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Test: BITSAT Past Year Paper- 2017 - Question 1

What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of  2R?

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 1

As we know,
Gravitational potential energy 

and orbital velocity





Therefore minimum required energy,

Test: BITSAT Past Year Paper- 2017 - Question 2

A mercury drop of radius 1 cm is sprayed into 106 drops of equal size. The energy expressed in joule is (surface tension of Mercury is 460 × 10–3 N/m)

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 2

W = TΔA = 4pR2T(n1/3 – 1)
= 4 × 3.14 × (10-2)2 × 460 × 10–3 [(106)1/3 –1]
= 0.057

Test: BITSAT Past Year Paper- 2017 - Question 3

Two plano-concave lenses (1 and 2) of glass of refractive index 1.5 have radii of curvature 25 cm and 20 cm. They are placed in contact with their curved surface towards each other and the space between them is filled with liquid of refractive index 4/3. Then the combination is

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 3



Now

Test: BITSAT Past Year Paper- 2017 - Question 4

A charged particle moves through a magnetic field perpendicular to its direction. Then

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 4

When a charged particle enters a magnetic field at a direction perpendicular to the direction of motion, the path of the motion is circular. In circular motion the direction of velocity changes at every point (the magnitude remains constant).

Therefore, the tangential momentum will change at every point. But kinetic energy will remain constant as it is given by 1/2mv2  and v2 is the square of the magnitude of velocity which does not change.

Test: BITSAT Past Year Paper- 2017 - Question 5

After two hours, one-sixteenth of the star ting amount of a certain radioactive isotope remained undecayed. The half life of the isotope is

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 5



Half life

Test: BITSAT Past Year Paper- 2017 - Question 6

A coil of inductance 300 mH and resistance 2Ω is connected to a source of voltage 2 V. The current reaches half of its steady state value in

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 6

The charging of inductance given by,


Taking log on both the sides,

Test: BITSAT Past Year Paper- 2017 - Question 7

Two concentric conducting thin spherical shells A, and B having radii rA and rB ((rB > rA) are charged to QA and –QB (|QB| > |QA|). The electric field along a line passing through the centre is

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 7

For, r < rA, E = 0



These values are correctly represent in option (a).

Test: BITSAT Past Year Paper- 2017 - Question 8

A capillary tube of radius R is immersed in water and water rises in it to a height H. Mass of water in the capillary tube is M. If the radius of the tube is doubled, mass of water that will rise in the capillary tube will now be :

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 8


and

Test: BITSAT Past Year Paper- 2017 - Question 9

A sonometer wire resonates with a given tuning fork forming standing waves with five antinodes between the two bridges when a mass of 9 kg is suspended from the wire. When this mass is replaced by a mass M,  the wire resonates with the same tuning fork forming three antinodes for the same positions of the bridges. The value of M is

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 9

From above equations, we get M = 25 kg.

Test: BITSAT Past Year Paper- 2017 - Question 10

When a metal surface is illuminated by light of wavelengths 400 nm and 250 nm, the maximum velocities of the photoelectrons ejected are v and 2v respectively. The work function of the metal is (h - Planck's constant, c = velocity of light in air)

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 10

We have,
E = W0 + K

On simplifying above equations, we get
W = 2 hc x 106 J

Test: BITSAT Past Year Paper- 2017 - Question 11

Two conductin g shells of radius a an d b are connected by conducting wire as shown in  figure.
The capacity of system is :

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 11

V = 0, and so 

Test: BITSAT Past Year Paper- 2017 - Question 12

When 92U235 under goes fission , 0.1% of its original mass is changed into energy. How much energy is released if 1 kg of 92U235 undergoes fission

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 12

Mass of uranium changed into energy

The energy released = mC2
= 10–3 x (3 x 108)2
= 9 x 1013 J.

Test: BITSAT Past Year Paper- 2017 - Question 13

One mole of an ideal gas is taken from state A to state B by three different processes,

(i) ACB  (ii) ADB (iii) AEB as shown in the P-V diagram. The heat absorbed by the gas is

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 13

The change in internal energy DU is same in all process.


Here WACB is positive and WAEB is negative.
Hence QACB > QADB > QAEB.

Test: BITSAT Past Year Paper- 2017 - Question 14

In the formula X = 3 YZ2, X and Z have dimensions of capacitance and magnetic induction respectively. The dimensions of Y in MKSA system are :

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 14

Test: BITSAT Past Year Paper- 2017 - Question 15

Two very long, straight, parallel wires carry steady currents I and -I respectively. The distance between the wires is d. At a certain instant of time, a point charge q is at a point equidistant from the two wires, in the plane of the wires. Its instantaneous velocity v is perpendicular to this plane. The magnitude of the force due to the magnetic field acting on the charge at this instant is

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 15

Net magnetic field due to the wires will be downward as shown below in the figure.
Since angle between  is 180°,

Test: BITSAT Past Year Paper- 2017 - Question 16

Two projectiles A and B thrown with speeds in the ratio  1: √2 acquired the same heights. If A is thrown at an angle of 45° with the horizontal, the angle of projection of B will be

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 16

For projectile A

For projectile B
Maximum height,

As we know, HA = HB




Test: BITSAT Past Year Paper- 2017 - Question 17

​A meter bridge is set up as shown, to determine an unknown resistance ‘X’ using a standard 10 ohm resistor. The galvanometer shows null point when tapping-key is at 52 cm mark. The endcorrections are 1 cm and 2 cm respectively for the ends A and B. The determined value of ‘X’ is

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 17

At Null point

Here ℓ1 = 52 + End correction
= 52 + 1 = 53 cm
2 = 48 + End correction = 48 + 2 = 50 cm

Test: BITSAT Past Year Paper- 2017 - Question 18

A dis k of r ad ius a/4 having a uniformly distributed charge 6 C is placed in the x - y plane with its centre at (–a / 2, 0, 0). A rod of length a carrying a uniformly distributed charge 8C is placed on the x-axis from x = a /4 to x = 5a / 4.
Two point charges – 7 C and 3 C are placed at (a/4, – a/4, 0) and (– 3a/4, 3a/4, 0), respectively. Consider a cubical surface formed by six surfaces x = ± a/2, y = ± a/2, z = ± a/2. The electric flux through this cubical surface is

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 18

Total flux through the cubical surface,

Test: BITSAT Past Year Paper- 2017 - Question 19

A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving in the y direction with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 19


Initial momentum of the system

Final momentum of the system = 3mV
By the law of conservation of momentum
2√2mv = 3mV

Loss in energy

Test: BITSAT Past Year Paper- 2017 - Question 20

A coil is suspended in a uniform magnetic field, with the plane of the coil parallel to the magnetic lines of force. When a current is passed through the coil it starts oscillating; It is very difficult to stop. But if an aluminium plate is placed near to the coil, it stops. This is due to :

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 20

Because of the Lenz's law of conservation
of energy.

Test: BITSAT Past Year Paper- 2017 - Question 21

A steel wire of length ‘L’ at 40°C is suspended from the ceiling and then a mass ‘m’ is hung from its free end. The wire is cooled down from 40°C to 30°C to regain its original length ‘L’. The coefficient of linear thermal expansion of the steel is 10–5/°C, Young’s modulus of steel is 1011 N/ m2 and radius of the wire is 1 mm. Assume that L >> diameter of the wire. Then the value of ‘m’ in kg is nearly

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 21

We know that

From (1) and (2)

Test: BITSAT Past Year Paper- 2017 - Question 22

On a hypotenuse of a right prism (30° – 60° – 90°) of refractive index 1.50, a drop of liquid is placed as shown in figure. Light is allowed to fall normally on the short face of the prism. In order that the ray of light may get totally reflected, the maximum value of refractive index is :

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 22

Cmax = 60°

Test: BITSAT Past Year Paper- 2017 - Question 23

A tuning fork of frequency 392 Hz, resonates with 50 cm length of a string under tension (T). If length of the string is decreased by 2%, keeping the tension constant, the number of beats heard when the string and the tuning fork made to vibrate simultaneously is :

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 23

The frequency of tuning fork, f = 392 Hz.
Also

After decreasing the length by 2%, we have

From above equations,
f' = 400 Hz.
∴ Beats frequency=8 Hz.

Test: BITSAT Past Year Paper- 2017 - Question 24

Hydrogen (H), deuterium (D), sin gly ionized helium (He+) and doubly ionized lithium (Li++) all have one electron around the nucleus. Consider n = 2 to n = 1 transition. The wavelengths of emitted radiations are λ1, λ2, λ3 and λ4 respectively.Then approximately  :

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 24

Z1 = 1, Z2 = 1, Z3 = 2 and Z4 = 3.

or λZ2 =constant
So 
λ1(1)2 = λ2 (1)2 = λ3(2)2 = λa(32 )
or λ1 = λ2 = 4λ3 = 9λ4.

Test: BITSAT Past Year Paper- 2017 - Question 25

The following figure depict a circular motion. The radius of the circle, the period of revolution, the initial position and the sense of revolution are indicated on the figure.

The simple harmonic motion of the x-projection of the radius vector of the rotating particle P can be shown as :

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 25

Test: BITSAT Past Year Paper- 2017 - Question 26

There are two sources kept at distances 2λ. A large screen is perpendicular to line joining the sources. Number of maximas on the screen in this case is (λ = wavelength of light)

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 26

Δxmax = 0 and Δxmax = 2λ
Theortical maximas are = 2n + 1 = 2 × 2 + 1 = 5
But on the screen there will be three maximas.

Test: BITSAT Past Year Paper- 2017 - Question 27

In the circuit shown in figure the current through

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 27

The net resistance of the circuit is 9Ω as shown in the following figures.

The flow of current in the circuit is as follows.

The current divides into two equal parts if passes through two equal resistances in parallel.
Thus current through 4Ω resistor is 0.25 A.

Test: BITSAT Past Year Paper- 2017 - Question 28

A telescope has an objective lens of 10 cm diameter and is situated at a distance of one kilometer from two objects. The minimum distance between these two objects, which can be resolved by the telescope, when the mean wavelength of light is 5000 Å, is of the order of

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 28


x is of the order of 5 mm.

Test: BITSAT Past Year Paper- 2017 - Question 29

During vapourisation
I. change of state from liquid to vapour state occurs.
II. temperature remains constant.
III. both liquid and vapour states coexist in equilibrium.
IV. specific heat of substance increases.

Correct statements are

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 29

The change of state from liquid to vapour (for gas) is called vapourisation. It is observed that when liquid is heated, the temperature remains constant untill the entire amount of the liquid is converted into vapour.
The temperature at which the liquid and the vapour states of the substance coexists is called its boiling point.

Test: BITSAT Past Year Paper- 2017 - Question 30

A wire is connected to a battery between the point M and N as shown in the figure (1). The same wire is bent in the form of a square and then connected to the battery between the points M and N as shown in the figure (2). Which of the following quantities increases ?

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 30

When the wire is bent in the form of a square and connected between M and N as shown in fig. (2), the effective resistance between M and N decreases to one fourth of the value in fig. (1). The current increases four times the initial value according to the relation  V = IR. Since H = I2Rt, the decrease in the value of resistance is more than compensated by the increases in the value of current. Hence heat produced increases.
Percentage loss in energy during the collision ; 56%

Test: BITSAT Past Year Paper- 2017 - Question 31

A body moves in a circular orbit of radius R under the action of a central force. Potential due to the central force is given by V(r) = kr (k is a positive constant). Period of revolution of the body is proportional to :

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 31

U = mV = kmr.

Test: BITSAT Past Year Paper- 2017 - Question 32

Two equal heavy spheres, each of radius r, are in equilibrium within a smooth cup of radius 3r. The ratio of reaction between the cup and one sphere and that between the two sphere is

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 32


sin θ = 1/2
Thus, N1 sin θ =N2

Test: BITSAT Past Year Paper- 2017 - Question 33

A long, hollow conducting cylinder is kept coaxially inside another long, hollow conducting cylinder of larger radius. Both the cylinders are initially electrically neutral

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 33

When charge is given to inner cylinder, an electric field will be in between the cylinders.
So there is potential difference between the cylinders.

Test: BITSAT Past Year Paper- 2017 - Question 34

A thin but rigid semicircular wire frame of radius r is hinged at O and can rotate in its own vertical plane. A smooth peg P starts from O and moves horizontally with constant speed v0, lifting the frame upward as shown in figure.

Find the angular velocity w of the frame when its diameter makes an angle of 60° with the vertical :

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 34

⇒ x  =2r sin θ

Test: BITSAT Past Year Paper- 2017 - Question 35

Given that A + B = R and A = B = R. What should be the angle between A and B ?

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 35

R2 = [A2 + B2 + 2AB cos θ]
R2 = R2 + R2 + 2R2 cos θ
- R2 = 2 R2 cos θ or cos θ = -1 / 2
or  θ = 2 π/3

Test: BITSAT Past Year Paper- 2017 - Question 36

The basic magnetization curve for a ferromagnetic material is shown in figure. Then, the value of relative permeability is highest for the point

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 36

According to the given graph, slope of the graph is highest at point Q.

Test: BITSAT Past Year Paper- 2017 - Question 37

Five gas molecules chosen at random are found to have speeds of 500, 600, 700, 800 and 900 m/s:

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 37


and

Thus vrms is greater than average speed by 14 m/s.

Test: BITSAT Past Year Paper- 2017 - Question 38

What is equivalent capacitance of circuit between points A and B?

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 38

The effective circuit is shown in figure.

The capacitance of upper series,

Now

Test: BITSAT Past Year Paper- 2017 - Question 39

A cyclic process ABCD is shown in the figure PV diagram. Which of the following curves represent the same process

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 39

Proce ss AB is is ob as i c and BC is isothermal, CD isochoric and DA isothermic compression.

Test: BITSAT Past Year Paper- 2017 - Question 40

In the circuit given below, V(t) is the sinusoidal voltage source, voltage drop VAB(t) across the resistance R is

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 40

During the operation, either of D1 and D2 be in forward bias. Also R1 and R2 are different, so output across R will have different peaks.

Test: BITSAT Past Year Paper- 2017 - Question 41

Which of the following can be repeatedly soften on heating?
(i) Polystyrene
(ii) Melamine
(iii) Polyesters
(iv) Polyethylene
(v) Neoprene

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 41

Polystyrene and polyethylene belong to the category of thermoplastic polymers which are capable of repeatedly softening on heating and harden on cooling.

Test: BITSAT Past Year Paper- 2017 - Question 42

Which one of the following complexes is an outer orbital complex ?

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 42

Hybridisation :

Hence [Ni(NH3)6]2+ is outer or bital complex.

Test: BITSAT Past Year Paper- 2017 - Question 43

For the reaction H2(g) + Br2 (g) → 2HBr (g), the experimental data suggest, rate = k[H2][Br2]1/2.
The molecularity and order of the reaction are respectively

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 43

The order of reaction is 3/2 and molecularity
is 2.

Test: BITSAT Past Year Paper- 2017 - Question 44

Dead burn plaster is

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 44

CaSO4

Test: BITSAT Past Year Paper- 2017 - Question 45

Stronger is oxidising agent, more is

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 45

More is  more is the tendency to get itself reduced or more is oxidising power.

Test: BITSAT Past Year Paper- 2017 - Question 46

Which of the following relation represents correct relation between standard electrode potential and equilibrium constant?

Choose the correct statement(s)

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 46

ΔG = –2.303 RT log K
-nFE° = –2.303 RT log K

Test: BITSAT Past Year Paper- 2017 - Question 47

Which of the following shows nitrogen with its increasing order of oxidation number?

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 47

Therefore increasing order of oxidation state of N is:
NH4+ < N2O < NO < NO2 < NO3

Test: BITSAT Past Year Paper- 2017 - Question 48

Raoult’s law becomes a special case of Henry’s law when

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 48

Raoult’s law becomes special case of Henry’s law when KH become equal to p1°.

Test: BITSAT Past Year Paper- 2017 - Question 49

E° for the cell, Zn | Zn2+ (aq) | | Cu2+ (aq) | Cu is 1.10 V at 25°C. The equilibrium constant for the cell reaction

is of the order of

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 49


∴ Kc = 1.9 x 1037

Test: BITSAT Past Year Paper- 2017 - Question 50

Which of the following represents Gay Lussac's law ?
I. P/T = constant
II. P1T2 = P2T1
III. P1V1 = P2V2

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 50

P/T = constant (Gay Lussac's law)

PV = constant
P1V1 = P2V[Boyle's law]

Test: BITSAT Past Year Paper- 2017 - Question 51

For the reaction
CO(g) + 1/2O2 (g) → CO2 (g)
Which one of the statement is correct at constant T and P ?

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 51

Test: BITSAT Past Year Paper- 2017 - Question 52

The energy of an electron in second Bohr orbit of hydrogen atom is :

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 52


For second orbit, n = 2
Z = At. no. = 1 (for hydrogen)

Test: BITSAT Past Year Paper- 2017 - Question 53

Which of the following order is wrong?

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 53

The right sequence of I.E1 of Li < B < Be < C.

Test: BITSAT Past Year Paper- 2017 - Question 54

Which of the following is not in volved in the formation of photochemical smog?

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 54

Photochemical smog does not involve SO2.

Test: BITSAT Past Year Paper- 2017 - Question 55

Which of the following is not present in Portland cement?

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 55

There are four chief miner als present in a Portland cement tricalcium silicate (Ca3SiO5), dicalcium silicate (Ca2SiO4), tricalcium aluminate (Ca3Al2O6) and calcium aluminoferrite (Ca4AlnFe2-nO7).

Test: BITSAT Past Year Paper- 2017 - Question 56

Which of the following can form buffer solution?

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 56

Ammonia is a weak base and a salt containing its conjugate acid, the ammonium cation, such as NH4OH functions as a buffer solution when they are present together in a solution.

Test: BITSAT Past Year Paper- 2017 - Question 57

Which of the following complex shows sp3d2 hybridization?

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 57

Among these ligands, ‘F’ is a weak field ligand, makes only high spin complexes which has sp3d2 hybridization.

Test: BITSAT Past Year Paper- 2017 - Question 58

Which has glycosidic linkage?

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 58

Glycosidic linkage is a type of covalent bond that joins either two carbohydrate (sugar) molecule or one carbohydrate to another group. All molecules show such type of linkages.

Test: BITSAT Past Year Paper- 2017 - Question 59

Which of the following represents Schotten Baumann reaction?

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 59

Schotten-Baumann Conditions

The use of added base to drive the equilibrium in the formation of amides from amines and acid chlorides.

Test: BITSAT Past Year Paper- 2017 - Question 60

In the following structures, which two forms are staggered conformations of ethane ?

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 60

Note that in structures 1 and 2, every two adjacent hydrogen atoms are at maximum possible distance from each other (staggered conformation).

Test: BITSAT Past Year Paper- 2017 - Question 61

Which of the following shows correct order of bond length?

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 61

Bond length decreases with an increase in bond order. Therefore, the order of bond length in these   species is O2+ < O2 > O2 < O22- (bond order - O2+ = 2.5, O2 =2, O2 =1.5, O22– =1)

Test: BITSAT Past Year Paper- 2017 - Question 62

The number of radial nodes of 3s and 2p orbitals are respectively

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 62

For a given orbital with principal quan tum number (n) and azimuthal quantum number (l)
number of radial nodes = (n – l – 1)  for 3s orbital: n = 3 and l = 0 therefore, number of radial nodes= 3 – 0 – 1 = 2
for 2p orbital: n = 2 and l = 1
therefore, number of radial nodes = 2 – 1 – 1 = 0

Test: BITSAT Past Year Paper- 2017 - Question 63

If a 25.0 mL sample of sulfuric acid is titrated with 50.0 mL of 0.025 M sodium hydroxide to a phenolphthalein endpoint, what is the molarity of the acid?

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 63

M1V1 = M2V2
(0.025 M) (0.050 L) = (M2) (0.025 L)
M2 = 0.05 M
but, there are 2 H’s per H2SO4 so [H2SO4]  = 0.025 M

Test: BITSAT Past Year Paper- 2017 - Question 64

Find which of the following compound can have mass ratios of C:H:O as 6:1:24

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 64

Given, mass ratio is C:H:O (6:1:24) so, molar ratio will be 6/12:1/1:24/16 = 1:2:3 therefore, HO-(C=O)-OH has molar ratio 1:2:3

Test: BITSAT Past Year Paper- 2017 - Question 65

The number of atoms per unit cell of bcc structure is

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 65

In bcc structure,
no. of atoms at corner = 1/8 × 8 = 1
no. of atom at body centre = 1
therefore, total no of atom per unit cell = 2.

Test: BITSAT Past Year Paper- 2017 - Question 66

Which of these doesn’t exist?

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 66

PH5 does not exist because d-or bital of ‘P’ interacts with s-orbital of H. Bond formed is not stable and not energetically favorable. It depends on size and orientation of interaction.

Test: BITSAT Past Year Paper- 2017 - Question 67

Which of these compounds are directional?

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 67

Ion ic bonding is non directional, wher eas covalent bonding is directional. So, CO2 is directional.

Test: BITSAT Past Year Paper- 2017 - Question 68

For a given reaction, ΔH = 35.5 kJ mol-1  and ΔS = 83.6 JK-1 mol-1. The reaction is spontaneous at : (Assume that ΔH and ΔS do not vary with tempearature)

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 68

Given ΔH  35.5 kJ mol–1 ΔS = 83.6 JK–1 mol–1
∴ ΔG = ΔH – TΔS
For a reaction to be spontaneous, ΔG = –ve
i.e., ΔH < TΔS

So, the given reaction will be spontaneous at T > 425 K

Test: BITSAT Past Year Paper- 2017 - Question 69

Specific conductance of 0.1 M HA is 3.75 x 10–4 ohm–1 cm–1. If λ (HA) = 250 ohm–1 cm2 mol–1, the dissociation constant Ka of HA is :

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 69

Test: BITSAT Past Year Paper- 2017 - Question 70

The rate of reaction between two reactants A and B decreases by a factor of 4 if the concentration of reactant B is doubled. The order of this reaction with respect to reactant B is:

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 70

Test: BITSAT Past Year Paper- 2017 - Question 71

A compound of molecular formula of C7H16 shows optical isomerism, compound will be

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 71

A compound is said to exhibit optical isomerism if it atleast contains one chiral carbon atom, which is an atom bonded to 4 different atoms or groups.

Test: BITSAT Past Year Paper- 2017 - Question 72

Which of the following does not contain Plane of symmetry?

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 72

Meso compounds ar e ch aracterized by an internal plane of symmetry that renders them achiral.

Test: BITSAT Past Year Paper- 2017 - Question 73

Cadmium is used in nuclear reactors for?

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 73

Control rods slowdown the motion of neutrons and help in controlling the rate of fission. Cadmium is efficient for this purpose.

Test: BITSAT Past Year Paper- 2017 - Question 74

Which reagent converts nitrobenzene to Nphenyl hydroxyamine?

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 74

Reducing reagent is needed, as shown in given reaction.

Test: BITSAT Past Year Paper- 2017 - Question 75

Which of the following can act as both Bronsted acid and Bronsted base?

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 75

Test: BITSAT Past Year Paper- 2017 - Question 76

Identify the structure of water in the gaseous phase.

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 76

Test: BITSAT Past Year Paper- 2017 - Question 77

Electrometallurgical process is used to extract

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 77

Because Na is very reactive and cannot be extracted by means of the reduction by C, CO etc. So it is extracted by electrolysis.

Test: BITSAT Past Year Paper- 2017 - Question 78

The correct statement about the compounds A, B, and C

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 78

Rotation of B through 180º within the plane of the paper gives D which is an enantiomer of A, hence A and B are enantiomers


Test: BITSAT Past Year Paper- 2017 - Question 79

Correct formula of the complex formed in the brown ring test for nitrates is

Test: BITSAT Past Year Paper- 2017 - Question 80

Which one of the following is an amine hormone ?

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 80

Thyroxine is an amine hormone.

Test: BITSAT Past Year Paper- 2017 - Question 81

choose the one which best expresses the meaning of the given word.

Loquacious

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 81

The word Loquacious (Adjective) means : talking a lot; talkative. Option (a) is the right synonym while others have different meanings.

Test: BITSAT Past Year Paper- 2017 - Question 82

Choose the word opposite in meaning to the given word.

Meticulous

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 82

The word Meticulous (Adjective) means : paying careful attention to every detail; fastidious; thorough.
Careless in option (c) is the correct antonomy.

Test: BITSAT Past Year Paper- 2017 - Question 83

Direction: Read the passage carefully and choose the best answer to each question out of the four alternatives.

PASSAGE
To write well you have to be able to write clearly and logically, and you cannot do this unless you can think clearly and logically. If you cannot do this yet you should train yourself to do it by taking particular problems and following them through, point by point, to a solution, without leaving anything out and without avoiding any difficulties that you meet.

At first you find clear, step-by-step thought very difficult. You may find that your mind is not able to concentrate. Several unconnected ideas may occur together. But practice will improve your ability to concentrate on a single idea and think about it clearly and logically. In order to increase your vocabulary and to improve your style, you should read widely and use a good dictionary to help you find the exact meanings and correct usages of words.

Always remember that regular and frequent practice is necessary if you want to learn to write well.

It is no good waiting until you have an inspiration before you write. Even with the most famous writers, inspiration is rare. Someone said that writing is ninetynine percent hard work and one percent inspiration, so the sooner you get into the habit of disciplining your-self to write, the better.

To write well, a person must train himself in

Test: BITSAT Past Year Paper- 2017 - Question 84

Direction: Read the passage carefully and choose the best answer to each question out of the four alternatives.

PASSAGE
To write well you have to be able to write clearly and logically, and you cannot do this unless you can think clearly and logically. If you cannot do this yet you should train yourself to do it by taking particular problems and following them through, point by point, to a solution, without leaving anything out and without avoiding any difficulties that you meet.

At first you find clear, step-by-step thought very difficult. You may find that your mind is not able to concentrate. Several unconnected ideas may occur together. But practice will improve your ability to concentrate on a single idea and think about it clearly and logically. In order to increase your vocabulary and to improve your style, you should read widely and use a good dictionary to help you find the exact meanings and correct usages of words.

Always remember that regular and frequent practice is necessary if you want to learn to write well.

It is no good waiting until you have an inspiration before you write. Even with the most famous writers, inspiration is rare. Someone said that writing is ninetynine percent hard work and one percent inspiration, so the sooner you get into the habit of disciplining your-self to write, the better.

Initially it is difficult to write because

Test: BITSAT Past Year Paper- 2017 - Question 85

Direction: Read the passage carefully and choose the best answer to each question out of the four alternatives.

PASSAGE
To write well you have to be able to write clearly and logically, and you cannot do this unless you can think clearly and logically. If you cannot do this yet you should train yourself to do it by taking particular problems and following them through, point by point, to a solution, without leaving anything out and without avoiding any difficulties that you meet.

At first you find clear, step-by-step thought very difficult. You may find that your mind is not able to concentrate. Several unconnected ideas may occur together. But practice will improve your ability to concentrate on a single idea and think about it clearly and logically. In order to increase your vocabulary and to improve your style, you should read widely and use a good dictionary to help you find the exact meanings and correct usages of words.

Always remember that regular and frequent practice is necessary if you want to learn to write well.

It is no good waiting until you have an inspiration before you write. Even with the most famous writers, inspiration is rare. Someone said that writing is ninetynine percent hard work and one percent inspiration, so the sooner you get into the habit of disciplining your-self to write, the better.

According to the passage, writing style can be improved by

Test: BITSAT Past Year Paper- 2017 - Question 86

Direction: Read the passage carefully and choose the best answer to each question out of the four alternatives.

PASSAGE
To write well you have to be able to write clearly and logically, and you cannot do this unless you can think clearly and logically. If you cannot do this yet you should train yourself to do it by taking particular problems and following them through, point by point, to a solution, without leaving anything out and without avoiding any difficulties that you meet.

At first you find clear, step-by-step thought very difficult. You may find that your mind is not able to concentrate. Several unconnected ideas may occur together. But practice will improve your ability to concentrate on a single idea and think about it clearly and logically. In order to increase your vocabulary and to improve your style, you should read widely and use a good dictionary to help you find the exact meanings and correct usages of words.

Always remember that regular and frequent practice is necessary if you want to learn to write well.

It is no good waiting until you have an inspiration before you write. Even with the most famous writers, inspiration is rare. Someone said that writing is ninetynine percent hard work and one percent inspiration, so the sooner you get into the habit of disciplining your-self to write, the better.

Famous writers have achieved success by

Test: BITSAT Past Year Paper- 2017 - Question 87

DIRECTIONS: In question below, sentences are given with blanks to be filled in with an appropriate word (s). Four alternatives are suggested for each question. Choose the correct alternative out of the four.

China is a big country, in area it is bigger than any other country _________ Russia.

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 87

China is a big country. In area it is bigger than any other country except Russia. [except means other than, accept means consent, expect means to anticipate and access means entrance].

Test: BITSAT Past Year Paper- 2017 - Question 88

DIRECTIONS: In question below, sentences are given with blanks to be filled in with an appropriate word (s). Four alternatives are suggested for each question. Choose the correct alternative out of the four.

The treasure was hidden ______ a big shore.

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 88

The treasure was hidden off the shore.
When something is hidden "off the shore," it just means that it's hidden somewhere near it.

Test: BITSAT Past Year Paper- 2017 - Question 89

DIRECTIONS: In question, some parts of the sentences have errors and some are correct.
Find  out which part of a sentence has an error. If a sentence is free from error, mark (d) in your Answer.

My father gave me (a) / a pair of binocular (b) / on my birthday. (c) / No error. (d)

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 89

Delete 'pair of' before binocular because the word 'binocular' itself suggests a pair.

Test: BITSAT Past Year Paper- 2017 - Question 90

DIRECTIONS: In question, some parts of the sentences have errors and some are correct.
Find  out which part of a sentence has an error. If a sentence is free from error, mark (d) in your Answer.

The teacher as well as his students, (a) / all left (b) / for the trip. (c) / No error. (d)

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 90

Delete 'all' before 'left'. Here the usage of 'all' is superfluous as 'the teacher as well as his students' itself signifies everyone.

Test: BITSAT Past Year Paper- 2017 - Question 91

Which answer figure complete the form in question figure ?

Test: BITSAT Past Year Paper- 2017 - Question 92

DIRECTIONS:  In the following question  which answer figure will complete the question figure?


Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 92

Option (d) will complete the question figure.

Test: BITSAT Past Year Paper- 2017 - Question 93

Which answer figure includes all the components given in the question figure ?

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 93

Test: BITSAT Past Year Paper- 2017 - Question 94

Which of the answer figures include the separate components found in the question figure?

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 94

All the components of question figure are present in Answer Figure (c)

Test: BITSAT Past Year Paper- 2017 - Question 95

Which of the answer figures include the separate components found in the question figure?

Question Figure

Answer Figures

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 95

In each row, the second figure is obtained from the first figure by adding two mutually perpendicular line segments at the centre and the third figure is obtained from the first figure by adding four circles outside the main figure.

Test: BITSAT Past Year Paper- 2017 - Question 96

Which of the answer figures include the separate components found in the question figure?

Question Figure



Answer Figures

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 96

In each row, the second figure is obtained by rotating the first figure through 90° CW or 90° ACW and adding a circle to it. Also, the third figure is obtained by adding two circles to the first figure (without rotating the figure).

Test: BITSAT Past Year Paper- 2017 - Question 97

M is the son of P. Q is the grand daughter of O who is the husband of P. How is M related to O?

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 97

O is  the husband  of P. M is the son of P.
Therefore , M is the son of O.

Test: BITSAT Past Year Paper- 2017 - Question 98

Vinod introduces Vishal as the son of the only brother of his father's wife. How is Vinod related to Vishal?

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 98

Wife of Vinod’s father means the mother of Vinod.
Only brother of Vinod’s mother means maternal uncle of Vinod.
Therefore, Vinod is cousin of Vishal.

Test: BITSAT Past Year Paper- 2017 - Question 99

AGMSY, CIOUA, EKQWC,  ?  IOUAG, KQWCI

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 99

The pattern is as follows:

Test: BITSAT Past Year Paper- 2017 - Question 100

(?), PSVYB, EHKNQ, TWZCF, ILORU

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 100

The pattern is as follows :

Therefore, the first term should be

Test: BITSAT Past Year Paper- 2017 - Question 101

In the following question, one statement is given followed by two assumptions I and II. You have to consider the statement to be true even if it seems to be at variance from commonly known facts. You have to decide which of the given assumptions, if any, follow from the given statement.

Statements : Politicians become rich by the votes of the people.
Assumptions :
I. People vote to make politicians rich.
II. Politicians become rich by their virtue.

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 101

The statemen t implies that politicians win elections by the votes of people. Therefore, neither of the assumptions is implicit in the statement.

Test: BITSAT Past Year Paper- 2017 - Question 102

Two statements are given followed by four conclusions, I, II, III and IV. You have to consider the statements to be true, even if they seem to be at variance from commonly known facts. You have to decide which of the given conclusions can definitely be drawn from the given statements. Indicate your answer.
Statements:
(A) No cow is a chair
​(B) All chairs are tables.
Conclusions:
I. Some tables are chairs.
II. Some tables are cows
III. Some chairs are cows
IV. No table is a cow

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 102

Test: BITSAT Past Year Paper- 2017 - Question 103

DIRECTIONS: In question one/two statements are given, followed by two conclusions I and II. You have to consider the statements to be true, even if they seem to be at variance from commonly known facts. You have to decide which of the given conclusions, if any follow from the given statement.

Statements :
1. Temple is a place of worship.
2. Church is also a place of worship.

Conclusions :
I. Hindus and Christians use the same place for worship.
II. All  churches are temples.

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 103

Temple and Church are places of worship. It does not imply that Hindus and Christians use the same place for worship. Church is different temple. Therefore, neither Conclusion I nor II follows.

Test: BITSAT Past Year Paper- 2017 - Question 104

DIRECTIONS: In question one/two statements are given, followed by two conclusions I and II. You have to consider the statements to be true, even if they seem to be at variance from commonly known facts. You have to decide which of the given conclusions, if any follow from the given statement.

Statement : The human organism grows and develops through stimulation and action.

Conclusions:
I. Inert human organism cannot grow and develop.
II. Human organisms do not react to stimulation and action.

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 104

Growth and development of human organism is a continuous process. Some changes take place in human body now and then.
Therefore, neither Conclusion I nor II follows.

Test: BITSAT Past Year Paper- 2017 - Question 105

Choose the set of figure which follows the given rule.
Rule: Closed figures gradually become open and open figures gradually become closed.


Test: BITSAT Past Year Paper- 2017 - Question 106

Let f and g be functions from  R to R defined as

​Then

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 106

We have
g (–3) = 0
⇒ f (g(–3)) = f (0) = 7 (0)2 +0 – 8  = – 8
∴ fog (–3) = –8 g (9) = 92 + 4 = 85
⇒ f (g(9))  = f (85) =  8(85) + 3= 683
∴ fog (9) = 683 f (0) = 7.02 + 0 – 8 = – 8
⇒ g (f (0)) = g (–8) = | –8 | = 8
∴ gof (0) = 8 f (6) = 4(6) +5 = 29
⇒ g (f (6)) = g (29) = (29)2 +  4 = 845
∴ gof (6) = 845

Test: BITSAT Past Year Paper- 2017 - Question 107

How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even positions ?

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 107

X - X - X - X - X. The four digits 3, 3, 5,5 can be arranged at (-) places in 
The five digits 2, 2, 8, 8, 8 can be arranged at (X) places in  
Total no. of ar rangements = 6 x 10 = 60 ways

Test: BITSAT Past Year Paper- 2017 - Question 108

if   where p, q, t and s are constants, then the value of s is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 108



Test: BITSAT Past Year Paper- 2017 - Question 109

The length of the semi-latus rectum of an ellipse is one thrid of its major axis, its eccentricity would be

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 109

Let eq. of ellipse be 
length of semi-latus rectum

Test: BITSAT Past Year Paper- 2017 - Question 110

If α and β are roots of the equation  such that  then p belongs to the set :

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 110

Given quadratic eqn. is

Now, given

Test: BITSAT Past Year Paper- 2017 - Question 111

Given the system of straight lines a(2x + y – 3) + b(3x + 2y – 5) = 0, the line of the system situated farthest from the point (4, –3) has the equation

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 111

The given system of lines passes through the point of intersection of the straight lines 2x + y – 3 = 0 and 3x + 2y – 5 = 0 [L1 + λL2 = 0 form], which is (1, 1).
The required line will also pass through this point. Further, the line will be farthest from point (4, –3) if it is in direction perpendicular to line joining (1, 1) and (4, –3).
∴ The equation of the required line is

Test: BITSAT Past Year Paper- 2017 - Question 112

One mapping is selected at random from all mappings of the set S = {1, 2, 3, ......n} into itself. The probability that it is one-one is 3/32. Then the value of n is 

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 112

Test: BITSAT Past Year Paper- 2017 - Question 113

The integer just greater than (3 + √5)2n is divisible by (n ∈ N)

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 113


Let [R] + 1 = I   ( ∵ [ . ] greatest integer function)
⇒ R + G = I ( ∵ 0 < G < 1)

seeing the option put n = 1
I = 28 is divisible by 4 i.e 2n+1

Test: BITSAT Past Year Paper- 2017 - Question 114

The domain of the function

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 114

For f (x) to be defined, we must have




Form (2) and (3), we get the domain of

Test: BITSAT Past Year Paper- 2017 - Question 115

The marks obtained by 60 students in a certain test are given below:

Median of the above data is

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 115

We construct the followin g table taking assumed mean a = 55 (step deviation method).




Here n = 60 ⇒ n/2 = 30 therefore, 60–70 is
the median class
Using the formula :

Test: BITSAT Past Year Paper- 2017 - Question 116

 If A, B, C are the angles of a triangle and eiA , eiB , eiC ar e in A.P. Then the trian gle must be

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 116


Squaring and adding we get
cos( A - C) = 1 ⇒ A - C = 0
∴ A = C, From (i) and (ii) cos B  = cos A and sin B = sin A
So, A = B.
⇒ A = B = C

Test: BITSAT Past Year Paper- 2017 - Question 117

An observer on the top of a tree, finds the angle of depression of a car moving towards the tree to be 30°. After 3 minutes this angle becomes 60°.After how much more time, the car will reach the tree?

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 117



It will travel distance h cot 60° in

Test: BITSAT Past Year Paper- 2017 - Question 118

After striking the floor a certain ball rebounds 4/5th of  its height from which it has fallen. Thetotal distance that the ball travels before coming to rest if it is gently released from a height of 120m is

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 118

Clearly, the total distance described

Except in the first fall the same ball will travel twice in each step the same distance one upward and second downward travel.

∴ Distance travelled

Test: BITSAT Past Year Paper- 2017 - Question 119

An equilateral triangle is inscribed in the circle x2 + y2 = a2 with one of the vertices at (a, 0). What is the equation of the side opposite to this vertex?

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 119

Since the equilateral triangle is inscribed in the circle with centre at the origin, centroid lies on the origin.

So, other vertices of triangle have coordinates,

∴ Equation of line BC is :

Test: BITSAT Past Year Paper- 2017 - Question 120

The function f(x) = x – | x – x2 |, –1≤ x ≤ 1 is continuous on the interval

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 120

we have,

∴ Continuity is to be checked at x = 0 and  x = 1. At x = 0


and f(0) = 0
Since LHL = RHL = f(0),
∴ f(x) is continuous at x = 0.
At x = 1


∴ f(x) is contin uous at x = 1
Hence f(x) is con tinuous for all x ∈ [-1, 1]

Test: BITSAT Past Year Paper- 2017 - Question 121

If  then P(n) is true for

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 121



which is true.
Let for n = m ≥ 2, P(m) is true.




Hence, for n ≥ 2, P(n) is true.

Test: BITSAT Past Year Paper- 2017 - Question 122

If a system of equation
– ax + y + z = 0
x – by + z = 0 x + y – cz = 0    (a, b, c≠ –1) 
has a non-zero solution then 

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 122

 for non-zero
solution
⇒  abc – a – b – c – 2 = 0
⇒  abc = a + b + c + 2

Test: BITSAT Past Year Paper- 2017 - Question 123

If f (x) = xx, then f (x) is increasing in interval :

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 123

A function f(x) is said to be increasing function in

Differentiate equation (i)
f '(x) = xx (1 + log x)
Put f '(x) = 0
0 = xx (1 + log x)


∴ f (x) is increasing in interval 

Test: BITSAT Past Year Paper- 2017 - Question 124

If x is real number, then  must lie between

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 124


⇒ x2y – (5y + 1)x + 9y = 0 for real x, Discriminant = b2 – 4ac ≥ 0
(5y + 1)2 – 36y2 ≥ 0
⇒ (5y + 1 – 6y) (5y + 1 + 6y) ≥ 0
⇒ (– y + 1) (11y + 1) ≥ 0

Test: BITSAT Past Year Paper- 2017 - Question 125

The value of 

ai > 0, i = 1,  2, ...... n, is

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 125





Test: BITSAT Past Year Paper- 2017 - Question 126

The value of cot–1 7 + cot–1 8 + cot–1 18 is

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 126

We have cot-1 7 + cot-1 8+ cot-1 18





Test: BITSAT Past Year Paper- 2017 - Question 127

If  is equal to :

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 127






Test: BITSAT Past Year Paper- 2017 - Question 128

A random variable X has the probability distribution

For the events E = {X is a prime number} and F = {X < 4} then P(E ∪ F) is

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 128

P(E) = P(2 or 3 or 5 or 7)
= 0.23 + 0.12 + 0.20 + 0.07 = 0.62

P( F ) =P(1 or2 or3)
= 0.15 + 0.23 + 0.12 = 0.50

P(E Ç F) = P(2 or 3)
= 0.23 + 0.12 = 0.35

∴ P(EUF ) = P(E) + P(F ) - P(E∩ F )
= 0.62 + 0.50 - 0.35 = 0.77

Test: BITSAT Past Year Paper- 2017 - Question 129

The number of roots of equation cos x + cos 2x + cos 3x = 0 is (0 ≤ x ≤ 2 π)

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 129

We have cos x + cos 2 x + cos 3x = 0
or (cos 3x + cos x ) + cos 2 x = 0
or 2 cos 2x. cos x + cos 2x = 0
or cos 2x (2 cos x + 1) = 0
We have  either cos 2x = 0 or 2 cos x + 1 = 0



Hence the required general solution are

Test: BITSAT Past Year Paper- 2017 - Question 130

The area under the curve y = |cos x – sin x|,   and above x-axis is :

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 130

y = | cos x – sin x |



Required area 

Test: BITSAT Past Year Paper- 2017 - Question 131

If 
 
then f(x) is 

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 131

we have




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

Test: BITSAT Past Year Paper- 2017 - Question 132

​The maximum value of z = 3x + 2y subject to x + 2y ≥ 2, x + 2y ≤ 8, x, y ≥ 0 is :

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 132


and x, y ≥ 0

For equation (1)

and for equation (2)

Given : z = 3x + 2y
At point (2, 0); z = 3 x 2 + 0 = 6
At point (0, 1); z = 3 x 0 + 2 x 1 = 2
At point (8, 0); z = 3 x 8 + 2 x 0 = 24
At point (0, 4); z = 3 x 0 +2 x 4 = 8
 ∴maximum value of z is 24 at point (8, 0).

Test: BITSAT Past Year Paper- 2017 - Question 133

A cylindircal gas container is closed at the top and open at the bottom. if the iron plate of the top is 5/4 time as thick as the plate forming thecylindrical sides. The ratio of the radius to the height of the cylinder using minimum material for the same capacity is

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 133

V = πr2h = constant. If k be the thickness of the sides then that of the top will be (5/4)k.
∴ S = (2πrh)k + (πr2). (5/4)k
(‘S’ is vol. of material used)



When r3 = 4V/5π or   5πr3 = 4πr2h.
∴ r/h = 5/4

Test: BITSAT Past Year Paper- 2017 - Question 134

Let A, B, C be finite sets. Suppose that n (A) = 10, n (B) = 15, n (C) = 20, n(A ∩ B) = 8 and  n(B ∩ C) = 9. Then the possible value of n (A ∪ B ∪ C) is

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 134

We have
n (A ∪ B ∪ C) = n (A) + n (B) + n (C) – n (A ∩ B) – n(B∩C) – n (C ∩ A) + n (A∩B ∩ C)
= 10 +15 + 20 – 8 – 9 – n  (C ∩ A) + n (A ∩ B ∩C)
= 28 – {n(C ∩ A) – n (A ∩ B ∩ C)}   ...(i)
Since n (C ∩ A) ≥  n (A ∩ B ∩ C)
We have
n (C ∩ A) – n (A ∩ B ∩ C) ≥ 0 ...(ii)
From (i) and (ii)
n (A ∪ B ∪ C) ≤ 28 ...(iii)
Now, n(A ∪ B) = n (A) +n (B) – n (A Ç B)  = 10 + 15 – 8 = 17
and   n (B ∪ C) = n (B) + n (C) – n (B Ç C)  = 15 + 20 – 9 = 26
Since, n (A ∪ B ∪ C) ≥ n (A∪C) and n (A∪B∪C) ≥ n (B∪C),
we have n (A∪B∪C) ≥ 17 and n (A∪B∪C) ≥ 26

Hence n (A∪B∪C) ≥ 26 ...(iv)
From (iii) and (iv)
we obtain 26 ≤ n (A∪B∪C)≤ 28
Also n (A∪B∪C) is a positive integer
∴ n(A∪B∪C) = 26 or 27 or 28

Test: BITSAT Past Year Paper- 2017 - Question 135

If  where z = 1 + 2i , then |f(z)| is equal to :
 

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 135



Test: BITSAT Past Year Paper- 2017 - Question 136

If   then the value of  f'(e) is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 136


Put log x = t  in f (x)

Now, put t = tan θ , we get

= cos–1 [cos 2 θ] =2θ = 2 tan–1 t = 2
tan–1 (log x)
Diff. both side w.r.t 'x', we get

Now,

Test: BITSAT Past Year Paper- 2017 - Question 137

Statement 1: A five digit number divisible by 3 is to be formed using the digits 0, 1, 2, 3, 4 and 5 with repetition. The total number formed are 216.
Statement 2 : If sum of digits of any number is divisible by 3 then the number must be divisible by 3.

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 137

Number form by using 1, 2, 3, 4, 5 = 5! = 120 Number formed by using 0, 1, 2, 4, 5

Total number formed, divisible by 3 (taking numbers without repetition) = 216
Statement 1 is false and statement 2 is true.

Test: BITSAT Past Year Paper- 2017 - Question 138

The equation of one of the common tangents to the parabola y2 = 8x and x2 + y2 - 12x + 4 = 0 is

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 138

Any tangent to parabola y2 = 8x is

It touches the circle x2 + y2 - 12x + 4 = 0 , if the length of perpendicular from the centre (6, 0) is equal to radius

⇒ (3m2 + 1)2 = 8(m4 + m2)
⇒ m4 – 2m2 + 1 = 0
⇒ m = ±1
Hence, the required tangents are y = x + 2 and y = –x – 2.

Test: BITSAT Past Year Paper- 2017 - Question 139

​If R , then R(s) R(t) equals 

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 139



Test: BITSAT Past Year Paper- 2017 - Question 140

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 140




Test: BITSAT Past Year Paper- 2017 - Question 141

Let  be non-coplanar unit vectors equally inclined to one another at an acute angle θ . Then  in terms of θ is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 141


are unit vectors and equally inclined to each other at an acute angle θ.

∴   ABC is an equilateral triangle and 


∴ Area of Δ ABC

If G is the centroid of the D ABC, then



Test: BITSAT Past Year Paper- 2017 - Question 142

21/4. 22/8. 23/16. 24/32......∞  is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 142

The given product




Test: BITSAT Past Year Paper- 2017 - Question 143

If  then a – n is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 143








Test: BITSAT Past Year Paper- 2017 - Question 144

if  then the value of  

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 144


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

[Expansion along first row]
⇒ xyr + xzq - xzy + yzp - zyx = 0

Test: BITSAT Past Year Paper- 2017 - Question 145

An urn contains five balls. Two balls are drawn and found to be white. The probability that all the balls are white is

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 145

Let Ai ( i = 2, 3, 4, 5) be the event that urn contains 2, 3, 4, 5 white balls and let B be the event that two white balls have been drawn then we have to find P (A5/B).
Since the four events A2, A3, A4 and A5 are equally likely we have P (A2) = P (A3) = P(A4) = P(A5) = 1/4
P(B/A2) is probability of event that the urn contains 2 white balls and both have been drawn.



By Baye’s theorem,

Test: BITSAT Past Year Paper- 2017 - Question 146

The ratio in which the join of ( 2, 1, 5) and (3, 4, 3) is divided by the plane (x + y – z) = 1/2 is:

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 146

As given plane  divides the linejoining the points A (2, 1, 5) and B (3, 4, 3) at a point C in the ratio k : 1.
Then coordinates of C

Point C lies on the plane,
⇒ Coordinates of C must satisfy the equation of plane.


Test: BITSAT Past Year Paper- 2017 - Question 147

Value of   is

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 147


then

Adding (i) and (ii), we get 

Test: BITSAT Past Year Paper- 2017 - Question 148

The dot product of a vector with the vectors  are 0, 5 and 8 respectively. The vector is

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 148

Let the required vector be


Subtracting (ii) from (i),we have
– 2y – z = –5 Þ 2y + z = 5 ..(iv)
Multiply (ii) by 2 and subtracting (iii) from it, we obtain
5y – 8z = 2 ...(v)
Multiply (iv) by 8 and adding (v) to it, we have 21y = 42
⇒ y = 2 ...(v)
Substituting y = 2 in (iv),
we get 2 × 2 + z = 5
⇒ z = 5 – 4 = 1
Substituting these values in (i),we get x + 2 – 3 = 0
⇒ x = 3 – 2 = 1
Hence, the required vector is

Test: BITSAT Past Year Paper- 2017 - Question 149

The angle between the lines whose intercepts on the axes are a, –b and b, –a respectively , is

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 149

Equation of lines are  and 

Test: BITSAT Past Year Paper- 2017 - Question 150

If the line through the points A (k, 1, –1) and B (2k, 0, 2) is perpendicular to the line through the points B and C (2 + 2k , k, 1), then what is the value of k?

Detailed Solution for Test: BITSAT Past Year Paper- 2017 - Question 150

Given points are A (k, 1, –1), B (2k, 0, 2) and C (2 + 2k, k, 1)
Let r1 = length of line

and r2 = length of line BC
Now, let ℓ1 , m1,n1 be direction-cosines of line AB and ℓ2, m2, n2 be the direction cosines of BC.
Since AB is perpendicular to BC




⇒ 2k -k - 3=0
⇒ k= 3 For k = 3,
AB is perpendicular to BC.

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