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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

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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

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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?

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
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)
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?
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.
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
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 25oC in kJ mol-1 is
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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
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
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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 27oC, in the presence of a catalyst, the activation energy of a reaction is lowered by 10 J mol-1.
The ratio  is
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
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 SN2 reaction.

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
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
Detailed Solution for KIITEE Mock Test - 5 - Question 15

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:

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:
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?
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 Cr2O72- to CrO5, which has two (-O-O-) bonds. The oxidation state of Cr in CrO5 is
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
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
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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
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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?
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
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 ofax sin  when (a > 1) is
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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 intercepts α and β on the coordinate axes, where α2 + β2 = 61, then the value of |a| is
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.
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KIITEE Mock Test - 5 - Question 28
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Evaluate
 .
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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
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KIITEE Mock Test - 5 - Question 30
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 is equal to
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