UGEE SUPR Mock Test-1 - JEE MCQ

Forward resistance

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.

This happen when vertical velocity of both are same.

W = F s cos θ = 10 × 2 cos 60° = 10J.

since both P and m are constants

In tunnel body will perform SHM at centre
V_max = Aω (see chapter on SHM)

PV^3/2 = K

Length of latus rectum = 2 × distance of focus from directrix

We know that the surface area of cube is given by S = 6x²   
∴ 
Δx = 5% of x = 05x
The surface area of' a cube is increased by

let sin x = t

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=1):

• 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).

(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

Given, |A| = 5


To solve for |A adj A|, we use the propeny of determinants and adjugates. For a square matrix
A of order n:
A · adj A = det (A) · I
Taking the determinant on both sides:
det(A · adj A) = det(det(A) · I)
For a matrix I of order 3 and det(A) = 5:
det(A · adj A) = (det(A))ⁿ = 5³ = 125
So, the correct answer is: 125

pH = 8, pOH = 6; [OH^-] = 10^-6 M;
Ionic product of Fe(OH)₂ = 0.2 × (1 × 10^-6)²
= 2 × 10^-13 > (K_sp = 8.1 × 10^-16)

Disproportionation involves simultaneous oxidation and reduction of the same atom in a molecule.

Nitrogen due to small size is able to show pπ-pπ lateral overlap forming N ≡ N, rest ele-ments due to bigger size are not able to show pπ-pπ lateral overlap.

The S₆ molecule has a chair-form hexagon ring with the approx same bond length as that in S₈, but with some what smaller bond angles i.e. bond lengths are approx same but bond angles are different.

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