JEE AMU B.Tech Entrance Exam Mock Test- 7 Free Online Test 2026

Energy of incident light E = 12375/2000 = 6.18 eV
According to relation E = W₀ + eV₀ 

T = 2u sin θ / g = 10 sec
⇒ u sin θ= 50
So, 
= 125 m

Moment of inertia

For the centre of rod

∴ The percentage error in density is

= 3% + 3(2%) = 3% + 6% = 9%

The fastest possible rate of rotation of a planet is that for which the gravitational force on material at the equator just barely provides the centripetal force needed for the rotation. Let M be the mass of the planet R its radius and m the mass of a particle on its surface. Then

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

The correct answer is Option a - V C₁/(C₁ + C₂)

Because the circuit is isolated when the battery is removed, the conservation of charge applies to the two-capacitor system.

The initial charge on the charged capacitor is Q = C₁V.

After connection, the two capacitors share charge and attain a common potential V' = Q / (C₁ + C₂), since total capacitance is C₁ + C₂.

Substituting Q = C₁V gives the common potential V' = (C₁V) / (C₁ + C₂), which is the potential difference across the combination.

Radius of electron orbit
r = 0.72Å = 0.72 × 10⁻¹⁰m
Frequency of revolution of electron in orbit of given atom
v = 9.4 × 10¹⁸ rev/s
(where T is time period of revolution of electron in orbit)
∴ Then equivalent current is
I = e/T = ev = 1.6 × 10⁻¹⁹ × 9.4 × 10¹⁸
= 1.504A

For an ideal solution, ΔH and Delta ΔV for mixing should be zero. P_Total = p_A + p_B and A − A, B − B and A − B interactions are nearly same.

Only (1) can show geometrical isomerism.

Bond order =

Bond order of

= 1

has lowest bond order among the given options hence the bond length will be longest.

Oxidation Number Transition in the Conversion of Br₂ to BrO₃

• The oxidation number of an element in its free or uncombined state is zero. This rule applies to bromine in Br₂, so the initial oxidation state of Br in Br₂ is zero.
• In BrO₃^-, the combined state of Bromine, Oxygen has an oxidation state of -2. However, the overall charge of the ion is -1. Since we have three Oxygen atoms, the total contribution of Oxygen is -6. To balance this, the Bromine must have an oxidation state of +5.
• Therefore, in the conversion of Br₂ to BrO₃^-, the oxidation state of Bromine changes from 0 in Br₂ to +5 in BrO₃^-
• Hence, the correct answer is Option A: zero to +5.
   

Molybdenum (Mo) is used in X-ray tubes because of its high melting point. This property makes it suitable for withstanding the extreme temperatures involved in operation.

Equation of any tangent to parabola  

If the tangent to parabola is also tangent to the circle (x-3)² + y² = 9 then:

9m⁴ + 6m² + 1 = 9m⁴ + 9m²
3m² – 1 = 0

From the figure it is evident that for the tangent above x-axis   Hence equation of required common tangent is:

= x + ln being a constant of integration.

• Given Vectors:
• A = 7i - 11j + k
• B = 5i + 3j - 2k
• C = 12i - 8j - k
• Step 1: Calculate Lengths of the Sides
• The lengths of the sides are calculated using the distance formula:
• Distance AB:
• AB = B - A = (5 - 7)i + (3 + 11)j + (-2 - 1)k = -2i + 14j - 3k
• |AB| = √((-2)² + (14)² + (-3)²) = √(4 + 196 + 9) = √209
• Distance BC:
• BC = C - B = (12 - 5)i + (-8 - 3)j + (-1 + 2)k = 7i - 11j + k
• |BC| = √((7)² + (-11)² + (1)²) = √(49 + 121 + 1) = √171
• Distance AC:
• AC = C - A = (12 - 7)i + (-8 + 11)j + (-1 - 1)k = 5i + 3j - 2k
• |AC| = √((5)² + (3)² + (-2)²) = √(25 + 9 + 4) = √38
• Step 2: Determine the Type of Triangle
• The lengths are:
• |AB| = √209
• |BC| = √171
• |AC| = √38
• Triangle Type Analysis:
• a) Equilateral Triangle: All sides equal. (Not applicable here)
• b)Isosceles Triangle: At least two sides equal. (Not applicable here)
• c)Right-Angled Triangle: Check if |AC|² + |BC|² = |AB|²: (38 + 171 = 209) (True)
• Therefore, it is a right-angled triangle.

Since mean (X + b) = E( X + b) = E(X) + b and Var (X + b) = Var X,
We get correct mean as 30 + 2 = 32 g and (s.d.) standard deviation as 2 g.

Hence the result.

Let A: First 3 cards are king

B: Fourth card is a king

P(A/B) = P(A ∩ B) / P(B)

Let k: drawing of a king

B can occur in four types:

Type 1: k̅ k̅ k̅ k → 1

Type 2: k k̅ k̅ k → 3

Type 3: k̅ k k̅ k → 3

Type 4: k k k k → 1

= 1 / 20825

∴ a = 5

Comparing on both sides of the equation


⇒ θ = π/3

If x ∈ (0, 1), we have [x] = 0
0 ≤ y ≤ 2 √x
And if x ∈ (1, 2), we have [x] = 1
(x - 1) ≤ y ≤ 2 √x

Tangent at (1, − 2) to x² + y² = 5 is
x − 2y − 5 = 0 ... (i).
Centre and radius of
x² + y² − 8x + 6y + 20 = 0 are
C (4, − 3) and radius r = √5
Perpendicular distance from
C (4, − 3) to (i) is radius.
∴ (i) is also a tangent to the second circle.
Let P (h, k) be the foot of the drawn circle from C (4, − 3) on (i)

∴ (h, k) = (3, −1)

The equation of the normal at (x₁, y₁) to the given ellipse is

Here x₁ = ae and y₁ = b²/a
So the equation of the normal at positive end of the latus rectum is