JEE AMU B.Tech Entrance Exam Mock Test- 5 Free Online Test 2026

Let us consider 1 mole of an ideal gas at kelvin temperature T. It has N molecules (Avogadro's number). The internal energy of an ideal gas is entirely kinetic. The average KE per molecule of a ideal gas is ½ ​nkT (k is boltzman constant), where n is degree of freedom. Therefore, the internal energy of one mole of a gas would be
 
E=N(1/2​nKT)=1/2​nRT             (∵k=R/N​)
 
Now, C_v​=dE/dT​=n/2 ​R
 
and Cp​=n/2 ​R+R=(n/2​+1)R
 
​C_p/ C_v ​​=​(n/2​+1)R/n/2​=(1+2/n​)

Velocity of A and Velocity of B relative to A are as shown.

and direction of is away from 

Calculation:

For Case 1 :

For Case 2 :

From (i) and (ii)

288 f_app = 240 × 240

f_app = 200 Hz

According to Newton’s Law of Cooling, the rate of loss of heat of a body at any instant is directly proportional to the difference in temperature between the body and the surroundings at that instant.

If the temperature of the body is T₂ and that of the surroundings is T₁, and T₂ > T₁​, then the temperature of the body decreases with time.
So, the mathematical form is:

The negative sign shows that the body cools (temperature falls) when its temperature is more than that of the surroundings.

Let the radii be r₁​ and r₂​ respectively.
Since there are two turns of radius r₂​, r₁​=2r_2​
Magnetic field B at the centre of  the coil of radius r_1​ B_1​=​μo​i/2r₁​=​μ_o​i​/4r₂
Magnetic field B at the center of the coil of radius r_2​ B_2​=2×​μ_o​i​/2r₂
∴ B₂/B₁ =(2× μ_o​i/2r₂​)/(μ_o​i /4r₂​)​ ​​=4
Hence the answer is option C, four times its initial value.
 

K = power × thickness (m) ÷ area × ∆T
K = (J s^-1) (m) ÷ (m²) (K)
K = J m^-1 s^-1 k^-1

From figure, F = 6 mg
As speed is constant, acceleration a = 0
∴ 6 mg = 6ma = 0 
F = 6 mg
T = 5 mg , T' = 3 mg T" = 0
F_net on block of mass 2 m = T – T' – 2 mg = (5 - 3 - 2)mg = 0
Alternate Method:
 v  = constant  so, a = 0, Hence, F_net = ma = 0

PV=nRT

Hence the value of PV/T where the curves meet on the y-axis is 0.26 jK^-1

Step 1: Understanding the problem

• Speed of boat in still water = 5 km/h

• Width of river = 1 km

• Time taken to cross the river (shortest path, straight across) = 15 minutes
  (15 min = 15/60 = 0.25 hours)

We need to find: Speed of river water

Step 2: Finding effective speed across the river

To cross the river via shortest possible path means going straight across the river without drifting downstream.

• Effective speed across the river = distance ÷ time

• Effective speed = 1 km ÷ 0.25 hr = 4 km/h

So, the boat moves straight across at a speed of 4 km/h.

Step 3: Using Pythagoras theorem

Since the boat's speed of 5 km/h is at an angle to counteract the river's current, we have a right triangle:

• Boat’s speed in still water (hypotenuse) = 5 km/h

• Effective speed across the river (vertical component) = 4 km/h

• Speed of river current (horizontal component) = ?

According to Pythagoras theorem:

Boat speed² = (effective speed)² + (river speed)²

Thus:

5² = 4² + (river speed)²
25 = 16 + (river speed)²

Step 4: Solving for the river speed

• (river speed)² = 25 – 16 = 9

• river speed = √9 = 3 km/h

Three 2Ω resistors are in series. Their total resistance =6Ω. Now it is in parallel with 2Ω resistor, so total resistance,
1/R​=1/2+1/6​=3+1/6​=4/6=2/3
R=3/2​
∴I=RV​=3/(3/2)​=3×2​/3=2A

For the reaction, A+B⟶C+D, the  variation of the concentration of the products is given by the curve Y.
Initially, the product concentration is 0, then it gradually rises and reaches a maximum value. After that it remains constant. 

This behavior is represented by the curve Y.

For a buffer, we need to have a weak base/acid and its salt. Here, we have a weak base and so we will need its salt. In option c, concentration of strong acid is less than base. So we will have both a weak base and its salt. Therefore, only option c gives us a buffer solution.

O₅ (Xe)₄f¹⁴ sd⁶ 6s²
O₅⁸⁺(Xe) 4f⁴

Oxidation at Cr electrode (anode)

Thus, oxidation half-cell is

reduction at Fe electrode (cathode)

Thus, reduction half-cell is

Two half-cells are joined by a salt-bridge

T₅₀ is independent of concentration of [A], Thus, it follows first order kinetics.

Fe 2⁺ and Cr³⁺ are oxidised (increase in oxidation number)
O₂, is reduced .

Since, F^- ion is a weak ligand hence, pairing of electrons does not occur and outer orbital complex is formed in [CoF₆]^3-.

Since E° _cell < 0, disproportionation of Sn²⁺ to Sn⁴⁺ (by oxidation) and Sn (by reduction) is not spontaneous.

0 < sin⁸θ ≤ sin²θ ...(i)
and 0 < cos¹⁴θ ≤ cos²θ ...(ii)
Adding (i) and (ii) ⇒ 0 < A ≤ 1

Let, the end points of the diameter of circle are
(a sec θ, b tan θ) & (−a sec θ, b tan θ)
⇒ Equation of circle is
(x − a sec θ)(x + a sec θ)+(y − b tan θ)² = 0
⇒ x² − a²sec²θ + y² + b² tan²θ − 2b tan θ y = 0
Because (2,0) satisfies above equation,
hence, 4 − a²(1 + tan²θ) + 0 + b² tan 2θ = 0
⇒(4 − a²) + (b² −a²)tan²θ = 0
Which is always true if |a| = 2 & |b| =2