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Test: BITSAT Past Year Paper- 2014 - Question 1

### A rifle man, who together with his rifle has a mass of 100 kg, stands on a smooth surface and fires 10 shots horizontally. Each bullet has a mass 10 g and a muzzle velocity of  800 ms–1. The velocity which the rifle man attains after  firing 10 shots is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 1

According to law of conservation of
momentum,
100v = -1/1000 x 10 x 800
i.e., v = 0.8 ms–1.

Test: BITSAT Past Year Paper- 2014 - Question 2

### A train accelerating uniformly from rest attains a maximum speed of 40 ms–1 in 20 s. It travels at the speed for 20 s and is brought to rest with uniform retardation in further 40 s. What is the average velocity during the period ?

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 2

(i) v = u + at1
40 = 0 + a × 20
a = 2 m/s2
v2 – u2 = 2as
402 – 0 = 2 × 2 s1
s1 = 400 m
(ii) s2 = v × t2 = 40 × 20 = 800 m
(iii) v = u + at
0 = 40 + a × 40
a = –1 m/s2
02 – 402 = 2(–1)s3
s3 = 800 m
Total distance travelled = s1 + s2 + s3
= 400 + 800 + 800 = 2000 m
Total time taken = 20 + 20 + 40 = 80 s
Average velocity 2000/80 = 25m/s

Test: BITSAT Past Year Paper- 2014 - Question 3

### A projectile is fired with a velocity u making an angle q with the horizontal. What is the magnitude of change in velocity when it is at the highest point –

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 3

Initially u = cosθ iˆ + usinθ ˆj .
At highest point v = u cosθ iˆ
∴ difference is u sin q.

Test: BITSAT Past Year Paper- 2014 - Question 4

For the equation F = Aavbdc, where F is the force, A is the area, v is the velocity and d is the density, the values of a, b and c are respectively

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 4

[MLT–2] = [L2a] × [LbT–b][McL–3c]
= [McL2a + b –3cT–b]
Comparing powers of M, L and T, on both
sides, we get
c = 1, 2a + b –3c = 1, –b = –2 or b = 2
Also, 2a + 2 – 3(1) =1 Þ 2a = 2 or a = 1
∴ This is 1, 2, 1

Test: BITSAT Past Year Paper- 2014 - Question 5

A person with his hand in his pocket is skating on ice at the rate of 10m/s and describes a circle of radius 50 m. What is his inclination to vertical: (g = 10 m/sec2)

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 5

Since surface (ice) is frictionless, so the
centripetal force required for skating will be
provided by inclination of boy with the
vertical and that angle is given as
tan θ = v2/rg
= where v is speed of skating &
r is radius of circle in which he moves.

Test: BITSAT Past Year Paper- 2014 - Question 6

A small block of mass m is kept on a rough inclined surface of inclination q fixed in a elevator. The elevator goes up with a uniform velocity v and the block does not slide on the wedge. The work done by the force of friction on the block in time t will be :

Test: BITSAT Past Year Paper- 2014 - Question 7

An equilateral prism of mass m rests on a rough horizontal surface with coefficient of friction µ.

A horizontal force F is applied on the prism as shown in the figure.
If the coefficient of  friction is sufficiently high so that the prism does not slide before toppling, then the minimum force required to topple the prism is –

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 7

The tendency of rotation will be about the point C.

For minimum force, the torque of F about C has to be equal to the torque of mg about C.
∴

Test: BITSAT Past Year Paper- 2014 - Question 8

A spherically symmetric gravitational system of particles has a mass density  where r0 is a constant. A test mass can undergo circular motion under the influence of the gravitational field of particles. Its speed V as a function of distance r (0 < r < ∞) from the centre of the system is represented by

Test: BITSAT Past Year Paper- 2014 - Question 9

The load versus elongation graph for four wires is shown. The thinnest wire is

Test: BITSAT Past Year Paper- 2014 - Question 10

Th e wor k done in blowin g a soap bubble of surface tension 0.06 × Nm–1 from 2 cm radius to 5 cm radius is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 10

= 0.003168 J

Test: BITSAT Past Year Paper- 2014 - Question 11

The wavelength of radiation emitted by a body depends upon

Test: BITSAT Past Year Paper- 2014 - Question 12

One mole of O2 gas having a volume equal to 22.4 Litres at 0°C and 1 atmospheric pressure in compressed isothermally so that its volume reduces to 11.2 litres. The work done in this process is

Test: BITSAT Past Year Paper- 2014 - Question 13

In a thermodynamic process, the pressure of a fixed mass of a gas is changed in such a manner that the gas releases 20 J of heat and 8 J of work is done on the gas. If the initial internal energy of the gas was 30 J, then the final internal energy will be

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 13

Given ΔQ = – 20 J, W = – 8 J
Using Ist law ΔQ = ΔU + ΔW
⇒ ΔQ = – 20 + 8 = – 12 J
Uf = – 12 + 30 = 18 J

Test: BITSAT Past Year Paper- 2014 - Question 14

In the kinetic theory of gases, which of these statements is/are true ?
(i) The pressure of a gas is proportional to the mean speed of the molecules.
(ii) The root mean square speed of the molecules is proportional to the pressure.
(iii) The rate of diffusion is proportional to the mean speed of the molecules.
(iv) The mean translational kinetic energy of a gas is proportional to its kelvin temperature.

Test: BITSAT Past Year Paper- 2014 - Question 15

Two balloons are filled one with pure he gas and other with air respectively. If the pressure and temperature of these balloons are same, then the number of molecules per unit volume is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 15

Assumingthe balloons have the same volume, as pV= nRT. If P, V and T are the same, n the number of moles present will be the same, whether it is He or air.
Hence, number of molecules per unit volume will be same in both the balloons.

Test: BITSAT Past Year Paper- 2014 - Question 16

Two particles P and Q describe S.H.M. of same amplitude a, same frequency f along the same straight line. The maximum distance between the two particles is a√2 . The initial phase difference between the particle is –

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 16

x1 = a sin(ωt + ϕ1), x2 = a sin(ωt + ϕ2 )

⇒

To maximize

⇒

⇒

⇒

Test: BITSAT Past Year Paper- 2014 - Question 17

A tunnel has been dug through the centre of the earth and a ball is released in it. It executes S.H.M. with time period

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 17

=  84.6 min

Test: BITSAT Past Year Paper- 2014 - Question 18

A soun d source, em itting sound of con stan t frequency, moves with a constant speed and crosses a stationary observer. The frequency (n) of sound heard by the observer  is plotted against time (t). Which of the following graphs represents the correct variation ?

Test: BITSAT Past Year Paper- 2014 - Question 19

When a string is divided into three segments of length l1, l2, and l3 the fundamental frequencies of these three segments are v1, v2 and v3 respectively. The original fundamental frequency (v) of the string is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 19

Fundamental frequency is given by

Since, P divided into l, land l3 segments
Herel = l1 + l2 + l3
So

Test: BITSAT Past Year Paper- 2014 - Question 20

Two point dipoles pkˆ and p/2 kˆ are located at(0, 0, 0) and (1m, 0, 2m) respectively. The resultant electric field due to the two dipoles at the point (1m, 0, 0) is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 20

The given point is at axis of
dipole and at equatorial line of dipole so that field at given point.

Test: BITSAT Past Year Paper- 2014 - Question 21

Electric field in the region is given by  then the correct expression for the potential in the region is [assume potential at infinity is zero]

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 21

Test: BITSAT Past Year Paper- 2014 - Question 22

Three capacitors C1 = 1 µF, C2 = 2 µF and C3 = 3 µF are connected as shown in figure, then the equivalent capacitance between points A and B is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 22

Ceq = (1 + 2 + 3)μF = 6 μF

Test: BITSAT Past Year Paper- 2014 - Question 23

Two long coaxial and conducting cylinders of radius a and b are separated by a material of conductivity s and a constant potential difference V is maintained between them, by a battery. Then the current, per unit length of the cylinder flowing from one cylinder to the other is –

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 23

Now

From (1) :

Test: BITSAT Past Year Paper- 2014 - Question 24

A wire X is half the diameter and half the length of a wire Y of similar material. The ratio of resistance of X to that of Y is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 24

Test: BITSAT Past Year Paper- 2014 - Question 25

A narrow beam of protons and deuterons, each having the same momentum, enters a region of uniform magnetic field directed perpendicular to their direction of momentum. The ratio of the radii of the circular paths described by them is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 25

Since, the radius of circular path of a charged particle in magnetic field is
Now, the radius of circular path of charged particle of given momentum r and magnetic field B is given by

Test: BITSAT Past Year Paper- 2014 - Question 26

For th e cir cuit (figur e), the cur rent is to be measured. The ammeter shown is a galvanometer with a resistance RG = 60.00W converted to an ammeter by a shunt resistance rs = 0.02W. The value of the current is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 26

RG = 60.00W, shunt resistance, rs = 0.02W
Total resistance in the circuit is RG + 3 = 63W
Hence, I = 3/63 = 0.048 A
Resistance of the galvanometer converted to an ammeter is,
Total resistance in the circuit = 0.02 + 3
= 3.02W
Hence, I = 3/3.02 = 0.99 A

Test: BITSAT Past Year Paper- 2014 - Question 27

The susceptibility of a magnetism at 300 K is 1.2 × 10–5. The temperature at which the susceptibility increases to 1.8 × 10–5 is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 27

⇒
⇒

Test: BITSAT Past Year Paper- 2014 - Question 28

A coil 10 turns and a resistance of 20W is connected in series with B.G. of resistance 30W. The coil is placed with its plane perpendicular to the direction of a uniform magnetic field of induction 10–2 T. If it is now turned through an angle of 60° about an axis in its plane. Find the charge induced in the coil. (Area of a coil = 10–2 m²)

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 28

Given : n = 10 turns, Rcoil = 20W, RG = 30W, Total resistance in the circuit = 20 + 30 = 50W.

= 1 × 10–5 C (Charge induced in a coil)

Test: BITSAT Past Year Paper- 2014 - Question 29

Voltage V and current i in AC circuit are given by V = 50 sin (50 t) volt, i = 50 sin. The power dissipated in the circuit is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 29

Given V = 50 sin (50 t) V
Maximum voltage, V0 = 50 V,

Maximum current, i0 = 50 mA = 50 × 10–3 A
Power dissipated,

Test: BITSAT Past Year Paper- 2014 - Question 30

Resolving power of the telescope will be more, if the diameter of the objective is

Test: BITSAT Past Year Paper- 2014 - Question 31

The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and the eye piece is found to be 20 cm. The focal length of lenses are

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 31

∴ f0 = 9fe
Also f0 + fe = 20 (Qfinal image is at infinity)
9 fe + fe = 20, fe = 2 cm, ∴ f0 = 18 cm

Test: BITSAT Past Year Paper- 2014 - Question 32

The angular size of the central maxima due to a single slit diffraction is (a ⇒ slit width)

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 32

Angular size of central maxima is

Test: BITSAT Past Year Paper- 2014 - Question 33

Find the final intensity of light (I"), if the angle between the axes of two polaroids is 60°.

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 33

From the first polaroid

From second polaroid

Test: BITSAT Past Year Paper- 2014 - Question 34

The threshold wavelength of the tungsten is 2300 Å. If ultraviolet light of wavelength 1800 Å is incident on it, then the maximum kinetic energy of photoelectrons would be about –

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 34

Test: BITSAT Past Year Paper- 2014 - Question 35

Gr aph betwen stopping potential for most energetic emitted photoelectrons (Vs) with frequency (u) of incident radiation on metal is given below. Value of AB/BC, in graph is [where h = plank’s constant, e = electronic charge]

Test: BITSAT Past Year Paper- 2014 - Question 36

If hydrogen atom, an electron jumps from bigger orbit to smaller orbit so that radius of smaller orbit is one-fourth of radius of bigger orbit. If speed of electron in bigger orbit was v, then speed in smaller orbit is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 36

Radius of the orbit , rn ∝ n2

⇒

⇒

Velocity of electron in nth orbit

⇒ vn small = 2(vn big) = 2v

Test: BITSAT Past Year Paper- 2014 - Question 37

A nucleus of uranium decays at rest into nuclei of thorium and helium. Then :

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 37

In an explosion a body breaks up into two pieces of unequal masses both part will have numerically equal momentum and lighter part will have more velocity.
U → Th + He

sinc mHe is less so KEHe will be more.

Test: BITSAT Past Year Paper- 2014 - Question 38

Let binding energy per nucleon of nucleus is denoted by Ebn and radius of nucleus is denoted as r. If mass number of nuclei A, B and 64 and 125 respectively then

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 38

r increases with increasing A mass number So, rA < rB as mass number of A is smaller Ebn decreases with increasing A for A > 56, 56Fe has highest Ebn value.
So, Ebn for A = 64 is larger as compared to Ebn for nucleus with A = 125

Test: BITSAT Past Year Paper- 2014 - Question 39

For a CE transistor amplifier, the audio signal voltage across the collector resistance of 2.0 kW is 2.0 V. Suppose the current amplification factor of the transistor is 100, What should be the value of RB in series with VBB supply of 2.0V if the dc base current has to be 10 times the signal current?

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 39

The output ac voltage is 2.0 V. So, the ac collector current iC = 2.0/2000 = 1.0 mA.
The signal current through the base is,
therefore given by
iB = iC /β = 1.0 mA/100 = 0.010 mA.
The dc base current has to be 10 × 0.010
= 0.10 mA.
RB = (VBB – VBE ) /IB.
Assuming VBE = 0.6 V, RB = (2.0 . 0.6 )/0.10
= 14 kW.

Test: BITSAT Past Year Paper- 2014 - Question 40

The combination of gates shown below yields

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 40

The final boolean expression is,

Test: BITSAT Past Year Paper- 2014 - Question 41

The formation of CO and CO2 illustrates the law of

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 41

Formation of CO and CO2 illustrates the law of multiple proportion that is constant mass of C reacts with different masses of oxygen. These masses here bears simple ratio of 1 : 2.

Test: BITSAT Past Year Paper- 2014 - Question 42

The wave number of the limiting line in Lyman series of hydrogen is 109678 cm–1. The wave number of the limiting line in Balmer series of He+ would be :

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 42

RH = 109678 cm–1
Wave number of the limiting line in
Balmer series of He+

= 109678 cm–1

Test: BITSAT Past Year Paper- 2014 - Question 43

Th e va len cy sh el l of elemen t A con ta in s 3 electrons while the valency shell of element B contains 6 electrons. If A combines with B, the probable formula of the compound formed will be

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 43

The element A is ns2p1 and B is ns2p4. They can form compound of the type A2B3.

Test: BITSAT Past Year Paper- 2014 - Question 44

The enthalpy of sublimation of aluminium is 330 kJ/mol. Its Ist, IInd and IIIrd ionization enthalpies are 580, 1820 and 2740 kJ respectively. How much heat has too be supplied (in kJ) to convert 13.5 g of aluminium into Al3+ ions and electrons at 298 k

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 44

Heat needed too be supplied
per mol = 330 + 580 + 1820 + 2740
= 5470 kJ
Heat required = 0.5 × 5470 kJ = 2735 kJ

Test: BITSAT Past Year Paper- 2014 - Question 45

Which one of the following pairs is isostructural (i.e., having the same shape and hybridization)?

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 45

BF4- hybridisation sp3, tetrahedral
structure.
NH4+ hybridisation sp3, tetrahedral
structure.

Test: BITSAT Past Year Paper- 2014 - Question 46

N2 and O2 are converted into mono anions, N2 and O2 respectively. Which of the following statements is wrong ?

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 46

We know that in O2 bond, the order is 2
and in O2 bond, the order is 1.5. Therefore the wrong statements is (b).

Test: BITSAT Past Year Paper- 2014 - Question 47

If the enthalpy of vaporization of water is 186.5 kJmol–1, the entropy if its vaporization will be :

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 47

Given enthalpy of vaporization,
ΔH = 186.5kJmol-1
Boiling point of water
= 100°C =100 + 273 = 373K
Entropy change,

= 0.5 k JK–1mol–1

Test: BITSAT Past Year Paper- 2014 - Question 48

Th e heats of neutra lisation of CH3CO OH, HCOOH, HCN and H2S are – 13.2, – 13.4, – 2.9 and – 3.8 kCal per equivalent respectively. Arrange the acids in increasing order of acidic strength.

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 48

The greater the (negative value) of heat of neutralisation, the more is the strength of  the acid. Hence,
HCOOH > CH3COOH > H2S > HCN

Test: BITSAT Past Year Paper- 2014 - Question 49

Kc for the the reaction, [Ag(CN)2] Ag+ + 2CN, the equillibrium constant at 25°C is 4.0 × 10–19, then the silver ion concentration in a solution which was originally 0.1 molar in KCN and 0.03 molar in AgNO3 is :

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 49

0.04 ≈0.04

∴

a = 7.5 × 10–18

Test: BITSAT Past Year Paper- 2014 - Question 50

The ratio of oxidation states of Cl in potassium chloride to that in potassium chlorate is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 50

KCl x = –1 KClO3

potassium chloride +1 + x – 6 = 0
x = + 5 potassium chlorate.
∴ Ratio of oxidation state of Cl = -1/5

Test: BITSAT Past Year Paper- 2014 - Question 51

Which of the following among alkali metal is most reactive ?

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 51

Amongst alkali metals, Cs is most reactive because of its lowest IE.

Test: BITSAT Past Year Paper- 2014 - Question 52

Which of the following compounds has wrong IUPAC name?

Test: BITSAT Past Year Paper- 2014 - Question 53

The compound which gives th e most stable carbonium ion on dehydration is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 53

Test: BITSAT Past Year Paper- 2014 - Question 54

The correct order of increasing C - O bond length CO, CO32- , CO2 is:

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 54

bond order. O = C = O has second lowest bond length due to double bond.

has highest has bond length due to lowest bond order which is due to resonance.

Test: BITSAT Past Year Paper- 2014 - Question 55

An organic compound A (C4H9Cl) on reaction with Na/diethyl ether gives a hydrocarbon which on monochlorination gives only one chloro derivative, then  A is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 55

Test: BITSAT Past Year Paper- 2014 - Question 56

When rain is accompanied by a t hunderstorm, the collected rain water will have a pH value:

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 56

Normal rain water has pH 5.6. Thunderstorm results in the formation NO and HNO3 which lowers the pH.

Test: BITSAT Past Year Paper- 2014 - Question 57

An elemental crystal has a density of 8570 kg/m3. The packing efficiency is 0.68. The closest distance of approach between neighbouring atom is 2.86 Å. What is the mass of one atom approximately?

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 57

The packing efficiency = 0.68, means the given lattice is BCC.
The closest distance of approach = 2r

Let atomic weight of the element = a
∴

a = 8.57 × 3 × (3.3)3 × 0.1
= 92.39 ≌ 93 amu

Test: BITSAT Past Year Paper- 2014 - Question 58

Identify the correct or der of solubilty of Na2S. CuS and ZnS in aqueous medium

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 58

The correct order of solublity of sulphides is Na2S > ZnS > CuS

Test: BITSAT Past Year Paper- 2014 - Question 59

In the cell reaction

E0cell = 0.46 V. By doubling the concentration of Cu2+, E0cell is

Test: BITSAT Past Year Paper- 2014 - Question 60

Cu+aq is un stable in solution a n d un der goes simultaneous oxidation and reduction according to the reaction :

choose correct Eº for above reaction if

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 60

Now

Again

= 0.53 – 0.15 = 0.38 V.

Test: BITSAT Past Year Paper- 2014 - Question 61

The reduction of peroxydisulphate ion by Iion is expressed by . If rate of disappearance of I is 9/2 × 10–3 mol lit–1 s–1, what is the rate of formation of 2SO42- during same time?

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 61

∴

= 3 × 10–3 mol Lit–1 s–1

Test: BITSAT Past Year Paper- 2014 - Question 62

A gaseous reaction
There is increase in pressure from 100 mm to 120 mm in 5 minutes. The rate of disappearance of X2 is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 62

The increase in pressure shows the increase in conc. of Z. Rate of appearance of
Z = 120-100/5 = 4 mm min–1
Rate of disappearance of X2 = 2 × rate of appearance of Z
= 2 × 4 mm min–1 = 8 mm min–1

Test: BITSAT Past Year Paper- 2014 - Question 63

Two substances R and S decompose in solution independently, both following first order kinetics.
The rate constant of R is twice that of S. In an experiment, the solution initially contained 0.5 millimoles of R and 0.25  of S. The molarities of R and S will be equal just at the end of time equal to

Test: BITSAT Past Year Paper- 2014 - Question 64

The isoelectric-point of a colloidially dispersed material is the pH value at which

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 64

At isoelectric point there is no migration of dispersed phase in an electric field.

Test: BITSAT Past Year Paper- 2014 - Question 65

Which of the following halogens exhibit only one oxidation state in its compounds ?

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 65

Fluorine, since it is the most electronegative element.

Test: BITSAT Past Year Paper- 2014 - Question 66

Star ch can be used as an indicator for the detection of traces of

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 66

I2 gives blue colour with starch.

Test: BITSAT Past Year Paper- 2014 - Question 67

Which one of t h e foll owing arr a n g em ents represents the correct order of electron gain enthalpy (with negative sign) of the given atomic species?

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 67

The amount of energy released when an electron is added to an isolated gaseous atom to produce a monovalent anion is called electron gain enthalpy.
Electron affinity value generally increase on moving from left to right in a period however there are exceptions of this rule in the case of those atoms which have stable configuration. These atoms resist the addition of extra electron, therefore the low value of electron affinity

On the other hand Cl, because of its
comparatively bigger size than F, allow the addition of an extra electron more easily.

Test: BITSAT Past Year Paper- 2014 - Question 68

Which form coloured salts :

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 68

Most of the transition metal compounds (ionic as well as covalent) are coloured both in the solid state and in aqueous so lution in contrast to the compounds of s andp- block elements due to the presence of incomplete d-subshell.

Test: BITSAT Past Year Paper- 2014 - Question 69

The correct order of magnetic moments (spin only values in B.M.) is:

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 69

Number of unpaired electrons in central
atom

The greater the number of unpaired electrons, the higher the value of magnetic
moment

Test: BITSAT Past Year Paper- 2014 - Question 70

The number of double bonds in gammexane is :

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 70

Gammexane is C6H6Cl6 or (6, 6, 6). It is a saturated compound so no double bond is there in it.

Test: BITSAT Past Year Paper- 2014 - Question 71

P and Q are isomers. Identify Q.

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 71

Test: BITSAT Past Year Paper- 2014 - Question 72

Consider the following phenols :

The decreasing order of acidity of the above phenols is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 72

Electron withdrawing group (–NO2)
increases the acidity while electron
releasing group (–CH3, –H) decreases
acidity.
Also effect will be more iffunctional group is present at para position than ortho
and meta position.

Test: BITSAT Past Year Paper- 2014 - Question 73

The ionization constant of phenol is higher than that of ethanol because :

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 73

The acidic nature of phenol is due to the
formation of stable phenoxide ion in
solution

The phenoxide ion is stable due to
resonance.

The negative charge is delocalized in the benzene ring which is a stabilizing  factor in the phenoxide ion and increase acidity of phenol. wheras no resonance is possible in alkoxide ions(RO) derived from alcohol. The negative charge is localized on oxygen atom. Thus, alcohols are not acidic.

Test: BITSAT Past Year Paper- 2014 - Question 74

The reaction,

is known as:

Test: BITSAT Past Year Paper- 2014 - Question 75

Aniline reacts with ph osgene and KOH to form

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 75

Test: BITSAT Past Year Paper- 2014 - Question 76

Which one of the following monomers gives the polymer neoprene on polymerization ?

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 76

Test: BITSAT Past Year Paper- 2014 - Question 77

Which of the following can possibly be used as analgesic without causing addiction and modification?

Test: BITSAT Past Year Paper- 2014 - Question 78

Which among the following is not an antibiotic?

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 78

Oxytocin is a hormone (nanopetide) which contracts uterus after the child birth and produces lactation in the mammary glands.

Test: BITSAT Past Year Paper- 2014 - Question 79

Which of the following ions can be separated by aq. NH4OH in presence of NH4Cl

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 79

Cu2+ is of group II and Al3+ is of group III of cation analysis.

Test: BITSAT Past Year Paper- 2014 - Question 80

3.92 g of fer rous ammon ium sulph ate react completely with 50 ml N/10 KMnO4 solution.  Thepercentage purity of the sample is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 80

Eq of KMnO4 used =

∴ Eq of FAS reacted = 0.005
∴ weight of FAS needed
= 0.005 × 392 = 1.96 g
Thus percentage purity of FAS is 50%

Test: BITSAT Past Year Paper- 2014 - Question 81

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

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 81

Augment means make greater, so increase is the correct option.

Test: BITSAT Past Year Paper- 2014 - Question 82

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

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 82

Consolation means ‘comfort received by a person after a loss or disappointment’, so comfort is correct option.

Test: BITSAT Past Year Paper- 2014 - Question 83

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

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 83

Auxiliary means ‘providing additional help’, so supplemental is correct option.

Test: BITSAT Past Year Paper- 2014 - Question 84

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

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 84

Auspicious means ‘favourable’, so ‘unfavourable’ is best opposite word for it.

Test: BITSAT Past Year Paper- 2014 - Question 85

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

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 85

Recompense means ‘reward given for loss, so ‘penalty’ is the correct opposite word for it.

Test: BITSAT Past Year Paper- 2014 - Question 86

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

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 86

Impede means ‘hinder’ or ‘obstruct’, so
‘push’ is correct opposite word for it.

Test: BITSAT Past Year Paper- 2014 - Question 87

DIRECTIONS: A part of sentence is underlined. Belence are given alternatives to the underlined part a, b, c and d which may improve the sentence. Choose the correct alternative.
Q. They requested me to follow them.

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 87

Here a sense of command is depicted insentence, sowe should use ‘ordered’ for proper meaning of sentence.

Test: BITSAT Past Year Paper- 2014 - Question 88

DIRECTIONS: A part of sentence is underlined. Belence are given alternatives to the underlined part a, b, c and d which may improve the sentence. Choose the correct alternative.
Q. She did not believed me.

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 88

Sentence is in past tense and V1 is used in those sentence which contain ‘did’, so option (c) is correct.

Test: BITSAT Past Year Paper- 2014 - Question 89

DIRECTIONS: A part of sentence is underlined. Belence are given alternatives to the underlined part a, b, c and d which may improve the sentence. Choose the correct alternative.
Q. I am fine, what about you?

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 89

No improvement is needed as sentence is right.

Test: BITSAT Past Year Paper- 2014 - Question 90

DIRECTIONS: Fill in the blanks.
Q. They were afraid ........... the lion, so they dropped the idea of hunting in jungle.

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 90

Afraid agrees with preposition ‘of’,so option (d) is correct.

Test: BITSAT Past Year Paper- 2014 - Question 91

DIRECTIONS: Fill in the blanks.
Q. Our company signed a profitable ...... last month.

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 91

Normally, company signs a contract or deal, so use of ‘deal’ is proper here.

Test: BITSAT Past Year Paper- 2014 - Question 92

DIRECTIONS: Fill in the blanks.
Q. What is your ......... for tonight?

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 92

The question gives a sense of query about normal routine of some special/specific day, so use of ‘plan’ is more proper  here.

Test: BITSAT Past Year Paper- 2014 - Question 93

DIRECTIONS: Arrange the following sentences in correct pattern and mark at the correct combination.
1. Today we live in modern technology era.
P. We have a log of problems now.
Q. We want to get everything in one day.
R. Ancient time was quite pleasant.
S. We has no problems then. 6. Perhaps greed is the main cause for this.

Test: BITSAT Past Year Paper- 2014 - Question 94

DIRECTIONS: Arrange the following sentences in correct pattern and mark at the correct combination.
1. He is a common man.
P. Yesterday our city saw a brutal crime.
Q. Police is trying to arrest innocent persons.
R. The criminals are well known.
S. Police as well as whole system in corrupt. 6. Police will arrest him as he is an easy target because of being a common man.

Test: BITSAT Past Year Paper- 2014 - Question 95

DIRECTIONS: Arrange the following sentences in correct pattern and mark at the correct combination.
1. I want to change the room.
P. Last month I got a job.
Q. I had been living there for six months.
R. The office is far from the room.
S. I want to cut expenses of travelling. 6. Hopefully I will do this next week.

Test: BITSAT Past Year Paper- 2014 - Question 96

In a certain code language, ‘SAFER’ is written as ‘5@3#2’ and ‘RIDE’ is written as ‘2©%#’, how would ‘FEDS’ be written in that code?

Test: BITSAT Past Year Paper- 2014 - Question 97

Find the missing number from the given response

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 97

From the given responses,
4 × 2 × 3 × 3 = 72
9 × 4 × 2 × 10 = 720
Similarly, 6 × 20 × 1 × 6 = 720

Test: BITSAT Past Year Paper- 2014 - Question 98

If the fir st an d second letters in the word DEPRESSION were interchanged, also the third and fourth letters, the fifth and the sixth letters and so on, then which of the following would be seventh letter from the right.

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 98

Since, consecutive two letters are interchanged. Therefore,

7th from Right.

Test: BITSAT Past Year Paper- 2014 - Question 99

Today is Thrusday. The day after 59 days will be

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 99

Every day of week repeats after seven days.
Hence, 59 = 7 × 8 + 3 = 56 + 3
It will be Thursday after 56 days.
∴ 57th day = Thursday ⇒ 58th day = Friday
59th day = Saturday ⇒ 60th day = Sunday
∴ It will be Sunday after 59 days.

Test: BITSAT Past Year Paper- 2014 - Question 100

Which of the following represents coal mines, factories and fields?

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 100

Both coal mines and factories are located in the fields.

Test: BITSAT Past Year Paper- 2014 - Question 101

Find out the missing term in the series. 1, 8, 27,  ?  , 125, 216

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 101

From the given series,

Test: BITSAT Past Year Paper- 2014 - Question 102

If ‘+’ means ‘×’, ‘–’ means ‘+’, ‘×’ means ‘÷’ and ‘÷’ means ‘–’, then 6 – 9 + 8 × 3 ÷ 20 = ?

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 102

Interchanging the symbols as given in the above question, the above equation
becomes.

= 6 + 24 – 20 = 10

Test: BITSAT Past Year Paper- 2014 - Question 103

Here are some words translated from an artificial language.
mallon piml means blue light
mallon tifl means blue berry
arpan tifl means rasp berry
Which word could means ‘light house’?

Test: BITSAT Past Year Paper- 2014 - Question 104

What is the water image of below figure?

Test: BITSAT Past Year Paper- 2014 - Question 105

A piece of paper is folded and penched as shown in the figure below

How will it appear when unfolded?

Test: BITSAT Past Year Paper- 2014 - Question 106

The set (A \ B) ∪ (B \ A) is equal to]

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 106

Given set can be written as
(A – B) ∪ (B – A) = (A ∪ B) – (A ∩ B)

(By definition of symmetric difference)
Hence, (A \ B) ∪ (B \ A) = (A ∪ B) \ (A ∩ B)

Test: BITSAT Past Year Paper- 2014 - Question 107

The domain of the function

is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 107

f (x) is defined if

⇒

⇒

⇒

⇒ 0 < x < 1

Test: BITSAT Past Year Paper- 2014 - Question 108

=

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 108

Test: BITSAT Past Year Paper- 2014 - Question 109

The solution of (2 cos x – 1) (3 + 2 cos x) = 0 in the interval 0 ≤ x ≤ 2π is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 109

We have (2cos x – 1) (3 + 2 cos x) = 0If 2 cos x – 1=0, then cos x = 1/2
∴x = π/3, πp/3
If 3 + 2 cos x = 0, the cos x = –3/2
which is not possible.

Test: BITSAT Past Year Paper- 2014 - Question 110

23n – 7n – 1 is divisible by

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 110

23n – 7n – 1 Taking n = 2;
26 – 7 × 2 – 1
= 64 – 15 = 49

Test: BITSAT Past Year Paper- 2014 - Question 111

The gr eatest positive in teger, wh ich divides n (n + 1)(n + 2)(n + 3) for all n ∈ N , is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 111

The product of r consecutive integers is divisible by r ! . Thus n (n+ 1 ) (n + 2) (n + 3)is divisible by 4 ! = 24.

Test: BITSAT Past Year Paper- 2014 - Question 112

If z = x + iy, z1/3 = a – ib, then

where k is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 112

Test: BITSAT Past Year Paper- 2014 - Question 113

when simplified has the value

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 113

= 0

Test: BITSAT Past Year Paper- 2014 - Question 114

The complex number z = z + iy which satisfies the equation  , lies on

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 114

[if | z-z1 |=| z+z2 | , then it is a perpendicular
bisector of z1 and z2]
Hence, perpendicular bisector of (0, 3) and (0, – 3) is X–axis.

Test: BITSAT Past Year Paper- 2014 - Question 115

The number of all three elements subsets of the set {a1, a2, a3 . . . an} which contain a3 is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 115

The number of three elements subsets containing ais equal to the number of ways of selecting 2 elements out of n –1 elements. So, the required number of  subsets is n –1C2.

Test: BITSAT Past Year Paper- 2014 - Question 116

In how many ways can a committee of 5 made out 6 men and 4 women containing atleast one woman?

Test: BITSAT Past Year Paper- 2014 - Question 117

The coefficient of x4 in the expansion of (1 + x + x2 + x3)11, is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 117

We have coefficient of x4 in (1 + x + x2 + x3)11

= coefficient of x4 in (1 + x2)11 (1 + x)11

= coefficient of x4 in (1 + x)11 + coefficient of x2

= 990

Test: BITSAT Past Year Paper- 2014 - Question 118

If T0, T1, T2.....Tn represent the terms in the expansion of  (x + a)n, then (T0 –T2 + T4 – .......)2 + (T1 – T3  + T5 – .....)2 =

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 118

From the given condition, replacing a by ai and – ai respectively, we get

and

Multiplying (ii) and (i) we get required result i.e.,

Test: BITSAT Past Year Paper- 2014 - Question 119

If the (2p)th term of a H.P. is q and the (2q)th term is p, then the 2(p + q)th term is-

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 119

If a is the first term and d is the common
difference of the associated A.P.

⇒

Test: BITSAT Past Year Paper- 2014 - Question 120

If  are in A. P., then  is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 120

∴

Test: BITSAT Past Year Paper- 2014 - Question 121

The product of n positive numbers is unity, then their sum is :

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 121

Since, product of n positive number is unity.
⇒ x1 x2 x3 ....... xn = 1 ..(i)
Using A.M. ≥ GM

⇒

Hence

Test: BITSAT Past Year Paper- 2014 - Question 122

If P1 and P2 be the length of perpendiculars from the origin upon the straight lines x secθ + y cosecθ= a  and x cosθ – y sinqθ = a cos2θ respectively, then the value of  4P12 + P22.

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 122

We have P1 = length of perpendicular from (0, 0) on x secθ + y cosecqθ = a

P2= Length of the perpendicular from (0, 0)on x cosθ – y sin q = a cos2θ

Now

Test: BITSAT Past Year Paper- 2014 - Question 123

The angle of intersection of the two circles x2 + y2 – 2x – 2y = 0 and x2 + y2 = 4, is

Test: BITSAT Past Year Paper- 2014 - Question 124

An arch of a bridge is semi-elliptical with major axis horizontal. If the length the base is 9 meter and the highest part of the bridge is 3 meter from the horizontal; the best approximation of the height of the arch. 2 meter from the centre of the base is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 124

The equation of the ellipse is

Where centre is assumed as origin and base as x-axis. Put x = 2, we get

(approximately)

Test: BITSAT Past Year Paper- 2014 - Question 125

is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 125

Let

Taking log on both sides, we get

log y = –1

y =  e-1 = 1/e

Test: BITSAT Past Year Paper- 2014 - Question 126

If M. D. is 12, the value of S.D. will be

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 126

We know that

∴

Test: BITSAT Past Year Paper- 2014 - Question 127

A bag contains 5 brown and 4 white socks. A man pulls out 2 socks. Find the probability that they are of the same colour.

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 127

Let A = event of two socks being brown.
B = event of two socks being white.
Then P (A) =

Now, since A and B are mutually exclusive events, so required probability

Test: BITSAT Past Year Paper- 2014 - Question 128

Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)} be a relation on the set A = {3, 6, 9, 12}. Then, the relation is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 128

(3, 3), (6, 6), (9, 9), (12, 12), ∈R. R is not symmetric as (6, 12) ∈R but (12, 6) ÎR.
R is transitive as the only pair which
needs verification is (3, 6) and (6, 12) ∈R.

⇒ (3, 12) ∈ R

Test: BITSAT Past Year Paper- 2014 - Question 129

Let f : R →R be a function defined by , where m ≠ n , then

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 129

Let f : R → R be a function defined by

For any (x, y) ∉ R

Let f (x) = f (y)
⇒
∴ f is one – one
Let α ∈ R such that f (x) = α

⇒

⇒

So, f is not onto.

Test: BITSAT Past Year Paper- 2014 - Question 130

Find the value of tan

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 130

Test: BITSAT Past Year Paper- 2014 - Question 131

If  is square root of identity matrix of order 2 then –

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 131

⇒ α2 + βγ = 1

Test: BITSAT Past Year Paper- 2014 - Question 132

The value of λ , for which the lines 3x - 4y= 13, 8x - 11y= 33 and 2 x - 3y + λ=0 are concurrent is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 132

For concurrency of 3 lines the determinant of coefficients of equations should be 0.

i.e.

⇒ 3(-11λ-99)+4(8λ+66) -13(-24+22) = 0
⇒ -33λ - 297 + 32λ + 264 + 312 - 286 = 0
⇒ -λ - 583+ 576 = 0

⇒ λ = -7

Test: BITSAT Past Year Paper- 2014 - Question 133

Let f(x) =
Then which one of the following is true?

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 133

We have;

∴ which does not exist.

∴ f is not differentiable at x = 1

Also

=  –sin 1 + cos 1

∴ f is differentiable at x = 0

Test: BITSAT Past Year Paper- 2014 - Question 134

The interval in which the function 2x3 + 15 increases less rapidly than the function 9x2 – 12x, is –

Test: BITSAT Past Year Paper- 2014 - Question 135

The fuel charges for running a train are proportional to the square of the speed generated in miles per hour and costs ' 48 per hour at 16 miles per hour. The most economical speed ifthe fixed charges i.e. salaries etc. amount to ' 300 per hour is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 135

Let the speed of the train be v and distance to be covered be s so that total time taken is s/v hours. Cost of fuel per hour = kv2 (k is constant)

Also 48 = k. 162 by given condition k = 3/16

∴ Cost to fuel per hour 3/16 v2

Other charges per hour are 300. Total running cost,

Test: BITSAT Past Year Paper- 2014 - Question 136

Evaluate:

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 136

Dividing the numerator and denominator by cos2x, we get

Putting tan x = t ⇒ sec2x dx = dt, we get

⇒

Test: BITSAT Past Year Paper- 2014 - Question 137

is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 137

Let

Adding (1) and (2), we get

Test: BITSAT Past Year Paper- 2014 - Question 138

The area bounded by the x-axis, the curve y = f(x) and the lines x =1, x =b, is equal to √b2 + 1 - 2 for all b > 1, then f(x) is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 138

Given

Differentiate with respect to b

Test: BITSAT Past Year Paper- 2014 - Question 139

Solution of differential equation

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 139

⇒

Test: BITSAT Past Year Paper- 2014 - Question 140

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

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 140

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

So, position vector of centre of ΔDEF

=

Test: BITSAT Past Year Paper- 2014 - Question 141

The angle between any two diagonal of a cube is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 141

for a unit cube unit vector along the diagonal

unit vector along the diagonal

∴

Test: BITSAT Past Year Paper- 2014 - Question 142

Find the angle between the line  and the plane 10x + 2y – 11z = 3.

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 142

Let q be the angle between the line and the normal to the plane. Converting the given equations into vector form, we have

Here

Test: BITSAT Past Year Paper- 2014 - Question 143

The equation of the right bisector plane of the segment joining (2, 3, 4) and (6, 7, 8) is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 143

If the given points be A (2, 3, 4) and B
(6, 7, 8), then their mid-point N(4, 5, 6) must lie on the plane. The direction ratios of AB are 4, 4, 4, i.e. 1, 1, 1.

The required plane passes through N (4, 5, 6) and is normal to AB. Thus its equation is

1(x - 4) +1(y - 5) +1(z - 6) = 0 ⇒ x + y + z = 15

Test: BITSAT Past Year Paper- 2014 - Question 144

A bag contains n + 1 coins. It is known that one of these coins shows heads on both sides, whereas the other coins are fair. One coin is selected at random and tossed. If the probability that toss results in heads is , 7/12 then the value of n is.

Test: BITSAT Past Year Paper- 2014 - Question 145

A coin is tossed 7 times. Each time a man calls head. Find the probability that he wins the toss on more occasions.

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 145

The man has to win at least 4 times.

Reqd. probability =

Test: BITSAT Past Year Paper- 2014 - Question 146

Consider  Then number of possible solutions are :

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 146

Consider

convert them into equation and solve them and draw the graph of these equations  we get y = 1 and x = 3/2

From graph region is finite but numbers of possible solutions are infinite because for different values of x and y we have different or infinite no. of solutions.

Test: BITSAT Past Year Paper- 2014 - Question 147

If A =  then A100 :

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 147

⇒ A100 = 2100-1A \ A100 = 299A

Test: BITSAT Past Year Paper- 2014 - Question 148

If , then the value of  is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 148

Apply R1 → R1 – R3 and R2 → R2 – R3, we get

⇒ x[yr + z(q - y)]- z[0 - y(p - x)] = 0
[Expansion along first row]

⇒

Test: BITSAT Past Year Paper- 2014 - Question 149

Through the vertex O of a parabola y2 = 4x, chords OP and OQ are drawn at right angles to one another. The locus of the middle point of PQ is

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 149

Given parabola is y2 = 4x ...(1)

Let

Slope of OP  =

and slope of OQ =

Since OP OQ,

Let R (h, k) be the middle point of PQ, then

and k = t1 + t2 ...(4)

From (4),

[ From(2) and (3)]

Hence locus of R (h, k) is y2 = 2x – 8.

Test: BITSAT Past Year Paper- 2014 - Question 150

Let f (x ) =
If f(x) is continuous at x = π/2 , (p, q) =

Detailed Solution for Test: BITSAT Past Year Paper- 2014 - Question 150

∴

q = 4.

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