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BITSAT Mock Test - 7 - JEE MCQ


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30 Questions MCQ Test - BITSAT Mock Test - 7

BITSAT Mock Test - 7 for JEE 2024 is part of JEE 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.
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BITSAT Mock Test - 7 - Question 1
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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?

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BITSAT Mock Test - 7 - Question 2
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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 .

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BITSAT Mock Test - 7 - Question 3
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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
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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?
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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
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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
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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 :
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BITSAT Mock Test - 7 - Question 7
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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:
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BITSAT Mock Test - 7 - Question 8
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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
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BITSAT Mock Test - 7 - Question 9
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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
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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
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Select the incorrect statement.
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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
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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
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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
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Which among the following molecules have the same number of lone pairs on Xe?
(i)
(ii)
(iii)

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

BITSAT Mock Test - 7 - Question 14
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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
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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
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x moles of potassium dichromate oxidise 1 mole of ferrous oxalate in the acidic medium. Here, x is
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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
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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.
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BITSAT Mock Test - 7 - Question 18
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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
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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
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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
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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
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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
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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
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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
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The degree of the differential equation
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BITSAT Mock Test - 7 - Question 26
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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
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BITSAT Mock Test - 7 - Question 27
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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
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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
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The sum of the series terms is
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BITSAT Mock Test - 7 - Question 30
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,
then is equal to
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