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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 2023 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 2023 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

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

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

Test: BITSAT Past Year Paper- 2009 - Question 81

In the following question, choose the alternatives which can be substituted for the given word.

Agnostic

Test: BITSAT Past Year Paper- 2009 - Question 82

In the following question, choose the alternatives which can be substituted for the given word.
Bohemian

Test: BITSAT Past Year Paper- 2009 - Question 83

In the following question, choose the alternatives which can be substituted for the given word.

Cacographist

Test: BITSAT Past Year Paper- 2009 - Question 84

DIRECTION: Which one of the following word is correctly spelt?

Spelling test-find correct spelling :

Test: BITSAT Past Year Paper- 2009 - Question 85

DIRECTION: Which one of the following word is correctly spelt?

Spelling test-find correct spelling:

Test: BITSAT Past Year Paper- 2009 - Question 86

DIRECTION: Which one of the following word is correctly spelt?

Spelling test-find correct spelling :

Test: BITSAT Past Year Paper- 2009 - Question 87

DIRECTIONS : In the following questions, choose the alternative which is opposite in meaning to the word given in capital letters.

REPRIMAND

Test: BITSAT Past Year Paper- 2009 - Question 88

DIRECTIONS : In the following questions, choose the alternative which is opposite in meaning to the word given in capital letters.

IMPERTINENT

Test: BITSAT Past Year Paper- 2009 - Question 89

DIRECTIONS : In the following questions, choose the alternative which is opposite in meaning to the word given in capital letters.

EQUIVOCAL

Test: BITSAT Past Year Paper- 2009 - Question 90

DIRECTION : In the following question, choose the most appropriate altarnative to fill in the blank.

It is difficult to believe what he tells us because his account of any event is always full of ............ of all sorts.

Test: BITSAT Past Year Paper- 2009 - Question 91

DIRECTION : In the following question, choose the most appropriate altarnative to fill in the blank.

The bank clerk tried to ............... money from his friend’s account.

Test: BITSAT Past Year Paper- 2009 - Question 92

DIRECTION : In the following question, choose the most appropriate altarnative to fill in the blank.

TEight scientists have ............. the national awards for outstanding contribution and dedication to the profession.

Test: BITSAT Past Year Paper- 2009 - Question 93

DIRECTION : In the following question, some parts have been jumbled up. You are required to rearrange these parts, which are labelled P, Q, R and S to produce the correct sentence.

Freedom, is the restricted kind in the sense/(P), the rich and poor woman/(Q), that a wide gulf separates/(R), which a modern woman enjoys (S)

Test: BITSAT Past Year Paper- 2009 - Question 94

DIRECTION : In the following question, some parts have been jumbled up. You are required to rearrange these parts, which are labelled P, Q, R and S to produce the correct sentence.

In life, some rules are/(P), as in business/(Q), they seem almost instinctive/(R), learnt so early that/ (S)

Test: BITSAT Past Year Paper- 2009 - Question 95

DIRECTION : In the following question, some parts have been jumbled up. You are required to rearrange these parts, which are labelled P, Q, R and S to produce the correct sentence.

Kapil, left in an aeroplane/(P), after reading a sailing magazine/(Q), had decided/(R), to build his own boat nine years earlier/(S)

Test: BITSAT Past Year Paper- 2009 - Question 96

DIRECTION : In the following question, select the related letter/word/number from the given alternatives.

Distance  : Odometer :: ? : Barometer

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 96

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

Test: BITSAT Past Year Paper- 2009 - Question 97

One of the, numbers does not fit into the series.

Find the wrong number 13, 16, 38, 124, 504, 2535

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 97

The number should be 123.
× 1 + 3, × 2 + 6, × 3 + 9……

Test: BITSAT Past Year Paper- 2009 - Question 98

DIRECTION: In the question below is given a statement followed by three assumptions numbered I, II and III. You have to consider the statement and the following assumptions, decide which of the assumptions is implicit in the statement and choose your answer accordingly.

Statement: In order to reduce the gap between income and expenditure, the company has decided to increase the price of its product from next month.

Assumptions: I. The rate will remain more or less same after the increase.
II. The expenditure will more or less remain the same in near future.
III. The rival companies will also increase the price of the similar product.

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 98

Clearly, the company intends to reduce the gap between income and expenditure by increasing the price of its product i.e. by keeping the expenditure unaltered and increasing the income only. So, II is implicit while I is not. However, the rival companies may or may not follow the same pursuit.
So, III is not implicit.

Test: BITSAT Past Year Paper- 2009 - Question 99

DIRECTION: In the following question, select the related letter/word/number from the given alternatives.

FLMO  :  ?  : :  BFEN  :  ARSO

Test: BITSAT Past Year Paper- 2009 - Question 100

If
A denotes '+'
B denotes '-'
C denotes '×'
Then what is the value of (10 C 4) A (4 C 4) B 6?

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 100

Using correct symbols we have:
(10 × 4) + (4 × 4) – 6 = 40 + 16 – 6 = 50

Test: BITSAT Past Year Paper- 2009 - Question 101

DIRECTION : In this questions, two figure/ words are given to the left of the sign and one  the sign:: with four alternatives under it out. of which one of the alternatives has the same relationship with the figures/words to the right of the sign:: as between the two figures/words to the left of the sign (::). Find the correct alternative.

Test: BITSAT Past Year Paper- 2009 - Question 102

Identify the missing part of the figure and select it from the given alternatives.

Test: BITSAT Past Year Paper- 2009 - Question 103

Figure (x) is embedded in anyone of the four alternative figures.

Choose the alternative which contains figure (x).

Test: BITSAT Past Year Paper- 2009 - Question 104

Which symbol will appear on the opposite surface to the symbol x?

Test: BITSAT Past Year Paper- 2009 - Question 105

The three figures marked X, Y, Z show the manner in which a paper is folded step by step and then cut. From the answer figures (a), (b), (c), (d), select the one, showing the unfolded position of the Paper after the cut.

Test: BITSAT Past Year Paper- 2009 - Question 106

If A = {1, 2, 3, 4, 5}, then the number of proper subsets of A is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 106

Number of proper subsets of A = 2n –1 Given : A = {1, 2, 3, 4, 5}
Here n = 5
∴ no. of proper subsets = 25 – 1

Test: BITSAT Past Year Paper- 2009 - Question 107

The range of the function  where x ∈ R, is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 107

y – 1 ≠ 0
⇒ y ≠ 1
If y = 1 then original equation gives x = 0, so taking y = 1
Also 3y2 – 10y + 3 ≤ 0
⇒ (3y – 1) (y – 3) ≤ 0

Test: BITSAT Past Year Paper- 2009 - Question 108

if  then value of  is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 108

Test: BITSAT Past Year Paper- 2009 - Question 109

Period of

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 109

Hence period = 2π/3

Test: BITSAT Past Year Paper- 2009 - Question 110

The general solution of  is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 110

We have,

⇒ 8 cos x - 8 cos2 x = (1 + cos x)2
⇒ 9 cos2 x - 6 cos x + 1 = 0
⇒ (3 cos x - 1) 2 = 0
⇒ 3 cos x -1 = 0

Test: BITSAT Past Year Paper- 2009 - Question 111

10n + 3(4n+2) + 5 is divisible by ( n ∈ N)

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 111

10n + 3(4n+ 2 )+5 Taking n = 2;
102 + 3 x 44+5 = 100 + 768 + 5 = 873
Therefore this is divisible by 9.

Test: BITSAT Past Year Paper- 2009 - Question 112

If the expression x2 – 11x + a  and x2 – 14x + 2a must have a common factor and a ≠ 0, then, the common factor is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 112

Here Let x – α is the common factor then x = α is root of the corresponding equation
∴ α2 – 11α + α = 0
α2  – 14α + 2α = 0
Subtracting 3α – α = 0
⇒  α = α/3

Test: BITSAT Past Year Paper- 2009 - Question 113

For the equation  if the product of roots is zero, then the sum of roots is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 113

or  x2 + (a + b) x + ab = (b – a) x + (b - a) c
or   x2 + 2ax + ab + ca – bc = 0
Since product of the roots = 0

Test: BITSAT Past Year Paper- 2009 - Question 114

If  , then

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 114

Test: BITSAT Past Year Paper- 2009 - Question 115

If  then x lies in the interval

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 115

Test: BITSAT Past Year Paper- 2009 - Question 116

The letters of the word TOUGH are written in all possible orders and these words are written out as in a dictionary, then the rank of the word TOUGH is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 116

Rank = (4!x 3) + (3!x 2) + (2!x 2)+ 1 = 72 + 12 + 4 + 1 = 89

Test: BITSAT Past Year Paper- 2009 - Question 117

If in the expansion of  and sum of the binomial coefficients of second and third terms is 36, then the value of x is –

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 117

Test: BITSAT Past Year Paper- 2009 - Question 118

The 100th term of the sequence 1, 2, 2, 3, 3, 3, 4, 4, 4, 4,... is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 118

1st term → 1, 2nd term = 2, 4th term → 3, 7th term → 4, 11th term → 5,...
Series is 1, 2, 4, 7, 11,...

If n = 14, then an = 92,If n = 15, then an = 106.

Test: BITSAT Past Year Paper- 2009 - Question 119

The line 3x – 4y + 7 = 0 is rotated through an angle π/4 in the clockwise direction about thepoint (–1, 1). The equation of the line in its new position is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 119

As (–1, 1) is a point on 3x – 4y + 7 = 0, the rotation is possible. Slope of the given line = 3/4. Slope of the line in its new position

The required equation is

Test: BITSAT Past Year Paper- 2009 - Question 120

Find the vertex of the parabola x2 – 8y – x + 19 = 0.

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 120

The given equation of Parabola can be written as

Test: BITSAT Past Year Paper- 2009 - Question 121

If  then f ' (1/t) is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 121

Test: BITSAT Past Year Paper- 2009 - Question 122

If : p Raju is tall and q: Raju is intelligent, then the symbolic statement ~ p v q means

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 122

~ p v q : Raju is not tall or he is intelligent.

Test: BITSAT Past Year Paper- 2009 - Question 123

Given below is a frequency distribution with median 46. In this distribution, some of the frequencies are missing : Determine the missing frequencies.

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 123

∴ Median class = 40 – 50
∴ l = 40 , c.f. = 42 + x,  f = 65,  h = 10

∴ x = 34 (∵ Number of students cannot be in fraction) Now Sfi = 29
∴ x + y + 150 = 229 x + y = 229 – 150 = 79 .......... (i)
Putting the value of x in (i), we get 34x + y = 79
⇒ y = 79 – 34 = 45
∴ x = 34, y = 45

Test: BITSAT Past Year Paper- 2009 - Question 124

If the function f : (– ∞ ,∞) → B defined by f (x) = – x2 + 6x – 8 is bijective, then B =

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 124

Since the function f is bijective, therefore f is onto. Therefore range of f = B.
Let y = – x2 + 6x – 8
⇒  x2 – 6x + (8 + y ) = 0

Test: BITSAT Past Year Paper- 2009 - Question 125

Find the value of

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 125

Test: BITSAT Past Year Paper- 2009 - Question 126

If A and B are 2 x 2 matrices, then which of the following is true ?

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 126

For two 2 x 2 matrices, A & B (A – B) × (A + B)
= A × A + A × B – B × A – B × B
= A2 – B2 + AB – BA
Hence, (A – B) (A + B) = A2 + AB – BA – B2

Test: BITSAT Past Year Paper- 2009 - Question 127

If a > 0, b > 0, c > 0 are respectively the pth, qth, rth terms of G.P., then the value of the determinant

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 127

Let A be the 1st term and R the common ratio of G.P., then
a = Tp = AR-1
∴ log a = log A + (p - 1) log R
Similarly, log b = log A + (q - 1) log R
and    log c = log A + (r - 1) log R

Split into two determinants and in the first take logA common and in the second take log R common

Test: BITSAT Past Year Paper- 2009 - Question 128

The digits A, B and C are such that the three digit numbers A88, 6B8, 86C are divisible by 72 then the determinant   is divisible by

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 128

which is divisible by 72.

Test: BITSAT Past Year Paper- 2009 - Question 129

If   then [M(α) M (β)]–1 is equal to -

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 129

Test: BITSAT Past Year Paper- 2009 - Question 130

If y = e–x cos x and y4 + ky = 0, where then k =

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 130

Test: BITSAT Past Year Paper- 2009 - Question 131

The set of points of discontinuity of the function 1/log | x | is –

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 131

Let
The points of disconinuity of f (x) are those points where f(x) is undefined or infinite. It is undefined when x = 0 and is infinite when
log | x | = 0, | x | = 1, i.e. x = ± 1.
∴ Set of points of discontinuity = {–1, 0, 1}.

Test: BITSAT Past Year Paper- 2009 - Question 132

The minimum value of the function y = x4 – 2x2 + 1 in the interval  is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 132

For max. or min,
4x (x2 – 1) = 0; either x = 0 or x = ± 1
x = 0 and x = – 1 does not belong to

∴ there is minimum value of function at x = 1
∴ minimum value is
y(1) = 14 – 2(1)2 + 1 = 1 – 2 + 1 = 0

Test: BITSAT Past Year Paper- 2009 - Question 133

The value of a in order that f (x) = sin x – cos x – ax + b decreases for all real values is given by

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 133

We have ;  f (x) = sin x – cos x – ax + b ⇒  f ' (x) = cos x + sin x – a
⇒  f ' (x) < 0  x ∈ R
⇒  (cos x + sin x) < a ∀ x ∈ R
As the max. value of (cos x + sin x)  is √2

Test: BITSAT Past Year Paper- 2009 - Question 134

The equation of tangent to the curve y = sin x at the point (π, 0) is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 134

Therefore the equation of tangent at ( π, 0) is given by
y – 0 = –1 (x – π)
⇒ x + y = π

Test: BITSAT Past Year Paper- 2009 - Question 135

If  = Aln | cos x + sin x -2 | +Bx+C.

Then the ordered triplet A, B, l is –

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 135

Test: BITSAT Past Year Paper- 2009 - Question 136

Evaluate:

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 136

Test: BITSAT Past Year Paper- 2009 - Question 137

Evaluate :

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 137

Test: BITSAT Past Year Paper- 2009 - Question 138

if

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 138

Test: BITSAT Past Year Paper- 2009 - Question 139

Area of the triangle formed by the line x + y = 3 and angle bisectors of the pair of straight lines x2 – y2 + 2y =1 is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 139

x2 – y2 + 2y = 1
⇒ x = ±( y - 1)
Bisectors of above line are x = 0 & y = 1

So area between x = 0, y = 1 & x + y = 3 is shaded Region shown in figure.

Area = ½ × 2 × 2 = 2 sq. units

Test: BITSAT Past Year Paper- 2009 - Question 140

Solution of the differential equation is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 140

= – xcosx + sinx + C
⇒ x (y + cosx) = sinx + C

Test: BITSAT Past Year Paper- 2009 - Question 141

If the line  lies in the plane 2x – 4y + z = 7, then the value of k is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 141

The point (4, 2, k) on the line also lies on the plane 2x – 4y + z = 7.
So,  8 – 8 + k = 7
⇒   k = 7

Test: BITSAT Past Year Paper- 2009 - Question 142

A line segment has length 63 and direction ratios are  3, –2, 6. If the line makes an obtuse angle with x–axis, the components of the line vector are

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 142

Let the components of the line vector be a, b, c.
Then  a2 + b2 + c2 = (63)2 ... (i)
Also

(say),  then a = 3λ ,
b = –2λ and c = 6λ  and from (i) we have

Since a = 3λ < 0 as the line makes an obtuse angle with x-axis, λ = –9 and the required components are –27, 18, –54.

Test: BITSAT Past Year Paper- 2009 - Question 143

It is given that the events A and B are such that

Then P(B) is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 143

By conditional probability,

Test: BITSAT Past Year Paper- 2009 - Question 144

The random variable X has the following probability distribution

Then the value of P(X ≥ 2) is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 144

Since, SPi(X = x) = 1

Test: BITSAT Past Year Paper- 2009 - Question 145

In a triangle ABC, ∠C = 90°, then   is equal to:

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 145

A + B = 180° – C = 90°
a = 2R sin A, b = 2R sin B, c = 2R sin C

Test: BITSAT Past Year Paper- 2009 - Question 146

A person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is 60º. when he retreats 20 feet from the bank, he finds the angle to be 30º. The breadth of the river in feet is :

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 146

Let h be the height of tree PQ and breadth of river PS be x ft. Angle of elevation subtended by a tree is 60°. Also, when he retreats 20 feet, the angle becomes 30°.
Also, in DPQS,

Test: BITSAT Past Year Paper- 2009 - Question 147

The minimum value of the function z = 4x + 3y subject to the constraints 3x + 2y ≥ 160, 5x + 2y ≥ 200, x + 2y ≥ 80, x ≥ 0, y ≥ 0 is

Test: BITSAT Past Year Paper- 2009 - Question 148

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 148

Test: BITSAT Past Year Paper- 2009 - Question 149

Two tangents PQ and PR drawn to the circle x2 + y2 – 2x – 4y – 20 = 0 from point P (16, 7). If the centre of the circle is C then the area of quadrilateral PQCR is

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 149

where L = length of tangent and r = radius of circle.

Hence the required  area = 75 sq. units.

Test: BITSAT Past Year Paper- 2009 - Question 150

The value of

Detailed Solution for Test: BITSAT Past Year Paper- 2009 - Question 150

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