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150 Questions MCQ Test BITSAT Mock Tests Series & Past Year Papers - Test: BITSAT Past Year Paper- 2016

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Test: BITSAT Past Year Paper- 2016 - Question 1

What should be the velocity of rotation of earth due to rotation about its own axis so that the weight of a person becomes 3/5 of the presentweight at the equator. Equatorial radius of the earth is 6400 km.

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 1

True weight at equator, W = mg
Observed weight at equator,
W ' = mg ' = 3/5 mg
At equator, latitude λ = 0 
Using the formula,
mg ' = mg - mRω2 cos2λ
3/5 mg = mg - mRω2 cos20 = mg - mRω
⇒ mRω = mg - 3/5 mg = 2/5 mg

Test: BITSAT Past Year Paper- 2016 - Question 2

Block A of mass m and block B of mass 2m are placed on a fixed triangular wedge by means of a massless, inextensible string and a frictionless pulley as shown in figure. 


The wedge is inclined at 45° to the horizontal on both the sides. If the coefficient of friction between the block A and the wedge is 2/3 and that between the block B and the wedge is 1/3 and both the blocks A and B are released from rest, the acceleration of A will be

Test: BITSAT Past Year Paper- 2016 - Question 3

The surface charge density of a thin charged disc of radius σ is s. The value of the electric field at the centre of the disc is σ/2∈0. With respect to the field at the centre, the electric field along the axis at a distance R from the centre of the disc

Test: BITSAT Past Year Paper- 2016 - Question 4

The molecules of a given mass of a gas have r.m.s. velocity of 200 ms–1 at 27°C and 1.0 × 105 Nm–2 pressure. When the temperature and pressure of the gas are respectively, 127°C and 0.05 × 105 Nm–2, the r.m.s. velocity of its molecules in ms–1 is :

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 4

Here v1 = 200 m/s;
temperature T1 = 27°C = 27 + 273 = 300 k
temperature T2 = 127° C = 127 + 273 = 400 k, V = ?
R.M.S. Velocity, V ∝ √T

Test: BITSAT Past Year Paper- 2016 - Question 5

An inductor of inductance L = 400 mH and resistors of resistance R1 = 2Ω and R2 = 2Ω are connected to a battery of emf 12 V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at t = 0. The potential drop across L as a function of time is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 5

Growth in current in LR2 branch when switch is closed is given by

Hence, potential drop across

Test: BITSAT Past Year Paper- 2016 - Question 6

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

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 6


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

For wire 2,

From (i) and (ii),

⇒ F ' = 9F

Test: BITSAT Past Year Paper- 2016 - Question 7

Two spheres of different materials one with double the radius and one-fourth wall thickness of the other are filled with ice. If the time taken for complete melting of ice in the larger sphere is 25 minute and for smaller one is 16 minute, the ratio of thermal conductivities of the materials of larger spheres to that of smaller sphere is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 7

Radius of small sphere = r Thickness of small sphere = t Radius of bigger sphere = 2r Thickness of bigger sphere = t/4 Mass of ice melted = (volume of sphere) × (density of ice)
Let K1 and K2 be the thermal conductivities of larger and smaller sphere.
For bigger sphere,

For smaller sphere,

Test: BITSAT Past Year Paper- 2016 - Question 8

A biconvex lens h as a radius of curvature of magnitude 20 cm. Which one of the following options best describe the image formed of an object of height 2 cm placed 30 cm from the lens?

Test: BITSAT Past Year Paper- 2016 - Question 9

In the figure below, what is the potential difference between the point A and B and between B and C respectively in steady state

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 9

The equivalent circuit is shown in figure.
V1 + V2 = 100
and   2V1 = 6V2

On solving above equations, we get
V1 = 75V, V2 = 25V

Test: BITSAT Past Year Paper- 2016 - Question 10

A radioactive element X converts into another stable element Y. Half life of X is 2 hrs. Initially only X is present. After time t, the ratio of atoms of X and Y is found to be 1 : 4, then t in hours is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 10

Let N0 be the number of atoms of X at time t = 0.
Then at t = 4 hrs (two half lives)

∴ Nx/Ny = 1/3
and at t = 6 hrs  (three half lives)


The given ratio 1/4 lies between 1/3 and 1/7.
Therefore, t lies between 4 hrs and 6 hrs.

Test: BITSAT Past Year Paper- 2016 - Question 11

The approximate depth of an ocean is 2700 m. The compressibility of water is 45.4 × 10–11 Pa–1 and density of water is 103 kg/m3.What fractional compression of water will be obtained at the bottom of the ocean?

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 11

Compressibility of water,
K = 45.4 × 10–11 Pa–1
density of water P = 103 kg/m3
depth of ocean, h = 2700 m
We have to find ΔV/V = ?
As we know, compressibility,

So,(ΔV/V) = KPgh
= 45.4 × 10–11 × 103 × 10 × 2700
= 1.2258 × 10–2

Test: BITSAT Past Year Paper- 2016 - Question 12

A frictionless wire AB is fixed  on a sphere of radius R. A very small spherical ball slips on this wire. The time taken by this ball to slip from A to B is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 12

Acceleration of body along AB is g cos q Distance travelled in time t sec =
From ΔABC, AB = 2R cos θ

Test: BITSAT Past Year Paper- 2016 - Question 13

A string of length ℓ is fixed at both ends. It is vibrating in its 3rd overtone with maximum amplitude 'a'.  The amplitude at a distance ℓ/3 from one end is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 13

For a string vibrating in its nth overtone (n + 1)th harmonic)


Test: BITSAT Past Year Paper- 2016 - Question 14

A deuteron of kinetic energy 50 keV is describing a circular orbit of radius 0.5 metre in a plane perpendicular to the magnetic field B. The kinetic energy of the proton that describes a circular orbit of radius 0.5 metre in the same plane with the same B is

Test: BITSAT Past Year Paper- 2016 - Question 15

In the circuit shown in the figure, find the current in 45 Ω.

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 15


Test: BITSAT Past Year Paper- 2016 - Question 16

Kepler's third law states that square of period of revolution (T) of a planet around the sun, is proportional to third power of average distance r between sun and planet i.e. T2 = Kr3 here K is constant. If the masses of sun and planet are M and m respectively then as per Newton's law of gravitation force of attraction between them is here G is gravitational constant. The relation between G and K is described as

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 16

As we know, orbital speed, 
Time period

Squarring both sides,

Test: BITSAT Past Year Paper- 2016 - Question 17

Find the number of photon emitted per second by a 25 watt source of monochromatic light of wavelength 6600 Å. What is the photoelectric current assuming 3% efficiency for photoelectric effect ?

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 17

Pin = 25W, λ = 6600 Å = 6600 × 10–10 m nhv = P
⇒ Number of photons emitted/sec,

3% of emitted photons are producing current

Test: BITSAT Past Year Paper- 2016 - Question 18

A ray of light of intensity I is incident on a parallel glass slab at point A as shown in diagram. It undergoes partial reflection and refraction. At each reflection, 25% of incident energy is reflected. The rays AB and A'B' undergo interference. The ratio of Imax and Imin is :

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 18


Test: BITSAT Past Year Paper- 2016 - Question 19

A capillary tube of radius r is immersed vertically in a liquid such that liquid rises in it to height h (less than the length of the tube).  Mass of liquid in the capillary tube is m. If radius of the capillary tube is increased by 50%, then mass of liquid that will rise in the tube, is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 19



New mass

Test: BITSAT Past Year Paper- 2016 - Question 20

The drift velocity of electrons in silver wire with cross-sectional area 3.14 × 10–6 m2 carrying a current of 20 A is. Given atomic weight of Ag = 108, density of silver = 10.5 × 103 kg/m3.

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 20

Number of electrons per kg of silver

Number of electrons per unit volume of silver

Test: BITSAT Past Year Paper- 2016 - Question 21

A parallel plate capacitor of area ‘A’ plate separation ‘d’ is filled with two dielectrics as shown. What is the capacitance of the arrangement?

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 21


(∵ C1 and C2 are in series and resultant of these two in parallel with C3)

Test: BITSAT Past Year Paper- 2016 - Question 22

In the Young’s double-slit experiment, the intensity of light at a point on the screen where the path difference is λ is K, (λ being the wave length of light used). The intensity at a point where the path difference is λ/4, will be :

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 22

For path differ ence λ, phase difference = 2π rad.
For path difference  λ/4, phase difference
= π/2 rad.
As K = 4I0  so intensity at given point where path difference is λ/4

Test: BITSAT Past Year Paper- 2016 - Question 23

The mass of 7N15 is 15.00011 amu, mass of 8O16 is 15.99492 amu and mp = 1.00783 amu. Determine binding energy of last proton of 8O16.

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 23

M(8O16) = M (7N15) + 1mP
binding energy of last proton
= M (N15) + mP – M (1O16)
= 15.00011 + 1.00783 – 15.99492
= 0.01302 amu = 12.13 MeV

Test: BITSAT Past Year Paper- 2016 - Question 24

A wire carrying current I has the shape as shown in adjoining figure. Linear parts of the wire are very long and parallel to X-axis while semicircular portion of radius R is lying in Y-Z plane. Magnetic field at point O is:

Test: BITSAT Past Year Paper- 2016 - Question 25

A stone projected with a velocity u at an angle θ with the horizontal reaches maximum height H1. When it is projected with velocity u at an angle with the horizontal, it reaches maximum height H2. The relation between the horizontal range R of the projectile, heights H1 and H2 is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 25


and

Test: BITSAT Past Year Paper- 2016 - Question 26

If the series limit wavelength of Lyman series for the hydrogen atom is 912 Å then the series limit wavelength for Balmer series of hydrogen atoms is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 26


For limiting wavelength of Lyman series

For limiting wavelength of Balmer series
n1 = 2, n2 = ∞

∴ λB = 4λL = 4 × 912 Å.

Test: BITSAT Past Year Paper- 2016 - Question 27

In the shown arrangement of the experiment of the meter bridge if AC corresponding to null deflection of galvanometer is x, what would be its value if the radius of the wire AB is doubled?

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 27


At null point 
If radius of the wire is doubled, then the resistance of AC will change and also the resistance of CB will change. But since R1/R2 does not change so, R3/R4 should also not change at null point. Therefore the point C does not change.

Test: BITSAT Past Year Paper- 2016 - Question 28

A 1 kg mass is attached to a spring of force constant 600 N/m and rests on a smooth horizontal surface with other end of the spring tied to wall as shown in figure. A second mass of 0.5 kg slides along the surface towards the first at 3m/s. If the masses make a perfectly inelastic collision, then find amplitude and time period of oscillation of combined mass.

Test: BITSAT Past Year Paper- 2016 - Question 29

The frequency of vibration of string is given by 

Here p is number of segments in the string and / is the length. The dimensional formula for m will be

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 29


Now, dimensional formula of R.H.S.

[p will have no dimension as it is an integer] = ML–1T0.
So, dimensions of m will be ML–1T0.

Test: BITSAT Past Year Paper- 2016 - Question 30

For the angle of minimum deviation of a prism to be equal to its refracting angle, the prism must be made of a material whose refractive index:

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 30


The angle of minimum deviation is given as
δmin = i + e–A
for minimum deviation
δmin = A then
 2A = i + e
in case of δmin i = e
2A = 2i r1 = r2 = A/2
i =  A = 90°
from smell’s law
1 sin i = n sin r1
sin A = n sin A/2


when  A = 90° = imin
then    nmin = √2
i = A = 0  nmax = 2

Test: BITSAT Past Year Paper- 2016 - Question 31

Consider elastic collision of a particle of mass m moving with a velocity u with another particle of the same mass at rest. After the collision, the projectile and the struck particle move in directions making angles θ1 and θ2 respectively with the initial direction of motion. The sum of the angles θ1 + θ2, is:

Test: BITSAT Past Year Paper- 2016 - Question 32

A conducting circular loop is placed in a uniform magnetic field of 0.04 T with its plane perpendicular to the magnetic field. The radius of the loop starts shrinking at 2 mm/s. The induced emf in the loop when the radius is 2 cm is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 32

Induced emf in th e loop is gi ven by
e = -B. dA/dt where A is the area of the loop.

r = 2cm = 2 × 10–2 m
dr = 2 mm = 2 × 10–3 m
dt = 1s

= 0.32 π × 10–5V
= 3.2 π ×10–6V
= 3.2 π μV

Test: BITSAT Past Year Paper- 2016 - Question 33

Figure below shows two paths that may be taken by a gas to go from a state A to a state (c) In process AB, 400 J of heat is added to the system and in process BC, 100 J of heat is added to the system. The heat absorbed by the system in the process AC will be

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 33

In cyclic process ABCA
Qcycle = Wcycle
QAB + QBC + QCA = ar. of  ΔABC
+ 400 + 100 + QC→A = 1/2 (2 × 10–3) (4 × 104)
⇒ QC→A  = – 460 J
⇒ QA→C = + 460 J

Test: BITSAT Past Year Paper- 2016 - Question 34

Two resistances at 0°C with temperature coefficient of resistance α1 and α2 joined in series act as a single resistance in a circuit. The temperature coefficient of their single resistance will be

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 34

R1 = R0(1 + α1t) + R0(1 + α2t)

Comparing with R = R0 (1+ αt)

Test: BITSAT Past Year Paper- 2016 - Question 35

Two identical charged spheres suspended from a common point by two massless strings of lengths l, are initially at a distance d (d << l ) apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity v. Then v varies as a function of the distance x between the spheres, as :

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 35

From figure tan θ = Fe/mg ≅ θ


Differentiating eq. (1) w.r.t. time

so x2(v) ∝ q Replace q from eq. (2)
x2(v) ∝ x3/2 or v ∝ x–1/2

Test: BITSAT Past Year Paper- 2016 - Question 36

A point particle of mass 0.1 kg is executing S.H.M. of amplitude of 0.1 m. When the particle passes through the mean position, its kinetic energy is 8 x 10-3 Joule. Obtain the equation of motion of this particle if this initial phase of oscillation is 45°.

Test: BITSAT Past Year Paper- 2016 - Question 37

A source of sound S emitting waves of frequency 100 Hz and an observor O are located at some distance from each other. The source is moving with a speed of 19.4 ms-1 at an angle of 60° with the source observer line as shown in the figure. The observor is at rest. The apparent frequency observed by the observer is (velocity of sound in air 330 ms-1)

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 37

Here, original frequency of sound, f0 = 100 Hz
Speed of source Vs = 19.4 cos 60° = 9.7

19.4 cos 60° = 9.7
From Doppler's formula

Apparent frequency f1 = 103 Hz

Test: BITSAT Past Year Paper- 2016 - Question 38

A resistor of resistance R, capacitor of capacitance C and inductor of inductance L are connected in parallel to AC power source of voltage ε0 sin ωt. The maximum current through the resistance is half of the maximum current through the power source. Then value of R is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 38


Test: BITSAT Past Year Paper- 2016 - Question 39

A lens having focal length f and aperture of diameter d forms an image of intensity I. Aperture of diameter d/2 in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively:

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 39

By covering aperture, focal length does not change. But intensity is reduced by 1/4 times, as aperture diameter d/2 is covered.

∴ New focal length = f and intensity = 3I/4.

Test: BITSAT Past Year Paper- 2016 - Question 40

A circular disc of radius R and thickness R/6 has moment inertia I about an axis passing through its centre perpendicular to its plane. It is melted and recasted into a solid sphere. The moment of inertia of the sphere about its diameter is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 40

According to problem disc is melted and recasted into a solid sphere so their volume will be same.
Moment of inertia of disc

Moment of inertia of sphere

Test: BITSAT Past Year Paper- 2016 - Question 41

In PO3-4, the formal charge on each oxygen atom and the P - O bond order respectively are

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 41


Bond order

Three unit negative charge is being shared by four O atoms. Formal charge = –3/4

Test: BITSAT Past Year Paper- 2016 - Question 42

The decreasing order of the ionization potential of the following elements is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 42

Closed shell (Ne), half filled (P) and completely filled configuration (Mg) are the cause of higher value of I.E.

Test: BITSAT Past Year Paper- 2016 - Question 43

Knowing that the chemistry of lanthanoids(Ln) is dominated by its + 3 oxidation state, which of the following statements is incorrect?

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 43

Most of the Ln3+ compounds except La3+ and Lu3+ are coloured due to the presence of f-electrons.

Test: BITSAT Past Year Paper- 2016 - Question 44

Which of the following arrangements does not represent the correct order of the property stated against it?

Test: BITSAT Past Year Paper- 2016 - Question 45

Which of the following is paramagnetic?

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 45

Fe2+ electr onic configura tion is [Ar] 3d6. Since CN is strong field ligand d electrons are paired . In Ni(CO)4 O. S. of Ni is zero electron ic configuration is [Ar}3d8 4s2. In presence of CO it is [Ar] 3d10 4s0, electrons are paired. Electronic configuration of Ni2+ [Ar]3d8 4s0, due to CN ligand all electrons are paired. Co3+ is [Ar] 3d6 since F is weak ligand hence paramagnetic.

Test: BITSAT Past Year Paper- 2016 - Question 46

The hypothetical complex chloro-diaquatriamminecobalt (III) chloride can be represented as

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 46

The complex chlorodiaquatriammine cobalt (III) chloride can have the structure [CoCl(NH3)3(H2O)2]Cl2

Test: BITSAT Past Year Paper- 2016 - Question 47

The normality of 26% (wt/vol) solution of ammonia (density = 0.855 ) is approximately :

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 47

Wt. of NH3 = 26g = 26/17 g eg = 1.53 g eg
Vol. of soln. = 100 mc= 0.1 L
∴ Normality = 1.53/0.1 = 15.3N

Test: BITSAT Past Year Paper- 2016 - Question 48

1.25 g of a sample of Na2CO3 and Na2SO4 is dissolved in 250 ml solution. 25 ml of this solution neutralises 20 ml of 0.1N H2SO4.The % of Na2CO3 in this sample is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 48

Let the amount of Na2CO3 present in the mixture be xg . Na2SO4 will not react with H2SO4. Then

Test: BITSAT Past Year Paper- 2016 - Question 49

Which of the following compound has all the four types (1°, 2°, 3° and 4°) of carbon atoms?

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 49


C1 and C5 are 1°, C3 is 2°, C4 is 3°, and C2 is 4°.

Test: BITSAT Past Year Paper- 2016 - Question 50

Which of the following has two stereoisomers?

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 50

A rapid umbrella type inversion rapidly converts the structure III to its enantiomer; hence the two enantiomers are not separable.

Test: BITSAT Past Year Paper- 2016 - Question 51

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 51

Test: BITSAT Past Year Paper- 2016 - Question 52


The compounds A and B, respectively are

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 52

Catalytic hydrogen ation of alkyn es gives cis-alkene which in turn adds deuterium atoms in presence of H2 again in cis-manner forming meso-2, 3-dideuterobutane.

Test: BITSAT Past Year Paper- 2016 - Question 53

Give the possible structure of X in the following reaction :

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 53

D2SO4 transfers D+, an electrophile, to form hexadeuterobenzene.

Test: BITSAT Past Year Paper- 2016 - Question 54

An aromatic compound has molecular formula C7H7Br. Give the possible isomers and the appropriate method to distinguish them.

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 54

Test: BITSAT Past Year Paper- 2016 - Question 55

Which of the following method gives better yield of p-nitrophenol?

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 55


Thus here, oxidation of phenol is minimised by forming p-nitrosophenol.

Test: BITSAT Past Year Paper- 2016 - Question 56

Formation of polyethylene from calcium carbide takes place as follows
CaC2 + 2H2O → Ca (OH)2 + C2H2
C2H2 + H2 → C2H4
nC2H4 → (-CH2 - CH2 -)n
The amount of polyethylene obtained from 64.1 kg of CaC2 is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 56

The concerned chemical reactions are

Thus 64 kg of CaC2 gives 26 kg of acetylene which in turn gives 28 kg of ethylene whose 28 kg gives 28 kg of the polymer, polythene.

Test: BITSAT Past Year Paper- 2016 - Question 57

The most likely acid-catalysed aldol condensation products of each of the two aldehydes I and II will respectively be

Test: BITSAT Past Year Paper- 2016 - Question 58

Sometimes, the colour observed in Lassaigne’s test for nitrogen is green. It is because

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 58

Although blue coloured ferric ferrocyanide is formed but due to the presence of yellow coloured Fe3+ salts, the blue colour gives the shade of green.

Test: BITSAT Past Year Paper- 2016 - Question 59

Fructose on reduction gives a mixture of two alcohols which are related as

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 59

Ketoses on reduction produce a new chiral carbon leading to the formation of two isomeric alcohols which are diastereomeric as well as C-2 epimers.

Test: BITSAT Past Year Paper- 2016 - Question 60

What will happen when D-(+)-glucose is treated with methanolic —HCl followed by Tollens’ reagent?

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 60

Reaction of D-(+)-glucose with methanolic —HCl leads to formation of methyl glucoside (C1—OH group is methylated) which, being acetal, is not hydrolysable by base, so it will not respond Tollens’ reagent.

Test: BITSAT Past Year Paper- 2016 - Question 61

Which of the followings forms the base of talcum powder?

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 61

Magnesium hydrosilicate forms base of Talcum powder.

Test: BITSAT Past Year Paper- 2016 - Question 62

The important antioxidant used in food is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 62

BHT is the important antioxidant used in food.

Test: BITSAT Past Year Paper- 2016 - Question 63

The first emission line in the atomic spectrum of hydrogen in the Balmer series appears at

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 63

For Balmer  n1 = 2 and n2 = 3;

Test: BITSAT Past Year Paper- 2016 - Question 64

An e has magnetic quantum number as –3, what is its principal quantum number?

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 64

When m = – 3, l = 3, ∴ n = 4 .

Test: BITSAT Past Year Paper- 2016 - Question 65

At what temperature, the rate of effusion of N2 would be 1.625 times than that of SO2 at 50oC?