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Test: BITSAT Past Year Paper- 2009 - JEE MCQ


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150 Questions MCQ Test BITSAT Mock Tests Series & Past Year Papers - Test: BITSAT Past Year Paper- 2009

Test: BITSAT Past Year Paper- 2009 for JEE 2024 is part of BITSAT Mock Tests Series & Past Year Papers preparation. The Test: BITSAT Past Year Paper- 2009 questions and answers have been prepared according to the JEE exam syllabus.The Test: BITSAT Past Year Paper- 2009 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: BITSAT Past Year Paper- 2009 below.
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Test: BITSAT Past Year Paper- 2009 - Question 1

Given that  and A2 + B2 = R2. The angle between  is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 1

Test: BITSAT Past Year Paper- 2009 - Question 2

In the relation :

P is pressure, Z is distance, k is Boltzmann constant and θ is the temperature. The dimensional formula of b will be​

Test: BITSAT Past Year Paper- 2009 - Question 3

Which of the following is most accurate?

Test: BITSAT Past Year Paper- 2009 - Question 4

A projectile projected at an angle 30º from the horizontal has a range R. If the angle of projection at the same initial velocity be 60º, then the range will be

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 4

If sum of angle of projection = 90° for given speed then range for that angle of projection is same.

Test: BITSAT Past Year Paper- 2009 - Question 5

A block of mass M is pulled along a horizontal frictionless surface by a rope of mass M/2. If a force 2Mg is applied at one end of the rope, the force which the rope exerts on the block is –​

Test: BITSAT Past Year Paper- 2009 - Question 6

A chain of mass M is placed on a smooth table with 1/n of its length L hanging over the edge.The work done in pulling the hanging portion of the chain back to the surface of the table is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 6

W = change in PE of COM of hanging part

Test: BITSAT Past Year Paper- 2009 - Question 7

A particle of mass 10 kg moving eastwards with a speed 5 ms–1 collides with another particle of the same mass moving north-wards with the same speed 5 ms–1. The two particles coalesce on collision. The new particle of mass 20 kg will move in the north-east direction with velocity

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 7

Here


Test: BITSAT Past Year Paper- 2009 - Question 8

A uniform cube of side a and mass m rests on a rough horizontal table. A horizontal force F is applied normal to one of the faces at a point that is directly above the centre of the face, at a height 3a/4 above the base. The minimum value of F for which the cube begins to topple an edge is (assume that cube does not slide)

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 8

For toppling about edge xx'
At the moment of toppling the normal force pass through axis xx'.

Test: BITSAT Past Year Paper- 2009 - Question 9

The rotation of the earth having radius R about its axis speeds upto a value such that a man at latitude angle 600 feels weightless. The duration of the day in such case will be :

Test: BITSAT Past Year Paper- 2009 - Question 10

A metallic rod breaks when strain produced is 0.2%. The Young’s modulus of the material of the rod is 7 × 109 N/m2. What should be its area of cross-section to support a load of 104 N ?

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 10

Maximum possible strain = 0.2/100

Test: BITSAT Past Year Paper- 2009 - Question 11

A liquid is flowing through a non-sectional tube with its axis horizontally. If two points X and Y on the axis of tube has a sectional area 2.0 cm3 and 25 mm2 respectively then find the flow velocity at Y when the flow velocity at X is 10m/s.

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 11

According to principle of continuity

Test: BITSAT Past Year Paper- 2009 - Question 12

A body of length 1m having cross-sectional area 0.75m2 has heat flow through it at the rate of 6000 Joule/sec. Then find the temperature difference if K = 200 Jm–1K–1.

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 12

Test: BITSAT Past Year Paper- 2009 - Question 13

Which of the following combinations of properties would be most desirable for a cooking pot?

Test: BITSAT Past Year Paper- 2009 - Question 14

A particle starts moving rectilinearly at time t = 0 such that its velocity v changes with time t according to the equation v = t2 – t where t is in seconds and v is in m/s. Find the time interval for which the particle retards.

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 14

Acceleration of the particle a = 2t – 1
The particle retards when acceleration is opposite to velocity.

Now t is always positive
∴  (2t – 1) (t – 1) < 0

This is not possible
or  
2t – 1 > 0 & t – 1 < 0 ⇒  1/2 < t < 1

Test: BITSAT Past Year Paper- 2009 - Question 15

A sample of gas expands from volume V1 to V2.
The amount of work done by the gas is greatest when the expansion is

Test: BITSAT Past Year Paper- 2009 - Question 16

A cyclic process is shown in the p-T diagram. Which of the curves show the same process on a P-V diagram ?

Test: BITSAT Past Year Paper- 2009 - Question 17

Which one the following graphs represents the behaviour of an ideal gas

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 17

For an ideal gas PV = constant i.e., PV does not vary with V.

Test: BITSAT Past Year Paper- 2009 - Question 18

In case of a forced vibration, the resonance wave becomes very sharp when the

Test: BITSAT Past Year Paper- 2009 - Question 19

A pendulum bob carries a +ve charge +q. A positive charge +q is held at the point of support.
Then the time period of the bob is – [where, L = length of pendulum, geff = effective value of g]

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 19

Test: BITSAT Past Year Paper- 2009 - Question 20

Two tuning forks A and  B sounded together give 6 beats per second. With an air resonance tube closed at one end, the two forks give resonance when the two air columns are 24 cm and 25 cm respectively. Calculate the frequencies of forks.

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 20

Let the frequency of the first fork be f1 and that of second be f2.
We then have,

We also see that f1 > f2

Solving (i) and (ii), we get
f1 = 150 Hz   and f2 = 144 Hz

Test: BITSAT Past Year Paper- 2009 - Question 21

If an electron has an initial velocity in a direction different from that of an electric field, the path of the electron is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 21

The path is a parabola, because initial velocity can be resolved into two rectangular components, one along  and other ⊥ to  . The former decreases at a constant rate and latter is unaffected. The resultant path is therefore a parabola.

Test: BITSAT Past Year Paper- 2009 - Question 22

If on combining two charged bodies, the current does not flow then

Test: BITSAT Past Year Paper- 2009 - Question 23

Calculate the area of the plates of a one farad parallel plate capacitor if separation between plates is 1 mm and plates are in vacuum

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 23

For a parallel plate capacitor


This corresponds to area of square of side 10.6 km which shows that one farad is very large unit of capacitance.

Test: BITSAT Past Year Paper- 2009 - Question 24

The length of a potentiometer wire is ℓ. A cell of emf E is balanced at a length ℓ/3 from the positive end of the wire. If the length of the wire is increased by ℓ/2. At what distance will be the same cell give a balance point.

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 24

Potential gradient in the first case 

Potential gradient in second case

From equations (i) and (ii),

Test: BITSAT Past Year Paper- 2009 - Question 25

A conducting circular loop  of radius r carries a constant current i. It is placed in a uniform magnetic field such that is perpendicular to the plane of the loop. The magnetic force acting on the loop is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 25

The magnetic field is perpendicular to the plane of the paper. Let us consider two diametrically opposite elements. By Fleming's Left hand rule on element AB the direction of force  will be Leftwards and the magnitude will be
dF = Idl B sin 90° = IdlB

On element CD, the direction of force will be towards right on the plane of the papper and the magnitude will be dF = IdlB.

Test: BITSAT Past Year Paper- 2009 - Question 26

An ammeter reads upto 1 ampere. Its internal resistance is 0.81ohm. To increase the range to 10 A the value of the required shunt is

Test: BITSAT Past Year Paper- 2009 - Question 27

At the magnetic north pole of the earth, the value of horizontal component of earth’s magnetic field and angle of dip are, respectively

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 27

At the magnetic north pole, the magnetic needle will point vertically. There is no component of earth’s magnetic field in the horizontal direction and the angle of dip (the angle  that the resultant magnetic field at the place makes with the horizontal) is 90°.
H = 0,   δ = 90° (maximum)

Test: BITSAT Past Year Paper- 2009 - Question 28

Lenz’s law is a consequence of the law of conservation of

Test: BITSAT Past Year Paper- 2009 - Question 29

The instantaneous current from an a.c. source is I = 6 sin 314 t. What is the rms value of the current?

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 29

Test: BITSAT Past Year Paper- 2009 - Question 30

A coil has resistance 30 ohm and inductive reactance 20 ohm at 50 Hz frequency. If an ac source, of 200 volt, 100 Hz, is connected across the coil, the current in the coil will be

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 30

If ω = 50 × 2π then wL = 20Ω If ω' = 100 × 2π then ω'L = 40Ω Current flowing in the coil is

Test: BITSAT Past Year Paper- 2009 - Question 31

The magnetic field in a travelling electromagnetic wave has a peak value of 20 nT. The peak value of electric field strength is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 31

Test: BITSAT Past Year Paper- 2009 - Question 32

A plano-convex lens of focal length 30 cm has its plane surface silvered. An object is placed 40 cm from the lens on the convex side. The distance of the image from the lens is

Test: BITSAT Past Year Paper- 2009 - Question 33

When a mica sheet of thickness 7 microns and μ = 1.6 is placed in the path of one of interfering beams in the biprism experiment then the central fringe gets at the position of seventh bright fringe.The wavelength of light used will be

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 33


According to question
n = 7 . m = 1.6,  t = 7 x 10–6 meter .....(2)
From eqs. (1) and (2),  l = 6 x 10–7 meter

Test: BITSAT Past Year Paper- 2009 - Question 34

In Young's double slit experiment, if the slit widths are in the ratio 1 : 2, the ratio of the intensities at minima and maxima will be

Test: BITSAT Past Year Paper- 2009 - Question 35

In a photoelectric experiment, with light of wavelength l, the fastest electron has speed v. If the exciting wavelength is changed to 3l/4, the speed of the fastest emitted electron will become

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 35


Test: BITSAT Past Year Paper- 2009 - Question 36

Taking Rydberg’s constant RH = 1.097 x 107m, first and second wavelength of Balmer series in hydrogen spectrum is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 36


For first wavelength, n1 = 2, n2 = 3
⇒ λ1 = 6563 Å. For second wavelength, n1 = 2, n2 = 4
⇒ λ2 = 4861 Å

Test: BITSAT Past Year Paper- 2009 - Question 37

An X-ray tube is operated at 15 kV. Calculate the upper limit of the speed of the electrons striking the target.

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 37

The maximum kinetic energy of an electron accelerated through a potential difference of V volt is 1/2mv= eV

Test: BITSAT Past Year Paper- 2009 - Question 38

Nuclear energy is released in fission since binding energy per nucleon is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 38

Nuclear energy is relased in fission because BE/nucleon is larger for fission fragments than for parent nucleus.

Test: BITSAT Past Year Paper- 2009 - Question 39

Assuming the diodes to be of silicon with forward resistance zero, the current I in the following circuit is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 39

Test: BITSAT Past Year Paper- 2009 - Question 40

The truth table given below corr espon d to the logic gate​

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 40

The given gate is a NOR gate. The output is high, when all inputs are low.


Boolean expression: 

Test: BITSAT Past Year Paper- 2009 - Question 41

Given the numbers : 161 cm, 0.161 cm, 0.0161 cm.The number of significant figures for the three numbers are

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 41

Each has three significant figures. When zero is used to locate the decimal point, it is not considered as significant figure.

Test: BITSAT Past Year Paper- 2009 - Question 42

Beryllium resembles much with :

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 42

Beryllium resembles with aluminium due to similarity in the size of ions and similarity in electropositive character. This type of resemblance between first element of a group in second period with second element of the next group is termed as diagonal relationship.

Test: BITSAT Past Year Paper- 2009 - Question 43

Which one of the following ions has the highest value of ionic radius?

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 43

The ionic radii follows the order
O2– > F > Li+ > B3+

Test: BITSAT Past Year Paper- 2009 - Question 44

Which of the following two are isostructural ?

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 44

XeF2 an d IF-2  both are lin ear and h ave hybridisation sp3d.

Test: BITSAT Past Year Paper- 2009 - Question 45

The cooking time in a pressure cooker is less because :

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 45

In pressure cooker, pressure is high thus, the boiling point of water increases, resulting cooking time is less than other open pots.

Test: BITSAT Past Year Paper- 2009 - Question 46

For the reaction : N2 + 3H2 ⇌ 2NH3
Which one of the following is correct regarding DH :

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 46


According to thermodynamics’s Ist law ΔH = ΔE + nRT
Where ΔH = enthalpy of reaction at constant pressure
ΔE = heat of reaction at constant volume
R = molar gas constant
T = temperature of the reaction
n = (no. of moles of product) – (no. of moles of reactant.)
From reaction, n = nP – nR = 2 – 4 = – 2 Hence, ΔH = ΔE - 2RT .

Test: BITSAT Past Year Paper- 2009 - Question 47

One mole of an ideal gas at 300 K is expanded isothermally from an initial volume of 1 litre to 10 litres. The ΔE for this process is (R = 2 cal mol–1 K–1)

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 47

For isothermal process, ΔE = 0

Test: BITSAT Past Year Paper- 2009 - Question 48

At 25°C and 1 bar which of the following has a non-zero ΔH°f ?

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 48

Ozone is allotropic form of oxygen and is of higher energy (by 68 K Cal mol–1) than O2.
Hence it can not be taken as the reference in standard state.

Test: BITSAT Past Year Paper- 2009 - Question 49

If the equilibrium constant of the reaction  is 0.25, then the equilibrium constant for the reaction ​ would be

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 49

When the reaction is reversed,

Test: BITSAT Past Year Paper- 2009 - Question 50

The oxidation states of sulphur in the anions  follow the order

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 50

According to chemical bond method:

Test: BITSAT Past Year Paper- 2009 - Question 51

The value of x is maximum for

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 51

Because of smaller size, Mg2+ ions are extensively hydrated.

Test: BITSAT Past Year Paper- 2009 - Question 52

For making good quality mirrors, plates of float glass are used. These are obtained by floating molten glass over a liquid metal which does not solidify before glass. The metal used can be

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 52

It is mercury, because mercury exists as liquid at room temperature.

Test: BITSAT Past Year Paper- 2009 - Question 53

The intermediate formed during the addition of HCl to propene in the presence of peroxide is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 53

The addition of HCl to propene proceeds by ionic mechanism and not by free radical mechanism. Hence it forms intermediate carbonium ion.

Test: BITSAT Past Year Paper- 2009 - Question 54

Which of the following has zero dipole moment?

Test: BITSAT Past Year Paper- 2009 - Question 55

Keto-enol tautomerism is observed in

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 55

Test: BITSAT Past Year Paper- 2009 - Question 56

Which one of the following contain isopropyl group?

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 56


It contains isopropyl group.

Test: BITSAT Past Year Paper- 2009 - Question 57

The statement which is not correct about control of particulate pollution is:

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 57

Par ticulates acquire n egative charge and are attracted by the positive electrode.

Test: BITSAT Past Year Paper- 2009 - Question 58

Chief source of soil and water pollution is:

Test: BITSAT Past Year Paper- 2009 - Question 59

The false statement among the followings:

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 59

The average residence time of NO is 4 days.

Test: BITSAT Past Year Paper- 2009 - Question 60

The atomic radius of atom is r.Total volume of atoms present in a fcc unit cell of an element is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 60

4 atom are present in fcc.

Test: BITSAT Past Year Paper- 2009 - Question 61

Which one of the following statements is false?

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 61

ΔTf = Kf x m x i. Since Kf has different values for different solvents, hence even if the m is the same ΔTf will be different.

Test: BITSAT Past Year Paper- 2009 - Question 62

The degree of dissociation of Ca(NO3)2 in dilute aq. solution containing 7.0 g of salt per 100 g of water at 100° C is 70%. If vapour pressure of water at 100° C is 760 mm Hg. The vapour pressure of solution is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 62



Test: BITSAT Past Year Paper- 2009 - Question 63

When the sample of copper with zinc impurity is to be purified by electrolysis, the appropriate electrodes are
 

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 63

In electrolytic purification cathode is of pure metal and anode is of impure metal.

The pure metal is thus deposited at cathode.

Test: BITSAT Past Year Paper- 2009 - Question 64

Th e con ductivity of a saturated solution of BaSO4 is 3.06 x 10–6 ohm–1 cm–1 and its equivalent conductance is 1.53 ohm–1 cm2 equiv–1. The Ksp for BaSO4 will be

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 64

Test: BITSAT Past Year Paper- 2009 - Question 65

In a cell that utilises the reaction 2
Zn (s) + 2H+ (aq) → Zn+ (aq) + H2 (g)  
addition of H2SO4 to cathode compartment, will

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 65


Addition of H2SO4 will increase [H+] and Ecell will also increase and the equilibrium will shift towards RHS.

Test: BITSAT Past Year Paper- 2009 - Question 66

The chemical reaction 2O3 → 3O2 proceeds as follows:

the rate law expression should be

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 66



Note: Intermediates are never represented in rate law equation.

Test: BITSAT Past Year Paper- 2009 - Question 67

Among th e following statements the incorr ect one is :

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 67

Cuprite is Cu2O and Argentite is Ag2S.

Test: BITSAT Past Year Paper- 2009 - Question 68

Cinnabar is an ore of

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 68

Cinnabar (HgS) is an ore of Hg.

Test: BITSAT Past Year Paper- 2009 - Question 69

Which of the following is used in the preparation of chlorine?

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 69

Both MnO2 an d KMnO4 used for the preparation of chlorine by the action of conc. HCl.

Test: BITSAT Past Year Paper- 2009 - Question 70

Which of the following elements does not belong to first transition series?

Test: BITSAT Past Year Paper- 2009 - Question 71

[EDTA]4– is a :

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 71

Test: BITSAT Past Year Paper- 2009 - Question 72

Which of the following order is not correct ?

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 72

The more is the stability of intermediate carbonium ion, the more is the chance of SN1 mechanism. The intermediates obtained will be

The stabilty is of the order iv > iii > ii > i.

Test: BITSAT Past Year Paper- 2009 - Question 73

When esters are hydrolysed the product gives hydrogen ions. The product which gives hydrogen ion is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 73

When esters are hydrolysed , then acid and alcohol are formed, where acid gives hydrogen ion.

Test: BITSAT Past Year Paper- 2009 - Question 74

Which of the following compound can not be used in preparation of iodoform?

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 74

Formaldehyde can not produce iodoform, as only those compound which contains either  group or
group on reaction with potassium iodide and sod. hypochlorite yield iodoform.

Test: BITSAT Past Year Paper- 2009 - Question 75

Which of the following compound is obtained by heating ammonium cyanate?

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 75

Urea is obtained by heating ammonium cyanate

Test: BITSAT Past Year Paper- 2009 - Question 76

Which of the following statements about vitamin B-12 is incorrect?

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 76

Vitamin B12 does not occur in plants.

Test: BITSAT Past Year Paper- 2009 - Question 77

Ammonia forms the complex ion [Cu(NH3)4]2+ with copper ions in alkaline solutions but not in acidic solutions. What is the reason for it ?

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 77

Test: BITSAT Past Year Paper- 2009 - Question 78

An aqueous solution of a substan ce gives a white precipitate on treatment with dil. HCl which dissolves on heating. When H2S is passed through the hot acidic solution, a black precipitate is obtained. The substance is a

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 78

PbCl2 is insoluble in cold water, soluble in hot water and PbS is black ppt in acidic medium.

Test: BITSAT Past Year Paper- 2009 - Question 79

The one which is least basic is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 79

More the electron density on N, higher will be the basicity. Density on N is influenced by the (i) nature of the group (+I or –I) present in alkyl group or benzene nucleus and (ii) resonance (delocalisation of the electron  present on N). In (C6H5)N: electron pair is delocalised to the maximum extent due to three benzene rings and hence least available for protonation, thus it will be least basic.

Test: BITSAT Past Year Paper- 2009 - Question 80

Interparticle forces present in n ylon 66 are