JEE AMU B.Tech Entrance Exam Mock Test- 6 Free Online Test 2026

∴

= 0.075 kg/m³

If the total energy is E that means that E = U(x) + KE and if KE = 0 then E = U(x)

From momentum conservation

M * 30 + m * 10 = Mv₁ + mv₂        ..(i)

e * velocity of approch = velocity of separation

on solving,

The correct option is C

Statement I is true, statement II is false

Assertion:

1. Forces can act on a system to conserve linear momentum, but their vector sum should be zero.
2. Hence if some external non-zero force is acting on the system, momentum changes. Statement I is true

Reason:

• For a system of moving particles, the law of conservation of linear momentum applies. It is not applicable to individual particles of a system.
• So the above statement II is not correct

Therefore, Statement I is true, and statement II is false. hence option (c) is the correct answer

PE ,

and K.E.

We have, m = ZIt
where, Z is the electrochemical equivalent
of copper.
⇒ m = 30 x10^-5 x1.5 x10 x 60
= 0.27 gm.

Step 1: Molar C_v values for gases

• Monoatomic gas (like He, Ne):
  C_v = (3/2)R = 12.5 J/mol·K

• Diatomic gas (like O₂, N₂ at room temp):
  C_v = (5/2)R = 20.8 J/mol·K

Let’s use R = 8.314 J/mol·K for accurate values:

• C_v (monoatomic) = 3/2 × 8.314 = 12.471 J/mol·K

• C_v (diatomic) = 5/2 × 8.314 = 20.785 J/mol·K

Step 2: Total heat capacity of the mixture

We have:

• 1 mole of monoatomic gas

• 3 moles of diatomic gas

Total heat capacity at constant volume:

C_v(total) = (1 mol × 12.471) + (3 mol × 20.785)
= 12.471 + 62.355 = 74.826 J/K

Step 3: Molar specific heat of the mixture

Total moles = 1 + 3 = 4 moles

So, C_v (mixture) = 74.826 / 4 = 18.7065 J/mol·K

= 10^-3 x 100[(32)² - (30)²] = 1240 cal

Hooke’s law: a law stating that the strain in a solid is proportional to the applied stress within the elastic limit of that solid.
In the OA line Hooke’s law is valid because stress is directly proportional to strain.

Explanation:

At const volume, 

Q = 500 J

Q=nCPΔT

500=1×2.5×8.31ΔT

ΔT=24.06

W=nRΔT=1×8.31×24.06=200J

Given,
 l=2m
m=1kg
ω=15rd//s
K.E.=1/2I ω
=1/2x(ml²/3) ω²
=1/2x(1x4/3)(15)²
=150J

∆H = ∆E + ∆n_gRT
176 = ∆E + 1×8.314×1240/1000
∆E = 176 - 10.30 = 165.69 kJ

For the strongest oxidizing agent, the oxidizing potential should be least. Here, the oxidizing potential of Fe⁺² is less than that of [Fe(CN)₆​]⁴⁻. Therefore, Fe⁺² is stronger oxidizing agent than [Fe(CN)₆​]⁴⁻. Also, the stronger oxidizing agent should easily reduce itself. Here, Fe⁺³ is easily reduced than Fe⁺². Therefore, among all the four, Fe⁺³ is the stronger oxidizing agent.

Angular nodes in 4s = 0

Only one pair of enantiomers is possible for cis-2,2-dibromobicyclo [2.2.1] heptane. The trans arrangement of one carbon bridge is structurally impossible. Such a molecule would have too much strain.

The addition in unsymmetrical alkenes follows the Markovnikov rule, according to which the negative part of the addendum goes to the carbon having lesser number of hydrogens.

Number of ten degrees in 
Increase in rate due to 10° rise in temperature = 2
Increase in rate due to 50^° rise in temperature = 2⁵ = 32 times

(iii) and (iv) will form ideal solutions hence do not form azeotropes. Azeotropes have same composition in liquid and vapour form when distilled.

To determine which binary mixtures will have the same composition in the liquid and vapor phases, we need to identify mixtures that form ideal solutions. In an ideal solution, the composition of the liquid phase and the vapor phase is the same at equilibrium.

Here’s a brief analysis of each mixture:

1. Ethanol + Chloroform:

   • This mixture does not behave ideally due to strong hydrogen bonding interactions between ethanol and chloroform, which can cause deviations from Raoult's Law.

2. Nitric Acid + Water:

   • Nitric acid and water form a non-ideal solution. Nitric acid forms strong hydrogen bonds with water, resulting in significant deviations from ideal behavior. Thus, the composition in the liquid and vapor phases will not be the same.

3. Benzene + Toluene:

   • Benzene and toluene form an ideal solution. The interactions between benzene and toluene are similar, and thus the composition of the liquid and vapor phases will be the same.

4. Ethyl Chloride + Ethyl Bromide:

   • Ethyl chloride and ethyl bromide also form an ideal solution. The interactions between these two similar substances lead to minimal deviations from ideal behavior.

Based on the above analyses, the mixtures that will have the same composition in the liquid and vapor phases are:

2. (iii) and (iv)

So the correct answer is:

2. (iii) and (iv)

Many trivalent lanthanide ions are coloured due to the presence of f-electrons, in solid state as well as in aqueous solutions.

The number of P-O-P bond in cyclic metaphosphoric acid is three.

Given, D.R.'s of OA are a 0, b - 0, c - 0 or a, b, c.

Equation of any plane through A, i.e. (a, b, c) is: A(x – a) + B(y - b) + C(z - c) = 0

But plane is at right angles to OA which therefore is normal and whose D.R.'s are a, b, c.

So, the plane is, a(x - a) + b(y - b) + c(z - c) = 0

ax + by + cz = a² + b² + c²

Equation of tangent for parabola y² = 8x:

n(A × A) = 9
Number of relations containing (1, 2), (2, 1), (a, a) = 2³
Number of relations containing (1, 2), (2, 1), (1, 3), (3, 1), (a, a) = 2³
Number of relations containing (1, 2), (2, 1), (2, 3), (3, 2), (a, a) = 2³
Number of relations containing (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2), (a, a) = 2³
⇒ Total number of symmetric relations = 32.

Mean, np = 25 and q < 1

⇒0 ≤ σ< 5

Total number of ways of choosing two numbers out of 30 is ³⁰C₂ We rewrite the first 30  natural numbers in three rows as follows:

Row I: 1, 4, 7, 10, 13, 16, 19, 22, 25, 28
Row II: 2, 5, 8, 11, 14, 17, 20, 23, 26, 29
Row III: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30

For x² - y² to be divisible by 3, either both x and y must be chosen from the same row or exactly one of x, y from Row I and the other from Row II.

Thus, the number of favourable ways

= 3 (¹⁰C₂) + 10 x 10 = 235

∴ probability of the required event

1. Express the three terms in A P
   Since a, b and c are in arithmetic progression, there is some common difference d so that
   a = b – d
   c = b + d

2. Write down the product condition
   We are given a·b·c = 64, so substitute a and c:
   (b – d) × b × (b + d) = 64

3. Expand the product
   First multiply (b – d)(b + d) = b² – d².
   So the left side becomes b × (b² – d²) = b³ – b·d².
   Hence the equation is
   b³ – b·d² = 64

4. Solve for d² in terms of b
   Rearrange to isolate d²:
   b·d² = b³ – 64
   d² = (b³ – 64) ÷ b = b² – (64 ÷ b)

5. Impose the reality condition
   Because a, b, c must all be real, d² cannot be negative. So
   b² – (64 ÷ b) ≥ 0

6. Find the range of b
   Multiply both sides by b (which is positive) to avoid fractions:
   b³ – 64 ≥ 0
   b³ ≥ 64
   Taking cube roots (remembering b > 0), we get
   b ≥ 4

7. Determine the minimum
   The smallest value of b satisfying b ≥ 4 is b = 4.
   Check that this works: if b = 4 then d² = 4² – (64 ÷ 4) = 16 – 16 = 0, so d = 0 and a = b = c = 4, giving product 4·4·4 = 64.

Therefore, the minimum possible value of b is 4.

The mean of a probability distribution is given as
M = Σ P_ix_i = (¼ × 0) + (0 × 1) + (¼ × 2 ) + (0 × 3) + (½ × 4) + (0 × 5)= 5/2

σ² = Σ P_i (M – x_i)²