JEE AMU B.Tech Entrance Exam Mock Test- 1 Free Online Test 2026

Since E and I are in the same phase.
Therefore, phase difference will be 0 and since power factor= cosx (where x= phase difference) and x =0
therefore, cos x or power factor will be =1

Mass of Iron = 100g
Water Eq of caloriemeter = 10g
Mass of water = 240g
Let the Temp. of surface = 0^ºC
S_iron = 470J/kg°C 

Total heat gained = Total heat lost.

So,100/1000× 470 × (θ – 60) = 250/1000 × 4200 × (60 – 20)
⇒ 47θ – 47 × 60 = 25 × 42 × 40
⇒  θ = 4200 + 2820/47= 44820/47 =953.61°C

Explanation: Total voltage= VR+VL+VC.
VR=1000x3x10-3=3V.
Therefore, total voltage = 30+18+3=51V.

Coefficient of performance(cop) of a refrigerator,
β= Q₂/Q₁-Q₂=T₂/T₁-T₂=277/300-277=277/23≈12
OR, Q₂/W=12. Therefore, W=1J.Q₂=12J
Also, Q1=Q2+W
=(12+1) J
=13J

The tangential acceleration is clearly gsinθ, by simply taking the components of the tension vector.
And the net for on the bob is
F = T – mgcosθ
And it will be equal to centripetal force, so,
T – mgcosθ = Mv²/l

Since the block is moving down with constant speed, the net force acting on it along the plane is zero. This means the component of gravitational force down the incline is balanced by the frictional force. The force exerted by the plane on the block is the normal reaction, which depends on the angle of inclination. The normal force is given by:
N = mg cosθ
Thus, the force on the block by the plane is determined by the angle of inclination.

We know that potential energy of discrete system of charges is given by

According to question,

For heat added at constant volume, Work doe = 0
∆U = _T1∫^T2 nC_v,mdT
Or ∆U = nC_v,m∆T
∆T = ∆U/nC_v,m = 100/1×33.33 = 3
So, the correct answer is 3

In order to obtain ethoxy ethane from ethanol, the ethanol must be subjected to the effect of concentrated sulphuric acid at 413 Kelvin or 140°C as shown below:

The correct IUPAC name of the given compound is 3-ethyl-4,4-dimethylheptane.

Sc³⁺ : 1s², 2s²p⁶, 3s²p⁶d⁰, 4s⁰; no unpaired electron.
Cu⁺ : 1s², 2s²p⁶, 3s2p⁶d¹⁰, 4s⁰; no unpaired electron.
Ni²⁺: 1s², 2s²p⁶, 3s²p⁶d⁸, 4s⁰; unpaired electron present.
Ti³⁺ : 1s², 2s²p⁶, 2s²p⁶d¹, 4s⁰; unpaired electron present Co²⁺ : 1s², 2s²p⁶, 3s²p⁶d⁷, 4s⁰; unpaired electron present So from the given option the only correct combination is Ni²⁺ and Ti³⁺

The formation of NCl₃ be like
½ N₂ + 3/2Cl₂ ⇋ NCl₃
We can see that for this setup, we need to have eqn (ii) divided by 2, reversing eqn (iii) and multiplying it by 3/2 and then adding all these to equation (i).
So option a is correct.

Given, pKa = 3.45
Concentration of HF = 0.1 M, concentration of KF = 0.300 M
For acidic buffer;
pH = pKa + log [salt of weak acid]/[weak acid]
= 3.45 + log0.3/0.1
= 3.45 + 0.48
= 3.93

Explanation of Chelating Ligand
A chelating ligand is a molecule that forms multiple bonds with a single metal ion. In other words, it can bind to a metal ion at more than one point, creating a ring-like structure. This process is known as chelation. Chelating ligands have two or more donor atoms, which allow them to form multiple bonds with a single metal ion. These ligands are used in various fields such as medicine, chemistry, and environmental science due to their ability to sequester metal ions.

Examination of Options
A: Cl^- Chloride ion (Cl-) is a monodentate ligand, meaning it can bind to a metal ion at only one point. Therefore, it is not a chelating ligand.
B: DMG DMG, or Dimethylglyoxime, is a type of ligand that can bind to a metal ion at two points, forming a five-membered ring. Thus, DMG is a chelating ligand.

C: H₂O Water (H₂O) is a monodentate ligand, meaning it can bind to a metal ion at only one point. Therefore, it is not a chelating ligand.
D: OH^- Hydroxide ion (OH^-) is a monodentate ligand, meaning it can bind to a metal ion at only one point. Therefore, it is not a chelating ligand.

Conclusion
Therefore, out of the given options, only DMG (Dimethylglyoxime) is a chelating ligand. Hence, the correct answer is B: DMG.

(i) CO₂ and OCS both are sp-hybrid ised and have linear geometry.

(ii)   both are sp³-hybridised and have tetrahedral geometry.

 

k + |k + z²| = |z|²
Hence k and |z²| are collinear.
Now k ∈ R⁻
Since, k and z² is collinear, then z² has to be purely real.
Thus, arg(z²) = (2n − 1)π
Now z  =re^iθ
z² = r²e^2iθ = me^2iθ
Hence arg(z²) = 2arg(z)
Therefore, arg(z) = 1/2 arg(z²)

Putting n = 1, we get
arg(z) = π/2
Hence, option 'C' is correct.

so it is reflexive

so, it is symmetric 

So, it is not transitive. 

f(x) = e^{{4{x}+3}=e{4(x}}=e{4x}}
because {4{x} + 3} = {4{x}} and 4{x} = 4x − 4[x]⟹{4{x}} = {4x}

Equation of the tangent to  whose slope 

From symmetry of ellipse it is obvious that area of the triangle ABC will be same with respect either tangent.
Let us consider 

Hence area Δ of the triangle ABC is:

Let equation of hyperbola and ellipse be

⇒ a₁e₁ = a₂e₂ (since, same pair of foci)
e₂ = 1/2 (given)
Both intersect at (2,2)

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

Given that, T_p = a + (p − 1)d = q …….(i) and

T_q = a + (q − 1)d = p ……. (ii)

From (i) and (ii), we get d = [−(p − q)] / [(p − q)] = −1

Putting value of d in equation (i), then a = p + q − 1

Now, r^th term is given by A.P. T_r = a + (r − 1)d

= (p + q − 1) + (r − 1) (−1)

= p + q − r