KIITEE Mock Test - 10 - JEE MCQ

As light moves from air to water, its speed decreases due to refractive index of water.

Fringe width, as λ decreases on entering in water, β also decreases.

λ = wavelength in medium,

D = separation between slit and screen,

d = separation between slits.

The situation can be shown below.

where, O is the centroid of the equivalent triangle.

The length AO can be found by

=

=

The moment of inertia about O is given by

I = MR²

…(i)

According to the law of conservation of angular momentum,

Iω = constant

l x (2π/T) = constant

⇒ T ∝ I

From Eq. (i), we get

T ∝ L²

Two blocks of masses M and m are connected by a string passing over a pulley. Let T is the tension in the string and a is the downward acceleration. Applying II law of motion for the block of mass M.
T = Ma ...(i)
For block of mass m
ma = mg - T ...(ii)

From Equations, (i) and (ii), we get
ma = mg - Ma

i.e …(i)

where, T = temperature and M = molecular mass. When the temperature is doubled, then O₂ dissociates into two atoms.

Hence, the atomic mass of the oxygen atom.

(M') = M/2

∴ rms velocity of the oxygen atom =

E_i = 2π² mr²₁f²_{1 ………..}(i)

Where, f₁ = frequency of revolution of the body.

When, r₂ = 2r₁ and f₂ = and 2f₁ , then

Change in kinetic energy,

ΔE = E_r -E_i

= 16E_i - E_i = 15 E_i

For steel wire: Y = 20 × 10¹⁰ Pa; L = 1.5 m ; d = 0.3 cm = 0.3 × 10⁻²m
Therefore, area of cross-section of the steel wire,

The stretching force for the steel wire,
F = 4.0 + 6.0 = 10 kgf = 10 × 9.8 N
If l is extension in the steel wire, then

For brass wire: Y = 9.0 × 10¹⁰ Pa ; L = 1m;

The stretching force for the brass wire,

F = 6 kgf = 6 × 9.8 N

where, T = tension in string

m = mass/length

and l = length of string.

New frequency,

⇒ 13v = 10v + 90

v = 30Hz

The reaction gives an α, β-unsaturated aromatic acid by the aldol condensation of an aromatic aldehyde and an acid anhydride, in the presence of an alkali salt of the acid. This reaction is known as Perkin's reaction.

(i) In nucleophilic substitution reaction more powerful nucleophile replaces weaker nucleophile.
(ii) In rearrangement reaction atoms replace their position within molecule.
(iii) In elimination reaction small molecules (e.g., H₂O, NH₃) are lost.

∵ KCl and H₂O molecules are lost during reaction.
∴ It is an elimination reaction.

K_p = K_c(RT)^Δn
For the reaction,
N₂ + O₂ ⇌ 2NO
Δn = 2 − 2 = 0
∴ K_p = K_c(RT)⁰
K_p = K_c

A is isopropyl benzene (Cumene)

When photo chlorination of cumene is carried out then chlorine radical get substituted to that carbon where free radical is most stable (major product) therefore, structure of B is

The shape of a molecule depends upon the hybridisation and number of lone pair(s) on the central atom of the molecule. The hybridisation in a molecule can be determined by the total steric number around the central atom, which is equal to the sum of sigma bonds and lone pair(s) of electrons on the central atom. 
The number of sigma bonds and lone pairs can be determined from the Lewis structure of the molecule.
OSF₂
The Lewis structure of the molecule can be represented as under :

Number of sigma bonds = 3
Number of lone pair = 1
Total steric number = 3 + 1 = 4
The hybridistion of a molecule with a steric effect of 4 is sp³. A hybridisation of sp³ is expected to have a tetrahedral geometry, but the presence of a lone pair on the sulphur atom comes into play (VSEPR theory). 
The geometry of a molecule with three sigma bond pairs and one lone pair, according to the VSEPR theory, can be represented as under:

Lone Pair − Bond Pair repulsion > Bond Pair − Bond Pair repulsion
Due to a greater repulsion between the lone pair and bond pair, the bond angle between the lone pair and each S−FS-F bond is more than the bond angles of the F−S−O and O−S−FF-S-O and O-S-F bonds. Thus, the molecule OSF2OSF2 has a trigonal pyramidal geometry.  

SO₂
The Lewis structure of the molecule can be represented as under :

Number of sigma bonds = 2
Number of lone pair = 1
Total steric count = 2 + 1 = 3
A steric count of 3 has a hybridisation of sp². So, the molecule is expected to have a trigonal planar shape, but due to the presence of a lone pair of electrons on the sulphur atom, the geometry changes to a bent structure as follows:

BrF₃
The Lewis structure of the molecule can be represented as under :

Number of sigma bonds = 3

Number of lone pairs = 2
Due to the presence of three sigma bonds, and two lone pairs of electrons, BrF3BrF3 shows a trigonal bipyramidal geometry with a distorted T−T-shaped structure.

SiO²⁻₃

The Lewis structure of the molecule can be represented as under :

Number of sigma bond pairs = 3

Numebr of lone pair = 0

Total steric number = 3 + 0 = 3

Hybridisation → sp²
Since there are no lone pairs on the silicon atom, the molecule shows a geometry of trigonal planar, as per its hybridisation. 

For the reaction

ΔG^o for the reaction
2ZnS + 3O₂ → 2ZnO + 2SO₂
Can be obtained by adding eqn. (1), (2) and (3)
⇒ ΔG^o = 293 − 480 − 544 = −731 kJ

Anisole reacts with HI to give phenol and methyl iodide. First, anisole on protonation forms methylphenyl oxonium ion. Since the bond between O-CH₃ is weaker than the bond between O-C₆H₅, (because of a partial double bond is character between oxygen and carbon of the benzene ring) it breaks more readily.
Hence, the I^- ion attacks and breaks the O-CH₃ bond to give CH₃I and phenol.

The phenols formed do not react with halides because nucleophilic substitution on aromatic ring is difficult.

Rate = k[X][Y₂]
But the rate law expression cannot be expressed in terms of a reaction intermediate.

Rate = 
Thus, order of the reaction 

Any redox reaction would occur spontaneously, if the cell emf is positive.

Addition polymerisation takes place via free radical mechanism. Out of the given polymers, Teflon is an addition polymer. Teflon is manufactured by heating tetrafluoroethene with a free radical or persulphate catalyst at high pressures.

The Cannizzaro reaction is a redox reaction in which two molecules of aldehyde having no α−hydrogen are reacted to produce an alcohol and a salt of carboxylic acid using a hydroxide base. For example formaldehyde, bezaldehyde.

2- methyl propanal exceptionally undergoes Cannizzaro reaction, although it has α −H atom.

Propanal does not undergo Cannizzaro reaction since it has two alpha hydrogen atoms.

Chain isomerism is type of structural isomerism. Chain isomers differ in the branching of carbon skeleton. Possible chain isomers of C₆H₁₄ are:

Hence, five isomers are possible.

dθ/dt = -k(θ - θ₀), where k is constant

⇒ (dθ/dt) + kθ = kθ₀

Which is linear differential equation in the form of

(dy/dx) + Py = Q

IF = e^∫kt = e^kt

before the required solution,

(θ)(e^kt) = ∫(e^kt x kθ₀)dt

⇒ θe^kt = e^kt θ₀ + a

⇒ θ = θ₀ + ae^-kt

⇒ sin A = 2Δ/bc …(i)

also, R = a/2sinA …(ii)

From Eqs. (i) and (ii), we have

⇒ R = abc/4Δ

⇒ Δ = A(ΔABC) = abc/4R

We have, A is non-singular matrix

∴ |A| = 0

and (A + l)(A - l) = 0

⇒ A² - I² = 0

⇒ A² = I² = I

⇒ A·A = I

⇒ A^-1 = A

∴ A + A^-1 = A + A = 2A

Equation of a line passing through 4(6, 1) and perpendicular to given line is y - 1 = (-2)(x - 6)

⇒ y - 1 = -2x + 12

⇒ 2x + y = 13

⇒

∴ the y-intercept of slope (i) is 13.

Let the first term of the G.P. be a and the common ratio be r.
The fourth term is ar³ = p.
The seventh term is ar⁶ = q.
The tenth term is ar⁹ = r.
Now, p² = (ar³)² = a²r⁶.
q × r = (ar⁶) × (ar⁹) = a²r¹⁵.
But if we compare ratios:
q = ar⁶ = (ar³)r³ = p r³
r = ar⁹ = (ar³) r⁶ = p r⁶
So, p² = (ar³)² = a²r⁶ = q × r / r⁶
But with the standard formula, for terms at positions n, n+3, n+6 (which this is, since 4, 7, 10 differ by 3), we always have (arⁿ)² = (ar^n-3)(arⁿ⁺³), so p² = q × r.
Thus, option b) p² = qr is correct.

Given, c = (x - 2)a + b and d = (2x + 1)a - b are collinear

∴ c = λd

⇒ (x - 2) a + b = λ[(2x + 1)a - b]

⇒ (x -2)/(2x + 1) = 1/-1 = λ

⇒ 2x + 1 = -x + 2

⇒ 3x = 1

⇒ x = 1/3

Let sin x - cos x = t

⇒ (cos x + sinx)dx = dt

Here,A=½ , B=½ so, A+B=½+½=1

Now,

put cosx + sinx = t

(-sinx + cosx)dx = dt

∴ p = 6