailed Solution: Question 3

BC, CD and BA are known resistances. The unknown resistance is connected between A1 and D.

BITSAT Mock Test - 9 - Question 4
Save

310 J of heat is required to raise the temperature of 2 moles of an ideal gas at constant pressure from 250C to 350C. The amount of heat required to raise the temperature of the gas through the same range at constant volume is

Detailed Solution: Question 4

BITSAT Mock Test - 9 - Question 5
Save

If a system changes from state (P1, V1) to (P2, V2) as shown below, the work done by the system is

Detailed Solution: Question 5

Work done = Area of graph ACD

BITSAT Mock Test - 9 - Question 6
Save

The resistance of an ammeter is 13Ω, and its scale is graduated for a current upto 100 A. After an additional shunt has been connected to this ammeter, it becomes possible to measure current upto 750 A by this meter. The value of shunt resistance is

Detailed Solution: Question 6

BITSAT Mock Test - 9 - Question 7
Save

A solution containing 0.319 g of CrCI3.6H2O was passed through a cation exchange resin and acid coming out of the cation exchange resin required 28.5 ml of 0.125 M NaOH. Determine the correct formula of the complex.
[Mol. wt. of the complex = 266.5 u]

Detailed Solution: Question 7

When the solution of the complex is passed through cation exchanger, nCI- ions will combine with H+ (of cation exchanger) to form HCI.
nCl- + nH+ nHCI
As 1 mole of the complex will form n moles of HCI, 1 mol of complex n mol of HCI n mol of NaOH
Mole of the complex = = 0.0012 mol
Mole of NaOH used = = 0.0036 mol
0.0012 mol of complex = 0.0036 mol of NaOH = 0.0036 mol of HCI
1 mol of complex = = 3 mol of HCI
Thus, all the CI- ions are outside the coordination sphere. Hence, complex is [Cr(H2O)6]CI3.

BITSAT Mock Test - 9 - Question 8
Save

In which of the following crystals of ionic compounds would you expect maximum distance between the centres of cations and anions?

Detailed Solution: Question 8

Ionic radius is the distance between the nucleus of an ion and a point up to which the nucleus has its influence on its electron cloud.
The size of ions increases on moving from top to bottom in a group. Hence, the maximum distance between the centres of cations and anions is in CsI because Cs is the largest cation and I is the largest anion.

BITSAT Mock Test - 9 - Question 9
Save

The conductivity of 0.001028 mol L-1 acetic acid is 4.95 × 10-5 Scm-1. Find out its dissociation constant, if for acetic acid is 390.5 Scm-1 mol-1.

Detailed Solution: Question 9



BITSAT Mock Test - 9 - Question 10
Save


The reagent X is

Detailed Solution: Question 10

The reaction is called Stephen reaction and is used for the preparation of aldehydes from alkylcyanides.

BITSAT Mock Test - 9 - Question 11
Save

Cd2+(aq) + Zn(s) → Zn2+(aq) + Cd(s)
If and
The equilibrium constant for the above reaction is

Detailed Solution: Question 11

BITSAT Mock Test - 9 - Question 12
Save

Directions: Find the odd one out.

Detailed Solution: Question 12

The rule here is as follows.
Third Letter = Position of first letter + Position of second letter
G(7) = B(2) + E(5)
U(21) = G(7) + N(14)
O(15) = D(4) + O(11)
The above rule is not followed in option 4.

BITSAT Mock Test - 9 - Question 13
Save

Directions: In the given question, a set of figures carrying certain characters is given. Assuming that the characters in each set follow a similar pattern, find the missing character in the given question.

Detailed Solution: Question 13

Column 1: 40 = 20 X 2
Column 2: 18 = 3 X 6
Column 3: 72 = x X 4
x = 72/4 = 18

BITSAT Mock Test - 9 - Question 14
Save

Directions: Find the missing term.

Detailed Solution: Question 14

Going by columns,
32 + 42 = 52
252 - 72 = 242
62 + 82 = 102

BITSAT Mock Test - 9 - Question 15
Save

Directions: In the following question, a series is given with one term missing. Choose the alternative that will complete the series.
18, 8, 10, 22, 73, ?

Detailed Solution: Question 15

The pattern of given series is:
⇒ 18 × 0.5 - 1 = 8
⇒ 8 × 1.5 - 2 = 10
⇒ 10 × 2.5 - 3 = 22
⇒ 22 × 3.5 - 4 = 73
⇒ 73 × 4.5 - 5 = 323.5
⇒ ? = 323.5

BITSAT Mock Test - 9 - Question 16
Save

Directions: Find the missing term in the following series.
?, 97, 167, 257, 367, 497, 647

Detailed Solution: Question 16

The pattern is the addition of 50, 70, 90, 110, 130, 150.... to the series 47, 97, 167, 257, 367, 497, 647.
Therefore, the missing number is 47.

BITSAT Mock Test - 9 - Question 17
Save

Directions: Read the following information carefully to answer the question that follows.
(A) Six flats on a floor in two rows facing north and south are allotted to P, Q, R, S, T and U.
(B) Q gets a north-facing flat and is not next to S.
(C) S and U get flats at diagonally opposite ends.
(D) R, next to U, gets a south-facing flat and T gets a north-facing flat.
Which of the following combinations gets south-facing flats?

Detailed Solution: Question 17

From (B) and (C), we get the arrangement as:

Combining the above arrangement with the information in (D), the arrangement becomes:

U, R and P get south-facing flats.

BITSAT Mock Test - 9 - Question 18
Save

If y = (x + 1)(x + 2)(x + 3), then dy/dx is equal to

Detailed Solution: Question 18

Since,



BITSAT Mock Test - 9 - Question 19
Save

How many four digit numbers that are divisible by 4 can be formed using 1, 2, 3, 4, 5, 6 and 7 in all possible ways without repetition?

Detailed Solution: Question 19

Last two digits of the four digit number can be 12, 16, 24, 32, 36, 52, 56, 64, 72 and 76 (10 ways).
Other two digits can be filled in 5 x 4 = 20 ways
Total number of ways = 20 x 10 = 200 ways

BITSAT Mock Test - 9 - Question 20
Save

The value of the integral is equal to

Detailed Solution: Question 20

BITSAT Mock Test - 9 - Question 21
Save

If a, b, c, d, e and f are in GP, then the value of depends on

Detailed Solution: Question 21

BITSAT Mock Test - 9 - Question 22
Save

If z = , then z69 is equal to

Detailed Solution: Question 22

BITSAT Mock Test - 9 - Question 23
Save

The amplitude of is

Detailed Solution: Question 23

Amplitude of is

=
=
=


=

BITSAT Mock Test - 9 - Question 24
Save

dx is equal to

Detailed Solution: Question 24

Let 1 + cos2 x = t
and dt = -2 cos x.sin x dx = -sin 2x dx
The given equation becomes

Therefore, the problem is reduced to −log(1 + cos2 (x)) + C

BITSAT Mock Test - 9 - Question 25
Save

The area bounded by y = sin x, y = cos x and the ordinates x = 0 and x = /4 is

Detailed Solution: Question 25

BITSAT Mock Test - 9 - Question 26
Save

A circle touches the x-axis and also touches the circle with centre (0, 3) and radius 2. The locus of the centre of the circle is

Detailed Solution: Question 26

BITSAT Mock Test - 9 - Question 27
Save

The maximum value of 5 + 4x − 4x2 is

Detailed Solution: Question 27

BITSAT Mock Test - 9 - Question 28
Save

If p is nearly equal to q and n > 1 such that = , then the value of k is

Detailed Solution: Question 28

Let p = q + h, where h is so small that its square and higher powers may be neglected.
Then, =
= = =
= [Expanding the second term and neglecting h2, h3, etc.]
= 1 + h - h [Neglecting h2]
= 1 + = =
∴ k =

BITSAT Mock Test - 9 - Question 29
Save

In an ellipse, if the lines joining a focus to the extremities of the minor axis make an equilateral triangle with the minor axis, then the eccentricity of the ellipse is

Detailed Solution: Question 29

According to the question we have,
Distance between focus and one end of minor axis is equal to its minor axis
[because triangle formed by the lines are equilateral.]
Then a2e2 + b2 = (2b)2
a2e2 = 3b2 = 3a2(1 - e2) [Since b2 = a2(1 - e2)]
e2 =
Hence, e =

BITSAT Mock Test - 9 - Question 30
Save

is equal to

Detailed Solution: Question 30

We know = log a
=
=
=
=

View more questions
2 videos|18 docs|70 tests
Join Course
Information about BITSAT Mock Test - 9 Page
In this test you can find the Exam questions for BITSAT Mock Test - 9 solved & explained in the simplest way possible. Besides giving Questions and answers for BITSAT Mock Test - 9, EduRev gives you an ample number of Online tests for practice
Download as PDF

About this JEE Mock Test

This mock test contains 130 questions designed to replicate the pattern, difficulty level and time pressure of the actual JEE exam. It is part of BITSAT Mock Tests Series & Past Year Papers 2026 on EduRev, covering all the topics that are being asked as per the current JEE syllabus. Attempting this test under timed conditions will help you assess your exam readiness, identify subject-wise weak areas, and improve your speed and accuracy before the actual exam.

Explore more JEE Resources

If you found this BITSAT Mock Test - 9 helpful, you can further enhance your JEE preparation with complete mock tests, previous year questions & practice tests available in BITSAT Mock Tests Series & Past Year Papers 2026 course on EduRev. The course offers:
  • Video Lectures: covering all the topics of BITSAT Mock Tests Series & Past Year Papers 2026
  • Notes for Revision: offering revision notes, important questions, summaries, flashcards, cheatsheets & more for JEE preparation
  • MCQ Test: variety of questions as per the pattern of JEE exam
Signup to see your scores go up
within 7 days!
Takes less than 10 seconds to signup