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Test: BITSAT Past Year Paper- 2011 - Question 1

A passenger in a open car travelling at 30 m/s throws a ball out over the bonnet. Relative to the car the initial velocity of the ball is 20 m/s at 60° to the horizontal. The angle of projection of the ball with respect to the horizontal road will be

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 1

Test: BITSAT Past Year Paper- 2011 - Question 2

A particle is moving in a straight line with initial velocity and uniform acceleration a. If the sum of the distance travelled in tth and (t + 1)th seconds is 100 cm, then its velocity after t seconds, in cm/s, is

Test: BITSAT Past Year Paper- 2011 - Question 3

The two vectors are drawn from a common point and , then angle between is –
(1) 90° if C2 = A2 + B2
(2) greater than  90° if C2 < A2 + B2
(3) greater than  90° if C2 > A2 + B2
(4) less than  90° if C2 > A2 + B2
Correct options are –

Test: BITSAT Past Year Paper- 2011 - Question 4

If then find the dimensions of q. Where T is the time period of bar of mass M, length L and Young modulus Y.

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 4

, writing dimensions of both the sides, we get 
or q = [L4]

Test: BITSAT Past Year Paper- 2011 - Question 5

An object experiences a net force and accelerates from rest to its final position in 16s. How long would the object take to reach the same final position from rest if the object's mass was four times larger ?

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 5

When the mass increases by a factor of 4 the acceleration must decrease by a factor of four if the same force is applied. The question asks about position so we need to relate acceleration and time to position. We can do this by the equation : xf – xi = vxi t + ½ ax t2 We want the change in position to stay the same. The initial velocity is zero so in order for the change in position to remain constant the term (1/2) at2 must remain the same. If the acceleration is reduced by a factor of 4 you can see that the time must be increased by a factor of 2 in order for the term to remain the same.

Test: BITSAT Past Year Paper- 2011 - Question 6

Three blocks of masses m1, m2 and m3 are connected by massless strings, as shown, on a frictionless table. They are pulled with a force T3 = 40 N. If m1 = 10 kg, m2 = 6 kg and m3 = 4kg, the tension T2 will be

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 6

For equilibrium of all 3 masses,

For equilibrium of m1 & m2
T2 = (m1 + m2) × a   or, 2
Given m1 = 10 kg, m2 = 6 kg, m3 = 4 kg, T3 = 40 N

Test: BITSAT Past Year Paper- 2011 - Question 7

A massless platform is kept on a light elastic spring as shown in fig. When a sand particle of mass 0.1 kg is dropped on the pan from a height of 0.24 m, the particle strikes the pan and spring is compressed by 0.01 m. From what height should the particle be dropped to cause a compression of 0.04 m.

Test: BITSAT Past Year Paper- 2011 - Question 8

A constant torque of 31.4 N-m is exerted on a pivoted wheel. If angular acceleration of wheel is 4 p rad/s2, then the moment of inertia of the wheel is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 8

Test: BITSAT Past Year Paper- 2011 - Question 9

A man of mass m starts falling towards a planet of mass M and radius R. As he reaches near to the surface, he realizes that he will pass through a small hole in the planet. As he enters the hole, he sees that the planet is really made of two pieces a spherical shell of negligible thickness of mass 2M/3 and a point mass M/3 at the centre.
Change in the force of gravity experienced by the man is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 9

Change in force of gravity

(only due to mass M/3 due to shell gravitational field is zero (inside the shell))
= 2GMm/3R2

Test: BITSAT Past Year Paper- 2011 - Question 10

Geo-stationary satellite is one which

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 10

Geo-stationary satellites are also called synchronous satellite. They always remain about the same path on equater, i.e., it has a period of exactly one day (86400 sec) So orbit radius comes out to be 42400 km, which is nearly equal to the circumference of earth. So height of Geostationary satellite from the earth surface is 42,400 – 6400 = 36,000 km.

Test: BITSAT Past Year Paper- 2011 - Question 11

Two wires are made of the same material and have the same volume. However wire 1 has crosssectional area A and wire 2 has cross-sectional area 3A. If the length of wire 1 increases by Dx on applying force F, how much force is needed to stretch wire 2 by the same amount?

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 11


As shown in the figure, the wires will have the same Young’s modulus (same material) and the length of the wire of area of crosssection 3A will be  l/3 (same volume as wire 1).

⇒ F' = 9F

Test: BITSAT Past Year Paper- 2011 - Question 12

An iron rod of length 2m and cross-sectional area of 50 mm2 stretched by 0.5 mm, when a mass of 250 kg is hung from its lower end. Young’s modulus of iron rod is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 12

Test: BITSAT Past Year Paper- 2011 - Question 13

Viscosity is the property of a liquid due to which it :

Test: BITSAT Past Year Paper- 2011 - Question 14

The radiation emitted by a perfectly black body is proportional to

Test: BITSAT Past Year Paper- 2011 - Question 15

A copper sphere cools from 62°C to 50°C in 10 minutes and to 42°C in the next 10 minutes.Calculate the temperature of the surroundings.

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 15

By Newton's law of cooling,

A sphere cools from 62°C to 50°C in 10 min.

Now, sphere cools from 50°C to 42°C in next 10 min.

Dividing eqn. (2) by (3) we get,

Hence θ0 = 26°C

Test: BITSAT Past Year Paper- 2011 - Question 16

An air bubble of volume v0 is released by a fish at a depth h in a lake. The bubble rises to the surface. Assume constant temperature and standard atmospheric pressure above the lake.
The volume of the bubble just before touching the surface will be (density) of water is ρ

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 16

As the bubble rises the pr essur e gets reduced for constant temperature, if P is the standard atmospheric pressure, then (P + ρgh ) V0 = PV
or, 

Test: BITSAT Past Year Paper- 2011 - Question 17

The molecules of a given mass of gas have a root mean square velocity of 200m s–1 at 27°C and 1.0 × 105 N m–2 pressure. When the temperature is 127°C and the pressure 0.5 × 105 Nm–2, the root mean square velocity in ms–1, is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 17

Test: BITSAT Past Year Paper- 2011 - Question 18

Which of the following expressions corresponds to simple harmonic motion along a straight line, where x is the displacement and a, b, c are positive constants?

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 18

In linear S.H.M., the restoring force acting on particle should always be proportional to the displacement of the particle and directed towards the equilibrium position. i.e., F ∝ x
or F = –bx where b is a positive constant.

Test: BITSAT Past Year Paper- 2011 - Question 19

A mass m is suspended from a spring of force constant k and just touches another identical spring fixed to the floor as shown in the figure. The time period of small oscillations is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 19

When the spring undergoes displacement in the downward direction it completes one half oscillation while it completes another half oscillation in the upward direction. The total time period is:

Test: BITSAT Past Year Paper- 2011 - Question 20

The fundamental frequency of an open organ pipe is 300 Hz. The first overtone of this pipe has same  frequency as first overtone of a closed organ pipe. If speed of sound is 330 m/s, then the length of closed organ pipe is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 20

For open pipe, n = V2l, where n0 is the fundamental frequency of open pipe.

As freq. of 1st overtone of open pipe = freq. of 1st overtone of closed pipe.

= 41.25 cm

Test: BITSAT Past Year Paper- 2011 - Question 21

In an uniformly charged sphere of total charge  Q and radius R, the electric field E is plotted as function of distance from the centre. The graph which would correspond to the above will be

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 21

Electric field inside the uniformly charged sphere varies linearly, while outside the sphere, it varies as inverse square of distance,which is correctly represented in option (c).

Test: BITSAT Past Year Paper- 2011 - Question 22

A charge Q1 exerts some force on a second charge Q2. If a 3rd charge Q3 is brought near, then the force of Q1 exerted on Q2

Test: BITSAT Past Year Paper- 2011 - Question 23

A hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 10 V. The potential at a distance of 2 cm from the centre of the sphere is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 23

Potential at any point inside the sphere = potential at the surface of the sphere = 10V.

Test: BITSAT Past Year Paper- 2011 - Question 24

If the potential of a capacitor having capacity 6 mF is increased from 10 V to 20 V, then increase in its energy will be

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 24

Capacitan ce of capacitor (C) = 6 m F = 6 × 10–6 F; Initial potential (V1) = 10 V and final potential (V2) = 20 V.
The increase in energy (ΔU)

= 1/2 x (6x10-6) x [(20)2 - (10)2]
= (3x10-6) x 300 = 9 x 10-4 J.

Test: BITSAT Past Year Paper- 2011 - Question 25

Calculate the effective resistance between A and B in following network.

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 25

Equivalent resistance = (5 + 10 + 15) | | (10 + 20 + 30)
So, 

Test: BITSAT Past Year Paper- 2011 - Question 26

A steady current is set up in a cubic network composed of wires of equal resistance and length d as shown in figure. What is the magnetic field at the centre P due to the cubic network

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 26

By symmetry, the magnetic field at the centre P is zero.

Test: BITSAT Past Year Paper- 2011 - Question 27

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- 2011 - Question 27

Torque, 

Test: BITSAT Past Year Paper- 2011 - Question 28

A metal rod of length 1 m is rotated about one of its ends in a plane right angles to a field of inductance 2.5 × 10–3 Wb/m². If it makes 1800 revolutions/min. Calculate induced e.m.f. between its ends.

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 28

Given: l = 1m, B = 5 × 10–3 Wb/m²
f = 1800/60 = 30 rotations/sec In one rotation, the moving rod of the metal traces a circle of radius r = l
∴ Area swept in one rotation = πr2

= Bfπr2 = (5 × 10–3) × 3.14 × 30 × 1 = 0.471 V

Test: BITSAT Past Year Paper- 2011 - Question 29

Which one of the following curves represents the variation of impedance (Z) with frequency f in series LCR circuit?

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 29

Test: BITSAT Past Year Paper- 2011 - Question 30

An electromagnetic wave passes through space and its equation is given by E = E0 sin (ωt – kx) where E is electric field. Energy density of electromagnetic wave in space is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 30

Energy density

Test: BITSAT Past Year Paper- 2011 - Question 31

A thin convergent glass lens (μg = 1.5) has a power of + 5.0 D. When this lens is immersed in a liquid of refractive index m, it acts as a divergent lens of focal length 100 cm. The value of m must be

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 31



Test: BITSAT Past Year Paper- 2011 - Question 32

A vessel of depth 2d cm. is half filled with a liquid of refractive index µ1 and the upper half with a liquid of refractive index µ2. The apparent depth of the vessel seen perpendicularly is –

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 32

Test: BITSAT Past Year Paper- 2011 - Question 33

If the distance between the first maxima and fifth minima of a double slit pattern is 7mm and the slits are separated by 0.15 mm with the screen 50 cm. from the slits, then the wavelength of the light used is :

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 33

There are three and a half fringes from first maxima to fifth minima as shown.

Test: BITSAT Past Year Paper- 2011 - Question 34

If the energy of a photon is 10 eV, then its momentum is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 34

Momentum of a photon 
= 5.33 × 10–27 kg ms–1

Test: BITSAT Past Year Paper- 2011 - Question 35

The energies of energy levels A, B and C for a given atom are in the sequence EA < EB < EC. If the radiations of wavelengths λ1, λ2 and λ3 are emitted due to the atomic transitions C to B, B to A and C to A respectively then which of the following relations is correct ?

Test: BITSAT Past Year Paper- 2011 - Question 36

Which one is correct about fission?

Test: BITSAT Past Year Paper- 2011 - Question 37

The output of an OR gate is connected to both the inputs of a NAND gate. The combination will serve as a:

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 37

= NOR gate When both inputs of NAND gate are connected, it behaves as NOT gate OR + NOT = NOR.

Test: BITSAT Past Year Paper- 2011 - Question 38

In a semiconductor diode, the barrier potential offers opposition to

Test: BITSAT Past Year Paper- 2011 - Question 39

An electron, in a hydrogen-like atom, is in an excited state. It has a total energy of –3.4 eV. The kinetic energy and the de-Broglie wavelength of the electron are respectively

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 39

En = – 3.4 eV
The kinetic energy is eq
∴ K.E. = + 3.4 eV
The de Broglie wavelength of electron

= 0.66 × 10–9 m

Test: BITSAT Past Year Paper- 2011 - Question 40

Light of wavelength 180 nm ejects photoelectron from a plate of a metal whose work function is 2 eV. If a uniform magnetic field of 5 × 10–5 T is applied parallel to plate, what would be the radius of the path followed by electrons ejected normally from the plate with maximum energy ?

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 40

If vmax is the speed of the fastest electron emitted from the metal surface, then

∴ v = 1.31 × 106 m/s
The radius of the electron is given by

Test: BITSAT Past Year Paper- 2011 - Question 41

The product of atomic weight and specific heat of any element is a constant, approximately 6.4.This is known as :

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 41

According to Dulong and Pettit’s law Atomic weight × Specific heat = 6.4 (approx) This law is applicable only to solid elements but it fails to explain very high specific heat of diamond.

Test: BITSAT Past Year Paper- 2011 - Question 42

1.520 g of hydroxide of a metal on ignition gave 0.995g of oxide. The equivalent weight of metal is:

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 42

Let E be the equivalent weight of the metal

[17 is equivalent weight of OH and 8 is equivalent weight of oxygen]
⇒ 0.995 E + 17 × 0.995  = E × 1.52 +  8 × 1.52
⇒ 0.525 E = 16. 915 – 12.16 = 4.755

Test: BITSAT Past Year Paper- 2011 - Question 43

The correct order of radii is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 43

Effective nuclear char ge (i,e. Z/e ratio) decreases from F to N3– hence the radii follows the order: F < O2– < N3–. Z/e for F = 9/10 = 0.9, for O2– = 8/10 = .8, for N3 – = 0.7

Test: BITSAT Past Year Paper- 2011 - Question 44

Beryllium and aluminium exhibit many properties which are similar. But, the two elements differ in

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 44

The valency of beryllium is +2 while that of aluminium is +3

Test: BITSAT Past Year Paper- 2011 - Question 45

Among Al2O3, SiO2, P2O3 and SO2 the correct order of acid strength is:

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 45

Test: BITSAT Past Year Paper- 2011 - Question 46

A σ bonded molecule MX3 is T-shaped. The number of non bonded pair of electrons is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 46

For T-shape geometry the molecule must have 3 bonded pair and 2 lone pair of electrons.

Test: BITSAT Past Year Paper- 2011 - Question 47

The correct bond order in the following species is:

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 47

ion - Total number of electrons (16 – 1) = 15.
E.C. :

(Super oxide ion): Total number of electrons (16 + 1) = 17 .
E.C. :

ion: Total number of electrons = (16 – 2) = 14
E.C:

Test: BITSAT Past Year Paper- 2011 - Question 48

What is the free energy change, DG , when 1.0 mole of water at 100º C and 1 atm pressure is converted into steam at 100°C and 1 atm. pressure?

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 48

Condition of equilibrium, hence ΔG = 0.

Test: BITSAT Past Year Paper- 2011 - Question 49

H2S gas wh en passed thr ough a solution of cations containing HCl precipitates the cations of second group of qualitative analysis but not those belonging to the fourth group. It is because

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 49

IVth group needs higher S2- ion concentration. In presence of HCl, the dissociation of H2S decreases hence produces less amount of sulphide ions due to common ion effect, thus HCl decreases the solubility of H2S which is sufficient to precipitate IInd group radicals.

Test: BITSAT Past Year Paper- 2011 - Question 50

The pH of a solution is increased from 3 to 6; its H+ ion concentration will be

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 50

pH = 3. ∴ [H+] = 10–3; pH = 6,
∴ [H+] = 10–6. Hence [H+] is reduced by 10–3 times

Test: BITSAT Past Year Paper- 2011 - Question 51

A gas X at 1 atm is bubbled through a solution containing a mixture of 1 M Y and 1 M Z at 25°C. If the reduction potential is Z > Y > X, then

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 51

The more the reduction potential, the more the oxidising power.

Test: BITSAT Past Year Paper- 2011 - Question 52

When a crystal of caustic soda is exposed to air, a liquid layer is deposited because :

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 52

It is hygroscopic and deliquescent and hence absorbs moisture and CO2 to form Na2CO3
2NaOH + CO2 → Na2CO3 + H2O

Test: BITSAT Past Year Paper- 2011 - Question 53

Which of the following compound is not chiral?

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 53

None of the carbon atoms in
DCH2CH2CH2Cl is chiral i.e., each carbon atom is achiral (symmetric).

Test: BITSAT Past Year Paper- 2011 - Question 54

C6H5C ≡ N and C6H5N C exhibit which type of isomerism?

Test: BITSAT Past Year Paper- 2011 - Question 55

The correct nucleophilicity order is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 55

Nucleophilicity increases with the decrease in electronegativity of the central atom.
Since electronegativity follows the order: F > O > N > C; nucleophilicity of the concerned group will follow the reverse order i.e.,
CH3- > NH2- > OH- > F-

Test: BITSAT Past Year Paper- 2011 - Question 56

In the anion HCOOthe two carbon-oxygen bonds are found to be of equal length. What is the reason for it ?

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 56

Test: BITSAT Past Year Paper- 2011 - Question 57

What will be the product in the following reaction?

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 57

Test: BITSAT Past Year Paper- 2011 - Question 58

The fraction of total volume occupied by the atoms present in a simple cube is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 58

Number of atoms per unit cell = 1 Atoms touch each other along edges.
Hence r = a/2
(r = radius of atom and a = edge length)
Therefore % fraction = 

Test: BITSAT Past Year Paper- 2011 - Question 59

1.00 g of a non-electrolyte solute (molar mass 250 g mol–1) was dissolved in 51.2 g of benzene.If the freezing point depression constant, Kf of benzene is 5.12 K kg mol–1, the freezing point of benzene will be lowered by

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 59

Test: BITSAT Past Year Paper- 2011 - Question 60

The number of coulombs required for the deposition of 108 g of silver is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 60

Amt. deposited = 

Test: BITSAT Past Year Paper- 2011 - Question 61

During the kinetic study of the reaction, 2A + B → C + D, following results were obtained:

Based on the above data which one of the following is correct?

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 61

In case of (II ) and (I II) Keeping concentration of [A] constant, when the concentration of [B] is doubled, the rate quadruples. Hence it is second order with respect to B. In case of I & IV Keeping the concentration of [B] constant, when the concentration of [A] is increased four times, rate also increases four times. Hence, the order with respect to A is one. Hence Rate = k [A] [B]2

Test: BITSAT Past Year Paper- 2011 - Question 62

Position of non-polar and polar part in micelle is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 62

Test: BITSAT Past Year Paper- 2011 - Question 63

For adsorption of a gas on a solid, the plot of log x/m vs log P is linear with slope equal to (n being whole number)

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 63

According to Freundlich adsorption
isotherm, at intermediate pressure, extent of adsorption

plot of log x/m  vs log P is linear with slope = 1/n

Test: BITSAT Past Year Paper- 2011 - Question 64

Calcination is used in metallurgy for removal of?

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 64

Calcination is used for removal of volatile impurities and decompose carbonates.

Test: BITSAT Past Year Paper- 2011 - Question 65

Phosphine is not obtained by the reaction

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 65

Red P does not react with NaOH to give PH3.

Test: BITSAT Past Year Paper- 2011 - Question 66

Which of the following halides is not oxidized by MnO2

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 66

F2 is strongest oxidising agent. F is not oxidised by MnO2

Test: BITSAT Past Year Paper- 2011 - Question 67

Which of the following exhibit only + 3 oxidation state?

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 67

Ac (89) = [Rn] [6d1] [7s2]

Test: BITSAT Past Year Paper- 2011 - Question 68

Which of the following pairs has the same size?

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 68

Due to lanthanide contraction, the size of Zr and Hf (atom and ions) become nearly similar.

Test: BITSAT Past Year Paper- 2011 - Question 69

Which of the following is not considered as an organometallic compound?

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 69

The structural formula of cis-platin is

Since no carbon is involved it is not a organometallic compound.

Test: BITSAT Past Year Paper- 2011 - Question 70

The most stable ion is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 70

A more basic ligand forms stable bond with metal ion, Cl is most basic amongst all.

Test: BITSAT Past Year Paper- 2011 - Question 71

A is an optically inactive alkyl chloride which on reaction with aqueous KOH gives B. B on heating with Cu at 300°C gives an alkene C, what are A and C

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 71

Test: BITSAT Past Year Paper- 2011 - Question 72

The reaction

is called:

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 72

Williamson syn thesis is on e of the best methods for the preparation of symmetrical and unsymmetrical ethers. In this method, an alkyl halide is allowed to react with sodium alkoxide.

Test: BITSAT Past Year Paper- 2011 - Question 73

Which of the following esters cannot undergo Claisen self condensation ?

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 73

Claisen conden sation is given by esters having two a-hydrogen atoms.

Test: BITSAT Past Year Paper- 2011 - Question 74

Schotten -Baumann reaction is a reaction of phenols with

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 74

Test: BITSAT Past Year Paper- 2011 - Question 75

Identify X,

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 75


Test: BITSAT Past Year Paper- 2011 - Question 76

The reagent (s) which can be used to distinguish acetophenone from benzophenone is (are)

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 76

I2 and NaOH react with acetophenone (C6H5COCH3) to give yellow ppt. of CHI3 but benzophenone (C6H5COC6H5) does not give haloform test.

Test: BITSAT Past Year Paper- 2011 - Question 77

Aniline reacts with nitrous acid to produce

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 77

When aniline is treated with nitrous acid in the presence of HCl, then benzene diazonium chloride is obtained.

Test: BITSAT Past Year Paper- 2011 - Question 78

Th e structural feature wh ich distinguish es proline from natural a-amino acids?

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 78

Proline contains imino (secondary amino),group.

Test: BITSAT Past Year Paper- 2011 - Question 79

Which of the following cann ot give iodometric titration?

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 79

There is no r eaction between I and Fe3+.

Test: BITSAT Past Year Paper- 2011 - Question 80

Acetaldehyde and acetone can be distinguished by :

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 80

Acetaldehyde is easily oxidised to acetic acid by a mild oxidising agent like Fehling solution. Acetone is not easily oxidised.
Both acetone and acetaldehyde give iodoform test. Other two conditions are not relevant to aldehydes and ketones.

Test: BITSAT Past Year Paper- 2011 - Question 81

DIRECTIONS: In the following questions, two sentences are given. There may be an error in the sentence(s). Mark as your correct answer.

I. Although he was innocent, baseless accusations were leveled at him.
II. Despite of repeated representations from the people, the authorities have failed to take any action.

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 81

Sentence I, accusations were leveled against him not at him. Sentence II, despite is not followed by of.

Test: BITSAT Past Year Paper- 2011 - Question 82

DIRECTIONS: In the following questions, two sentences are given. There may be an error in the sentence(s). Mark as your correct answer.

I. I deem it as a privilege to address the gathering.
II. Perfection can be achieved with practice.

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 82

Sentence I: I deem it a privilege not as a privilege.
Sentence II: …… achieved through practice not with practice.

Test: BITSAT Past Year Paper- 2011 - Question 83

DIRECTIONS: For each of the following questions, select the option which is CLOSEST in meaning to the capitalized word.

TURBULENCE

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 83

Commotion means an disorderly outburst or tumult. Its most close to tubulence which means unstable flow of a liquid or gas.
Turbulence also refers to a state of disturbance.

Test: BITSAT Past Year Paper- 2011 - Question 84

DIRECTIONS: For each of the following questions, select the option which is CLOSEST in meaning to the capitalized word.

DEFER

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 84

Other synonyms are prorogue, put off, set back, shelve.

Test: BITSAT Past Year Paper- 2011 - Question 85

DIRECTIONS: For each of the following questions, select the option which is CLOSEST in meaning to the capitalized word.

ADAGE

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 85

An adage is a proverb or byword.

Test: BITSAT Past Year Paper- 2011 - Question 86

DIRECTIONS: Choose the word, which is most OPPOSITE in meaning as the word given in bold.

FRAGRANCE

Test: BITSAT Past Year Paper- 2011 - Question 87

DIRECTIONS: Choose the word, which is most OPPOSITE in meaning as the word given in bold.

PECULIAR

Test: BITSAT Past Year Paper- 2011 - Question 88

DIRECTIONS: Choose the word, which is most OPPOSITE in meaning as the word given in bold.

ETERNAL

Test: BITSAT Past Year Paper- 2011 - Question 89

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.

______ to popular belief that red meat makes human aggressive, scientist have found that it actually has a calming effect.

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 89

Sticking-extending out above a surface or boundary contrary-very opposite in nature or character or purpose.

Test: BITSAT Past Year Paper- 2011 - Question 90

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.

From its ______ opening sequence, it is clear that we are in the grip of a delicious new voice, a voice of breathtaking.

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 90

Evocative-recreate strong feelings, memory etc.
Mesmerizing-attract strongly.

Test: BITSAT Past Year Paper- 2011 - Question 91

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

1. making ourselves
P. our language
Q. part of growing into
R. Masters of
S. is an important
6. full manhood or womanhood

Test: BITSAT Past Year Paper- 2011 - Question 92

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

1. The very first battle they fought
P. and they had to fall back
Q. cross the border
R. was lost
S. letting the enemy
6. an enter the country

Test: BITSAT Past Year Paper- 2011 - Question 93

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

1. A nation
P. the material assets it possesses
Q. is not made by
R. and collective determination
S. but by the will
6. of the people

Test: BITSAT Past Year Paper- 2011 - Question 94

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

1. When the Governor
P. the bell had rung
Q. justice should be immediately
R. he ordered that
S. found out why
6. done to the horse

Test: BITSAT Past Year Paper- 2011 - Question 95

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

1. When you ponder over
P. that the only hope
Q. you will realize
R. of world peace lies
S. the question deeply
6. in the United Nations

Test: BITSAT Past Year Paper- 2011 - Question 96

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

One of the, numbers does not fit into the series. Find the wrong number.
15, 20, 45, 145, 565, 2830

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 96

The number should be 140.
× 1 + 5, × 2 + 5, × 3 + 5……

Test: BITSAT Past Year Paper- 2011 - Question 97

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

VWX,    BCD,    HIJ,    ?

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 97

The pattern is:

Test: BITSAT Past Year Paper- 2011 - Question 98

In a code lan guage, if TARGET is coded as 201187520, then the word WILLIUM will be coded as

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 98

Test: BITSAT Past Year Paper- 2011 - Question 99

Sanjay is taller than Suresh but sh orter than Rakesh. Rakesh is taller than Harish but shorter than Binit. Who among is the tallest?

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 99

From the question we get, Rakesh > Sanjay > Suresh Binit > Rakesh > Harish So, Binit is the tallest among them.

Test: BITSAT Past Year Paper- 2011 - Question 100

In a row of 62 persons. Rahul is 36th from left side of the row and Nitesh is 29th form the right side of the row. Find out the number of persons sitting between them?

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 100

No. of Persons between Rahul and Nitesh = (36 + 29) - 62-2 = 65 - 62-2 = 1

Test: BITSAT Past Year Paper- 2011 - Question 101

The missing number in the given figure is

Test: BITSAT Past Year Paper- 2011 - Question 102

Select the combination of number so that the letters arranged will from a meaningful word.

Test: BITSAT Past Year Paper- 2011 - Question 103

Which of the given Venn diagrams out of (a), (b), (c) or (d) correctly represents the relationship among the following classes?
Rose, Flower, Lotus

Test: BITSAT Past Year Paper- 2011 - Question 104

A piece of paper is folded and a cut is made as shown below. From the given responses indicate how it will appear when opened?

Test: BITSAT Past Year Paper- 2011 - Question 105

Which answer figure will complete the question figure?
Question figure

Test: BITSAT Past Year Paper- 2011 - Question 106

If f(x) is a function that is odd and even simultaneously, then f(3) - f(2) is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 106

f (x) = 0 x ∈ R  ⇒  f (3) – f (2) = 0

Test: BITSAT Past Year Paper- 2011 - Question 107

If, tan A = 1/2 and tan B = 1/3, then find the valueof A + B

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 107


∴ A + B = 45° = π/4

Test: BITSAT Past Year Paper- 2011 - Question 108

If sin θ = – 1/2 and tan θ = 1/3 , then θ =

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 108

We shall first consider values of θ between 0 and 2π sin θ = 
or sin (2π – π/6)
∴ θ = 7π/6 ; 11π/6
tan θ = 1 /3 = tan (π/6) or tan (π + π/6)
∴ θ = π/6, 7π/6
The value of θ which satisfies both the equations is 7π/6
Hence the general value of θ is 2nπ + 7π/6 where n ∈ I.

Test: BITSAT Past Year Paper- 2011 - Question 109

is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 109

Test: BITSAT Past Year Paper- 2011 - Question 110

For n ∈ N, xn+1 + (x + 1)2n–1 is divisible by

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 110

For n = 1, we have;
xn+1 + (x + 1)2n–1 = x2 + (x + 1) = x2 + x + 1,
which is divisible by x2 + x + 1
For n = 2, we have; xn+1 + (x + 1)2n–1
= x3 + (x + 1)3 = (2x + 1) (x2 + x + 1),
which is divisible by x2 + x + 1.

Test: BITSAT Past Year Paper- 2011 - Question 111

If a, b are the roots of the equation ax2 + bx + c = 0, then the roots of the equation ax2 + bx (x + 1) + c (x + 1)2 = 0 are

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 111

Putting z = a + 2i in the given equation and comparing imaginary parts, we get a2 + 4 = a2, which is not possible.

Test: BITSAT Past Year Paper- 2011 - Question 112

If a > 0, a ∈ R, z = a + 2i and z |z| – az + 1 = 0 then

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 112

Putting z = a + 2i in the given equation and comparing imaginary parts, we get a2 + 4 = a2, which is not possible.

Test: BITSAT Past Year Paper- 2011 - Question 113

Which of the following is not a vertex of the positive region bounded by the inqualities 2x + 3y ≤ 6, 5x + 3y ≤ 15 and x, y ≥ 0

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 113

Here (0, 2), (0, 0) and (3, 0) all are vertices of feasible region.

Test: BITSAT Past Year Paper- 2011 - Question 114

If 20Cr = 20Cr – 10 then 18Cr is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 114

20Cr = 20Cr–10 ⇒ r + (r – 10) = 20 ⇒ r = 15
∴ 18Cr = 18C15 = 18C3 = 18.17.16/1.2.3 = 816

Test: BITSAT Past Year Paper- 2011 - Question 115

The term independent of x in the expansion of ,  is a times the corresponding binomial coefficient. Then a is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 115


is independent of x provided r = 12 and then a = 1.

Test: BITSAT Past Year Paper- 2011 - Question 116

In the binomial (21/3 + 3-1/3)n,  if  the  ratio  of the seventh term from the beginning of the expansion to the seventh term from its end is 1/6, then n equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 116

Tr + 1 = nCr an - r . br where  a = 21/3  and
b = 3-1/3
T7  from  beginning = nC6 an - 6 b6 and
T7 from end  = nC6  bn - 6  a6

⇒  n - 12 = - 3   ⇒   n = 9

Test: BITSAT Past Year Paper- 2011 - Question 117

If pth,qth and rth terms of H.P. are u,v,w respectively, then find the value of the expression (q - r) vw + (r - p) wu + (p - q) uv.

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 117

Let H.P. be 

Test: BITSAT Past Year Paper- 2011 - Question 118

If the sum of the first 2n terms of 2, 5, 8, ....... is equal to the sum of the first n terms of 57, 59, 61......., then n is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 118

Given,
2n/2 {2.2 + (2n - 1)3} = n/2 {2.57 + (n - 1)2}
or  2 (6n + 1) = 112 + 2n   or  10n = 110, ∴  n = 11

Test: BITSAT Past Year Paper- 2011 - Question 119

The distance of the point (–1, 1) from the line 12(x + 6) = 5 (y – 2) is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 119

The given line is 12 (x + 6) = 5(y – 2)
⇒ 12x + 72 = 5y – 10 or 12x – 5y + 72 + 10 = 0
⇒ 12x – 5y + 82 = 0 The perpendicular distance from (x1, y1) to the line  ax + by + c = 0 is 
The point (x1, y1) is (–1, 1), therefore, perpendicular distance from (–1, 1) to the line 12x – 5y + 82 = 0 is

Test: BITSAT Past Year Paper- 2011 - Question 120

The family of straight lines (2a + 3b) x + (a – b) y + 2a – 4b = 0 is concurrent at the point

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 120

Rewriting the equation
(2x + y + 2) a +  (3x – y – 4)b = 0 and for all a, b the straight lines pass through the intersection of  2x + y + 2 = 0 and 3x – y – 4 = 0
i.e., the point 

Test: BITSAT Past Year Paper- 2011 - Question 121

The length of the latus-rectum of the parabola whose focus is and directrix is  is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 121

According to the figure, the length of latus rectum is

Test: BITSAT Past Year Paper- 2011 - Question 122

The equation of the ellipse with focus at (±5, 0) and x =36/5 as one directrix is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 122

We have ae = 5 [Since focus is (±ae, 0)] and a/e = 36/5 
On solving we get a = 6 
Thus, the required equation of the ellipse is

Test: BITSAT Past Year Paper- 2011 - Question 123

For what value of k the circles x2 + y2 + 5x + 3y + 7 = 0 and x2 + y2 – 8x + 6y + k = 0 cuts orthogonally

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 123

Let the two circles be x2 + y2+ 2g1 x + 2f1y + c1 = 0 and  x2 + y2 + 2g2 x + 2f2y + c2 = 0 where g1 = 5/2, f1 = 3/2, c1 =7, g2 =–4, f2 =3 and c2 = k
If the two circles intersects orthogonally, then
⇒ 11 = 7 + k ⇒ k = – 18

Test: BITSAT Past Year Paper- 2011 - Question 124

If the lines 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0 are tangents to a circle, then the radius of the circle is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 124

The diameter of the circle is perpendicular distance between the parallel lines (tangents) 3x – 4y + 4 = 0 and 3x – 4y – 7/2 = 0 and so it is equal to Hence radius is 3/4.

Test: BITSAT Past Year Paper- 2011 - Question 125

Evaluate 

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 125


Test: BITSAT Past Year Paper- 2011 - Question 126

Negation of “Paris in France and London is in England” is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 126

Let p : Paris is in France, q : London is in England
∴ i.e., Paris is not in France or London is not in England..

Test: BITSAT Past Year Paper- 2011 - Question 127

Find the A.M. of the first ten odd numbers.

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 127

First ten odd numbers are 1, 3, 5, 7, 11, 13, 15, 17, 19 respectively. So A.M.

Test: BITSAT Past Year Paper- 2011 - Question 128

If A and B are mutually exclusive events and if P(B) = 1/3, P(A ∪ B) = 13/21 then P(A) is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 128

For mutually exclusive events
P(A ∪ B) = P(A) + P(B) ⇒ P(A) = 2/7

Test: BITSAT Past Year Paper- 2011 - Question 129

A die is loaded such that th e pr obability of throwing the number i is proportional to its reciprocal. The probability that 3 appears in a single throw is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 129

Test: BITSAT Past Year Paper- 2011 - Question 130

If is given as

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 130

Test: BITSAT Past Year Paper- 2011 - Question 131

If the domain of f–1 (x) is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 131


Clearly, f –1 (x) is defined for 1 + x ≠ 0.
Hence, domain of f –1 (x) is R – {– 1}

Test: BITSAT Past Year Paper- 2011 - Question 132

The value of

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 132


The given expression = 0

Test: BITSAT Past Year Paper- 2011 - Question 133

The matrix A2 + 4A – 5I, where I is identity matrix and equals:

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 133

A2 + 4A – 5 I = A × A + 4A – 5I

Test: BITSAT Past Year Paper- 2011 - Question 134

If A = then adj ( adj  A) is equal to -

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 134

Test: BITSAT Past Year Paper- 2011 - Question 135

If and then dy/dx is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 135

Test: BITSAT Past Year Paper- 2011 - Question 136

The function f(x) = is at x = 1

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 136

Con tinuous as well as differ entiable, so f '(1) = 0

Test: BITSAT Past Year Paper- 2011 - Question 137

The function f(x) = sin x – kx – c, where k and c are constants, decreases always when

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 137

Let f (x) = sin x – kx – c where k and c are constants. f'(x) = cos x – k
∴ f decreases if cos x ≤ k
Thus, f (x) = sin x – kx – c decrease always when k ≥ 1.

Test: BITSAT Past Year Paper- 2011 - Question 138

The minimum value of f (x) = sin4 x + cos4 x in the interval is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 138

Let y = sin4x  +  cos4x
dy/dx = 4 sin3 x cos x 4 cos3 x ( -sin x)
= 4 sin xcox (sin2 x - cos2x)
= (2 sin 2x) (– cos 2x) = – sin 4x
∴ dy/dx = 0 ⇒ sin 4x = 0 ⇒ 4x = 0, π, 2π​, 3π​

Test: BITSAT Past Year Paper- 2011 - Question 139

The curve y –exy+ x = 0 has a vertical tangent at

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 139

y – exy + x = 0

for the vertical tangents
1 – x (x + y) = 0 i.e., 
∴ x = 1 and y = 0

Test: BITSAT Past Year Paper- 2011 - Question 140

The function f(x) = 2x3 – 3x2 – 12x + 4, has

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 140

f(x) = 2x3 – 3x2 – 12x + 4
⇒ f '(x) = 6x2 – 6x – 12 = 6(x2 – x – 2)
= 6(x – 2) (x + 1)
For maxima and minima f '(x) = 0
∴ 6(x – 2)(x + 1) = 0 Þ x = 2, – 1
Now, f ''(x) = 12x-6
At x = 2; f ''(x) = 24 - 6 = 18> 0
∴ x = 2 , local min. point
At x = – 1; f ''(x) = 12 ( -1) - 6 = -18<0
∴ x = –1 local max. point

Test: BITSAT Past Year Paper- 2011 - Question 141

Evaluate 

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 141

Given integral  

Test: BITSAT Past Year Paper- 2011 - Question 142

 Find the value of dx

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 142

We know that |sinx| is a periodic function of π

Test: BITSAT Past Year Paper- 2011 - Question 143

Let  then

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 143

Test: BITSAT Past Year Paper- 2011 - Question 144

What is the area bounded by y = tan x , y = 0 and  x = π/4?

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 144

 Required area = tan x dx

Test: BITSAT Past Year Paper- 2011 - Question 145

The degree of the differential equation

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 145


It is a differential equation of degree 2.

Test: BITSAT Past Year Paper- 2011 - Question 146

Two vectors are such that The angle between the two vectors will be–

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 146

Squaring both the sides, we get
or dot product is zero) . Therefore angle between is 90°

Test: BITSAT Past Year Paper- 2011 - Question 147

Gives the line L: and theplane π : x – 2y – z = 0. Of the following assertions, the only one that is always true is 

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 147

Since 3(1) + 2(– 2) + (–1) (–1) = 3 – 4 + 1 = 0
∴ Given line is ^ to the normal to the planei.e., given line is parallel to the given plane.
Also, (1, –1, 3) lies on the plane x – 2y – z = 0 if 1 – 2 (–1) – 3 = 0   i.e.,   1 + 2 – 3 = 0 which is true  ∴ L lies in plane π.

Test: BITSAT Past Year Paper- 2011 - Question 148

A ladder rests against a wall so th at its top touches the roof of the house. If the ladder makes an angle of 60° with the horizontal and height of the house be 63 meters, then the length of the ladder in meters is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 148

Length of ladder = 

Test: BITSAT Past Year Paper- 2011 - Question 149

In an equilateral triangle, the in radius, circumradius and one of the ex-radii are in the ratio

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 149

Test: BITSAT Past Year Paper- 2011 - Question 150

For the constraints of a L.P. Problem given by x1 + 2x2 ≤  2000, x1 + x2 ≤ 1500  and x2 ≤ 600 and x1, x2 ≥ 0, which one of the following points does not lie in the positive bounded region

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 150

Clearly point (2000, 0) is outside.

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