Test: BITSAT Past Year Paper- 2019 - JEE MCQ

# Test: BITSAT Past Year Paper- 2019 - JEE MCQ

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## 150 Questions MCQ Test - 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.

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

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

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.                      log(17)  =  1.23044892138

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