JEE Exam  >  JEE Tests  >  BITSAT Mock Test - 8 - JEE MCQ

BITSAT Mock Test - 8 - JEE MCQ


Test Description

30 Questions MCQ Test - BITSAT Mock Test - 8

BITSAT Mock Test - 8 for JEE 2024 is part of JEE preparation. The BITSAT Mock Test - 8 questions and answers have been prepared according to the JEE exam syllabus.The BITSAT Mock Test - 8 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for BITSAT Mock Test - 8 below.
Solutions of BITSAT Mock Test - 8 questions in English are available as part of our course for JEE & BITSAT Mock Test - 8 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 - 8 | 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 - 8 - Question 1

If Earth is supposed to be a sphere of radius 'R' and if '' is the value of acceleration due to gravity at a latitude of 30°, while 'g' is at the equator, the value of '(g - )' is

Detailed Solution for BITSAT Mock Test - 8 - Question 1

The variation of acceleration due to gravity with latitude is given by,

BITSAT Mock Test - 8 - Question 2

The pressure and density of a diatomic gas change adiabatically from to . If , then what should be the value of ?

Detailed Solution for BITSAT Mock Test - 8 - Question 2

In an adiabatic process.
= constant
or
or ...(i)
Volume of gas =
i.e.,
= 32
Thus, from Eq. (i), we have
= 27 = 128

1 Crore+ students have signed up on EduRev. Have you? Download the App
BITSAT Mock Test - 8 - Question 3

A solid cylinder of mass M and radius R rolls down an inclined plane of height 'h'. The angular velocity of the cylinder when it reaches the bottom of the plane will be

Detailed Solution for BITSAT Mock Test - 8 - Question 3

BITSAT Mock Test - 8 - Question 4

When the potential function at any point is given by the expression , the magnitude of the electric field at any point is given by

Detailed Solution for BITSAT Mock Test - 8 - Question 4

By definition,

BITSAT Mock Test - 8 - Question 5
1 mole of a gas with = 7/5 is mixed with 1 mole of a gas with = 5/3, then the value of for the resulting mixture is
Detailed Solution for BITSAT Mock Test - 8 - Question 5
BITSAT Mock Test - 8 - Question 6

The potential energy of a 1 kg particle, free to move along the x-axis, is given by The total mechanical energy of the particle is 2 J. Then, the maximum speed (in m/s) is

Detailed Solution for BITSAT Mock Test - 8 - Question 6

Velocity of a particle will be maximum when its kinetic energy is maximum and corresponding to this, potential energy will be minimum.
J

For minimum value,


Hence, minimum potential energy of the particle is
J


Hence, maximum velocity of the particle
m/s

BITSAT Mock Test - 8 - Question 7

A hollow sphere of mass M and radius R is initially at rest on a horizontal rough surface. It moves under the action of a constant horizontal force F as shown in the figure.

The frictional force between the sphere and the surface is

Detailed Solution for BITSAT Mock Test - 8 - Question 7


Let a and α be the linear and angular accelerations of the sphere respectively.
For translational motion,
F + f = Ma ...(1)
The magnitude of the net torque acting on the sphere = FR - fR.
Hence, for rotational motion the equation is
FR - fR = Iα = (∵ a = αR)
For a hollow sphere, I = MR2. Hence

FR - fR = MR2 × MRa
F - f = Ma ...(2)
From Eqs. (1) and (2) we get f = .
Hence the correct choice is (4).

BITSAT Mock Test - 8 - Question 8

A 2 F, capacitor C1 is charged to a voltage 100 V and a 4 capacitor C2 is charged to a voltage 50 V. The capacitors are then connected in parallel. What is the loss of energy due to parallel connection?

Detailed Solution for BITSAT Mock Test - 8 - Question 8

Loss in energy when two capacitors are connected in parallel is


BITSAT Mock Test - 8 - Question 9
When a thin transparent plate of thickness t and refractive index m is placed in the path of one of the two interfering waves of light, then the path difference changes by
Detailed Solution for BITSAT Mock Test - 8 - Question 9
BITSAT Mock Test - 8 - Question 10

A mark is made on the surface of a glass sphere of diameter 10 cm and refractive index 1.5. It is viewed through the glass from a portion directly opposite. The distance of the image of the mark from the centre of the sphere will be

Detailed Solution for BITSAT Mock Test - 8 - Question 10

Let P be the position of the mark. Q is the position of its image.
Since the incident ray PA lies in a medium of refractive index μ2, and is refracted into a medium of refractive index μ1, our formula becomes


where u = -2 R = -10 cm,
R = -5 cm, μ2 = 1.5 and μ1 = 1
Putting these values in the above formula, we have

which gives v = -20 cm
∴ Distance of image Q from O = 20 - 5 = 15 cm

BITSAT Mock Test - 8 - Question 11
If x is the amount of adsorbate and m is the amount of adsorbent, then which of the following relations is not related to the adsorption process?
Detailed Solution for BITSAT Mock Test - 8 - Question 11
According to Freundlich, the adsorption isotherms and adsorption isobars are plotted when temperature and pressure are constant respectively.
The extent of adsorption is directly proportional to the pressure but inversely proportional to the temperature.
Therefore, option (3) is incorrect and the correct relation is .
BITSAT Mock Test - 8 - Question 12

The compound which gives an oily nitrosamine on reaction with nitrous acid at low temperature, is

Detailed Solution for BITSAT Mock Test - 8 - Question 12

Secondary amines, on reaction with nitrous acid at low temperature, give an oily nitrosamine.

BITSAT Mock Test - 8 - Question 13

Energy of an electron in the first Bohr orbit for a hydrogen atom is -13.6 eV. Which of the following is the possible excited state for an electron in a Bohr orbit of a hydrogen atom?

Detailed Solution for BITSAT Mock Test - 8 - Question 13

En =
Or, = n2
n2 = = 4 = 22
n2 = 22
Since, n has an integral value, this can be a possible excited state in a hydrogen atom.

BITSAT Mock Test - 8 - Question 14

Which of the following statements is true in context of the solutions of alkali metals?

Detailed Solution for BITSAT Mock Test - 8 - Question 14

Solutions of alkali metals in liquid ammonia are blue coloured due to ammoniated electrons.

These ammoniated electrons absorb red colour from the visible light and so the transmitted light is blue.
Hence, option (4) is correct.
Option (1) is incorrect because the reducing power of the alkali metals increases from sodium to cesium. However, Li has the highest reducing power because of its highest heat of hydration.
Option (2) is incorrect because Li+, being highly hydrated, has the least ion mobility and the least value of ionic velocity.
Option (3) is incorrect because the metal ions are extensively hydrated in their aqueous solutions and the trend of the hydrated ionic radius is as follows.
Li+(aq) > Na+(aq) > K+(aq) > Rb+(aq) > Cs+(aq)
[Note: In solid ionic crystals, the trend of the radii of metal ions is the same as that for the atomic radii, which is Li+ < Na+ < K+ < Rb+ < Cs+]

BITSAT Mock Test - 8 - Question 15
A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressures of the pure hydrocarbons at 20oC are 440 mm of Hg for pentane, and 120 mm of Hg for hexane. The mole fraction of pentane in the vapour phase would be
Detailed Solution for BITSAT Mock Test - 8 - Question 15
Total vapour pressure (VP) of mixture = VP of pentane in mixture + VP of hexane in mixture
PT = (Mole fraction of pentane × VP of pentane) + (Mole fraction of hexane × VP of hexane)
PT =
VP of pentane in mixture = VP of mixture × Mole fraction of pentane in vapour
88 = 184 x Mole fraction of pentane in vapour
Or, mole fraction of pentane in the vapour phase =
BITSAT Mock Test - 8 - Question 16

Directions: The passage given below is followed by four alternative summaries. Choose the option that best captures the essence of the passage.
With two hands, how can you receive the abundance which God can bestow on you with His infinite hands? And why do you cling to petty things? If He starts drawing everything from you with His infinite hands, how much can you grab? He will suck everything if He wants. So there is no use clinging to small things, and also no use craving too much as if we are poverty-stricken, uncared for. Nothing of the kind. We have become poverty-stricken due to our own making.

Detailed Solution for BITSAT Mock Test - 8 - Question 16

1. Not true. The author has talked about both greed and self-pity; however, has not mentioned whether these have any limits or not.
2. True. This captures the essence of the paragraph. The ideas of greed and need are within us and these are misleading.
3. Not true. This is not meant by the paragraph. God, though, wants us to be free from greed and self-pity.
4. Not true. Nowhere in the paragraph has attachment or detachment to material things been talked of, or the outcomes thereof (rich or poor).

BITSAT Mock Test - 8 - Question 17

Directions: Choose the set of letters which when sequentially placed in the gaps in the following letter series will complete it.
_ ab _ b _ aba _ _ abab

Detailed Solution for BITSAT Mock Test - 8 - Question 17

The series is aabab | aabab | aabab.
Hence, option 4 is correct.

BITSAT Mock Test - 8 - Question 18

War is the negation of truth means

Detailed Solution for BITSAT Mock Test - 8 - Question 18

According to the passage, the meaning of the sentence, war is the negation of truth is that wars always spread and give rise to falsehood. Therefore, (4) is correct.

BITSAT Mock Test - 8 - Question 19

Hatred and aversions are unwholesome as they are

Detailed Solution for BITSAT Mock Test - 8 - Question 19

According to the author, hatred and aversions are unwholesome as they create problems in the visibility of the truth. Therefore, (4) is correct.

BITSAT Mock Test - 8 - Question 20

Directions: In the following question, choose the word with similar meaning of the given word out of the given alternatives.
Moot

Detailed Solution for BITSAT Mock Test - 8 - Question 20

Moot' means open for debate or discussion and 'controversial' means disputed or debatable. Therefore, option (4) is correct.

BITSAT Mock Test - 8 - Question 21
Directions: Find the odd one out.
Detailed Solution for BITSAT Mock Test - 8 - Question 21
Except option (3), in all the other options, both the numbers are successive multiples of the same number.
Option (1): 21 : 24, i.e. 3 × 7 : 3 × 8
Option (2): 28 : 32, i.e. 4 × 7 : 4 × 8
Option (3): 54 : 62, but 9 × 6 : 9 × 7 = 54 : 63
Option (4); 70 : 80, i.e. 10 × 7 : 10 × 8
Option (3) does not follow the pattern.
BITSAT Mock Test - 8 - Question 22

Five men and four women are to be seated on 9 adjacent seats in a cinema hall in such a way that no two women sit together. Find the number of ways in which they can be arranged.

Detailed Solution for BITSAT Mock Test - 8 - Question 22

* M1 * M2 * M3 * M4 * M5 *
Five men can be seated in 5! ways.
The women can occupy any of the places marked *.
Thus, the women can be seated in 6P4 = 360 ways.
Total number of ways = 5! × 360 = 43200

BITSAT Mock Test - 8 - Question 23

The value of the integral log tan xdx is equal to

Detailed Solution for BITSAT Mock Test - 8 - Question 23

BITSAT Mock Test - 8 - Question 24

The product of (32)(32)1/6(32)1/36 ..... up to is

Detailed Solution for BITSAT Mock Test - 8 - Question 24

(32)(32)1/6(32)1/36 … up to can be written as

BITSAT Mock Test - 8 - Question 25

The number of solutions of y = ex and y = sin x, where x > 0, is

Detailed Solution for BITSAT Mock Test - 8 - Question 25

y = ex and y = sin x
We can solve this question by graphing.



From the graph and the tables, it is clear that for x > 0, there is no solution for y = ex and y = sin x.

BITSAT Mock Test - 8 - Question 26

The eccentricity of the hyperbola in the standard form passing through (3, 0) and is

Detailed Solution for BITSAT Mock Test - 8 - Question 26

passes through (3, 0) and (3, 2)

and

BITSAT Mock Test - 8 - Question 27
The function f(x) = log (x + ) is
Detailed Solution for BITSAT Mock Test - 8 - Question 27
BITSAT Mock Test - 8 - Question 28
The vertices of a triangle are A (0, 0), B (0, 2) and C (2, 0). The distance between circumcentre and orthocentre is
Detailed Solution for BITSAT Mock Test - 8 - Question 28
BITSAT Mock Test - 8 - Question 29
If tann x dx, n N, then ln + 2 + ln equals
Detailed Solution for BITSAT Mock Test - 8 - Question 29
BITSAT Mock Test - 8 - Question 30
The equations of two lines are 3y - 4x = 5 and 4y + 6 = 3x. What is the relation between these lines?
Detailed Solution for BITSAT Mock Test - 8 - Question 30
The slope of the first line is and the slope of the second line is .
Clearly, these slopes are neither negative reciprocal of each other nor equal.
Hence, they are neither parallel nor perpendicular.
View more questions
Information about BITSAT Mock Test - 8 Page
In this test you can find the Exam questions for BITSAT Mock Test - 8 solved & explained in the simplest way possible. Besides giving Questions and answers for BITSAT Mock Test - 8, EduRev gives you an ample number of Online tests for practice

Top Courses for JEE

Download as PDF

Top Courses for JEE