When a mixture of 1-hexanol and hexanoic acid in diethyl ether is shaken with an aqueous NaHCO 3 solution, then which of the following is the correct composition of the mixture after the reaction?

Sodium hexanoate + Hexanoic acid

Sodium hexoxide + Hexanoic acid

Sodium hexanoate + Sodium Hexoxide

CH 3 - CH 2 - CH = CH 2 + HBr X (major) + Y (minor) X and Y are

BrCH 2 - CH 2 - CH = CH 2 and C 2 H 5 - CHBr - CH 3

C 2 H 5 - CH 2 - CH 2 Br and Br - CH 2 - CH 2 - CH = CH 2

C 2 H 5 - CH 2 - CH 2 Br and C 2 H 5 - CHBr - CH 3

C 2 H 5 CHBr - CH 3 and C 2 H 5 - CH 2 - CH 2 Br

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JEE Exam > JEE Tests > KIITEE Mock Test - 5 - JEE MCQ

KIITEE Mock Test - 5 - JEE MCQ

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30 Questions MCQ Test - KIITEE Mock Test - 5

KIITEE Mock Test - 5 for JEE 2024 is part of JEE preparation. The KIITEE Mock Test - 5 questions and answers have been prepared according to the JEE exam syllabus.The KIITEE Mock Test - 5 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for KIITEE Mock Test - 5 below.

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KIITEE Mock Test - 5 - Question 1

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A transverse sinusoidal wave of amplitude a, wavelength λ and frequency f is travelling on a stretched string. The maximum speed at any point on the string is v/10, where v is the speed of propagation of the wave. If a = 10^-3 m and v = 10 ms^-1, then λ and f are given by

A.
B.
λ = 10^-2 m, f = 10³ Hz
C.
λ = 10^-3 m, f = 10⁴ Hz
D.

Detailed Solution for KIITEE Mock Test - 5 - Question 1

KIITEE Mock Test - 5 - Question 2

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An observer moves towards a stationary source of sound with a velocity one-tenth the velocity of sound. The apparent increase in frequency is

A.
zero
B.
5%
C.
10%
D.
0.1%

Detailed Solution for KIITEE Mock Test - 5 - Question 2

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KIITEE Mock Test - 5 - Question 3

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A 2 μF, capacitor C₁ is charged to a voltage 100 V and a 4 μF 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?

A.
1.7 J
B.
1.7 x 10^-1 J
C.
1.7 x 10^-2 J
D.
1.7 x 10^-3 J

Detailed Solution for KIITEE Mock Test - 5 - Question 3

Loss in energy when two capacitors are connected in parallel is

KIITEE Mock Test - 5 - Question 4

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The electric potential V (in volts) varies with x (in metres) according to the relation V = 5 + 4x². The force experienced by a negative charge of 2 x 10^-6 C, located at x = 0.5 m, is

A.
2 x 10^-⁶ N
B.
4 x 10^-⁶ N
C.
6 x 10^-⁶ N
D.
8 x 10^-⁶ N

Detailed Solution for KIITEE Mock Test - 5 - Question 4

Electric field E = -
Force on charge (-q) = -qE = +8qx
At x = 0.5 m, force = 8 x 2 x 10^-⁶ x 0.5 = 8 x 10^-⁶ N
Hence, the correct option is (d).

KIITEE Mock Test - 5 - Question 5

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Directions: The question below is based on the following passage:

Two light springs of force constants k₁ = 1.8 Nm^-1 and k₂ = 3.2 Nm^-1 and a block of mass m = 200 g are arranged on a horizontal frictionless table as shown in the figure. One end of each spring is fixed on rigid supports and the other end is free. The distance CD between the free ends of the springs is 60 cm and the block moves with a velocity 'v' = 120 cms^-1 between the springs.

When the block moves towards the spring k₂, the time taken by it to move from D up to the maximum compression of spring k₂ is (in seconds)

A.
π
B.
π/2
C.
π/3
D.
π/4

Detailed Solution for KIITEE Mock Test - 5 - Question 5

Time taken by block to move from D upto the maximum compression of spring k₂ = half of the time period of oscillation of the block if it were attached to the free end of k₂, i.e.,

The correct choice is (4).

KIITEE Mock Test - 5 - Question 6

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The figure below shows the P - V diagram for a fixed mass of an ideal gas undergoing cyclic process ABCA. If the temperature at A is T, then what is the temperature at C?

A.
4T
B.
2T
C.
T
D.
T/2

Detailed Solution for KIITEE Mock Test - 5 - Question 6

Volume at A = Volume at C = V, but pressure at A = 2P and pressure at C = P. According to Charles' law, at constant volume, P is directly proportional to T. Hence, the temperature at C will become half of temperature at A, i.e. T/2, which is choice (4).

KIITEE Mock Test - 5 - Question 7

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When sounded together, a column of air and a tuning fork produce 4 beats per second. The tuning fork gives the lower note. The temperature of air is 15°C. When the temperature falls to 10°C, the two produce 3 beats per second. Find the frequency of the fork.

A.
100 Hz
B.
110 Hz
C.
120 Hz
D.
130 Hz

Detailed Solution for KIITEE Mock Test - 5 - Question 7

Let v = Frequency of the tuning fork
v₁ = Frequency of the air column
At 15°C, v₁ = v + 4
At 10°C, v₁ = v + 3
We know, velocity u = v
u₁₅ = (v + 4)
and u₁₀ = (v + 3)

Also, …(2)
From (1) and (2), we get

Using binomial expansion and neglecting terms, we get

Or, 1 +
Or, (v + 3) = 113.2
v = 110.2 Hz 110 Hz

KIITEE Mock Test - 5 - Question 8

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A particle executes SHM on a linear path with an amplitude of 2 cm. When the displacement is 1 cm, the magnitude of its velocity is equal to that of its acceleration. The frequency of the oscillations is

A.
3 s^-1
B.
√3 s^-1
C.
D.

Detailed Solution for KIITEE Mock Test - 5 - Question 8

Velocity of the particle executing SHM at position y is

Magnitude of the acceleration of the particle is

Hence, the frequency of the oscillation is

KIITEE Mock Test - 5 - Question 9

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The difference between heat of reaction at constant pressure and constant volume for the reaction 2C₆H₆ (l) + 15SO₂ (g) 12 CO₂ (g) + 6H₂O (l) + 15S (s) at 25^oC in kJ mol^-1 is

A.
-7.43
B.
+3.72
C.
-3.72
D.
+7.43

Detailed Solution for KIITEE Mock Test - 5 - Question 9

KIITEE Mock Test - 5 - Question 10

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A molecule contains two σ^_- and two π-bonds and one lone pair of electrons in the valence shell of central atom. The arrangement of lone pair as well as bond pairs is

A.
square planar
B.
trigonal planar
C.
linear
D.
tetrahedral

Detailed Solution for KIITEE Mock Test - 5 - Question 10

Since there are three sigma bond equivalents, therefore hybridisation should be sp²(lone pair = 1 sigma bond equivalent, and two sigma bonds). So, the structure should be trigonal planar, but bond angle will be a little contracted due to lone pair and bond pair repulsion.

KIITEE Mock Test - 5 - Question 11

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On reaction with HNO₂, C₂H₅NH₂ produces

A.
C₂H₅OH
B.
C₂H₅N-N=O
C.
CH₃CHO
D.
CH₃COOH

Detailed Solution for KIITEE Mock Test - 5 - Question 11

Nitrous acid is a source of electrophilic nitrosonium ion (O = N⁺), which reacts with primary aliphatic amines to form diazonium salts, which are unstable and liberate nitrogen to form a carbonium ion, which in aqueous reaction medium forms an alcohol.

KIITEE Mock Test - 5 - Question 12

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At 27^oC, in the presence of a catalyst, the activation energy of a reaction is lowered by 10 J mol^-1.
The ratio is

A.
191.47x 10^-3
B.
1.741 x 10^-3
C.
0.034 x 10^-3
D.
0.0145 x 10^-3

Detailed Solution for KIITEE Mock Test - 5 - Question 12

For uncatalysed reaction:
log k = log A -
For catalysed reaction:
log k' = log A -
Or, log k' = log A -
log k' = log k +
log k' - log k =
= 1.741 x 10^-3

KIITEE Mock Test - 5 - Question 13

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t-butyl methyl ether is obtained in

A.
CH₃OH + HO - CH₂ - CH₃
B.
C.
D.

Detailed Solution for KIITEE Mock Test - 5 - Question 13

t-butyl methyl ether is prepared by Williamson synthesis process. It involves the reaction of an alkoxide ion with a primary alkyl halide via an S_N2reaction.

KIITEE Mock Test - 5 - Question 14

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At 25°C, the dissociation constant of a base, BOH, is 1.0 x 10^-12. The concentration of hydroxyl ions in 0.01 M aqueous solution of the base would be

A.
2.0 x 10^-6 mol L^-1
B.
1.0 x 10^-7 mol L^-1
C.
1.0 x 10^-6 mol L^-1
D.
1.0 x 10^-5 mol L^-1

Detailed Solution for KIITEE Mock Test - 5 - Question 14

Base BOH is dissociated as follows.

So, dissociation constant of BOH base,

At equilibrium, [B⁺] = [OH]^-

KIITEE Mock Test - 5 - Question 15

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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

A.
MO
B.
M₂O₃
C.
M₂O
D.
M₂O₅

Detailed Solution for KIITEE Mock Test - 5 - Question 15

Let the formula of oxide = M_xO_y
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:

Thus, formula of metal oxide = M₂O

KIITEE Mock Test - 5 - Question 16

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The most suitable reagent for the following conversion, is:

A.
Na/liquid NH₃
B.
H₂, Pd/C, quinoline
C.
Zn/HCl
D.
Hg²⁺/ H⁺, H₂O

Detailed Solution for KIITEE Mock Test - 5 - Question 16

(i) Na/ Liquid NH₃ (Birch reduction) it reduces non-terminal alkynes into alkene (trans/anti addition)
(ii) H₂/Pd−C, quinoline (lindlar catalyst) reduces alkyne into alkene by syn addition. This will give the required product.
(iii) Zn/HCl is Clemmensen reduction which reduced carbonyl into an alkene.
(iv) Hg⁺²/H⊕, H₂O is an electrophilic addition reaction, which converts alkyne into ketone.

KIITEE Mock Test - 5 - Question 17

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When a mixture of 1-hexanol and hexanoic acid in diethyl ether is shaken with an aqueous NaHCO₃ solution, then which of the following is the correct composition of the mixture after the reaction?

A.
Sodium hexanoate + Hexanoic acid
B.
Sodium hexoxide + Hexanoic acid
C.
Sodium hexanoate + Hexan-1-ol
D.
Sodium hexanoate + Sodium Hexoxide

Detailed Solution for KIITEE Mock Test - 5 - Question 17

Both hexanol and hexanoic acid are insoluble in ether, but hexanoic acid reacts with NaHCO3 to form sodium hexanoate, carbon dioxide and water.

CH₃-(CH₂)₄-COOH(aq) + NaHCO₃(aq) → CH₃-(CH₂)-COONa + CO₂ + H₂O

KIITEE Mock Test - 5 - Question 18

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In the acidic medium, hydrogen peroxide changes Cr₂O₇^2- to CrO₅, which has two (-O-O-) bonds. The oxidation state of Cr in CrO₅ is

A.
+1
B.
+3
C.
+6
D.
-5

Detailed Solution for KIITEE Mock Test - 5 - Question 18

This option is correct because CrO₅ has a butterfly structure having two peroxo bonds as shown below:

In the peroxo bonds, the oxidation state of each oxygen is -1. Now, let the oxidation state of Cr in CrO₅ be x.
Therefore, x + 4(-1) + 1(-2) = 0
x = +6

KIITEE Mock Test - 5 - Question 19

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CH₃ - CH₂ - CH = CH₂ + HBr X (major) + Y (minor)
X and Y are

A.
BrCH₂ - CH₂ - CH = CH₂ and C₂H₅ - CHBr - CH₃
B.
C₂H₅ - CH₂ - CH₂Br and Br - CH₂ - CH₂ - CH = CH₂
C.
C₂H₅ - CH₂ - CH₂Br and C₂H₅ - CHBr - CH₃
D.
C₂H₅CHBr - CH₃ and C₂H₅ - CH₂ - CH₂Br

Detailed Solution for KIITEE Mock Test - 5 - Question 19

According to anti-Markovnikov's rule, X is 1-bromobutane and Y is 2-bromobutane.

In this reaction, organic peroxide dissociates butene into free radicals. There is the formation of free radicals as intermediates.

More stable free radicals give major products according to anti-Markovnikov's rule.

KIITEE Mock Test - 5 - Question 20

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The differential equation of all non-vertical lines in a plane is

Detailed Solution for KIITEE Mock Test - 5 - Question 20

The general equation of all non-vertical lines in a plane is: ax + by = 1, where b ≠0.
a + b = 0
b = 0
= 0
[ b 0]
This is the required differential equation.

KIITEE Mock Test - 5 - Question 21

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The arithmetic mean of two positive numbers is

A.
greater than the geometric mean
B.
equal to the geometric mean
C.
less than the geometric mean
D.
greater than or equal to the geometric mean

Detailed Solution for KIITEE Mock Test - 5 - Question 21

KIITEE Mock Test - 5 - Question 22

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The number of integral values of 'a' for which the equation cos 2x + a sin x = 2a – 7 possesses possible solutions is

A.
2
B.
3
C.
4
D.
5

Detailed Solution for KIITEE Mock Test - 5 - Question 22

KIITEE Mock Test - 5 - Question 23

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For what value of 'a' will the lines represented by the equation ax² - 6xy + 9y² + 3x - 9y - 4 = 0 be perpendicular?

A.
0
B.
1/2
C.
-9
D.
9

Detailed Solution for KIITEE Mock Test - 5 - Question 23

ax² - 6xy + 9y² + 3x - 9y - 4 = 0
This is a homogeneous second degree equation.
It represents a pair of perpendicular lines if
a + b = 0, where a is coefficient of x² and b is coefficient of y²
Here, b = 9
a = a
⇒ a + 9 = 0
Or, a = -9

KIITEE Mock Test - 5 - Question 24

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The equation of smallest degree with real coefficients having 2 + 3i as one of the roots is

A.
x² − 4x + 13 = 0
B.
x² + 5x + 6 = 0
C.
x² − 2x + 1 = 0
D.
x² + 2x + 1 = 0

Detailed Solution for KIITEE Mock Test - 5 - Question 24

Since 2+ 3i is a root
∴ 2-3i is also a root
Hence, required equation is
x²-(sum of roots) x + (product of roots) = 0
Sum of roots=2+3i+2-3i=4
Product of roots
=(2+31) (2-3)=4-9i²-13 [∵ i²= -1]
So, equation is x²- 4x + 13 = 0.

KIITEE Mock Test - 5 - Question 25

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The value ofa^xsinwhen (a > 1) is

A.
b log a
B.
log b
C.
b
D.
None of these

Detailed Solution for KIITEE Mock Test - 5 - Question 25

KIITEE Mock Test - 5 - Question 26

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If the tangent to the curve 2y³ = ax² + x³ at the point (a, a) cuts off interceptsα and β on the coordinate axes, where α² + β² = 61, then the value of |a| is

A.
16
B.
28
C.
30
D.
31

Detailed Solution for KIITEE Mock Test - 5 - Question 26

KIITEE Mock Test - 5 - Question 27

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Integrate log⁡(tan⁡x)sec²⁡x dx.

A.
log⁡(tan⁡x) (tan⁡x-1)+C
B.
log⁡(tan⁡x) (sec⁡x+1)+C
C.
tan⁡x (log⁡(tan⁡x)-1)+C
D.
tan⁡x (log⁡sec⁡x +1)+C

Detailed Solution for KIITEE Mock Test - 5 - Question 27

KIITEE Mock Test - 5 - Question 28

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Evaluate
.

A.
1
B.
2
C.
0
D.
√2

Detailed Solution for KIITEE Mock Test - 5 - Question 28

KIITEE Mock Test - 5 - Question 29

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Let A and B be two sets such that n(A) = 70, n(B) = 60 and n(A ∪ B) = 110. Then, n(A ∩ B) is equal to

A.
120
B.
100
C.
20
D.
240

Detailed Solution for KIITEE Mock Test - 5 - Question 29

KIITEE Mock Test - 5 - Question 30

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is equal to

A.
4
B.
8
C.
10
D.
None of these

Detailed Solution for KIITEE Mock Test - 5 - Question 30

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KIITEE Mock Test - 5 - JEE MCQ

30 Questions MCQ Test - KIITEE Mock Test - 5

A transverse sinusoidal wave of amplitude a, wavelength λ and frequency f is travelling on a stretched string. The maximum speed at any point on the string is v/10, where v is the speed of propagation of the wave. If a = 10-3 m and v = 10 ms-1, then λ and f are given by

An observer moves towards a stationary source of sound with a velocity one-tenth the velocity of sound. The apparent increase in frequency is

A 2 μF, capacitor C1 is charged to a voltage 100 V and a 4 μF 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?

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Important Questions for KIITEE Mock Test - 5

KIITEE Mock Test - 5 MCQs with Answers

Online Tests for KIITEE Mock Test - 5

A transverse sinusoidal wave of amplitude a, wavelength λ and frequency f is travelling on a stretched string. The maximum speed at any point on the string is v/10, where v is the speed of propagation of the wave. If a = 10^-3 m and v = 10 ms^-1, then λ and f are given by

A 2 μF, capacitor C₁ is charged to a voltage 100 V and a 4 μF 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?