Test: BITSAT Past Year Paper- 2014 - JEE MCQ

# Test: BITSAT Past Year Paper- 2014 - JEE MCQ

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## 150 Questions MCQ Test BITSAT Mock Tests Series & Past Year Papers - Test: BITSAT Past Year Paper- 2014

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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