JEE Exam  >  JEE Tests  >  BITSAT Mock Tests Series & Past Year Papers  >  Test: BITSAT Past Year Paper- 2020 - JEE MCQ Download as PDF

Test: BITSAT Past Year Paper- 2020 - JEE MCQ


Test Description

150 Questions MCQ Test BITSAT Mock Tests Series & Past Year Papers - Test: BITSAT Past Year Paper- 2020

Test: BITSAT Past Year Paper- 2020 for JEE 2024 is part of BITSAT Mock Tests Series & Past Year Papers preparation. The Test: BITSAT Past Year Paper- 2020 questions and answers have been prepared according to the JEE exam syllabus.The Test: BITSAT Past Year Paper- 2020 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: BITSAT Past Year Paper- 2020 below.
Solutions of Test: BITSAT Past Year Paper- 2020 questions in English are available as part of our BITSAT Mock Tests Series & Past Year Papers for JEE & Test: BITSAT Past Year Paper- 2020 solutions in Hindi for BITSAT Mock Tests Series & Past Year Papers course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt Test: BITSAT Past Year Paper- 2020 | 150 questions in 180 minutes | Mock test for JEE preparation | Free important questions MCQ to study BITSAT Mock Tests Series & Past Year Papers for JEE Exam | Download free PDF with solutions
1 Crore+ students have signed up on EduRev. Have you? Download the App
Test: BITSAT Past Year Paper- 2020 - Question 1

An organ pipe, open from both end produces 5 beats per second when vibrated with a source of frequency 200 Hz. The second harmonic of the same pipes produces 10 beats per second with a source of frequency 420 Hz. The fundamental frequency of organ pipe is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 1

Let the fundamental frequency of organ pipe be f
Case I :  f  = 200 ± 5 =  205 Hz or 195 Hz

Case II : frequency of 2nd harmonic of organ pipe = 2f (as is clear from the second figure) 2f  = 420 ± 10 or f  = 210 ± 5 or f  = 205 or 215
Hence fundamental frequency of organ pipe  = 205 Hz

Test: BITSAT Past Year Paper- 2020 - Question 2

A vessel of depth 2d cm is half filled with a liquid of refractive index μ1 and the upper half with a liquid of refractive index μ2. The apparent depth of the vessel seen perpendicularly is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 2

Apparent depth = d/μ1 + d/μ2

Test: BITSAT Past Year Paper- 2020 - Question 3

The upper half of an inclined plane with inclination f is perfectly smooth while the lower half is rough.A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is given by

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 3

Acceleration of block while sliding down upper half = g sin ф;
retardation of block while sliding down lower half = – (g sin ф - mg cos ф) For the block to come to rest at the bottom, acceleration in I half = retardation in II half. g sin ф = -(g sin ф - mg cos ф)
Þ m = 2 tan ф
Alternative method : According to work-energy theorem, W = ΔK = 0 (Since initial and final speeds are zero)
∴ Work done by friction + Work done by gravity = 0
i.e., 
or 

Test: BITSAT Past Year Paper- 2020 - Question 4

A body of density ρ' is dropped from rest at a height h into a lake of density r where ρ > ρ' neglecting all dissipative forces, calculate the maximum depth to which the body sinks before returning to float on the surface :

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 4

The effective acceleration of the body 

Now, the depth to which the body sinks

Test: BITSAT Past Year Paper- 2020 - Question 5

If the forward voltage in a semiconductor diode is changed from 0.5V to 0.7 V, then the forward current changes by 1.0 mA. The forward resistance of diode junction will be

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 5

Forward resistance

Test: BITSAT Past Year Paper- 2020 - Question 6

The heat generated in a cir cuit is given by Q = I2 Rt,  where I is current, R is resistance and t is time. If the percentage errors in measuring I, R and t are 2%, 1% and 1% respectively, then the maximum error in measuring heat will be

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 6


= 2 x 2% + 1% + 1% = 6%.

Test: BITSAT Past Year Paper- 2020 - Question 7

The r.m.s. velocity of oxygen molecule at 16ºC is 474 m/sec. The r.m.s. velocity in m/s of hydrogen molecule at 127ºC is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 7


Test: BITSAT Past Year Paper- 2020 - Question 8

A projectile A is thrown at an angle of 30° to the horizontal from point P. At the same time, another projectile B is thrown with velocity v2 upwards from the point Q vertically below the highest point. For B to collide with A, should be

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 8

This happen when vertical velocity of both are same.

Test: BITSAT Past Year Paper- 2020 - Question 9

The coefficient of friction between the rubber tyres and the road way is 0.25. The maximum speed with which a car can be driven round a curve of radius  20 m without skidding is (g = 9.8 m/s2)

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 9

μ mg = mv2 / r or v = 
or v = = 7 m / s

Test: BITSAT Past Year Paper- 2020 - Question 10

A boy pushes a toy box 2.0 m along the floor by means of a force of 10 N directed downward at an angle of 60º to the horizontal. The work done by the boy is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 10

W = F s cos θ = 10 × 2 cos 60º = 10 J.

Test: BITSAT Past Year Paper- 2020 - Question 11

The engine of a truck moving  along a straight road delivers constant power. The distance travelled by the truck in time t is proportional to

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 11


since both P and m are constants

Test: BITSAT Past Year Paper- 2020 - Question 12

The escape velocity from a planet is ve. A tunnel is dug along a diameter of the planet and a small body is dropped into it at the surface. When the body reaches the centre of the planet, its speed will be

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 12


In tunnel body will perform SHM at centre Vmax = Aω (see chapter on SHM)

Test: BITSAT Past Year Paper- 2020 - Question 13

If the the earth is at one-fourth of its present distance from the sun, the duration of year will be

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 13

The duration of a year is determined by the time it takes for a planet to complete one orbit around the sun. This is governed by Kepler's Third Law of Planetary Motion, which states that the square of the orbital period (T) of a planet is directly proportional to the cube of the semi-major axis (a) of its orbit.

Mathematically, this can be written as:

T² ∝ a³

Let's consider two cases: the current Earth-sun distance and the new distance where Earth is one-fourth of its present distance from the sun.

Case 1 (current distance):
T1² ∝ a1³

Case 2 (new distance, one-fourth the current distance):
T2² ∝ a2³

Since the new distance is one-fourth of the current distance, we can write a2 = 1/4 * a1. Now, we need to find the ratio of T2 to T1.

Divide the equations for Case 2 by Case 1:

(T2² / T1²) = (a2³ / a1³)
(T2² / T1²) = ((1/4 * a1)³ / a1³)
(T2² / T1²) = (1/64)

Now, take the square root of both sides:

T2 / T1 = √(1/64)
T2 / T1 = 1/8

So, the duration of the year when the Earth is at one-fourth of its present distance from the sun would be one-eighth of the present year

Test: BITSAT Past Year Paper- 2020 - Question 14

A vessel with water is placed on a weighing pan and it reads 600 g. Now a ball of mass 40 g and density 0.80 g cm–3 is sunk into the water with a pin of negligible volume, as shown in figure keeping it sunk. The weighing pan will show a reading

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 14

Volume of ball = 
Downthrust on water = 50 g.
Therefore reading is 650 g.

Test: BITSAT Past Year Paper- 2020 - Question 15

In an adiabatic process, the pressure is increased by If γ = 3/2, then the volume decreases by
nearly 

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 15

PV3/2 = K

Test: BITSAT Past Year Paper- 2020 - Question 16

The equation of a projectile is The angle of projection is given by

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 16

Comparing the given equation with we get tan θ = √3

Test: BITSAT Past Year Paper- 2020 - Question 17

Frequency of oscillation  is proportional to

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 17


Let mass is displaced towards left by x then force on mass = – kx – 2kx = – 3kx [negative sign is taken because force is opposite to the direction of motion]


Thus it is propotional to 

Test: BITSAT Past Year Paper- 2020 - Question 18

Two wires A and B of the same material, having radii in the ratio 1 : 2 and carry currents in the ratio 4 : 1. The ratio of drift speed of electrons in A and B is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 18

Current flowing through the conductor, I = n e v A. Hence

Test: BITSAT Past Year Paper- 2020 - Question 19

An instantaneous displacemen t of a simple harmonic oscillator is x = A cos (ωt + π/4). Its speed will be maximum at time

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 19

 Velocity, = -Aω sin (ωt + π/4)
Velocity will be maximum, when
ωt + π/4 = π/2 or ωt = π/2 – π/4 = π/4 or t = π/4ω

Test: BITSAT Past Year Paper- 2020 - Question 20

The energy of electron in the n th orbit of hydrogen atom is expressed as 
The shortest and longest wavelength of Lyman series will be

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 20



and

Test: BITSAT Past Year Paper- 2020 - Question 21

In the circuit given below, the charge in μC, on the capacitor  having 5 μF is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 21

Potential difference across the branch de is 6 V. Net capacitance of de branch is 2.1 µF
So, q = CV
⇒ q = 2.1 × 6 µC ⇒ q = 12.6 µ C Potential across 3 µF capacitance is

Potential across 2 and 5 combination in parallel is 6 – 4.2 = 1.8 V
So, q' = (1.8) (5) = 9 µC

Test: BITSAT Past Year Paper- 2020 - Question 22

A crystal has a coefficient of expansion 13×10– 7 in one direction and 231 × 10–7 in every direction at right angles to it. Then the cubical coefficient of expansion is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 22

γ = ∝1 + ∝2 + ∝3
= 13 x 10-7 + 231 x 10-7 + 231 x 10-7 = 475 x l0-7

Test: BITSAT Past Year Paper- 2020 - Question 23

A solid cylinder and a hollow cylinder both of the same mass and same external diameter are released from the same height at the same time on an inclined plane. Both roll down without slipping. Which one will reach the bottom first?

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 23

Solid cylin der reaches the bottom first because for solid cylinder and for hollow cylinder,  Acceleration down the inclined plane Solid cylin der h as greater acceleration. It reaches the bottom first.

Test: BITSAT Past Year Paper- 2020 - Question 24

A whistle of frequency 1000 Hz is sounded on a car travelling towards a cliff with velocity of 18 m s–1 normal to the cliff. If c = 330 m s–1, then the apparent frequency of the echo as heard by the car driver is nearly

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 24

(a) By the concept of accoustic, the observer and source are moving towards each other, each with a velocity of 18 m s–1.

Test: BITSAT Past Year Paper- 2020 - Question 25

A thin sheet of glass (μ = 1.5) of thickness 6 micron introduced in the path of one of the interfering beams in a double slit experiment shifts the central fringe to a position previously occupied by fifth bright fringe. Then the wavelength of light used is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 25

nλ = (μ - 1)t;

Test: BITSAT Past Year Paper- 2020 - Question 26

M.I of a circular loop of radius R about the axis in figure is​

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 26

Use theorem of parallel axes.

Test: BITSAT Past Year Paper- 2020 - Question 27

Three charge q, Q and 4q are placed in a straight line of length l at points distant 0, 1/2
 and l respectively from one end. In order to make the net froce on q zero, the charge Q must be equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 27

(Fnet )q =0

Test: BITSAT Past Year Paper- 2020 - Question 28

In series combination of R, L and C with an A.C. source at resonance, if R = 20 ohm, then impedence Z of the combination is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 28

We know at resonance, reactance (resistance due to inductor and capacitor) be zero (0).
At resonance, Impedance (Z) = Resistance (R)
Therefore, Z=20 ohm

Test: BITSAT Past Year Paper- 2020 - Question 29

An electron moves in a circular arc of radius 10 m at a contant speed of 2 × 107 ms–1 with its plane of motion normal to a magnetic flux density of 10–5 T. What will be the value of specific charge of the electron?

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 29

Bqv = mv2 /r  or q/m = v /rB.

Test: BITSAT Past Year Paper- 2020 - Question 30

From a 200 m high tower, one ball is thrown upwards with speed of 10 ms-1 and another is thrown vertically downwards at the same speeds simultaneously. The time difference of their reaching the ground will be nearest to

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 30

The ball thrown upward will lose velocity in 1s. It return back to thrown point in another 1 s with the same velocity as second. Thus the difference will be 2 s.

Test: BITSAT Past Year Paper- 2020 - Question 31

A wheel is rotating at 900 r.p.m. about its axis.When power is cut off it comes to rest in 1 minute.The angular retardation in rad/s2 is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 31

Angular retardation,

Test: BITSAT Past Year Paper- 2020 - Question 32

 A particle executing simple harmonic motion along y-axis has its motion described by the equation y = A sin(ω t)+B . The amplitude of the simple harmonic motion is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 32

The  amplitude is a maximum displacement from the mean position.

Test: BITSAT Past Year Paper- 2020 - Question 33

A conducting square loop of side L and resistance R moves in its plane with a uniform velocity v perpendicular to one of its side. A magnetic induction B constant in time and space, pointing perpendicular and into the plane of the loop exists everywhere.

The current induced in the loop is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 33

Since the magnetic field is uniform the flux f through the square loop at any time t is constant, because
f  = B × A = B × L2 = constant

Test: BITSAT Past Year Paper- 2020 - Question 34

A body of mass 2  kg is placed on a horizontal surface having kinetic friction 0.4 and static friction 0.5. If the force applied on the body is 2.5 N, then the frictional force acting on the body will be [g = 10 ms–2]

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 34

Limiting friction = 0.5 x 2 x 10= 10 N
The applied force is less than force of friction, therefore the force of friction is equal to the applied force.

Test: BITSAT Past Year Paper- 2020 - Question 35

A nucleus splits into two nuclear parts which have their velocity ratio equal to 2 : 1. What will be the ratio of their nuclear radius?

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 35

As momentum is conserved, therefore

Test: BITSAT Past Year Paper- 2020 - Question 36

A charge +q is at a distance L/2 above a square of side L. Then what is the flux linked with the surface?

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 36


The given square of side L may be considered as one of the faces of a cube with edge L. Then given charge q will be considered to be placed at the centre of the cube. Then according to Gauss's theorem, the magnitude of the electric flux through the faces (six) of the cube is given by
ф = q/ε0
Hence, electric flux through one face of the cube for the given square will be

Test: BITSAT Past Year Paper- 2020 - Question 37

A plane electromagnetic wave is incident on a plane surface of area A, normally and is perfectly reflected. If energy E strikes the surface in time t then average pressure exerted on the surface is (c = speed of light)

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 37

Incident momentum, 
For perfectly reflecting surface with normal incidence

Test: BITSAT Past Year Paper- 2020 - Question 38

There are two wire of same material and same length while the diameter of second wire is two times the diameter of first wire, then the ratio of extension produced in the wires by applying same load will be

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 38

∵ Both wires are same materials so  both will have same Young’s modulus, and let it be Y.

A = area of cross-section of wire Now,

Since load and length are same for both

ΔL1 : ΔL2= 4 :1

Test: BITSAT Past Year Paper- 2020 - Question 39

Determine the current in 2Ω resistor.​

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 39

At steady state the capacitor will be fully charged and thus there will be no current in the 1W resistance. So the effective circuit becomes

Net current from the 6V battery,

Between A and B, voltage is same in both resistances,  2I1 = 3I 2  where
I1 + I 2 = I = 1.5
⇒ 2I1 = 3(1.5 - I1 ) ⇒ I1 = 0.9A

Test: BITSAT Past Year Paper- 2020 - Question 40

The potential energy of a satellite of mass m and revolving at a height Re above the surface of earth where Re = radius of earth, is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 40

At a height h above the surface of earth the gravitational potential energy of the particle of mass m is

Where Me & Re are the mass & radius of earth respectively.
In this question, since h = Re

Test: BITSAT Past Year Paper- 2020 - Question 41

If 0.2 gram of an organic compound containing carbon, hydrogen and oxygen on combustion, yielded 0.147 gram carbon dioxide and 0.12 gram water. What will be the content of oxygen in the substance ?

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 41



% of O = 100 - (20.045 + 6.666) = 73.29 %

Test: BITSAT Past Year Paper- 2020 - Question 42

The Lassaigne’s extract is boiled with dil. HNO3 before testing for halogens because

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 42

Na2S and NaCN, formed during fusion with metallic sodium, must be removed before adding AgNO3, otherwise black ppt. due to Na2S or white precipitate due to AgCN will be formed and thus white precipitate of AgCl will not be identified easily.

Test: BITSAT Past Year Paper- 2020 - Question 43

What is X in the following conversion ?

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 43

Pyridinium chlorochromate selectively oxidises a primary alcohol to an aldehyde
e.g.  

Test: BITSAT Past Year Paper- 2020 - Question 44

Maleic acid and fumaric acids are

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 44

Maleic acid and fumaric acids are geometrical isomers.

Test: BITSAT Past Year Paper- 2020 - Question 45

For which one of the processes represented by the following equations the enthalpy (heat) change is likely to be negative

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 45

Gaseous ions, when dissolved in water, get hydrated and heat is evolved (heat of hydration).
Cl- (g) + aq → Cl- (aq) is such reaction .

Test: BITSAT Past Year Paper- 2020 - Question 46

A cyclic process ABCD is shown in P–V diagram for an ideal gas. Which of the following diagram represents the same process?

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 46

Correct option is (C)

Solution: (C)

In the given fig.,

Process A→B⇒ Isobaric

Process B→C⇒ Isothermal

Process C→D⇒ Isochoric

Process D→A⇒ Isothermal

Hence the process is correctly represented in fig. (C).

Test: BITSAT Past Year Paper- 2020 - Question 47

In a monoclinic unit cell, the relation of sides and angles are respectively:

Test: BITSAT Past Year Paper- 2020 - Question 48

Phosphine is not obtained by which of the following reaction?

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 48

Red P  does not react with NaOH to give PH3.

Test: BITSAT Past Year Paper- 2020 - Question 49

An ideal gaseous mixture of ethane (C2H6) and ethene (C2H4) occupies 28 litre at 1 atm and 273 K. The mixture reacts completely with 128 g O2 to produce CO2 and H2O. Mole fraction at C2H6 in the mixture is:

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 49

C2H6 + 3.5O2 → 2CO2 + 3H2O;
C2H4 + 3O2 → 2CO2 + 2H2O
Let volume of ethane is x litre,
22.4 × 4 = 3.5x + 3(28 – x)
⇒  x = 11.2 litre
at constant T and P, V αx n;
⇒ Mole fraction of C2H6 in mixture = 

Test: BITSAT Past Year Paper- 2020 - Question 50

What is the maximum wavelength line in the Lyman series of He+ ion?

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 50


Test: BITSAT Past Year Paper- 2020 - Question 51

The correct order of acidic strength :

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 51

Acidic character of oxide ∝ Non-metallic nature of element.
Non-metallic character increases along the period. Hence order of acidic character is
Cl2O7 > SO2 > P4O10.

Test: BITSAT Past Year Paper- 2020 - Question 52

In the reaction 2PCl5 PCl4+ + PCl6- , the change in hybridisation is from

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 52

Test: BITSAT Past Year Paper- 2020 - Question 53

Arrange the following ions in the order of decreasing X – O bond length, where X is the central atom

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 53

More will be the electronegativity of X, lesser will be the bond length of X-O bond.

Test: BITSAT Past Year Paper- 2020 - Question 54

Two vessels of volumes 16.4 L and 5 L contain two ideal gases of molecular existence at the respective temperature of 27 °C and 227 °C and exert 1.5 and 4.1 atmospheres respectively. The ratio of the number of molecules of the former to that of the later is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 54

Given conditions
V1 = 16.4 L, V2 = 5 L
P1 = 1.5 atm, P2 = 4.1 atm
T1 = 273 + 27 = 300 K,
T2 = 273 + 227 = 500 K
Applying gas equation, 

Test: BITSAT Past Year Paper- 2020 - Question 55

For the combustion reaction at 298 K
2Ag(s) + 1 / 2O2 (g ) → 2Ag2O(s)
Which of the following alternatives is correct?

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 55

ΔH = ΔU + ΔnRT
Δn = nP – nR
Now, 

Thus, 
∴  ΔU > ΔH

Test: BITSAT Past Year Paper- 2020 - Question 56

The ratio for the reaction

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 56

KP = KC ( RT)Δng
For the reaction

Test: BITSAT Past Year Paper- 2020 - Question 57

A solution of NH 4Cl and NH 3 has pH = 8.0.Which of the following hydroxides may be precipitated when this solution is mixed with equal volume of 0.2 M of metal ion.

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 57

pH = 8, pOH = 6; [OH - ] = 10 -6 M;
Ionic product of Fe(OH)2 = 0.2 × (1 x10 -6 ) 2
 2 x 10 -13 > K sp (= 8.1 x 10-16 )

Test: BITSAT Past Year Paper- 2020 - Question 58

Which of the following is not a disproportionation reaction?

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 58

Disproportionation involves simultaneous oxidation and reduction of the same atom in a molecule.

Test: BITSAT Past Year Paper- 2020 - Question 59

The amount of H2O2 present in 1 litre of 1.5 N H2O2 solution, is :

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 59

Molecular weight of H2O2 = 34
Equivalent weight of H2O2 = 17
∴ 1 L of 1 N H2O2 has = 17 g of H2O
∴ 1 L of 1.5 N H2O2 has = 1.5 × 17 = 25.5 g of H2O2

Test: BITSAT Past Year Paper- 2020 - Question 60

BeF2 is soluble in water whereas fluorides of other alkaline earth metals are insoluble because of

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 60

Be2+ being small in size is heavily hydrated and heat of hydration exceeds the lattice energy.
Hence BeF2 is soluble in water.

Test: BITSAT Past Year Paper- 2020 - Question 61

Identify the incorrect statement :

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 61

The hydrolysis of Trialkylchlorosilane R3 SiCl yields dimer :

Test: BITSAT Past Year Paper- 2020 - Question 62

The alcohol product(s) of the reduction of 2-methyl-3-pentanone with LiAlH4 is (are)

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 62

The resulting compound is C3H7CH(OH)C2H5 which is optically inactive and reaction leads the racemisation

Test: BITSAT Past Year Paper- 2020 - Question 63


Br will abstract which of the hydrogen most readily?

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 63

Bromine is more selective
∴ abstract that hydrogen which forms stable free- radical
∴ 

Test: BITSAT Past Year Paper- 2020 - Question 64

The vapour pressure of two pure liquids A and B that form an ideal solution, are 400 and 800 mm of Hg respectively at a temperature t°C. The mole fraction of A in a solution of A and B whose boiling point is t°C will be (a) 0.4 (b) 0.8 (c) 0.1 (d) 0.2

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 64

V.P. of solution at t°C = 760 mm
[at b.p., V.P. of solution =atompheric pressure]
Thus  = PA°.xA + PB°.xB
or P = PA°.xA + PB°.(1 – xA)  [∵xA + xB = 1]
or 760 = 400XA + 800(1 – XA) [∵ P = 760 mm of Hg]
or – 800 + 760 = – 400 xA
or  – 40 = – 400 xA

Thus mole fraction in solution is 0.1

Test: BITSAT Past Year Paper- 2020 - Question 65

On reaction with sodium, 1 mol of a compound X gives 1 mol of H2. Which one of the following compounds might be X?

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 65

Since 1 mole of compound X reacts with Na to evolve 1 mole of H2 gas, therefore the compound should have 2 active hydrogen atoms per mole which is possible only in option d.
CH 2OHCH 2CH2CH2OH + 2 Na → NaOCH 2CH2CH2CH2ONa+ 2H

Test: BITSAT Past Year Paper- 2020 - Question 66

The following change can be carried out with

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 66

LiAlH4 will give the desired compound.

Test: BITSAT Past Year Paper- 2020 - Question 67

The greenhouse effect is because of the

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 67

Green house gases such as CO2, ozone, methane, the chlorofluorocarbon compounds and water vapour form a thick cover around the earth which prevents the IR rays emitted by the earth to escape. It gradually leads to increase in temperature of atmosphere.

Test: BITSAT Past Year Paper- 2020 - Question 68

Methylene blue, from its aqueous solution, is adsorbed on activated charcoal at 25°C. For this process, which of the following statement is correct?