JEE AMU B.Tech Entrance Exam Mock Test- 10 Free Online Test 2026

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We know that,
I=V/R​
∴​I₁/I₂ ​​= ​R₂/ R₁
​​R₁​=90Ω​
Current flowing through 90Ω resistance is reduced by 90%
∴ Current ratio =10%:90%
=1:9
∴1/9​=R₂​R₁
​​⇒1/9​= R₂​/90
​⇒R₂​=90/9​=10
∴R₂​=shunt=10Ω

Given,
λ= 4000 Å ;W=3.2× 10^-19 j
as we know that
K.E = (hc/ λ) - W
= (6.6× 10^-34 × 3 × 10⁸/ 4000×10^-10 )- 3.20× 10^-19 
= 1.75 × 10^-19
 

P=P1​+P2​=+2−1=+1 dioptre, lens behaves as convergent
F=1​/P=1/1​=1m=100cm

In an ideal environment where there is no resistance to oscillatory motion, that is, damping is zero, when we oscillate a system at its resonant frequency, since there is no opposition to oscillation, a very large value of amplitude will be recorded. Forced oscillation is when you apply an external oscillating force.

The cyclic process 1 is clockwise and the process 2 is anticlockwise. Therefore w₁ will be positive and w₂ will be negative area₂> area₁. Hence, the network will be negative.

Given : Mass, m = 3 kg;

Initial Velocity, u = 2m/s;

Final velocity, v = 3.5 m/s

Time taken, t = 10 seconds

v = u+at

3.5 = 2 +a x 10

F = ma = 3 x 0.15 = 0.45 N

Since the applied force increase the speed of the body, it acts along the direction of motion.

Φ= hνλ => 2eV= 6.626 x 10^-34 x ν

2 x 1.6 x 10^-19= 6.626 x 10^-34 x ν

On solving we get ν = 4.6 x 10¹⁴ Hz.

When Young's double-slit set up for interference is shifted from air to within water then the fringe width decreases because the refractive index of water is more than that of the air.
Originally the fringe width is given by:
β1​=λD/2d​
The new fringe width within water will be given by 
β2​= λD​/2nd
So, β2​= β1/n​​
Here, n is the refractive index of medium.
 

Given Data,
Length of the piece of copper = l = 19.1 mm = 19.1 × 10^-3m
Breadth of the piece of copper = b = 15.2 mm = 15.2× 10^-3m
Tension force applied on the piece of cooper, F = 44500N
Area of rectangular cross section of copper piece,
Area = l× b
⇒ Area = (19.1 × 10-3m) × (15.2× 10-3m)
⇒ Area = 2.9 × 10^-4 m²
Modulus of elasticity of copper from standard list, η = 42× 10⁹ N/m2
By definition, Modulus of elasticity, η = stress/strain

⇒ Strain = F/Aη

⇒ Strain = 3.65 × 10^-3
Hence, the resulting strain is 3.65 × 10^-3

Name of ligands name first in alphabetical order followed by anme of central ion.

The correct answer is option D
By treating diazonium salts with cuprous cyanide or KCN and copper powder it forms aryl nitrile. Illustrative is the preparation of benzonitrile using the reagent cuprous cyanide:

From the structure of three species we can determine the number of lone pair electrons on central atom (i.e. N atom) and thus the bond angle.

We know that higher the number of lone pair of electron on central atom, greater is the lp – lp repulsion between Nitrogen and oxygen atoms. Thus smaller is bond angle..
The correct order of bond angle is

NO₂^- <NO₂<NO₂⁺  i .e. opti on (b) is correct.

The correct answer is Option C.
The ionization equilibrium of silver (I) sulfide is 
Ag₂S⇌2Ag⁺ +S2⁻
The solubility product expression for silver (I) sulfide is KSP[Ag⁺]2 [S2⁻].
But [Ag⁺]² = x and [S²⁻] = y.
Hence the expression for the solubility product becomes K_SP[Ag⁺]² [S²⁻] = x²y.

The correct answer is Option B.

Ca_(s) + O2_(g) + H2_(g) → Ca(OH)₂ , ΔHf = ?

Desired equation = eq (iii) + eq(i) - eq (ii)

ΔH_f = (−151.80)+(−15.26)−(−68.37)
ΔH_f = (-151.80)+(-15.26)-(-68.37)
ΔH_f = −235.43KCalmol⁻¹

When acetone and chloroform are mixed together, a hydrogen bond is formed between them which increases intermolecular interactions. Hence, A − B interactions are stronger than A − A and A − B interactions.

Oxidation number of Cr changes from +6 to +3.

This reaction is called Gattermann reaction. In this reaction, Cl, Br and CN can be introduced into the benzene ring by simply treating diazonium salts with HCl, HBr, KCN. Respectively in presnce of copper powder instead of using Cu(I) salts.

The carbon atom involved in the formation of 1 π bond is sp² hybridised, and the C-atom involved in the formation of 2- bonds is sp hybridised.
In the formation of CH₂=C=CH₂ compound, only sp and sp² hybrid orbitals are involved,
i.e. sp² sp sp²
CH₂=C=CH₂

Zn reacts with 2,3 - dibromobutane to give alkene. As trans 2-butene is more stable than cis 2-butene(owing tosterric hinderancce), former is major product.

₂₈ Ni⁰ :

Remember CO is a strong ligand

Remember CN^– is also a strong ligand

Cl^– is a weak ligand, hence no pairing of electrons takes place.

Alkenes exhibit geometrical isomerism due to restricted rotation around the caron-carbon double bond. However, the alkene molecule should not have the same groups attached to the double bonded carbon atom.
Hence, but-2-ene (CH₃CH=CHCH₃) is the smallest alkene to exhibit geometrical isomerism.

3Mn²⁺(aq) + 2MnO₄^-(aq)+ 2H₂O (/) → 5MnO₂(s) + 4H⁺  E°_cell = 0.47 V
Since E°_cell > 0, hence spontaneous.
Thus, Mn²⁺ is oxidised to MnO₂ by MnO₄^- in acidic medium.

The C-C chain is zigzag and not linear. Hence, statement C is incorrect.
Some examples of alkanes are methane, ethane and propane. First three members do not exhibit isomerism, but the higher members (butane and onwards) exhibit chain isomerism. Hence, statement D is incorrect.

w (reversible) < w (irreversible) (for compression process)

• Work done is the area under the P−V curve. It can be seen in the curve above that the area under the P−V curve for irreversible compression of the gas is more than the area under the curve for reversible compression.
• Thus, work done for irreversible compression is more than that for reversible compression.

Ammonia is a Lewis base. It forms complexes with cations of transition elements. Pb²⁺ does not form a complex with ammonia as lead (Pb) is not a transition element. Lead (Pb) is a p - block element.

The given two vertices of the triangle are (3, -2) and (-2, 3), and its orthocenter is (-6, 1).

Let the third vertex be A(α, β).

Using the condition OA ⊥ BC:

⇒ Slope of OA × Slope of BC = -1

⇒ (β - 1) / (α + 6) × (3 + 2) / (-2 - 3) = -1

⇒ (β - 1) / (α + 6) × (5 / -5) = -1

⇒ (β - 1) / (α + 6) × -1 = -1

⇒ β - 1 = α + 6

⇒ α + 7 = β ....(i)

Similarly, using OB ⊥ AC:

⇒ Slope of OB × Slope of AC = -1

⇒ (-2 - 1) / (3 + 6) × (β - 3) / (α + 2) = -1

⇒ (-3 / 9) × (β - 3) / (α + 2) = -1

⇒ (-1 / 3) × (β - 3) / (α + 2) = -1

⇒ β - 3 = 3(α + 2)

⇒ β - 3 = 3α + 6

⇒ 3α - β + 9 = 0

⇒ 3α + 9 = β ....(ii)

Using equations (i) and (ii):

⇒ 3α + 9 = α + 7

⇒ 3α - α = 7 - 9

⇒ 2α = -2

⇒ α = -1

Substituting α = -1 in equation (i):

⇒ β = α + 7 = -1 + 7 = 6

Thus, the third vertex is (-1, 6).