JEE Main Mock Test - 11 - JEE MCQ

Pitch = 3/6 = 0.5 mm
L.C. = 0.5 mm  / 50 = 1 / 100 mm = 0.01 mm
=0.001 cm

When an ideal gas is compressed adiabatically, its temperature and the average kinetic energy of the gas molecule increases because of collision of molecules with wall. Hence, Both A and R are true and R is the correct explanation of A.

Given inclination of plane,

Mass of block, m = 20 g = 0.02 kg
charge on the block, q = 4mC = 4 × 10⁻³C
Magnetic field, B = 1 T
Magnetic force on the charge particles, F = Bqv
Particle will leave the inclined plane, when
F = mgcosθ ⇒ Bqv = mgcosθ ⇒ v= mgcosθ / qB
Time taken to reached at the velocity v is given by
v = 0 + g sin θt [∴ u = 0, a = g sin θ]
t = v / g sin θ = mg cos θ / qB · g sin θ = m cot θ / qB
t = 0.02 cot 45° / 4 × 10⁻³ × 1 = 5 s [∴ cot 45° = 1]

P (in dioptre) = 100 / f (in cm)

∴ f = 100 / 10 = 10 cm

According to lens maker's formula,

1 / f = (μ - 1) (1 / R₁ - 1 / R₂)

For a biconvex lens, R₁ = +R, R₂ = -R

∴ 1 / f = (μ - 1) (1 / R + 1 / R)

⇒ 1 / 10 = (μ - 1) (2 / 10)

∴ μ = 1/2 + 1 = 3/2

Given that

Force 

Initial point (0, 0, 0)

Final point (2, 4, 0)

Work done by the variable force is,

y = x²

dy = 2x dx

dw = Fx dx + Fy dy
= 0 + (3xy) dy
= 3x × x² × 2x dx

= 192 / 5 J

Wavelength of the photon in terms of energy will be given by,

λ₂ = hc / E

De-Broglie of the proton will be,

λ₁ = h / p = h / √(2mₚE)

(where p is the momentum of the proton and mₚ is the mass of the proton)

From the above two equations,

Binding Energy (B.E) is given by:
B.E = Δm * c²
Δm = (50 × 1.00783) + (70 × 1.00867) - 119.902199
= (120.9984 - 119.902199) u
= 1.0962 u
B.E = 1.0962 × 931
= 1020.5622 MeV
B.E per nucleon ≈ 1020.5622 / 120
≈ 8.5 MeV

As the condensers are connected in parallel, therefore potential difference across both the condensers remains the same.

Q₁ = CV; Q₂ = C₂V
Also, Q = Q₁ + Q₂ = CV + C₂V = (3/2) CV
Work done in charging fully both the condensers is given by
W = (1/2) QV = (1/2) × (3/2 CV) V = (3/4) CV²

The circuit given in the problem can be redrawn as shown below. 

The three branches are connected between the same points P and Q and hence they are in parallel. For such a circuit the potential difference between the points P and Q is

Where E₁, E₂ and E₃ are the emf of the batteries present in the first, second and the third branch.
Similarly, R₁, R₂ and R₃ are the values of resistance in each of the respective branches.

The value of current through the 1 Ω resistance is

It can be observed from figure that P and Q shall collide if the initial component of velocity of P and Q on inclined plane i.e along incline should be equal. That is particle is projected perpendicular to incline.

∴∴ Time of flight 

(A) Atomic radius does not decrease regularly down the group due to d - and f -block contraction. Hence, this statement is incorrect.

(B) & (C) check video solution for details.

(D) Compounds of boron, like boric acid (H₃BO₃), exhibit significant pπ−pπ character due to boron's small size and its ability to form strong π-overlap. This statement is correct.

(E) Boron shows a diagonal relationship with silicon, resulting in similar chemical behavior, such as the formation of covalent compounds, acidic oxides, and tetrahedral structures like BF₄ and SiF₄. This statement is correct.

Correct Answer: (C), (D), and (E) only.

Molecular forces of attraction get stronger as molecules get bigger in size. Further, as the branching increases, the surface area of the molecule decreases. Because of this, the Van der Waal's force of attraction between the molecule decreases and consequently boiling point decreases.
In the above three isomers given Structure (b) is a straight-chain with no branching, structure (c) is branched but has less surface area compared to the structure (a). Hence, the increasing order of their boiling points is, (c) < (a)<(b)

The inverse of resistivity is called conductivity or specific conductance. It is represented by the symbol, κ. It may be defined as the conductance of a solution of 1 cm length and having 1 square centimetre as the area of cross-section. In other words, conductivity is the conductance of one cm³ of a solution of an electrolyte. The mathematical expression for conductivity is:

Therefore, the units are: ohm⁻¹ cm⁻¹ = Ω⁻¹cm⁻¹

(A) Bi³⁺ + 3NH₄OH ⟶ Bi(OH)₃↓ (white) + 3NH₄⁺
(B) Bi³⁺ + C₆H₃(OH)₃ ⟶ Bi(C₆H₃O₃)↓ (yellow) + 3H⁺
(C) Bi³⁺ + 3I⁻ ⟶ BiI₃↓ (black); BiI₃ + I⁻ ⟶ [BiI₄]⁻ (orange solution)=
(D) Bi³⁺ + 3OH⁻ ⟶ Bi(OH)₃↓ (white); 2Bi(OH)₃↓ + 3[Sn(OH)₄]²⁻ ⟶ 2Bi↓ (black) + 3[Sn(OH)₆]²⁻

K₂Cr₂O₇ contains Cr⁶⁺ (3d⁰), which is diamagnetic but colored due to charge transfer spectra. Charge transfer spectra in highly ionic crystals correspond to electron transfer between neighboring atoms. They can be classified as donor or acceptor, depending on whether the metal atom donates or accepts an electron.
The orange color of the K₂Cr₂O₇ solution is due to the transfer of an electron from the oxygen lone pair to a low-lying chromate ion orbital.
(NH₄)₂[TiCl₆] contains Ti⁴⁺ (3d⁰), which is diamagnetic and colorless.
VOSO₄ contains V⁴⁺ (3d¹), which is paramagnetic and colored.
K₃[Cu(CN)₄] contains Cu⁺ (3d¹⁰), which is diamagnetic and colorless.
Hence, option C is the correct answer.

Acid-hydrolysis of benzonitrile forms benzoic acid.

Note: Alkaline-hydrolysis of cyanide forms benzoate and NH3NH3 .

From two data, (for zero order kinetics)

K_I = x / t = (0.25 / 0.05) = 5
⇒ K_II = x / t = (0.60 / 0.12) = 5

Cr has a 3d⁵4s¹ configuration. But in [Cr(CO)₆], it adopts a 3d⁶ configuration with no unpaired electrons because CO is a strong field ligand that causes electron pairing.
Cr = 24
Cr = 3d⁵4s¹
The charge on Cr:
⇒ x + 0 × 6 = 0
⇒ x = 0
So, Cr = 3d⁵4s¹

For f to be one-one, f'(x) > 0 and f'(x) < 0 for all x.

Clearly, f is continuous at x = 0 and f(0) = -1.

For x ≤ 0, f'(x) = 2(x + m) for x < 0.

f'(x) cannot be > 0 for all x < 0 if m > 0.

Thus, f'(x) < 0 for all m ≤ 0.

But m ≠ 0, as for x > 0, f is constant and for all m < 0, f'(x) < 0 for all x > 0.

⇒ a₁a₂ + b₁b₂ + c₁c₂ = 0

Case-1: a₁a₂ = 4  
⇒ (a₁, a₂) = (-2, -2), (2,2)  
⇒ b₁b₂ = -2  
(b₁, b₂) = (1, -2), (-1, 2), (2, -1), (-2, 1)  
⇒ c₁c₂ = -2 (4 cases)  
Total number of cases = 2 × 4 × 4 × 3 = 96  

Case-2: a₁a₂ = -4  
⇒ (a₁, a₂) = (-2, 2) or (2, -2)  
b₁b₂ = 2  
(b₁, b₂) = (1, 2), (2, 1), (-1, -2), (-2, -1)  
Similarly, c₁c₂ = 2 gives 4 cases  
Total number of cases = 2 × 4 × 4 × 3 = 96  

Case-3: a₁a₂ = 2  
⇒ (a₁, a₂) = (2,1), (1,2), (-2,-1), (-1,-2)  
b₁b₂ = -1  
(b₁, b₂) = (1, -1), (-1, 1)  
c₁c₂ = -1 (2 cases)  
Total number of cases = 4 × 2 × 2 × 3 = 48  

Case-4: a₁a₂ = -2  
⇒ (a₁, a₂) = (1, -2), (-1, 2), (2, -1), (-2, 1)  
b₁b₂ = 1  
(b₁, b₂) = (-1, -1), (1,1)  
c₁c₂ = 1 (2 cases)  
Total number of cases = 48  

Total number of cases = 96 + 96 + 48 + 48 = 288  

r / (a/2) = tan(30°)

2r / a = 1 / √3

⇒ r = a / (2√3)

∴ Area of incircle, A = πr² = (πa²) / 12

∴ dA/dt = (2πa / 12) * (da/dt)

= (π/6) * (6/π) * 2

= 2 cm²/sec.

As, a̅ lies in the plane of the vectors b̅ and c̅, So, 

Compare with  a̅ = αî + 2ĵ + βk̂

Not in option
So, now consider


Compare with a̅ = αî + 2ĵ + βk̂

Given, 2x² (dy/dx) - 2xy + 3y² = 0
Dividing by 2x²y², we get

General solution will be

 is the particular solution.

when x = 1;  y(1) = 2/3

∵y = e^x,  y = e^−x will meet, when e^x = e^−x
⇒ e^2x = 1,
∴ x = 0,y = 1
∴ A and B meet on (0, 1), 
∴ A ∩ B ≠ ϕ

Let A = {a₁, a₂, a₃, ......, aₙ}
P and Q are subsets of A. Elements in P and Q can be chosen in the following ways:
Case I: aᵢ ∈ P, aᵢ ∈ Q for i = 1, 2, ..., n
Case II: aᵢ ∈ P, aᵢ ∉ Q for i = 1, 2, ..., n
Case III: aᵢ ∉ P, aᵢ ∈ Q for i = 1, 2, ..., n
Case IV: aᵢ ∉ P, aᵢ ∉ Q for i = 1, 2, ..., n
Since P and Q are non-intersecting sets, every element can be chosen as in Case II, Case III, or Case IV, i.e., each element can be chosen in 3 ways.
Total number of ways = 3 × 3 × ... (n times) = 3ⁿ
However, one case needs to be excluded when P = ∅ and Q = ∅.
∴ The number of valid ways = 3ⁿ − 1

Any point on the first line is (4r + k, 2r + 1, r − 1), and any point on the second line is (r′ + k + 1, −r′, 2r′ + 1) for some values of r and r′. The lines are intersecting if these two points coincide, i.e.,
4r + k = r′ + k + 1,
2r + 1 = −r′,
r − 1 = 2r′ + 1,
for some r and r′.
⇒ 4r − r′ = 1,
2r + r′ = −1,
r − 2r′ = 2.
Now, solving 4r − r′ = 1 and 2r + r′ = −1:
⇒ r = 0, r′ = −1,
which satisfies r − 2r′ = 2.
⇒ The given lines are intersecting for all real values of k.

∵ a₁ + a₂ + a₃ = 1 ...(i)
and due to symmetry a₁ = a₃ ...(ii)

Now,
= 

So, a₁ + a₃ + 1/3 = 1

a₁ + a₃ = 2/3  ..........(ii)

From Eqs. (ii) and (iii),

a₁ = a₃ = 1/3 = a₂

So, a₁ + 2a₂ + 3a₃ = 1/3 + 2/3 + 3/3 = 2

T₁ = 2014
T₂ = (2)³ + (0)³ + (1)³ + (4)³ = 73
T₃ = (7)³ + (3)³ = 370
T₄ = (3)³ + (7)³ + (0)³ = 370
T₅ = (3)³ + (7)³ + (0)³ = 370
T₂₀₁₄ = (3)³ + (7)³ + (0)³ = 370

|x − 1| + |x − 2| + |x − 3| = k
To find the number of solution we draw the graph of
y = |x − 1| + |x − 2| + |x − 3| and y = k

In the graph of above curve can be drawn as following

Clearly the number of solution of the equation
(i) for k < 2 is zero.