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150 Questions MCQ Test BITSAT Mock Tests Series & Past Year Papers | Test: BITSAT Past Year Paper- 2019

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Test: BITSAT Past Year Paper- 2019 - Question 1

Which one of the following graphs represents the variation of electric field with distance r from the centre of a charged spherical conductor of radius R?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 1

The charged sphere is a conductor.
Therefore the field inside is zero and outside it is proportional to 1/r2.

Test: BITSAT Past Year Paper- 2019 - Question 2

 are the electric and magnetic field vectors of e.m. waves then the direction of propagation of e.m. wave is along the direction of

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 2

The direction of propagation of electromagnetic wave is perpendicular to the variation of electric field  as well as to the magnetic field 

Test: BITSAT Past Year Paper- 2019 - Question 3

The young's modulus of a wire of length L and radius r is Y N/m2. If the length and radius are reduced to L/2 and r/2, then its young's modulus will be

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 3

Young's modulus of wire does not vary with dimention of wire. It is a constant quantity.

Test: BITSAT Past Year Paper- 2019 - Question 4

Twelve resistors each of resistance 16 Ω are connected in the circuit as shown. The net resistance between A and B is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 4

Redraw the given circuit,

where, R = 16 Ω
Rnet = 4 Ω

Test: BITSAT Past Year Paper- 2019 - Question 5

The time period of a satellite of earth is 5 hours.If the separation between the earth and the satellite is increased to 4 times the previous value, the new time period will become

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 5

Accor ding to Kepler ’s law of plan etary motion, T2 ∝ R 

Test: BITSAT Past Year Paper- 2019 - Question 6

Two trains are moving towards each other with speeds of 20 m/s and 15 m/s relative to the ground. The first train sounds a whistle of frequency 600 Hz. The frequency of the whistle heard by a passenger in the second train before the train meets, is (the speed of sound in air is 340 m/s)

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 6

Test: BITSAT Past Year Paper- 2019 - Question 7

You are asked to design a shaving mirror assuming that a person keeps it 10 cm from his face and views the magnified image of the face at the closest comfortable distance of 25 cm. The radius of curvature of the mirror would then be :

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 7

Concave morror is used as a shaving mirror.



Therefore radius of curvature,
R = 2f = – 60 cm

Test: BITSAT Past Year Paper- 2019 - Question 8

A block is kept on a frictionless inclined surface with angle of inclination ‘α’. The incline is given an acceleration ‘a’ to keep the block stationary.
Then ‘a’  is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 8

From free body diagram, For block to remain stationary,


⇒ a = g tan α

Test: BITSAT Past Year Paper- 2019 - Question 9

With the increase in temperature, the angle of contact

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 9

On increasing the temperature, angle of contact decreases.

Test: BITSAT Past Year Paper- 2019 - Question 10

Forward biasing is that in which applied voltage

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 10

Forward bias opposes the potential barrier and if the applied voltage is more than knee voltage it cancels the potential barrier.

Test: BITSAT Past Year Paper- 2019 - Question 11

Number of significant figures in expression 

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 11

In multiplication or division the final result should return as many significant figures as there are in the original number with the least significant figures. (Rounding off to three significant digits)

Test: BITSAT Past Year Paper- 2019 - Question 12

The ratio of the specific heats Cp/Cv = γ in terms of degrees of freedom (n) is given by

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 12

Let ‘n’ be the degree of freedom

Test: BITSAT Past Year Paper- 2019 - Question 13

A stone is thrown with a velocity u making an angle θ with the horizontal. The horizontal distance covered by its fall to ground is maximum when the angle θ is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 13

Since range on horizontal plane is

so it is maximum when, sin 2θ = 1
θ = π/4

Test: BITSAT Past Year Paper- 2019 - Question 14

A ball of mass 150 g, moving with an acceleration 20 m/s2, is hit by a force, which acts on it for 0.1 sec. The impulsive force is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 14

Force = Mass × acceleration

Impulsive force = F.Δt = 3 × 0.1 = 0.3 N

Test: BITSAT Past Year Paper- 2019 - Question 15

A man drags a block through 10 m on rough surface (µ = 0.5). A force of √3 kN acting at 30° to the horizontal. The work done by applied force is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 15

Given,  d = 10 m
θ = 30°
μ = 0.5


W = Fsdcosθ
Where,
Fs = μF
Fs = 0.5 × √3 kN
Fs = 0.866 kN
Fs = 866 N
So, W = 866 × 10 × cos 30°

Test: BITSAT Past Year Paper- 2019 - Question 16

A force of  acts on a body for 4 second, produces a displacement of  The power used is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 16

Test: BITSAT Past Year Paper- 2019 - Question 17

The Earth is assumed to be a sphere of radius R. A platform is arranged at a height R from the surface of the Earth. The escape velocity of a body from this platform is fv, where v is its escape velocity from the surface of the Earth. The value of f is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 17

Test: BITSAT Past Year Paper- 2019 - Question 18

Kepler ’s second law regarding constancy of areal velocity of a planet is a consequence of the law of conservation of

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 18

dA/dt = L/2m = Constant

Test: BITSAT Past Year Paper- 2019 - Question 19

Water is flowing through a horizonal tube having cross-sectional areas of its two ends being A and A' such that the ratio A/A' is 5. If the pressure difference of water between the two ends is 3 × 105 N m–2, the velocity of water with which it enters the tube will be (neglect gravity effects)

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 19

According to Bernoulli’s theorem

According to the condition,


From equation of continuity,

From equation (i)

Test: BITSAT Past Year Paper- 2019 - Question 20

A thermodynamic system is taken from state A to B along ACB and is brought back to A along BDA  as shown in the PV diagram. The net work done during the complete cycle is given by the area

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 20

Work done = Area under curve ACBDA

Test: BITSAT Past Year Paper- 2019 - Question 21

A boat crosses a river from port A to port B, which are just on the opposite side. The speed of the water is Vw and that of boat is VB relative to still water. Assume Vw = 2Vw. What is the time taken by the boat, if it has to cross the river directly on the AB line [D = width of the river]

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 21

From figure, VB sin θ = VW 

Time taken to cross the river.

Test: BITSAT Past Year Paper- 2019 - Question 22

Two springs, of force constants k1 and k2 are connected to a mass m as shown. The frequency of oscillation of the mass is f. If both k1 and k2 are made four times their original values, the frequency of oscillation becomes

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 22

The two springs are in parallel.
∴ Effective spring constant, k = k1 + k2 Now, frequency of oscillation is given by


When both k1 and k2 are made four times their original values, the new frequency is given by


Test: BITSAT Past Year Paper- 2019 - Question 23

When a potential difference V is applied across a conductor at a temperature T, the drift velocity of electrons is proportional to

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 23

Drift velocity,

Test: BITSAT Past Year Paper- 2019 - Question 24

The amplitude of a damped oscillator becomes (1/3)rd in 2 secon ds. If its amplitude after 6 seconds is 1/n times the original amplitude, the value of n is 

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 24

Amplitude of a damped oscillator
A = A0e–bt/2m
Case 1 :-

Test: BITSAT Past Year Paper- 2019 - Question 25

The angular speed of the electron in the nth orbit of Bohr hydrogen atom is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 25

Angular speed of electron in the nth orbit of Bohr H-atom is inversely proportional to n3 

Test: BITSAT Past Year Paper- 2019 - Question 26

In the given figure, the charge on 3 µF capacitor is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 26

C = equivalent capacitance

Charge on each capacitor in series circuit will be same.
∴ q = CV = (1 × 10-6) × 10 = 10μC
∴ Charge across 3µF capacitor will be 10µC.

Test: BITSAT Past Year Paper- 2019 - Question 27

Two bodies A and B are placed in an evacuated vessel maintained at a temperature of 27ºC. The temperature of A is 327ºC and that of B is 227ºC.
The ratio of heat loss from A and B is about

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 27

Test: BITSAT Past Year Paper- 2019 - Question 28

If a rigid body is rotating about an axis with a constant velocity, then

Test: BITSAT Past Year Paper- 2019 - Question 29

The fundamental frequency of an open organ pipe is 300 Hz. The first overtone of this pipe has same  frequency as first overtone of a closed organ pipe. If speed of sound is 330 m/s, then the length of closed organ pipe is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 29

For open pipe, n1 = v/2ℓ , where n1 is the fundamental frequency of open pipe. length of open pipe is,

Ist overtone of open pipe, 
Ist overtone of closed pipe

where, ℓ’ = length of closed pipe
As freq. of 1st overtone of open pipe = freq. of 1st overtone of closed pipe

Test: BITSAT Past Year Paper- 2019 - Question 30

In Young's experiment, the distance between the slits is reduced to half and the distance between the slit and screen is doubled, then the fringe width

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 30

Test: BITSAT Past Year Paper- 2019 - Question 31

If a rolling body’s angular momentum changes by 20 Sl units in 3 seconds, by a constant torque.Then find the torque on the body

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 31

As we know, τ is change in angular momentum.

Test: BITSAT Past Year Paper- 2019 - Question 32

Charge Q is distributed to two different metallic spheres having radii x and 2x such that both spheres have equal surface charge density, then charge on large sphere is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 32

Let q and q' be the  charges on spheres of radii x and 2x respectively.
Given, q + q' = Q …(i)
Surface charge densities are


From eq. (i), q'  = Q – q  or,  4q = Q – q
or, Q = 5q …(ii)

Test: BITSAT Past Year Paper- 2019 - Question 33

In an LR circuit f = 50 Hz, L = 2 H, E = 5 volts, R = 1 Ω then energy stored in inductor is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 33

L = 2 H, E = 5 volts, R = 1Ω

Test: BITSAT Past Year Paper- 2019 - Question 34

A straight wire of length 0.5 metre and carrying a current of 1.2 ampere is placed in uniform magnetic field of induction 2 tesla. The magnetic field is perpendicular to the length of the wire.The force on the wire is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 34

F = Biℓ = 2 ×1.2 × 0.5 = 1.2 N

Test: BITSAT Past Year Paper- 2019 - Question 35

A man drives a car from station B towards station A at speed 60 km/h. A car leaves station A for station B every 10 min. The distance between A and B is 60 km. The car travels at the speed of 60 km/h. A man drives a car from B towards A at speed of 60 km/h. If he starts at the moment when first car leaves the station B, then how many cars would be meet on the route ?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 35

Distance between two cars leaving from the station A is,

Man meets the first car after time,


He will meet the next car after time,

In the remaining half an hour, the number of cars he will meet again is,


∴  Total number of cars would be meet on route will be 7.

Test: BITSAT Past Year Paper- 2019 - Question 36

In rotatory motion, linear velocities of all the particles of the body are

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 36

From v = r ω, linear velocities (v) for particles at different distances (r) from the axis of rotation are different.

Test: BITSAT Past Year Paper- 2019 - Question 37

If x, v and a denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period T, then, which of the following does not change with time?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 37

For an SHM, the acceleration a = -ω2x where, ω is a constant = 2π/T

The period of oscilation T is a constant.
∴ aT/x is a constant.

Test: BITSAT Past Year Paper- 2019 - Question 38

A conducting wire frame is placed in a magnetic field which is directed into the paper. The magnetic field is increasing at a constant rate.
The directions of induced current in wires AB and CD are

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 38

As the inward magnetic field increases, its flux also increases into the page and so induced current in bigger loop will be anticlockwise. i.e., from D to C in bigger loop and then from B to A in smaller loop.

Test: BITSAT Past Year Paper- 2019 - Question 39

Find the acceleration of block A and B. Assume pulley is massless.

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 39

Since A moves twice the distance moved by B.
If acceleration of B is ‘a’, then acceleration of A is ‘2a’.
T' – (T + T) = 0 (since pulley is massless)
⇒  T' = 2T ....(i)

For 5 kg block
5g – T' = 5a
for 2 kg block
⇒  5g – 2T = 5a ....(ii) [T’ = 2T]
T = 2 × (2a) = 4a ....(iii)
From equations (ii) and (iii),
5g – (2 × 4a) = 5a
5g – 8a = 5a
5g = 13a
a  = 5g/13

Test: BITSAT Past Year Paper- 2019 - Question 40

The nuclei of which one of the following pairs of nuclei are isotones?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 40

Isotones means equal number of neutrons
i.e.,   (A–Z) = 74 – 34 = 71 – 31 = 40

Test: BITSAT Past Year Paper- 2019 - Question 41

Plots showing the variation of the rate constant (k) with temperature (T) are given below. The plot that follows Arrhenius equation is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 41

As per Arrhenius equation (k = Ae - Ea / RT) , the rate constant increases exponentially with temperature.

Test: BITSAT Past Year Paper- 2019 - Question 42

3.6 g of oxygen is adsorbed on 1.2 g of metal powder. What volume of oxygen adsorbed per gram of the adsorbent at 1 atm and 273 K?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 42

Mass of O2 absorbed per gram of adsorbent = 3.6/1.2 = 3
No. of moles of O2 absorbed per gram of adsorbent = 3/32
Volume of O2 absorbed per gram of adsorbent
PV = nRT

Test: BITSAT Past Year Paper- 2019 - Question 43

In the purification of impure nickel by Mond's process, metal is purified by :

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 43

Test: BITSAT Past Year Paper- 2019 - Question 44

When chlorine water is added to an aqueous solution of sodium iodide in the presence of chloroform, a violet colouration is obtained. On adding more of chlorine water and vigorous shaking, the violet colour disappears. This shows the conversion of ...... into ......

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 44

3Cl2 + 2NaI  → 2NaCl + I2
I2 gives violet colouration in CHCl3.

Test: BITSAT Past Year Paper- 2019 - Question 45

In the clathrates of xenon with water, the nature of bonding between xenon and water molecule is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 45

Clathrate formation in volves dipole– induced dipole interaction.

Test: BITSAT Past Year Paper- 2019 - Question 46

The electronic configurations of Eu(Atomic No. 63), Gd(Atomic No. 64) and Tb (Atomic No. 65) are

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 46

Eu (63) = [Xe] 4 f7 6s2
Gd (64) = [Xe] 4 f7 5d1 6s2
Tb (65) = [Xe] 4 f9 6s2 

Test: BITSAT Past Year Paper- 2019 - Question 47

Which of the following carbonyls will have the strongest C – O bond ?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 47

As positive charge on the central metal atom increases, the less readily the metal can donate electron density into the π* orbitals of CO ligand (donation of electron density into π* orbitals of CO result in weakening of C – O bond). Hence, the C – O bond would be strongest in [Mn(CO)6]+.

Test: BITSAT Past Year Paper- 2019 - Question 48

How many chiral compounds are possible on monochlorination of 2- methyl butane ?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 48


Four monochloro derivatives are chiral.

Test: BITSAT Past Year Paper- 2019 - Question 49

Which of the following are intermediates in the reaction of excess of CH3MgBr with C6H5COOC2H5 to make 2-phenyl - 2-propanol?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 49


Test: BITSAT Past Year Paper- 2019 - Question 50

What is Z?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 50

Test: BITSAT Past Year Paper- 2019 - Question 51

Which of the following is the strongest base?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 51

In compound (b), the lone pair of nitrogen is not involved in resonance therefore it is the strongest base.

Test: BITSAT Past Year Paper- 2019 - Question 52

Which of the following does not reduce Benedict’s solution?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 52

Sucrose, being a non-reducing sugar, does not reduce Benedict’s solution. Remember that fructose has an α-hydroxy ketonic group, which is also reducing group (different from ordinary ketonic group)

Test: BITSAT Past Year Paper- 2019 - Question 53

General formula of solid in zinc blende structure is:

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 53

Zn+2 present in alternate tetrahedral void

S2– present in ccp = 4
∴ Zn4S4 = ZnS i.e., AB type compound.

Test: BITSAT Past Year Paper- 2019 - Question 54

Glycine in alkaline solution exists as ______ and migrates to __________.

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 54

Glycine in alkaline solution exists as anion and migrates to anode.

Due to internal proton tranfer of H+ from the –COOH group to the –NH2, the amino acid exists as an ion with both a negative charge and a positive charge, called a Zwitter ion

Adding an alkali to glycine

Now, during electrophoresis, glycine moves towards anode.

Test: BITSAT Past Year Paper- 2019 - Question 55

Product on reaction of ethanamide with phosphorus pentoxide is:

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 55

Test: BITSAT Past Year Paper- 2019 - Question 56

Ka of HX is 10–5, then find concentration of H3O+ when equal volumes of 0.25M HX and 0.05 M NaOH are mixed.

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 56


Test: BITSAT Past Year Paper- 2019 - Question 57

Net cell reaction of Pt | H2 (640 mm) | HCl | H2 (510 mm) | Pt.

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 57

Pt | H2 (640 mm) | HCl | H2 (510 mm) | Pt
E°cell = 0 
(p1)H2(g) → 2H++2e
2H+ + 2e → H2(g)(p2)

Test: BITSAT Past Year Paper- 2019 - Question 58

Which of the following has zero net dipole moment?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 58


XeF4 has zero net dipole moment

Test: BITSAT Past Year Paper- 2019 - Question 59

Which of the following element has the highest ionisation enthalpy?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 59

Boron has the highest ionisation enthalpy amongst the following.
Ionisation enthalpy decreases down the group and increases across the period.

Test: BITSAT Past Year Paper- 2019 - Question 60

Out of the elements with atomic number 7, 8, 9, 13 which has the smallest size and highest ionization enthalpy?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 60

Elemen t with atomic number 7 has the smallest size and highest ionization enthalpy Nitrogen – Atomic Number 7

N has a stable half-filled electronic configuration therefore it is difficult to remove electron and hence it has a high ionization enthalpy.

Test: BITSAT Past Year Paper- 2019 - Question 61

Which one is classified as a condensation polymer?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 61

Except dacron all are additive polymers.
Terephthalic acid condenses with ethylene glycol to give dacron.

Test: BITSAT Past Year Paper- 2019 - Question 62

Which of the following compounds is not an antacid?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 62

Phenelzine is an antidepressant, while others are antacids.

Test: BITSAT Past Year Paper- 2019 - Question 63

Mole fraction of the solute in a 1.00 molal aqueous solution is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 63

1 molal solution means 1 mole of solute dissolved in 1000 g solvent.
∴  nsolute = 1 wsolvent = 1000 g

Test: BITSAT Past Year Paper- 2019 - Question 64

The IUPAC name of the following compound is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 64

Test: BITSAT Past Year Paper- 2019 - Question 65

Most stable carbocation among the following is:

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 65

Stability of carbocation ∝ no. of α–H present on carbocation.

Test: BITSAT Past Year Paper- 2019 - Question 66

Which is correct for the following changes?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 66

Only Lindlar ’s catalyst converts alkyne to alkene (cis addition) and alkenes with Baeyer’s reagent give cis glycols.

Test: BITSAT Past Year Paper- 2019 - Question 67

The stability of +1 oxidation state among Al, Ga, In and Tl increases in the sequence :

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 67

Lower oxidation state become more stable on moving down the group
Al < Ga < ln < Tl

Test: BITSAT Past Year Paper- 2019 - Question 68

Which of the following alkaline earth metal hydroxides is amphoteric in character?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 68

Be(OH)2 is amphoteric while Ca(OH)2, Sr(OH)2 and Ba(OH)2 are all basic.

Test: BITSAT Past Year Paper- 2019 - Question 69

Which reaction shows oxidising nature of H2O2?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 69

H2O2 + 2KI → I2, O.S. of I (–1) changes to I2 (Zero) There is increase in oxidation number, hence oxidation.

Test: BITSAT Past Year Paper- 2019 - Question 70

aK2Cr2O7 + bKCl + cH2SO4 → xCrO2Cl2 + yKHSO4 + zH2O
The above equation balances when

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 70

The balanced equation is
K2Cr2O+ 4KCl + 6H2SO4 → 2CrO2Cl+ 6KHSO4 + 3H2O

Test: BITSAT Past Year Paper- 2019 - Question 71

For the reactions 
A ⇌ B ; Kc = 2
B ⇌ C ; Kc = 4
C ⇌ D ; Kc = 6
Kc for the reaction A ⇌ D is 

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 71


Multiply the three equations,

Test: BITSAT Past Year Paper- 2019 - Question 72

Which of the following will always lead to a non-spontaneous change?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 72

ΔG = ΔH – TΔS = +ve for spontaneous change, ΔH < 0, ΔS > 0
for non-spontaneous change, ΔH > 0, ΔS < 0

Test: BITSAT Past Year Paper- 2019 - Question 73

The densities of two gasses are in the ratio of 1: 16. The ratio of their rates of diffusion is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 73

Test: BITSAT Past Year Paper- 2019 - Question 74

In the reaction the change in hybridisation is from

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 74

Test: BITSAT Past Year Paper- 2019 - Question 75

The group having isoelectronic species is:

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 75

Isoelectronic species have same no. of electrons.


∴ O2–, F, Na+ , Mg+2 are isoelectronic

Test: BITSAT Past Year Paper- 2019 - Question 76

100 mL O2 and H2 kept at same temperature and pressure. What is true about their number of molecules

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 76

This is Avogadro’s hypothesis. According to this, equal volume of all gases contain equal no. of molecules under similar condition of temperature and pressure.

Test: BITSAT Past Year Paper- 2019 - Question 77

If mA gram of a metal A displaces mB gram of another metal B from its salt solution and if the equivalent mass are EA and EB respectively then equivalent mass of A can be expressed as:

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 77

Eq. of A = Eq. of B

Test: BITSAT Past Year Paper- 2019 - Question 78

Which one of the following set of quantum numbers is not possible for 4p electron?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 78

For 4p electron n = 4, l = 1, m = –1, 0, + 1 and s = +½ or –½

Test: BITSAT Past Year Paper- 2019 - Question 79

Which of the following radial distribution graphs correspond to l = 2 for the H atom?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 79

l = 2 represent d orbital for which

Test: BITSAT Past Year Paper- 2019 - Question 80

Which of the following is paramagnetic?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 80

B2 is paramagnetic due to the presence of unpaired electron in π2px = π2py orbital.


M.O diagram for B2 molecule

Test: BITSAT Past Year Paper- 2019 - Question 81

Direction: In the following questions below, out of the four alternatives, choose the one which best expresses the meaning of the given word.

Garrulous

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 81

The word Garrulous (Adjective) means : talkative; talking a lot.

Test: BITSAT Past Year Paper- 2019 - Question 82

Direction: In the following questions below, out of the four alternatives, choose the one which best expresses the meaning of the given word.

Tinsel

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 82

The word Tinsel (Noun/Adjective) means : strips of shiny material like metal used as decorations.

Test: BITSAT Past Year Paper- 2019 - Question 83

Direction: In the following questions below, out of the four alternatives, choose the one which best expresses the meaning of the given word.

Labyrinth

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 83

The word Labyrinth (Noun) means : a place that has many confusing paths or passage.
The correct synonym  will be 'meandering' which means, 'to have a lot of curves on a path'.

Test: BITSAT Past Year Paper- 2019 - Question 84

Direction: In the following questions, choose the word opposite in meaning to the given word.

Knack :

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 84

Knack means a clever way of doing something.

Test: BITSAT Past Year Paper- 2019 - Question 85

Direction: In the following questions, choose the word opposite in meaning to the given word.

Pernicious :

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 85

Pernicious means highly injurious  or destructive.

Test: BITSAT Past Year Paper- 2019 - Question 86

Direction: In the following questions, choose the word opposite in meaning to the given word.

Opulence :

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 86

Opulence means wealthy.

Test: BITSAT Past Year Paper- 2019 - Question 87

Direction: Read the passage carefully and choose the best answer to each question out of the four alternatives and mark it by blackening the appropriate circle [·].

Like watering a plant, we grow our friendships [and all our relationships) by running them. Friendships need the same attention as other relationships. If they are to continue. These relationships can be delightfully non-judgemental, supportive, understanding and fun.
Sometimes a friendship can  bring out the positive side that you never show in any other relationship. This may be because the pressure of playing a 'role' (daughter, partner or child) is removed.
With a friend you are to be yourself and free to change.
Of course, you are free  to do this in all other relationships as well, but in friendships you get to have lats of rehearsals and discussion about changes as you experience them. It is an unconditional experience where you receive as much as you give.
You can explain yourself to a friend openly without the fear of hurting a family member. How do friendships grow ? The answer is simple. By revealing yourself; being attentive: remembering what is most showing empathy; seeing the world through the eyes of your friend, you will understand the value of friendship. All this means learning to accept a person from a completely different family to your own or perhaps someone from a completely different cultural background. This is the way we learn tolerance. In turn we gain tolerance and acceptance for our own differences.

Q. In good friendships, we

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 87

In good friendships, we receive as much as we give.

Test: BITSAT Past Year Paper- 2019 - Question 88

Direction: Read the passage carefully and choose the best answer to each question out of the four alternatives and mark it by blackening the appropriate circle [·].

Like watering a plant, we grow our friendships [and all our relationships) by running them. Friendships need the same attention as other relationships. If they are to continue. These relationships can be delightfully non-judgemental, supportive, understanding and fun.
Sometimes a friendship can  bring out the positive side that you never show in any other relationship. This may be because the pressure of playing a 'role' (daughter, partner or child) is removed.
With a friend you are to be yourself and free to change.
Of course, you are free  to do this in all other relationships as well, but in friendships you get to have lats of rehearsals and discussion about changes as you experience them. It is an unconditional experience where you receive as much as you give.
You can explain yourself to a friend openly without the fear of hurting a family member. How do friendships grow ? The answer is simple. By revealing yourself; being attentive: remembering what is most showing empathy; seeing the world through the eyes of your friend, you will understand the value of friendship. All this means learning to accept a person from a completely different family to your own or perhaps someone from a completely different cultural background. This is the way we learn tolerance. In turn we gain tolerance and acceptance for our own differences.

Q. Empathy means

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 88

Empathy means the ability to show and understand the feelings of others.

Test: BITSAT Past Year Paper- 2019 - Question 89

Direction: Read the passage carefully and choose the best answer to each question out of the four alternatives and mark it by blackening the appropriate circle [·].

Like watering a plant, we grow our friendships [and all our relationships) by running them. Friendships need the same attention as other relationships. If they are to continue. These relationships can be delightfully non-judgemental, supportive, understanding and fun.
Sometimes a friendship can  bring out the positive side that you never show in any other relationship. This may be because the pressure of playing a 'role' (daughter, partner or child) is removed.
With a friend you are to be yourself and free to change.
Of course, you are free  to do this in all other relationships as well, but in friendships you get to have lats of rehearsals and discussion about changes as you experience them. It is an unconditional experience where you receive as much as you give.
You can explain yourself to a friend openly without the fear of hurting a family member. How do friendships grow ? The answer is simple. By revealing yourself; being attentive: remembering what is most showing empathy; seeing the world through the eyes of your friend, you will understand the value of friendship. All this means learning to accept a person from a completely different family to your own or perhaps someone from a completely different cultural background. This is the way we learn tolerance. In turn we gain tolerance and acceptance for our own differences.

Q. Through strong friendships, we gain

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 89

A strong friendship helps us gain acceptance and tolerance.

Test: BITSAT Past Year Paper- 2019 - Question 90

Direction: Read the passage carefully and choose the best answer to each question out of the four alternatives and mark it by blackening the appropriate circle [·].

Like watering a plant, we grow our friendships [and all our relationships) by running them. Friendships need the same attention as other relationships. If they are to continue. These relationships can be delightfully non-judgemental, supportive, understanding and fun.
Sometimes a friendship can  bring out the positive side that you never show in any other relationship. This may be because the pressure of playing a 'role' (daughter, partner or child) is removed.
With a friend you are to be yourself and free to change.
Of course, you are free  to do this in all other relationships as well, but in friendships you get to have lats of rehearsals and discussion about changes as you experience them. It is an unconditional experience where you receive as much as you give.
You can explain yourself to a friend openly without the fear of hurting a family member. How do friendships grow ? The answer is simple. By revealing yourself; being attentive: remembering what is most showing empathy; seeing the world through the eyes of your friend, you will understand the value of friendship. All this means learning to accept a person from a completely different family to your own or perhaps someone from a completely different cultural background. This is the way we learn tolerance. In turn we gain tolerance and acceptance for our own differences.

Q. Friendships and relationships grow when they are

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 90

The very first line of the passage states that friendships  and relationships grow when they are nurtured just like nurturing a plant.

Test: BITSAT Past Year Paper- 2019 - Question 91

Direction: In the following questions, sentences are given with blanks to be filled with an appropriate word(s). Four alternatives are suggested for each question. Choose the correct alternative out of the four as your answer.

Q. There are not solitary, free-living creatures ; every form of life is ______ other forms.

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 91

Dependent on = needing som ebod y / something in order to survive or be successful; affected or decided by something.

Test: BITSAT Past Year Paper- 2019 - Question 92

Direction: In the following questions, sentences are given with blanks to be filled with an appropriate word(s). Four alternatives are suggested for each question. Choose the correct alternative out of the four as your answer.

Q. I'll take ______ now as I have another's appointment some where else.

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 92

Take your leave = to say good bye.

Test: BITSAT Past Year Paper- 2019 - Question 93

Direction: In the following questions, some parts of the sentences have errors and some are correct. Find out which part of a sentence has an error. The number of that part is the answer. If a sentence is free from error, then your answer is (d). i.e., No error.

When one hears of the incident (a)/about the plane crash (b)/ he feels very sorry. (c)/ No error (d)

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 93

Here, indefinite article i.e., 'about a plane crash' should be used. No particular incident is evident here.

Test: BITSAT Past Year Paper- 2019 - Question 94

Direction: In the following questions, some parts of the sentences have errors and some are correct. Find out which part of a sentence has an error. The number of that part is the answer. If a sentence is free from error, then your answer is (d). i.e., No error.

I went there (a)/ with a view to survey (b)/ the entire procedure. (c)/ No error (d)

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 94

‘With a View to’ should be followed by gerund i.e., surveying.

Test: BITSAT Past Year Paper- 2019 - Question 95

Direction: In the following questions, some parts of the sentences have errors and some are correct. Find out which part of a sentence has an error. The number of that part is the answer. If a sentence is free from error, then your answer is (d). i.e., No error.

It had laid (a)/ in the closet (b)/ for a week before we found it. (c)/ No error (d)

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 95

Here, time period is given. Hence, Past Perfect Continuous i.e., 'It had been lying' ....should be used.

Test: BITSAT Past Year Paper- 2019 - Question 96

Direction: In the following questions, which answer figure will complete the question figure?

Question Figures :

Test: BITSAT Past Year Paper- 2019 - Question 97

Direction: In the following questions, which answer figure will complete the question figure?

Question Figure:

Test: BITSAT Past Year Paper- 2019 - Question 98

A piece of paper is folded and cut/punched as shown below in the question figures. From the given answer figures, indicate how it will A appear when opened.

Question figures:

Test: BITSAT Past Year Paper- 2019 - Question 99

Select there lated word from the given alternatives:

Medicine : Patient : : Education : ?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 99

Medicine is given to patient. Similar ly, Education is given to student.

Test: BITSAT Past Year Paper- 2019 - Question 100

Choose the correct alternative from the given ones that will complete the series.
A3E, F5J, K7O,  _____?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 100

Test: BITSAT Past Year Paper- 2019 - Question 101

Which one of the following numbers lacks the common property in the series?
81, 36, 25, 9, 5, 16

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 101

Except  5, all numbers are perfect square numbers.

Test: BITSAT Past Year Paper- 2019 - Question 102

In a certain code language, "TIRED" is written as "56" and "BRAIN" is written as "44". How is 'LAZY" written in that code language?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 102

As,
TIRED = 20 + 9 + 18 + 5 + 4 = 56
BRAIN = 2 + 18 + 1 + 9 + 14 = 44
Similarly,
LAZY = 12 + 1 + 26 + 25 = 64.

Test: BITSAT Past Year Paper- 2019 - Question 103

Select the missing number from the given response.

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 103

8 × 8 × 88 = 5632
7 × 7 × 77 = 3773
Similarly, 6 × 6 × ? = 3132

Test: BITSAT Past Year Paper- 2019 - Question 104

Which one of the following diagrams best depicts the relationship among Human Society - Youth Club, Political Party and Youths?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 104

Test: BITSAT Past Year Paper- 2019 - Question 105

Among her children, Ganga's favourites are Ram and Rekha. Rekha is the mother of Sharat, who is loved most by his uncle Mithun. The head of the family is Ram Lal, who is succeeded by his sons Gopal and Mohan. Gopal and Ganga have been married for 35 years and have 3 children.What is the relation between Mithun and Mohan?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 105

Mohan is son of Ram Lal and uncle of Ram and Rekha. Mithun is uncle of Sharat who is son of Rekha. Rekha is niece of Mohan.
Therefore, Mithun is brother of Rekha's husband.

Test: BITSAT Past Year Paper- 2019 - Question 106

If x cos α + y sin α = P is a tangent to the ellipse

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 106

Given line is x cos α + y sin α = P ....(1)
Any tangent to the ellipse is


Comparing (1) and (2)


Eliminate θ, cos2θ + sin2θ

or a2 cos2 α + b2 sin2 α = P2 

Test: BITSAT Past Year Paper- 2019 - Question 107

If a1 , a2 , a3........,an are in A.P. where ai > 0 for all i, then

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 107

As a1, a2 , a3 , .......,an , are in A.P. we get, a2 - a1 = a3 - a2 = ............. = an - an-1 = d (say)


 [formula for nth term]

Test: BITSAT Past Year Paper- 2019 - Question 108

In order to solve the differential equation the integrating factor is:

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 108

Given differential equation is :
x cos x dy/dx + y (x sin x + cos x) = 1 Dividing both the sides by x cos x,

which is of the form 


Integrating factor 

= e(log sec x + log x) = elog (sec x . x) = x sec x

Test: BITSAT Past Year Paper- 2019 - Question 109

Equation of two straight lines are


Then

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 109

Equation of the first line L1 is (x-1)/2 = (y-2)/3 = (z-3)/4 and that of the second line


Clearly, these lines are not parallel (the ratios of D.R. are not equal).
Any point P on the first line is (1 + 2λ, 2 + 3λ, 3 + 4λ) and any point Q on the second line is (4 + 5μ, 1 + 2μ, μ). If these two points P and Q are identical then.
1 + 2λ =  4 + 5μ ...(1)
2 + 3λ =  1 + 2μ ...(2)
3 + 4λ =  μ ...(3)
From (2) and (3), we get λ = μ = –1, which also satisfies (1). Thus the two lines L1 and L2 ; entersect and the coordinates of the point of intersection are (– 1, – 1, – 1).

Test: BITSAT Past Year Paper- 2019 - Question 110

The equation of the curve  passing through the point  and satisfying the differential equation 

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 110

⇒ ydx - xdy = ay2 dx + ady
⇒ y(1 - ay)dx = (x + a)dy

Integrating, we get
log(x + a) - log y + log(1 - ay) = log C


Since the curve passes through 

So, (x + a )(1 - ay) = -4a2y

Test: BITSAT Past Year Paper- 2019 - Question 111

The locus of the mid-point of a chord of the circle x2 + y2 = 4 , which subtends a right angle at the origin is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 111


Equation of given circle is  x2 +y= 4 Its centre , O =   (0, 0) and radius, r = 2 Draw OM ⊥ AB
Clearly M is the mid-point of AB which subtends a right angle at O.
In ΔAOB, OA = OB radius
∴ ∠A = ∠B = π/4
and in ΔOMA, sin A = OM/OA

⇒ OM = √2    ...(1)
Let M = (x, y) then OM   ....(2)
From (1) and (2), x2 + y= 2
This is the required equation of locus.

Test: BITSAT Past Year Paper- 2019 - Question 112

With the usual notation  is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 112


Test: BITSAT Past Year Paper- 2019 - Question 113

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 113


Test: BITSAT Past Year Paper- 2019 - Question 114

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 114



Squaring both sides, we get
x2(1+ y) = y2(1 + x)
⇒ x2 – y2 + x2y – xy2 = 0
⇒ (x – y) (x + y + xy) = 0
⇒ y = x or y(1 + x) = – x ⇒ y = x or 

Test: BITSAT Past Year Paper- 2019 - Question 115

If f (x) = 3x4 + 4x3 – 12x2 + 12, then f (x) is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 115

Given :  f(x) = 3x4 + 4x3 – 12x2 + 12
Differentiating with respect to x, we get
f'(x) = 12x3+ 12x2– 24x
For f (x) to be increasing
f '(x) > 0 ⇒ 12x3 + 12x2 – 24x  >  0
⇒  12x (x2 + x – 2)  >  0
⇒  12x (x – 1) (x + 2)  >  0
⇒  x (x – 1) (x + 2) > 0
⇒  – 2 < x < 0     or       x > 1

It means x ∈ ( -2, 0) ∪ (1,∞).
Hence f (x) is increasing in (– 2, 0) and (1,∞)

Test: BITSAT Past Year Paper- 2019 - Question 116

Consider 
Then number of possible solutions are :

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 116


x, y ≥ 0 convert them into equation and solve them and draw the graph of these equations we get y = 1 and x = 3/2

Test: BITSAT Past Year Paper- 2019 - Question 117

The distance of a point (2, 5, –3) from the plane 

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 117

Here, and d = 4.
Therefore, the distance of the point (2, 5, –3) from the given plane is

Test: BITSAT Past Year Paper- 2019 - Question 118

The value of definite integral  

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 118


Test: BITSAT Past Year Paper- 2019 - Question 119

For the following feasible region, the linear constraints are

Test: BITSAT Past Year Paper- 2019 - Question 120

The general solution of differential equation (ex + 1) ydy = (y + 1)ex dx is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 120

Since, (ex + 1) ydy = (y + 1)ex dx


After integrating on both sides, we have

Hence y = log [k (1 + y) (1 + ex)]

Test: BITSAT Past Year Paper- 2019 - Question 121

What is the slope of the normal at the point (at2, 2at) of the parabola y2 = 4ax?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 121

Equation of parabola is y2 = 4ax

 (On differentiating w.r.t ‘x’)
∴ dy/dx = 2a/y , [slope of tangent]
So, slope of normal

Test: BITSAT Past Year Paper- 2019 - Question 122

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 122


From the definite integral property


we have


By adding (i) and (ii)



[∵ sin 2x = 2 sin x cos x]



Test: BITSAT Past Year Paper- 2019 - Question 123

 then what is x equal to ?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 123


= 6i [3i2 + 3] + 3i [4i + 20] + 1 [12 – 60i]
= 6i [–3 + 3] + 12i2 + 60i + 12 – 60i
= –12 + 12 = 0 = x + iy
∴ x = 0

Test: BITSAT Past Year Paper- 2019 - Question 124

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 124

Test: BITSAT Past Year Paper- 2019 - Question 125

If 2 cos2 x + 3 sin x – 3 = 0, 0 ≤ x ≤ 180°, then x =

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 125

2 cos2 x + 3 sin x – 3 = 0
2 – 2 sin2 x + 3 sin x – 3 = 0
⇒ (2 sin x – 1) (sin x – 1) = 0
⇒ sin x = 1/2 or sin x = 1

Test: BITSAT Past Year Paper- 2019 - Question 126

If the number of available constraints is 3 and the number of parameters to be optimized is 4, then

Test: BITSAT Past Year Paper- 2019 - Question 127

 then y'(1) is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 127

On differentiating w.r.t. x, we get

Test: BITSAT Past Year Paper- 2019 - Question 128

The maximum area of rectangle inscribed in a circle of diameter R is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 128

The diagonal = R
Thus the area of rectangle

Test: BITSAT Past Year Paper- 2019 - Question 129

If A and B are two events, such that P(A∪B) = 3/4, P(A∩B) = 1/4 , P(Ac) = 2/3
where Ac stands for the complementary event of A, then P(B) is given by:

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 129

From the given problem: P(A ∪ B) = 3/4 , P(A ∩ B) = 1/4

Test: BITSAT Past Year Paper- 2019 - Question 130

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 130



Limit doesn't exist, so f (x) is not continuous at 0.

Test: BITSAT Past Year Paper- 2019 - Question 131

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 131

Put x = cos 2θ

Test: BITSAT Past Year Paper- 2019 - Question 132

The equation of chord of the circle x2 + y2 = 8x bisected at the point (4, 3) is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 132

T = S1 ⇒  x(4)  + y(3) – 4 (x + 4) = 16 + 9 – 32
⇒ 3y – 9 = 0  ⇒ y = 3

Test: BITSAT Past Year Paper- 2019 - Question 133

x and y are positive number. Let g and a be G. M. and AM of these numbers. Also let G be G. M. of x + 1 and y + 1. If G and g are roots of equation x2 – 5x + 6 = 0, then

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 133

The roots of equation are 2 and 3


∴  x = y = 2

Test: BITSAT Past Year Paper- 2019 - Question 134

The co-efficient of xn in the expansion of

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 134


Test: BITSAT Past Year Paper- 2019 - Question 135

A pair of tangents are drawn from the origin to the circle x2 + y2+ 20 (x + y) + 20 = 0, then the equation of the pair of tangent are

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 135

Equation of pair of tangents is given by SS1 = T2,
or S = x2 + y2 + 20 (x + y) + 20, S, = 20,
T = 10 (x + y) + 20 = 0
∴ SS1 = T2
⇒ 20 (x2 + y2 + 20 (x + y) + 20)
= 102 (x + y + 2)2
⇒ 4x2 + 4y2 + 10xy = 0
⇒ 2x2 + 2y2 + 5xy = 0

Test: BITSAT Past Year Paper- 2019 - Question 136

If the sum of a certain number of terms of the A.P. 25, 22, 19, ........ is 116. then the last term is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 136

a = 25, d = 22 – 25 = –3.
Let n be the no. of terms

or  232 = n[50 – 3n + 3] = n[53- 3n]
= –3n2 + 53n
⇒ 3n2 – 53 + 232 = 0  ⇒ (n – 8) (3n – 29) = 0

∴ Now, T8 = a + (8 – 1)d = 25 + 7 × (–3)
= 25 – 21
∴ Last term = 4

Test: BITSAT Past Year Paper- 2019 - Question 137

If 1, a and P are in A. P. and 1, g and P are in G. P., then

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 137

2a = 1 + P and g2 = P
⇒ g2 = 2a -1 ⇒ 1- 2a +g2 = 0

Test: BITSAT Past Year Paper- 2019 - Question 138

If y = sin x + ex, then d2x/dy2 is equal to 

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 138

y = sin x = ex 

Test: BITSAT Past Year Paper- 2019 - Question 139

The foci of the hyperbola 4x2 – 9y2 – 1 = 0 are

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 139

4x2 – 9y2 = 1



Test: BITSAT Past Year Paper- 2019 - Question 140

From the top of a cliff 50 m high, the angles of depression of the top and bottom of a tower are observed to be 30° and 45°. The height of tower is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 140

Let height of the tower be h m and distance between tower and cliff be x m.

∴ CD = h, BD = x
In ΔABD,  tan 45° = AB/BD or 1 = 50/x  
x = 50      ... (i)
In ΔAEC

or  50 = 50√3- h√3
[From (i), x = 50]
or  h√3 = 50√3 - 50

Test: BITSAT Past Year Paper- 2019 - Question 141

The coefficient of x2 term in the binomial expansion of 

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 141

General term of the given binomial series is given by:

Put r = 4, we get

Thus coefficient of x2 = 70/243.

Test: BITSAT Past Year Paper- 2019 - Question 142

The value of λ, for which the circle x2 + y2 + 2λx + 6y + 1 = 0 intersects the circle x2 + y2 + 4x + 2y = 0 orthogonally, is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 142

Two circles x2 + y2 + 2g1x + 2f1y + c1 = 0 and x2 + y2 + 2g2x + 2f2y + c2 = 0  cuts orthogonally if 2g1g2 + 2f1 f2 = c1 + c2
Given equations of two circles are
x2 + y2 + 2λx + 6y + 1 = 0 .... (i)
x2 + y2 +4x + 2y = 0 .... (ii)
On comparing (i) and (ii) with original equation, we get g1 = λ, f1 = 3, c1 = 1 and g2 = 2,f2 = 1, c2 = 0
So, from orthogonality condition, we have 4λ + 6 = 1 ⇒ 4λ = -5
∴ λ = -5/4

Test: BITSAT Past Year Paper- 2019 - Question 143

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 143

We know, scalar triple product


(By definition of scalar triple product)

Test: BITSAT Past Year Paper- 2019 - Question 144

If f (x) = (a - xn)1/n , where a > 0 and n ∈ N , then fof (x) is equal to :

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 144

Given that f(x) = (a - xn)1/n
∴  fof (x ) = [a - {(a - xn)1/n}n]1/n
= [a- (a- xn)]1/n
= [xn]1/n = x

Test: BITSAT Past Year Paper- 2019 - Question 145

Sum of n terms of the series 8 + 88 + 888 + .... equals

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 145

Test: BITSAT Past Year Paper- 2019 - Question 146

The modulus of the complex number z such that | z + 3 – i | = 1 and arg(z) = π is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 146

Let z = x + iy
∴ | z + 3 – i | = | (x + 3) + i(y – 1) | = 1

From equations (i) and (ii), we get x = –3, y = 0 ∴ z = –3
⇒ | z | = | –3 | = 3

Test: BITSAT Past Year Paper- 2019 - Question 147

Bag P contains 6 red and 4 blue balls and bag Q contains 5 red and 6 blue balls. A ball is transferred from bag P to bag Q and then a ball is drawn from bag Q. What is the probability that the ball drawn is blue?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 147

Let E1, E2 and A be the even ts defined as follows:
E1 = red ball is transferred from bag P to bag Q
E2 = blue ball is transferred from bag P to bag Q
A = the ball drawn from bag Q is blue
As the bag P contains 6 red and 4 blue balls,

Note that E1 and E2 are mutually exclusive and exhaustive events.
When E1 has occurred i.e., a red ball has already been transferred from bag P to Q, then bag Q will contain 6 red and 6 blue balls, So, P(A|E1) = 6/12 = 1/2
When E2 has occurred i.e., a blue ball has already been transferred from bag P to Q, then bag Q will contain 5 red and 7 blue balls, So, P(A|E2) = 7/12
By using law of total probability, we get P(A) = P(E1) P(A|E1) + P(E2) P(A|E2)

Test: BITSAT Past Year Paper- 2019 - Question 148

The number of 4-digit numbers that can be formed with the digits 1, 2, 3, 4 and 5 in which at least 2 digits are identical, is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 148

Total number of 4-digit numbers = 5 × 5 × 5 × 5 = 625
(as each place can be filled by anyone of the numbers 1, 2, 3, 4 and 5)
Numbers in which no two digits are identical = 5 × 4 × 3 × 2 = 120 (i.e. repetition not allowed) (as 1st place can be filled in 5 different ways, 2nd place can be filled in 4 different ways and so on)
Number of 4-digits numbers in which at least 2 digits are identical
= 625 – 120 = 505

Test: BITSAT Past Year Paper- 2019 - Question 149

Consider the system of linear equations;
x1 + 2x2 + x3 = 3
2x1 + 3x2 + x3 = 3
3x1 + 5x2 + 2x3 = 1
The system has

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 149


⇒  Given system, does not have any solution.
⇒  No solution

Test: BITSAT Past Year Paper- 2019 - Question 150

What is the value of y so that the line through (3, y) and (2, 7) is parallel to the line through (– 1, 4) and (0, 6)?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 150

Let A(3, y), B(2, 7), C(–1, 4) and D(0, 6) be the given points.


Since AB and CD are parallel.  
∴ m1 = m2 ⇒ y = 9.

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