COMEDK Mock Test - 1 - JEE MCQ

For the first case, when the current is increasing di/dt = 1A / s

8 = 2R + L …..(1)

For the first case, when the current is increasing di/dt = -1A / s

4 = 2R - L …..(2)

From 1 and 2, R = 3Ω, L = 2H

Two beams of red and violet colors are made to pass separately through a prism (angle of the prism is 60°).
We have to find the angle of refraction at the position of minimum deviation.
In the position of minimum deviation, light passes symmetrically through the prism irrespective of the light wavelength used.

In the position of minimum deviation, r + r = 2r = A[ as deviation δ = 0]
  for both the colour.

B = B₀sin(π × 10⁷C)t + B₀sin (2π × 10⁷C)t
Since there are two EM waves with different frequency, to get maximum kinetic energy, we take the photon with higher frequency.
B₁ = B₀sin(π × 10⁷C)t  v₁ = 10⁷/2 x c 
B₂ = B₀sin(2π × 10⁷C)t v₂ = 10⁷C
where C is speed of light C = 3 × 10⁸ m/s
v₂ > v₁
so KE of photoelectron will be maximum for photon of higher energy.
v₂ = 10⁷C Hz
hv = φ + KE_max
energy of photon
E_ph = hv = 6.6 × 10^-34 × 10⁷ × 3 × 10⁹
E_ph = 6.6 × 3 × 10^-19J

KE_max = E_ph - φ
= 12.375 - 4.7 = 7.675 eV ^_≈ 7.72 eV

We know that PV=nRT also PM=dRT
So in the equation The value of R depends on P , V , n , T , d , M
except atomicity 
 so the ans is A

B = μ₀ni
Now i → 2i
And n  → n/2

• When an external electric field is applied to the conductor, the free electrons in the conductor move in an opposite direction to that of the applied electric field.
• This movement of electrons induces another electric field inside the conductor which opposes the original external electric field.
• This continues until the induced electric field cancels out the external field. 

Freundlich gave an empirical relationship between quantity of gas adsorbed by unit mass of solid adsorbent and pressure at a particular temperature.

(x is mass of gas adsorbed on a mass mm of the adsorbent at a pressure p, k and n are constant, which depends on nature of adsorbent and gas at a particular temperature)
From above equation of Freundlich adsorption isotherm

Taking log both sides 

Now if 1/n = 0, so the above equation will be independent of p and, we will get a straight line parallel to x-axis as required. 

From the given equations, we have
K₁ = 
K₂ = 
K₃ = 
For the given reaction,
K = 

K = 

The Newman projection formula using C₂ and C₃ is:

Hence, option (d) is correct.

Mn in [Mn(H₂O)₆]²⁺ is present as Mn²⁺(3d⁵).

sp³d² hybridisation occurs in [Mn(H₂O)₆]²⁺
Due to the presence of a weak field ligand, pairing of electrons do not take place. Hence, the number of unpaired electrons is five.

Metallic lustre is the property of metals which describes the amount of light that reflects by metals.

The reflection of light happens due to oscillation of free electron. Because loosely held electrons oscillate in all directions so, when light hits the metals, it gets reflected.

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

The Gibbs free energy change at a constant temperature and pressure is given as:

where, ΔH is the change in the enthalpy of the process and ΔS is the change in the entropy of the process.

For the process to be spontaneous, ΔG must be negative.

In the lewis structure of disulfate  ion, sulfur atoms are located as center atoms and oxygen atoms are located around sulfur atoms. There are two double bonds around a sulfur atom in the lewis structure. Oxidation number of sulfur is six in  ion.

has no s−s linkage :

Acid hydrolysis of an ester gives two different organic compounds.

According to Huckle, an aromatic compound must be cyclic planar having (4n + 2)π electrons.

At constant pressure, heat given by one mole gas (qp) = nCpΔT

Isopropylbenzene is cumene. Its air oxidation and then, acidic hydrolysis gives phenol.

Enthalpy of neutralisation is defined as the amount of heat evolved when 1 equivalent of H⁺ ions from an acid is completely neutralised by 1 equivalent of HO^- ions from a base.
For a strong base and a strong acid, which are strong electrolytes, this value is fixed under standard conditions at -57.62 kJ mol^-1.
But for a weak acid like HCN or a weak base, the value is less as some of the heat liberated is utilised to ionise the weak acid or the weak base.

f : N → R, f(x + y) = 2 f(x) f(y) ...(1)
f (1) = 2,

= 2f(α)(f(1) + f(2) + ... + f(10)) ...(2)
From (1),
f(2) = 2f²(1) = 2³
f(3) = 2f(2) f(1) = 2⁵
---------------------------------
f(10) = 2⁹ f¹⁰(1) = 2¹⁹
f (α) = 2^2α-1; α ∈ N
From (2),


Hence, α = 4

∵ Option (3) is correct.

P1 : ~(p → ~q)
P2 : (p ∧ ~q) ∧ ((~p) ∨ q)
if p →  ((~p) ∨ q) is false
Then p is ture and ~p v q is false
∴ q  is false
⇒ P1 is false and P2 is also false

Let the equation of the required plane be:

∵ (2, 1, -2) lies on it, 2 + λ (-2) = 0
 ⇒ λ = 1
Hence, π : 3x + 2y - 8 = 0

Clearly, X and Y are on the opposite sides of plane .

Let P₁(x, y, z) be any point on plane P.
Then (x + 4)² + (y - 2)² + (z - 1)² = (x - 2)² + (y + 2)² + (z - 3)²
⇒ 12x − 8y + 4z + 4 = 0
⇒ 3x - 2y + z + 1 = 0
And P₂ : 2x + y + 3z = 1
∴ Angle between P₁ and P₂ 

sin 12° + sin12° - sin 72°
= sin 12° - 2cos 42° sin30°
= sin 12° - sin48°
= -2cos30° sin 18°

x, y > 0 and x³y² = 215
Now, 3x + 2y = (x + x + x) + (y + y)
So, by A.M ≥ G.M inequality

∴ Least value of 3x + 2y = 40