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150 Questions MCQ Test BITSAT Mock Tests Series & Past Year Papers - Test: BITSAT Past Year Paper- 2017

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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