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Test: BITSAT Past Year Paper- 2010 - Question 1

If P represents radiation pressure, c represents speed of light and Q represents radiation energy striking a unit area per second, the non-zero integers x, y and z such that PxQycz is dimensionless, are.

Test: BITSAT Past Year Paper- 2010 - Question 2

The position x of a particle varies with time (t) as x = A t2 – B t3. The acceleration at time t of the particle will be equal to zero. What is the value of t?

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 2

Given that x = A t2 – B t3
∴ velocity dx/dy = 2A t - 3 Bt 2
and  acceleration =   = 2 A - 6Bt
For acceleration to be zero 2A – 6Bt = 0.

Test: BITSAT Past Year Paper- 2010 - Question 3

Two projectiles A and B are thrown with the same speed but angles are 40º and 50º with the horizontal. Then

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 3

lesser is the value of θ, lesser is sinq and hence lesser will be the time taken. Hence A will fall earlier.

Test: BITSAT Past Year Paper- 2010 - Question 4

A body is travelling in a circle at a constant speed. It

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 4

Body moves with constant speed it means that tangential acceleration aT = 0 & only centripetal acceleration aC exists whose direction is always towards the centre or inward (along the radius of the circle).

Test: BITSAT Past Year Paper- 2010 - Question 5

Two blocks are connected over a massless pulley as shown in fig. The mass of block A is 10 kg and the coefficient of kinetic friction is 0.2. Block A slides down the incline at constant speed. The mass of block B in kg is:

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 5

Considering the equilibrium of A, we get 10a = 10g sin 30º – T – mN
where N = 10g cos 30°


but a = 0, T = mBg

⇒ mB = 3.268 ≈ 3.3 kg

Test: BITSAT Past Year Paper- 2010 - Question 6

A spring is compressed between two toy carts of mass m1 and m2. When the toy carts are released, the springs exert equal and opposite average forces for the same time on each toy cart. If v1 and v2 are the velocities of the toy carts and there is no friction between the toy carts and the ground, then :

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 6

Applyin g law of conservation of lin ear momentum m1v1 + m2v2 = 0,

 

Test: BITSAT Past Year Paper- 2010 - Question 7

The potential energy for a force field is given by U (x, y) = cos (x + y). The force acting on a particle at position given by coordinates (0, π/4) is –

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 7

Test: BITSAT Past Year Paper- 2010 - Question 8

A long string is stretched by 2 cm and the potential energy is V. If the spring is stretched by 10 cm, its potential energy will be

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 8

Test: BITSAT Past Year Paper- 2010 - Question 9

The ratio of the accelerations for a solid sphere (mass ‘m’ and radius ‘R’) rolling down an incline of angle ‘θ’ without slipping and slipping down the incline without rolling is:

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 9

For solid sphere rolling without slipping on inclined plane, acceleration

For solid sphere slipping on inclined plane without rolling, acceleration
a2 = g sin θ
Therefore required ratio = 

Test: BITSAT Past Year Paper- 2010 - Question 10

A system consists of three particles, each of mass m and located at (1, 1), (2, 2) and (3, 3). The co-ordinates of the centre of mass are

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 10

The coordinates of C.M of three particle are

so coordinates of C.M. of three particle are (2, 2) V = 300 m/s

Test: BITSAT Past Year Paper- 2010 - Question 11

Suppose the gravitational force varies inversely as the nth power of distance. Then the time period of a planet in circular orbit of radius ‘R’ around the sun will be proportional to

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 11

F = KR- n = MRω2 ⇒ ω2 = KR- (n +1)

[where K' = K1/2, a constant]

Test: BITSAT Past Year Paper- 2010 - Question 12

Two planets A and B have the same material density. If the radius of A is twice that of B, then the ratio of the escape velocity vA/vB is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 12


as ve ∝ R for same density, 

Test: BITSAT Past Year Paper- 2010 - Question 13

The upper end of a wire of diameter 12mm and length 1m is clamped and its other end is twisted through an angle of 30°. The angle of shear is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 13

Test: BITSAT Past Year Paper- 2010 - Question 14

A spherical ball is dropped in a long column of a viscous liquid. The speed (v) of the ball as a function of time (t) may be best represented by

Test: BITSAT Past Year Paper- 2010 - Question 15

Two mercury drops (each of radius r) merge to form a bigger drop. The surface energy of the bigger drop, if T is the surface tension, is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 15


Surface energy of bigger drop,
E = 4πR2T = 4 x 22/3 πr2T = 28/3 πr2T

Test: BITSAT Past Year Paper- 2010 - Question 16

Two circular plates of radius 5 cm each, have a 0.01 mm thick water film between them. Then what will be the force required to separate these plate (S.T. of water = 73 dyne/cm) ?

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 16


    = 36.5 π ≈ 115 newton

Test: BITSAT Past Year Paper- 2010 - Question 17

One kilogram of ice at 0°C is mixed with one kilogram of water at 80°C. The final temperature of the mixture is (Take specific heat of water = 4200 kJ/kg-°C, Latent heat of ice = 336 kJ/kg)

Test: BITSAT Past Year Paper- 2010 - Question 18

In the equation PVγ = constant, the value of γ is unity. Then the  process is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 18

PV = constant represents isothermal process.

Test: BITSAT Past Year Paper- 2010 - Question 19

An ideal refrigerator has a freezer at a temperature of 13ºC. The coefficient of performance of the engine is 5. The temperature of the air (to which heat is rejected) is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 19

T2 = 273 – 13 = 260,

or T1 – 260 = 52; T1 = 312 K,
T2 = 312 – 273 = 39°C

Test: BITSAT Past Year Paper- 2010 - Question 20

3 moles of an ideal gas at a temperature of 27°C are mixed with 2 moles of an ideal gas at a temperature 227°C, determine the equilibrium temperature of the mixture, assuming no loss of energy.

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 20

Energy possessed by the ideal gas at 27°C is

Energy possessed by the ideal gas at 227°C is

If T be the equilibrium temperature, of the mixture, then its energy will be

Since, energy remains conserved,
Em = E+ E2

or T = 380 K  or 107°C

Test: BITSAT Past Year Paper- 2010 - Question 21

A simple pendulum has time period 't'. Its time period in a lift which is moving upwards with acceleration 3 ms–2 is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 21

Test: BITSAT Past Year Paper- 2010 - Question 22

A wave y = a sin (ωt – kx) on a string meets with another wave producing a node at x = 0. Then the equation of the unknown wave is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 22

Equation of a wave
y1 = a sin (ωt – kx)         ....(i)
Let equations of another wave may be,
y2 = a sin (ωt + kx)         ....(ii)
y3 = –a sin (ωt + kx)      ....(iii)
If Eq. (i) propagate with Eq. (ii), we get y = 2a cos kx sin ωt
If Eq. (i), propagate with Eq. (iii), we get y = –2a sin kx cos ωt
At x = 0, y = 0, wave produce node
So, Eq.(iii) is the equation of unknown wave

Test: BITSAT Past Year Paper- 2010 - Question 23

A source of sound produces waves of wavelength 60 cm when it is stationary. If the speed of sound in air is 320 m s-1 and source moves with speed 20 m s-1, the wavelength of sound in the forward direction will be nearest to

Test: BITSAT Past Year Paper- 2010 - Question 24

A charge +q is at a distance L/2 above a square of side L. Then what is the flux linked with the surface?

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 24


The given square of side L may be considered as one of the faces of a cube with edge L. Then given charge q will be considered to be placed at the centre of the cube. Then according to Gauss's theorem, the magnitude of the electric flux through the faces (six) of the cube is given by
φ = q/ε0
Hence, electric flux through one face of the cube for the given square will be

Test: BITSAT Past Year Paper- 2010 - Question 25

Two metallic spheres of radii 1 cm and 3 cm are given charges of -1 × 10-2 C and 5 × 10-2 C, respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 25

Test: BITSAT Past Year Paper- 2010 - Question 26

In a region, the potential is represented by V(x, y, z) = 6x - 8xy - 8y + 6yz, where V is in volts and x, y, z are in metres. The electric force experienced by a charge of 2 coulomb situated at point (1, 1, 1) is :

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 26

Test: BITSAT Past Year Paper- 2010 - Question 27

The power dissipated in the circuit shown in the figure is 30 Watts. The value of R is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 27

     ...(i)

Substituting the values in equation (i)

Test: BITSAT Past Year Paper- 2010 - Question 28

Which of the following quantities do not change when a resistor connected to a battery is heated due to the current?

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 28

Only number of free electrons is constant, other factors are temperature dependent.

Test: BITSAT Past Year Paper- 2010 - Question 29

The magnetic field at the origin due to the current flowing in the wire is –

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 29

Test: BITSAT Past Year Paper- 2010 - Question 30

The back emf induced in a coil, when current changes from 1 ampere to zero in one milli-second, is 4 volts, the self inductance of the coil is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 30

Test: BITSAT Past Year Paper- 2010 - Question 31

Two solenoids of same cross-sectional area have their lengths and number of turns in ratio of 1 : 2.
The ratio of self-inductance of two solenoids is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 31

From

we get,

Test: BITSAT Past Year Paper- 2010 - Question 32

An alternating voltage V = V0 sin ωt is applied across a circuit. As a result, a current I = I0 sin (ωt – π/2) flows in it. The power consumed per cycle is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 32

The phase angle between voltage V and current I is π/2.

Test: BITSAT Past Year Paper- 2010 - Question 33

A resistance R and inductance L and a capacitor C all are connected in series with an AC supply. The resistance of R is 16 ohm and for a given frequency, the inductive reactance of L is 24 ohm and capacitive reactance of C is 12 ohm. If the current in the circuit is 5 amp., find the potential difference across R, L and C.

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 33

VR = iR = 5 × 16 = 80 Volt
VL = i × (ωL) = 5 × 24 = 120 Volt
VC = i × (l /ωC) = 5 × 12 = 60 Volt

Test: BITSAT Past Year Paper- 2010 - Question 34

The diameter of the objective of a telescope is a, its magnifying power is m and wavelength of light is λ, The resolving power of the telescope is:

Test: BITSAT Past Year Paper- 2010 - Question 35

The photoelectric threshold of metal is 2000Å. The energy of the electrons ejected from the surface by ultraviolet light of wavelength 1500Å is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 35

Test: BITSAT Past Year Paper- 2010 - Question 36

A material particle with a rest mass m0 is moving with a velocity of light c. Then the wavelength of the de Broglie wave associated with it is :

Test: BITSAT Past Year Paper- 2010 - Question 37

Hydrogen atom in ground state is excited by a monochromatic radiation of λ = 975 Å. Number of spectral lines in the resulting spectrum emitted will be

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 37

Number of spectral lines = 

Test: BITSAT Past Year Paper- 2010 - Question 38

Which of the following is best nuclear fuel

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 38

Pluton ium 239 is pr ocessed by br eeder mechanism to be used as nuclear feul.

Test: BITSAT Past Year Paper- 2010 - Question 39

A transistor has a base current of 1 mA and emitter current 90 mA. The collector current will

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 39

IC = IE –IB = 90 – 1 = 89 m A

Test: BITSAT Past Year Paper- 2010 - Question 40

A d.c. battery of V volt is connected to a series combination of a resistor R and an ideal diode D as shown in the figure below. The potential difference across R will be

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 40

In forward biasing, the diode conducts. For ideal junction diode, the forward resistance is zero; therefore, entire applied voltage occurs across external resistance R i.e., there occurs no potential drop, so potential across R is V in forward biased.

Test: BITSAT Past Year Paper- 2010 - Question 41

The vapour density of ozone is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 41

We know that,
Molecular weight of compound or molecules = 2 × V. D.
Vapour density (V. D.) of ozone molecules

Hence, V. D. of O3 is 24.

Test: BITSAT Past Year Paper- 2010 - Question 42

In redox reaction 1 g-eq of reducing agent requires P gm-eq. of oxidising agent. The value of P is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 42

In redox reaction,
g equivalent of reducing agent = g. equivalent of oxidising agent
Hence 1g equ. of reducing agent = P g equ. of oxidising agent.

Test: BITSAT Past Year Paper- 2010 - Question 43

Chloride ion and potassium ion are isoelectronic. Then :

Test: BITSAT Past Year Paper- 2010 - Question 44

Which of the following pairs has both members from the same period of periodic table :

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 44

11Na ⇒ 2, 8, 1; 17Cl ⇒ 2, 8, 7
These have same number of shells. Hence, they are the elements of the same period.

Test: BITSAT Past Year Paper- 2010 - Question 45

In the periodic table, with the increase in atomic number, the metallic character of an element

Test: BITSAT Past Year Paper- 2010 - Question 46

Which of the following statements is/are true?
1. PH5 and BiCl5 do not exist.
2. pπ – dπ bond is present in SO2.
3. I3+ has bent geometry.
4. SeF4 and CH4 have same shape.

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 46

All the three statements are correct.

Test: BITSAT Past Year Paper- 2010 - Question 47

When the temperature is raised, the viscosity of liquid decreases this is because

Test: BITSAT Past Year Paper- 2010 - Question 48

At a constant volume the specific heat of a gas is 0.075 and its molecular weight is 40. The gas is:

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 48


⇒ CV = q × m
= 0.075 × 40
= 3.0 cal
CP – Cv = R
CP = CV + R
= 3 + 2 = 5

Monoatomic gas.

Test: BITSAT Past Year Paper- 2010 - Question 49

Which of these is least likely to act as Lewis base?

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 49

BF3 is Lewis acid (e pair acceptor)

Test: BITSAT Past Year Paper- 2010 - Question 50

The Ksp of CuS, Ag2S and HgS are 10–31,10–44 and 10–54 respectively. The solubility of these sulphides are in the order :

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 50

For CuS, solubility = (10–31)1/2;
for  HgS = 

Test: BITSAT Past Year Paper- 2010 - Question 51

In which of the following reactions, H2O2 is acting as a reducing agent?

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 51

SO2 changes to H2SO4 (O.N. changes from +4 to +6 oxidation)
2KI →  I2 (O.S. changes from –1 to 0 oxidation)
PbS → PbSO4 (O.S. changes from –2 to +6 oxidation)
Ag2O → 2Ag (O.S. changes from +1 to 0 reduction)

Test: BITSAT Past Year Paper- 2010 - Question 52

Sodium peroxide in contact with moist air turns white due to formation of :

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 52

Na2O2 + H2O → 2NaOH + 1/2 O2

Test: BITSAT Past Year Paper- 2010 - Question 53

Which of the following is similar to graphite

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 53

Graphite and boron nitride have similar structure.

Test: BITSAT Past Year Paper- 2010 - Question 54

The number of geometrical isomers of CH3 – CH = CH – CH = CH – CH = CHCl is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 54

The given structure has three double bonds whose each carbon atom is differently substituted hence number of geometrical isomers will be 2n = 23’ = 8, where n is the number of double bonds whose each carbon atom is differently substituted.

Test: BITSAT Past Year Paper- 2010 - Question 55

According to IUPAC system, the correct name of the compound having the formula

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 55

Test: BITSAT Past Year Paper- 2010 - Question 56

Liebig’s method is used for the estimation of

Test: BITSAT Past Year Paper- 2010 - Question 57

Hyperconjugation involves

Test: BITSAT Past Year Paper- 2010 - Question 58

Name of following reaction is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 58

The given reaction is Diel’s Alder reaction.

Test: BITSAT Past Year Paper- 2010 - Question 59

The unsaturated hydrocarbon which on ozonolysis gives one mole each of formaldehyde, acetaldehyde and methyl glyoxal (CH3COCHO) is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 59

Test: BITSAT Past Year Paper- 2010 - Question 60

Minamata disease is due to pollution of

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 60

Minamata is caused by Hg poisoning.

Test: BITSAT Past Year Paper- 2010 - Question 61

Phosphate pollution is caused by:

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 61

Phosphate pollution is caused by sewage and agricultural fertilizers.

Test: BITSAT Past Year Paper- 2010 - Question 62

Eutrophication causes reduction in

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 62

Eutrophication causes reduction in dissolved oxygen.

Test: BITSAT Past Year Paper- 2010 - Question 63

Coolant used in car radiator is aqueous solution of ethylene glycol. In order to prevent the solution from freezing at – 0.3°C. How much ethylene glycol must be added to 5 kg of water?
(Kf = 1.86 K Kg mol–1)

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 63

ΔTf = 0.3 ° C

∴ WB = 50 g
The amount used should be more than 50 g.

Test: BITSAT Past Year Paper- 2010 - Question 64

Which of the following will form the cathode with respect to iron anode in an electrolyte cell?

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 64

In electrolytic cell the cathode is of higher reduction potential.

Test: BITSAT Past Year Paper- 2010 - Question 65

The activation energy for a simple chemical reaction A → B is Ea in forward direction. The activation energy for reverse reaction

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 65

Since the nature of reaction (i.e. exothermic or endothermic) is not given, Ea for reverse reaction can be more or less.

Test: BITSAT Past Year Paper- 2010 - Question 66

The following data are for the decomposition of ammonium nitrite in aqueous solution:

The order of reaction is:

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 66

NH4 NO2 → N2 + 2H2O
Volume of N2 formed in successive five minutes are 2.75 cc, 2.40 cc and 2.25 cc which is in decreasing order. So rate of reaction is dependent on  concentration of NH4 NO2. As decrease is not very fast so it will be first order reaction.

Test: BITSAT Past Year Paper- 2010 - Question 67

Which liberates ammonia when treated with water?

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 67

All nitrides react with H2O to give NH3 and CaCN2 also react with H2O
CaNCN + 3H2O → CaCO3 + NH3

Test: BITSAT Past Year Paper- 2010 - Question 68

The correct order of reactivity of halogens with alkalies is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 68

Reactivity follows the order F > Cl > Br > I

Test: BITSAT Past Year Paper- 2010 - Question 69

In the manufacture of iron from haematite, limestone is added to act as:

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 69

Limestone (CaCO3) is mixed with Fe2O3 and it acts as flux to form slag (CaSiO3).

Test: BITSAT Past Year Paper- 2010 - Question 70

Which of the following has square planar geometry?

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 70

[PtCl4]2– has square planar geometry. Pt : 5d96s1

Two electrons are removed from 5d shell and 6s shell. So, hybridisation takes place is dsp2 i.e. square planar geometry.

Test: BITSAT Past Year Paper- 2010 - Question 71

In which of the following conversions, phosphorus pentachloride is used as the reagent?

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 71

When ethyl alcohol is treated with PCl 5 , then ethyl chloride is formed.
CH3CH2 - OH + PCl5 CH3CH2 - Cl + HCl + POCl3

Test: BITSAT Past Year Paper- 2010 - Question 72

Match List I (Reaction) with List II (Reagent) and select the correct answer using the codes given below the lists:

Test: BITSAT Past Year Paper- 2010 - Question 73

Which of the following will not form a yellow precipitate on heating with an alkaline solution of iodine?

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 73

CH3OH does not have – CH(OH)CH3 group hence it will not form yellow precipitate  with an alkaline solution of iodine (haloform reaction).

Test: BITSAT Past Year Paper- 2010 - Question 74

Formic acid and acetic acid can be distinguished by:

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 74

Formic acid(HCOOH) has aldehydic group.

Test: BITSAT Past Year Paper- 2010 - Question 75

When ethylamine react with sodium metal, the gas evolved is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 75

When ethylamine is heated with sodium metal, then hydrogen gas is evolved.
2C2H5NH2 + 2Na  →  2C2H5NHNa+ + H2 ­↑

Test: BITSAT Past Year Paper- 2010 - Question 76

The secondary structure of a protein refers to

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 76

The secondary structure of a protein refers to the shape in which a long peptide chain can exist. There are two different conformations of the peptide linkage present in protein are α-helix and β-conformation.
The α-helix always has a right handed arrangement.
In β-conformation all peptide chains are streched out to nearly maximum extension and then laid side by side held together by intermolecular hydrogen bonds. The structure resembles the pleated folds of drapery and therefore is known as β-pleated shelt.

Test: BITSAT Past Year Paper- 2010 - Question 77

When H2S gas is passed through the HCl containing aqueous solution of CuCl2, HgCl2, BiCl3 and CoCl2 it does not precipitate out

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 77

CoS is not precipitated in acidic medium.

Test: BITSAT Past Year Paper- 2010 - Question 78

Which one of the following statements is correct?

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 78

Ammonia can dissolve ppt. of AgCl only due to formation of complex as given below:
AgCl + 2NH3 → [Ag (NH3)2]Cl

Test: BITSAT Past Year Paper- 2010 - Question 79

Three separate samples of a solution of a single salt gave these results. One formed a white precipitate with excess ammonia solution, one formed a white precipitate with dil. HCl solution and one formed a black precipitate with H2S. The salt could be

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 79

Pb(NO3)2 + 2NH4OH → Pb(OH)2 ↓ + 2NH4NO3 (white ppt)
Pb(NO3)2 + 2HCl → PbCl2 ↓ + 2HNO3 (white ppt)
Pb(NO3)2 + H2S → PbS ↓ + 2HNO3 (black)

Test: BITSAT Past Year Paper- 2010 - Question 80

Experiment to study kinetics of the dissociation of hydrogen peroxide must be performed by group of two or three so that–

Test: BITSAT Past Year Paper- 2010 - Question 81

Directions: In the following questions, choose the alternative which can most appropriately replace the group of words italicised in the sentence.

Despite being in the career of singing for the last 10 yr, he has not been able to earn fame on account of his practice of borrowing ideas and words from others and using them as his own.

Test: BITSAT Past Year Paper- 2010 - Question 82

Directions: In the following questions, choose the alternative which can most appropriately replace the group of words italicised in the sentence.

Every person is not allowed to enter the place where public, government or historical records are  kept

Test: BITSAT Past Year Paper- 2010 - Question 83

Directions: In the following questions, choose the alternative which can most appropriately replace the group of words italicised in the sentence.

The advertisement assured the public that the medicine would give back to the users, their youthful vigour and appearance.

Test: BITSAT Past Year Paper- 2010 - Question 84

Directions: Choose the alternative which is most similar in meaning to the word given in capital letters.

PARAMOUR

Test: BITSAT Past Year Paper- 2010 - Question 85

Directions: Choose the alternative which is most similar in meaning to the word given in capital letters.

REFECTORY

Test: BITSAT Past Year Paper- 2010 - Question 86

Directions: Choose the alternative which is most similar in meaning to the word given in capital letters.

ASSENT

Test: BITSAT Past Year Paper- 2010 - Question 87

Directions: Choose the alternative which best expresses the meaning of the given idiom/ phrase.

To show one’s teeth

Test: BITSAT Past Year Paper- 2010 - Question 88

Directions: Choose the alternative which best expresses the meaning of the given idiom/ phrase.

To pour oil in troubled water

Test: BITSAT Past Year Paper- 2010 - Question 89

Directions: Rearrange the following six sentences (A), (B), (C), (D), (E), (F) in the proper sequence to form a meaningful paragraph; then answer the questions given below them.

(A) This is being done to ensure that when the cess is removed or merged ,the flexibility to impose higher rate on luxury goods is not taken away ,a senior finance ministry official told ET.
(B) This is only an enabling provision and the highest rate levied on goods will still be 28% (14% central GST and 14% state GST).The demerit and luxury goods will attract higher 28% rate plus cess.
(C) India has decided to peg the peak goods and services tax (GST) rate at 40% in the legislation instead of 28%, giving it the flexibility to raise rates without having to reach out to parliament.
(D) "Some members of the council felt such an enabling provision was needed." the official privy to the development said.
(E) The GST council has proposed to peg the peak tax rate at 40% (20% central GST and 20% state GST) in the model GST law to preclude the requirement of approaching parliament or state assemblies for any change in future.
(F) This enabling provision will also allow the government to remove the cess at some stage and instead have a higher GST rate only, which will make for a neater GST.

Q. Which sentence should come second in the paragraph?

Test: BITSAT Past Year Paper- 2010 - Question 90

Directions: Rearrange the following six sentences (A), (B), (C), (D), (E), (F) in the proper sequence to form a meaningful paragraph; then answer the questions given below them.

(A) This is being done to ensure that when the cess is removed or merged ,the flexibility to impose higher rate on luxury goods is not taken away ,a senior finance ministry official told ET.
(B) This is only an enabling provision and the highest rate levied on goods will still be 28% (14% central GST and 14% state GST).The demerit and luxury goods will attract higher 28% rate plus cess.
(C) India has decided to peg the peak goods and services tax (GST) rate at 40% in the legislation instead of 28%, giving it the flexibility to raise rates without having to reach out to parliament.
(D) "Some members of the council felt such an enabling provision was needed." the official privy to the development said.
(E) The GST council has proposed to peg the peak tax rate at 40% (20% central GST and 20% state GST) in the model GST law to preclude the requirement of approaching parliament or state assemblies for any change in future.
(F) This enabling provision will also allow the government to remove the cess at some stage and instead have a higher GST rate only, which will make for a neater GST.

Q. Which sentence shouldcome before the last?

Test: BITSAT Past Year Paper- 2010 - Question 91

Directions: Rearrange the following six sentences (A), (B), (C), (D), (E), (F) in the proper sequence to form a meaningful paragraph; then answer the questions given below them.

(A) This is being done to ensure that when the cess is removed or merged ,the flexibility to impose higher rate on luxury goods is not taken away ,a senior finance ministry official told ET.
(B) This is only an enabling provision and the highest rate levied on goods will still be 28% (14% central GST and 14% state GST).The demerit and luxury goods will attract higher 28% rate plus cess.
(C) India has decided to peg the peak goods and services tax (GST) rate at 40% in the legislation instead of 28%, giving it the flexibility to raise rates without having to reach out to parliament.
(D) "Some members of the council felt such an enabling provision was needed." the official privy to the development said.
(E) The GST council has proposed to peg the peak tax rate at 40% (20% central GST and 20% state GST) in the model GST law to preclude the requirement of approaching parliament or state assemblies for any change in future.
(F) This enabling provision will also allow the government to remove the cess at some stage and instead have a higher GST rate only, which will make for a neater GST.

Q. Which sentence will come complete the passage?

Test: BITSAT Past Year Paper- 2010 - Question 92

Directions: Rearrange the following six sentences (A), (B), (C), (D), (E), (F) in the proper sequence to form a meaningful paragraph; then answer the questions given below them.

(A) This is being done to ensure that when the cess is removed or merged ,the flexibility to impose higher rate on luxury goods is not taken away ,a senior finance ministry official told ET.
(B) This is only an enabling provision and the highest rate levied on goods will still be 28% (14% central GST and 14% state GST).The demerit and luxury goods will attract higher 28% rate plus cess.
(C) India has decided to peg the peak goods and services tax (GST) rate at 40% in the legislation instead of 28%, giving it the flexibility to raise rates without having to reach out to parliament.
(D) "Some members of the council felt such an enabling provision was needed." the official privy to the development said.
(E) The GST council has proposed to peg the peak tax rate at 40% (20% central GST and 20% state GST) in the model GST law to preclude the requirement of approaching parliament or state assemblies for any change in future.
(F) This enabling provision will also allow the government to remove the cess at some stage and instead have a higher GST rate only, which will make for a neater GST.

Q. Which sentence will be third after the rearrangement?

Test: BITSAT Past Year Paper- 2010 - Question 93

Directions: Rearrange the following six sentences (A), (B), (C), (D), (E), (F) in the proper sequence to form a meaningful paragraph; then answer the questions given below them.

(A) This is being done to ensure that when the cess is removed or merged ,the flexibility to impose higher rate on luxury goods is not taken away ,a senior finance ministry official told ET.
(B) This is only an enabling provision and the highest rate levied on goods will still be 28% (14% central GST and 14% state GST).The demerit and luxury goods will attract higher 28% rate plus cess.
(C) India has decided to peg the peak goods and services tax (GST) rate at 40% in the legislation instead of 28%, giving it the flexibility to raise rates without having to reach out to parliament.
(D) "Some members of the council felt such an enabling provision was needed." the official privy to the development said.
(E) The GST council has proposed to peg the peak tax rate at 40% (20% central GST and 20% state GST) in the model GST law to preclude the requirement of approaching parliament or state assemblies for any change in future.
(F) This enabling provision will also allow the government to remove the cess at some stage and instead have a higher GST rate only, which will make for a neater GST.

Q. Which sentence will start the passage?

Test: BITSAT Past Year Paper- 2010 - Question 94

Directions: Pick out the most effective word from the given words to fill in the blanks to make the sentence meaningfully complete in the contest of the sentence.

A novel of real_____________ must invent its own language, and this one does,

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 94

Ambition-a cherished desire Abstruse-hard to understand intricate complex

Test: BITSAT Past Year Paper- 2010 - Question 95

Directions: Pick out the most effective word from the given words to fill in the blanks to make the sentence meaningfully complete in the contest of the sentence.

Information technology, and the hardware and software______________with the IT industry.

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 95

Amalgamation-joining of two organization to form one large organization Regulated means controlled.

Test: BITSAT Past Year Paper- 2010 - Question 96

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

EFJK : LMQR :: LMQR : ?

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 96

Test: BITSAT Past Year Paper- 2010 - Question 97

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

Mahatma Gandhi :  Porbandar  : :  Pt. Jawaharlal Nehru  :  ?

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 97

As, the birth place of Mahatma Gandhi was Porbandar. Similarly, the birth place of Pt. Jawaharlal Nehru was Allahabad.

Test: BITSAT Past Year Paper- 2010 - Question 98

Directions: 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: The education of a student at collegiate level, not taking into account maintenance expenses, costs four hundred rupees a year. Collegiate education is thus drawing heavily upon the national resources of an impoverished community. So college education should be restricted to a brilliant few.
Assumptions:
I. Our resources are very limited.
II. Only a few students should be admitted to the colleges.

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 98

The use of the words 'impoverished community' in the statement makes I implicit while the phrase 'college education should be restricted to a brilliant few' makes II implicit.

Test: BITSAT Past Year Paper- 2010 - Question 99

In a code language, if BANGED is coded as JJKQCC, then the word STRAY will be coded as

Test: BITSAT Past Year Paper- 2010 - Question 100

Directions: In the following question, a series is given with one terms missing. Choose the correct alternative from the given ones that will complete the series:

2,    3,    7,    22, 155,   ?

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 100

Test: BITSAT Past Year Paper- 2010 - Question 101

Which one of the following diagram represents the correct relationship among Colour, Black and White.

Test: BITSAT Past Year Paper- 2010 - Question 102

Find out the alternative figure which contains figure (X) as its part.

Test: BITSAT Past Year Paper- 2010 - Question 103

A piece of paper is folded and cut. From the figures given, indicate how it will appear when opened?

Test: BITSAT Past Year Paper- 2010 - Question 104

Identify the figure that completes the pattern.

Test: BITSAT Past Year Paper- 2010 - Question 105

Find out which of the figures (a), (b), (c) and (d) can be formed from the pieces given in figure (X).

Test: BITSAT Past Year Paper- 2010 - Question 106

Let A = {x : x ∈ R |x| < 1};
B = {x : x ∈ R, |x - 1| ≥ 1} and A ∪ B = R- D, then the set D is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 106

A = [ x : x ∈ R, -1 < x < 1]
B = [x : x ∈ R : x -1 ≤ -1 or x -1 ≥ 1]
= [x: x ∈ R : x ≤ 0 or x ≥ 2]
∴ A ∪ B = R - D,
where  D = [ x : x ∈ R,1 ≤ x < 2]

Test: BITSAT Past Year Paper- 2010 - Question 107

If 12 cot2θ – 31 coses θ + 32 = 0, then the value of sin θ is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 107

12 cot2θ - 31cosec θ + 32 = 0
⇒ 12(cosec2θ - 1) - 31cosec θ + 32 = 0
⇒ 12cosec2θ - 31cosec θ + 20 = 0
⇒ 12cosec2θ - 16 cosec θ - 15 cosecθ + 20 = 0
⇒ (4cosec θ - 5)(3cosec θ - 4) = 0
⇒ 

Test: BITSAT Past Year Paper- 2010 - Question 108

tan 20° + tan 40° + √3 tan 20° tan 40° is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 108

√3 = tan 60° = tan (40° + 20°)

∴ √3 – √3 tan 40° tan 20°
= tan 40° + tan 20°
Hence tan 40° + tan 20° + √3 tan 40° tan20°
= √3

Test: BITSAT Past Year Paper- 2010 - Question 109

The roots of the equation x2 – 2 √2x + 1 = 0 are

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 109

The discriminant of the equation
( -2√2)2 – 4 (1) (1) = 8 – 4 = 4 > 0 and a perfect square, so roots are real and different but we can't say that roots are rational because coefficients are not rational therefore.

this is irrational
∴ the roots are real and different.

Test: BITSAT Past Year Paper- 2010 - Question 110

If  = A + iB, then A2 + B2 equals to

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 110

Test: BITSAT Past Year Paper- 2010 - Question 111

In a polygon no three diagonals are concurrent. If the total number of points of intersection of diagonals interior to the polygon be 70 then the number of diagonals of the polygon is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 111

A combination of four vertices is equivalent to one interior poin of intersection of diagonals.

∴ No. of interior points of intersection
nC4 = 70
⇒ n (n – 1) (n – 2) (n – 3) = 5. 6. 7. 8
∴ n = 8
So, number of diagonals  = 8C2 – 8 = 20

Test: BITSAT Past Year Paper- 2010 - Question 112

With 17 consonants and 5 vowels the number of words of four letters that can be formed having two different vowels in the middle and one consonant, repeated or different at each end is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 112

The two letters, th e first and the last of the four lettered word can be chosen in (17)2 ways, as repetition is allowed for consonants. The two vowels in the middle are distinct so that the number of ways of filling up the two places is 5P2 = 20.
The no. of different words = (17)2 . 20 = 5780.

Test: BITSAT Past Year Paper- 2010 - Question 113

The coefficient of x32 in the expansion of is:

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 113

We know by Binomial expansion, that (x + a)

Given expansion is 
On comparing we get n = 15, x = x4,

⇒ x60 –7r = x32 ⇒  60 – 7r = 32
⇒ 7r = 28  ⇒  r = 4
So, 5th term, contains x32

15C4X32.
Thus, coefficient of  x32 = 15C4.

Test: BITSAT Past Year Paper- 2010 - Question 114

If m arithmetic means are inserted between 1 and 31 so that the ratio of the 7th and (m - 1)th means is 5 : 9, then find the value of m.

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 114

Let the means be x1, x2,.....xm so that 1, x1, x2,....xm, 31 is an A.P. of (m + 2) terms.
Now, 31 = Tm+2 = a + (m + 1)d = 1 + (m + 1)d


⇒ 9a + 63d = 5a + (5m – 5)d

Test: BITSAT Past Year Paper- 2010 - Question 115

The reflection of the point (4, –13) in the line 5x + y + 6 = 0 , is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 115

Let Q(a, b) be the reflection of P(4, – 13) in the line 5x + y + 6 = 0
Then the mid-point lies on
5x + y + 6 = 0

⇒ 5a + b + 19 = 0 ...(i)
Also PQ is perpendicular to 5x + y + 6 = 0
Therefore

⇒ a – 5b – 69 = 0 ..(ii)
Solving (i) and (ii), we get a = – 1, b = – 14

Test: BITSAT Past Year Paper- 2010 - Question 116

If the equations of the pairs of opposite sides of a parallelogram are x2 – 5x + 6 = 0 and y2  – 6y + 5 = 0, then equations of its diagonals are

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 116

Equations of the sides of the parallelogram are
(x – 3) (x – 2) = 0 and (y – 5) (y – 1) = 0
i.e. x = 3, x = 2; y = 5, y = 1
Hence its vertices are : A (2 ,1 ); B (3, 1); C (3, 5); D (2, 5)
Equation of the diagonal AC is
y – 1 = 4/1 (x – 2) ⇒ y = 4x – 7
Equation of the diagonal BD is

Test: BITSAT Past Year Paper- 2010 - Question 117

If the line 2x – 1 = 0 is the directrix of the parabola y2 – kx + 6 = 0, then one of the values of k is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 117

Given eqn of parabola is y2 – kx + 6 = 0

Test: BITSAT Past Year Paper- 2010 - Question 118

The line ax + by = 1 cuts ellipse cx2 + dy2 = 1 only once if

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 118

Clearly ax + by = 1

Test: BITSAT Past Year Paper- 2010 - Question 119

Find the equation of chord of the circle x2 + y2 = 8x bisected at the point (4, 3)

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 119

T = S1 ⇒ x (4)  + y (3) – 4 (x + 4) = 16 + 9 – 32
⇒ 3y – 9 = 0 ⇒ y = 3

Test: BITSAT Past Year Paper- 2010 - Question 120

Find the value of 

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 120

Test: BITSAT Past Year Paper- 2010 - Question 121

Mean of 25 observations was found to be 78.4. But later on it was found that 96 was misread 69. The correct mean is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 121

Mean

∑x = 25 × 78.4 = 1960
But this ∑x is incorrect as 96 was misread as 69.
∴ correct ∑x = 1960 + (96 – 69) = 1987
∴ correct mean = 1987/25 = 79.48

Test: BITSAT Past Year Paper- 2010 - Question 122

If the mean, mode and S.D. of a fr equency distribution are 41, 45 and 8 respectively, then its Pearson’s coefficient of skewness is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 122

Test: BITSAT Past Year Paper- 2010 - Question 123

A black die and a white die are rolled. Find the probability that the number shown by the black die will be more than twice that shown by the white die.

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 123

The number of favourable cases are shown below:

There are 6 favourable cases in which the number on black die is more than twice the number on the white die.
∴  m = 6
n = Total number of cases = 6 × 6
(∴ with each die there are six possibilities)
∴ Probability

Test: BITSAT Past Year Paper- 2010 - Question 124

Let E = {1, 2, 3, 4} and F = {1, 2}. Then the number of onto functions from E to F is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 124

If set A has m elements and set B has n elements then number of onto functions from A to B is

Here E = {1, 2, 3, 4}, F = {1, 2}
m = 4, n = 2
∴ No. of onto functions from E to F

Test: BITSAT Past Year Paper- 2010 - Question 125

If , then (fof of) (x) is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 125


Test: BITSAT Past Year Paper- 2010 - Question 126

The value of  cos–1x + cos–1 is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 126

Let cos–1x = y

Test: BITSAT Past Year Paper- 2010 - Question 127

If  and A + B – D = 0 (zero matrix), then D matrix will be -

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 127



 

Test: BITSAT Past Year Paper- 2010 - Question 128

The value of  is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 128


= 1 (3 × 9 – 6 (–7)) –2 (–4 × 9 –2 × 6) +3 ((–4) (–7) – 3 × 2)
= (27 + 42) –2 (–36 – 12)+3 (28 – 6) = 231

Test: BITSAT Past Year Paper- 2010 - Question 129

Let Then f (x) is derivable at x = 1, if

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 129

Test: BITSAT Past Year Paper- 2010 - Question 130

If a circular plate is heated uniformly, its area expands 3c times as fast as its radius, then the value of c when the radius is 6 units, is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 130

Let A sq. units in the area measure when the radius is r units.
their A = πr2
Differentiate both side w.r.t ‘t’

We have,

From eqn (i), we get

Test: BITSAT Past Year Paper- 2010 - Question 131

The function f(x) = tan x – 4x is strictly decreasing on

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 131

f(x) = tanx – 4x ⇒ f'(x) = sec2x – 4
When

Therefore, 1 < sec2x < 4
⇒ – 3 < (sec2x – 4) < 0
Thus, for 
Hence, f is strictly decreasing on 

Test: BITSAT Past Year Paper- 2010 - Question 132

The slope of the tangent to the hyperbola 2x2 – 3y2 = 6 at (3, 2) is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 132

Differentiating the given equation of the curve
4x – 6y. (dy/dx) = 0 ∴ dy/dx = 2x / 3y

Test: BITSAT Past Year Paper- 2010 - Question 133

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 133

Test: BITSAT Past Year Paper- 2010 - Question 134

If , where m ∈ N, then I10 + 10 I9 is equal to -

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 134

Test: BITSAT Past Year Paper- 2010 - Question 135

The area of the region bounded by the curve y = x | x |, x-axis and the ordinates x = 1, x = –1 is given by:

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 135

The area of the region bounded by the curve y = f (x) and the ordinates x = a, x = b is given by

According to the question,

Required area
= area of region OAB + area of region OCD
= 2 × Area of region OAB

Test: BITSAT Past Year Paper- 2010 - Question 136

What is the solution of dy/dx + 2y = 1 satisfying y(0) = 0 ?

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 136

Test: BITSAT Past Year Paper- 2010 - Question 137

The solution of differential equation  – y = 3 represents a family of

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 137

We have, 

integrating, 2 ln (y + 3) = ln x + ln c = ln cx
⇒ ln (y + 3)2 = ln cx  ⇒   (y + 3)2 = cx
which is a family of parabolas.

Test: BITSAT Past Year Paper- 2010 - Question 138

If = 676 and  = 2 then  is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 138

Test: BITSAT Past Year Paper- 2010 - Question 139

Which one of the following is the unit vector perpendicular to both  and 

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 139

Test: BITSAT Past Year Paper- 2010 - Question 140

With respect to a rectangular cartesian coordinate system, three vectors are expressed as:
and where are unit vectors, along the X, Y and Z-axis respectively. The unit vector along the direction of sum of these vector is –

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 140

Test: BITSAT Past Year Paper- 2010 - Question 141

If the middle points of sides BC, CA & AB of triangle ABC are respectively D, E, F then position vector of centre of triangle DEF, when position vector of A, B, C are respectively 

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 141

The position vector of points D, E, F are respectively

So, position vector of centre of ΔDEF

Test: BITSAT Past Year Paper- 2010 - Question 142

The perpendicular distance of  P(1, 2, 3) from the line is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 142

The point A (6, 7, 7) is on the line . Let the perpendicular from P meet the line in L. Then AP2 = (6 - 1) 2 + (7 - 2) 2 + (7 - 3) 2 = 66

∴ ⊥ distance d of P from the line is given by
d2 = AP2 - AL2 = 66 - 17 = 49 so that d = 7

Test: BITSAT Past Year Paper- 2010 - Question 143

The equation of the plane containing the line is a (x – x1) + b (y – y1) + c (z – z1) = 0, then

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 143

If the given plane contains the given line then the normal to the plane, must be perpendicular to the line and the condition for the same is aℓ + bm + cn = 0.

Test: BITSAT Past Year Paper- 2010 - Question 144

If mean of a poisson distribution of a random variable X is 2, then the value of P(X > 1.5) is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 144

Test: BITSAT Past Year Paper- 2010 - Question 145

If P(A ∪ B) = 2/3 , P(A ∪ B) = 1/6 and P(A) = 1/3 then

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 145


Now, P(AB) = P(A) P(B), A and B are independent events.

Test: BITSAT Past Year Paper- 2010 - Question 146

If a flagstaff of 6 metres high placed on the top of a tower throws a shadow of 2√3 metres along the ground, then the angle (in degrees) that the sun makes with the ground is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 146

Accordingly,

Test: BITSAT Past Year Paper- 2010 - Question 147

A wholesale merchant wants to start the business of cereal with ₹ 24000. Wheat is ₹ 400 per quintal and rice is ₹ 600 per quintal. He has capacity to store 200 quintal cereal. He earns the profit ₹ 25 per quintal on wheat and ₹ 40 per quintal on rice. If he store x quintal rice and y quintal wheat, then for maximum profit the objective function is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 147

For maximum profit, z = 40x + 25y.

Test: BITSAT Past Year Paper- 2010 - Question 148

The minimum value of  for positive real  number x, y, z is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 148

By A.M. ≥ G.M.

Test: BITSAT Past Year Paper- 2010 - Question 149

Let and f(0) = 12. If f is continuous at x = 0, then the value of a is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 149

Test: BITSAT Past Year Paper- 2010 - Question 150

Which of the following functions is differentiable at x = 0?

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 150

|x| is non-differentiable function at x = 0 as L.H.D = –1 and R.H. D = 1

But cos |h| is differentiable
∴ Any combination of two such functions will be non-differentiable. Hence option (a) and (b) are ruled out.
Now, consider sin |x| + |x|


Hence, sin |x| - |x| is diffble at x = 0.

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