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

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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

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.

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

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?

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 68

The adsorption of methyene blue on activated charcoal is an example of physiosorption which is exothermic, multilayer and does not have energy barrier.

Test: BITSAT Past Year Paper- 2020 - Question 69

The final product obtained in the reaction

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 69

Test: BITSAT Past Year Paper- 2020 - Question 70

What is the product of the following reaction ?

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 70

Only –CHO group reacts with CN– ion and the reaction is

Test: BITSAT Past Year Paper- 2020 - Question 71

The d-electron configurations of Cr2+, Mn2+, Fe2+ and Co2+ are d 4, d 5, d 6 and d 7 respectively.Which one of the following will exhibit the lowest paramagnetic behaviour?|
(Atomic no. Cr = 24, Mn = 25, Fe = 26, Co = 27).

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 71

Electronic No. of unpaired configuration electrons

∴ Sin ce Co2+ ha s lowest no. of unpaired electrons hence lowest paramagnetic behaviour is shown by [Co(H2O)6]2+

Test: BITSAT Past Year Paper- 2020 - Question 72

An organic compound with the formula C6H12O6 forms  a yellow crystalline solid with phenylhydrazine and gives a mixture of sorbitol and mannitol when reduced with sodium. Which among the following could be the compound?

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 72

Since the compound forms a yellow crystalline solid, i.e. osazone with phenylhydrazine, it may be an aldohexose or a ketohexose. Further, since on reduction, compound forms a  mixture of sorbitol and mannitol, it must be a ketohexose, i.e. fructose.
Recall that glucose on reduction gives only one alcohol glucitol (Sorbitol)

Test: BITSAT Past Year Paper- 2020 - Question 73

Compounds I and II may be grouped as

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 73

When structures I and II are C–2 epimers, it implies that these are epimers and diastereomers too.

Test: BITSAT Past Year Paper- 2020 - Question 74

Which is not the correct statement?
(At. nos. Ce = 58, Lu = 71, La = 57, Yb = 70)

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 74

Option (a) is incorrect as Yb3+ is colorless.

Test: BITSAT Past Year Paper- 2020 - Question 75

Which of the following reactions will not give N, N- dimethyl benzamide ?

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 75

(a) C6 H5COOC 2 H5 + (CH3 )2NH → C6H5CON(CH 3 ) 2 + C2H 5OH
(b) C6H5CONH 2 + CH3MgI → C6H5CO NH MgI + CH 4
(c) C6 H 5COCl + (CH3 )2NH → C6H5CON(CH3)2 + HCl
(d) C6H5CO.O.COC6H5 +
(CH3 )2NH → C6H5CON(CH 3 ) 2
+ C6H5COOH

Test: BITSAT Past Year Paper- 2020 - Question 76

Here, X is :

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 76

Here X is polytetrafluoroethylene. So, none of these i.e., option (d) is correct choice.

Test: BITSAT Past Year Paper- 2020 - Question 77

Select the incorrect statement.

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 77

Bacteriostatic drugs inhibit the growth of organisms while bactericidal drugs kill the microorganisms.

Test: BITSAT Past Year Paper- 2020 - Question 78

Product (D) in above reaction is:

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 78

Test: BITSAT Past Year Paper- 2020 - Question 79

ΔfG vs T plot in the Ellingham diagram slopes downward for the reaction

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 79

Hence, as the temperature increases, TΔS increases and hence ΔG (ΔH – TΔS) decreases.
In other words, the slope of the curve for formation of CO decreases. However, for all other oxides, it increases.

Test: BITSAT Past Year Paper- 2020 - Question 80

A radioactive isotope having a half - life period of 3 days was received after 12 days. If 3g of the isotope is left in the container, what would be the initial mass of the isotope?

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 80

Given t1/2 = 3
Total time T = 12
No. of half lives (n) =

N = 48 g

Test: BITSAT Past Year Paper- 2020 - Question 81

DIRECTIONS: Select the most appropriate option to fill in the blank.​

One of them would keep a look-out on the road behind to warn us _______ approaching vehicles.

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 81

Warn a person of possible danger or a potential problem.
Warn a person against a fault.
Hence out of all four prepositions in options, ‘of’ is the correct choice.

Test: BITSAT Past Year Paper- 2020 - Question 82

DIRECTIONS: Select the most appropriate option to fill in the blank.​

The top -ranking manager _________  his success in the profession to his managing director’s guidance.

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 82

Attributes means regard something as being caused by.

Test: BITSAT Past Year Paper- 2020 - Question 83

DIRECTIONS: Some parts of the sentences have errors and some are correct. Find out which part of a sentence has an error.

Mother Teresa asked a building where she and her workers could care for the poor people always.

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 83

Test: BITSAT Past Year Paper- 2020 - Question 84

DIRECTIONS: Some parts of the sentences have errors and some are correct. Find out which part of a sentence has an error.

You may not realize it but the weather in Barbados during Christmas is like New York in June.

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 84

Write ‘that in’ after ‘like’.

Test: BITSAT Past Year Paper- 2020 - Question 85

DIRECTIONS : Some parts of the sentences have errors and some are correct. Find out which part of a sentence has an error.

The parents decided it was worth the risk because these children would have succumbed their disease before adulthood.

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 85

Use preposition ‘to’ after ‘succumbed’.
Verb ‘Succumb’ is characterized by preposition ‘to’. Moreover, ‘Succumb to something’ is an idiomatic expression which means to yield to something, especially a temptation, fatal disease, a human weakness, etc.

Test: BITSAT Past Year Paper- 2020 - Question 86

DIRECTIONS : Read the following passage carefully and choose the best answer to each question out of the four given alternatives.

Harold, a professional man who had worked in an office for many years, had a fearful dream. In it, he found himself in a land where small slug-like animals with slimy tentacles lived on people’s bodies. The people tolerated the loathsome creatures because after many years they grew into elephants which then became the nation’s system of transport, carrying everyone wherever he wanted to go.
Harold suddenly realized that he himself was covered with these things, and he woke up screaming. In a vivid sequence of pictures his dream dramatized for Harold what he had never been able to put into words; he saw himself as letting society feed on his body in his early years so that it would carry him when he retired. He later threw off the ‘security bug’ and took up freelance work.

Q. The statement that ‘he later threw off the security bug’ means that

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 86

Harold succeeded in overcoming the need for security

Test: BITSAT Past Year Paper- 2020 - Question 87

DIRECTIONS : Read the following passage carefully and choose the best answer to each question out of the four given alternatives.

Harold, a professional man who had worked in an office for many years, had a fearful dream. In it, he found himself in a land where small slug-like animals with slimy tentacles lived on people’s bodies. The people tolerated the loathsome creatures because after many years they grew into elephants which then became the nation’s system of transport, carrying everyone wherever he wanted to go.
Harold suddenly realized that he himself was covered with these things, and he woke up screaming. In a vivid sequence of pictures his dream dramatized for Harold what he had never been able to put into words; he saw himself as letting society feed on his body in his early years so that it would carry him when he retired. He later threw off the ‘security bug’ and took up freelance work.

Q. Which one of the following phrases best helps to bring out the precise meaning of ‘loathsome creatures’?

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 87

Slimy tentacles and slug-like animals

Test: BITSAT Past Year Paper- 2020 - Question 88

DIRECTIONS : Read the following passage carefully and choose the best answer to each question out of the four given alternatives.

Harold, a professional man who had worked in an office for many years, had a fearful dream. In it, he found himself in a land where small slug-like animals with slimy tentacles lived on people’s bodies. The people tolerated the loathsome creatures because after many years they grew into elephants which then became the nation’s system of transport, carrying everyone wherever he wanted to go.
Harold suddenly realized that he himself was covered with these things, and he woke up screaming. In a vivid sequence of pictures his dream dramatized for Harold what he had never been able to put into words; he saw himself as letting society feed on his body in his early years so that it would carry him when he retired. He later threw off the ‘security bug’ and took up freelance work.

Q. In his dream, Harold found the loathsome creatures

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 88

In a different land

Test: BITSAT Past Year Paper- 2020 - Question 89

DIRECTIONS : Read the following passage carefully and choose the best answer to each question out of the four given alternatives.

Harold, a professional man who had worked in an office for many years, had a fearful dream. In it, he found himself in a land where small slug-like animals with slimy tentacles lived on people’s bodies. The people tolerated the loathsome creatures because after many years they grew into elephants which then became the nation’s system of transport, carrying everyone wherever he wanted to go.
Harold suddenly realized that he himself was covered with these things, and he woke up screaming. In a vivid sequence of pictures his dream dramatized for Harold what he had never been able to put into words; he saw himself as letting society feed on his body in his early years so that it would carry him when he retired. He later threw off the ‘security bug’ and took up freelance work.

Q. Harold’s dream was fearful because

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 89

In it he saw slimy creatures feeding on people’s bodies

Test: BITSAT Past Year Paper- 2020 - Question 90

DIRECTIONS: In questions, choose the word opposite in meaning to the given word and mark it in.

Q. Subdued

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 90

Subdued Lacking in vitality, intensity, or strength

Test: BITSAT Past Year Paper- 2020 - Question 91

DIRECTIONS: In questions, choose the word opposite in meaning to the given word and mark it in.

Q. Fervent

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 91

Fervent having or displaying a passionate intensity. Dispassionate not influenced by strong emotion, and so able to be rational and impartial.

Test: BITSAT Past Year Paper- 2020 - Question 92

DIRECTIONS: In questions, choose the word opposite in meaning to the given word and mark it in.

Q. Scrupulous

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 92

Scrupulous (of a person or process) diligent, thorough, and extremely attentive to details.

Test: BITSAT Past Year Paper- 2020 - Question 93

DIRECTIONS: In questions no. 26 to 28, out of the four alternatives, choose the one which best expresses the meaning of the given word and mark it.

Q. Onslaught

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 93

Onslaught a fierce or destructive attack.
Invasion an instance of invading a country or region with an armed force.

Test: BITSAT Past Year Paper- 2020 - Question 94

DIRECTIONS: In questions no. 26 to 28, out of the four alternatives, choose the one which best expresses the meaning of the given word and mark it.

Q. Ignominy

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 94

Ignominy public shame or disgrace.

Test: BITSAT Past Year Paper- 2020 - Question 95

DIRECTIONS: In questions no. 26 to 28, out of the four alternatives, choose the one which best expresses the meaning of the given word and mark it.

Q. Tryst

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 95

Tryst An agreement (as between lovers) to meet

Test: BITSAT Past Year Paper- 2020 - Question 96

Which of the option figures bear s the closest resemblance to the question figure?
Question Figure:

Option Figures:

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 96

Closest Resemblance is mirror image of question figure.

Test: BITSAT Past Year Paper- 2020 - Question 97

In the following questions a piece of paper is folded and punched as shown in problem figures and four answer figure marked (a), (b), (c), (d) are given. Select the answer figure which indicates how the paper will appear when open (unfolded).
Question Figure:

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 97

Test: BITSAT Past Year Paper- 2020 - Question 98

Which answer figure complete th e form in question figure ?
Question Figure:

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 98

Test: BITSAT Past Year Paper- 2020 - Question 99

From the answer figure choose the box that is similar to the box formed by folding the question figure.
Question figure:

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 99

Only cube (4) can be formed.

Test: BITSAT Past Year Paper- 2020 - Question 100

In each of the following question select the missing number from the given responses.

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 100

(100  + 12) – (28 + 25) = 59
Similarly,
(102 + 52) – (36 + 28) = 90

Test: BITSAT Past Year Paper- 2020 - Question 101

A clock with only dot markings 3, 6, 9 and 12 positions has been kept upside down in front of a mirror. A person reads the time in the reflections of the clock as 12 : 30 the actual that will be

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 101

In this time 12 : 30 we subtract that time from 17 : 90 – 12 : 30 = 5 : 60 = 6 : 00

Test: BITSAT Past Year Paper- 2020 - Question 102

Six friends - A, B, C, D, E and F are sitting in a circle facing center. F is to the immediate left of A and B is sitting  opposite to F, E and D are sitting opposite to each other. Who is sitting third right to A?

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 102

So, C is sitting third right to A

Test: BITSAT Past Year Paper- 2020 - Question 103

Ganesh cycles towards South-West a distance of 8 m. Then he moves towards East a distance of 20 m. From there he moves towards North-East a distance of 8 m, then he moves towards West a distance of 6m. From there he moves towards North-East a distance of 2 m. Then he moves towards West a distance of 4 m and then towards South-West 2 m and stops at that point. How far is he from the starting point ?

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 103

Ganesh is 10m. far from the starting point

Test: BITSAT Past Year Paper- 2020 - Question 104

If in a code language, LAPTOP is written as PNSOZL and NOTEBOOK and KNNADSNN, then which letter will be  there in first and seventh letter from left after coding MEDICINE in the same way?

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 104

Test: BITSAT Past Year Paper- 2020 - Question 105

Find the next term in the following series: X24C, V22E, T20G, ______

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 105

Test: BITSAT Past Year Paper- 2020 - Question 106

Let [x] denote the greatest integer £ x. If f (x) = [x] and g (x) = |x|, then the value of

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 106

Given that, f (x) = [x] and g (x) = |x|

Test: BITSAT Past Year Paper- 2020 - Question 107

If a chord of the circle x2 + y2 = 8 makes equal intercets of length a on the coordinate axes, then

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 107

Since the chord makes equal intercepts of length a on the coordinate axes. So, its equation can be written as x ± y = ±a . This line meets the given circle at two distinct points.
So, length of the perpendicular from the centre (0, 0) of the given circle must be less than the radius.

Test: BITSAT Past Year Paper- 2020 - Question 108

In the given figure, the equation of the larger circle is x2 + y2 + 4 y - 5 = 0 and the distance between centres is 4. Then the equation of smaller circle is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 108

We have x2 + y2 + 4y – 5 = 0. Its centre is C1 (0, – 2) ,
r1 = Let C2 (h, k) be the centre of the smaller circle and its radius r2. Then C1C2 = 4.

⇒ r2 = 1
But k = r2 = 1 [it touches x-axis]
∴ From eq (1), 4 =
⇒ 16 = h2 + 9 ⇒ h2 = 7 ⇒ h = ± √7
Since h > 0     ∴ h = √7
Hence, required circle is
(x - √7 )2 + (y -1)2 = 1

Test: BITSAT Past Year Paper- 2020 - Question 109

The equation sin 4 x - (k + 2) sin 2 x - (k + 3) = 0 possesses a solution if

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 109

We have,
sin 4 x - (k + 2) sin 2 x - (k + 3) = 0

⇒ sin 2 x = k + 3 (∵ sin2x = – 1 is not possible)
Since 0 < sin 2 x ,< 1,
∴ 0 < k +3 < 1or - 3 < k < -2

Test: BITSAT Past Year Paper- 2020 - Question 110

If 16 identical pencils are distributed among 4 children such that each gets at least 3 pencils.The number of ways of distributing the pencils is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 110

Required number of ways
= coeff. of x16 in (x3 + x4 + x5 + x6 +  x7)4

Test: BITSAT Past Year Paper- 2020 - Question 111

If a, b, c are positive numbers, then least value of (a + b + c) is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 111

AM > HM

Test: BITSAT Past Year Paper- 2020 - Question 112

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 112

Test: BITSAT Past Year Paper- 2020 - Question 113

If and are non-zero constant vectors and the scalar b is chosen such that | + b| is minimum, then the value of | b |2 + | + b|2 is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 113

For minimum value | + b |= 0.
Let and sr are anti-parallel so bsr =-
∴| b |2 + | + b |2 =| - |2 + | - |2=| |2 .

Test: BITSAT Past Year Paper- 2020 - Question 114

Statement-1 : If the general equation x2 + y2 + 2xy + 2gx + 2fy + 4 = 0 represents a pair of real lines then | g |>2 .
Statement-2 : The equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents pair of real lines if abc + 2fgh – af2 – bg2 – ch2 = 0.

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 114

The equation represents a pair of lines if 1×1×4 + 2× f ×g×1-1×f 2 -1×g 2 - 4×12 = 0
⇒ (f - g) 2 = 0 ⇒ f=g
The equation becomes (x + y)2 + 2g(x + y) + 4 = 0 Which represents pair of parallel lines, which are real provided (2g) 2 - 4 × 4 ×1 > 0 Þ| g | > 2.

Test: BITSAT Past Year Paper- 2020 - Question 115

If | z1 | = | z2 | =    | zn | = 1. then the
value of | z1 | + | z2 | +    zn I -

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 115

Test: BITSAT Past Year Paper- 2020 - Question 116

In a statistical investigation of 1003 families of Calcutta, it was found that 63 families has neither a radio nor a T.V, 794 families has a radio and 187 has T.V. The number of families in that group having both a radio and a T.V is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 116

Let R be the set of families having a radio and T the set families having a T.V., then n (R ∪ T) = The number of families having at least on of the radio and T.V. = 1003 – 63 = 940 n(R) = 794 and n(T) = 187 Let x families have both a radio and a T.V.
Then number of families who have only radio = 794 – x
And the number of families who have only T.V. = 187 – x From Venn diagram, 794 – x + x – 187 – x = 940
⇒ 981 – x = 940 or x = 981 – 940 = 41 Hence, the required number of families having both a radio and a T.V. = 41

Test: BITSAT Past Year Paper- 2020 - Question 117

The domain of the function

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 117

For f (x) to be defined, we must have

∴ x2 > 1 – x2 or
Also, 1 – x2 > 0 or x2 < 1.
Now,

Also, x2 < 1 ⇒ (x – 1) ( x + 1) < 0
⇒ –1 < x < 1
Thus, x > 0, x2 >  and x2 < 1

Test: BITSAT Past Year Paper- 2020 - Question 118

If the coordinates at one end of a diameter of the circle x2 + y2 – 8x – 4y + c = 0 are (–3, 2), then the coordinates at the other end are

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 118

The centre of the given circle is C º (4, 2) Let A ≡ (-3, 2)

If (α, β) are the coordinates of the other end of the diameter, then, as the middle ploint of the diameter is the centre,

Thus, the coordinates of the other end of diameter are (11, 2)

Test: BITSAT Past Year Paper- 2020 - Question 119

The value of is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 119

[∵ sin(nπ + θ) = (-1) n sin θ]

∴ sin[(-1) n θ] = (-1) n sin θ]

Test: BITSAT Past Year Paper- 2020 - Question 120

The number of possible outcomes in a throw of n ordinary dice in which at least one of the dice shows an odd number is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 120

Total number of ways = 6 × 6 × ..... to n times = 6n.
Total number of ways to show only even number = 3 × 3 × ...... to n times = 3n.
∴ required number of ways = 6n – 3n.

Test: BITSAT Past Year Paper- 2020 - Question 121

The sum of the infinite series is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 121

We know that

keeping x = 2, we get

Test: BITSAT Past Year Paper- 2020 - Question 122

The solution of the differential equation

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 122

The given equation is

The solution is

Test: BITSAT Past Year Paper- 2020 - Question 123

Consider the function f(x) = |x – 1|/x2, then f (x) is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 123

Clearly,  f (x) is continuous for all x ∈ R except at x = 0.

f '(x) > 0 ⇒ x< 0 or 1 < x < 2
f '(x) < 0 ⇒ 0 < x <1 or x > 2
Hence, f (x) is increasing in (-∞, 0)∪ (1, 2) and decreasing in (0, 1) ∪ (2,∞).

Test: BITSAT Past Year Paper- 2020 - Question 124

A straight rod of length 9 units slides with its ends A, B always on the X and Y-axis respectively.Then the locus of the centroid of Δ OAB is :

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 124

Let end A is (a, 0) and end B is (0,  b) then a 2 + b 2 = 81 …(1) If centroid is (x, y) then

putting in (1) we get the locus as x2 + y2  = 9

Test: BITSAT Past Year Paper- 2020 - Question 125

If xy + y2 = tan x + y, then find is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 125

The given relation is xy + y2 = tan x + y.
Differentiating both sides with respect to x, we get

Test: BITSAT Past Year Paper- 2020 - Question 126

Th e probability th at certain electr on ic component fails when first used is 0.10. If it does not fail immediately, the probability that it lasts for one year is 0.99. The probability that a new component will last for one year is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 126

Probability that the electronic component fails when first used is P(F) = 0.10. Therefore, P(F') = 1 – P(F) = 0.90
Let E be the event that a new component will last for one year.
[total probability theorem]
= 0.10 × 0 + 0.90 × 0.99 = 0.891.

Test: BITSAT Past Year Paper- 2020 - Question 127

Let x +y = 3 - cos4θ and x -y = 4 sin2θ then the greatest of xy is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 127

∴ xy = (1 - sin 2 2θ)2 = cos4< 1

Test: BITSAT Past Year Paper- 2020 - Question 128

Total number of ways of selecting 2 white squares on a normal chess board if the squares are not from the same row or column is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 128

2 white squares can be selected in 32C2 ways If they belong to some row or same column then 2(8.4C2) ways.

Test: BITSAT Past Year Paper- 2020 - Question 129

The length of the latus rectum of the parabola which has focus at (–1, 1) and the directrix is 4x + 3y – 24 = 0 is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 129

Length of latus rectum = 2 × distance of focus from directrix

Test: BITSAT Past Year Paper- 2020 - Question 130

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 130

In the interval

Test: BITSAT Past Year Paper- 2020 - Question 131

A student read common difference of an A. P. as – 2 instead of 2 and got the sum of first 5 terms as – 5. Actual sum of first five terms is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 131

∴ Actual sum =

Test: BITSAT Past Year Paper- 2020 - Question 132

The number of real solution s of the equation ( x – 1) 2 + ( x – 2) 2 + ( x – 3) 2 = 0 is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 132

Equation holds if and only if x – 1 = 0, x – 2 = 0 and x – 3 = 0 simultanesouly, which is not possible.

Test: BITSAT Past Year Paper- 2020 - Question 133

The least positive non-integral solution of the equation sin p(x 2 + x) = sin πx 2 is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 133

We have, sin p( x 2 + x) = sin πx 2
⇒ p(x 2 + x ) = np + (-1) n πx 2
∴ Either x 2 + x = 2m + x 2 ⇒ x = 2m ∈ I
or x 2 + x = k - x 2 , where k is an odd integer
⇒ 2x2 +x -k =0 ⇒ x =
For least positive non-integral solution is x 1/2, when k = 1

Test: BITSAT Past Year Paper- 2020 - Question 134

ABC is a triangular park with AB = AC = 100 m. A TV tower stands at the mid-point of BC. The angles of elevation of the top of the tower at A, B, C are 45°, 60°, 60° respectively. The height of the tower is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 134

Where PQ is tower and ABC is the park, with Q being mid point of the side BC and PQ = h
Also, AQ2 + BQ2 = 1002
⇒ h 2 + h 2 cot 2 60° = 1002

Test: BITSAT Past Year Paper- 2020 - Question 135

If x, y and z are real numbers, then
x 2 + 4 y 2 + 9 z 2 – 6yz –3zx – 2xy is always

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 135

x2 + 4y2 + 9z2 – 6yz – 3zx – 2xy
= x2 + (2y)2 + (3z)2 – (2y) (3z) – (3z)(x) – x (2 y) > 0 x, y, z
[∵ a2 + b2 + c2 - ab - bc - ca > 0]

Test: BITSAT Past Year Paper- 2020 - Question 136

A person invites a party of 10 friends at dinner and place them so that 4 are on one round table and 6 on the other round table. The number of ways in which he can arrange the guests is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 136

Selection of 6 guests = 10C6 Permutation of 6 on round table = 5!
Permutation of 4 on round table = 3!
Then, total number of arrangements = 10C6.5!.3!

Test: BITSAT Past Year Paper- 2020 - Question 137

If a > 0, b > 0, c > 0 and a, b, c are distinct, then (a + b) (b + c) (c + a) is greater than

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 137

AM > GM

So, (a + b)(b + c)(c + a ) > 8abc

Test: BITSAT Past Year Paper- 2020 - Question 138

The equation of the tangent to the curve y = be– x/a at the point where it crosses the y-axis is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 138

y = be- x /a meets the y-axis at (0, b). Again,

Therefore, required tangent is

Test: BITSAT Past Year Paper- 2020 - Question 139

The solutions of (x + y + 1) dy = dx are

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 139

Putting x + y + 1 = u, we have du = dx + dy and the given equation reduces to u(du – dx) = dx

⇒ log(x + y + 2) = y + constant
⇒ x + y + 2 = Ce y

Test: BITSAT Past Year Paper- 2020 - Question 140

If y = 3x + 6x2 + 10x3 + ........ ∞, then
is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 140

We have y = 3x + 6x 2 + 10x 3 + .......
⇒ 1 + y = 1 + 3x + 6x 2 + 10x 3 + .....
⇒ 1+ y = (1- x)-3 ⇒ 1- x = (1+ y)-1/3
⇒ x = 1 - (1 + y) -1 / 3

Test: BITSAT Past Year Paper- 2020 - Question 141

Suppose p, q, r ≠ 0 and system of equation
(p + a)x + by + cz = 0,
ax + (q + b) y + cz = 0 , ax + by + (r + c)z = 0 has a non-trivial solution,
then value of is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 141

Since the system is homogeneous and has a non-trivial solution

using R1 → R1 - R2 and R2 → R2 - R3 .
Expanding along C1 we get
⇒ p[q(r + c) + br ] + aqr = 0
⇒ pqr + pqc + prb + qra = 0
⇒

Test: BITSAT Past Year Paper- 2020 - Question 142

If a, c, b are in G.P., then the area of the triangle formed by the lines ax + by + c = 0 with the coordinates axes is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 142

Area of the triangle

[∵ a, c, b are in G.P. ⇒ c2 = ab]

Test: BITSAT Past Year Paper- 2020 - Question 143

Suppose that f(0) = –3 and f '(x) £5 for all values of x.
Then, the largest value which f(2) can attain is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 143

Using LMVT in [0, 2]

Test: BITSAT Past Year Paper- 2020 - Question 144

The maximum value of  z = 6x + 8y  subject to constraints 2x + y < 30, x + 2y < 24 and x > 0, y > 0 is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 144

Here, 2x + y < 30, x + 2y < 24, x, y > 0 The shaded region represents the feasible region, hence z = 6x + 8y. Obviously it is maximum at (12, 6).
Hence z = 12 × 6 + 8 × 6 = 120

Test: BITSAT Past Year Paper- 2020 - Question 145

If a, b, c are three natural numbers in AP and a + b + c = 21 then the possible number of values of the ordered triplet (a, b, c) is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 145

Let a = b – d and c = b + d, then a + b + c = 21
⇒ b = 7 So, the equation is a + c = 14
∴ No. of solution = coeff. of x14 in (x + x2 + .....) = 13C12 = 13

Test: BITSAT Past Year Paper- 2020 - Question 146

If touches the ellipse

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 146

The line is It will touch the ellipse if

Test: BITSAT Past Year Paper- 2020 - Question 147

Let f be the function defined by

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 147

∴ The function is not continuous at x = 1.

Test: BITSAT Past Year Paper- 2020 - Question 148

The line y = mx bisects the area enclosed  by lines x =0, y = 0 and x = 3/2 and the curve y = 1 + 4x – x2.
Then the value of m is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 148

y = 1 + 4x - x 2 = 5 - (x - 2) 2

We have

On solving we get

Test: BITSAT Past Year Paper- 2020 - Question 149

There are four numbers of which the first three are in G.P. and the last three are in A.P., whose common difference is 6. If the first and the last numbers are equal then two other numbers are

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 149

Let the last three numbers in A.P. be a, a + 6, a + 12, then the first term is also a + 12.
But  a + 12, a, a + 6 are in G.P.
∴ a 2 = (a +12) (a + 6) ⇒ a 2 = a 2 +18a + 72
∴ a  =  –4.
∴ The numbers are 8, –4, 2, 8.

Test: BITSAT Past Year Paper- 2020 - Question 150

The distance from the point (3, 4, 5) to the point where the line meets the
plane x + y + z = 17 is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 150

Any point on the line is ( r + 3, 2r + 4, 2r + 5).
It lies on the plane x + y + z = 17,
∴ (r + 3) + (2r + 4) + (2r + 5) = 17    i.e   r = 1
Thus the point of intersection of the plane and the line is (4, 6, 7) Required distance = distance between (3, 4, 5) and (4, 6, 7)

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