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## 150 Questions MCQ Test BITSAT Mock Tests Series & Past Year Papers | Test: BITSAT Past Year Paper- 2012

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Test: BITSAT Past Year Paper- 2012 - Question 1

### What is the moment of inertia of a solid sphere of density r and radius ρ about its diameter?

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 1

For solid sphere

Test: BITSAT Past Year Paper- 2012 - Question 2

### A body moves with uniform acceleration, then which of the following graph is correct ?

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 2

An object is said to be moving with a uniform acceleration, if its velocity changes by equal amount in equal intervals of time. The velocity-time graph of uniformly accelerated motion is a straight line inclined to time axis.
Acceleration of an object in a uniformly accelerated motion in one dimension is equal to the slope of the velocity-time graph with time axis.

Test: BITSAT Past Year Paper- 2012 - Question 3

### A projectile can have the same range R for two angles of projection. If t1 and t2 be the times of flight in two cases, then what is the product of two times of flight?

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 3

where R is the range.
Hence t1 t2 ∝ R

Test: BITSAT Past Year Paper- 2012 - Question 4

A horizontal overhead powerline is at height of 4m from the ground and carries a current of 100A from east to west. The magnetic field directly below it on the ground is (μ0 = 4π × 10–7 Tm A–1)

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 4

The magnetic field is

According to right hand palm rule, the magnetic field is directed towards south.

Test: BITSAT Past Year Paper- 2012 - Question 5

A man of mass 100 kg. is standing on a platform of mass 200 kg. which is kept on a smooth ice surface. If the man starts moving on the platform with a speed 30 m/sec relative to the platform then calculate with what velocity relative to the ice the platform will recoil?

Test: BITSAT Past Year Paper- 2012 - Question 6

If the unit of force and length be each increased by four times, then the unit of energy is increased by

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 6

Since unit of energy = (unit of force).(unit of length) so if we increase unit of length and force, each by four times, then unit of energy will increase by sixteen times.

Test: BITSAT Past Year Paper- 2012 - Question 7

Which of the following must be known in order to determine the power output of an automobile?

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 7

Power is defined as the rate of doing work.
For the automobile, the power output is the amount of work done (overcoming friction) divided by the length of time in which the work was done.

Test: BITSAT Past Year Paper- 2012 - Question 8

If the force is given by F = at + bt2 with t as time. The dimensions of a and b are

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 8

Dimension of at = Dimension of F

Dimension of bt2 = Dimension of F

Test: BITSAT Past Year Paper- 2012 - Question 9

A wheel of radius R rolls on the ground with a uniform velocity v. The relative acceleration of topmost point of the wheel with respect to the bottom most point is

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 9

As  aCM = 0   [vCM = constant],
Tangential acceleration of each point

Test: BITSAT Past Year Paper- 2012 - Question 10

If the radius of the earth were to shrink by one per cent, its mass remaining the same, the value of g on the earth’s surface would

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 10

Test: BITSAT Past Year Paper- 2012 - Question 11

The Young’s modulus of a perfectly rigid body is

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 11

For a perfectly rigid body strain produced is zero for the given force applied, so Y = stress/strain = ∞

Test: BITSAT Past Year Paper- 2012 - Question 12

An ice block floats in a liquid whose density is less than water. A part of block is outside the liquid. When whole of ice has melted, the liquid level will

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 12

Ice is lighter than water. When ice melts, the volume occupied by water is less than that of ice. Due to which the level of water goes down.

Test: BITSAT Past Year Paper- 2012 - Question 13

A large drop of oil (density 0.8 g/cm3 and viscosity η0) floats up through a column of another liquid (density 1.2 g/cm3 and viscosity ηL).
Assuming that the two liquids do not mix, the velocity with which the oil drop rises will depend on:

Test: BITSAT Past Year Paper- 2012 - Question 14

A solid body of constant heat capacity 1 J/°C is being heated by keeping it in contact with reservoirs in two ways:
(i) Sequen tially keepin g in con tact with 2 reservoirs such that each reservoir supplies same amount of heat.
(ii) Sequen tially keepin g in con tact with 8 reservoirs such that each reservoir supplies same amount of heat.
In both the cases body is brought from initial temperature 100°C to final temperature 200°C.
Entropy change of the body in the two cases respectively is :

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 14

The entropy change of the body in the two cases is same as entropy is a state function.

Test: BITSAT Past Year Paper- 2012 - Question 15

Which of the following process is possible according to the first law of thermodynamics?

Test: BITSAT Past Year Paper- 2012 - Question 16

For an isothermal expansion of a perfect gas, the value of ΔP/P is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 16

Differentiate PV = constant w.r.t V

Test: BITSAT Past Year Paper- 2012 - Question 17

A sample of ideal monoatomic gas is taken round the cycle ABCA as shown in the figure. The work done during the cycle is

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 17

ΔW = area under the p – V curve

Test: BITSAT Past Year Paper- 2012 - Question 18

The average translational kinetic energy of O2 (molar mass 32) molecules at a particular temperature is 0.048 eV. The translational kinetic energy of N2 (molar mass 28) molecules in eV at the same temperature is

Test: BITSAT Past Year Paper- 2012 - Question 19

For a gas if ratio of specific heats at constant pressure and volume is γ then value of degrees of freedom is

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 19

Test: BITSAT Past Year Paper- 2012 - Question 20

One end of a long metallic wire of length L tied to the ceiling. The other end is tied with a massless spring of spring constant K. A mass hangs freely from the free end of the spring. The area of cross section and the young’s modulus of the wire are A and Y respectively. If the mass slightly pulled down and released, it will oscillate with a time period T equal to :

Test: BITSAT Past Year Paper- 2012 - Question 21

The transverse displacement y(x, t) of a wave on a string is given by

This represents a

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 21

It is a function of type  y = f (ωt + kx)
∴ y (x, t) represents wave travelling along –x direction.

Test: BITSAT Past Year Paper- 2012 - Question 22

A sound source is moving towards stationary listener with 1/10th of the speed of sound. The ratio of apparent to read frequency is

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 22

Test: BITSAT Past Year Paper- 2012 - Question 23

In a region of space having a uniform electric field E, a hemispherical bowl of radius r is placed.The electric flux ϕ through the bowl is

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 23

ϕ = E(ds) cos θ = E(2πr2) cos 0° = 2πr2 E.

Test: BITSAT Past Year Paper- 2012 - Question 24

The electric field intensity just sufficient to balance the earth’s gravitational attraction on an electron will be: (given mass and charge of an electron respectively are 9.1 × 10–31 kg and 1.6 × 10–19 C.)

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 24

– eE = mg

Test: BITSAT Past Year Paper- 2012 - Question 25

Two capacitors C1 and C2 are charged to 120 V and 200 V respectively. It is found that by connecting them together the potential on each one can be made zero. Then

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 25

For potential to be made zero, after connection

⇒ 3C1 = 5C2

Test: BITSAT Past Year Paper- 2012 - Question 26

Three voltmeters A, B and C having resistances R, 1.5 R and 3R, respectively, are connected as shown. When some potential difference is applied between X and Y, the voltmeter readings are VA, VB and VC respectively. Then –

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 26

VA = IR

∴ VA = V= VC

Test: BITSAT Past Year Paper- 2012 - Question 27

The range of the particle when launched at an angle of 15º with the horizontal is 1.5 km. What is the range of the projectile when launched at an angle of 45º to the horizontal.

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 27

Test: BITSAT Past Year Paper- 2012 - Question 28

If m is magnetic moment and B is the magnetic field, then the torque is given by

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 28

Test: BITSAT Past Year Paper- 2012 - Question 29

Magnetic moment of bar magnet is M. The work done to turn the magnet by 90° of magnet in direction of magnetic field B will be

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 29

Work done, W = MB (1 – cos θ)
θ = 90°
W = MB

Test: BITSAT Past Year Paper- 2012 - Question 30

The laws of electromagnetic induction have been used in the construction of a

Test: BITSAT Past Year Paper- 2012 - Question 31

The impedance of a circuit consists of 3 Ω resistance and 4Ω reactance. The power factor of the circuit is

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 31

Power = cos ϕ = 3/5 = 0.6

Test: BITSAT Past Year Paper- 2012 - Question 32

The r.m.s. value of potential difference V shown in the figure is

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 32

Test: BITSAT Past Year Paper- 2012 - Question 33

A ray of light is incident at the glass-water interface at an angle i, it emerges finally parallel to the surface of water, then the value of μg would be

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 33

Test: BITSAT Past Year Paper- 2012 - Question 34

A mica slit of thickness t and refractive index µ is introduced in the ray from the first source S1. By how much distance of fringes pattern will be displaced?

Test: BITSAT Past Year Paper- 2012 - Question 35

In a Young’s double slit experiment the angular width of a fringe formed on a distant screen is 1°. The wavelength fo the light used is 6280 Å. What is the distance between the two coherent sources?

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 35

The angular fringe width is given by α = λ/d
where λ is wavelength and d is the distance between two coherent sources. Thus d =  λ/α

Test: BITSAT Past Year Paper- 2012 - Question 36

A light having wavelength 300 nm fall on a metal surface. The work function of metal is 2.54 eV, what is stopping potential ?

Test: BITSAT Past Year Paper- 2012 - Question 37

If the total binding energies of  nuclei are 2.22, 28.3, 492 and 1786 MeV respectively, identify the most stable nucleus of the following.

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 37

is most stable as it has maximum binding energy per nucleon.

Test: BITSAT Past Year Paper- 2012 - Question 38

An oscillator is nothing but an amplifier with

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 38

A positive feedback from output to input in an amplifier provides oscillations of constant amplitude.

Test: BITSAT Past Year Paper- 2012 - Question 39

In an experiment on photoelectric effect photons of wavelength 300 nm eject electrons from a metal of work function 2.25eV. A photon of energy equal to that of the most energetic electron corresponds to the following transition in the hydrogen atom:

Test: BITSAT Past Year Paper- 2012 - Question 40

A letter 'A' is constructed of a uniform wire with resistance 1.0 Ω per cm, The sides of the letter are 20 cm and the cross piece in the middle is 10 cm long. The apex angle is 60. The resistance between the  ends of the legs is close to:

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 40

Solving we get
x = 10 Ω
Putting the value of x = 10 Ω in equation (i)
We get

Test: BITSAT Past Year Paper- 2012 - Question 41

Number of atoms of He in 100 amu of He (atomic wt. of He is 4) are :

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 41

100 amu of He = 100/4 atoms of He
= 25 atoms.
[1 a.m.u. = mass of one proton (approx.)]

Test: BITSAT Past Year Paper- 2012 - Question 42

If the radius of H is 0.53 Å, then what will be the radius of 3Li2+ ?

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 42

Test: BITSAT Past Year Paper- 2012 - Question 43

Which of the following does not have valence electron in 3d-subshell?

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 43

P (At no. 15) has electronic configuration 1s2, 2s2 p6, 3s2 p3, hence no electron in d-subshell.

Test: BITSAT Past Year Paper- 2012 - Question 44

The vapour pressure of

due to

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 44

Ortho-nitrophenol has intramolecular

and paranitrophenol has intermolecular H-bonding.

Hence former is more volatile than latter.

Test: BITSAT Past Year Paper- 2012 - Question 45

An ideal gas can’t be liquefied because

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 45

In an ideal gas, the intermolecular forces of attraction are negligible and hence it cannot be liquefied.

Test: BITSAT Past Year Paper- 2012 - Question 46

In which of the following reactions, standard entropy change (ΔS°) is positive and standard Gibb’s energy change (ΔG°) decreases sharply with increasing temperature ?

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 46

Since, in the first reaction gaseous products are forming from solid carbon hence entropy will increase i.e. ΔS = +ve.

Since, ΔG° = ΔH° – TΔS hence the value of ΔG decrease on increasing temperature.

Test: BITSAT Past Year Paper- 2012 - Question 47

Bond enthalpies of H2, X2 and HX are in the ratio 2 : 1 : 2. If enthalpy of formation of HX is –50 kJ mol–1, the bond enthalpy of X2 is

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 47

Let the bond enthalpy of X – X  bond be x.
ΔHf (HX) = – 50

∴ x = 50 × 2 = 100 kJ mol–1

Test: BITSAT Past Year Paper- 2012 - Question 48

The pOH value of a solution whose hydroxide ion concentration is 6.2 × 10–9 mol/litre is

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 48

–log (OH) = pOH; – log 6.2 × 10–9 = pOH;
∴ pOH = 8.21

Test: BITSAT Past Year Paper- 2012 - Question 49

Which of the following combinations would not result in the formation of a buffer solution?

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 49

Combination of NaOH and CH3COOH is the mixture of alkali and acetic acid. Therefore this combination can not be buffer forming solution.

Test: BITSAT Past Year Paper- 2012 - Question 50

The reaction, SO2 + Cl2 → SO2Cl2 is exothermic and reversible. A mixture of SO2 (g), Cl2 (g) and SO2Cl2 (l) is at equilibrium in a closed container. Now a certain quantity of extra SO2 is introduced into the container, the volume remaining the same. Which of the following is/ are true?

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 50

By addition of SO2, equilibrium will shift to RHS which is exothermic. Hence temp, will increase.

Test: BITSAT Past Year Paper- 2012 - Question 51

In the reaction

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 51

O.N. of Br2 changes from 0 to –1 and +5, hence it is reduced as well as oxidised.

Test: BITSAT Past Year Paper- 2012 - Question 52

The boiling point of water is exceptionally high because

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 52

The high boiling point of water is due to H-bonding.

Test: BITSAT Past Year Paper- 2012 - Question 53

Which of the following has correct increasing basic strength?

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 53

The basic character of oxides increases down the group.

Test: BITSAT Past Year Paper- 2012 - Question 54

The following two compounds are

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 54

The given two structures are optical isomers but as these are mirror image of each other, hence they represent enantiomers of each other.

Test: BITSAT Past Year Paper- 2012 - Question 55

In paper chromatography :

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 55

Paper chromatography is a special case of partition chromatography where the special quality paper containing water trapped in it acts as a stationary phase and solvent as a mobile phase. Thus, both phases are liquids.

Test: BITSAT Past Year Paper- 2012 - Question 56

In which case the NO2 will attack at the meta position

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 56

m-directing in nature.

Test: BITSAT Past Year Paper- 2012 - Question 57

Which alkene on ozonolysis gives CH3CH2CHO and

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 57

Test: BITSAT Past Year Paper- 2012 - Question 58

Formation of ozone in the upper atmosphere from oxygen takes place by the action of

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 58

In presence of U.V. rays O2 is converted into O3.

Test: BITSAT Past Year Paper- 2012 - Question 59

CO2 goes to air, causes green house effect and gets dissolved in water. What will be the effect on soil fertility and pH of the water?

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 59

Here [H+] increases hence, pH decreases due to which soil fertility will also decreases.

Test: BITSAT Past Year Paper- 2012 - Question 60

The van’t Hoff factor i for an electrolyte which undergoes dissociation and association in solvents are respectively

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 60

When an electrolyte dissociates van’t Hoff factor i is greater than 1 and when it associates the i is less than 1.

Test: BITSAT Past Year Paper- 2012 - Question 61

If the elevation in boiling point of a solution of 10 g of solute (mol. wt. = 100) in 100 g of water is ΔTb, the ebullioscopic constant of water is

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 61

Test: BITSAT Past Year Paper- 2012 - Question 62

The ionic conductance of Ba2+ and Cl respectively are 127 and 76Ω–1cm2 at infinite dilution.The equivalent conductance of BaCl2 at infinite dilution will be

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 62

Test: BITSAT Past Year Paper- 2012 - Question 63

2N2O5 ⇌ 4NO2 + O2
If rate and rate constant for above reaction are 2.40 × 10–5 mol L–1 s–1 and 3 × 10–5 s–1 respectively, then calculate the concentration of N2O5.

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 63

The reaction is of first and for a first order reaction, rate, R = k [N2O5] 2.4 × 10–5 = 3 × 10–5 × [N2O5]

Test: BITSAT Past Year Paper- 2012 - Question 64

Which of the following gas molecules have maximum value of enthalpy of physisorption?

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 64

The more the liquifiable nature of a gas, the more is the enthalpy of adsorption. Water is more liquifiable.

Test: BITSAT Past Year Paper- 2012 - Question 65

Which of the following will be the most effective in the coagulation of Fe(OH)3 soil?

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 65

According to Hardy-Schulze rule, coagulation power of ions is directly proportional to charge on ion.
∵ Fe(OH)3 is positively charged colloid.
∴ It will be coagulated by anion.

Because  has highest charge among the given anions, therefore, Mg3(PO4)2 is the most effective in the coagulation of Fe(OH)3 solution.

Test: BITSAT Past Year Paper- 2012 - Question 66

When chlorine water is exposed to sunlight, O2 is liberated. Hence,

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 66

Hydrogen has more affinity for chlorine.

Test: BITSAT Past Year Paper- 2012 - Question 67

An extremely hot copper wire reacts with steam to give

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 67

Test: BITSAT Past Year Paper- 2012 - Question 68

Among the following the lowest degree of paramagnetism per mole of the compound at 298 K will be shown by

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 68

Hence lowest paramagnetism is shown by CuSO4.5H2O

Test: BITSAT Past Year Paper- 2012 - Question 69

The following reaction is known as

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 69

At 120-140°C temperature and 1.5 atm pressure, sodium phenoxide reacts with CO2 to yield sodium salicylate which on further hydrolysis give to salicylic acid.
This reaction is known as Kolbe’s reaction.

Test: BITSAT Past Year Paper- 2012 - Question 70

Which of the following is process used for the preparation of acetone?

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 70

In Wacker process, when mixture of propene and air is passed through mixture of Pd and CuCl2 at high pressure acetone is formed.

Test: BITSAT Past Year Paper- 2012 - Question 71

The preparation of ethyl acetoacetate involves:

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 71

In Claisen con den sation inter molecular condensation of esters containing a-hydrogen atom in presence of strong base produce β-keto ester.

Test: BITSAT Past Year Paper- 2012 - Question 72

Which one of the following pairs is not correctly matched?

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 72

Like clemmensen reduction, Wolf-Kishner reduction involves reduction of  > C = O to > CH2 , of course by different reagent.

Test: BITSAT Past Year Paper- 2012 - Question 73

Identify ‘C’ in the following reaction:

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 73

Test: BITSAT Past Year Paper- 2012 - Question 74

The helical structure of protein is stabilised by

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 74

Fibrous proteins have thread like molecules which lie side by side to form fibres. The various molecules are held together by hydrogen bonds.

Test: BITSAT Past Year Paper- 2012 - Question 75

Complete hydrolysis of cellulose gives

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 75

Test: BITSAT Past Year Paper- 2012 - Question 76

Alizarin is an example of

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 76

Alizarin is an anthraquinone dye. It gives a bright red colour with aluminium and a blue colour with barium.

Test: BITSAT Past Year Paper- 2012 - Question 77

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 77

2,4-dichlorophenoxyacetic acid is used as a herbicide.

Test: BITSAT Past Year Paper- 2012 - Question 78

0.45 g of acid  molecular weight 90 is neutralised by 20 ml of 0.5N caustic potash. The basicity of acid is

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 78

Eq. of acid = Eq of base,

Test: BITSAT Past Year Paper- 2012 - Question 79

In the reaction of KMnO4 with an oxalate in acidic medium, MnO-4 is reduced to Mn2+ and  is oxidised to CO2. Hence, 50 mL of 0.02 M KMnO4 is equivalent to

Test: BITSAT Past Year Paper- 2012 - Question 80

Which of the following is soluble in yellow ammonium sulphide?

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 80

SnS +(NH4)2 S2 → (NH4)2 SnS3 soluble

Test: BITSAT Past Year Paper- 2012 - Question 81

Direction: In the following questions choose the word opposite in meaning to the given word.

Florid

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 81

The word Flor id (Adjective) means : rosy; gaudy; ornated; red; having too much decoration or detail.
The word Pale (Adjective) means : light in colour; not strong or bright; having skin that is almost white because of illness.
Hence, the words florid and pale are antonymous.

Test: BITSAT Past Year Paper- 2012 - Question 82

Direction: In the following questions choose the word opposite in meaning to the given word.

Verity

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 82

The word Verity (Noun) means : a belief or principle about life that is accepted as true; truth.
Hence, the words verity and falsehood are antonymous.

Test: BITSAT Past Year Paper- 2012 - Question 83

Direction: In the following questions choose the word opposite in meaning to the given word.

Perspicuity

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 83

The word Perspicuity (Noun) means : clarity.
The word vagueness (Noun) means : no clarity in a person’s mind.
Hence, the words perspicuity and Vagueness are antonymous.

Test: BITSAT Past Year Paper- 2012 - Question 84

Direction: In question  out of the four alternative, choose the one which best expresses the meaning of the given word.

Disgrace

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 84

Disgrace means a state of shame.

Test: BITSAT Past Year Paper- 2012 - Question 85

Direction: In question  out of the four alternative, choose the one which best expresses the meaning of the given word.

Striking

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 85

Striking means extraordinary, attractive.

Test: BITSAT Past Year Paper- 2012 - Question 86

Direction: In question  out of the four alternative, choose the one which best expresses the meaning of the given word.

Fiasco

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 86

Fiasco means a complete failure.

Test: BITSAT Past Year Paper- 2012 - Question 87

Direction: In the following questions a part of the sentence is bold. Below are given alternatives to the Underline part at (a), (b) and (c) which may improve the sentence. Choose the correct alternative. In case no improvement is needed, your answer is (d).

Q. Power got with money is the most craved for today.

Test: BITSAT Past Year Paper- 2012 - Question 88

Direction: In the following questions a part of the sentence is bold. Below are given alternatives to the Underline part at (a), (b) and (c) which may improve the sentence. Choose the correct alternative. In case no improvement is needed, your answer is (d).

Q. You are asked to copy this letter word by word.

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 88

Word for word means : in exactly the same words or when translated exactly equivalent words.

Test: BITSAT Past Year Paper- 2012 - Question 89

Direction: Sentences are given with blanks to be filled in with an appropriate word(s). Four alternatives are suggested for each question. Choose the correct alternative out of the four:

Q. Let us quickly __________.

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 89

Huddle : come close in a gr oup

Test: BITSAT Past Year Paper- 2012 - Question 90

Direction: Sentences are given with blanks to be filled in with an appropriate word(s). Four alternatives are suggested for each question. Choose the correct alternative out of the four:

Q. Rajesh’s car wasn’t __________ Ramesh’s, so we were too exhausted by the time we reached home.

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 90

Right use of as - as comparison

Test: BITSAT Past Year Paper- 2012 - Question 91

Direction: In the following questions, the 1st and the last sentences of the passage are numbered 1 and 6. The rest of the passage is split into four parts and named P, Q, R and S. These four parts are not given in their proper order. Read the sentence and find out which of the four combinations is correct. Then find the correct answer.

1. The most vulnerable section of the society are the students.
P. Revolutionary and new fledged ideas have a great appeal to them.
Q. Agitations may be non-violent methods of protest.
R. They cannot resist the charm of persuasion.
S. They are to be taught that without discipline they cannot get proper education.
6. However if th ese become violent , the antisocial elements get encouraged and they put all proper working out of gear.

Test: BITSAT Past Year Paper- 2012 - Question 92

Direction: In the following questions, the 1st and the last sentences of the passage are numbered 1 and 6. The rest of the passage is split into four parts and named P, Q, R and S. These four parts are not given in their proper order. Read the sentence and find out which of the four combinations is correct. Then find the correct answer.

1. Venice is a strange city.
P. There are about 400 odd bridges connecting the islands of Venice.
Q. There are no motor cars, no horses and no buses there.
R. Th ese small islands ar e close to on e another.
S. It is not one island but a hundred islands.
6. This is because Venice has no streets.

Test: BITSAT Past Year Paper- 2012 - Question 93

The World health Organisation is briefly called W.H.O. It is a specialised agency of the United Nations and was established in 1948.
International health workers can be seen working in all kinds of surroundings in deserts, jungles, mountains, coconut groves, and rice fields. They help the sick to attain health and the healthy to maintain their health.
This global health team assists the local health workers in stopping the spread of what are called communicable diseases, like cholera. These diseases can spread from one country to another and so can be a threat to world health.
W.H.O. assists different national health authorities not only in controlling diseases but also in preventing them altogether. Total prevention of diseases is possible in a number so ways. Everyone knows how people, particularly children, are vaccinated against one disease or another. Similarly, most people are familiar with the spraying of houses with poisonous substances which kill disease-carrying insects.

Q. "It is a specialised agency of the United Nations and was established in 1948". Here specialised means :

Test: BITSAT Past Year Paper- 2012 - Question 94

The World health Organisation is briefly called W.H.O. It is a specialised agency of the United Nations and was established in 1948.
International health workers can be seen working in all kinds of surroundings in deserts, jungles, mountains, coconut groves, and rice fields. They help the sick to attain health and the healthy to maintain their health.
This global health team assists the local health workers in stopping the spread of what are called communicable diseases, like cholera. These diseases can spread from one country to another and so can be a threat to world health.
W.H.O. assists different national health authorities not only in controlling diseases but also in preventing them altogether. Total prevention of diseases is possible in a number so ways. Everyone knows how people, particularly children, are vaccinated against one disease or another. Similarly, most people are familiar with the spraying of houses with poisonous substances which kill disease-carrying insects.

Q. "International health workers can be seen working in all kinds of surroundings: in deserts, jungles, mountains, coconout groves, and rice fields". Here International means

Test: BITSAT Past Year Paper- 2012 - Question 95

The World health Organisation is briefly called W.H.O. It is a specialised agency of the United Nations and was established in 1948.
International health workers can be seen working in all kinds of surroundings in deserts, jungles, mountains, coconut groves, and rice fields. They help the sick to attain health and the healthy to maintain their health.
This global health team assists the local health workers in stopping the spread of what are called communicable diseases, like cholera. These diseases can spread from one country to another and so can be a threat to world health.
W.H.O. assists different national health authorities not only in controlling diseases but also in preventing them altogether. Total prevention of diseases is possible in a number so ways. Everyone knows how people, particularly children, are vaccinated against one disease or another. Similarly, most people are familiar with the spraying of houses with poisonous substances which kill disease-carrying insects.

Q. They help the sick to attain health and the healthy to maintain their health. here they stands for:

Test: BITSAT Past Year Paper- 2012 - Question 96

In a code language, if SUMMER is coded as SDNLVR, then the word WINTER will be coded as:

Test: BITSAT Past Year Paper- 2012 - Question 97

Direction: In question number, select the missing number from the given responses.

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 97

(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- 2012 - Question 98

Today is Monday. After 61 days, it will be:

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 98

Each day of the week is repeated after 7 days.
So, after 63 days, it will be Monday.
After 61 days, it will be Saturday.

Test: BITSAT Past Year Paper- 2012 - Question 99

Rahul and Nitesh are standing in a row of persons. Rahul is 12th from left side and Nitesh is 18th from the right side of the row. If they interchanged their positions Rahul becomes 25th from left. Find the new position of Nitesh from right side?

Test: BITSAT Past Year Paper- 2012 - Question 100

One of the numbers does not fit into the series. Find the wrong number.
52, 152, 414, 1312, 5348, 26840

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 100

The number should be 404. × 1 + 100, × 2 + 100, × 3 + 100……

Test: BITSAT Past Year Paper- 2012 - Question 101

In the following question and Δ stands for any of Mathematical signs at different places, which are given as choices under each question. Select the choice with the correct sequence of signs which when substituted makes the question as correct equation?  24 Δ 4 Δ 5 Δ 4

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 101

After putting sign 24 = 4 × 5 + 4
24 = 24
Hence, (b) is correct choice.

Test: BITSAT Past Year Paper- 2012 - Question 102

Which represents carrot, food, vegetable?

Test: BITSAT Past Year Paper- 2012 - Question 103

"All the members of the Tennis club are members of the badminton club too". No woman plays badminton?

Test: BITSAT Past Year Paper- 2012 - Question 104

Test: BITSAT Past Year Paper- 2012 - Question 105

Which answer figure is the exact mirror image of the given figure when the mirror held form the right at PQ?

Test: BITSAT Past Year Paper- 2012 - Question 106

Let A and B be two sets then (A ∪ B)‘∪ (A ‘∩ B) is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 106

From Venn-Euler ’s Diagram.

∴ (A ∪ B)' ∪(A'∩ B) = A'

Test: BITSAT Past Year Paper- 2012 - Question 107

Let x and y be two natural numbers such that xy = 12(x + y) and x ≤ y. Then the total number of pairs (x, y) is

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 107

xy – 12x – 12y = 0  ⇒ (x – 12) (y – 12) = 144
Now 144 can be factorised into two factors x and y where x ≤ y and the factors are (1, 144), (2, 72), (3, 48), (4, 36), (6, 24), (8, 18),
(9, 16), (12, 12).
Thus there are eight solutions.

Test: BITSAT Past Year Paper- 2012 - Question 108

If sin2θ + sin2ϕ = 1/2, cos2θ +cos2ϕ = 3/2, then cos2 (q – ϕ) is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 108

Using cosine formula
2 sin (θ + ϕ) cos (θ – ϕ) = 1/2 .....(i)
2 cos (θ + ϕ) cos (θ – ϕ) =3/2 .....(ii)
Squaring (1) and (2) and then adding

Test: BITSAT Past Year Paper- 2012 - Question 109

Let T(k) be the statement 1 + 3 + 5 + ... +  (2k – 1) = k2 +10
Which of the following is correct?

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 109

When k = 1, LHS = 1 but RHS = 1 + 10 = 11
∴ T(1) is not true
Let T(k) is true. That is
1+ 3+ 5+ ..... + (2k - 1) = k2 + 10
Now, 1+ 3 + 5 + ..... + (2k-1) + (2k+1)
= k2 + 10 + 2k +1 = (k + 1)2 + 10
∴ T(k+1) is true.
That is T(k) is true ⇒ T(k+ 1) is true.
But T(n) is not true for all n ∈ N , as T(1) is not true.

Test: BITSAT Past Year Paper- 2012 - Question 110

The amplitude of

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 110

Test: BITSAT Past Year Paper- 2012 - Question 111

If x = ω – ω2 –2, then the value of x4 + 3x3 + 2x2 – 11x – 6 is

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 111

We have, x = ω – ω2 –2 or x + 2 = ω – ω2
Squaring, x2 + 4x + 4  = ω2 + ω4 – 2ω3
= ω2 + ω3ω. –2ω3  = ω2 + ω – 2 [ω3 = 1]
= –1 – 2 = – 3  ⇒ x2 + 4x + 7 = 0
Dividing x4 + 3x3 + 2x2 –11x – 6  by x2 + 4x + 7, we get
x4 + 3x3 + 2x2 – 11x – 6 = (x2 + 4x + 7)(x2 – x – 1) + 1
= (0) (x2 – x – 1)+1 = 0 + 1 = 1

Test: BITSAT Past Year Paper- 2012 - Question 112

In how many ways can 5 prizes be distributed among 4 boys when every boy can take one or more prizes ?

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 112

First prize may be given to any one of the 4 boys, hence first prize can be distributed in 4 ways. similarly every one of second, third, fourth and fifth prizes can also be given in 4 ways.
∴ the number of ways of their distribution = 4 × 4 × 4 × 4 × 4 = 45 = 1024

Test: BITSAT Past Year Paper- 2012 - Question 113

The number of positive integral solution of abc = 30 is

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 113

We have : 30 = 2 × 3 × 5. So, 2 can be assigned to either a or b or c i.e. 2 can be assigned in 3 ways. Similarly, each of 3 and 5 can be assigned in 3 ways. Thus, the number of solutions is 3 × 3 × 3 = 27.

Test: BITSAT Past Year Paper- 2012 - Question 114

The coefficient of x20 in the expansion of

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 114

= (1 + x2)30. x10
The coefficient of x20 in x10 (1 + x2)30
= the coefficient of x10 in (1 + x2)30
= 30C5 = 30C30–5 = 30C25

Test: BITSAT Past Year Paper- 2012 - Question 115

If x is positive then the sum to infinity of the series

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 115

The  series is a G.P. with common ratio  is less than 1 since x is
positive

Test: BITSAT Past Year Paper- 2012 - Question 116

The nearest point on the line 3x + 4y = 12 from the origin is

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 116

If ‘D’ be the foot of altitude, drawn from origin to the given line, then ‘D’ is the required point.

Let ∠OBA = θ
⇒ tan q = 4/3
⇒ ∠ DOA = θ we have OD = 12/5.
If D is (h, k) then h = OD cosθ, k = OD sinθ
⇒ h = 36/25, k  = 48/25.

Test: BITSAT Past Year Paper- 2012 - Question 117

The length of the tangent drawn from any point on the circle x2 + y2 + 2fy + λ = 0 to the circle x2 + y2 + 2fy + μ = 0, where μ > λ  > 0, is

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 117

Let the radius of the first circle be CT = r1.
Also, let the radius of the second circle be CP = r2.
In the triangle PCT,  T is a right angle

Test: BITSAT Past Year Paper- 2012 - Question 118

Find the eccentricity of the conic represented by x2 – y2 – 4x + 4y + 16 = 0

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 118

We have x2 – y2 – 4x + 4y + 16 = 0
⇒ (x2 – 4x) – (y2 – 4y) = 16
⇒ (x2 – 4x + 4) – (y2 – 4y + 4) = – 16
⇒ (x – 2)2 – (y – 2)2 = – 16

This is rectangular hyperbola, whose eccentricity is always √2.

Test: BITSAT Past Year Paper- 2012 - Question 119

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 119

∴ Given limit

Test: BITSAT Past Year Paper- 2012 - Question 120

Let f (x + y) = f (x) . f (y) for all x, y where f (0) ≠  0. If f (5) = 2 and f ' (0) = 3, then f ' (5) is equal to –

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 120

Test: BITSAT Past Year Paper- 2012 - Question 121

If sample A contains 100 observations 101, 102, .... 200 and sample B contains 100 obsections 151, 152, .......... 250, then ratio of variance vA/vB =

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 121

But both A and B have 100 observations, then both the sample A and B have same standard deviation and the same variance.
Hence, VA/VB = 1

Test: BITSAT Past Year Paper- 2012 - Question 122

The probability of simultaneous occurrence of atleast one of two events A and B is p. If the probability that exactly one of A, B occurs is q, then P(A’) + P(B’) is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 122

Since, P (exactly one of A, B occurs) = q.
∴ P (A∪B)- P (A∩B) = q
⇒ p - P (A∩B)= q ⇒ P ( A∩B)= p - q
⇒ 1- P ( A'∪B') = p-q ⇒ P (A'∪B') =1- p+q
⇒ P (A') + P(B') - P (A'∩B') =1- p+ q
⇒ P (A') + P( B') =(1- p+ q) + [1- P(A ∪B)]
= (1 – p + q) + (1 – p) = 2 – 2p + q

Test: BITSAT Past Year Paper- 2012 - Question 123

If f is an even function and g is an odd function, then the function fog is

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 123

We have, fog (–x) = f [g (–x)] = f [–g(x)
(∵ g is odd)
= f[g (x)]      (∵ f is even)
= fog (x) ∀ x ∈ R.
∴ fog is an even function.

Test: BITSAT Past Year Paper- 2012 - Question 124

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 124

Test: BITSAT Past Year Paper- 2012 - Question 125

If k ≤ sin–1 x + cos–1 x + tan–1 x ≤ K, then –

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 125

Since domain of the function x ∈ [-1,1]

Test: BITSAT Past Year Paper- 2012 - Question 126

The equations 2x + 3y + 4 = 0; 3x + 4y + 6 = 0 and 4x + 5y  + 8 = 0 are

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 126

Consider first two equations :
2x + 3y = –4   and  3x + 4y = –6

∴ x = –2 and y = 0
Now this solution satisfies the third, so the equations are consistent with unique solution.

Test: BITSAT Past Year Paper- 2012 - Question 127

The value of the determinant

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 127

Applying C1 – C2 and C2 – C3, we get

[by C1 – C2, C3 –10C2]

= 4 (180 – 180) = 0

Test: BITSAT Past Year Paper- 2012 - Question 128

If x = a sin θ and y = b cos θ, then d2y/dx2 is

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 128

Given x = a sin θ and y = b cos θ
⇒ dx/dθ = a cos θ and dy/dθ = -b sin θ

Test: BITSAT Past Year Paper- 2012 - Question 129

If f(x) = xα log x and f(0) = 0, then the value of a for which Rolle’s theorem can be applied in [0, 1] is

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 129

For Rolle’s theorem in [a, b],    f(a) = f(b),
In [0, 1] ⇒ f(0) = f(1) = 0
∵ the function has to be continuous in [0, 1]

Test: BITSAT Past Year Paper- 2012 - Question 130

If the function

is continuous at x = 2 and 4, then the values of a and b are

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 130

Since f (x) is continuous at x = 2

∴  1 = 2a + b ..... (1)
Again f(x) is continuous at x = 4,

Solving (1) and (2), we get a = 3, b = – 5

Test: BITSAT Past Year Paper- 2012 - Question 131

is a decreasing function of x in R, then the set of possible values of a (independent of x) is

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 131

f'(x) < 0 for all x if a2 -1 ≤ 0 ⇒ -1 ≤ a ≤ 1

Test: BITSAT Past Year Paper- 2012 - Question 132

The diagonal of a square is changing at the rate of 0.5 cm/sec. Then the rate of change of area, when the area is 400 cm2, is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 132

when area A is 400 cm2 then a = 20

Test: BITSAT Past Year Paper- 2012 - Question 133

If the normal to the curve y = f (x) at the point (3, 4) makes an angle 3π/4 with the positive x-axis, then f'(3) =

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 133

Slope of normal to y = f(x) at (3, 4) is -1/f'(3)

Test: BITSAT Past Year Paper- 2012 - Question 134

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 134

Test: BITSAT Past Year Paper- 2012 - Question 135

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 135

Test: BITSAT Past Year Paper- 2012 - Question 136

The area bounded by the curve y = sinx, x-axis and the ordinates x = 0 and x = π/2 is

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 136

Test: BITSAT Past Year Paper- 2012 - Question 137

The differential equation whose solution is Ax2 + By2 = 1 where A and B are arbitrary constants is of

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 137

Ax2 + By2 = 1    ......(1)

From (2) and (3)

Dividing both sides by –B, we get

Which is a DE of order 2 and degree 1

Test: BITSAT Past Year Paper- 2012 - Question 138

The unit vector perpendicular to the vectors

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 138

Unit vector perpendicular to both the given vectors is,

Test: BITSAT Past Year Paper- 2012 - Question 139

If a.b = a.c and a × b = a × c, then correct statement is

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 139

a. b. = a.c ⇒ a.(b – c) = 0
⇒ a = 0 or b – c = 0 or a ⊥ (b – c)
⇒ a = 0 or b = c or a ⊥ (b – c) ...(1)
Also a x b = a x c ⇒ a × (b – c) = 0
⇒ a = 0 or b – c = 0 or a || (b – c)
⇒ a = 0 or b = c or a || (b – c) ...(2) Observing to (1) and (2) we find that a = 0 or b = c

Test: BITSAT Past Year Paper- 2012 - Question 140

What is the value of n so that the angle between the lines having direction ratios (1, 1, 1) and (1, –1, n) is 60°?

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 140

If (l1, m1, n1) and (l2, m2, n2) are the direction ratios then angle between the lines is

Test: BITSAT Past Year Paper- 2012 - Question 141

The foot of the perpen dicular from the point (7, 14, 5) to the plane 2x + 4y – z = 2 are

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 141

We know that the length of the per pendicular from the point (x1 , y1, z1) to the plane ax + by + cz + d = 0 is

and the co-or dinate (α,β,γ) of the foot of the ⊥ are given by

In the given ques,, x1 = 7, y1 = 14, z1 = 5, a = 2 b = 4, c = -1, d = -2
By putting these values in (1), we get

Hence, foot of ⊥ is (1, 2, 8)

Test: BITSAT Past Year Paper- 2012 - Question 142

Find the coordinates of the point where the line joining the points (2, –3, 1) and (3, – 4, – 5) cuts the plane 2x + y + z = 7.

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 142

The direction ratios of the line are 3 – 2, – 4 – (–3), – 5 –1  i.e. 1, –1, – 6
Hence equation of the line joining the given points is

Coordinates of any point on this line are (r + 2, – r – 3, – 6r + 1)
If this point lies on the given plane 2x + y + z = 7, then 2(r + 2) + (– r – 3) + (– 6r + 1) = 7
⇒  r = – 1
Coordinates of any point on this line are (– 1 + 2, – (– 1) – 3, – 6 (–1) + 1) i.e. (1, – 2, 7)

Test: BITSAT Past Year Paper- 2012 - Question 143

A boy is throwing stones at a target. The probability of hitting the target at any trial is 1/2. The probability of hitting the target 5th time at the 10th throw is :

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 143

The probability of hitting the target 5th time at the 10th throw = P(the probability of hitting the target 4 times in the first 9 throws) × P(the probability of hitting the target at the 10th throw) =

Test: BITSAT Past Year Paper- 2012 - Question 144

Two dice are thrown together 4 times. The probability that both dice will show same numbers twice is -

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 144

The probability of showing same number by both dice p = 6/36 = 1/6
In binomial distribution here n = 4, r = 2, p = 1/6 , q = 5/6

Test: BITSAT Past Year Paper- 2012 - Question 145

In a triangle ABC, if a = 2, B = 60° and C = 75°, then b equals

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 145

A = 180° – 60° – 75° = 180° – 135° = 45°

Test: BITSAT Past Year Paper- 2012 - Question 146

Prabh at wan ts to invest the total amount of ₹ 15,000 in saving certificates and national saving bonds. According to rules, he has to invest at least ₹ 2000 in saving certificates and ₹ 2500 in national saving bonds. The interest rate is 8% on saving certificate and 10% on national saving bonds per annum. He invest ₹ x in saving certificate and ₹ y in national saving bonds. Then the objective function for this problem is

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 146

The function is given by profit function

Test: BITSAT Past Year Paper- 2012 - Question 147

For the function

f '(1) = mf ' (0), where m is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 147

Given,

⇒ f ' (1)  = 100   ...(ii)
Again, putting x = 0, we get
f ' (0) = 0 + 0 + ... + 0 + 1
⇒ f ' (0) = 1    ...(iii)
From eqs. (ii) and (iii), we get; f ' (1) = 100f ' (0) Hence, m = 100

Test: BITSAT Past Year Paper- 2012 - Question 148

find abc + abd + bcd + acd

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 148

As A2 = 0, Ak = 0 ∀ k ≥ 2.
Thus, (A + I)50 = I + 50A ⇒ (A + I)50 – 50A = I
∴ a = 1, b = 0, c = 0, d = 1
abc + abd + bcd + acd = 0

Test: BITSAT Past Year Paper- 2012 - Question 149

If the line x cos α + y sin α = p represents the common chord of the circles x2 + y2 = a2 and x2 + y2 + b2 (a > b), where  A and B lie on the first circle and P and Q lie on the second circle, then AP is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 149

The given circles are concentric with centre at (0, 0) and the length of the perpendicular from (0, 0) on the given line is p. Let OL = p

Test: BITSAT Past Year Paper- 2012 - Question 150

Let a1, a2, a3............ be terms on A.P. If

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 150

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