JEE Exam  >  JEE Tests  >  BITSAT Mock Test - 4 - JEE MCQ

BITSAT Mock Test - 4 - JEE MCQ


Test Description

30 Questions MCQ Test - BITSAT Mock Test - 4

BITSAT Mock Test - 4 for JEE 2025 is part of JEE preparation. The BITSAT Mock Test - 4 questions and answers have been prepared according to the JEE exam syllabus.The BITSAT Mock Test - 4 MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for BITSAT Mock Test - 4 below.
Solutions of BITSAT Mock Test - 4 questions in English are available as part of our course for JEE & BITSAT Mock Test - 4 solutions in Hindi for JEE course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt BITSAT Mock Test - 4 | 130 questions in 180 minutes | Mock test for JEE preparation | Free important questions MCQ to study for JEE Exam | Download free PDF with solutions
BITSAT Mock Test - 4 - Question 1

Bullets of mass 0.03 kg each hit a plate at the rate of 200 bullets per second with a velocity of 50 ms-1 and reflect back with a velocity of 30 ms-1. The average force (in Newton) acting on the plate is

Detailed Solution for BITSAT Mock Test - 4 - Question 1
Change of momentum of one bullet = m(v - u)
= 0.03 × {50 - (-30)
= 2.4 kg ms-1
Average force = rate of change of momentum of 200 bullets
= 200 × 2.4 = 480 N,
which is choice (4).
BITSAT Mock Test - 4 - Question 2

A circular coil of 5 turns and 10 cm mean diameter is connected to a voltage source. If the resistance of the coil is 10Ω , what should be the voltage of the source, so as to nullify the horizontal component of Earth's magnetic field of 30 A turn m–1 at the centre of the coil?

Detailed Solution for BITSAT Mock Test - 4 - Question 2

Magnetic field of 1 A turn/m = 4π × 10-7 T

or V = 6 volt
To nullify the horizontal component of magnetic field of earth, plane of the coil should be normal to magnetic meridian.

BITSAT Mock Test - 4 - Question 3

A horizontal platform is rotating with uniform angular velocity around the vertical axis passing through its centre. At some instant of time, a viscous fluid of mass 'm' is dropped at the centre and is allowed to spread out and finally fall. The angular velocity during this period

Detailed Solution for BITSAT Mock Test - 4 - Question 3

According to the law of conservation of momentum,
lω = constant
When viscous fluid of mass m is dropped and starts spreading out, then its moment of inertia increases and angular velocity decreases. But when it starts falling, then its moment of inertia again starts decreasing and angular velocity increases.

BITSAT Mock Test - 4 - Question 4

The figure shows three spherical and equipotential surfaces A, B and C around a point charge q. The potential difference VA - VB = VB - VC. If t1 and t2 are the distances between them, then

Detailed Solution for BITSAT Mock Test - 4 - Question 4

Potential difference between two equipotential surfaces A and B,


or 

or t1 ∝ rA rB    
Similarly, t₂ ∝ rB rC
Since rA < rB < rC, therefore rA rB < rB rC,
∴  t₁ < t₂.

BITSAT Mock Test - 4 - Question 5

Two point charges, each of charge + q, are fixed at (+ a, 0) and (– a, 0). Another positive point charge q placed at the origin is free to move along the x-axis. The charge q at the origin in equilibrium will have

Detailed Solution for BITSAT Mock Test - 4 - Question 5

When considering the behaviour of the point charges, it's important to understand the concepts of force and potential energy.

  • The point charges at (a, 0) and (-a, 0) exert forces on the charge at the origin.
  • In equilibrium, there is a balance of forces acting on the charge at the origin.
  • This balance results in the charge experiencing minimum force because it is not accelerating.
  • At this equilibrium point, the potential energy is also at a minimum level, as potential energy is lowest when forces are balanced.

Thus, the charge at the origin in equilibrium will have minimum force and minimum potential energy.

BITSAT Mock Test - 4 - Question 6

Match the following:

Detailed Solution for BITSAT Mock Test - 4 - Question 6

Unit of magnetic field intensity is Am-1.
Unit of magnetic flux is Wb (weber)
Unit of magnetic potential is Wbm-1.
Unit of magnetic induction is Wbm-2.

BITSAT Mock Test - 4 - Question 7

Weight of a body of mass decreases by 1% when it is raised to height 'h' above Earth's surface. If the body is taken to a depth 'h' in a mine, the change in its weight is

Detailed Solution for BITSAT Mock Test - 4 - Question 7

Weight decreases by 1% at height h 

For height: 

thus

At depth h:
 
Weight decreases by 0.5%.

BITSAT Mock Test - 4 - Question 8

ΔH and ΔS for the reaction Ag2O(s) →2Ag(s) + 1/2O2(g), are 30.56 kJ mol-1 and 66.00 JK-1 mol-1, respectively. The temperature at which the free energy change for the reaction will be zero, is

Detailed Solution for BITSAT Mock Test - 4 - Question 8

From the Gibb's-Helmholtz equation
ΔG = ΔH - TΔS, at equilibrium Δ​​​​​​​H = T Δ S
Hence, ΔG - 0
T = ΔH / Δ S
Δ H = 30.56 kJ mol-1= 30,560 J mol-1
Δ S = 66.0 JK-1 mol-1
∴ T = 30560 / 66.0 = 463 K

BITSAT Mock Test - 4 - Question 9

In diborane, two H-B-H angles are nearly

Detailed Solution for BITSAT Mock Test - 4 - Question 9

The structure of diborane is as follows.

Thus, H-B-H angles are nearly 97° and 120°.

BITSAT Mock Test - 4 - Question 10

Given below are some catalysts and corresponding processes/reactions. Which of the following is incorrectly matched?

Detailed Solution for BITSAT Mock Test - 4 - Question 10


Ni and RhCI(PPH3)2 are used as catalysts in hydrogenation.
TiCI4 + AI(C2H5)3 is a Ziegler-Natta catalyst used in polymerisation reaction.
In Haber-Bosch process, for the manufacture of NH3, finely divided iron + molybdenum (as promoter) is used as catalyst.
N2 + 3H2 ⇌ 2NH3
V2O5 is used in contact process as a catalyst for the conversion of SO2 into SO3.
2SO2 + O2 ⇌ 2SO3

BITSAT Mock Test - 4 - Question 11

On reduction with hydrogen, 3.6 g of an oxide of metal left 3.2 g of metal. If the vapour density of the metal is 32 g/L, the simplest formula of the oxide would be

Detailed Solution for BITSAT Mock Test - 4 - Question 11

Let the formula of oxide = MxOy
Molar mass of metal oxide = 2 x 32 = 64 g
Mass of metal = 3.2g
Mass of oxygen = 3.6 - 3.2 = 0.4 g
According to law of constant proportions:
(64 X x) / (16 X y) = 3.2 / 0.4;
= x/y = 2/1
Thus, formula of metal oxide = M2O

BITSAT Mock Test - 4 - Question 12
On adding a few drops of dilute HCl to a freshly precipitated ferric hydroxide, a red-coloured colloidal solution is obtained. This phenomenon is called
Detailed Solution for BITSAT Mock Test - 4 - Question 12
Peptisation may be defined as the process of converting a precipitate into colloidal solution by shaking it with dispersion medium, in the presence of a small amount of electrolyte.
BITSAT Mock Test - 4 - Question 13

Consider the following reaction:
CaCO3(s) ⇌ CaO(s) + CO2(g)
The decomposition of limestone is non-spontaneous at 298 K.
The ΔHº and ΔSºvalues for the reaction are 176.0 kJ and 160 JK-1, respectively.
At what temperature will the decomposition become spontaneous?

Detailed Solution for BITSAT Mock Test - 4 - Question 13

According to Gibb's equation:
ΔGº = ΔHº - T Δ Sº
The reaction becomes spontaneous at a temperature at which ΔGº is a negative value.
At 298K:
ΔGº = 176000 - 298 x 160 = 128320 J mol-1
Since ΔGº is positive, the reaction is not spontaneous at this temperature.
Let ΔGº = 0; at T K
0 = 176000 - T x 160
T = 176000 / 160 = 1100 K
Hence, the reaction is spontaneous above 1100 K or 827oC.

BITSAT Mock Test - 4 - Question 14

The oxidation state of sulphur in Na2S4O6 is

Detailed Solution for BITSAT Mock Test - 4 - Question 14

The oxidation state of sulphur in Na2S4O6 is 5/2.
Let the oxidation state of sulphur be x.
2 + 4x – 12 = 0
4x = +10
x = 10/4
x = 5/2

BITSAT Mock Test - 4 - Question 15

Directions: In the following question, choose the option which best expresses the meaning of the underlined part of the sentence.
The inspector was a vigilant young man.

Detailed Solution for BITSAT Mock Test - 4 - Question 15
The word "vigilant" means being watchful and alert, especially to avoid danger or problems. Therefore, the best alternative that captures this meaning is "watchful."
BITSAT Mock Test - 4 - Question 16

Directions: The following sentence has a blank, which can be filled in from the given options. Choose the option that is correct semantically and syntactically.
Only ____ boys in our class were interested in playing chess.

Detailed Solution for BITSAT Mock Test - 4 - Question 16

Few' means next to none; 'a few' means some; 'the few' is specific to the group already referred to. As there is no comparison and no specification, option 3 is the correct usage.

BITSAT Mock Test - 4 - Question 17

Directions: This problem consists of four 'Question Figures' followed by four 'Answer Figures' (A, B, C and D). Select the Answer Figure which continues the same pattern as established by the Question Figures.

Detailed Solution for BITSAT Mock Test - 4 - Question 17

There are 10 straight lines in the first Question Figure, 9 straight lines in the second, 8 straight lines in the third and 7 straight lines in the fourth. The number of lines is decreasing by one in every step, so the next figure must have 6 straight lines.
Hence, option (1) is correct.

BITSAT Mock Test - 4 - Question 18

Why do HEVs use two types of propulsion?

Detailed Solution for BITSAT Mock Test - 4 - Question 18

The following sentence, 'Hybrids use two types of propulsion in order to use gasoline more efficiently than conventional vehicles do' justifies the correct choice.
Thus, option (2) is the correct answer.

BITSAT Mock Test - 4 - Question 19

Directions: In the following question, select the word or group of words that is most opposite in meaning to the underlined word or group of words in the given sentence.
The fallen trees blocked our passage to freedom from the woods.

Detailed Solution for BITSAT Mock Test - 4 - Question 19

Block' means to obstruct (someone or something) by placing obstacles in the way. 'Facilitate' means to assist the progress of a person. The sentence mentions that the fallen trees obstructed/blocked the passage. 'Facilitate' implies that instead of being blocked, the subjects were assisted to pass through. Thus, option 1 is the answer.

BITSAT Mock Test - 4 - Question 20

Directions: In this question, Problem figures 1 and 2 are related in some way. Similarly, Problem figures 3 and 4 are related, but Problem figure 4 is missing. Find out the correct alternative from the given options for that.

Detailed Solution for BITSAT Mock Test - 4 - Question 20

Problem figure 2 is the water image of Problem figure 1. Similarly, Answer figure (b) is the water image of Problem figure (3). Hence, Answer figure (b) is correct.

BITSAT Mock Test - 4 - Question 21
Directions: Find the odd one out.
Detailed Solution for BITSAT Mock Test - 4 - Question 21
In this question, four consecutive letters are arranged in some random order in all the options, except option (2). In option (2), the letters are not consecutive letters in the English alphabet. Therefore, it is the odd one out.
BITSAT Mock Test - 4 - Question 22

Directions: In the following question, give the antonym of the underlined word in the sentence.
A zealous and hardworking person can accomplish any task he/she undertakes.

Detailed Solution for BITSAT Mock Test - 4 - Question 22

Zealous' means someone who works with enthusiasm, and 'indifferent' is someone marked by lack of interest. Hence, it would be the correct answer.

BITSAT Mock Test - 4 - Question 23

The centres of a set of circles, each of radius 3, lie on the circle x2 + y2 = 25. The locus of any point in the set is

Detailed Solution for BITSAT Mock Test - 4 - Question 23

Distance can be calculated by using formula

Let (h, k) be any point in the set, then the equation of the circle is:

Since (h, k) lies on x2 + y2 = 25
we have:

Thus, the distance between the centers of the two circles satisfies:

BITSAT Mock Test - 4 - Question 24

If (z - 1) / (z + 1) is purely imaginary, then | z | = ?

Detailed Solution for BITSAT Mock Test - 4 - Question 24

Given (z - 1) / (z + 1) is purely imaginary, the real part must be zero:
Re (z - 1) / (z + 1) = 0

Multiply by the conjugate of the denominator:

The numerator must be zero (denominator ≠ 0 since x ≠ -1):

Expand:

BITSAT Mock Test - 4 - Question 25

If the roots of the equation x2 - bx + c = 0 are two consecutive integers, then b2 - 4c is equal to

Detailed Solution for BITSAT Mock Test - 4 - Question 25

Let α, α + 1 be roots
α + α + 1 = b
α(α + 1) = c
∴ b2 - 4c = (2α + 1)2 - 4α(α + 1) = 1.

BITSAT Mock Test - 4 - Question 26

is equal to

Detailed Solution for BITSAT Mock Test - 4 - Question 26

BITSAT Mock Test - 4 - Question 27

The position vector of the centre of the sphere is

Detailed Solution for BITSAT Mock Test - 4 - Question 27

Let

The equation of sphere =
Centre of sphere = (-1/2, -1/2, 1/2)
The position vector of centre of sphere =

BITSAT Mock Test - 4 - Question 28

The two circles x2 + y2 = r2 and x2 + y2 – 10x + 16 = 0 intersect at real and distinct points. What will be the possible values of r?

Detailed Solution for BITSAT Mock Test - 4 - Question 28

C₁(0,0) R₁ = r
C₂(5,0) R₂ = 3
The two circles intersect at real and distinct points
|R₁ - R₂| < C₁C₂ < |R₁ + R₂|
r - 3 < 5 < r + 3
2 < r & r < 8
2 < r < 8

BITSAT Mock Test - 4 - Question 29

One of the equations of lines passing through the point (3, − 2) and inclined at 60° to the line√3x + y = 1 is:

Detailed Solution for BITSAT Mock Test - 4 - Question 29

Let slope of required line L be m.
Given line is L₁: √3x + y = 1
Slope of line (m₁) = - Coefficient of x / Coefficient of v = √3 / 1 = -√3
Angle between L and L₁ is 60°.


As line passes through (3, -2), hence equation of L is given by:

BITSAT Mock Test - 4 - Question 30

The probability that a man will live 10 more years is 1/4 and the probability that his wife will live 10 more years is 1/3. The probability that neither of them will be alive in 10 years is

Detailed Solution for BITSAT Mock Test - 4 - Question 30

Probability that a man will live 10 more years = 1/4
Probability that man will not live 10 more years = 1 - 1/4 = 3/4
Probability that his wife will live 10 more years = 1/3
Probability that his wife will not live 10 more years = 1 - 1/3 = 2/3
Then, probability that neither will be alive in 10 years = (3/4) x (2/3) = 1/2

View more questions
Information about BITSAT Mock Test - 4 Page
In this test you can find the Exam questions for BITSAT Mock Test - 4 solved & explained in the simplest way possible. Besides giving Questions and answers for BITSAT Mock Test - 4, EduRev gives you an ample number of Online tests for practice
Download as PDF