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150 Questions MCQ Test BITSAT Mock Tests Series & Past Year Papers - Test: BITSAT Past Year Paper- 2018

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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

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 16

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.

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 27

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)