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BITSAT Mock Test - 10 - JEE MCQ


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30 Questions MCQ Test - BITSAT Mock Test - 10

BITSAT Mock Test - 10 for JEE 2024 is part of JEE preparation. The BITSAT Mock Test - 10 questions and answers have been prepared according to the JEE exam syllabus.The BITSAT Mock Test - 10 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for BITSAT Mock Test - 10 below.
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BITSAT Mock Test - 10 - Question 1
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Suppose the gravitational force varies inversely as the nth power of the distance. The time period of a planet in circular orbit of radius R around sun will be proportional to:

Detailed Solution for BITSAT Mock Test - 10 - Question 1

Necessary centripetal force is provided by the gravitational force,


Time period of revolution is

or

BITSAT Mock Test - 10 - Question 2
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Consider a thin square sheet of side L and thickness t, made of a material of resistivity ρ. The resistance between two opposite faces shown by the shaded areas in the figure is

Detailed Solution for BITSAT Mock Test - 10 - Question 2

Resistance is given by
Here, area of cross section A = L × t
∴
Hence, resistance is inversely proportional to the thickness 't' and is independent of L.

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BITSAT Mock Test - 10 - Question 3
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The figure below shows a circuit with two cells in opposition to each other. One cell has an emf of 6 V and internal resistance of 2Ω and the other cell has an emf of 4 V and internal resistance of 8Ω. The potential difference across the terminals X and Y is

Detailed Solution for BITSAT Mock Test - 10 - Question 3

Since the two cells are in opposition, the effective voltage = 6 - 4 = 2 V. The current in the circuit is

Terminal voltage of 6 V cell = 6 - 2 x 0.2 = 5.6 V
Terminal voltage of 4 V cell = 4 + 8 x 0.2 = 5.6 V
Therefore, the potential difference across terminals X and Y is 5.6 V. Hence, the correct option is (2).

BITSAT Mock Test - 10 - Question 4
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A body is projected vertically upwards from the surface of a planet of radius R, with a velocity equal to one-third of the escape velocity of the planet. The maximum height attained by the body is
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BITSAT Mock Test - 10 - Question 5
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The angular momentum of the Earth revolving around the Sun is proportional to Rn, where R is the distance between the Earth and the Sun. The value of n is
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BITSAT Mock Test - 10 - Question 6
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A radioactive sample S1 having an activity of 5 has twice the number of nuclei as another sample S2, which has an activity of 10 The half lives of S1 and S2 can be
Detailed Solution for BITSAT Mock Test - 10 - Question 6

Activity, R =
Then, 5 = But, N1 = 2N2 (given)

5 =
And 10 =

i.e.
Of the given options, option (1) satisfies the above condition.

BITSAT Mock Test - 10 - Question 7
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The resistance of a coil is 4.2 at 100oC and the temperature coefficient of resistance of its material is 0.004oC. Its resistance will be 4 at
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BITSAT Mock Test - 10 - Question 8
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The formula mass of Mohr's salt is 392. The iron present in it is oxidised by KMnO4 in acidic medium. The equivalent mass of Mohr's salt is
Detailed Solution for BITSAT Mock Test - 10 - Question 8

Mohr's salt is (NH4)2SO4.FeSO4.6H2O.
The equation is 5Fe2+ + MnO-4 + 8H+ 5Fe3+ + Mn2+ + 4H2O
Total change in oxidation number of iron = (+3) - (+2)
So, equivalent weight of Mohr's salt
=
=
= 392

BITSAT Mock Test - 10 - Question 9
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For a Van der Waals gas, the expression for Boyle's temperature is
Detailed Solution for BITSAT Mock Test - 10 - Question 9

Then,

Since

Thus, on rearrangement,

includes only the first two terms of the expansion in x.

At Boyle's temperature, T = TB and z = 1

BITSAT Mock Test - 10 - Question 10
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Which of the following amines will not undergo carbylamine reaction?
Detailed Solution for BITSAT Mock Test - 10 - Question 10

Carbylamine test is given by aliphatic as well as aromatic primary amines, i.e. amines having --NH2 group.

Hence, dimethyl amine, being a secondary amine, does not undergo carbylamine reaction.

BITSAT Mock Test - 10 - Question 11
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At 25°C, the pH of a solution containing 0.10 M sodium acetate and 0.03 M acetic acid is
[pKa value of CH3COOH = 4.57]
Detailed Solution for BITSAT Mock Test - 10 - Question 11

The mixture consists of weak acid and its salt with strong base, so it acts as an acidic buffer solution.
According to Henderson Hasselbalch equation,
pH = pKa + log
= 4.57 + log
= 4.57 + (log 10 - log 3) [log 3 = 0.4771]
pH = 4.57 + 0.52
= 5.09

BITSAT Mock Test - 10 - Question 12
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The resistance of 0.01 N solution of an electrolyte was found to be 210 ohm at 298 K, using a conductivity cell of cell constant 0.66 cm−1. The equivalent conductance of the solution is
Detailed Solution for BITSAT Mock Test - 10 - Question 12





Hence, the equivalent conductance of the solution is 314.29 Scm2eq−1 or 314.29 mho cm2eq−1.

BITSAT Mock Test - 10 - Question 13
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500 ml of a sample of water required 19.6 mg of K2Cr2O7 for the oxidation of dissolved organic matter in it in the presence of H2SO4. The COD of water sample is
Detailed Solution for BITSAT Mock Test - 10 - Question 13





Amount of oxygen in 1 L of water = 6.4 x 10-3 g
Therefore, COD of water = 6.4 ppm

BITSAT Mock Test - 10 - Question 14
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The absolute enthalpy of neutralisation of the reaction will be
Detailed Solution for BITSAT Mock Test - 10 - Question 14
The heat of neutralisation of strong acid and strong base is -57.33 kJ. MgO is a weak base, while HCl is a strong acid. So, the heat of neutralisation of MgO and HCl is less than -57.33 kJ because MgO requires some heat in ionisation. Thus, the net released amount of heat is decreased.
BITSAT Mock Test - 10 - Question 15
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The chemical added to leaded petrol to prevent the deposition of lead in the combustion chamber is
Detailed Solution for BITSAT Mock Test - 10 - Question 15
1,2-dibromoethane (ethylene dibromide) is added to petrol to prevent lead deposition, which occurs by decomposition of the anti-knock lead tetraethyl bromomethane.
BITSAT Mock Test - 10 - Question 16
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Excess of iron is added to a 0.1 M Cd2+ solution and the system is allowed to attain equilibrium.
Cd2+ (aq) + Fe (s) → Cd (s) + Fe2+ (aq)
The concentration of Cd2+ ions, if Eo = 0.037 V, is
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BITSAT Mock Test - 10 - Question 17
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Directions: In this question, the second figure of the Problem figures bears a certain relationship to the first figure. Similarly, one of the figures in the Answer figures bears the same relationship to the third figure. You have to select the figure from the set of answer figures which will replace the sign of question mark (?).
Detailed Solution for BITSAT Mock Test - 10 - Question 17

Figures in the top corners are rotated by 180° and then, moved to the diagonally opposite corners. The figures in the bottom corners are rotated by 180° and then, moved to their respective top corners.

BITSAT Mock Test - 10 - Question 18
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Directions: A sentence has been given in Active/Passive Voice. Out of the four alternatives suggested, select the one which best expresses the same sentence in Passive/Active Voice.
Somebody should have cleaned the windows yesterday.
Detailed Solution for BITSAT Mock Test - 10 - Question 18

The context sentence is in active voice. A sentence is in active form, when the doer is in subject place. But in passive form, the subject (person/place/thing or idea) receives the action.
The sentence has 'should have + V3' construction. We add 'been' in the passive form of such sentence (should have + been + cleaned).
The doer of the action may or may not be mentioned at the end of the sentence especially when 'someone, somebody' etc. are used.
Option 3 has all the modifications in the sentence as per the rules of active/passive voice. Thus, it is the correct answer.

BITSAT Mock Test - 10 - Question 19
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Directions: There is some relationship between the two terms to the left of the sign (: :). The same relationship exists between the two terms to its right, out of which one is missing. Find the missing one from the given alternatives.
FILM : ADGH : : MILK : ?
Detailed Solution for BITSAT Mock Test - 10 - Question 19

Each letter of the first group is replaced by its fifth predecessor in the English alphabet to get the second group.
So,
M - 5 = H
I - 5 = D
L - 5 = G
K - 5 = F
Hence, HDGF is the correct answer.

BITSAT Mock Test - 10 - Question 20
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Directions: The question is followed by four options. Choose the best completion for the given proverb.
All is well _________.
Detailed Solution for BITSAT Mock Test - 10 - Question 20

The proverb 'All is well that ends well' is used to say that a person can forget about how unpleasant or difficult something was because everything ended in a good way. Thus, option 2 is the answer.

BITSAT Mock Test - 10 - Question 21
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Directions: There are two sets of figures namely the Problem figures, containing five figures 1, 2, 3, 4 and 5, and the Answer figures (a), (b), (c), (d). You have to 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 - 10 - Question 21

The dot rotates 90°, 135°, 90°, 135° ..... clockwise direction in subsequent figures. Hence, the correct option is (4).

BITSAT Mock Test - 10 - Question 22
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Directions: In the following question, a part of the figure is missing. From the given options (a), (b), (c) and (d), find the correct figure to fit in the missing figure.
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BITSAT Mock Test - 10 - Question 23
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Directions: Out of the given four figures, three are similar in a certain way and form a group, while one figure is not like the other three. Choose the one that does not belong to this group.
Detailed Solution for BITSAT Mock Test - 10 - Question 23

In figure (3), both the circles are shaded, while in the rest of them, only one circle is shaded.
Hence, figure (3) does not belong to the group.

BITSAT Mock Test - 10 - Question 24
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In India, the concept of agriculture insurance
Detailed Solution for BITSAT Mock Test - 10 - Question 24

This suggests that not too much headway has been made in agriculture insurance since its initiation.

BITSAT Mock Test - 10 - Question 25
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Most of the major food crop-producing states
Detailed Solution for BITSAT Mock Test - 10 - Question 25

Some of the bigger crop-producing states like Punjab, Haryana and Rajasthan did not participate in the insurance scheme.

BITSAT Mock Test - 10 - Question 26
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Who among them plays Basketball?
Detailed Solution for BITSAT Mock Test - 10 - Question 26

In terms of height, we have:
T, who is taller than P and S, plays Tennis.
T > P, T > S
Only T is between Q, who plays Football, and P in order of height.
Q > T > P
The shortest among them plays Volleyball.
R plays Volleyball, so R is the shortest.
S plays neither Volleyball nor Basketball.
So, Q is not the tallest.
Thus, U is the tallest.
So, the sequence becomes:
U > Q > T > P > S > R
Now, T plays Tennis. U, being the tallest, plays Basketball. R plays Volleyball. Q plays Football. Both P and S play either Cricket or Badminton.
So, we get the final arrangement as given below:

U plays Basketball.

BITSAT Mock Test - 10 - Question 27
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Which of the following will replace the question mark (?) in the grid given below?
Detailed Solution for BITSAT Mock Test - 10 - Question 27

Column-wise relation:
Column 3: = 9
Column 2: = 8
Column 1: The missing number is = = 10.

BITSAT Mock Test - 10 - Question 28
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The area of the region satisfying and is
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BITSAT Mock Test - 10 - Question 29
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Assuming the variance of four numbers a, b, c and d as 9, the variance of 5a, 5b, 5c and 5d would be
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BITSAT Mock Test - 10 - Question 30
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One factor of the determinant
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