JEE Main Practice Mock Test - 9 Free Online Test 2026

We are given a circuit, we need to find the current across  1 Ω resistance.

In the given figure, upper diode is in forward bias and lower diode is in reverse bias.

Thus, the equivalent circuit is,

In forward bias mode, diode behaves like  on state and in reverse bias mode, diode behaves like  off state.
Current through the circuit ,  I = V / R 
⇒I = 6 / 2 + 1 = 6 / 3 = 2 A

N = m(g − a),N < mg if a(↓) and N > mg if a (↑)
Reading of spring balance is less than m if a(↓) and reading of spring balance is

greater than m if a(↑)

n = 5 moles, T₁ = 100°C

T₂ = 120°C, ΔU = 80 J

Rise in temperature Δt = 120 – 100 = 20°C

ΔU = msΔt

80 / 5 = 1 × s × 20

s = 0.8 J

∴ For 5 moles, s = 0.8 × 5 J/K

= 4 J/K

Given, MI = 2.5 kg m⁻²

W = 40 rad s⁻¹

T = 10 Nm

As T = Iα

10 = 2.5α

α = 4 rad s⁻²

Now, ω = ω₀ + αt

60 = 40 + 4 × t

20 = 4t

t = 5 s

F.B.D. of man and plank are

For plank to be at rest, applying Newton's second law to plank along the incline
Mgsinα = f                   ...(1)
and applying Newton's second law to man along the incline.
mgsinα + f = ma           ...(2)
a = gsinα(1 + M/m)down the incline

Here, the point of suspension has an acceleration.  (down the plane). Further,  can be resolved into two components gsinθ (along the plane) and gcosθ (perpendicular to the plane).

= gcosθ (perpendicular to plane)

PR = d
∴ PO = d sec θ
and CO = PO cos 2θ = d sec θ cos 2θ
Path difference between the two rays is,
Δx = PO + OC = (d sec θ + d sec θ cos 2θ)
Phase difference between the two rays is
Δϕ = π (one is reflected, while another is direct)
Therefore, the condition for constructive interference should be,

The beat frequency is given as |f₁ - f₂|.
With 250 Hz, x produces 8 beats per second.
⇒ |250 - x| = 8
⇒ 250 - x = ±8
⇒ x = 242 Hz or 258 Hz ...(1)
Also, x produces 12 beats per second with 270 Hz.
⇒ |270 - x| = 12
⇒ 270 - x = ±12
⇒ x = 258 Hz or 282 Hz ...(2)
From (1) and (2), we get:
x = 258 Hz

Before the slab is introduced:
f = 40 cm, u = -10 cm
1/v - 1/u = 1/f
1/v + 1/20 = 1/40
v = -40 cm
The shift in the image of O due to the slab:
du = t (1 - 1/μs)
= 2 (1 - 3/4)
= 12 mm = 0.5 mm
Also,
dv/du = v²/u²
dv = (v²/u²) * du
(40/20)² * du = 4 × 0.5 mm
= 2 mm

Given:

• Mass (m) = 2 kg
• Initial radius of the path (r₁) = 0.8 m
• Initial angular velocity (ω₁) = 44 rad/s
• Final radius of the path (r₂) = 1 m

Moment of inertia calculations:

• I₁ = m r₁² = 2 × (0.8)² = 1.28 kg·m²
• I₂ = m r₂² = 2 × (1)² = 2 kg·m²

Using the law of conservation of angular momentum:
I₁ω₁ = I₂ω₂
Solving for ω₂:
ω₂ = (I₁ × ω₁) / I₂ = (1.28 × 44) / 2
ω₂ = 28.16 rad/s^-1

Let us take wedge + spring + block as a system. The forces responsible for performing work are spring force kx and the external force F.

Work Energy theorem for block + plank relative to ground :
Applying work-energy theorem, we have W_ext + W_sp = ΔK
where W_sp the total work done by the spring on wedge and block −1/2kx² and ΔK = change in KE of the block (because the plank does not change its kinetic energy)
Then,

As the block was initially stationary and it will acquire a velocity v₀ equal to that of the plank at the time of maximum compression of the spring, the change in kinetic energy of the block relative to ground is

Substituting ΔK in the above equation, we have

W-E theorem for block+ plank relative to the plank W_ext + W_sp = DK The plank moves with constant velocity, there is no pseudo-force acting on the block. W_ext = O Then the net work done on the system (block + plank), due to the spring, can be given as

As the relative velocity between the observer (plank) and block decreases from v₀ to zero at the time of maximum compression of the spring, the change in kinetic energy of the block is 
Substituting W_sP and ΔK in above equation

Let time, T ∝ c^X G^Y h^Z

⇒ T = k c^X G^Y h^Z

Taking dimensions on both sides:

[M⁰ L⁰ T¹] = [L T⁻¹]^X [M⁻¹ L³ T⁻²]^Y [M L² T⁻¹]^Z

i.e.,

[M⁰ L⁰ T¹] = [M^Y+Z L^(X+3Y+2Z) T^(-X-2Y-Z)]

Equating powers of M, L, and T on both sides, we get:

1. -Y + Z = 0 ...(1)
2. X + 3Y + 2Z = 0 ...(2)
3. -X - 2Y - Z = 1 ...(3)

From equation (1):
Y = Z

Adding (2) and (3):
Y + Z = 1

Since Z = Y, we get:
2Y = 1
∴ Z = Y = 1/2

Substituting these values in (2):
X + 3(1/2) + 1 = 0
X + 3/2 + 1 = 0
X = -5/2
Hence, 

In the saturated solution of a sparingly soluble electrolyte, the ionic product of ions is constant at constant temperature and it is known as the Solubility Product.

Here,
Mg(OH)₂ → [Mg²⁺] + [OH⁻]²

Given,
Kₛₚ for Mg(OH)₂ = 5.6 × 10⁻¹²

[Mg²⁺] = 10⁻¹⁰ M

Substituting the value, we get:

Kₛₚ = [Mg²⁺] [OH⁻]² ⇒ 5.6 × 10⁻¹² = [10⁻¹⁰] [OH⁻]² ⇒ OH⁻ = 0.24 M

Hence, the concentration of [OH⁻] = 0.24 M

ΔT_f = K_f × (w₁ × 1000) / (w₂ × m₁)

w₁ = weight of solute  
w₂ = weight of solvent  
m₁ = molar mass of solute  
K_f = 4.3 K kg mol⁻¹  

Now,  
ΔT_f = 4.3 × (2 × 1000) / (57.3 × 500)  

= 17.2 / 57.3 = 0.30 K

B atom after losing outermost electron acquires noble gas configuration (stable configuration). It is difficult to remove the next electron from B⁺(1s², 2s² 2p⁶) ion.

In HClO₄,Cl is at maximum oxidation state so can't act as reducing agent.

Actinoids show different oxidation states such as +2, +3, +4, +5, +6 and +7. However, +3 oxidation state is most common among all the  lanthanoids. The wide range of oxidation states of actinoids is attributed to the fact that the 5f ,6d and 7s energy levels are of comparable energies. Therefore, all these three subshells can participate.

Boiling point of Cis−2−butene is more than trans−2−butene

Cis −2− butene is polar so boiling point is increase.
trans−2− butene is non polar because dipole moment is zero so boiling point is less.

A graph plotted between  log k versus 1/T for calculating activation energy is shown as:

From the Arrhenius equation

Comparing with the equation of straight line, y = mx + c, we get a straight line having positive intercept on y-axis and negative slope.

nₑ = (5.79 × 20000) / 96500 = 1.2 mole

Cathode:

Cu⁺² + 2e⁻ → Cu
0.3 mole ⇒ 0.6 mole

2H₂O(l) + 2e⁻ → H₂ ↑ + 2OH⁻
0.6 mole ⇒ 0.3 mole

Anode:

2H₂O(l) → O₂ + 4H⁺ + 4e⁻
← 1.2 mole

V_O₂ = (1.2 / 4) = 0.3 mole

V_H₂ / Vₒ₂ = 0.3 / 0.3 = 1

The sum of partial pressures is given by:
Σ_{partial pressures} = p₀ − x + 2x + x + 28(Vapor pressure of moisture)
= p₀ + 2x + 28 = 188
⇒p₀ + 2x = 160 mm Hg   (i)
After a long time, when P_A = 0, we have:
P_B = 2P₀ and P_C = P₀
So,

From equation (i) & (ii), we get,
∴  x = 20 mm Hg

Point on the line = (2t, 2 + 3t, 3 + 4t)

Equating to B(q, 5, 7), we get t = 1 and hence q = 2

∴ B = (2, 5, 7)

D.R.s of AB are (p - 1, -6, 4)

AB is perpendicular to the line ⇒ (p - 1)(2) + (-6)(3) + (4)(4) = 0

∴ p = 2 and p - q = 0

∴ (x, y) ∈ R ⇒ xʸ = yˣ

∴ (x, x) ∈ R as xˣ = xˣ ∀ x ∈ I - {0}

∴ R is reflexive

Now, (x, y) ∈ R ⇒ xʸ = yˣ ⇒ yˣ = xʸ

⇒ (y, x) ∈ R ∴ R is symmetric

Now, (x, y) ∈ R ⇒ xʸ = yˣ and (y, z) ∈ R ⇒ yᶻ = zʸ

∴ xʸ = yˣ ⇒ y = xʸ / x and yᶻ = zʸ

⇒ yᶻ = zʸ ⇒ (xʸ / x) ᶻ = zʸ

⇒ xʸᶻ = z ʸᶻ / x = zˣ

⇒ xʸᶻ ≠ zˣ² ⇒ (x, z) ∉ R

R is not transitive.

R = {(1, 1),(1, 2),(1, 3),(1, 4),
(2, 1),(2, 2),(2, 3),(2, 4),
(3, 2),(3, 3),(3, 4)
(4, 2),(4, 3),(4, 4)}
So, m = 14
To make R symmetric, we need to add (3, 1) and (4, 1).
So, n = 2
Since, R is already reflexive, p = 0

Let a, b, c > 1, and a³, b³, c³ be in A.P.
This implies:
2b³ = a³ + c³
Also, given that logₐb, logcₐ, and logbₐ are in G.P., we set:
(logₐb)(logbₐ) = (logcₐ)²
Now, the sum of the first 20 terms of an A.P. is given by:
S₂₀ = (20/2) [2(a + 4b + c/3) + 19(a - 8b + c/10)]
Setting this equal to -444 and solving for abc, we get:
abc = 216
Thus, the correct answer is B) 216.

Let the sides of the right-angle triangle be coinciding with the coordinate axes and the vertices of △OAB be O(0, 0), A(a, 0) & B(0, b).
Also, let BD & AE are the medians cutting the sides at D & E.
So, coordinates of D = (a/2, 0) & E = (0, b/2).
Now Slope of (let say)
Also, slope of (let say)

Since, by Pythagoras theorem in △OAB,

Putting equations (ii) in equation (i), we get

Hence, area of triangle is √3 Sq. units

Given, liquid evaporates at a rate proportional to its surface are.

We know, volume of cone = 13πr²h

The surface area is given by:
S = πr²
The volume is given by:
V = (1/3) πr²h
Also, we have:
tanθ = R / H
and r / h = tanθ
From equations (2) and (3), we get:
V = (1/3) πr³ cotθ
and S = πr³
On substituting Eq. (4) into Eq. (1), we get:

∴ Required time after which the cone is empty, T = H/k

Let, x be the length of an edge, V be the volume and S be the surface area of the cube.

S = 6x²