JEE Main Mock Test - 14 - JEE MCQ

Let Y = f(V, F, A)

Y = K V^x F^y A^z, where K is unitless

[Y] = [V]^x [F]^y [A]^z

[ML^-1T^-2] = [LT^-1]^x [MLT^-2]^y [LT^-2]^z

[M] [L^-1] [T^-2] = [M]^y [L^(x+y+z)] [T^(-2y-2z)]

From the equations:
y = 1, x + y + z = -1; x - 2y - 2z = -2

x + z = -2
x + 2y + 2z = 2

z = 2, x = -4
x + 2z = 0

|Y| = V^-4 A² F

Diameter of wire,
d = MSR + CSR × LC
= 0 + 52 × (1/100)
= 0.52 mm = 0.052 cm.

For A to B
u = 0
s = -(H - h)
a = -g

Using s = ut + (1/2) at², we get:
-(H - h) = (1/2)(-g)t²
⇒ t = √(2(H - h) / g)

For B to C Vertical Motion
uᵧ = 0
sᵧ = -h
aᵧ = -g
t = t'

Using s = ut + (1/2) at², we get:
-h = -(1/2) gt² ⇒ t' = √(2h / g)

∴ Total time of fall T = t + t'

or T = [ (2(H - h)) / g ]^(1/2) + [ (2h) / g ]^(1/2)

For finding the maximum time using the concept of differentiation:

dT/dh = 0

⇒ d/dt [ (2(H - h)) / g ]^(1/2) + d/dt [ (2h) / g ]^(1/2) = 0

⇒ (1/2) [ (2(H - h)) / g ]^(1/2) × (-2g) + (1/2) [ (2h) / g ]^(1/2) × (2g) = 0

⇒ [ (2(H - h)) / g ]^(1/2) = [ 2h / g ]^(1/2)

⇒ H - h = h

⇒ H = 2h

⇒ h = H / 2

C₁, C₂ and C₃ are in series

or 

or 

All the capacitors in upper branch are in series so the charge on each capacitor is Q′ = 6 / 11CV

Also charge on capacitor C₄ is Q = 4CV

∴ 

Net electric field will be zero at origin.
At any co-ordinate (0, y)


For maximum electric field, dE / dy = 0
Solving, 

τ = Iα
iAB = Iα
iπR²B = (1/2)MR²α (∵ Ring is rotating about its diameter)
α = (2πiB) / M = (2π × 4 × 10) / 2 = 40π rad/s²

In Statement I:
The decrease in mechanical energy in Case I will be:
ΔME₁ = (1/2) mv²
However, the decrease in mechanical energy in Case II will be:
ΔME₂ = (1/2) mv² - mgh
∴ ΔU₂ < ΔU₁, meaning Statement I is correct.
In Statement II:
The coefficient of friction will not change, which makes this statement incorrect.

Let F ∝ Pˣ Vʸ Tᶻ
By substituting the following dimensions:
[P] = ML⁻¹T⁻², [V] = LT⁻¹, [T] = T
And [F] = [MLT⁻²]
Therefore,
[MLT⁻²] = [ML⁻¹T⁻²]ˣ [LT⁻¹]ʸ [T]ᶻ
Comparing the dimensions on both sides, we get:
x = 1,
−x + y = 1 ⇒ y = 2,
−2x − y + z = −2 ⇒ z = 2
Hence, [F] = P V² T²

The pressure exerted by the photons is given by:
P = 0.75 × (I/c) + 0.25 × (2I/c)
(P = pressure, c = speed of light, I = intensity)
P = 1.25 × (I/c)
We know that force can be computed using:
F = P × A
(F = force, P = pressure, A = area)
F = 1.25 × (50 × 1 / (3 × 10⁸))
F = 20.83 × 10⁻⁸ N

The de Broglie wavelength is given by

So if the velocity of the electron increases, the de Broglie wavelength decreases.

Magnetic fields at P :

In case I. B₁ = 

In case II. B₂ = 

If R₁ and R₂ be the corresponding radii,

It is an example of internal Cannizzaro reaction; where -CHO is oxidized while ketonic group of the same molecule is reduced. Since keto group can't be oxidized, alcohol corresponding to -CHO group will not be formed.

The boiling point of an azeotropic mixture of water and ethyl alcohol is less than that of theoretical value of water and alcohol mixture. Hence, the mixture shows positive deviation from Raoult's law.
Positive deviations from Raoult's law are noticed when
(i) Exp. vaue of vapour pressure of mixture is more than calculated value.
(ii) Exp. value of boiling point of mixture is less than calculated value.
(iii) ΔH_mixing = +ve
(iv) ΔV_mixing = +ve

(i, ii, iii and iv) all are false statement
Correct Option: A, B, C, D (All Options are correct)

Here X is 2, 4, 6-tribromofluorobenzene

When an organic compound is present in an aqueous medium, it is separated by shaking it with an organic solvent in which it is more soluble than in water. The organic solvent and the aqueous solution should be immiscible with each other so that they form two distinct layers which can be separated by a separatory funnel.

The hybridisation of the central atom can be decided based on the number of atomic orbitals used in the hybridisation. This can be found using the formula:
H = 1 / 2 [V.E. + M − C + A]
Where,
H = Number of hybrid orbitals;
V.E.= Number of valence electrons on the central atom;
M = Number of monovalent atoms attached to the central atom;
C = Magnitude of charge if the given species is a cation;
A = Magnitude of charge if the given species is an anion.
Based on this, the hybridisation of NO⁻³, NO⁺²  and NH⁺⁴ can be found out as:

For an n^th order reaction,
When, 
When, 

= 2^{n - 1} + 1

The elevation in the boiling point is given by:
T′_b − T_b = ΔT_b = i × K_b × m
Where:

• i = Van't Hoff factor for the solute
• K_b = Molal elevation constant
• m = Molality of the solution

Now,
100.16 − 100 = 0.16 = 1 × K_b × 0.1 ...(1)
And
T′_b − 100 = ΔT_b = 1 × Kb × 0.5 ...(2)
From equations (1) and (2):
T′_b − 100 = 0.8
T′_b = 100.80°C

Δ_rG^∘ = 0−77.1 × 2 = −154.2 kJ/mol

= 
ΔG = Δ_rG^∘ + RT ln Q

= −129.5 kJ/mol

If the increase in molecular mass is 42, the reactant has one NH₂ group.
Actual increase in molecular mass = 348−180 = 168
Number of NH₂ groups = 168/42 = 4

Benzoic acid is monobasic acid, so its valence factor is 1 and molar conductivity of benzoic acid is equal to its equivalent conductivity.
By Kohlrausch's law:

We have 2[x] = x + {x} = [x] + 2{x}

⇒ {x} = [x] / 2

Now, 0 ≤ {x} < 1

⇒ 0 ≤ [x] / 2 < 1

⇒ 0 ≤ [x] < 2 ⇒ [x] = 0, 1

For [x] = 0, {x} = 0 ⇒ x = 0

For [x] = 1, {x} = 1/2 ⇒ x = 3/2

Thus, x = 0 and 3/2

The shortest distance (SD)

= 

= 1 / √78

So, 1 / SD = √78

f(x) = |x - 1| + 2|x - 1/2| + ... + 119|x - 1/119|

Minimum value occurs at the median.

Total number of terms:

= 1 + 2 + ... + 119 = 7140

n(n + 1) / 2 = 3570

n = 84

Minimum occurs at x = 1/84

EHINTZ
words starting with E is 5 !
words starting with H is 5 !
words starting with I is 5 !
words starting with N is 5 !
words starting with T is 5 !

The position of word ZENITH is 616.

In 3n consecutive natural numbers, either:
(i) n numbers are of the form 3P
(ii) n numbers are of the form 3P + 1
(iii) n numbers are of the form 3P + 2
Here, the number of favorable cases = Either we can select three numbers from any one of the sets or we can select one from each set.

Total number of selections = ³ⁿC₃
∴ Required probability

A² − 4A + 10I = A

⇒ 9 − 3k = −3, −6 + 2k = 2
And 4 + k² − 4k = k
⇒ k2 − 5k + 4 = 0 ⇒ k = 4, 1
But, k = 1 does not satisfy the Eq (1)1.

Suppose M divides BD in ratio K : 1




K = 2
Ratio 2 : 1

Given, f(x + y, x − y) = xy
Let, x + y = p, x − y = q
Then, 


= 
= 0