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Test: BITSAT Past Year Paper- 2013 - Question 1

The velocity and acceleration vectors of a particle undergoing circular motion are and respectively at an in stant of time. The radius of the circle is –

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 1

It can be observed th at component of acceleration perpendicular to velocity is a = 4 m/s2

Test: BITSAT Past Year Paper- 2013 - Question 2

Wave pulse can travel along a tense string like a violin spring. A series of experiments showed that the wave velocity V of a pulse depends on the following quantities, the tension T of the string, the cross-section area A of the string and then as per unit volume ρ of the string. Obtain an expression for V in terms of the T, A and r using dimensional analysis.

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 2

Let  V = kTa Abρc,
k = dimensional constant
Writing dimension on both we side
[LT–1] = [MLT–2]a [L2]b [ML–3]c              = [Ma+cLa+2b–3cT–2a]Comparing power on both sides we have a + c = 0,   a + 2b – 3c = 1,    –2a = –1

Test: BITSAT Past Year Paper- 2013 - Question 3

A man runs at a speed of 4 m/s to overtake a standing bus. When he is 6 m behind the door at t = 0, the bus moves forward and continuous with a constant acceleration of 1.2 m/s2. The man reaches the door in time t. Then,

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 3

Let us draw the figure for given situation,


⇒ 4 t = 6 + 0.6 t2

Test: BITSAT Past Year Paper- 2013 - Question 4

A body is projected, making an acute angle with the horizontal. If angle between velocity and acceleration is θ, then

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 4

Here velocity is acting upwards wh en projectile is going upwards and acceleration is downwards. The angle θ between and ar is more than 0º and less than 180º.

Test: BITSAT Past Year Paper- 2013 - Question 5

The minimum velocity (in ms–1) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction 0.6 to avoid skidding is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 5

The condition to avoid skidding,

Test: BITSAT Past Year Paper- 2013 - Question 6

A bob is hanging over a pulley inside a car through, a string. The second end of the string is in the hand of a person standing in the car. The car is moving with constant acceleration 'a' directed horizontally as shown in figure. Other end of the string is pulled with constant acceleration ‘a’ vertically. The tension in the string is equal to –

Test: BITSAT Past Year Paper- 2013 - Question 7

A block of mass m is placed on a smooth inclined wedge ABC of inclination q as shown in the figure. The wedge is given an acceleration ‘a’ towards the right. The relation between a and θ for the block to remain stationary on the wedge is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 7

Let the mass of block is m. It will remains stationary if forces acting on it are in equilibrium. i.e., ma cos θ = mg sin θ
⇒ a = g tan θ

Here ma = Pseudo force on block, mg = weight.

Test: BITSAT Past Year Paper- 2013 - Question 8

A 3.628 kg freight car moving along a horizontal rail road spur track at 7.2 km/hour strikes a bumper whose coil springs experiences a maximum compression of 30 cm in stopping the car. The elastic potential energy of the springs at the instant when they are compressed 15 cm is

Test: BITSAT Past Year Paper- 2013 - Question 9

A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses 0.36 kg and 0.72 kg. Taking g = 10 m/s2, find the work done (in joules) by the string on the block of mass 0.36 kg during the first second after the system is released from rest.

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 9

Given m = 0.36 kg, M = 0.72 kg.
The figure shows the forces on m and M.
When the system is released, let the acceleration be a. Then

T – mg = ma
Mg – T = Ma
∴ 
and T =  4 mg/3
For block m :
u = 0, a = g/3, t = 1, s = ?

∴ Work done by the string on m is

Test: BITSAT Past Year Paper- 2013 - Question 10

Two rings of radius R and nR made of same material have the ratio of moment of inertia about an axis passing through centre is 1 : 8. The value of n is 

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 10

Ratio of moment of inertia of the rings

(λ = linear density of wire = constant)

∴ n3 = 8 ⇒ n = 2

Test: BITSAT Past Year Paper- 2013 - Question 11

A particle of mass ‘m’ is projected with a velocity v making an angle of 30° with the horizontal. The magnitude  of angular momentum of the projectile about the point of projection when the particle is at its maximum  height ‘h’ is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 11

Angular momentum L0 = pr ⊥
(∴ linear momentum p = mv cos θ and r = H)
⇒ L0 = mv cos θH


Test: BITSAT Past Year Paper- 2013 - Question 12

There is a shell of mass M and density of the shell is uniform. The work done to take a point mass from point A to B is  [AB = r]

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 12

Gravitaional field inside the shell is zero, so no work required.

Test: BITSAT Past Year Paper- 2013 - Question 13

A disc is performing pure rolling on a smooth stationary surface with constant angular velocity as shown in figure. At any instant, for the lower most point of the disc –

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 13

As the disc is in combined rotation and translation, each point has a tangential velocity and a linear velocity in the forward direction. From figure

vnet (for lowest point) = v – Rω = v – v = 0 and acceleration

(since linear speed is constant)

Test: BITSAT Past Year Paper- 2013 - Question 14

A cube is subjected to a uniform volume compression. If the side of the cube decreases by 2% the bulk strain is

Test: BITSAT Past Year Paper- 2013 - Question 15

A ball whose density is 0.4 × 103 kg/m3 falls into water from a height of 9 cm. To what depth does the ball sink ?

Test: BITSAT Past Year Paper- 2013 - Question 16

Figure shows a copper rod joined to a steel rod.
The rods have equal length and equal crosssectional area. The free end of the copper rod is kept at 0ºC and that of steel rod is kept at 100ºC.
Find the temperature of the junction of the rod.
Conductivity of copper = 390 W/mºC.
Conductivity of steel = 46 W/m ºC

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 16

Heat current in first rod (copper)

Here q is ltemperature of the junction and A & l are area and length of copper rod.
Heat current in second rod (steel)

In series clombination, heat current remains same. So,

– 390 θ = = 46 θ – 4600 436 θ = 4600  ⇒  θ = 10.6ºC

Test: BITSAT Past Year Paper- 2013 - Question 17

If the radius of a star is R and it acts as a black body, what would be the temperature of the star, in which the rate of energy production is Q ?

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 17

Stefan’s law for black body radiation
Q = σe AT4

Here e = 1
A = 4πR2

Test: BITSAT Past Year Paper- 2013 - Question 18

A thermodynamical system is changed from state (P1, V1) to (P2, V2) by two different process, the quantity which will remain same will be

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 18

For all process
ΔU = ΔQ – ΔW
does not change as it depends on initial final states.

Test: BITSAT Past Year Paper- 2013 - Question 19

A Car not’s heat engine works between the temperatures 427°C and 27°C. What amount of heat should it consume per second to deliver mechanical work at the rate of 1.0 kW?

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 19

The efficiency of the heat engine is

But 

Thus, the engine would require 417 cal of heat per second, to deliver the requisite amount of work.

Test: BITSAT Past Year Paper- 2013 - Question 20

A vessel containing 1 more of O2 gas (molar mass 32) at temperature T. The pressure of the gas is p.An identical vessel containing one mole of he gas (molar mass 4) at temperature 2T has a pressure of

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 20

Applying gas equation, pV = nRT We can write, p1V = n1RT1 and p2V = n2RT2

Test: BITSAT Past Year Paper- 2013 - Question 21

The temperature of an ideal gas is increased from 27°C to 127°C, then percentage increase in vrms is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 21

We know, 
⇒ % increase in

Test: BITSAT Past Year Paper- 2013 - Question 22

Two gases occupy two containers A and B the gas in A, of volume 0.10m3, exerts a pressure of 1.40 MPa and that in B of volume 0.15m3 exerts a pressure 0.7 MPa. The two containers are united by a tube of negligible volume and the gases are allowed to intermingle. Then if the temperature remains constant, the final pressure in the container will be (in MPa)

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 22

We know that
PAVA = nA RT, PBVB = nBRT and Pf (VA + VB) = (nA + nB) RT
Pf (VA + VB) = PAVA + PBVB

Test: BITSAT Past Year Paper- 2013 - Question 23

An instan tan eous displacemen t of a simple harmonic oscillator is x = A cos (ωt + p/4). Its speed will be maximum at time

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 23

Velocity, 
Velocity will be maximum, when wt + ω/4 = π/2 or ωt = π/2 – π/4 = π/4
or t = π/4ω

Test: BITSAT Past Year Paper- 2013 - Question 24

Two waves of wavelengths 99 cm and 100 cm both travelling with velocity 396 m/s are made to interfere. The number of beats produced by them per second is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 24

Velocity of wave v = nλ
where n = frequency of wave 

no. of beats = n1 – n2 = 4

Test: BITSAT Past Year Paper- 2013 - Question 25

If equation of transverse wave is y = x0 cos Maximum velocity of particle istwice of wave velocity, if l is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 25


Test: BITSAT Past Year Paper- 2013 - Question 26

Three equal charges (q) are placed at corners of an equilateral triangle of side a. The force on any charge is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 26

Test: BITSAT Past Year Paper- 2013 - Question 27

Two identical capacitors, have the same capacitance C. One of them is charged to potential V1 and the other to V2. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is –

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 27

Initial energy of combined system

Final common potential, 
Final energy of system,

Hence loss of energy = U1 – U2

Test: BITSAT Past Year Paper- 2013 - Question 28

What should be the characteristic of fuse wire?

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 28

Fuse wire should be such that it melts immediatley when strong current flows through the circuit. The same is possible if its melting point is low and resistivity is high.

Test: BITSAT Past Year Paper- 2013 - Question 29

In the circuit shown in figure potential difference between points A and B is 16 V. the current passing through 2Ω resistance will be​

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 29

∴ 4i1 + 2(i1 + i2) – 3 + 4i1 = 16V ...(i) Using Kirchhoff’s second law in the closed loop we have
9 – i2 – 2(i1 + i2) = 0 …(ii)
Solving equations (i) and (ii), we get
i1 = 1.5 A   and i2 = 2 A
∴ current through 2W resistor = 2 + 1.5 = 3.5 A.

Test: BITSAT Past Year Paper- 2013 - Question 30

Two parallel conductors carry current in opposite directions as shown in figure. One conductor carries a current of 10.0 A. Point C is a distance d/2 to the right of the 10.0 A current. If d = 18 cmand I is adjusted so that the magnetic field at C is zero, the value of the current I is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 30

The magnetic field at C due to first conductor is (since, point C is separated by from 1st conductor). The direction of field is perpendicular to the plane of paper and directed outwards.
The magnetic field at C due to second conductor is (since, point C is separated by d/2 from 2nd conductor)
The direction of field is perpendicular to the plane of paper and directed inwards.
Since, direction of B1 and B2 at point C is in opposite direction and the magnetic field at C is zero, therefore,
 B1 = B2

On solving I = 30.0 A

Test: BITSAT Past Year Paper- 2013 - Question 31

A uniform electric field and uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the direction of fields, then the electron

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 31

and  are in same direction so th at magnetic force on electron becomes zero, only electric force acts. But force on electron due to electric field is opposite to the direction of velocity.

Test: BITSAT Past Year Paper- 2013 - Question 32

Eddy currents are produced when

Test: BITSAT Past Year Paper- 2013 - Question 33

Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area A = 10 cm2 and length = 20 cm. If one of the solenoid has 300 turns and the other 400 turns, their mutual inductance is
0 = 4p × 10 –7 Tm A–1)

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 33

Test: BITSAT Past Year Paper- 2013 - Question 34

The ratio of secondary and primary turns of step-up transformer is 4 : 1. If a current of 4 A is applied to the primary, the induced current in secondary will be

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 34

Test: BITSAT Past Year Paper- 2013 - Question 35

Which of the following electromagnetic radiations has the smallest wavelength?

Test: BITSAT Past Year Paper- 2013 - Question 36

When light is refracted, which of the following does not change ?

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 36

Frequency does not change on refraction.

Test: BITSAT Past Year Paper- 2013 - Question 37

The given len s is broken into four parts and rearranged as shown. If the initial focal length is f then after rearrangement the equivalent focal length is –

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 37

Cutting a lens in tra nsver se direction doubles their focal length i.e. 2f.
Using the formula of equivalent focal length,

We get equivalent focal length as f/2.

Test: BITSAT Past Year Paper- 2013 - Question 38

In Young's double slit experiment 10th order maximum is obtained at the point of observation in the interference pattern for λ = 7000 Å. If the source is replaced by another one of wavelength 5000 Å then the order of maximum at the same point will be –

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 38

n1 λ1 = n2 λ2
10 × 7000  = n2 × 5000 ⇒ n2 = 14

Test: BITSAT Past Year Paper- 2013 - Question 39

Transfer characteristics [output voltage (V0) vs input voltage (V1)] for a base biased transistor in CE configuration is as shown in the figure. For using transistor as a switch, it is used

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 39

I  → ON
II → OFF
In IInd state it is used as a amplifier it is active region.

Test: BITSAT Past Year Paper- 2013 - Question 40

The circuit is equivalent to​

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 40

Y = A + (.B) = (A + ).(A+ B) = A + B
⇒ OR gate

Test: BITSAT Past Year Paper- 2013 - Question 41

How many grams of con centrated nitric acid solution should be used to prepare 250 mL of 2.0M HNO3? The concentrated acid is 70% HNO3.

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 41

Molarity (M) = 


wt. of 70% acid = 

Test: BITSAT Past Year Paper- 2013 - Question 42

The Bohr orbit radius for the hydrogen atom (n = 1) is approximately 0.530 Å. The radius for the first excited state (n = 2) orbit is (in Å)

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 42

Radius of hydrogen atom = 0.530 Å, Number of excited state (n) = 2 and atomic number of hydrogen atom (Z) = 1. We know that the Bohr radius.

= 4 × 0.530 = 2.12 Å

Test: BITSAT Past Year Paper- 2013 - Question 43

The screening effect of d-electrons is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 43

The screening effect of inner electron of the nucleus causes the decrease in ionization potential, therefore the order of the screening effect is
f < d < p < s
Hence, the screening effect of d-electron is less than p-electron.

Test: BITSAT Past Year Paper- 2013 - Question 44

When the first ionisation energies are plotted against atomic number, the peaks are occupied by

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 44

Rare gases; as the e is to be removed from stable configuration.

Test: BITSAT Past Year Paper- 2013 - Question 45

The ions O2–, F, Na+, Mg 2+ a nd Al3+ are isoelectronic. Their ionic radii show :

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 45

Amongst isoelectronic species, ionic radii of anion is more than that of cations. Further size of anion increase with increase in –ve charge and size of cation decrease with increase in + ve charge. Hence ionic radii decreases from O to Al3+.

Test: BITSAT Past Year Paper- 2013 - Question 46

Using MOT, which of the following pairs denote paramagnetic species?

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 46

B2 and O2 are paramagnetic due to presence of unpaired electron. MO electronic configuration of B2 is :

MO electorn i.e. configuration of O2 is :

Test: BITSAT Past Year Paper- 2013 - Question 47

Increasing order of rms velocities of H2, O2, N2 and HBr is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 47

RMS velocity of molecules depends on mass. If mol. wt. increases, rms velocity of melocules decreases.

The order of increasing m. wt. is
H 2 < N 2 < O 2 < HBr
Order of Vrms of molecules.
HBr < O 2 < N 2 < H 2 .

Test: BITSAT Past Year Paper- 2013 - Question 48

For the dissociation reaction,
H2 (g) → 2H (g) ΔH = 162 Kcal,
heat of atomisation of H is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 48

ΔH = ΔH(product) – DH(reactant)
162 = 2 × ΔHH -DHH2

ΔHH = 81 Kcal

Test: BITSAT Past Year Paper- 2013 - Question 49

The enthalpy of combustion of 2 moles ofbenzene at 27°C differs from the value determinedin bomb calorimeter by

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 49

By bomb calorimeter we get DE . 2C6 H 6 (l) + 15 O2 (g) → 12CO2(g) + 6H2O (l)
ΔH – ΔE = ΔnRT
= (12 – 15) × 8.314 × 300 = – 7.483 kJ

Test: BITSAT Past Year Paper- 2013 - Question 50

If 1.0 mole of I2 is introduced into 1.0 litre flask at 1000 K, at equilibrium (Kc = 10–6), which one is correct?

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 50



It shows that (1 – x) < 2x

Test: BITSAT Past Year Paper- 2013 - Question 51

For the reaction CO (g) + (1/2) O2 (g) → CO2 (g), Kp / Kc is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 51

Kp = Kc(RT)Δn;

Test: BITSAT Past Year Paper- 2013 - Question 52

The oxidation state of sulphur in Na2S4O6 is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 52

Na2S4O6
2 + 4x – 12 = 0
4 x – 10 = 0

Oxidation state of S is = 

Test: BITSAT Past Year Paper- 2013 - Question 53

When same amount of  zinc is treated separately with excess of sulphuric acid and excess of sodium hydroxide solution, the ratio of volumes of hydrogen evolved is:

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 53

Zn + H2SO4 → ZnSO4 + H2
Zn + 2NaOH → Na2ZnO2 + H2
[ Ratio of volumes of H2 evolved is 1 : 1

Test: BITSAT Past Year Paper- 2013 - Question 54

The alkali metals form salt-like hydrides by the direct synthesis at elevated temperature. The thermal stability of these hydrides decreases in which of the following orders ?

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 54

The s tability of alkali metal h ydr ides decreases from Li to Cs. It is due to the  fact that M–H bonds becomes weaker with increase in size of alkali metals as we move down the group from Li to Cs. Thus the order of stability of hydrides is LiH > NaH > KH > RbH > CsH

Test: BITSAT Past Year Paper- 2013 - Question 55

The most stable carbanion among the following is :

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 55

-NO2 group, being strong electron withdrawing, disperses the -ve charge, hence stabilizes the concerned carbanion.

Test: BITSAT Past Year Paper- 2013 - Question 56

Among the following four structures I to IV,

it is true that

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 56

A chiral object or structure has four different groups attached to the carbon.

Test: BITSAT Past Year Paper- 2013 - Question 57

The number of enan tiomers of the compound CH3CHBrCHBrCOOH is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 57

The structure 
has two different chiral carbon atoms, hence number of enantiomers (optically active forms) is 2n = 22 = 4

Test: BITSAT Past Year Paper- 2013 - Question 58

Which one of the following reactions is expected to readily give a hydrocarbon product in good yields ?

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 58

Electrolysis of a concentrated aqueous solution of either sodium or potassium salts of saturated carboxylic acids yields higher alkane at anode.

Test: BITSAT Past Year Paper- 2013 - Question 59

What will be the main product when acetylene reacts with hypochlorous acid?

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 59



Test: BITSAT Past Year Paper- 2013 - Question 60

The greenhouse effect is because of the

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 60

Green house gases such as CO2, ozon e, methane, the chlorofluoro carbon compounds and water vapour form a thick cover around the earth which prevents the IR rays emitted by the earth to escape. It gradually leads to increase in temperature of atmosphere.

Test: BITSAT Past Year Paper- 2013 - Question 61

Due to Frenkel defect, the density of ionic solids

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 61

No change in density.

Test: BITSAT Past Year Paper- 2013 - Question 62

Equal weights of NaCl and KCl are dissolved separately in equal volumes of solutions. Molarity of the two solutions will be :

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 62

When equal weights of different solutes are present in equal volumes of solution the molarity is inversely related to molecular mass of the solute. Mol. mass of NaCl is less than KCl. Hence, molarity of NaCl solution will be more.

Test: BITSAT Past Year Paper- 2013 - Question 63

A current of 2.0 A passed for 5 hours through a molten metal salt deposits 22.2 g of metal (At wt. = 177). The oxidation state of the metal in the metal salt is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 63

Test: BITSAT Past Year Paper- 2013 - Question 64

The electrolytic cells, one containing acidified ferrous chloride and another acidified ferric chloride are connected in series. The ratio of iron deposited at cathodes in the two cells when electricity is passed through the cells will be

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 64

Test: BITSAT Past Year Paper- 2013 - Question 65

Velocity constant of a reaction at 290 K was found to be 3.2×10–3. At 300 K it will be

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 65

Th e velocity con stant doubles for every 10°C  rise in temperature.

Test: BITSAT Past Year Paper- 2013 - Question 66

At high pressure, the entire surface gets covered by a monomolecular layer of the gas follows

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 66

At high pressure the extent of adsorption follows zero order kinetics.

Test: BITSAT Past Year Paper- 2013 - Question 67

Which of the following is incorrect with respect to property indicated ?

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 67

F2 is expected to have highest bond en ergy but the correct decreasing order is Cl2 > Br2 > F2 because of fluorine atom has very small size due to which there is a high inter electronic repulsion between two fluorine  atoms so the bond between two fluorine gets weaker and   need less energy.

Test: BITSAT Past Year Paper- 2013 - Question 68

Strong reducing behaviour of H3PO2 is due to

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 68

The acids which contain P–H bond have strong reducing properties. Thus H3PO2 acid is good reducing agent as it contains two P–H  bonds. For example, it reduces AgNO3 to metallic silver.
4 AgNO3 + 2H2O + H3PO2 → 4Ag + 4HNO3 + H3PO4

Test: BITSAT Past Year Paper- 2013 - Question 69

The pair in which both species have same magnetic moment (spin only value) is :

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 69

[Cr(H2O)6]2+ Cr is in Cr2+ form

In [Fe(H2O)]2+ Fe2+ form. Both will have 4 unpaired electrons.

Test: BITSAT Past Year Paper- 2013 - Question 70

Which of the following is less acidic among the given halogen compounds?

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 70

Due to stronger–I-effect of F than that of Cl, CHF3 should be more acidic than CHCl3. But actually reverse is true.
This is due to : CCl3- left after the removal of a proton from CHCl3 is stabilised due to presence of d-orbitals in Cl than: CF3 left after the removal of a proton from CHF3 which is not stabilised due to the absence of d-orbitals on F.

Test: BITSAT Past Year Paper- 2013 - Question 71

In a SN2 substitution reaction of the type

which one of the following has the highest relative rate ?

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 71

SN2 mech an ism is followed in case of primary and secondary halides i.e., SN2 reaction is favoured by small groups on the carbon atom attached to halogens so

Test: BITSAT Past Year Paper- 2013 - Question 72

Which of th e following is not the product of dehydration of 

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 72


 

Test: BITSAT Past Year Paper- 2013 - Question 73

What will be the correct structural formula of product for the following reaction?

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 73


Test: BITSAT Past Year Paper- 2013 - Question 74

Nucleophilic addition reaction will be most favoured in

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 74

Aldehydes are more reactive than ketones due to +I effect of –CH3 group. There are two – CH3 group in acetone which reduces +ve charge density on carbon atom of carbonyl group. More hindered carbonyl group becomes less reactive. So in the given case CH3CHO is the right choice.

Test: BITSAT Past Year Paper- 2013 - Question 75

Identify the prdouct C in the series

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 75

Test: BITSAT Past Year Paper- 2013 - Question 76

When NH3 is passed over heated metal A, its amide is formed. The metal is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 76

When Potassium is treated with ammonia, then potassium amide is obtained.

Test: BITSAT Past Year Paper- 2013 - Question 77

Insulin production and its action in human body are responsible for the level of diabetes. This compound belongs to which of the following categories ?

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 77

Insulin is a biochemically active peptide harmone secreted by pancreas.

Test: BITSAT Past Year Paper- 2013 - Question 78

Which statement is incorrect about peptide bond?

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 78

Due to resonance,

bond acquires some double bond character, hence shorter in length.

Test: BITSAT Past Year Paper- 2013 - Question 79

A mixture of chlorides of copper, cadmium, chromium, iron and aluminium was dissolved in water acidified with HCl and hydrogen sulphide gas was passed for sufficient time. It was filtered, boiled and a few drops of nitric acid were added while boiling. To this solution ammonium chloride and sodium hydroxide were added and filtered. The filterate shall give test for.

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 79

CuS and CdS are precipitated by H2S. Hydroxide of A1 will pass into the solution in the form ofNaAlO2, being amphoteric in nature. Hence filtrate will give test for sodium and aluminium.

Test: BITSAT Past Year Paper- 2013 - Question 80

Volume of 3% solution of sodium car bonate necessary to neutralise a litre of 0.1 N sulphuric acid​

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 80

Normality of 3% Na2CO3.

For H2SO4 sol. N1 = 0.1, V1 = 100 mL For Na2CO3 sol. N2 = 0.566.
Now apply  N1V1  = N2V2

Test: BITSAT Past Year Paper- 2013 - Question 81

DIRECTIONS: Out of the four alternatives, choose the one which express the correct meaning of the word.​

Q. SAGACIOUS

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 81

Sagacious means ‘judicious’, so ‘wise’ is correct answer.

Test: BITSAT Past Year Paper- 2013 - Question 82

DIRECTIONS: Out of the four alternatives, choose the one which express the correct meaning of the word.​

Q. REMEDIAL

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 82

Remedial means ‘reformative’, so ‘corrective’ is correct answer.

Test: BITSAT Past Year Paper- 2013 - Question 83

DIRECTIONS: Out of the four alternatives, choose the one which express the correct meaning of the word.

Q. RETICENT

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 83

Reticent mean s ‘quiet’ so ‘secretive’ is correct answer

Test: BITSAT Past Year Paper- 2013 - Question 84

DIRECTIONS: Choose the word opposite is meaning to the given word.

Q. FIDELITY

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 84

Fidelity means ‘faithfulness in relation s’, so ‘treachery’ is correct antonym.

Test: BITSAT Past Year Paper- 2013 - Question 85

DIRECTIONS Choose the word opposite is meaning to the given word.

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 85

Infrangible means ‘strong’, so ‘breakable’ is correct antonym.

Test: BITSAT Past Year Paper- 2013 - Question 86

DIRECTIONS: Choose the word opposite is meaning to the given word.

Q. PROGENY

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 86

Progeny means ‘child’, so ‘parent’ is correct antonym.

Test: BITSAT Past Year Paper- 2013 - Question 87

DIRECTIONS: A part of sentence is underlined. Below are given alternatives to the underlined part (a), (b), (c) and (d) which may improve the sentence. Choose the correct alternative.

Q. It was not possible to drag any conclusion so he left the case.

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 87

Use of ‘draw’ is more suitable for using before word ‘conclusion’, so option (c) is correct.

Test: BITSAT Past Year Paper- 2013 - Question 88

DIRECTIONS: A part of sentence is underlined. Below are given alternatives to the underlined part (a), (b), (c) and (d) which may improve the sentence. Choose the correct alternative.

Q. I am looking after my pen which is missing.

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 88

Use of ‘looking for’ is proper because look for means ‘to search for something’ which suits here.

Test: BITSAT Past Year Paper- 2013 - Question 89

DIRECTIONS: A part of sentence is underlined. Below are given alternatives to the underlined part (a), (b), (c) and (d) which may improve the sentence. Choose the correct alternative.

Q. “Mind  your language !” he shouted.

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 89

‘Mind your language’ is proper to use here because it gives proper sense of sentence.

Test: BITSAT Past Year Paper- 2013 - Question 90

DIRECTIONS: Fill in the blanks.

Q. I ..... to go there when I was student.

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 90

‘Use to’ is used when any h abit is to be shown, so use of option (b) is proper.

Test: BITSAT Past Year Paper- 2013 - Question 91

DIRECTIONS: Fill in the blanks.

Q. She was angry ...... me

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 91

‘Angry’ agrees with preposition ‘with’, so use of option (c) is correct here.

Test: BITSAT Past Year Paper- 2013 - Question 92

DIRECTIONS: Fill in the blanks.

Q. You should not laugh ...... the poor.

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 92

Laugh agrees with preposition ‘at’, so use of option (b) is correct here.

Test: BITSAT Past Year Paper- 2013 - Question 93

DIRECTIONS: In the questions below, each passage consists of six sentences. The first and sixth sentence are given in the beginning. The middle four sentences in each have been removed and jumbled up. These are labeled as P, Q, R and S. Find out people order for the four sentences.

Q. 1. He is a famous doctor.
P. Once I had to consult with him.
Q. I never believed him.
R. He suggested me a proper remedy.
S. I become completely fine. 6. Now I also admit this fact.

Test: BITSAT Past Year Paper- 2013 - Question 94

DIRECTIONS: In the questions below, each passage consists of six sentences. The first and sixth sentence are given in the beginning. The middle four sentences in each have been removed and jumbled up. These are labeled as P, Q, R and S. Find out people order for the four sentences.

Q. 1. We don’t know the plan of Ram.
P. He cares for his friends.
Q. He is a complete person.
R. We want some help and advice.
S. As we are in a trouble. 6. We hope he will do his best for us.

Test: BITSAT Past Year Paper- 2013 - Question 95

DIRECTIONS: In the questions below, each passage consists of six sentences. The first and sixth sentence are given in the beginning. The middle four sentences in each have been removed and jumbled up. These are labeled as P, Q, R and S. Find out people order for the four sentences.

Q. 1. It is not my problem.
P. All residents of this society are careless.
Q. I am unable to convince anyone.
R. They don’t want to do some good.
S. Every one seems to be unwise here. 6. We all have to suffer one day.

Test: BITSAT Past Year Paper- 2013 - Question 96

In a certain code language ‘DOME’ is written as ‘8943’ and ‘MEAL’ is written as ‘4321’. What group of letters can be formed for the code ‘38249’?

Test: BITSAT Past Year Paper- 2013 - Question 97

Find the missing number from the given response.

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 97

From fig. 1, 93 – (27 + 63) = 93 – 90 = 3 From fig. 2, 79 – (38 + 37) = 79 – 75 = 4 From fig. 3, 67 – (16 +42) = 67 – 58 = 9 Hence, option (d) is correct.

Test: BITSAT Past Year Paper- 2013 - Question 98

Which of the following correctly represents the relationship among illiterates, poor people and unemployed?

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 98


Hence, option (b) is correct.

Test: BITSAT Past Year Paper- 2013 - Question 99

Sushma walks 20m towards north. Then she turns right and walks 30m. Now, she turns right and walks 35m. Now turning left, she walks 15m.Again, she turns left and moves 15m. Finally, she turns left and walks 15m. In which direction and how far is she from her original position.

Test: BITSAT Past Year Paper- 2013 - Question 100

In a classroom, there are 5 rows and 5 children A, B, C, D and E are seated one behind the other in 5 separate rows as follows.

  • A is sitting behind C but in front of B.
  • C is sitting behind E and D is sitting in front of E.
  • The order in which they are sitting from the first row to the last is
Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 100

From the information given in the question the arrangement of students is

Test: BITSAT Past Year Paper- 2013 - Question 101

Which of the following will fill the series?
2, 9, 28,  ?  , 126

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 101

The given series follows the pattern
13 + 1 = 2
23 + 1 = 8 + 1 = 9
33 + 1 = 27 + 1 = 28
43 + 1 = 64 + 1 = 65
53 + 1 = 125 + 1 = 126

Test: BITSAT Past Year Paper- 2013 - Question 102

Two signs in the equations have been interchanged, find out the two signs to make equation correct.
3 ÷ 5 x 8 + 2 - 10 = 13

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 102

Interchanging symbols ÷ and + as given in option (d), we get 3 + 5 × 8 ÷ 2 – 10

Test: BITSAT Past Year Paper- 2013 - Question 103

Assertion: [A] India is a democratic country.
Reason: [R] India has a constitution of its own.Choose the correct alternative from the given options.

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 103

Both Assertion and Reason ar e cor rect but India is a democratic country because the government is elected by its citizens and not because India has its own constitution.

Test: BITSAT Past Year Paper- 2013 - Question 104

Which one of the following figures completes the original figure?

Test: BITSAT Past Year Paper- 2013 - Question 105

How many squares are there in the following figure?

Test: BITSAT Past Year Paper- 2013 - Question 106

A class has 175 students. The following data shows the number of students obtaining one or more subjects. Mathematics 100, Physics 70, Chemistry 40; Mathematics and Physics 30, Mathematics and Chemistry 28, Physics and Chemistry 23; Mathematics, Physics and Chemistry 18. How many students have offered Mathematics alone?

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 106

n (M alone)
= n(M) – n(M ∩ C) – n(M ∩ P) + n(M ∩ C ∩ P) = 100 – 28 – 30 + 18 = 60

Test: BITSAT Past Year Paper- 2013 - Question 107

If x sin3 θ + y cos3 θ = sin θ cos θ and x sin θ = y cos θ, then x2 + y2 =

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 107

x sin 3 θ + y cos 3 θ = sin θ cos θ       ....(i) and x sin θ = y cos θ                 ...(ii) Equation (i) may be written as x sin q.sin 2 θ + y cos 3 θ = sin θ cos θ
⇒ y cos θ sin 2 θ + y cos 3 θ = sin θ cos θ
⇒ y cos θ(sin 2 θ + cos 2 θ) = sin θ cos θ
⇒ y cos θ = sin q cos θ
∴ y = sin θ    ...(iii) Putting the value of y from  (iii) in (ii), we get x sin θ = sin θ. cos θ
⇒ x = cos θ   ...(iv) Squaring (iii) and (iv) and adding , we get
x2 + y2 = cos2 θ + sin2 θ = 1

Test: BITSAT Past Year Paper- 2013 - Question 108

If cos 7θ = cos q – sin 4θ, then the general value of θ is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 108

cos 7θ = cos θ - sin4θ
⇒ sin 4θ = cos θ - cos 7θ
⇒ sin 4θ = 2 sin 4θ sin 3θ
⇒ sin 4θ (1 - 2 sin 3θ)= 0
∴ sin 4θ = 0  or
 

Test: BITSAT Past Year Paper- 2013 - Question 109

If the real part of is 4, z ≠ 1, then the locus of the point representing z in the complex plane is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 109

Real part of is given by

Comparing with the equation

= 2. Thus, the locus of z given by the equation
(i) is a circle with centre 3/2 and radius = 1/2

Test: BITSAT Past Year Paper- 2013 - Question 110

If α and β are the roots of  x2 – x + 1 = 0, then the equation whose roots are α100 and β100  are

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 110

We have x3 + 1 ≡ ( x + 1) ( x 2 - x + 1) .
Therefore, α and β are the complex cube roots of –1 so that we may take a = –ω and b =–ω2, where w ≠ 1 is a cube root of unity..
Thus α100 = (- ω)100 = ω and β100 = (–ω2)100 = ω2, so that the  required equation is x2 + x + 1 = 0.

Test: BITSAT Past Year Paper- 2013 - Question 111

The set of all real x satisfying the inequality 

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 111

Given, 
⇒ 3-| x | < 0 and 4 -|x | < 0
or   3-| x | > 0 and 4 -|x | < 0
⇒ | x | > 3 and |x |>4
or   | x | < 3 and |x |<4
⇒ | x|> 4 or|x| < 3
⇒ x∈(-∞,-4) ∪[-3, 3]∪ (4, ∞)

Test: BITSAT Past Year Paper- 2013 - Question 112

If x satisfies | 3 x – 2 | + | 3x – 4 | + | 3x – 6 | > 12, then

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 112

Dividing R at analyse 4 cases.
When x the inequality becomes
2 – 3x + 4 – 3x + 6 – 3x > 12.
implying – 9x > 0 ⇒ x < 0.
when x > 2 the ineqality becomes
3x – 2 + 3x – 4 + 3x – 6 > 12,
Implying 9x > 24 Þ x > 8/3 The inequality in invalid in the other two sections.
∴ either x < 0 or x > 8/3

Test: BITSAT Past Year Paper- 2013 - Question 113

In how many ways can 5 boys and 5 girls be seated at a round table so that no two girls may be together ?

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 113

Leaving one seat vacant between two boys, 5 boys may be seated in 4! ways. Then at remaining 5 seats, 5 girls any sit in 5! ways.
Hence the required number = 4! × 5!

Test: BITSAT Past Year Paper- 2013 - Question 114

A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box if at least one black ball is to be included in the draw?

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 114

Atleast one black ball can be drawn in the following ways (i) one black and two other colour balls = 3C1 × 6C2 = 3 × 15 = 45
(ii) two black and one other colour balls = 3C2 × 6C1 = 3 × 6= 18
(iii) All the three are black = 3C3 × 6C0 = 1
∴ Req. no. of ways = 45 + 18 + 1 = 64

Test: BITSAT Past Year Paper- 2013 - Question 115

The coefficient of th e middle ter m in the expansion of (2 + 3x)4 is :

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 115

When exponent is n then total number of terms are n + 1. So, total number of terms in (2 + 3x)4 = 5
Middle term is 3rd. ⇒  T3 = 4C2(2)2 . (3x)2

Test: BITSAT Past Year Paper- 2013 - Question 116

If C0, C1, C2, .............. Cn denote the binomial coefficients in the expansion of (1 + x)n, then the value of
C0 + (C0 + C1) + (C0+ C1 + C2) + ....+ (C0+ C1 + .....+ Cn–1)

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 116

C0 + (C0 + C1) + (C0 + C1 + C2) + ................ + (C0 + C1 + ............  Cn–1) = nC0 + (n - 1)C1 + (n - 2)C2+ .......Cn -1 = C1 + 2C2 + 3C3 + 4C4 .......nCn= n.2n-1 

Test: BITSAT Past Year Paper- 2013 - Question 117

The sum of the series 1 + 2 .2 + 3 .22 + 4 .23 + .... + 100 .299 is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 117

Let S = 1 + 2. 2 + 3. 22 +4. 22 +  .... + 100 .299 ....(i) It is an arithmetico - geometric series.
On multiplying Eq. (i) by 2 and then subtracting it from Eq. (i), we get

Test: BITSAT Past Year Paper- 2013 - Question 118

The quadratic equation whose roots are the x and y intercepts of the line passing through (1, 1) and making a triangle of area A with the co-ordinate axes is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 118

Equation of the line making intercepts a and b on the axes is 
Since, it passes through (1, 1)

Also the area of the triangle formed by the line and the axes is A.

From eqs. (i) and (ii), we get, a + b = 2A Hence, a and b are the roots of the eq.
x2 -(a +b)x +ab = 0 ⇒ x2 -2Ax +2A = 0

Test: BITSAT Past Year Paper- 2013 - Question 119

If 4a2 + b2 + 2c2 + 4ab – 6ac – 3bc = 0, the family of lines ax + by + c = 0 is concurrent at one or the other of the two points-

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 119

4a2 + b2 + 2c2 + 4ab – 6ac – 3bc ≡ (2a + b)2 – 3(2a + b) c + 2c2 = 0
⇒ (2a + b – 2c) (2a + b – c) = 0 ⇒ c = 2a + b or c = 
The equation of the family of lines is
a(x + 2) + b(y + 1) = 0 or a(x + 1) + b giving the point of concurrence (–2, –1) or 

Test: BITSAT Past Year Paper- 2013 - Question 120

A pair of tangents are drawn from the origin to the circle  x2 + y2+ 20 (x + y) + 20 = 0, then the equation of the pair of tangent are

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 120

Equation of pair of tangents is given by SS1 = T2,
or S = x2 + y2 + 20 (x + y ) + 20, S1 = 20,
T = 10 (x + y) + 20 = 0
∴ SS1 = T2
⇒ 20 (x2 + y2 + 20 (x + y ) + 20) = 102 (x + y + 2)2
⇒ 4x2 + 4y2 + 10xy = 0 ⇒ 2x2 + 2y2 + 5xy = 0

Test: BITSAT Past Year Paper- 2013 - Question 121

An ellipse has OB as semi minor axis, F and F ' its focii and the angle FBF ' is a right angle.
Then the eccentricity of the ellipse is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 121

∵ ∠FBF ' = 90° ⇒ FB 2 + F'B 2= FF'2


Also, e2 = 1- b 2 / a 2 =1-e2
(By using equation (i))

Test: BITSAT Past Year Paper- 2013 - Question 122

If the lin e 2x – 3y = k touches th e parabola y2 = 6x, then find the value of  k.

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 122


⇒ y2 = 3 (3y + k)  ⇒ y2 – 9y – 3k = 0   ........(3)
If line (1) touches parabola (2) then roots of quadratic equation (3) is equal
∴ (–9)2 = 4 × 1  × (– 3k) ⇒ k = – 27/4

Test: BITSAT Past Year Paper- 2013 - Question 123

S and T are the foci of an ellipse and B is an end of the minor axis. If STB is an equilateral triangle, then the eccentricity of the ellipse is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 123

Let eq. of ellipse be 
S is (–ae, 0), T is (ae, 0)  and B is (0, b).


Also  SB2 = ST2 ⇒ 4a 2e 2 = a 2 e 2 + b 2

Test: BITSAT Past Year Paper- 2013 - Question 124

Let f(x) = (x5 – 1)(x3 + 1), g(x) = (x2 – 1)(x2 – x + 1) and let h(x) be such that f(x) = g(x)h(x). Then 

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 124

Given f (x) = g(x) h(x)

Test: BITSAT Past Year Paper- 2013 - Question 125

In the truth table for the statement (p ∧ q) ® (q ν ~ p), the last column has the truth value in the following order is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 125

T T T T

Test: BITSAT Past Year Paper- 2013 - Question 126

If the value of mode and mean is 60 and 66 respectively, then the value of median is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 126

Mode = 3 Median – 2 Mean
∴ 

Test: BITSAT Past Year Paper- 2013 - Question 127

Find the variance of the data given below

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 127


Variance 
= 1.32

Test: BITSAT Past Year Paper- 2013 - Question 128

Let R be the relation on the set R of all real numbers, defined by aRb If  |a – b| < 1. Then, R is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 128

Since, | a - a | = 0 < 1, so aRa,
∴ R is reflexive.
Now, aRb ⇒ | a - b | < 1⇒ | b - a | 1⇒ bRa
∴ R is symmetric.
But R is not transitive as
1R2, 2R3 but 1 3
[∵|1 – 3| = 2 > 1]

Test: BITSAT Past Year Paper- 2013 - Question 129

The greatest and least values of (sin–1 x)2 + (cos–1 x)2 are respectively

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 129

We have, (sin–1 x)2 + (cos–1 x)2 = (sin–1 x + cos –1 x)2 – 2 sin–1 x. cos–1 x

Thus, the least value is 2 
and the greatest value is 

Test: BITSAT Past Year Paper- 2013 - Question 130

The value of 

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 130

The given trigonometric ratio

Test: BITSAT Past Year Paper- 2013 - Question 131

The determinant vanishes for

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 131

The given determinant  vanishes, i.e.,

Expanding along C1, we get (x – 4)(x – 5)2 – (x – 5)(x – 4)2 – {(x – 3)(x –5)2 – (x – 5)(x – 3)2} + (x – 3)(x – 4)2 – (x – 4)(x – 3)2 = 0 Þ (x – 4)(x – 5)(x – 5 – x + 4) – (x – 3)(x – 5)(x – 5 – x + 3) +(x – 3)(x – 4) (x – 4 –x + 3) = 0
⇒ – (x – 4)(x – 5) + 2(x – 3)(x – 5) – (x – 3) (x – 4) = 0
⇒ – x2 + 9x – 20 + 2x2 – 16x + 30 – x2 + 7x – 12 = 0
⇒ – 32 + 30 = 0 Þ  –2 = 0
Which is not possible, hence no value of x satisfies the given condition.

Test: BITSAT Past Year Paper- 2013 - Question 132

If the lines lx + my + n = 0, mx +ny + l = 0 and nx + ly + m = 0 are concurrent then

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 132

Since the lines are concurrent, so

⇒ (l + m + n) (l2 + m2 + n2 – lm – mn – nl) = 0
⇒ l + m + n = 0 [∵ l2 + m2  + n2 > lm + mn + n]

Test: BITSAT Past Year Paper- 2013 - Question 133

If y =1 + x + +....∞, then is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 133

Test: BITSAT Past Year Paper- 2013 - Question 134


is continuous at x = –5, then the value of ‘a’ will be

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 134

Test: BITSAT Past Year Paper- 2013 - Question 135

The equation of all lines having slope 2 which are tangent to the curve 

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 135

The equation of the given curve is

The slope of the tangent to the given curve at any point (x, y) is given by For tangent having slope 2, we must have 

which is not possible as square of a real number cannot be negative.
Hence, there is no tangent to the given curve having slope 2.

Test: BITSAT Past Year Paper- 2013 - Question 136

The function f (x) = (x(x–2))2 is increasing in the set

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 136

Here, f(x) = (x(x – 2))2
⇒  f '(x) = 4x (x – 2) (x – 1) For f(x) as increasing, f '(x) > 0 So, 4x (x – 1) (x – 2) > 0 ⇒  x(x – 1) (x – 2) > 0

From the above figure required interval is, (0, 1) ∪ (2, ∞)

Test: BITSAT Past Year Paper- 2013 - Question 137

If a2 x4 + b2 y4 = c4, then the maximum value of xy is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 137

If the sum of two positive quantities is a constant, then their product is maximum, when they are equal. ∴
a2 x4. b2y2 is maximum when

∴ maximum value of

Maximum value of xy = 

Test: BITSAT Past Year Paper- 2013 - Question 138

is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 138

Test: BITSAT Past Year Paper- 2013 - Question 139

Evaluate 

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 139

Let 
Let cos x = t and – sin x dx = dt.
Now, x = 0 ⇒ t = cos 0 = 1 and

Test: BITSAT Past Year Paper- 2013 - Question 140

Area intercepted by the curves y = cos x, x ∈ [0, π] and y = cos 2x, x ∈ [0, π], is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 140


Test: BITSAT Past Year Paper- 2013 - Question 141

 The general solution of the differential equation

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 141

The equation is,

Test: BITSAT Past Year Paper- 2013 - Question 142

The solution to the differential equation
where f (x) is a given function is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 142


Test: BITSAT Past Year Paper- 2013 - Question 143

are three unit vectors such that where is null vector, then 

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 143


Test: BITSAT Past Year Paper- 2013 - Question 144

If vectors 2i – j + k, i + 2j – 3k and 3i + aj + 5k are coplanar, then the value of a is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 144

If given vectors are coplanar, then there exists two scalar quantities x and y such that

Comparing coefficient of and  on both sides of (1) we get x + 3y = 2 , 2x + ay = –1 , –3x + 5y = 1 ...(2)
Solving first and third equations, we get x = 1/2, y = 1/2 Since the vectors are coplanar, therefore these values of x and y will satisfy the equation   2x + ay = –1
∴ 2 (1/2) + a (1/2) = – 1 ⇒ a = –4

Test: BITSAT Past Year Paper- 2013 - Question 145

The coordinates of the poin t wh ere the lin e through the points A (3, 4, 1) and B (5, 1, 6) crosses the XY-plane are

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 145

Equation of the line through the given points is  

Any point on this line can be taken as (3 + 2λ, 4 – 3λ, 1 + 5λ) If this point lies on XY-plane then the z-coordinate is zero

Thus the required coordinates of the point are

Test: BITSAT Past Year Paper- 2013 - Question 146

Find the angle between the two planes 2x + y – 2z = 5 and 3x – 6y – 2z = 7.

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 146

The angle between two planes is the angle between their normals. From the equation of the planes, the normal vectors are 
Therefore, 

Hence, 

Test: BITSAT Past Year Paper- 2013 - Question 147

For k = 1, 2, 3 the box Bk contains k red balls and (k + 1) white balls. Let and A box is selected at random and a ball is drawn from it. If a red ball is drawn, then the probability that it has come from box B2, is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 147

In a box, B1 = 1R, 2W; B2 = 2R, 3W and B3 = 3R, 4W
Also, given that, and 

Test: BITSAT Past Year Paper- 2013 - Question 148

The probability of India winning a test match against West Indies is 1/2.  Assuming independence from match to match, the probability that in a 5 match series India’s second win occurs at the third test, is –

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 148

The sample space is [LWW, WLW]
∴   P (LWW) + P (WLW)
= Probability that in 5 match series, it is India’s second win = P (L) P (W) P (W) + P (W) P (L) P (W)

Test: BITSAT Past Year Paper- 2013 - Question 149

An object is obseved from the points A, B and C lying in a horizontal straight line which passes directly underneath the object. The angular elevation at B is twice that at A and at C three times that at A. It AB = a, BC = b, then the height of the object is

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 149


Let ED = h, ∠EAB = a ∠EBD = 2α, ∠ECD = 2α Now, ∠DBE = ∠EAB + ∠BEA
⇒ 2α = α + ∠BEA
⇒∠BEA = α = ÐEAB
⇒ AB = EB = α Similarly, ∠EBC = α

Test: BITSAT Past Year Paper- 2013 - Question 150

A shopkeeper wants to purchase two articles A and B of cost price ` 4 and ` 3 respectively. He thought that he may earn 30 paise by selling article A and 10 paise by selling article B. He has not to purchase total articles worth more than ` 24. If he purchases the number of articles of A and B, x and y respectively, then linear constraints are

Detailed Solution for Test: BITSAT Past Year Paper- 2013 - Question 150

x, y > 0 and 4x + 3y < 24

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