JEE AMU B.Tech Entrance Exam Mock Test- 3 Free Online Test 2026

According to Kepler's Law, the square of the time period of revolution of a planet around the sun is directly proportional to the cube of semi major axis of its elliptical orbit.

T²αR³

At infinity, image of B will form at focus since it is at infinity. Image of A will be at A which can be calculated by mirror formula.
1/v​+(1/−u)​=(1/−f)
⟹v= fu /(f−u)​=−( fu/(u−f)​)
Image length = v−f=(fu/(u−f)​)−(f)

= f^2​/(u−f)
(We take the absolute values of the distances to calculate the rod length)

Translational (KE) per unit volume

T_Initial  = t K
Work, W = 9R
Ratio of specific heats, γ = C_p / C_v = 4/3
In an adiabatic process, we have
W = R(T_Final – T_initial) / (1-γ)
9R = R (T_Final – t) / (1 – 4/3)
T_Final – t = 9 (-1/3) = -3
T_Final  = (t-3) K

Frequency of the echo detected by the driver of the train is (According to Doppler effect in sound)

where f = original frequency of source of sound f' = Apparent frequency of source because of the relative motion between source and observer.

Apparent frequency of S₁ and S₂ heard by observer is 

∴ 

For the process C → A, ΔU = UA - UC = 0 - 80
⇒ ΔU = -80 J
and W = area ACED = area ACB + area ABED
∴ W = (1/2 × AB × BC) + (DE × DA)
∴ W = (1/2 × 2 × 60) + (2 × 30) = 120 J
Since, in this process the volume decreases, the work will be negative (w = 120 Joule). That is, the work will be done on the system. Now, by the first law of thermodynamics, we will have
Q = ΔU + W = -80 - 120 = -200 J
Since it is negative, this heat is given out by the system and work done in the whole cycle
= area ABC = (1/2 × 2 × 60) = 60 J
Since the cyclic process is traced anticlockwise, the net work will be done on the system.

The angle of incidence = 90^∘ - 47.5^∘ = 42.5^∘
Now by snells’ laws we get 
sin r = sin 42.5^∘ / 1.66^∘
Thus we get r = 24.01^∘
Thus the angle with surface = 90^∘ -24.01^∘ = 65.99^∘

The velocity of the body at any point during its motion is composed of:

1. Horizontal Component (v_x): This remains constant throughout the motion because there’s no horizontal acceleration.
   
   Using cos ⁡60^∘ = 1/2
   
2. Vertical Component (v_y): This changes with time due to the acceleration due to gravity (g = 9.8 m/s²). At the point where the angle with the horizontal is 30^∘, the relation between the components of velocity is:
    ​ ​
   Using tan ⁡ 
    ​ ​
   Solve for v_y​ :
   
3. Magnitude of the Velocity (v): The total velocity is the vector sum of v_x and v_y​ :
   
   Substitute
   
   

Final Answer:
The speed of the body at the point where it makes an angle of 30^∘ with the horizontal is: 20 m/s

The momentum of the initial particle remains conserved. Hence the emitted particles’ COM will remain along +ve x-axis.

Just before Releasing:

(a) Velocity of stone = Velocity of balloon
(b) Acceleration of stone = Acceleration of balloon
Just After Releasing:
(a) Velocity of stone = velocity of balloon
(b) Acceleration of stone = 9.8 m/s² downwards
Height at which stone is released = H
= 
Velocity with which stone is released = V
= 0 + 4.9 × 2 = 9.8 m/sec
∴ H_max above ground level = 
= 14.7 m

The equation of NAND gate is 
Y=Aˉ.Bˉ
When both the terminals are connected together.
A=B=A
Y=Aˉ.Aˉ=Aˉ.Aˉ=Aˉ
Which is nothing but the equation of NOT gate.
Y=Aˉ
Option A is the correct answer.

Output X = A.B

As,
B=μ_onI/2a
a ->radius
B ∝1/a
B₁/B₂=a₂/a₁
B₁=2B₂
B₂=(1/2) x B₁,
Magnetic field is halved.

Key Points

•  C₂H₅Br + KOH → C₂H₄ + KBr +H₂O
• The above reaction is the reaction between alkyl halide with alcoholic alkali.
• When ethyl bromide reacts with alcoholic potassium hydroxide (KOH), ethene (C₂H₄) is formed.
• This reaction is called an Elimination or dehydrohalogenation reaction.

Additional Information C₂H₂

• C₂H₂ is the chemical formula of acetylene.
• It contains a triple bond between two carbon atoms.

C₂H₆

• C₂H₆ is the chemical formula of ethane.
• It is the second member of the alkane group.
• It is a straight-chain alkane containing two carbon atoms having single covalent bonds.

CH₄

• CH₄ is the chemical formula of methane.
• It is the first member of the alkane group.
• It is a gaseous compound.

C₃H₆

• C₃H₆ is the chemical formula of propene and cyclopropane.
• Propene is an alkene having double bonds.
• Cyclopropane is a cyclic ring alkane containing 3 carbon atoms.

The correct answer is option C
Conformers are isomers having the same bond connectivity sequence and can be interconverted by rotation around one or more single (σ) bonds. Hence, the two structures represent conformers.

Let's check statement (a) 
Statement(a) : ΔrH° for the reaction H₂O(g) → 2H(g) + O(g) is 925 kJ/mol
For this, we need - (II) + (III) + ½ (IV)
We get, H₂O(g) → 2H(g) + O(g) - (-242)+436+½ 495 = 925.5 kJ mol^-1
So it is true.
Let's check statement (b)
Statement(b) : ΔrH° for the reaction OH(g) → H(g) + O(g) is 502 kJ/mol
For this we need -(I)+½ (III)+½ (IV)
We get OH(g)   →   H(g)  + O(g)       -(42) + ½ (436) + ½ (495) = 423.5 kJ mol-1
So statement (b) is wrong.
Let's check statement (c)
Statement(c) : Enthalpy of formation of H(g) is -218 kJ/mol
We can see that for enthalpy of formation, we need to divide eqn (III) by 2
So, it would become :- 
½ H₂(g) → H(g)
436/2 = 218 kJ
So, statement (c) is wrong.
Let's check statement (d)
Statement(d) : Enthalpy of formation of OH(g) is 42 kJ/mol
For that, we have eqn (I) as it is. So, statement (d) is correct.

(1) Reaction (i) is stereospecific. In this reaction, inversion of configuration occurs due to S_N² mechanism.

(2) Reaction (ii) is non-stereospecific. There is change in configuration due to racemisation.

(3) Reaction (iii) is stereospecific. There is no change in configuration as this reaction takes place through S_N2 mechanism.

• Addition through B₂H₆ follows the anti-Markovnikov rule.
• A pair of diastereomers are formed.

Any atom exhibit oxidation state in which it is found to be stable. It is second element of Lanthanide series having atomic number 58. General electronic configuration is
[Xe] 4f¹ 5d¹ 6s².
Lanthanide are very less electronegative. Their electronegative value is nearly equal to s-block elements. So they can lose electrons to form cation. When Cerium loses 4 e-, it will acquire fully field electronic configuration of Xenon (2,8,18,18,8). As fully filed or half filed electronic configuration possesses extra stability.
Therefore, Cerium exhibits +4 oxidation state.

Δ_rH=Δ_rH_(product)-ΔrH_(reactant)
=3×-393+81+3×110-9.7
=-777.7kJ
ΔH = ΔU+Δn_gRT
AS Δn_g = 0, ΔH = ΔU
So, ΔU = -777.7 kJ

Because this is a gas-phase reaction at constant temperature, concentration is proportional to pressure. For an n-th order reaction (n≠1),

Doubling the pressure from 50 to 100 mm Hg doubles the initial concentration, so

Required Probability = 5/6 x 1/6 + (5/6)³ x 1/6 + (5/6)⁵ x 1/6 + ....
= 1/6 x 5/6 / (1-25/36)
= 5/11

The given equation is:

Its solution is:

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