UGEE SUPR Mock Test-1 - JEE MCQ

When viewing a stack of transparent layers perpendicularly, the total apparent depth is the sum of each layer’s actual depth divided by its refractive index.

• The vessel has a total depth of 2d, with the bottom half (depth d) filled with a liquid of refractive index μ₁​ and the top half (depth d) with μ₂.
• Apparent depth of the bottom layer = d/μ₁​.
• Apparent depth of the top layer = d/μ₂​.
• Total apparent depth , which matches option B.

As the body falls from height h, it gains velocity

On entering the lake, it experiences an upward net force due to the difference between the buoyant force and its weight.

The net upward acceleration is

Using the equation v² = 2as, we calculate the maximum depth s: 

The r.m.s. velocity of a gas molecule can be calculated using the formula:

• v_rms = √(3RT/M)

Where:

• R is the universal gas constant
• T is the temperature in Kelvin
• M is the molar mass

To find the r.m.s. velocity of hydrogen at 127°C:

• Convert 127°C to Kelvin: T = 127 + 273 = 400 K
• The molar mass of hydrogen (H₂) is 2 g/mol or 0.002 kg/mol
• Comparing the r.m.s. velocities of hydrogen and oxygen, use the formula:
• (v_rms of H₂ / v_rms of O₂) = √(M_O2 / M_H2) × √(T_H2 / T_O2)
• Given: v_rms of O₂ = 474 m/s at 16°C (289 K)
• Molar mass of oxygen (O₂) is 32 g/mol or 0.032 kg/mol
• Calculate:
• v_rms of H₂ = 474 × √(0.032 / 0.002) × √(400 / 289)
• v_rms of H₂ ≈ 2230.59 m/s

Thus, the r.m.s. velocity of hydrogen at 127°C is 2230.59 m/s.

The coefficient of friction determines the maximum speed a car can safely travel around a curve without skidding.

• The given coefficient of friction is 0.25.
• The radius of the curve is 20 m.
• Gravitational acceleration (g) is 9.8 m/s².

To find the maximum speed, use the formula for the maximum speed on a curve:

• v = √(μ × g × r)
• Here, μ = 0.25, g = 9.8 m/s², and r = 20 m.

Substitute the values into the formula:

• v = √(0.25 × 9.8 × 20)
• Calculate the result: v = √(49)
• Therefore, v = 7 m/s.

The maximum speed without skidding is 7 m/s.

The work done W is calculated using the formula:

W = F · s · cos(θ)

• F = 10 N
• s = 2 m
• θ = 60°

Since cos(60°) = 0.5, the calculation is as follows:

W = 10 × 2 × 0.5 = 10 J

Thus, the work done by the boy is 10 J.

Upthrust on ball = weight of displaced water

= Vρg = (m/ρ)ρg = (40/0.8) × 1 × g = 50g

As the ball is sunk into the water with a pin by applying downward force equal to upthrust on it. So reading of weighing pan

= weight of water + downward force equal to upthrust = 600 + 50

= 650gm

let sin x = t

1. Use partial fractions:

2. Set up:

3. Coefficients:

4. Solve: Test D: 2, -7, 5:

5. Conclusion: A = 2, B = −7, C = 5 satisfies all, so the answer is D: 2, -7, 5.

1. Identify Intersection Points: Set √(8x) = 2x. Square both sides:

• 8x = 4x²
• 4x² - 8x = 0
• x(4x - 8) = 0

Solutions: x = 0 and x = 2.

2. Determine Upper and Lower Functions: For 0 < x="" />< 2,="" test="" a="" value="" (e.g.,="" x="" />

• y_parabola = √(8(1)) ≈ 2.828
• y_line = 2(1) = 2

Thus, y = √(8x) is above y = 2x.

3. Set Up the Integral: The area A is given by:

A = ∫₀² (√(8x) - 2x) dx.

4. Compute Each Part of the Integral:

• Integrate √(8x) to get:
• (8)^1/2 x^1/2 = 2√(2) (2/3) x^3/2 = (4√(2)/3) x^3/2
• Integrate 2x to get:
• x²

5. Evaluate the Integral from 0 to 2:

• Calculate ∫₀² √(8x) dx = [(4√(2)/3) x^3/2] from 0 to 2 = (4√(2)/3) (2)^3/2 = (4√(2)/3) * 2.828 ≈ (16/3).
• Calculate ∫₀² 2x dx = [x²] from 0 to 2 = 4.

6. Subtract the Two Results: A = (16/3) - 4 = (16/3) - (12/3) = (4/3).

We have,

Let us use the identity:

So,

Since cos⁡ (−x) = cosx,

The given equation c₁y = (c₂ + c₃)e^{x + c₄} can be simplified using properties of exponents to (c₂ + c₃)e^c₄ as a new constant, say C, making the equation c₁y = Ce^x. This results in only one arbitrary constant, leading to a first-order differential equation. Thus, the correct order is 1

(a) We have
x²dy - 2xydx = x⁴ cos x dx
⇒ 

∴ The solution of the given differential

Given, equation of line y = 2x + 1   ....(i)
Eq. (i) passing through the points  and (0, 1).

∴ Required area of shaded region

We know that, the sum of three vectors of triangle is zero.

∴ AB + BC + CA = 0
⇒ BC = AC - AB
 (since, M is a mid-point of BC)
And also, AB + BM + MA = 0

For a square matrix A of order 3, |A| = 5.
|adj A| = |A|^(n-1) = 5² = 25.
|A adj A| = |A| · |adj A| = 5 · 25 = 125.
Alternatively, A · adj A = |A|I, so |A adj A| = ||A|I| = |A|³ = 5³ = 125.
The correct answer is B: 125

In a first-order reaction, the concentration of a reactant decreases over time in a specific pattern. The key characteristics are:

• The rate of reaction depends directly on the concentration of the reactant.
• Logarithm of concentration (log C) decreases linearly with time.
• The expression for a first-order reaction is given by the formula: log C = log C0 - kt, where:
  • C is the concentration at time t.
  • C0 is the initial concentration.
  • k is the rate constant.
  • t is time.

This means that the graph of log C vs. time is a straight line with a negative slope.

BF₃, BCl₃, BBr₃ are sp² hybridised.
So, all have same and bond angle i.e. 120⁰.

Completing the reaction, we get

No. of lattice points = No. of octahedral
voids = 1/2 × No. of tetrahedral voids in ccp
structure
∴ No. of atoms of B = 4
No. of atoms of A = 1/2 × No. ofoctahedral voids

No. of atoms of O = All tetrahedral voids
= 2 × No. of lattice points = 2 × 4 = 8
Hence, A: B: O = 1 : 2 : 4
Therefore, the formula of the compound is
AB₂O₄

Let the densities of metal and water be ρ and ρ₀ respectively and let x be the length of the rod immersed in water at an instant of time t.
Then, acceleration at that instant = apparent weight divided by mass of the rod, i.e.