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Test: BITSAT Past Year Paper- 2018 - Question 1

Four point charges –Q, –q, 2q and 2Q are placed, one at each corner of the square. The relation between Q and q for which the potential at the centre of the square is zero is :​

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 1

Let the side length of square be 'a' then potential at centre O is


= – Q – q + 2q + 2Q = 0 ⇒ Q + q = 0 (Given) Q
= – q

Test: BITSAT Past Year Paper- 2018 - Question 2

Two long parallel wires carry equal current i flowing in the same direction are at a distance 2d apart. The magnetic field B at a point lying on the perpendicular line joining the wires and at a distance x from the midpoint is –

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 2

The magnetic field due to two wires at P


Both the magnetic fields act in opposite direction.

Test: BITSAT Past Year Paper- 2018 - Question 3

In the circuit shown, the symbols have their usual meanings. The cell has emf E. X is initially joined to Y for a long time. Then, X is joined to Z. The maximum charge on C at any later time will be

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 3

Current in inductor = E/R
∴  its energy = 1/2 LE2/R2
Same energy is later stored in capacitor

Test: BITSAT Past Year Paper- 2018 - Question 4

A point object O is placed in front of a glass rod having spherical end of radius of curvature 30 cm. The image would be formed at

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 4

Using, μ/v - 1/u = μ - 1/R

Or 1.5/v - 1/-15 = 1.5 - 1/+30
∴ v = - 30cm

Test: BITSAT Past Year Paper- 2018 - Question 5

In Young’s double slit exper iment, λ = 500n m, d = 1mm, D = 1m. Minimum distance from the central maximum for which intensity is half of the maximum intensity is

Test: BITSAT Past Year Paper- 2018 - Question 6

What is the voltage gain in a common emitter amplifier, where input resistance is 3 Ω and load resistance 24 Ω, b = 0.6 ?

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 6

Voltage gain,  Av = β RL/Ri = 0.6 x 24/3 = 4.8

Test: BITSAT Past Year Paper- 2018 - Question 7

The acceleration due to gravity on the surface of the moon is 1/6 that on the surface of earth and the diameter of the moon is one-fourth that of earth. The ratio of escape velocities on earth and moon will be​

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 7

Test: BITSAT Past Year Paper- 2018 - Question 8

Given The magnitude of their resultant is

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 8

Test: BITSAT Past Year Paper- 2018 - Question 9

A particle of mass m executes simple harmonic motion with amplitude a and frequency n. The average kinetic energy during its motion from the position of equilibrium to the end is

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 9

The kinetic energy of a particle executing S.H.M. is given by
K = 1/2 ma2 ω2 sin2ωt
= 1/2 mω2a2(1/2) (∵ < sin2 θ > = 1/2)
= 1/4 mω2a2 = 1/4 ma2 (2πv)2 (∵ ω = 2πv)
or, <K> = πma2v2

Test: BITSAT Past Year Paper- 2018 - Question 10

The dipole moment of the given charge distribution is

Test: BITSAT Past Year Paper- 2018 - Question 11

At a place, if the earth's horizontal and vertical components of magnetic fields are equal, then the angle of dip will be

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 11

tan θ = By/B= 1 ∴ θ = 45°

Test: BITSAT Past Year Paper- 2018 - Question 12

The third line of Balmer series of an ion equivalnet to hydrogen atom has wavelength of 108.5 nm.The ground state energy of an electron of this ion will be

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 12

For third line of Balmer series n1 = 2, n2 = 5

On putting values Z = 2 From
E

Test: BITSAT Past Year Paper- 2018 - Question 13

The binding energy per nucleon of 10X is 9 MeV and that of 11X is 7.5 MeV where X represents an element. The minimum energy required to remove a neutron from 11X is

Test: BITSAT Past Year Paper- 2018 - Question 14

If C, the velocity of light, g the acceleration due to gravity and P the atmospheric pressure be the fundamental quantities in MKS system, then the dimensions of length will be same as that of​

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 14

Test: BITSAT Past Year Paper- 2018 - Question 15

Figure shows a capillary rise H. If the air is blown through the horizontal tube in the direction as shown then rise in capillary tube will be​

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 15

Due to increase in velocity, pressure will be low above the surface of water.

Test: BITSAT Past Year Paper- 2018 - Question 16

A boy running on a horizontal road at 8 km/h finds the rain falling vertically. He increases his speed to 12 km/h and finds that the drops makes 30° with the vertical. The speed of rain with respect to the road is

Test: BITSAT Past Year Paper- 2018 - Question 17

A hunter aims his gun and fires a bullet directly at a monkey on a tree. At the instant the bullet leaves the barrel of the gun, the monkey drops.
Pick the correct statement regarding the situation.

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 17


t = OC/u cos θ = x/u cos θ

AC = x tan q
BC = distance travelled by bullet in time t, vertically.
y = u sin θ t – 1/2 gt2
AB = x tan θ – (u sin θt – 1/2 gt2)
= x tan θ – (usinθ × x/u cos θ  – 1/2 gt2)
(∴ bullet will always hit the monkey)

Test: BITSAT Past Year Paper- 2018 - Question 18

A particle of mass m1 moving with velocity v collides with a mass m2 at rest, then they get embedded. Just after  collision, velocity of the system

Test: BITSAT Past Year Paper- 2018 - Question 19

The ratio of the specific heats of a gas is Cp/Cv = 1.66, then the gas may be

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 19

Let ‘n’ be the degree of freedom

⇒ n = 3 ⇒ gas must be monoatomic.

Test: BITSAT Past Year Paper- 2018 - Question 20

Two oscillators ar e started simultaneously in same phase. After 50 oscillations of one, they get out of phase by p, that is half oscillation. The percentage difference of frequencies of the two oscillators is nearest to

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 20

Phase change π in 50 oscillations.
Phase change 2π in 100 oscillations.
So frequency different ~ 1 in 100.

Test: BITSAT Past Year Paper- 2018 - Question 21

A juggler keeps on moving four balls in the air throwing the balls after intervals. When one ball leaves his hand (speed = 20 ms–1) the position of other balls (height in m) will be (Take g = 10 ms–2)

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 21

Time taken by same ball to return to the hands of juggler = 2u/g = 2 x 20/10 = 4 S. So he is throwing the balls after each 1 s. Let at some instant he is throwing ball number 4. Before 1 s of it he throws ball. So height of ball 3:
h3 = 20 × 1 – 1/2 10(1)2 = 15
Before 2s, he throws ball 2. So height of ball 2:
h2 = 20 × 2 – 1/2 10(2)2 = 20 m
Before 3 s, he throws ball 1. So height of ball 1:
h1 = 20 × 3 – 1/2 10(3)2 = 15 m

Test: BITSAT Past Year Paper- 2018 - Question 22

If a stone of mass 0.05 kg is thrown out a window of a train moving at a constant speed of 100 km/ h then magnitude of the net force acting on the stone is

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 22

After the stone is thrown out of the moving train, the only force acting on it is the force of gravity i.e. its weight.
∴ F = mg = 0.05 × 10 = 0.5 N.

Test: BITSAT Past Year Paper- 2018 - Question 23

A body of mass M hits normally a rigid wall with velocity V and bounces back with the same velocity. The impulse experienced by the body is

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 23

Impulse experienced by the body = MV – (–MV) = 2MV.

Test: BITSAT Past Year Paper- 2018 - Question 24

A hoop rolls down an inclined plane. The fraction of its total kinetic energy that is associated with rotational motion is

Test: BITSAT Past Year Paper- 2018 - Question 25

Infinite number of masses, each 1 kg are placed along the x-axis at x = ± 1m, ± 2m, ± 4m, ± 8m, ± 16m ..... the magnitude of the resultant gravitational potential in terms of gravitational constant G at the orgin (x = 0) is

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 25

Test: BITSAT Past Year Paper- 2018 - Question 26

Water of volume 2 litre in a container is heated with a coil of 1 kW at 27°C. The lid of the container is open and energy dissipates at rate of 160 J/s.
In how much time temperature will rise from 27°C to 77°C? [Given specific heat of water is 4.2 kJ/kg]

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 26

Heat gained by the water = (Heat supplied by the coil) – (Heat dissipated to environment)
⇒ mc Δθ = PCoil t – PLoss t
⇒ 2 × 4.2 × 103 × (77 – 27) = 1000 t – 160 t
⇒ t = 4.2x105/840 = 500 s = 8 min 20 s

Test: BITSAT Past Year Paper- 2018 - Question 27

In the following P-V diagram of an ideal gas, two adiabates cut two isotherms at T1 = 300K and T2 = 200K. The value of VA = 2 unit, VB = 8 unit, VC = 16 unit. Find the value of VD.

Test: BITSAT Past Year Paper- 2018 - Question 28

The mass of H2 molecule is 3.32 × 10–24 g. If 1023 hydrogen molecules per second strike 2 cm2 of wall at an angle of 45o with the normal, while moving with a speed of 105 cm/s, the pressure exterted on the wall is nearly.

Test: BITSAT Past Year Paper- 2018 - Question 29

The wavelength of two waves are 50 and 51 cm respectively. If the temperature of the room is 20°C then what will be the number of beats produced per second by these waves, when the speed of sound at 0°C is 332 m/s?

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 29

λ1= 50 cm. λ2= 51 cm.

⇒ v2 = 319.23.
v= v2= 319.23/0.50 = 640 Hz.
v2 = v22 = 319.23/51x10-2 = 625.94 =  62 HZ
No. of beats = v2 - v= 14 Hz

Test: BITSAT Past Year Paper- 2018 - Question 30

The figure shows the interference pattern obtained in a double-slit experiment using light of wavelength 600nm. 1, 2, 3, 4 and 5 are marked on five fringes.The third order bright fringe is

Test: BITSAT Past Year Paper- 2018 - Question 31

Electric potential at any point is V = –5x + 3y + √15z , then the magnitude of the electric field is

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 31

Ex = - rv/rx = -r/rx(-5x + 5y + √15z) = 5

Ey = rv/ry = -3, Ez = -√15
Now E = 

Test: BITSAT Past Year Paper- 2018 - Question 32

Seven resistan ces, each of value 20 Ω , are connected to a 2 V battery as shown in the figure.
The ammeter reading will be

Test: BITSAT Past Year Paper- 2018 - Question 33

The variation of magnetic susceptibility (c) with temperature for a diamagnetic substance is best represented by

Test: BITSAT Past Year Paper- 2018 - Question 34

A copper rod of length l rotates about its end with angular velocity ω in uniform magnetic field B. The emf developed between the ends of the rod if the field is normal to the plane of rotation is​

Test: BITSAT Past Year Paper- 2018 - Question 35

A 10V battery with internal resistance 1W and a 15V battery with internal resistance 0.6 W are connected in parallel to a voltmeter (see figure).
The reading in the voltmeter will be close to:

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 35

As the two cells oppose each other hence, the effective emf in closed circuit is 15 – 10 = 5 V and net resistance is 1 + 0.6 = 1.6 Ω (because in the closed circuit the internal resistance of two cells are in series.
Current in the circuit, I = effective emf/total resistance = 5/1.6A
The potential difference across voltmeter will be same as the terminal voltage of either cell.
Since the current is drawn from the cell of 15 V
∴ V1 = E1 – Ir1 = 15 – 5/1.6 x 0.6 = 13.1 V

Test: BITSAT Past Year Paper- 2018 - Question 36

10 for ks are arranged in in creasing or der of frequency in such a way that any two nearest tuning forks produce 4 beats/sec. The highest frequency is twice of the lowest. Possible highest and the lowest frequencies (in Hz) are

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 36


Using  nLast = nFirst + (N – 1)x
where N = Number of tuning forks in series
x = beat frequency between two successive forks
⇒ 2n = n + (10 – 1) × 4 ⇒ n = 36 Hz

Test: BITSAT Past Year Paper- 2018 - Question 37

A charged particle enters in a uniform magnetic field with a certain velocity. The power delivered to the particle by the magnetic field depends on

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 37

Power = work done/time
As no work is done by magnetic force on the charged particle because magnetic force is perpendicular to velocity, hence power delivered is zero.

Test: BITSAT Past Year Paper- 2018 - Question 38

A r esistor an d an inductor a re connected to an ac supply of 120 V and 50 Hz. The current in the circuit is 3 A. If the power consumed in the circuit is 108 W, then the resistance in the circuit is​

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 38

In an ac cir cuit, a pure indcutor does not consume any power. Therefore, power is consumed by the resistor only.
∴ P = I2vR
or 108 = (3)2 R or R = 12 Ω

Test: BITSAT Past Year Paper- 2018 - Question 39

In an electron gun, the poten tial differ en ce between the filament and plate is 3000 V. What will be the velocity of electron emitting from the gun?

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 39

V = 3000 volt.
1/2 mv2 = eV ⇒ v = 

= 32.6 × 106 = 3.26 × 107 m/s.

Test: BITSAT Past Year Paper- 2018 - Question 40

A radioactive substance with decay constant of 0.5s–1 is being produced at a constant rate of 50 nuclei  per second. If there are no nuclei present initially, the time (in second) after which 25 nuclei will be present is

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 40

N = (100 (1 – e–t/2)) = 25
t = 2 ln (4/3)

Test: BITSAT Past Year Paper- 2018 - Question 41

The 25 mL of a 0.15 M solution of lead n itrate, Pb(NO3)2 reacts with all of the aluminium sulphate, Al2(SO4)3, present in 20 mL of a solution. What is the molar concentration of the Al2(SO4)3?​
3Pb(NO3)2 (aq) + Al2(SO4)3 (aq) → 3PbSO4(s)  + 2Al(NO3)3 (aq)

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 41

Molar mass of Pb(NO3)2 = 25 × 0.15 = 3.75 m. moles Molar mass of Al2 (SO4)3 = 1/3 x 3.75 M x 20
M = 0.0625 = 6.25 × 10–2 M

Test: BITSAT Past Year Paper- 2018 - Question 42

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- 2018 - Question 42

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- 2018 - Question 43

If the Planck’s constant h = 6.6×10–34 Js, the de Broglie wavelength of a particle having momentum of 3.3 × 10–24 kg ms –1 will be

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 43

Test: BITSAT Past Year Paper- 2018 - Question 44

Amongst the elements with following electronic configurations, which one of them may have the highest ionization energy?

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 44

The smaller the atomic size, larger is the value of ionisation potential. Further the atoms having half filled or fully filled orbitals are comparatively more stable, hence more energy is required to remove the electron from such atoms.

Test: BITSAT Past Year Paper- 2018 - Question 45

Wh ich of t h e followin g is th e cor r ect a n d increasing order of lone pair of electrons on the central atom?

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 45

The number of lone pairs of electrons on central atom in various given species are
Thus the correct increasing order is

Test: BITSAT Past Year Paper- 2018 - Question 46

According to molecular orbital theory which of the following statement about the magnetic character and bond order is correct regarding O+2

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 46

O2 : σ1s2, σ*1s2 , σ2s2, σ*2s2, σ2p2z,

Bond order 10 - 6/2 = 2
(two unpaired electrons in antibonding molecular orbital)
Bond order 10 -5/2 = 2.5
(One unpaired electron in antibonding molecular orbital).
Hence O2 as well as O+2 both are paramagnetic, and bond order of O+2 is greater than that of O2.

Test: BITSAT Past Year Paper- 2018 - Question 47

If V is the volume of one molecule of gas under given conditions, the van der Waal’s constant b is

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 47

van der Waals’s constant b = 4 times the actual volume of 1 mole molecules = 4VN0

Test: BITSAT Past Year Paper- 2018 - Question 48

For vapor ization of water at 1 atmosph er ic pressure, the values of DH and DS are 40.63 kJmol–1 and 108.8 JK–1 mol–1, respectively. The temperature when Gibbs energy change (DG) for this transformation will be zero, is:

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 48


ΔH = 40630 J mol –1
ΔS = 108.8 JK–1 mol–1
ΔG = ΔH - TΔS When ΔG = 0, ΔH - TΔS = 0
T = ΔH/ΔS
 

Test: BITSAT Past Year Paper- 2018 - Question 49

For the react ion taking place at certain temperature
NH2 COONH4 (s) ⇌ 2 NH3 (g) + CO2 (g), if equilibrium pressure is 3X bar then ΔrG° would be

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 49

ΔG° = – RT ln KP; KP = (2X)2 X = 4X3
ΔG° = – RT ln (4X3)
ΔG° = – RT ln 4 – 3RT ln X

Test: BITSAT Past Year Paper- 2018 - Question 50

The pH of 0.1 M solution of the following salts increases in the order :

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 50

(i) 
∴ [H+] = 0.1 M
pH = – log [H+] = – log 0.1 = 1
(ii) NaCl is a salt of strong acid and strong base so it is not hydrolysed and hence its pH is 7.
(iii) NH4Cl + H2O ⇌ NH4OH + HCl
∴ The solution is acidic and its pH is less than that of 0.1 M HCl.
(iv) NaCN + H2O ⇌ NaOH + HCN
∴ The solution is basic and its pH is more than that of 0.1 M HCl.
∴ Correct order for increase in pH is
HCl < NH4Cl < NaCl < NaCN.

Test: BITSAT Past Year Paper- 2018 - Question 51

When N2O5 is heated at certain temperature, it dissociates as N2O5 (g) ⇌ N2O3 (g) + O2 (g); Kc = 2.5. At the same time N2O3 also decomposes as:
N2O3 (g) ⇌ N2O (g) + O2 (g).
If initially 4.0 moles of N2O5 are taken in 1.0 litre flask and allowed to dissociate. Concentration of O2 at equilibrium is 2.5 M. Equilibrium concentration of N2O5 is :

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 51


Q [O2] = x + y = 2.5
for N2O5, Kc = [N2O5] [O2]/ [N2O5]
and  2.5 = (x +y (x -y)/4-x
∴ x = 2.166
[N2O5] = 4 – x = 1.846

Test: BITSAT Past Year Paper- 2018 - Question 52

Consider the reactions
(A) H2O2 + 2HI → I2 + 2H2O
(B) HOCl + H2O2 → H3O+ + Cl + O2
Which of the following statements is correct about H2O2 with reference to these reactions?
Hydrogen peroxide is ______.

Test: BITSAT Past Year Paper- 2018 - Question 53

Following are colours shown by some alkaline earth metals in flame test. Which of the following are not correctly matched?

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 53

Calcium gives brick red colour and barium gives apple green colour in flame test.

Test: BITSAT Past Year Paper- 2018 - Question 54

Ber yllium shows diagonal relationship with aluminium. Which of the following similarity is incorrect ?

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 54

The Be(OH)2 and Al(OH)3 are amphoteric in nature.

Test: BITSAT Past Year Paper- 2018 - Question 55

An elemen t X occurs in shor t period having configuration ns2 np1. The formula and nature of its oxide is

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 55

ns2 np1 is the electronic configuration of III A period. X2O3(Al2O3) is an amphoteric oxide.

Test: BITSAT Past Year Paper- 2018 - Question 56

Which of the following is strongest nucleophile

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 56

The strength of nucleophile depends upon the nature of alkyl group R on which nucleophile has to attack and also on the nature of solvent.
The order of strength of nucleophiles follows the order:

CN > I > C6H5O > OH > Br > Cl

Test: BITSAT Past Year Paper- 2018 - Question 57

The IUPAC name of the compound is

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 57


IUPAC name –  3, 3-Dimethyl -1-cyclohexanol

Test: BITSAT Past Year Paper- 2018 - Question 58

Which of the following will have a meso-isomer also?

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 58

The compound has two similar assymmetric C-atoms. It has plane of symmetry and exists in meso form.

Meso - 2, 3 dichlorobutane

Test: BITSAT Past Year Paper- 2018 - Question 59

In a set of reactions, ethylbenzene yielded a product D.

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 59


Test: BITSAT Past Year Paper- 2018 - Question 60

Identify the incorrect statement from thefollowing:

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 60

The ozone layer, existing between 20 to 35 km above the earth’s surface, shield the earth from the harmful U. V. radiations from the sun.
Depletion of ozone is caused by oxides of nitrogen
N2O + hn → NO + N
NO + O3 → NO2 + O2
O3 + hn → O2 + O
NO2 + O → NO + O2
2O3 + hn → O2 (Net reaction)
The presence of oxides of nitrogen increase the decomposition of O3.

Test: BITSAT Past Year Paper- 2018 - Question 61

Each edge of a cubic unit cell is 400 pm long. If atomic mass of the element is 120 and its density is 6.25 g/cm3, the crystal lattice is : (use NA = 6 × 1023)

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 61

Test: BITSAT Past Year Paper- 2018 - Question 62

Chloroform, CHCl3, boils at 61.7 °C. If the Kb for chloroform is 3.63°C/molal, what is the boiling point of a solution of 15.0 kg of CHCl3 and 0.616 kg of acenaphthalene, C12H10?

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 62

ΔTb = Kb.m ⇒ 
Tb = 61.7 + 0.968
= 62.67° C

Test: BITSAT Past Year Paper- 2018 - Question 63

pH of a 0.1 M monobasic acid is found to be 2.Hence,  its osmotic pressure at a given temperature TK is

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 63

pH = 2
[H+] = 0.01 M = Cx = 0.1x
x = 0.1
i = 1 + x = 1.1
π = i n/V RT = iMRT = 1.1 x 0.1RT = 0.11RT

Test: BITSAT Past Year Paper- 2018 - Question 64

On passing a current of 1.0 ampere for 16 min and 5 sec through one litre solution of CuCl2, all copper of the solution was deposited at cathode. The strength of CuCl2 solution was (Molar mass of Cu = 63.5; Faraday constant = 96,500 Cmol–1)

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 64

By Faraday's Ist Law, W/e = q/96500
(where q = it = charge of ion)
we know that no of equivalent

(where i = 1 A, t = 16×60+5 = 965 sec.) Since, we know that
Normality = no. of equivalent/Volume (in litre)

Test: BITSAT Past Year Paper- 2018 - Question 65

A 100.0 mL dillute solution of Ag+ is electrolysed for 15.0 minutes with a current of 1.25 mA and the silver is removed completely. What was the initial [Ag+]?

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 65

No. of moles of

Test: BITSAT Past Year Paper- 2018 - Question 66

The accompanying figure depicts a change in concentration of species A and B for the reaction A → B, as a function of time. The point of inter section of the two curves represents

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 66

The intersection point indicates the half life of the reactant A when it is converted to B.

Test: BITSAT Past Year Paper- 2018 - Question 67

The rate constant of a reaction is 1.5 × 10–3 at 25°C and 2.1 × 10–2 at 60°C. The activation energy is

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 67

T1 = 273 + 25 = 298 K
T2 = 273 + 60 = 333 K

Test: BITSAT Past Year Paper- 2018 - Question 68

Freundlich equation for adsorption of gases (in amount of x g) on a solid (in amount of m g) at constant temperature can be expressed as

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 68

According to Freundlich equation,
x/m ∝ p1/n or x/m = Kp1/n
or log x/m = log Kp1/n or log x/m = log K + 1/n log K + 1/n log p

Test: BITSAT Past Year Paper- 2018 - Question 69

Which of the following feature of catalysts is described in reactions given below?
(i) CO (g) + 2H2 (g) CH3OH (g)
(ii) CO (g) + H2 (g ) HCHO (g)
​(iii) CO (g) + 3H2 (g)  CH4 (g)+ H2O (g)

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 69

Given reactions shows that the selectivity of different catalysts for some reactants is different.

Test: BITSAT Past Year Paper- 2018 - Question 70

Which of the following is not a member of chalcogens?

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 70

chalcogens are defined as ore-forming elements.

Test: BITSAT Past Year Paper- 2018 - Question 71

Pick out the wrong statement.

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 71

Catenation tendency is higher in
phosphorus when compared with other elements of same group.

Test: BITSAT Past Year Paper- 2018 - Question 72

Which of the following element do not form complex with EDTA?

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 72

Be is the only group 2 element that does not form a stable complex with [EDTA]4–. Mg2+ and Ca2+ have the greatest tendency to form complexes  with [EDTA]4–.

Test: BITSAT Past Year Paper- 2018 - Question 73

Which one of the following cyano complexes would exhibit the lowest value of paramagnetic behaviour ?

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 73


CN is a strong field ligand and it causes pairing of electrons; as a result number of unpaired electrons in Co3+ becomes zero and hence it has lowest value of paramagnetic behaviour.

Test: BITSAT Past Year Paper- 2018 - Question 74

When an aqueous solution of copper (II) sulphate is saturated with ammonia, the blue compound crystallises on evaporation. The formula of this blue compound is:

Test: BITSAT Past Year Paper- 2018 - Question 75


Here [Y] is a

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 75



Test: BITSAT Past Year Paper- 2018 - Question 76

Following compounds are given:
(1) CH3CH2OH
(2) CH3COCH3
(3) 
(4) CH3OH
Which of the above compound(s), on being warmed with iodine solution and NaOH, will give iodoform?

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 76

Among the given compounds only CH3OH does not give iodoform reaction.

Test: BITSAT Past Year Paper- 2018 - Question 77

Arrange the following alcohols in increasing order of their reactivity towards the reaction with HCl.
(CH3)2CH-OH (1), (CH3)3C-OH (2), (C6H5)3C-OH (3)

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 77

Alkylhalide formation in the reaction of alcohol with HCl undergoes SN1 reaction in which formation of the carbocation as intermediate occurs. Stability of carbocation is greatest for (C6H5)3C+ due to resonance effect, and stability of tertiary carbocation is greater than the secondary carbocation hence the option (a) shows the correct order.

Test: BITSAT Past Year Paper- 2018 - Question 78

Thirty percent of the bases in a sample of DNA extracted from eukaryotic cells is adenine. What percentage of cytosine is present in this DNA?

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 78

If 30 percent of DNA is adenine, then by Chargaff’s rule 30 percent will be thymine. The remaining 40 percent of the DNA is cytosine and guanine. Since the ratio of cytosine to guanine must be equal, then each accounts for 20 percent of the bases.

Test: BITSAT Past Year Paper- 2018 - Question 79

The blue colour of snail is due to presence of

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 79

Most snail blood is blueish in color. This is because their blood cells use haemocyanin, which gets its blue color from the copper that is part of its structure.

Test: BITSAT Past Year Paper- 2018 - Question 80

Which of the following is a diamine?

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 80

Diamines are those compounds which contain two amino groups.

Test: BITSAT Past Year Paper- 2018 - Question 81

Choose the word which is most similar in meaning to the word 'Optimistic'.

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 81

Optimistic means hopeful and confident about the future.

Test: BITSAT Past Year Paper- 2018 - Question 82

Choose the word which is most opposite in meaning to the word 'Drowsy'.

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 82

Drowsy means sleepy and lethargic.
Therefore, option (d) is the correct antonym of it.
Rest of the options are its synonyms.

Test: BITSAT Past Year Paper- 2018 - Question 83

Direction: Which of the following phrases (I), (II), and (III) given below each sentence should replace the phrase printed in bold letters to make the sentence grammatically correct? Choose the best option among the five given alternatives that reflect the correct use of phrase in the context of the grammatically correct sentence. If the sentence is correct as it is, mark (d) i.e., "No correction required" as the answer.

He is really feeling under the weather today; he has a terrible cold.
(I) feeling like the weather
(II) feeling over the weather
(III) feeling in the weather

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 83

The phrase used in the sentence is grammatically correct hence, it doesn't require any correction. The meaning of the phrase' under the weather' is to feel ill.

Test: BITSAT Past Year Paper- 2018 - Question 84

Direction: Which of the following phrases (I), (II), and (III) given below each sentence should replace the phrase printed in bold letters to make the sentence grammatically correct? Choose the best option among the five given alternatives that reflect the correct use of phrase in the context of the grammatically correct sentence. If the sentence is correct as it is, mark (d) i.e., "No correction required" as the answer.

By working part-time and looking after his old mother, he managed to get the best for both worlds.
(I) the best at both worlds
(II) the best of both worlds
(III) the best on both worlds

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 84

The correct phrase to be used is 'get the best of both worlds' which means a win-win situation.

Test: BITSAT Past Year Paper- 2018 - Question 85

Direction: Which of the following phrases (I), (II), and (III) given below each sentence should replace the phrase printed in bold letters to make the sentence grammatically correct? Choose the best option among the five given alternatives that reflect the correct use of phrase in the context of the grammatically correct sentence. If the sentence is correct as it is, mark (d) i.e., "No correction required" as the answer.

Hey, Nanny, speak about the devil and you are here.
(I) speak at the devil
(II) speak on the devil
(III) speak of the devil

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 85

The correct phrase to be used is 'speak of the devil'. This phrase is said when a person appears just after being mentioned.

Test: BITSAT Past Year Paper- 2018 - Question 86

DIRECTION: Read the following passage carefully and answer the questions given below it.

The likelihood of at least 600,000 deaths being caused annually in India by fine particulate matter pollution in the air is cause for worry, even if the data released by the World Health Organisation are only a modelled estimate. The conclusion that so many deaths could be attributed to particulate matter 2.5 micrometres or less in size is, of course, caveated, since comprehensive measurement of PM2.5 is not yet being done and the linkages between pollution, disease and deaths need further study. What is not in doubt is that residents in many urban areas are forced to breathe unhealthy levels of particulates, and the smallest of these - PM10 and less - can penetrate and get lodged deep in the lungs.
The WHO Global Burden of Disease study has been working to estimate pollution-linked health impacts, such as stroke and ischaemic heart disease, acute lower respiratory infection and chronic obstructive pulmonary disease. Data on fine particulates in India show that in several locations the pollutants come from burning of biomass, such as coal, fuel wood, farm litter and cow dung cakes. In highly built-up areas, construction debris, road dust and vehicular exhaust add to the problem. The Prime Minister launched an Air Quality Index last year aimed at improving pollution control. The new data, which the WHO says provide the best evidence available on the terrible toll taken by particulates, should lead to intensified action. A neglected aspect of urban air pollution control is the virtual discarding of the Construction and Demolition Waste Management Rules, notified to sustainably manage debris that is dumped in the cities, creating severe particulate pollution.
The Environment Ministry has highlighted the role that debris can play as a resource. Municipal and government contracts are, under the rules, required to utilise up to 20 per cent materials made from construction and demolition waste, and local authorities must place containers to hold debris. This must be implemented without delay. Providing cleaner fuels and scientifically designed cookstoves to those who have no option but to burn biomass, would have a big impact on reducing particulate matter in the northern and eastern States, which are the worst-hit during winter, when biomass is also used for heating.
Greening the cities could be made a mission, involving civil society, with a focus on landscaping open spaces and paving all public areas to reduce dust. These measures can result in lower PM10 and PM2.5 levels.
Comprehensive measurement of these particulates is currently absent in many cities, a lacuna that needs to be addressed.

Q. According to the WHO Global Burden of Disease study which of the following is/are pollution linked health impacts?
(I) Infection of the lower respiratory system
(II) Chronic obstructive pulmonary disease
(III) Stroke and ischaemic heart disease

Test: BITSAT Past Year Paper- 2018 - Question 87

DIRECTION: Read the following passage carefully and answer the questions given below it.

The likelihood of at least 600,000 deaths being caused annually in India by fine particulate matter pollution in the air is cause for worry, even if the data released by the World Health Organisation are only a modelled estimate. The conclusion that so many deaths could be attributed to particulate matter 2.5 micrometres or less in size is, of course, caveated, since comprehensive measurement of PM2.5 is not yet being done and the linkages between pollution, disease and deaths need further study. What is not in doubt is that residents in many urban areas are forced to breathe unhealthy levels of particulates, and the smallest of these - PM10 and less - can penetrate and get lodged deep in the lungs.
The WHO Global Burden of Disease study has been working to estimate pollution-linked health impacts, such as stroke and ischaemic heart disease, acute lower respiratory infection and chronic obstructive pulmonary disease. Data on fine particulates in India show that in several locations the pollutants come from burning of biomass, such as coal, fuel wood, farm litter and cow dung cakes. In highly built-up areas, construction debris, road dust and vehicular exhaust add to the problem. The Prime Minister launched an Air Quality Index last year aimed at improving pollution control. The new data, which the WHO says provide the best evidence available on the terrible toll taken by particulates, should lead to intensified action. A neglected aspect of urban air pollution control is the virtual discarding of the Construction and Demolition Waste Management Rules, notified to sustainably manage debris that is dumped in the cities, creating severe particulate pollution.
The Environment Ministry has highlighted the role that debris can play as a resource. Municipal and government contracts are, under the rules, required to utilise up to 20 per cent materials made from construction and demolition waste, and local authorities must place containers to hold debris. This must be implemented without delay. Providing cleaner fuels and scientifically designed cookstoves to those who have no option but to burn biomass, would have a big impact on reducing particulate matter in the northern and eastern States, which are the worst-hit during winter, when biomass is also used for heating.
Greening the cities could be made a mission, involving civil society, with a focus on landscaping open spaces and paving all public areas to reduce dust. These measures can result in lower PM10 and PM2.5 levels.
Comprehensive measurement of these particulates is currently absent in many cities, a lacuna that needs to be addressed.

Q. The conclusion regarding the deaths attributed to particulate matter 2.5 micrometers is considered to be caveated because

Test: BITSAT Past Year Paper- 2018 - Question 88

DIRECTION: Read the following passage carefully and answer the questions given below it.

The likelihood of at least 600,000 deaths being caused annually in India by fine particulate matter pollution in the air is cause for worry, even if the data released by the World Health Organisation are only a modelled estimate. The conclusion that so many deaths could be attributed to particulate matter 2.5 micrometres or less in size is, of course, caveated, since comprehensive measurement of PM2.5 is not yet being done and the linkages between pollution, disease and deaths need further study. What is not in doubt is that residents in many urban areas are forced to breathe unhealthy levels of particulates, and the smallest of these - PM10 and less - can penetrate and get lodged deep in the lungs.
The WHO Global Burden of Disease study has been working to estimate pollution-linked health impacts, such as stroke and ischaemic heart disease, acute lower respiratory infection and chronic obstructive pulmonary disease. Data on fine particulates in India show that in several locations the pollutants come from burning of biomass, such as coal, fuel wood, farm litter and cow dung cakes. In highly built-up areas, construction debris, road dust and vehicular exhaust add to the problem. The Prime Minister launched an Air Quality Index last year aimed at improving pollution control. The new data, which the WHO says provide the best evidence available on the terrible toll taken by particulates, should lead to intensified action. A neglected aspect of urban air pollution control is the virtual discarding of the Construction and Demolition Waste Management Rules, notified to sustainably manage debris that is dumped in the cities, creating severe particulate pollution.
The Environment Ministry has highlighted the role that debris can play as a resource. Municipal and government contracts are, under the rules, required to utilise up to 20 per cent materials made from construction and demolition waste, and local authorities must place containers to hold debris. This must be implemented without delay. Providing cleaner fuels and scientifically designed cookstoves to those who have no option but to burn biomass, would have a big impact on reducing particulate matter in the northern and eastern States, which are the worst-hit during winter, when biomass is also used for heating.
Greening the cities could be made a mission, involving civil society, with a focus on landscaping open spaces and paving all public areas to reduce dust. These measures can result in lower PM10 and PM2.5 levels.
Comprehensive measurement of these particulates is currently absent in many cities, a lacuna that needs to be addressed.

Q. Which of the following is/are not true in the context of the passage?

Test: BITSAT Past Year Paper- 2018 - Question 89

DIRECTION: Read the following passage carefully and answer the questions given below it.

The likelihood of at least 600,000 deaths being caused annually in India by fine particulate matter pollution in the air is cause for worry, even if the data released by the World Health Organisation are only a modelled estimate. The conclusion that so many deaths could be attributed to particulate matter 2.5 micrometres or less in size is, of course, caveated, since comprehensive measurement of PM2.5 is not yet being done and the linkages between pollution, disease and deaths need further study. What is not in doubt is that residents in many urban areas are forced to breathe unhealthy levels of particulates, and the smallest of these - PM10 and less - can penetrate and get lodged deep in the lungs.
The WHO Global Burden of Disease study has been working to estimate pollution-linked health impacts, such as stroke and ischaemic heart disease, acute lower respiratory infection and chronic obstructive pulmonary disease. Data on fine particulates in India show that in several locations the pollutants come from burning of biomass, such as coal, fuel wood, farm litter and cow dung cakes. In highly built-up areas, construction debris, road dust and vehicular exhaust add to the problem. The Prime Minister launched an Air Quality Index last year aimed at improving pollution control. The new data, which the WHO says provide the best evidence available on the terrible toll taken by particulates, should lead to intensified action. A neglected aspect of urban air pollution control is the virtual discarding of the Construction and Demolition Waste Management Rules, notified to sustainably manage debris that is dumped in the cities, creating severe particulate pollution.
The Environment Ministry has highlighted the role that debris can play as a resource. Municipal and government contracts are, under the rules, required to utilise up to 20 per cent materials made from construction and demolition waste, and local authorities must place containers to hold debris. This must be implemented without delay. Providing cleaner fuels and scientifically designed cookstoves to those who have no option but to burn biomass, would have a big impact on reducing particulate matter in the northern and eastern States, which are the worst-hit during winter, when biomass is also used for heating.
Greening the cities could be made a mission, involving civil society, with a focus on landscaping open spaces and paving all public areas to reduce dust. These measures can result in lower PM10 and PM2.5 levels.
Comprehensive measurement of these particulates is currently absent in many cities, a lacuna that needs to be addressed.

Q. As per the given passage, which of the following is/are the measures for lowering particulate matter in the atmosphere?
(I) Making cleaner fuels available
(II) Landscaping open areas
(III) Providing cooking stoves designed scientifically

Test: BITSAT Past Year Paper- 2018 - Question 90

If sentence (B) "The Finance Ministry's warning to potential investors in bitcoin and other cryptocurrencies has come at a time when a new, seemingly attractive investment area has opened up that few have enough information about." is the first sentence, what is the order of other sentences after rearrangement?
(A) One of the main reasons for this volatility is speculation and the entry into the market of a large number of people lured by the prospect of quick and easy profits.
(B) The Finance Ministry's warning to potential investors in bitcoin and other cryptocurrencies has come at a time when a new, seemingly attractive investment area has opened up that few have enough information about.
(C) A number of investors, daunted by the high price of bitcoin, have put their money into less well-established and often spurious cryptocurrencies, only to lose it all.
(D) Investment in bitcoin and other cryptocurrencies increased tremendously in India over the past year, but most new users know close to nothing of the technology, or how to verify the genuineness of a particular cryptocurrency.
(E) The price of bitcoin, the most popular of all cryptocurrencies, not only shot up by well over 1000% over the course of the last year but also fluctuated wildly.
(F) The government's caution comes on top of three warnings issued by the Reserve Bank of India since 2013.

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 90

The first sentence talks about the fact that only few investors have idea about bitcoins and other cryptocurrencies (which seems an attractive investment area), so, the finance ministry has warned the potential investors about it. Sentence E will follow the first sentence because it says that 'bitcoin not only shot up well over by 1000%......' which justifies 'attractive investment area' and forms a link. Now, we are left with only option (b) and (d) to choose from.
When we consider the sentence F, we can see that this line seems to be a part somewhere in the middle of the paragraph, also, the first line starts with a warning, therefore, it must justify the consequences of the investment in bitcoins and other cryptocurrencies which is justified by sentence C. Hence, option (b) is the correct choice.

Test: BITSAT Past Year Paper- 2018 - Question 91

If sentence (C) "Clinical trials involving human subjects have long been a flashpoint between bioethicists and clinical research organisations (CROs) in India." is the first sentence, what is the order of other sentences after rearrangement?
(A) Such over-volunteering occurs more frequently in bioequivalence studies, which test the metabolism of generics in healthy subjects.
(B) Landmark amendments to the Drugs and Cosmetics Act in 2013 led to better protection of vulnerable groups such as illiterate people, but more regulation is needed to ensure truly ethical research.
(C) Clinical trials involving human subjects have long been a flashpoint between bioethicists and clinical research organisations (CROs) in India.
(D) The big problem plaguing clinical research is an over-representation of low-income groups among trial subjects.
(E) While CROs have argued that more rules will stifle the industry, the truth is that ethical science is often better science.
(F) Sometimes CROs recruit them selectively, exploiting financial need and medical ignorance; at other times people overvolunteer for the money.

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 91

After reading all the sentences carefully, we see that sentence A and F should go one after another as both talk about 'over-volunteer'.
Moreover, sentence A will follow sentence F because of the presence of the word 'such' which signifies that the subject of the sentence has already been discussed in the previous sentence.
So, we have option (c), (d) and (e) to choose from. Considering sentence D which talks about 'a big problem', we find that it can't be the second sentence as no problem of any sort has been dealt in the first sentence, so, option (c) and (e) gets eliminated. Hence, by elimination method, we can conclude that option (d) is the correct choice.

Test: BITSAT Past Year Paper- 2018 - Question 92

DIRECTION: Read the sentence to find out whether there is any grammatical error or idiomatic error in it. The error, if any, will be in one part of the sentence.
The letter of that part is the answer. If there is no error, the answer is (d). (Ignore errors of punctuation, if any.)

Despite being (a)/ a good teacher, (b)/ he has no influence on his pupil. (c)/ No error (d)

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 92

Replace the preposition 'on' with 'over' to make the sentence grammatically correct.

Test: BITSAT Past Year Paper- 2018 - Question 93

DIRECTION: Read the sentence to find out whether there is any grammatical error or idiomatic error in it. The error, if any, will be in one part of the sentence.
The letter of that part is the answer. If there is no error, the answer is (d). (Ignore errors of punctuation, if any.)

Yesterday, when we were returning from the party, (a)/ our car met with an accident, (b)/ but we were fortunate to reach our home safely. (c)/ No error (d)

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 93

Replace the adverb 'safely' with the adjective 'safe' to make the sentence.
Grammatically correct.

Test: BITSAT Past Year Paper- 2018 - Question 94

A group of sheep is known as :

Test: BITSAT Past Year Paper- 2018 - Question 95

A group of trees is known as:

Test: BITSAT Past Year Paper- 2018 - Question 96

In a code language, if REGAINS is coded as QDFZHMR, then the word PERIODS will be coded as -

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 96

According to question,

Test: BITSAT Past Year Paper- 2018 - Question 97

If 5 # 6 = 121 and 10 # 8 = 324, then find the value of 23 # 14 = ?

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 97

According to question, (5 + 6)2 = 121
(10 + 8)2 = 324
(23 + 14)2 = 1369

Test: BITSAT Past Year Paper- 2018 - Question 98

Which of the following cube in the answer figure cannot be made based on the unfolded cube in the question figure?

Test: BITSAT Past Year Paper- 2018 - Question 99

Which one of the following diagram represents the correct relationship among Professor,
Male and Female.

Test: BITSAT Past Year Paper- 2018 - Question 100

Select the related letter/word/ number from the given alternatives.

Distance : Odometer :: ? : Barometer

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 100

Distance is measured by odometer. Similarly, Pressure is measured by Barometer.

Test: BITSAT Past Year Paper- 2018 - Question 101

Find the odd word/letters/ number pair/number from the given alternatives.

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 101

According to question, 24 = 1 × 6 ×1 × 4
270 = 5 × 6 × 9
120 = 4 × 3 × 2 × 5
162 ¹ 6 × 9 × 3 × 0 = 0

Test: BITSAT Past Year Paper- 2018 - Question 102

Choose the correct alternatives from the given ones that will complete the series.
L_N O_ _MLLM_ OO_ML

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 102

According to question, L M N O /O N M L/ L M N O/ O N M L

Test: BITSAT Past Year Paper- 2018 - Question 103

Choose the correct alternatives from the given ones that will complete the series.

22, 26, 53, 69, 194, ?

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 103

The pattern is :

Test: BITSAT Past Year Paper- 2018 - Question 104

Select the missing number from the given responses.

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 104

The pattern is : (1 × 2 × 3 × 5) + (1 + 2 + 3 + 5) = 41
(3 × 4 × 2 × 6) + (3 + 4 + 2 + 6) = 159
(9 × 8 × 3 × 4) + (9 + 8 + 3 + 4) = 888

Test: BITSAT Past Year Paper- 2018 - Question 105

Identify the figure that will complete the pattern.

Test: BITSAT Past Year Paper- 2018 - Question 106

The domain of the function f (x) = , where [x] denotes the greatest integer less than or equal to x, is

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 106

f(x) is defined if x2 - [x2] ≥ 0 ⇒ x2 ≥ [x]2, which is true for all positive real x and all negative integers x.

Test: BITSAT Past Year Paper- 2018 - Question 107

If m sin θ = n sin(θ + 2α) then tan(θ + α) is

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 107


 = tan(θ + α) cot α

Test: BITSAT Past Year Paper- 2018 - Question 108

Number of solutions of equation sin 99 = sin θ in the interval [0,2π] is

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 108

sin 9θ = sin θ ⇒  9θ = nπ + (-1)n θ
If n = 2m then 9θ = 2mπ + θ ⇒ θ = mπ/4
If n = 2m +1 then 9θ = (2m +1)π - θ
⇒ θ =  (2m +1) π/10
The values belonging to [0, π] are

Test: BITSAT Past Year Paper- 2018 - Question 109

A pole stands vertically inside a triangular park ABC. If the angle of elevation of the top of the pole from each corner of the park is same, then the foot of the pole is at the

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 109


The foot of the pole is at the centroid.
Because centroid is the point of intersection of medians AD, BE and CF, which are the lines joining a vertex with the mid point of opposite side.

Test: BITSAT Past Year Paper- 2018 - Question 110

Let A, B, and C are the angles of a plain triangle and  tan A/2 = 1/3, tan B/2 = 2/3 Then tan C/2 is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 110

A + B + C = π

∴ tan C/2 = 7/9.

Test: BITSAT Past Year Paper- 2018 - Question 111

If the amplitude of z – 2 – 3i is p/4, then the locus of  z = x + iy is

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 111

z - 2-3i = x+ iy - 2-3i = (x - 2)+i(y -3)
⇒ x – y + 1 = 0

Test: BITSAT Past Year Paper- 2018 - Question 112

The roots of the equation x4 - 2x3 + x = 380 are: 

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 112

Given equation is x4 – 2x3 + x – 380 = 0
⇒  (x2 -x - 20)(x2 -x +19)= 0
⇒ (x - 5)(x + 4)(x2 -x + 19)= 0 Hence, the required roots of the equation are

Test: BITSAT Past Year Paper- 2018 - Question 113

Roots of the equation x2 + bx - c = 0(b,c > 0) are

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 113

Since b, c > 0 Therefore a + b = -b < 0 and ab = -c < 0 Since product of the roots is –ve therefore roots must be of opposite sign.

Test: BITSAT Past Year Paper- 2018 - Question 114

In how many ways can 12 gentlemen sit around a round table so that three specified gentlemen are always together?

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 114

It is obvious by fundamental property of circular permutations.

Test: BITSAT Past Year Paper- 2018 - Question 115

The number of ways in which first, second and third prizes can be given to 5 competitors is

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 115

First prize can be given in 5 ways. Then second prize can be given in 4 ways and the third prize in 3 ways (Since a competitior cannot get two prizes) and hence the no. of ways. = 5 × 4 × 3 = 60 ways

Test: BITSAT Past Year Paper- 2018 - Question 116

The coefficient of x3 in the expansion of is:

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 116

Given, and the (r + 1)th term in the expansion of (x + a)n is T(r + 1) = nCr(x)n – r
αr
∴ (r + 1)th term in expansion of

 =  7Cr(x)7 – 2r (– 1)r
Since  x3 occurs in Tr + 1
∴ 7 – 2r = 3 ⇒ r  = 2
thus the coefficient of x3 = 7C2 ( – 1)2
= 7 x 6/2 x 1 = 21.

Test: BITSAT Past Year Paper- 2018 - Question 117

If x > 0, the 1 + 
 

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 117

Test: BITSAT Past Year Paper- 2018 - Question 118

If a, b, c are in G.P., then

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 118

Q a, b, c are in G.P.
 ⇒ a2, b2, care in G.P.

Test: BITSAT Past Year Paper- 2018 - Question 119

The locus of the point of intersection of the lines represent (t being a parameter)

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 119

To eliminate the parameter t, square and add the equations, we have


Which is the equation of a circle.

Test: BITSAT Past Year Paper- 2018 - Question 120

The equation of the circle which passes through the point (4, 5) and has its centre at (2, 2) is

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 120

As the circle is passing through the point (4, 5) and its centre is (2, 2) so its radius is

∴ The required equation is:
(x – 2)2 + (y – 2)2 = 13

Test: BITSAT Past Year Paper- 2018 - Question 121

Eccentricity of ellipse x2 + α2 + y2/b2 = 1 if it passes through point (9, 5) and (12, 4) is

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 121

We have ....(1)
 ......(2)
From eq. (2) – eq. (1):

Test: BITSAT Past Year Paper- 2018 - Question 122

Consider the equation of a parabola  y2 + 4ax = 0, where a > 0 which of the following is/are correct?

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 122

Equation of parabola is y2 = – 4ax. Its focus is at(– a, 0).

Test: BITSAT Past Year Paper- 2018 - Question 123

The value of is equal to:

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 123


(By using sum of n natural number 1+ 2 + 3 + .... + n = n(n + 1)/2)
Take n2 common from Nr and Dr.

Test: BITSAT Past Year Paper- 2018 - Question 124

is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 124


Test: BITSAT Past Year Paper- 2018 - Question 125

 The probability of getting 10 in a single throw of three fair dice is :

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 125

Exhaustive no. of cases = 63
10 can appear on three dice either as distinct number as following (1, 3, 6) ; (1, 4, 5); (2, 3, 5) and each can occur in 3! ways. Or 10 can appear on three dice as repeated digits as following (2, 2, 6),
(2, 4, 4), (3, 3, 4) and each can occur in 3!/2!
 ways.
∴ No. of favourable cases = 3 x 3! + 3 x 3!/2! = 27

Test: BITSAT Past Year Paper- 2018 - Question 126

Number of solutions of the equation tan-1 (1 + x) + tan-1 (1 -x) π/2 are

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 126

tan-1 (1 + x) + tan-1 (1 -x) π/2
⇒ tan-1 (1 + x) =  π/2 - tan-1 (1 - x)
⇒ tan-1 (1 + x)= cot-1(1 - x)
⇒ tan-1 (1 + x) = tan-1(1/1-x)
⇒ 1 + x = 1/1-x
⇒ 1 - x2 = 1 ⇒ x = 0

Test: BITSAT Past Year Paper- 2018 - Question 127

 If A = is an orthogonal matrix, then 

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 127

As A is an orthogonal matrix, AAT = I

⇒ a + 4 + 2b = 0, 2a + 2 – 2b = 0 and a2 + 4 + b2 = 9
⇒ a + 2b + 4 = 0 , a – b + 1 = 0 and a2 + b2 = 5
⇒ a = – 2, b = – 1

Test: BITSAT Past Year Paper- 2018 - Question 128

 The points represented by the complex numbers 1 + i, -2 + 3i, 5/3 i on the argand plane are

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 128

Let z1 = 1 + i, z2 = – 2 + 3i and z3 = 0 + 5/3 i

Test: BITSAT Past Year Paper- 2018 - Question 129

 If matrix  and A-1 = 1/k adj(A) , then k is

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 129


A-1 = 1/k adj(A) .....(i) 
Also, we know  .....(ii)
∴ By comparing (i) and (ii) | A | = k

= 3 (2 + 1) + 2 (1 + 0) + 4 (1 – 0)  = 9 + 2 + 4 = 15

Test: BITSAT Past Year Paper- 2018 - Question 130

If x, y, z are complex numbers, and then Δ is

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 130

We have

[Interchanging rows and columns]

[Taking –1 common from each row]
∴  + Δ = 0 ⇒ 2Re(Δ) = 0
∴ Δ is purely imaginary.

Test: BITSAT Past Year Paper- 2018 - Question 131

If 

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 131

The function can be contin uous only at those points for which
sin x = cos x ⇒ x =  nπ + π/4

Test: BITSAT Past Year Paper- 2018 - Question 132

If 

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 132


Hence f (x) is continuous at x = 3π/4

Test: BITSAT Past Year Paper- 2018 - Question 133

The value of c in (0, 2) satisfying the mean value theorem for the function f(x) = x(x – 1)2, x ∈ [0, 2] is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 133

f(x) = x (x – 1)2 ; x ∈ [0, 2]

f '(x) = 3x2 – 4x + 1 ⇒ f '(c) = 3c2 – 4c + 1

Test: BITSAT Past Year Paper- 2018 - Question 134

If at x = 1 is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 134

Given expression can be written as

Test: BITSAT Past Year Paper- 2018 - Question 135

Let y = e2x. Then  is

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 135

Test: BITSAT Past Year Paper- 2018 - Question 136

A ball is dropped from a platform 19.6m high. Its position function is –

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 136

We have,  a = d2x/dt2 = 9.8
The initial conditions are x (0) = 19.6 and v (0) = 0
So,  v = dx/dt =  9.8t v(0) 9.8t
∴ x = – 4.9t2 + x (0) = – 4.9t2 + 19.6
Now, the domain of the function is restricted since the ball hits the ground after a certain time. To find this time we set x = 0 and solve for t. 0 = – 4.9t2 + 19.6 ⇒ t = 2

Test: BITSAT Past Year Paper- 2018 - Question 137

The value of the integral   is:

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 137


Test: BITSAT Past Year Paper- 2018 - Question 138

dx is equal to :

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 138

Put x2 = t ⇒ 2 x dx = dt

Test: BITSAT Past Year Paper- 2018 - Question 139

If f (2a - x )dx = m and f (x)dx= n, then  f (x)dx is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 139

Put x = 2a – t
so that dx = – dt
when x = a, t = a and when x = 2a, t = 0

Test: BITSAT Past Year Paper- 2018 - Question 140

An integrating factor of the differential equation sin x dx/dy  + 2 y cos x = 1 is

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 140

Given differential equation is

Test: BITSAT Past Year Paper- 2018 - Question 141

The expression satisfying the differ ential equation (x2 - 1) dy/dx + 2xy = 1 is

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 141

Rewrite the given differential equation as follows :
which is a linear form The integrating factor I.F.

Thus  multiplying the given equation by (x2 - 1),
we get (x2 - 1) dy/dx + 2xy = 1
⇒ d/dx [y(x2 - 1)] = 1
On integrating we get  y(x2 - 1) = x + c.

Test: BITSAT Past Year Paper- 2018 - Question 142

Let  and depends on

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 142



Hence is independent of x and y both.

Test: BITSAT Past Year Paper- 2018 - Question 143

If  are the position vectors of the vertices of a triangle ABC taken in order, then ∠A is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 143

Test: BITSAT Past Year Paper- 2018 - Question 144

The projection of line joining (3, 4, 5) and (4, 6, 3) on the line joining (–1, 2, 4) and (1, 0, 5) is

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 144

Let  A = (3, 4, 5), P = (–1, 2, 4) B = (4, 6, 3)   and    Q = (1, 0, 5)
∴ Dr's of line AB are (4 – 3), (6 – 4), (3 – 5) = 1, 2, –2
and Dr's of line PQ are (1 + 1), (0 – 2), (5 – 4) = 2, –2, 1
∴ Dr's of line PQ =  

∴ Projection of line segment AB on the line PQ is 
 

Test: BITSAT Past Year Paper- 2018 - Question 145

Which of the following statements is correct?

Test: BITSAT Past Year Paper- 2018 - Question 146

If the constraints in a linear programming problem are changed then

Test: BITSAT Past Year Paper- 2018 - Question 147

In a binomial distribution, the mean is 4 and variance is 3. Then its mode is :

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 147

In Bin omi al di str ibuti on , Mea n = np, Variance = npq and the mode is r if for x = r, the probability function p(x) is maximum.
Given np = 4 and npq = 3

⇒ The distribution will have unique mode (unimodal) & the mode = 4

Test: BITSAT Past Year Paper- 2018 - Question 148

The sum is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 148

The given series is


Test: BITSAT Past Year Paper- 2018 - Question 149

The Boolean expression is equivalent to :

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 149

Test: BITSAT Past Year Paper- 2018 - Question 150

If in a frequency distribution, the mean and median are 21 and 22 respectively, then its mode is approximately

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 150

We know that,
Mode = 3 Median – 2 Mean = 3(22) –2(21)
= 66 – 42 = 24

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