JEE Advanced Mix Test 51

Solution:

Applying the lens maker's formula for lens 1:

1 / f₁ = (μ - 1) × (1 / R₁ - 1 / R₂)

1 / f₁ = (1.5 - 1) × (1 / 14 - 0)

1 / f₁ = 0.5 / 14

For lens 2:

1 / f₂ = (μ - 1) × (1 / R₁ - 1 / R₂)

1 / f₂ = (1.2 - 1) × (0 - (1 / -14))

1 / f₂ = 0.2 / 14

Now, using the lens formula:

1 / f = 1 / f₁ + 1 / f₂

1 / f = (0.5 + 0.2) / 14

Therefore, from the lens formula:

1 / v = 7 / 140 - 1 / 40

v = 40 cm

Solution:

Given, the second harmonic of the first string is the fundamental of the second string.

Solution:

Given:

E = 81π/7 × 10⁵ V/m

v = 2 × 10^-3 m/s

η = 1.8 × 10^-5 N·s/m²

ρ = 900 kg/m³

We know that:

qE = mg (Equation i)

Also:

6πηrv = mg (Equation ii)

From this we get:

r = ^{3mg / 4πρg}_1/3 (Equation iii)

Substitute the value of r from Equation iii into Equation ii:

6πηv ^{(3mg / 4πρg)}_1/3 = mg

Or:

(6πηv)³ (3mg / 4πρg) = (mg)³

Next, substitute mg = qE, we get:

(qE)² = (3 / 4πρg) (6πηv)³

Or:

qE = ^{(3 / 4πρg)}_1/2 (6πηv)^3/2

Therefore, we get:

q = 1 / E × ^{(3 / 4πρg)}_1/2 (6πηv)^3/2

Substitute the values:

q = ⁷ / (81π × 10⁵) × ^{4 / 3} π × 900 × 9.8 × ^216π3

Multiplying and solving, we get:

q = 8.0 × 10^-19 C

The correct answer is: option 4) 8.0 × 10^-19 C

CONCEPT:

Electrophilic Aromatic Substitution and Ortho/Para Directing Groups

• In aromatic substitution reactions, groups attached to the benzene ring can influence where an electrophile will attack. Ortho/para directing groups, such as –OH (hydroxyl) and –CH₃ (methyl), push the incoming electrophile to the ortho and para positions relative to themselves on the ring.
• The presence of –OH and –CH₃ groups on a benzene ring leads to substitution occurring at the ortho and para positions with respect to these groups. Therefore, the positions where the substitution occurs can be predicted based on their electronic effects.
• When the compound reacts with bromine water (Br₂/H₂O), bromine will substitute at the positions directed by these groups. If a compound contains both –OH and –CH₃ groups, they will guide the bromine substitution to the ortho and para positions relative to them.

• The compound Y has a formula C₇H₈O and is insoluble in water, but dissolves in NaOH. This suggests that Y has a phenol group (–OH), as phenols are soluble in NaOH but insoluble in water.
• When Y reacts with bromine water, it is converted into a compound with the formula C₇H₅OBr₃. This indicates that three bromine atoms have substituted into the ring.
• Since both –OH and –CH₃ are ortho/para directing groups, they will direct the substitution to the ortho and para positions relative to each group. This means that the three bromine atoms will be positioned at the ortho and para positions relative to the –OH and –CH₃ groups.
• The correct structure that fits this pattern is given in option B, where the bromines are substituted at the positions that are ortho and para with respect to both –OH and –CH₃.

Therefore, the structure of compound Y is the one shown in option B, as it is the only one where the substitution pattern is consistent with the ortho/para directing effects of –OH and –CH₃ groups.

CONCEPT:

Deviation from Ideal Gas Behavior

• In real gases, deviations from ideal gas behavior occur due to intermolecular forces and the volume occupied by gas molecules.
• The ideal gas law assumes no intermolecular interactions and that the volume of the gas molecules is negligible. However, real gases experience attractive and repulsive forces, leading to deviations from this behavior, especially at low pressures and high temperatures.
• The PV vs. P curve for a gas shows its deviation from the ideal gas law. Positive deviation occurs when the actual volume is greater than the ideal volume, while negative deviation occurs when the actual volume is smaller than the ideal volume.
• At low pressure, gases like hydrogen (H₂) and helium (He) show a positive deviation, meaning they are less compressible than ideal gases. Conversely, gases like CO₂, CH₄, and O₂ show negative deviations at low pressure due to intermolecular attractions.

EXPLANATION:

• The graph shows the behavior of various gases (H₂, He, CO₂, CH₄, O₂) at a particular temperature.
• • Carbon dioxide (CO₂), Methane (CH₄), and Oxygen (O₂) initially show a negative deviation at lower pressures and then transition to a positive deviation as pressure increases. This suggests that the attractive forces between molecules dominate at low pressures, leading to a negative deviation, but as the pressure increases, repulsive forces start to take over, leading to a positive deviation.
  Hydrogen (H₂) and Helium (He) show a positive deviation from the ideal gas law, meaning they are less compressible than an ideal gas. This indicates that the intermolecular forces are relatively weak, leading to a behavior that deviates positively from the ideal gas equation.
• Option 3 incorrectly states that H₂ and He show negative deviation, which is not supported by the graph. They actually show positive deviation. It also wrongly states that CO₂, CH₄, and O₂ show positive deviation, which is true only at higher pressures. Hence, option 3 is the incorrect statement.

Therefore, option C is the correct answer, which states that H₂ and He show positive deviation while CO₂, CH₄, and O₂ show negative deviation at low pressure.

CONCEPT:

Work Done in a Cyclic Process

W = nR * (T_f - T_i) / (K - 1)

• In a cyclic process, the gas goes through a series of transformations that eventually bring it back to its original state. The total work done by the gas is the area enclosed by the path on the PV diagram or can be calculated from the individual segments of the cycle in the V-T diagram.
• For an ideal gas undergoing a polytropic process, where VT = constant, the work done in each segment can be calculated using the formula:
  • T_f and T_i are the final and initial temperatures for the process.
  • R is the ideal gas constant, n is the number of moles of the gas, and K is the polytropic index (which is 2 in this case for the given process).

EXPLANATION:

W_BC = nR * (T_f - T_i) / (K - 1)

W_BC = 2 * R * (1200 - 600) / (2 - 1) = 1200 * R

-1200R ln 2 + 1200R + 0 = 1200R (1 - ln 2)

• From the diagram, the process AB and BC are part of the cyclic process, with the condition VT = constant for the segment BC. This indicates that the gas undergoes a polytropic process between points B and C, where the work done can be calculated using the formula mentioned above.
• For segment BC, the work done is given by:
• Now, the total work done by the gas during the entire cycle is the sum of the work done during each segment:
  • W_AB + W_BC + W_CA
• Therefore, the work done during the entire cycle is 1200R (1 - ln 2).

Hence, the correct answer is option 3) 1200R(1 - ln 2).

In the given figure,

C is the incircle of △ABC.

Tangent t₁ is parallel to side a of triangle, tangent t₂ is parallel to side b of triangle and tangent t₃ is parallel to side c of triangle.