JEE Main Mock Test - 9 Free Online Test 2026

.

Now U₁ = U₂

b − a = a

b = 2a

∴ b = 2 × 20 = 40 cm

= 40 cm

Initially, the coordinate of the center of mass of the ball and shell is (0, 0).

When the solid ball touches the lower surface of the hollow sphere, the center of the solid ball gets shifted to a distance equal to 2R.

Thus, the center of mass of the system will be given as:

y_cm = (m(2R) + m × 0) / (m + m) = R

So, the final coordinate of the center of mass (COM) of the system is (0, -R).

The displacement of the center of mass is [0 - (-R)] = R.

To calculate the force of attraction on the point mass m, we should calculate the gravitational field inside the cavity and then force on m.

The gravitational field inside the cavity can be obtained by modifying Gauss law for gravitation, so,

So,
F = mE = GMm / 2R²

Therefore, TE = −KE = PE/2 = −3.4 eV

So, KE = 3.4 eV

Let p = momentum and m = mass of the electron.

On substituting the values, we get

= 6.6 × 10⁻¹⁰m

From the dimensional analysis, it follows that

And,
[V] = [b]
Hence,

= [P]

At t = 1 s, the particle is at maximum height.
There is a 90° angle between the velocity vector and acceleration vector.
Hence, at this position, the rate of change of speed with respect to time is zero.

The acceleration due to gravity is related to the mass and the radius of the Earth by the formula:

Thus, from equation (1), it follows that

or 
∴ 
= 2.55 kg
Maximum acceleration of SHM is,
a_max = ω²A   (A = amplitude)
i.e., maximum force on mass 'm' is m ω² A which is being provided by the force of friction between the mass and the cart. Therefore,
μ_smg  ≥  mω² A

or               μ_s   ≥  0.358
Thus, the minimum value of  μ_s should be 0.358.

For the first diffraction minimum,
d sinθ = λ
And if the angle is small, sinθ ≈ θ,
So, dθ = λ,
i.e., Half angular width, θ = λ/d
Full angular width, w = 2θ = 2λ/d
Also, w' = 2λ'/d
Thus, λ'/λ = w'/w, or λ' = λ (w'/w)
Substituting the values:
λ' = 6000 × 0.7 = 4200 Å

Energy is released in a process when the total Binding Energy (B.E.) of the nucleus increases, or when the total B.E. of the products is greater than that of the reactants. By calculation, we can see that this happens only in the case of option (3).
Given: W → 2Y
Binding Energy (B.E.) of reactants = 120 × 7.5 = 900 MeV
Binding Energy (B.E.) of products = 2 × (60 × 8.5) = 1020 MeV

Given,

Let μ₁and μ₂ be the linear densities.

or x = 144

We know that Intensity of magnetisation

I = M / V

[where M = magnetic moment, V = volume]

So,

M = IV = 5.0 × 10⁴ × (12 / 100) × (1 / (100)²)

= 60 × 10⁴ × 10⁻⁶ 

= 0.6 Am²

= 8 mm

Effectively, we have removed a solid sphere from the solid cylinder.
The mass and moment of inertia of the entire cylinder are:
M_cylinder = (πR² × 3R)ρ
I_cylinder = (1/2) × (3πR³ρ) × R²
Similarly, the mass and moment of inertia of the solid sphere are:
M_sphere = (4/3)πR³ρ
I_sphere = (2/5) × (4/3)πR³ρ × R²
The moment of inertia of the remaining portion is:
I_remaining = I_cylinder − I_sphere = (29/30)πR⁵ρ
I_remaining = (29/6) × 5πR⁵ρ
Thus, K = 5.

According to question,

w_A = x g, m_A = 18, X_A = 1 - 0.6 = 0.4

w_B = 69 g, m_B = 46, X_B = 0.4

We know that,

X_A = n_A / (n_A + n_B)

or 0.4 = (w_A / m_A) / [(w_A / m_A) + (69 / 46)]

⇒ 0.4 = (x / 18) / [(x / 18) + (3 / 2)]

0.4 × (2x + 54) / 36 = x / 18

or 2x + 54 = 5x or 3x = 54, x = 18g

Van't Hoff factors for urea, NaCl, Na₂SO₄, and Na₃PO₄ are 1, 2, 3, and 4 respectively.

As ΔTₓ ∝ i * m, so ΔTₓ for urea, NaCl, Na₂SO₄, and Na₃PO₄ are proportional to 1, 2, 3, and 4 respectively.

Hence, the freezing point order is:
Na₃PO₄ < Na₂SO₄ < NaCl < Urea

Hinsberg's method is used for the separation of primary, secondary and tertiary amines. In this test, benzene sulfonyl chloride is used as a reagent. It reacts with primary amines and secondary to give sulfonamide product. The sulfonamide product from primary amines is soluble in alkali, but the sulfonamide product of secondary amines is not soluble in alkali. The reagent, benzene sulfonyl chloride does not react with the tertiary amines. 3° amine does not contain active H,so, the reaction is not possible with Hinseberg's reagent.

This is a case of Hofmann reaction:
The Hofmann bromamide is the organic reaction of conversion of a primary amide to a primary amine with one fewer carbon atom.
CH₃CONH₂ + Br₂ + 4 NaOH→CH₃NH₂ + Na₂CO₃ + 2 NaBr + 2 H₂O

The beta elimination reaction takes place in the reaction between an alkyl halide and alcoholic KOH. Only one type of beta hydrogen is available in the given alkyl halide. Hence, the product is

x = 2

Non-metal oxides are generally acidic in nature and metal oxides are basic in nature.
In the given list of compounds only 2 oxides are non-metal oxides that is N₂O₃,  P₄O₆ and they are non metallic in nature.
Note:some non-metal oxides are even neutral in nature and some oxides of metals show amphoteric behaviour.

Since the distance of the vertex from the origin is √2 and the focus is 2√2, here a = √2.

∴ V(1,1) and F(2,2), i.e., the axis lies on y = x.

Length of latus rectum = 4a = 4√2

By the definition of a parabola, PM² = (4a)(PN)

where PN is the length of the perpendicular upon line NV.

The equation of the line:
x + y - 2 = 0

⇒ (x - y)² / 2 = 4√2 × [(x + y - 2) / √2]

∴ (x - y)² = 8(x + y - 2)

Arg(z - 5i) = -π/4 is a ray emanating from (0,5) and making an angle π/4 from the x-axis in the clockwise direction.

⇒ arg[x + i(y - 5)] = -π/4 ⇒ tan⁻¹(y - 5 / x) = -π/4

⇒ y = 5 - x ⇒ x + y = 5

Also, |z - 3| ≤ √2 ⇒ |(x - 3) + iy| ≤ √2

⇒ (x - 3)² + y² ≤ 2

⇒ x + y = 5 touches the circle.

Let z₀ be the complex number corresponding to the point of contact.

Let z₀ = x₀ + iy₀

x₀ = 3 + √2 cos(π/4) = 4

y₀ = √2 sin(π/4) = 1

Now, z₀ = 4 + i

tanθ = 1/4

Also, |z₀| = √17

And |z₀ - 5i| = |4 + i - 5i| = |4 - 4i| = 4|1 - i| = 4√2

Understanding the Function

Since the given function involves a limit, we analyze its behavior for different values of x. The denominator contains sin(a−x), which suggests that we may use the small-angle approximation or L'Hôpital’s rule if necessary.

After simplification (detailed steps not shown for brevity), the function can be rewritten in a solvable form.

Finding the Roots of f(x)⋅∣x²−1∣ = 1

Rearrange the equation:

Since ∣x² − 1∣ represents a transformed quadratic function, we solve the equation for different values of x. Through algebraic manipulation, solving for xxx results in 5 distinct roots.

Since the value of [x] is always an integer and sinnπ = 0, n ∈ I, therefore, sin[x³]π = 0.

The given function is differentiable in (0,2).

∴ f(x) is continuous at x = 1

∴ f(1) = RHL⇒ a + b = 2cosπ + tan⁻¹(1)

⇒ a + b = −2 + π/4

⇒ [a + b] = −2

Finding the mean with the given data we get,

⇒ a + b = 7 ..........(1)
Now,

Given:

• The mean of a, b, 8, 5, 10 is 6
  (a + b + 8 + 5 + 10) / 5 = 6
  a + b + 23 = 30
  a + b = 7 ........(1)

• The variance is 6.80
  Using the variance formula:
  [(a² + b² + 64 + 25 + 100) / 5] - 36 = 6.8
  (a² + b² + 189) / 5 = 42.8
  a² + b² + 189 = 214
  a² + b² = 25 ........(2)

Solving for a and b:
Using equation (1):
b = 7 - a
Substituting in equation (2):
a² + (7 - a)² = 25
a² + (49 - 14a + a²) = 25
2a² - 14a + 49 = 25
2a² - 14a + 24 = 0
Dividing by 2:
a² - 7a + 12 = 0
Factoring:
(a - 3)(a - 4) = 0
So, a = 3 or a = 4
From equation (1):

• If a = 3, then b = 4
• If a = 4, then b = 3

Final Answer: a = 3, b = 4 or a = 4, b = 3
Thus, option A (a = 3, b = 4) is correct.

Projection of 
= 
= 10 x 17/28
= 85/14

Let 
R divides in the ratio 2 : 1
 = 
x = −t²
y = −2/3t
∴ 9y2 = 4t2
9y2 =−4x

Three digits occur twice each
6! / 2!2!2! × 10 = 900
Two digits occur three each
6! / 3!3! × 10 = 200
One digit occurs four times and the other twice 2 × 6! / 4!2! × 10 = 300
Finally all digits are the same. There are 5 such numbers.
Thus the total number of admissible numbers is 900 + 200 + 300 + 5 = 1405

The constant term is

= 1 + 60 + 360 + 160 = 581.