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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?

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 65


Test: BITSAT Past Year Paper- 2016 - Question 66

The average kinetic energy of an ideal gas per molecule in SI unit at 25°C will be

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 66

Test: BITSAT Past Year Paper- 2016 - Question 67

The degree of dissociation of PCl5 (α) obeying the equilibrium PCl5 ⇌ PCl3 + Cl2 is related to the equilibrium pressure by

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 67

Test: BITSAT Past Year Paper- 2016 - Question 68

In a closed system, A(s) ⇌ 2B(g) + 3C(g), if partial pressure of C is doubled, then partial pressure of B will be

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 68


Test: BITSAT Past Year Paper- 2016 - Question 69

For a particular reversible reaction at temperature T, ΔH and ΔS were found to be both +ve. If Te is the temperature at equilibrium, the reaction would be spontaneous when

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 69

At equilibrium  ΔG = 0
Hence, ΔG = ΔH – TeΔS = 0

For a spontaneous reaction
ΔG must be negative which is possible only
if ΔH – TΔS < 0
∴ ΔH < TΔS

Test: BITSAT Past Year Paper- 2016 - Question 70

Given

Based on data provided, the value of electron gain enthalpy of fluorine would be:

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 70

Applying Hess’s Law

-617= 161+ 520 + 77 + E.A. + (–1047)
E.A. = –617 + 289 = –328 kJ mol–1
∴ electron affinity of fluorine = –328 kJ mol–1

Test: BITSAT Past Year Paper- 2016 - Question 71

The percentage hydrolysis of 0.15 M solution of ammonium acetate, Ka for CH3COOH is 1.8 × 10–5 and Kb for NH3 is 1.8 × 10–5

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 71

Test: BITSAT Past Year Paper- 2016 - Question 72

For a sparingly soluble salt ApBq, the relationship of its solubility product Ls → Ksp with its solubility (S) is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 72


Let the solubility be S mol/liter Thus,
Ksp = [A +]p [B]q
= [Sp]p [Sq]q = p p.qq (S)p+q .

Test: BITSAT Past Year Paper- 2016 - Question 73

Consider the reaction :
Cl2(aq) + H2S(aq) → S(s) + 2H+ (aq) + 2Cl (aq)
The rate equation for this reaction is rate = k[Cl2][H2S]
Which of these mechanisms is/are consistent with this rate equation?
A. Cl2 + H2S → H+ + Cl + Cl+ + HS (slow)
Cl+ + HS → H+ + Cl + S (fast)
B. H2S ⇌ H+ + HS- (fast equilibrium)
Cl2 + HS- → 2Cl- + H+ S (Slow )

Test: BITSAT Past Year Paper- 2016 - Question 74

In the reaction, P + Q  → ? R + S
The time taken for 75% reaction of P is twice the time taken for 50% reaction of P. The concentration of Q varies with reaction time as shown in the figure. The overall order of the reaction is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 74

For P, if t50% = x
then t75% = 2x
This is true only for first order reaction. So, order with respect to P is 1.
Further the graph shows that concentration of Q decreases with time. So rate, with respect to Q, remains constant. Hence, it is zero order wrt Q.
So, overall order is 1+ 0 = 1

Test: BITSAT Past Year Paper- 2016 - Question 75

The EMF of th e cell Tl/Tl+ (0.001M) || Cu2+ (0.01M) /Cu is 0.83. The cell EMF can be increased by

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 75

The oxidation potential
and reduction potential ∝ concentration of ions. The cell voltage can be increased by decreasing the concentration of ions around anode or by increasing the concentration of ions around cathode

Test: BITSAT Past Year Paper- 2016 - Question 76

Electrolysis is carried out in three cells
(A) 1.0 M CuSO4 Pt electrode
(B) 1.0 M CuSO4 copper electrodes
(C) 1.0 M KCl Pt electrodes
If volume of electrolytic solution is maintained constant in each of the cell, which is correct set of pH changes in (A), (B) and (C) cell respectively?

Test: BITSAT Past Year Paper- 2016 - Question 77

The equilibrium constant for the disproportionation reaction 2Cu+ (aq) → Cu(s) + Cu2+ (ag) at 25°C
(Eo Cu+ / Cu = 0.52V, EoCu2+ / Cu= 0.16V) is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 77

The reaction

At equilibrium Ecell = 0

∴ Kc = 1.2 x 106

Test: BITSAT Past Year Paper- 2016 - Question 78

The non stoichiometric compound Fe0.94O is formed when x % of Fe2+ ions are replaced by as many 2/3 Fe3+ ions, x is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 78

The number of Fe3+ ions replacing x Fe2+ ions = 2x/3 vacancies of cations

But x/3 = 1 – 0.94 = 0.06,  x = 0.06 × 3
= 0.18 = 18%

Test: BITSAT Past Year Paper- 2016 - Question 79

Al (at. wt 27) crystallizes in the cubic system with a cell edge of 4.05 Å. Its density is 2.7 g per cm3.
Determine the unit cell type calculate the radius of the Al atom

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 79


Hence it is face centred cubic unit lattice.
Again 4r = a√2 = 5.727 Å
∴ r = 1.432 Å

Test: BITSAT Past Year Paper- 2016 - Question 80

A compound of Xe and F is found to have 53.5% of Xe. What is oxidation number of Xe in this compound ?

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 80

Xe = 53.5 % ∴ F = 46.5%
Relative number of atoms Xe

Simple ratio Xe = 1 and F = 6 ; Molecular formula is XeF6
O.N.of Xe is +6

Test: BITSAT Past Year Paper- 2016 - Question 81

Directions: Choose the word which best expresses the meaning of the given word.

CORPULENT

Test: BITSAT Past Year Paper- 2016 - Question 82

Directions: Choose the word which best expresses the meaning of the given word.

EMBEZZLE

Test: BITSAT Past Year Paper- 2016 - Question 83

Directions: Choose the word which is the exact OPPOSITE of the given words.

ARROGANT

Test: BITSAT Past Year Paper- 2016 - Question 84

Directions: Choose the word which is the exact OPPOSITE of the given words.

EXODUS

Test: BITSAT Past Year Paper- 2016 - Question 85

Directions: Read the following passage and answer the questions that follows.

At this stage of civilisation, when many nations are brought in to close and vital contact for good and evil, it is essential, as never before, that their gross ignorance of one another should be diminished, that they should begin to understand a little of one another's historical experience and resulting mentality. It is the fault of the English to expect the people of other countries to react as they do, to political and international situations. Our genuine goodwill and good intentions are often brought to nothing, because we expect other people to be like us. This would be corrected if we knew the history, not necessarily in detail but in broad outlines, of the social and political conditions which have given to each nation its present character.

Q. According to the author of 'Mentality' of a nation is mainly product of its

Test: BITSAT Past Year Paper- 2016 - Question 86

Directions: Read the following passage and answer the questions that follows.

At this stage of civilisation, when many nations are brought in to close and vital contact for good and evil, it is essential, as never before, that their gross ignorance of one another should be diminished, that they should begin to understand a little of one another's historical experience and resulting mentality. It is the fault of the English to expect the people of other countries to react as they do, to political and international situations. Our genuine goodwill and good intentions are often brought to nothing, because we expect other people to be like us. This would be corrected if we knew the history, not necessarily in detail but in broad outlines, of the social and political conditions which have given to each nation its present character.

Q. The need for a greater understanding between nations

Test: BITSAT Past Year Paper- 2016 - Question 87

Directions: Read the following passage and answer the questions that follows.

At this stage of civilisation, when many nations are brought in to close and vital contact for good and evil, it is essential, as never before, that their gross ignorance of one another should be diminished, that they should begin to understand a little of one another's historical experience and resulting mentality. It is the fault of the English to expect the people of other countries to react as they do, to political and international situations. Our genuine goodwill and good intentions are often brought to nothing, because we expect other people to be like us. This would be corrected if we knew the history, not necessarily in detail but in broad outlines, of the social and political conditions which have given to each nation its present character.

Q. The character of a nation is the result of its

Test: BITSAT Past Year Paper- 2016 - Question 88

Directions: Read the following passage and answer the questions that follows.

At this stage of civilisation, when many nations are brought in to close and vital contact for good and evil, it is essential, as never before, that their gross ignorance of one another should be diminished, that they should begin to understand a little of one another's historical experience and resulting mentality. It is the fault of the English to expect the people of other countries to react as they do, to political and international situations. Our genuine goodwill and good intentions are often brought to nothing, because we expect other people to be like us. This would be corrected if we knew the history, not necessarily in detail but in broad outlines, of the social and political conditions which have given to each nation its present character.

Q. According to the author his countrymen should

Test: BITSAT Past Year Paper- 2016 - Question 89

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

S1: A force of exists between everybody in the universe.
P: Normally it is very small but when the one of the bodies is a planet, like earth, the force is considerable.
Q: It has been investigated by many scientists including Galileo and Newton.
R: Everything on or near the surface of the earth is attracted by the mass of earth.
S: This gravitational force depends on the mass of the bodies involved.
S6: The greater the ma ss, the greater is the earth's force of attraction on it. We can call this force of attraction gravity.

Q. The Proper sequence should be:

Test: BITSAT Past Year Paper- 2016 - Question 90

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

S1: Calcutta unlike other cities kepts its trams.
P: As a result there horrendous congestion.
Q: It was going to be the first in South Asia.
R: They run down the centre of the road
S: To ease in the city decided to build an underground railway line.
S6: The foundation stone was laid in 1972.

Q. The Proper sequence should be:

Test: BITSAT Past Year Paper- 2016 - Question 91

Directions: Pick out the most effective word from the given words to fill in the blank to make the sentence meaningfully complete.

Q. The miser gazed ...... at the pile of gold coins in front of him.

Test: BITSAT Past Year Paper- 2016 - Question 92

Directions: Pick out the most effective word from the given words to fill in the blank to make the sentence meaningfully complete.

Q. I saw a ...... of cows in the field.

Test: BITSAT Past Year Paper- 2016 - Question 93

Directions: Read the each sentence to find out whether there is any grammatical error in it. The error, if any will be in one part of the sentence. The letter of that part is the answer. If there is no error, the answer is 'd'. (Ignore the errors of punctuation, if any).

Test: BITSAT Past Year Paper- 2016 - Question 94

Directions: Read the each sentence to find out whether there is any grammatical error in it. The error, if any will be in one part of the sentence. The letter of that part is the answer. If there is no error, the answer is 'd'. (Ignore the errors of punctuation, if any).

Test: BITSAT Past Year Paper- 2016 - Question 95

Directions: Read the each sentence to find out whether there is any grammatical error in it. The error, if any will be in one part of the sentence. The letter of that part is the answer. If there is no error, the answer is 'd'. (Ignore the errors of punctuation, if any).

Test: BITSAT Past Year Paper- 2016 - Question 96

Select a suitable figure from the four alternatives that would complete the figure matrix.

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 96

In each row, the third figure comprises of a black circle and only those line segments which are not common to the first and the second figures.

Test: BITSAT Past Year Paper- 2016 - Question 97

Select a suitable figure from the four alternatives that would complete the figure matrix.

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 97

In each row (as well as each column), the third figure is a combination of all the, elements of the first and the second figures.

Test: BITSAT Past Year Paper- 2016 - Question 98

Select a suitable figure from the four alternatives that would complete the figure matrix.

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 98

In each row, the third figure is a collection of the common elements (line segments) of the first and the second figures.

Test: BITSAT Past Year Paper- 2016 - Question 99

Direction: Choose the correct alternative that will continue the same pattern and replace the question mark in the given series.

3, 4, 7, 7, 13, 13, 21, 22, 31, 34, ?

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 99

The given sequence is a combination of two series :
I.  3, 7, 13, 21, 31, ? and  II.  4, 7, 13, 22, 34
The pattern in I is + 4, + 6, + 8, + 10,.....
The pattern in II is + 3, + 6, + 9, + 12,.....
So, missing term = 31 + 12 = 43.

Test: BITSAT Past Year Paper- 2016 - Question 100

Introducing a boy, a girl said, "He is the son of the daughter of the father of my uncle." How is the boy related to the girl?

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 100

The father of the boy's uncle is the grandfather of the boy and daughter of the grandfather is sister of father.

Test: BITSAT Past Year Paper- 2016 - Question 101

Direction: In these series, you will be looking at both the letter pattern and the number pattern. Fill the blank in the middle of the series or end of the series.

QAR, RAS, SAT, TAU, _____

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 101

In this series, the third letter is repeated as the first letter of the next segment. The middle letter, A, remains static. The third letters are in alphabetical order, beginning with R.

Test: BITSAT Past Year Paper- 2016 - Question 102

Direction: In these series, you will be looking at both the letter pattern and the number pattern. Fill the blank in the middle of the series or end of the series.

DEF, DEF2, DE2F2, _____, D2E2F3

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 102

In this series, the letters remain the same: DEF.
The subscript numbers follow this series: 111, 112, 122, 222, 223, 233, 333, ...

Test: BITSAT Past Year Paper- 2016 - Question 103

Direction: In each question below are given two statements followed by two conclusions numbered I and II. You have to take the given two statements to be true even if they seem to be at variance from commonly known facts. Read the conclusion and then decide which of the given conclusions logically follows from the two given statements, disregarding commonly known facts.

Statements: Raman is always successful. No fool is always successful.
Conclusions:
I. Raman is a fool.
II. Raman is not a fool.

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 103

Since both the premises are universal and one premise is negative, the conclusion must be universal negative and should not contain the middle term. So, only II follows.

Test: BITSAT Past Year Paper- 2016 - Question 104

Direction: In each question below are given two statements followed by two conclusions numbered I and II. You have to take the given two statements to be true even if they seem to be at variance from commonly known facts. Read the conclusion and then decide which of the given conclusions logically follows from the two given statements, disregarding commonly known facts.

Statements: Some desks are caps. No cap is red.
Conclusions:
I. Some caps are desks.
II. No desk is red.

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 104

Since one premise is particular and the other premise is negative, the conclusion must be particular negative and should not contain the middle term. So, it follows that 'Some desks are not red'. However, I is the converse of the first premise and thus it holds.

Test: BITSAT Past Year Paper- 2016 - Question 105

Choose the set of figures which follows the given rule.
Rule: Closed figures losing their sides and open figures gaining their sides.

Test: BITSAT Past Year Paper- 2016 - Question 106

Let f (x) = , then fof (x) = x, provided that:

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 106



⇒ (ac + dc)x2 + (bc + d2 - bc - a2) x – ab – bd = 0, ∀ x ∈ R
⇒ (a + d)c = 0, d2 - a2 = 0 and  (a + d)b = 0
⇒ a + d = 0

Test: BITSAT Past Year Paper- 2016 - Question 107

Two finite sets have m and n elements. The number of subsets of the first set is 112 more than that of the second set. The values of m and n respectively are,

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 107

2m - 2n = 112 ⇒ 2n (2m-n - 1)= 16.7
∴ 2n (2m-n - 1) = 24 (23- 1)
Comparing we get n = 4 and m – n = 3
⇒ n = 4 and m = 7

Test: BITSAT Past Year Paper- 2016 - Question 108

If A and B are positive acute angles satisfying 3 cos 2 A + 2 cos 2 B = 4 and , Then the value of A + 2B is equal to :

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 108

Given, 3 cos2A +2 cos2 B = 4
⇒ 2 cos2 B -1 = 4 - 3cos2 A -1
⇒ cos 2B = 3 (1 - cos 2 A) = 3 sin2A...(1)
and 2 cos B sin B = 3 sin A cos A
sin 2B = 3 sin A cos A ...(2)
Now,  cos (A + 2B) = cos A cos 2B – sin A sin 2B
= cos A (3 sin2 A) – sin A (3 sin A cos A) = 0 [using eqs. (1) and (2)]
⇒ A +2B = π/2

Test: BITSAT Past Year Paper- 2016 - Question 109

If sin θ1 + sin θ2 + sin θ3 = 3, then cos θ1 + cos θ2 + cos θ3 =

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 109

Since, sin θ1 + sin θ2 + sin θ3 = 3
∴ sin θ1 = sin θ2 = sin θ3 = 1 ⇒ θ1 = θ2 = θ3 = π/2
∴ cos θ1 + cos θ2 + cos θ3 = 0

Test: BITSAT Past Year Paper- 2016 - Question 110

If tan (cot x) = cot (tan x), then sin 2x is equal to :

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 110

Given,

Test: BITSAT Past Year Paper- 2016 - Question 111

The general solution of the equation sin 2x + 2sin x +2 cos x+ 1 = 0 is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 111

Given, sin 2x + 2 sin x + 2 cos x + 1 = 0
⇒ 1 + sin 2x + 2(sin x + cos x ) = 0
⇒ (sin x + cos x ) 2 + 2(sin x + cos x) = 0
⇒ (sin x + cos x ) (sin x + cos x + 2) = 0
∴ sin x + cos x = 0 or sin x + cos x = -2
But, sin x + cos x = -2 is inadmissible.
Since, | sin x | ≤ 1, | cos x | ≤ 1

Test: BITSAT Past Year Paper- 2016 - Question 112

In a ΔABC, if  and the side a = 2, then area of the triangle is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 112



⇒ cot A = cot B = cot C
⇒ A = B = C = 60° ⇒ ΔABC is equilateral
Hence, area of 

Test: BITSAT Past Year Paper- 2016 - Question 113

If  , then what is the value of x?

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 113

Given,

∴ 2tan–1 a – 2tan–1b = 2 tan–1 x
⇒ tan–1 a – tan–1 b = tan–1 x

Test: BITSAT Past Year Paper- 2016 - Question 114

The arithmetic mean of numbers a, b, c, d, e is M. What is the value of (a – M) + (b – M) + (c – M) + (d – M) + (e – M) ?

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 114

Given

⇒ a + b + c + d + e = 5 M
⇒ a + b + c + d + e – 5 M = 0
⇒ (a – M) + (b – M) + (c – M) + (d – M) + (e – M) = 0
Hence, required value = 0

Test: BITSAT Past Year Paper- 2016 - Question 115

The fourth term of an A.P. is three times of the first term and the seventh term exceeds the twice of the third term by one, then the common difference of the progression is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 115

Let the progression be a, a + d, a + 2d,
Then x 4 = 3x1 ⇒ a + 3d = 3a ⇒ 3d = 2a ...(i)
Again x7 = 2x3 + 1
⇒ a + 6d = 2(a + 2d) + 1 ⇒ 2d = a + 1 ...(ii)
Solving (i) and (ii), we get
a = 3,  d = 2

Test: BITSAT Past Year Paper- 2016 - Question 116

The sum to n terms of the series is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 116

Test: BITSAT Past Year Paper- 2016 - Question 117

If log a, log b, and log c are in A.P. and also log a – log 2b, log 2b – log 3c, log 3c – log a are in A.P., then

Test: BITSAT Past Year Paper- 2016 - Question 118

....upto n terms is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 118

The series is

Test: BITSAT Past Year Paper- 2016 - Question 119

If z1 = √3 + i√3 an d z2 = √3 + i, then the complex number lies in the: 

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 119




Hence, lies in the first quadrant asboth real and imaginary parts of this number are positive.

Test: BITSAT Past Year Paper- 2016 - Question 120

If the matrix is singular, then λ  =

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 120

| A | = 0 as the matrix A is singular

Apply R2 → R2 – 2R1 and R3 → R3 – 3R1 and expand.
–2(4 – 3λ) + 4(4 – 2λ) = 0
⇒ 8 – 2λ = 0 ⇒ λ = 4
For λ = 4, the second and the third column are proportional.

Test: BITSAT Past Year Paper- 2016 - Question 121

Let α1, α2 and β1, β2 be the roots of ax2 + bx + c = 0 and px2 + qx + r = 0 respectively. If the system of equations α1y + α2z = 0 and β1y + β2z = 0 has a non-trivial solution, then

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 121

Since α1, α2 and β1, β2 are the roots of ax2 + bx + c = 0 and px2 + qx + r = 0 respectively, therefore

Since the given system of equation has a non-trivial solution

Test: BITSAT Past Year Paper- 2016 - Question 122

If [ ] denotes the greatest integer less than or equal to the real number under consideration and –1 ≤ x < 0; 0 ≤ y < 1; 1 ≤ z < 2 , then the value of the determinant is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 122

Since,  –1 ≤ x < 0
∴ [x] = –1
0 ≤ y < 1    ∴  [y] = 0
1 ≤ z < 2      ∴  [z] = 1
∴ Given determinant =

Test: BITSAT Past Year Paper- 2016 - Question 123

If α, β are the roots of the equations x2 – 2x– 1 = 0, then what is the value of α2 β–2+ α –2 β2

Test: BITSAT Past Year Paper- 2016 - Question 124

If a, b and c are real numbers then the roots of the equation (x – a) (x – b) + (x – b) (x – c) + (x – c) (x – a) = 0 are always

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 124

Given equation is (x – a) (x – b) + (x – b) (x – c) + (x – c) (x – a) = 0
⇒ 3x2 – 2(b + a + c) x + ab + bc + ca = 0
Now, here A = 3, B = – 2 (a + b + c)
C = ab + bc + ca

Test: BITSAT Past Year Paper- 2016 - Question 125

, where a > b > 1, is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 125


Test: BITSAT Past Year Paper- 2016 - Question 126

The number of points at which the function is discontinuous is:

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 126

The function log|x| is not defined at x = 0, so, x = 0 is a point of discontinuity. Also for f(x) to be defined, log |x| ≠ 0 ⇒ x ≠ ± 1.
Hence, 0, 1, –1 are three points of discontinuity.

Test: BITSAT Past Year Paper- 2016 - Question 127

If then f(x) is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 127

We have,


Since  Lf'(0) = Rf'(0), therefore f(x)  is differentiable at x = 0
Since differentiability ⇒ continuity, therefore f(x) is continuous at x = 0.

Test: BITSAT Past Year Paper- 2016 - Question 128

For any differentiable function y of x,

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 128


Test: BITSAT Past Year Paper- 2016 - Question 129

The set of all values of a for which the function f(x) = (a2 – 3a + 2) (cos2x/4 –sin2x/4) + (a –1) x + sin 1 does not possess critical points is

Test: BITSAT Past Year Paper- 2016 - Question 130

Match List I with List II and select the correct answer using the code given below the lists:

Codes:

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 130

(A) Graph of f(x) = cos x cuts x-axis at infinite number of points. (5 of list II)
(B) Graph of f(x) = In x cuts x-axis in only one point. (4 of list II)
(C) Graph of f(x) = x2 – 5x + 4 cuts x axis in two points (2 of list II)
(D) Graph of f(x) = ex cuts y-axis in only one point. (3 of list II)

Test: BITSAT Past Year Paper- 2016 - Question 131

What is the x-coordinate of the point on the curve  f (x) = √x (7x –  6), where the tangent is parallel to x-axis?

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 131

f(x) = √x (7x – 6) = 7x3/2 – 6x1/2

When tangent is parallel to x axis f '(x) = 0

Test: BITSAT Past Year Paper- 2016 - Question 132

A wire 34 cm long is to be bent in the form of a quadrilateral of which each angle is 90°. What is the maximum area which can be enclosed inside the quadrilateral?

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 132

Let one side of quadrilateral be x an d another side be y
so, 2(x + y) = 34
or, (x + y) = 17 ...(i)
We know from the basic principle that for a given perimeter square has the maximum area, so, x = y and putting this value in equation (i)
x = y = 17/2
Area = x . y = 

Test: BITSAT Past Year Paper- 2016 - Question 133

Consider the following statements in respect of the function f (x) = x3 – 1, x ∈ [–1, 1]
I. f (x) is increasing in [– 1, 1]
II. f (x) has no root in (– 1, 1).
Which of the statements given above is/are correct?

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 133

Since f (x) is an increasin g function in  [– 1, 1] and it has a root in (– 1, 1).
∴ Only statement I is correct.

Test: BITSAT Past Year Paper- 2016 - Question 134

At an extreme point of a function f (x), the tangent to the curve is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 134

At an extreme point of a function f (x), slope is always zero.
Thus, At an extreme point of a function f (x), the tangent to the curve is parallel to the x-axis.

Test: BITSAT Past Year Paper- 2016 - Question 135

The curve y = xex has minimum value equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 135

Let y = xex.
Differentiate both side w.r.t. ‘x’.

Hence, y = xex is minimum function and ymin =

Test: BITSAT Past Year Paper- 2016 - Question 136

A ray of light coming from the point (1, 2) is reflected at a point A on the x-axis and then passes through the point (5, 3). The co-ordinates of the point A is

Test: BITSAT Past Year Paper- 2016 - Question 137

The equation x2 - 2√3xy + 3y2 - 3x + 3√3y - 4 = 0  represents

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 137

We have


Thus abc + 2fgh - af2 - bg2 - ch2 = 0
Hence the equation represents a pair of straight lines.

∴ the lines are parallel. The distance between them

Test: BITSAT Past Year Paper- 2016 - Question 138

The line joining (5, 0) to ( (10 cos θ, 10 sin θ) is divided internally in the ratio 2 : 3 at P. If θ varies, then the locus of P is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 138

Let P(x, y) be the point dividing the join of A and B in the ratio 2 : 3 internally, then

Squaring and adding (i) and (ii), we get the required locus ( x - 3)2 + y2 = 16, which is a circle.

Test: BITSAT Past Year Paper- 2016 - Question 139

The number of integral values of λ for which x2 + y2 + λx + (1 – λ)y + 5 = 0 is the equation of a circle whose radius cannot exceed 5, is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 139

Radius ≤ 5

∴ λ = -7, - 6, - 5, .......,7, 8 , in all 16 values

Test: BITSAT Past Year Paper- 2016 - Question 140

The lengths of the tangent drawn from any point on the circle 15x2 + 15y2 - 48x + 64y = 0 to the two circles 5x2 + 5y2 – 24x + 32y + 75 = 0 and 5x2 + 5y2 – 48x + 64y + 300 = 0 are in the ratio of

Test: BITSAT Past Year Paper- 2016 - Question 141

The length of the chord x + y = 3 intercepted by the circle x2 + y2 - 2x - 2y - 2 = 0 is

Test: BITSAT Past Year Paper- 2016 - Question 142

The locus of the point of intersection of two tangents to the parabola y2 = 4ax, which are at right angle to one another is

Test: BITSAT Past Year Paper- 2016 - Question 143

The parabola having its focus at (3, 2) and directrix along the y-axis has its vertex at

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 143

Vertex of the parabola is a point which lies on the axis of the parabola, which is a line ⊥ to the directrix through the focus, i.e., y = 2 and equidistant from the focus and directrix x = 0, so that the vertex is  

Test: BITSAT Past Year Paper- 2016 - Question 144

The number of values of r satisfying the equation
39C3r-1 - 39Cr2 = 39Cr2-1 - 39C3r is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 144

39C3r-1 - 39Cr2 = 39Cr2-1 - 39C3r
⇒ 39C3r-1 + 39C3r = 39Cr2-1 + 39Cr2
40C3r = 40Cr
⇒ r2 = 3r or r2 = 40 - 3r
⇒ r = 0, 3 or – 8, 5
3 and 5 are the values as the given equation is not defined by r = 0 and r = –8. Hence, the number of values of r  is 2.

Test: BITSAT Past Year Paper- 2016 - Question 145

If , then n =

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 145




Comparing we get n + 1 = 6 ⇒ n = 5

Test: BITSAT Past Year Paper- 2016 - Question 146

All the words that can be formed using alphabets A, H, L, U and R are written as in a dictionary (no alphabet is repeated). Rank of the word RAHUL is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 146

No. of words starting with A are 4 ! = 24
No. of words starting with H are 4 ! = 24
No. of words starting with L are 4 ! = 24
These account for 72 words
Next word is RAHLU and the 74th word RAHUL.

Test: BITSAT Past Year Paper- 2016 - Question 147

If the sum of odd numbered terms and the sum of even numbered terms in the expansion of (x + a)n are A and B respectively, then the value of (x2 – a2)n is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 147


= A + B ....(1)
Similarly,  (x – a)n = A – B .....(2)
Multiplying eqns. (1) and (2), we get
(x2 - a2 )n = A2 - B2

Test: BITSAT Past Year Paper- 2016 - Question 148

If the third term in the expansion of [ x + x log10 x ]5 is 106, then x may be

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 148

Put log 10 x = y, th e given expr ession becomes (x + xy)5.
T3 = 5C2 . x3 ( xy)2 = 10x 3+2y = 106 (given)
⇒ (3 + 2y) log10 x = 5 log10 10 = 5
⇒ (3 + 2y)y = 5

∴ x = 10 or x = (10)–5/2

Test: BITSAT Past Year Paper- 2016 - Question 149

If three vertices of a regular hexagon are chosen at random, then the chance that they form an equilateral triangle is:

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 149

Three vertices can be selected in 6C3 ways.

The only equilateral triangles possible are A1A3A5 and A2A4A

Test: BITSAT Past Year Paper- 2016 - Question 150

A man takes a step forward with probability 0.4 and backward with probability 0.6. The probability that at the end of eleven steps he is one step away from the starting point is

Detailed Solution for Test: BITSAT Past Year Paper- 2016 - Question 150

As 0.4 + 0.6 = 1, the man either takes a step forward or a step backward. Let a step forward be a success and a step backward be a failure.
Then, the probability of success in one step 
The probability of failure in one step

In 11 steps he will be one step away from the starting point if the numbers of successes and failures differ by 1.
So, the number of successes = 6
The number of failures = 5 or the number of successes = 5,
The number of failures = 6
∴ the required probability

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