UGEE SUPR Mock Test- 3 - JEE MCQ

The de Broglie wavelength is given by λ = h/p, where p is the momentum.
Momentum where m is the mass and K is the kinetic energy.
When the kinetic energy doubles, K' = 2K, so the new momentum is: p' = 
New wavelength: 
Thus, the de Broglie wavelength changes by a factor of 1/√2.

1st excited state corresponds to n = 2
2nd excited state corresponds to n = 3

(A) is a NAND gate so output is

(B) is a NOR gate so output is

(C) is a NAND gate so output is

(D) is a XOR gate so output is

Equivalent capacitance of two parallel capacitors 10 μF and 6 μF = (10 + 6) μF = 16 μF
This 16 μF capacitor is in series combination with 4 μF capacitor,
∴ Equivalent capacitance of the entire combination 

Magnetic field due to long wire,

When r is doubled, the magnetic field becomes half, i.e., now the magnetic field will be 0.2 T.

Total distance covered in 10 s = Area 1 + Area 2 + Area 3

Total displacement in 10 s = Area 1 - Area 2 + Area 3

In a uniform electric field, the electric dipole experiences no net force because the forces on the two charges cancel each other out. However, it does experience a torque due to the electric field exerting a turning effect, trying to align the dipole with the field direction.

Therefore, the correct answer is that:

• The dipole experiences a torque.
• The dipole experiences no net force.

where n is a non-square natural number so 1 − n ≠ 0. Hence sec2θ is a rational number 
 

Thus, k = 4, matching option D.

Let the centre of the circle be (h, k).
Since the centre lies on the line y = x - 1,
∴ k = h - 1 …(i)
Given the circle passes through (7, 3) and has a radius of 3,
Using the distance formula: (7 - h)² + (3 - k)² = 3²

(7 - h)² + (3 - (h - 1))² = 3²  (∴ k = h-1)
(7 - h)² + (4 - h)² = 9.
Solving these, h = 4, k = 3.
Thus, the circle equation is: (x - 4)² + (y - 3)² = 9
Expanding this, x² - 8x + 16 + y² - 6y + 9 = 9
x² + y² - 8x - 6y + 16 = 0

Let u = log (sec θ + tan θ) and v = sec θ
After differentiating on both sides w.r.t. θ, we get

The equation of the bisector of the angle between the lines (x = 5) and (y = 3) is

⇒ x − 5 = +(y − 3) and x − 5 = −(y−3)
⇒ (x − y − 2) = 0 and (x + y − 8) = 0
Hence, combined equation of bisectors of angle between given lines is (x − y − 2)(x + y − 8) = 0
⇒ x² + xy − 8x − xy − y² + 8y − 2x − 2y − 16 = 0
⇒ x² − y² − 10x + 6y + 16 = 0

Given, 6y = x³ + 2…(i)
and Δy = 8Δx
On differentiating both sides of Eq. (i) w.r.t. x, we get

⇒ x² = 16
⇒ x = ±4
When x = 4,6y = (4)³ + 2
⇒ 6y = 66
⇒ y = 11
Hence, required point is (4, 11)

As, f(x) is continuous at x = 0
So, LHL = RHL = f(0) ......(i)

Now, from eq. (i), LHL = f(0)
⇒ 0 = k
Hence, k = 0

After differentiating on both sides, we get

The number of optical isomers depends upon the number of asymmetric centres (n). Possible number of optical isomers of the compound is 2ⁿ.

Octahedral sulphur (rhombic or α-sulphur) is the most stable allotrope of sulphur.

Thermoplastic polymers are the linear or slightly branched polymers in which the intermolecular forces of attraction are intermediate between elastomers and fibres.

We require nF to deposit 1 mol or 40 g of Ca.
n = 2 (no. of e⁻ involved)
∵ 10 g Ca is deposited by 0.5 F.

Paracetamol is classified as an analgesic because it is primarily used to relieve pain and reduce fever. In contrast, the other options are:

• Ofloxacin - an antibiotic used to treat bacterial infections
• Penicillin - another antibiotic for bacterial infections
• Aminoglycosides - a class of antibiotics used in treating serious infections

Thus, unlike the other medications listed, Paracetamol does not treat infections but instead alleviates discomfort and fever.

In the Wurtz reaction, two alkyl halides react with sodium metal in dry ether to form alkanes. When a mixture of n-butyl bromide (C₄H₉Br, 4 carbons) and ethyl bromide (C₂H₅Br, 2 carbons) is used, three possible products are formed:

1. Octane (C₈H₁₈): This results from the self-coupling of two n-butyl bromide molecules (2 × 4 = 8 carbons).

   • Reaction: C₂H₃CH₂CH₂CH₂Br + 2Na → CH₃CH₂CH₂CH₂CH₂CH₂CH₂CH₃ + 2NaBr

2. Butane (C₄H₁₀): This is formed from the self-coupling of two ethyl bromide molecules (2 × 2 = 4 carbons).

   • Reaction: C₂H₃CH₂Br + 2Na → CH₃CH₂CH₂CH₃ + 2NaBr

3. Hexane (C₆H₁₄): This is formed from the cross-coupling of n-butyl bromide and ethyl bromide (4 + 2 = 6 carbons).

   • Reaction: C₂H₃CH₂CH₂CH₂Br + CH₃CH₂Br + 2Na → CH₃CH₂CH₂CH₂CH₂CH₃ + 2NaBr

Note: Ethane (C₂H₆, 2 carbons) would require the coupling of two methyl bromide (CH₃Br) molecules, but methyl bromide is not present in the mixture. Thus, ethane is not formed, making option D incorrect.

Butylated hydroxy anisole is an antioxidant. The conjugated aromatic ring of BHA is able to stabilise free radicals.

Polonium crystallises in a simple cubic structure under certain conditions, known as the α-phase. The other metals listed have different crystal structures:

• Copper: exhibits a face-centred cubic (FCC) structure
• Nickel: exhibits a face-centred cubic (FCC) structure
• Iron (at room temperature): exhibits either a face-centred cubic (FCC) or body-centred cubic (BCC) structure

When amines react with nitrous acid (HNO₂) at low temperatures:

• Primary aliphatic amines (e.g., RNH₂) produce alcohols and nitrogen gas.

• Primary aromatic amines (e.g., aniline, C₆H₅NH₂), which are not yellow oils but can couple with phenols to form azo dyes (colored compounds).

• Secondary amines (e.g., RR'NH) form yellow oily nitrosoamines (RR'NNO).

• Tertiary amines (e.g., R₃N) may form nitrite salts or not react significantly, depending on conditions, but do not produce yellow oils.

Analyzing the options:

• A: Triethylamine (C₂H₅)₃N – Tertiary, no yellow oil.

• B: Trimethylamine (CH₃)₃N – Tertiary, no yellow oil.

• C: Aniline (C₆H₅NH₂) – Primary aromatic, forms diazonium salt, not a yellow oil directly.

• D: Methylphenylamine (C₆H₅NCH₃) – Secondary, forms C₆H₅N(CH₃)NO, a yellow oily nitrosoamine.

Thus, the amine 'D' that produces a yellow oily substance with nitrous acid is methylphenylamine, making option D correct.

The acidic oxides are formed by the non-metals (group 14-17) whereas basic oxides are formed by the metals of group 1 and group 2 elements. Ba belongs to group 2.