MHT CET 2024 PCM Question Paper(Available) : Download PCM Question Paper with Answers Key PDF

The electrostatic force between two point charges in vacuum is:

F = (1 / 4πε₀) × (q₁q₂ / r²).

In a medium with dielectric constant K, the force becomes:

F' = (1 / 4πKε₀) × (q₁q₂ / r'²).

If K = 5 and r' = r / 5, substitute these values:

F' = (1 / 4π(5)ε₀) × (q₁q₂ / (r / 5)²).

Simplify:

F' = (25 / 5) × (1 / 4πε₀) × (q₁q₂ / r²) = 5F.

Thus, the new force is 5F.

Using conservation of angular momentum:

I₁ω₁ = I₂ω₂.

Initial moment of inertia: I₁ = (1/2)MR².

Final moment of inertia after adding the second disc:

I₂ = (1/2)MR² + (1/2)(M/2)R² = (3/4)MR².

Substituting into the conservation equation:

(1/2)MR²ω = (3/4)MR²ω₂.

Solve for ω₂:

ω₂ = (1/2) / (3/4) ω = (2/3)ω.

Thus, the new angular velocity is (2/3)ω.

Bernoulli’s theorem states that for an incompressible, non-viscous fluid in steady flow, the sum of pressure energy, potential energy, and kinetic energy per unit volume is constant:

P + ρgh + (1/2)ρv² = constant.

Here:

- P is the pressure.

- ρ is the density of the fluid.

- g is the acceleration due to gravity.

- h is the height above a reference point.

- v is the velocity of the fluid.

Other options incorrectly use mass instead of density or modify the energy terms improperly.

The maximum height attained by a projectile is given by:

H = (u² × sin²θ) / (2g),

where:

- u is the initial velocity,

- θ is the angle of projection,

- g is the acceleration due to gravity.

Let the two projectiles have angles of projection θ₁ = 30° and θ₂ = 60°.

For the first projectile: H₁ = (u² × sin²30°) / (2g).

For the second projectile: H₂ = (u² × sin²60°) / (2g).

The ratio of their maximum heights is:

H₁ / H₂ = (sin²30°) / (sin²60°).

Substitute sin30° = 1/2 and sin60° = √3/2:

H₁ / H₂ = ((1/2)²) / ((√3/2)²) = (1/4) / (3/4) = 1 / 3.

Thus, the ratio of the maximum heights is 1 : 3.

The cyclic process consists of three steps: AC (isochoric), BC (isobaric), and AB (PV³ = RT).

Step 1: Isochoric process (AC)

Work done (W) = 0 (no change in volume).

Step 2: Isobaric process (BC)

Work done: W = P × ΔV.

W = 10 × (2 - 4) = -20 J.

Step 3: Path AB (PV³ = RT)

Work done: W = ∫ P dV = ∫ (RT / V³) dV.

W = -RT / 2 [(1/V²)] from V₁ to V₂.

Substitute V₁ = 2, V₂ = 4, R = 8 J/mol·K, and T = 300 K:

W = (8 × 300) / 2 × (1/16 - 1/4) = 225 J.

Total Work Done:

W = W(AC) + W(BC) + W(AB) = 0 - 20 + 225 = 205 J.

Step 1: Climbing upward
The tension in the rope while climbing upward is given by: T = m(g + a), where:
m = 50 kg (mass of the monkey),
g = 10 m/s² (gravitational acceleration),
a = 5 m/s² (acceleration while climbing upward).
Substitute values:
T = 50 × (10 + 5) = 750 N.
Since T > 350 N, the rope will break while climbing upward.

Step 2: Climbing downward
The tension in the rope while climbing downward is given by: T = m(g - a).
Substitute values (a = 4 m/s²):
T = 50 × (10 - 4) = 300 N.
Since T < 350 N, the rope will not break while climbing downward.

Step 1: Apply Ohm's Law
The voltage difference across the 20 Ω resistor is given by V = V_b - V_c.
Substitute the values:
V = 12 V - 2 V = 10 V.
Using Ohm's Law (V = IR), the current is:
I = V / R = 10 / 20 = 0.4 A.

Step 2: Verify using Kirchhoff's Laws
Calculate the currents through other resistors using Ohm's Law:
Current through the 10 Ω resistor (I₁):
I₁ = (V_a - V_b) / 10 = (30 - 12) / 10 = 1.8 A.
Current through the 30 Ω resistor (I₃):
I₃ = (V_b - V_c) / 30 = (12 - 2) / 30 = 0.333 A.
At the junction, I₁ = I₂ + I₃, which verifies I₂ = 0.4 A.

The deflection in a moving coil galvanometer is directly proportional to the current:

I ∝ θ.

Let:

I₁ = 200 μA, θ₁ = 60°, and θ₂ = π/10 radians.

Convert 60° to radians:

θ₁ = 60° = π/3 radians.

Using the proportionality relation:

I₂ / I₁ = θ₂ / θ₁.

Substitute the values:

I₂ / 200 = (π/10) / (π/3).

Simplify:

I₂ / 200 = 3 / 10.

Solve for I₂:

I₂ = 200 × (3 / 10) = 60 μA.

The acceleration due to gravity at an altitude h is given by:

gₕ = g / (1 + h / R)².

Given:

gₕ = g / 2.

Substitute into the formula:

g / 2 = g / (1 + h / R)².

Simplify:

(1 + h / R)² = 2.

Take the square root:

1 + h / R = √2.

Solve for h:

h / R = √2 - 1.

Substitute R = 6.4 × 10⁶ m:

h = (√2 - 1) × 6.4 × 10⁶.

Simplify:

h = (1.414 - 1) × 6.4 × 10⁶ = 0.414 × 6.4 × 10⁶ = 2.6 × 10⁶ m.

For a diamagnetic material:

The relative permeability μᵣ is slightly less than 1, as diamagnetic materials repel magnetic fields.

The relative permittivity εᵣ is always greater than 1, as it represents the material's ability to polarize in response to an electric field.

Among the given options, εᵣ = 1.5 and μᵣ = 0.5 satisfies these conditions.

Solution:

The induced EMF (e) is given by Faraday’s Law:

e = |dφ/dt|

Given:

φ = 5t³ + 4t² + 2t - 5

Differentiate φ with respect to t:

e = |15t² + 8t + 2|

At t = 2 seconds:

e = 15(2)² + 8(2) + 2 = 60 + 16 + 2 = 78 V

The induced current (i) is given by Ohm’s Law:

i = e / R

Substitute R = 5 Ω:

i = 78 / 5 = 15.6 A

Solution:

The increase in volume is given by:

ΔV = V₀ γ ΔT,

where:

• V₀ = a³ (initial volume of the cube),
• γ = 3α (coefficient of volumetric expansion),
• ΔT = 10°C (temperature change).

The surface area of a cube is:

Total Surface Area = 6a²

Given 6a² = 24 m², solve for a:

a² = 24 / 6 = 4 m² → a = 2 m

The initial volume is:

V₀ = a³ = 2³ = 8 m³ = 8 × 10⁶ cm³

Substitute into the formula for ΔV:

ΔV = 8 × 10⁶ × (3 × 5 × 10⁻⁴) × 10

ΔV = 8 × 10⁶ × 15 × 10⁻³ = 1.2 × 10⁵ cm³

Solution:

The mutual inductance between two coils is defined as:

M = Φ / I,

where Φ is the magnetic flux in one coil due to the current I in the other coil.

The mutual inductance depends on:

• The relative position and orientation of the two coils (determines the coupling of magnetic fields).
• The number of turns and dimensions of the coils.
• The magnetic permeability of the medium between the coils.

It does not depend directly on the current or the material of the coil wires.

Solution:

The de-Broglie wavelength is given by:

λ = h / √(2mE),

where h is Planck's constant, m is the mass of the particle, and E is its energy.

For particles with the same energy, λ ∝ 1 / √m.

The masses of the particles are:

• Electron (m_e): least mass,
• Proton (m_p): greater mass,
• Alpha particle (m_α = 4m_p): greatest mass.

Since λ ∝ 1 / √m, the order of wavelengths is:

λ_e > λ_p > λ_α.

Solution:

The refractive index (μ) is given by:

μ = Real Depth / Apparent Depth.

Let the real depth of the bubble be x cm. When viewed from one side, the apparent depth is 12 cm, and when viewed from the opposite side, the apparent depth is 24 - x cm.

From the refractive index formula:

μ = x / 12 (from one side),

μ = (24 - x) / 4 (from the opposite side).

Equate the two expressions for μ:

x / 12 = (24 - x) / 4.

Simplify:

4x = 12(24 - x).

4x = 288 - 12x.

16x = 288 → x = 18 cm.

Substitute x = 18 cm into μ = x / 12:

μ = 18 / 12 = 3/2.

Solution:

The wavelength for the Paschen series is given by:

1/λ = R_H (1/n₁² - 1/n₂²),

where:

• n₁ = 3 (Paschen series),
• n₂ = 4 (for the longest wavelength, n₂ is the next level above n₁).

Substitute:

1/λ = R_H (1/3² - 1/4²).

Simplify:

1/λ = R_H (1/9 - 1/16).

1/λ = R_H (16 - 9) / 144.

1/λ = R_H × 7/144.

Substitute R_H = 1.097 × 10⁷:

1/λ = 1.097 × 10⁷ × 7/144.

1/λ = 7.679 × 10⁷ / 144.

λ = 144 / 7.679 × 10⁷ ≈ 1.876 × 10^-6 m.

Solution:

Nuclear density (ρ) is nearly constant for all nuclei, regardless of their atomic number or mass number. This is because both the mass and the volume of the nucleus are proportional to the mass number (A).

1. Density of ⁴⁰Ca:

For ⁴⁰Ca, mass number A = 40. The nuclear density remains constant:

ρ_Ca = constant.

2. Density of ¹⁶O:

For ¹⁶O, mass number A = 16. The nuclear density is the same as for ⁴⁰Ca:

ρ_O = constant.

3. Ratio of Densities:

Since nuclear density is constant:

ρ_Ca / ρ_O = 1.

Solution:

To determine the location x₀ where the second charge (40 μC) should be placed such that the net electric field is zero at x = 2 cm:

Step 1: Electric Field Balance

At x = 2 cm, the electric field contributions from the two charges must cancel each other:

E_net = 0 → k × (10 × 10^-6) / (2²) = k × (40 × 10^-6) / (x₀ - 2)².

Step 2: Simplify the Relation

Cancel k and simplify:

(10 / 4) = (40 / (x₀ - 2)²).

1 / 2 = 2 / (x₀ - 2)².

Step 3: Solve for x₀

(x₀ - 2)² = 4.

x₀ - 2 = ±2.

x₀ = 4 cm or x₀ = 6 cm.

Step 4: Select the Correct Value of x₀

Since the second charge is placed to the right of x = 2 cm, the correct position is:

x₀ = 6 cm.

Solution:

A magnetic needle kept in a non-uniform magnetic field experiences both a force and a torque due to unequal forces acting on its poles.

The non-uniform magnetic field exerts different magnitudes of forces on the north and south poles of the needle, resulting in a net force on the needle.

Additionally, these forces create a torque that tends to align the needle with the magnetic field lines.

Solution:

The magnetic force per unit length F/ℓ between two parallel current-carrying wires is given by:

F/ℓ = (μ₀ I₁ I₂) / (4πr),

where:

• μ₀ is the permeability of free space,
• I₁ and I₂ are the currents in the two wires,
• r is the distance between the wires.

Step 1: Proportionality for F/ℓ

The magnetic force is directly proportional to the product of the currents:

F₁ ∝ I² (when the current is I).

When the current changes to 2I:

F₂ ∝ (2I)² = 4I².

Step 2: Ratio of Forces

The ratio of the magnetic forces is:

F₁ / F₂ = I² / 4I² = 1:4.

Solution:

The r.m.s. velocity \( v_{\text{rms}} \) of a gas molecule is given by:

v_rms = √(3RT / M),

where:

• R is the gas constant,
• T is the temperature (in Kelvin),
• M is the molar mass of the gas.

Step 1: Equating the r.m.s. velocities

For hydrogen (H₂) and oxygen (O₂):

v_rms(H₂) = v_rms(O₂).

√(3RT_H2 / M_H2) = √(3RT_O2 / M_O2).

Square both sides:

T_H2 / M_H2 = T_O2 / M_O2.

Rearrange for T_H2:

T_H2 = T_O2 × (M_H2 / M_O2).

Step 2: Substitute values

For oxygen:

T_O2 = 47°C + 273 = 320 K.

The molar masses are:

M_H2 = 2, M_O2 = 32.

Substitute into the equation:

T_H2 = 320 × (2 / 32).

Simplify:

T_H2 = 320 × 1 / 16 = 20 K.

Solution:

The wavelength of emitted light is related to the band gap energy E_g by the equation:

E_g (eV) = 1240 / λ (nm).

Rearrange to find the wavelength:

λ = 1240 / E_g.

Step 1: Substitute the given values

Given:

E_g = 1.42 eV.

Substitute:

λ = 1240 / 1.42.

λ ≈ 875 nm.

Solution:

By the ideal gas equation:

P × V = n × R × T, which implies P ∝ n,

where n is the number of moles.

Step 1: Calculate the number of moles

For hydrogen (H₂):

M_H2 = 2 g/mol, n_A = Mass / Molar Mass = 1 / 2 = 0.5 mol.

For oxygen (O₂):

M_O2 = 32 g/mol, n_B = Mass / Molar Mass = 1 / 32 = 0.03125 mol.

Step 2: Ratio of pressures

P_A / P_B = n_A / n_B = 0.5 / 0.03125.

Simplify:

P_A / P_B = 16.

Solution:

The induced emf ε in a moving conductor is given by:

ε = B × v × l,

where:

• B = 2 T is the magnetic field strength,
• v = 8 m/s is the velocity of the wire,
• l = 1 m is the length of the wire.

Step 1: Substitute the given values

Substitute B = 2 T, v = 8 m/s, and l = 1 m:

ε = 2 × 8 × 1.

Step 2: Simplify the calculation

ε = 16 V.

Solution:

The gravitational force provides the centripetal force for circular motion:

F_gravity = F_centripetal.

The gravitational force between two particles is:

F = (G × M₁ × M₂) / d²,

where:

• G is the gravitational constant,
• M₁ = M₂ = m,
• d = 2a is the distance between the two particles.

Substitute:

F = (G × m²) / (2a)² = Gm² / 4a².

For circular motion:

F = m × ω² × r,

where r = a. Substitute F = Gm² / 4a²:

(Gm²) / (4a²) = m × ω² × a.

Simplify:

ω² = Gm / 4a³.

Take the square root:

ω = √(Gm / 4a³).

Solution:

The total energy in SHM is given by:

E_total = (1/2) k A²,

where k is the spring constant and A is the amplitude of oscillation.

Step 1: Potential and Kinetic Energy

The potential energy at displacement x is:

PE = (1/2) k x².

The kinetic energy is:

KE = E_total - PE = (1/2) k A² - (1/2) k x².

Step 2: Displacement is half the amplitude

Substitute x = A/2:

PE = (1/2) k (A/2)² = (1/2) k (A² / 4) = (1/8) k A².

KE = (1/2) k A² - (1/8) k A² = (4/8) k A² - (1/8) k A² = (3/8) k A².

Step 3: Ratio of PE and KE

PE / KE = (1/8) k A² / (3/8) k A² = 1/3.

Solution:

When the drops coalesce, the volume remains constant. Let:

r be the radius of each small drop, R be the radius of the combined drop.

Step 1: Volume conservation

The total volume of the eight small drops is:

8 × (4/3) π r³ = (4/3) π R³.

Simplify:

R³ = 8r³ → R = 2r.

Step 2: Relationship between terminal velocity and radius

The terminal velocity v_T is proportional to the square of the radius:

v_T ∝ r².

Let v₁ = 10 cm/s be the terminal velocity of the small drops, and v₂ be the terminal velocity of the combined drop. Then:

v₁ / v₂ = (r / R)².

Substitute R = 2r:

v₁ / v₂ = (r / 2r)² = 1/4.

Step 3: Solve for v₂

v₂ = 4v₁ = 4 × 10 = 40 cm/s.

Solution:

The emf induced in a coil is given by:

ε = L × (ΔI / Δt),

where:

• ε = 0.1 V is the induced emf,
• ΔI = I_final - I_initial = 2 - (-2) = 4 A is the change in current,
• Δt = 0.2 s is the time interval,
• L is the self-inductance.

Step 1: Solve for L

L = ε × Δt / ΔI.

Substitute the values:

L = (0.1 × 0.2) / 4.

Simplify:

L = 0.02 / 4 = 0.005 H.

Convert to millihenries:

L = 5 mH.

Solution:

The moment of inertia (I_O) of the rectangular sheet about an axis passing through O (center of mass) is given by:

I_O = (M/12) × (a² + b²),

where a = 80 cm and b = 60 cm are the dimensions of the rectangle, and M is the mass of the sheet.

Substitute the values:

I_O = (M/12) × (80² + 60²).

Calculate:

I_O = (M/12) × (6400 + 3600) = (M/12) × 10000 = 10000M / 12.

Using the parallel axis theorem, the moment of inertia about O' is:

I_O' = I_O + M × d²,

where d is the perpendicular distance between O and O', given by:

d = √((a/2)² + (b/2)²) = √(40² + 30²) = 50 cm.

Substitute:

I_O' = (10000M / 12) + M × 50².

Simplify:

I_O' = (10000M / 12) + 2500M = (10000M / 12) + (30000M / 12) = 40000M / 12.

The ratio of moments of inertia is:

(I_O / I_O') = (10000M / 12) / (40000M / 12) = 10000 / 40000 = 1/4.

Solution:

The mean free path (λ) is the average distance a molecule travels between two successive collisions. It is derived from the kinetic theory of gases and is given by the formula:

λ = 1 / √(2πnd²),

where:

• n is the number density of molecules (number of molecules per unit volume),
• d is the diameter of a molecule,
• π is the constant representing the geometric cross-sectional area of a circle.

Final Answer: λ = 1 / √(2πnd²).

Solution:

For a monoatomic gas, the specific heat at constant volume (C_V) is:

C_V = (3/2)R.

For a diatomic gas (rigid), the specific heat at constant volume (C'_V) is:

C'_V = (5/2)R.

The ratio of specific heats is:

(C_V / C'_V) = ((3/2)R) / ((5/2)R) = 3/5.

Solution:

The average power in one cycle is given by:

P_avg = V_rms × I_rms × cos(Δφ).

Here, V_rms = V_0 / √2 = 100 / √2, I_rms = I_0 / √2 = (100 × 10⁻³) / √2, and Δφ = π/3.

Substitute:

P_avg = (100 / √2) × (100 × 10⁻³ / √2) × cos(π/3).

Simplify:

P_avg = (100 × 0.1) / 2 × (1/2) = 2.5 W.

Solution:

The wavelength (λ) is related to the work function (ϕ) by:

λ = (hc) / ϕ.

For ϕ_A = 9 eV:

λ_A = 1242 / 9 = 138 nm.

For ϕ_B = 4.5 eV:

λ_B = 1242 / 4.5 = 276 nm.

The difference is:

Δλ = λ_B - λ_A = 276 - 138 = 138 nm.

Solution:

The circuit performs the following operations:

1. The first two NOT gates take inputs a and b, and output their complements ¬a and ¬b.
2. The OR gate combines these complements: ¬a + ¬b.
3. The NOT gate takes the complement of the OR gate’s output: ¬(¬a + ¬b).

Using De Morgan’s law:

¬(¬a + ¬b) = a ∧ b (AND).

The circuit performs the logic of an AND gate.

Solution:

The velocity of a particle in SHM is:

v = ω√(A² - x²),

where A is the amplitude and x is the displacement.

Initially, at x = (2A/3), the velocity is:

v = ω√(A² - (2A/3)²).

Substitute:

v = ω√(A² - (4A²/9)) = ω√(5A²/9) = (ωA√5)/3.

When the speed is tripled, the velocity becomes:

3v = ω√(A'² - (2A/3)²),

where A' is the new amplitude.

Substitute for v:

3(ωA√5)/3 = ω√(A'² - (4A²/9)).

Simplify:

A√5 = √(A'² - (4A²/9)).

Square both sides:

5A² = A'² - (4A²/9).

Rearrange:

A'² = 5A² + (4A²/9) = (45A²/9) + (4A²/9) = (49A²/9).

Simplify:

A' = (7A/3).

Solution:

The binding energy (BE) of a nucleus is calculated using:

BE = Δm × c²

Here, Δm is the mass defect. For helium nucleus:

Δm = (2 × mass of proton + 2 × mass of neutron) - mass of helium nucleus

Δm = (2 × 1.0073 + 2 × 1.0087) - 4.0015

Δm = 4.0310 - 4.0015 = 0.0295 u

Convert mass defect to energy using 1 u = 931.5 MeV/c²:

BE = 0.0295 × 931.5 = 28.4 MeV

Solution:

The voltage across the inductor is given by:

V_L = I × X_L

where X_L is the inductive reactance and is calculated as:

X_L = ωL

Step 1: Calculate angular frequency:

ω = 2πf = 2 × 3.14 × 50 = 314 rad/s

Step 2: Calculate X_L:

X_L = 314 × 10 × 10⁻³ = 3.14 Ω

Step 3: Calculate current:

I = V_L / X_L = 31.4 / 3.14 = 10 A

Solution:

The pressure inside a soap bubble is given by:

P = P₀ + 4T / R

For two bubbles with radii a and b:

P₁ = P₀ + 4T / a, and P₂ = P₀ + 4T / b

The pressure difference across the common surface is:

P₁ - P₂ = 4T / R

Substitute P₁ and P₂:

(4T / a) - (4T / b) = 4T / R

Simplify:

1 / a - 1 / b = 1 / R

R = ab / (b - a)

Solution:

From the principle of continuity:

A₁v₁ = A₂v₂

Here, A₁ and A₂ are the cross-sectional areas, and v₁ and v₂ are the velocities of the liquid at the wider and narrower ends, respectively.

Since A₁ > A₂, the velocity at the wider end (v₁) is lower than the velocity at the narrow end (v₂).

The beat frequency is given by the formula:

f_beat = f₁ - f₂,

where f₁ and f₂ are the frequencies corresponding to the given wavelengths λ₁ and λ₂.

Step 1: Calculate the beat frequency
The beat frequency is:

f_beat = Number of beats ÷ Time = 40 ÷ 12 Hz.
Simplify:
f_beat = 10 ÷ 3 Hz.

Step 2: Relating wavelength to velocity and frequency
The frequency of a wave is related to its velocity by:

f = v ÷ λ.

Using the wavelengths λ₁ = 4.08 m and λ₂ = 4.16 m:

f₁ = v ÷ 4.08, f₂ = v ÷ 4.16.

Step 3: Solve for v
The beat frequency is:

f_beat = f₁ - f₂ = (v ÷ 4.08) - (v ÷ 4.16).

Simplify:
10 ÷ 3 = v × [(1 ÷ 4.08) - (1 ÷ 4.16)].

Calculate the difference in reciprocals:
(1 ÷ 4.08) - (1 ÷ 4.16) = (4.16 - 4.08) ÷ (4.08 × 4.16) = 0.08 ÷ 16.9728.

Substitute:
10 ÷ 3 = v × (0.08 ÷ 16.9728).

Solve for v:
v = [(10 ÷ 3) × 16.9728] ÷ 0.08 = 707.2 ms⁻¹.

Final Answer: 707.2 ms⁻¹

Using Gauss's law:

Electric flux = E × Area = Q / ε₀.

Step 1: Relate charge to surface charge density

The total charge Q on the spherical shell is:

Q = σ × 4πR².

Step 2: Simplify the electric field

The electric flux through a Gaussian surface of radius R is:

E × 4πR² = Q / ε₀.

Substitute Q = σ × 4πR² into the equation:

E × 4πR² = (σ × 4πR²) / ε₀.

Cancel 4πR² on both sides:

E = σ / ε₀.

Final Answer: σ / ε₀.

The electric flux is calculated as:

Flux = E · A.

Here, A = 30i m² represents the area vector.

Substitute E = (6i + 5j + 3k):

Flux = (6i + 5j + 3k) · (30i).

Only the i-components contribute to the dot product:

Flux = 6 × 30 = 180.

Final Answer: 180.

Step 1: Use volume conservation

(4/3)πR³ = 1000 × (4/3)πr³,

where R is the radius of the big drop, and r is the radius of each small drop.

Cancel (4/3)π:

R³ = 1000 × r³ → R = 10r.

Step 2: Compare surface energies

The initial surface energy is proportional to the surface area of 1000 small drops:

E₁ = 1000 × 4πr².

The final surface energy is proportional to the surface area of the big drop:

E₂ = 4πR² = 4π(10r)² = 100 × 4πr².

The ratio is:

E₂ / E₁ = (100 × 4πr²) / (1000 × 4πr²) = 1/10.

Final Answer: 1/10th.

The height of water in a capillary tube is given by:

h = (2T cos θ) / (r × ρ × g),

where T is surface tension, θ is the contact angle, ρ is the density of water, and g is gravitational acceleration.

The volume of water is:

V = πr² × h = πr² × [(2T cos θ) / (r × ρ × g)].

Simplify:

V ∝ r.

The mass of water is:

m = ρ × V ∝ r.

For a capillary of radius 2r:

m₂ = 2 × m₁.

Thus, the new mass is:

2M.

For a closed organ pipe, the length of the pipe for the n-th harmonic is given by:

L = [(2n - 1) × λ] / 4,

where λ is the wavelength.

For the third overtone:

n = 4.

The pipe supports four full loops plus a half loop, corresponding to:

4 nodes and 4 antinodes.

The electrostatic potential V due to an electric dipole at a point at distance r from its center is proportional to:

V ∝ 1 / r³.

This is because the electric potential of a dipole decreases as the cube of the distance due to the nature of its charge distribution.

The circuit contains a balanced Wheatstone bridge. The 5 Ω resistor does not contribute to the current as no current flows through it.

Step 1: Simplify the circuit

Remove the 5 Ω resistor and combine the remaining resistors.

• The two 3 Ω resistors in the top branch are in series: R_top = 3 + 3 = 6 Ω.
• The two 6 Ω resistors in the bottom branch are in parallel: R_bottom = (6 × 6) / (6 + 6) = 3 Ω.

Combine R_top and R_bottom in parallel:

R_eq = (6 × 3) / (6 + 3) = 18 / 9 = 2 Ω.

Add the 2 Ω resistor in series:

R_total = R_eq + 2 = 2 + 2 = 4 Ω.

Step 2: Apply Ohm's Law

Total current: I = V / R_total.

Substitute V = 6 V and R_total = 6 Ω:

I = 6 / 6 = 1 A.

In a FCC unit cell, octahedral voids are present at the body center and edge centers. The distance between two octahedral voids is:

1. The body diagonal length is x = √3 × a, where a is the edge length.
2. Edge length a = x / √3.
3. The distance between octahedral voids is a / √2.

Substitute a = x / √3:

Distance = (x / √3) / √2 = x / √6.

Ortho-sulphobenzimide is the chemical name for saccharin. It is widely used as a non-caloric artificial sweetener.

At NTP, 22,400 cm³ of gas contains 6.02 × 10²³ molecules.

For a volume of 1.12 × 10⁻⁷ cm³:

Number of molecules = (6.02 × 10²³ / 22,400) × 1.12 × 10⁻⁷.

Number of molecules = (6.02 × 1.12 × 10²³ × 10⁻⁷) / 22,400.

Final result = 3.01 × 10¹² molecules.

The wavelength λ is calculated using the Rydberg formula:

1 / λ = R × (1 / n₁² - 1 / n₂²), where n₁ = 1 (final state) and n₂ = infinity (initial state).

1. 1 / λ = R × (1 / 1²).
2. Substitute R = 1.097 × 10⁷: 1 / λ = 1.097 × 10⁷.
3. λ = 1 / (1.097 × 10⁷) = 91 nm.

The structure of NO₃⁻ ion is:

O = N⁺ - O⁻ - O.

In NO₃⁻:

• The nitrogen atom forms 4 bond pairs (3 sigma bonds and 1 pi bond).
• The nitrogen atom has no lone pairs.

In K₂MnO₄, the oxidation number of Mn can be calculated as:

2(+1) + x + 4(-2) = 0, where x is the oxidation number of Mn.

Simplify: 2 + x - 8 = 0 → x = +6.

In the other compounds:

• MnO₂: Mn = +4
• Mn₃O₄: Mn = +8/3 (average)
• MnSO₄: Mn = +2

Thus, the highest oxidation number is in K₂MnO₄.

Alkali metals have a single valence electron, and their low ionisation energy makes it easy to lose this electron and get oxidised. This property makes them strong reducing agents.

From Charles's law:

V₁ / T₁ = V₂ / T₂.

Substitute the known values:

3 / 308 = 2.5 / T₂, where T₁ = 35°C = 273 + 35 = 308 K.

Solve for T₂:

T₂ = (2.5 × 308) / 3 = 256.6 K.

Convert to Celsius:

T₂ = 256.6 - 273 = -16.4°C ≈ -16°C.

Silver sol can be used as an eye lotion because of its antibacterial and healing properties. It helps in treating eye infections effectively.

Electron-donating groups like -NHCOCH₃ and -OCH₃ are ortho and para directing because they increase electron density at the ortho and para positions of the benzene ring through resonance or inductive effects.

Electron-withdrawing groups like -CHO and -SO₃H are meta directing.

Carbanion stability depends on the electron-withdrawing nature of attached groups:

• -NO₂ and -CHO are strong electron-withdrawing groups, stabilising the negative charge.
• -CH₃ is electron-donating, which destabilises the negative charge.

Thus, CH₂-CH₃ is the least stable as it lacks electron-withdrawing groups.

The molecular orbital electronic configurations are:

• O₂⁻: π(2pₓ)² π(2pᵧ)² π*(2pₓ)¹ (7 antibonding electrons)
• O₂: π(2pₓ)² π(2pᵧ)² (6 antibonding electrons)
• O₂²⁻: π(2pₓ)² π(2pᵧ)² π*(2pₓ)² (8 antibonding electrons)

At high altitudes, the partial pressure of oxygen is lower, leading to reduced oxygen availability in the blood. This condition, called hypoxia, causes symptoms like weakness and difficulty in thinking, known as anoxia.

Molar conductance is given by:

Λ_m = κ × (1000 / C),

where:

• κ = 6.3 × 10⁻² ohm⁻¹ cm⁻¹,
• C = 0.1 M.

Substitute:

Λ_m = 6.3 × 10⁻² × (1000 / 0.1) = 6.3 × 10² = 630 ohm⁻¹ cm² mol⁻¹.

For a first-order reaction, the rate constant k is related to the half-life t_1/2 as:

k = 0.693 / t_1/2.

Substitute t_1/2 = 480 s:

k = 0.693 / 480 = 1.44 × 10⁻³ s⁻¹.

The enthalpy change ΔH for a reaction is related to the activation energies of the forward reaction (E_a^f) and the reverse reaction (E_a^r) as:

ΔH = E_a^f - E_a^r.

Substitute the given values:

ΔH = 150 - 260 = -110 kJ/mol.

The coagulation of a negatively charged sol like As₂S₃ is most effective with a highly positively charged ion, as stated by the Hardy-Schulze rule.

Among the given options:

• FeCl₃ produces Fe³⁺, which has the highest positive charge (+3).
• Higher charges neutralise the negative charge of the sol more effectively, causing coagulation.

Other options, such as CaCO₃ (Ca²⁺) and NaCl (Na⁺), have lower positive charges and are less effective.

Fluorine is the most electronegative element and does not exhibit variable oxidation states because:

• It cannot expand its octet (no available d-orbitals).
• Fluorine always has an oxidation state of -1 in its compounds due to its high electronegativity.

Other halogens, such as bromine, iodine, and chlorine, can show variable oxidation states due to the presence of vacant d-orbitals.

Molar conductance is given by:

Λ_m = κ × (1000 / C),

where:

• κ = 6.3 × 10⁻² ohm⁻¹ cm⁻¹,
• C = 0.1 M.

Substitute:

Λ_m = 6.3 × 10⁻² × (1000 / 0.1) = 6.3 × 10² = 630 ohm⁻¹ cm² mol⁻¹.

For a first-order reaction, the rate constant k is related to the half-life t_1/2 as:

k = 0.693 / t_1/2.

Substitute t_1/2 = 480 s:

k = 0.693 / 480 = 1.44 × 10⁻³ s⁻¹.

The enthalpy change ΔH for a reaction is related to the activation energies of the forward reaction (E_a^f) and the reverse reaction (E_a^r) as:

ΔH = E_a^f - E_a^r.

Substitute the given values:

ΔH = 150 - 260 = -110 kJ/mol.

The coagulation of a negatively charged sol like As₂S₃ is most effective with a highly positively charged ion, as stated by the Hardy-Schulze rule.

Among the given options:

• FeCl₃ produces Fe³⁺, which has the highest positive charge (+3).
• Higher charges neutralise the negative charge of the sol more effectively, causing coagulation.

Other options, such as CaCO₃ (Ca²⁺) and NaCl (Na⁺), have lower positive charges and are less effective.

Fluorine is the most electronegative element and does not exhibit variable oxidation states because:

• It cannot expand its octet (no available d-orbitals).
• Fluorine always has an oxidation state of -1 in its compounds due to its high electronegativity.

Other halogens, such as bromine, iodine, and chlorine, can show variable oxidation states due to the presence of vacant d-orbitals.

Paramagnetic behaviour depends on the number of unpaired electrons:

• V²⁺: 3 unpaired electrons (3d³)
• Cr²⁺: 4 unpaired electrons (3d⁴)
• Mn²⁺: 5 unpaired electrons (3d⁵)
• Fe²⁺: 4 unpaired electrons (3d⁶)

The correct order is:

V²⁺ < Cr²⁺ < Fe²⁺ < Mn²⁺.

The given order in Option (1) is incorrect.

The complex contains:

• ONO (nitrito ligand coordinated through oxygen)
• NH₃ (neutral ammine ligands)
• Cl₂ (counter ions)

The oxidation state of cobalt is:

Co + 5(0) + (-1) + 2(-1) = 0 → Co = +3.

The name is:

Pentaammine-nitrito-O-cobalt(III) chloride.

Compounds undergoing the S_N1 mechanism form a planar carbocation intermediate, allowing nucleophilic attack from both sides, resulting in a racemic mixture.

• Compound (i): Secondary carbon, follows S_N1 mechanism.
• Compound (ii): Tertiary carbon, strongly favours S_N1 mechanism.
• Compound (iii): Secondary carbon, follows S_N1 mechanism.

All three compounds produce a racemic mixture.

Properties of an ideal solution:

• Obeys Raoult's law.
• No heat change during mixing (ΔH = 0).
• No volume change during mixing (ΔV = 0).

Option (3), stating ΔH = ΔV ≠ 0, is incorrect because both ΔH and ΔV must be zero for an ideal solution.

For E_cell = E^°_cell, the Nernst equation is:

E_cell = E^°_cell - (RT/nF) ln Q,

where Q is the reaction quotient.

For E_cell = E^°_cell, ln Q must be zero, which occurs when:

Q = 1 → Concentration of all reactant and product species is unity.

The divalent state stability depends on the electronic configuration:

• Ce²⁺ → 4f¹
• Sm²⁺ → 4f⁶
• Eu²⁺ → 4f⁷ (half-filled, very stable)
• Yb²⁺ → 4f¹⁴ (fully filled, stable but less so than Eu²⁺)

Europium (Eu²⁺) is more stable due to its half-filled configuration.

The spin-only magnetic moment is given by:

μ = √(n(n + 2)) BM, where n is the number of unpaired electrons.

Calculations for configurations:

• d⁵ (strong field): n = 1, μ = √3 = 1.73 BM
• d³ (any field): n = 3, μ = √15 = 3.87 BM
• d⁴ (weak field): n = 4, μ = √24 = 4.89 BM
• d⁴ (strong field): n = 2, μ = √8 = 2.84 BM

The correct configuration is d⁴ in strong ligand fields.

Step 1: Chlorobenzene (C₆H₅Cl) reacts with Mg in dry ether to form phenyl magnesium chloride (C₆H₅MgCl).

Step 2: Phenyl magnesium chloride reacts with ethanol (C₂H₅OH) to produce benzene (C₆H₆) and magnesium ethoxide (C₂H₅OMgCl).

Acidity in carboxylic acids is increased by electron-withdrawing (-I) groups.

Order of electronegativity: F > Cl > Br > H.

Fluorine has the strongest -I effect, stabilizing the carboxylate ion and increasing acidity, followed by chlorine, bromine, and hydrogen.

Grignard reagents react with carbonyl compounds to form alcohols. Formaldehyde (HCHO) reacts with Grignard reagents to produce primary alcohols like ethyl alcohol (C₂H₅OH).

Step 1: Reaction mechanism.

1. The nucleophilic CH₃MgBr attacks the electrophilic carbon in formaldehyde:

HCHO + CH₃MgBr → CH₃CH₂OMgBr.

2. The intermediate alcoholate ion (CH₃CH₂OMgBr) is hydrolyzed to form ethyl alcohol:

CH₃CH₂OMgBr + H₃O⁺ → CH₃CH₂OH.

Phenols are more acidic than alcohols because the phenoxide ion formed after losing H⁺ is resonance stabilized. Among phenols, electron-withdrawing substituents like -Cl increase acidity by stabilizing the negative charge on the oxygen atom.

Step 1: Compare structures.

Benzyl alcohol: Less acidic because the -OH group is attached to a saturated carbon.

Cyclohexanol: Similar to benzyl alcohol, it lacks resonance stabilization.

Phenol: Acidic due to resonance stabilization of the phenoxide ion.

m-Chlorophenol: Most acidic because the -Cl group has an electron-withdrawing (-I) effect, enhancing resonance stabilization.

The compound undergoes Cannizzaro reaction but not aldol condensation, indicating the absence of alpha-hydrogens. This suggests that the compound is an aldehyde. Upon nitration, benzaldehyde forms 3-nitrobenzaldehyde.

Step 1: Reaction mechanism.

CHO → NO₂-CHO (meta-nitrobenzaldehyde).

The nitration occurs at the meta-position relative to the aldehyde group due to the electron-withdrawing nature of the -CHO group.

The reactions and their respective products are:

(a) Hofmann Bromamide Reaction: CH₃CONH₂ + Br₂ + KOH → CH₃NH₂ (Primary Amine)

(b) Reduction of Nitriles: CH₃CN + LiAlH₄ → CH₃CH₂NH₂ (Primary Amine)

(c) Reduction of Isocyanides: CH₃NC + LiAlH₄ → CH₃NHCH₃ (Secondary Amine)

(d) Reduction of Amides: CH₃CONH₂ + LiAlH₄ → CH₃CH₂NH₂ (Primary Amine)

Conclusion: Only reaction (3) produces a secondary amine and does not yield a primary amine.

Nucleophilic addition reactivity depends on steric hindrance and the electrophilicity of the carbonyl carbon.

• Aldehydes: Less steric hindrance and more electrophilic than ketones.
• Acetaldehyde: Minimal steric hindrance and no resonance effects, making it the most reactive.
• Ketones and benzophenone: Steric hindrance from two substituents reduces reactivity.
• Benzaldehyde: Resonance of the phenyl group decreases electrophilicity.

Conclusion: Acetaldehyde (CH₃CHO) is the most reactive compound.

A molecule is chiral if it has at least one carbon atom bonded to four different groups.

• 2-Bromobutane: Contains a chiral carbon bonded to bromine, hydrogen, a methyl group, and an ethyl group.
• 1-Bromobutane: The first carbon is bonded to two identical hydrogen atoms, making it achiral.
• 2-Bromopropane: The second carbon is bonded to two identical methyl groups, making it achiral.
• 2-Bromopropan-2-ol: The second carbon is bonded to two identical methyl groups, making it achiral.

Conclusion: Only 2-bromobutane is chiral.

The basicity of amines depends on the inductive effect and steric hindrance:

• Methylamine: A primary amine with moderate basicity.
• Dimethylamine: A secondary amine with strong inductive effects and minimal steric hindrance, making it the most basic.
• Trimethylamine: A tertiary amine with strong inductive effects but steric hindrance reduces reactivity.
• Aniline: Resonance of the aromatic ring reduces its basicity, making it the least reactive.

Conclusion: Dimethylamine reacts most readily with dilute hydrochloric acid.

Essential amino acids cannot be synthesized by the human body and must be obtained through diet. Lysine is one such essential amino acid critical for protein synthesis and other metabolic functions.

Serine, glycine, and proline are non-essential amino acids because they can be synthesized by the body.

A peptide bond is covalent and planar due to resonance. It can form hydrogen bonds because the NH group acts as a donor and the CO group acts as an acceptor.

• Option (1): Incorrect as the peptide bond is planar.
• Option (2): Correct because of hydrogen bond capability.
• Option (3): Incorrect, as the trans configuration is more stable due to reduced steric hindrance.
• Option (4): Incorrect, as rotation is restricted by partial double-bond character.

Chain-growth polymerization involves the successive addition of monomers, typically through free radicals or ions. Teflon is formed by the polymerization of tetrafluoroethylene, making it a chain-growth polymer.

• Bakelite, Nylon, and Terylene are step-growth polymers, formed by condensation reactions.

Buna-S is formed by the co-polymerization of 1,3-butadiene and phenylethene (styrene). Other polymers are:

• Buna-N: Co-polymer of 1,3-butadiene and acrylonitrile.
• Neoprene: Polymer of chloroprene.
• Novolac: Phenol-formaldehyde resin.

The freezing point depends on the number of particles (i) formed upon dissociation:

• (1) Dissociates into 4 particles.
• (2) Dissociates into 3 particles.
• (3) Dissociates into 2 particles.
• (4) Does not dissociate (1 particle).

The compound with the fewest particles ([Co(H₂O)₃Cl₃]·3H₂O) will have the highest freezing point.

Molar conductivity (Λ_m) is defined as the conductance of all ions in 1 mole of solution, placed between two electrodes of unit area separated by unit distance. All given formulas are valid representations:

• Λ_m = K / C: Relates molar conductivity to conductivity (K) and molar concentration (C).
• Λ_m = (K × A) / l: Derived from conductance G and the relationship to molar conductivity, where A is the area, and l is the electrode separation.
• Λ_m = K × V: Relates molar conductivity to conductivity and solution volume V.

For a zero-order reaction, the rate law is r = k, and the integrated rate equation is:

[R] = [R]₀ - kt.

This equation represents a straight line with a negative slope (-k) when [R] is plotted against t. Thus, the correct graph is a linear plot with a downward slope.

In acidic conditions:

• MnO₄^- retains its purple color.
• MnO₄^2- disproportionates to MnO₄^- and Mn²⁺.
• CrO₄^2- converts to Cr₂O₇^2-, changing from yellow to orange.
• FeO₄^2- decomposes to Fe³⁺, losing its color.

Thus, MnO₄^- does not change its color in acidic medium.

Heliox (a mixture of helium and oxygen) is used to treat asthma patients. Helium reduces airway resistance due to its low density, making it easier for patients to breathe.

These compounds differ in the exchange of anions between the inner and outer spheres:

• [PtCl₂(NH₃)₄]Br₂: Chloride ions are in the coordination sphere, bromide ions in the outer sphere.
• [PtBr₂(NH₃)₄]Cl₂: Bromide ions are in the coordination sphere, chloride ions in the outer sphere.

These are ionization isomers as they differ in the anion present in the coordination sphere.

Tincture of iodine is a solution of iodine (2-3%) in alcohol and water, commonly used as an antiseptic. The alcohol acts as a solvent, and the iodine provides antimicrobial properties. Other options like iodoform, 2-iodopropane, and iodobenzene are unrelated to tincture.

Buna-S rubber is formed by the copolymerization of butadiene (CH₂=CH-CH=CH₂) and styrene (C₆H₅-CH=CH₂). It is widely used in making automobile tires due to its durability and resistance to abrasion.

In a Body-Centered Cubic (BCC) structure, the atom at the center touches the eight corner atoms, making the coordination number 8. Other coordination numbers are incorrect for BCC: 12 applies to FCC, 6 to simple cubic, and 4 does not apply.

Using the Nernst equation:

E_cell = E⁰_cell - (0.059/n) log Q.

After calculating with Q = (0.01)²/(0.1)³, the cell potential is approximately 0.51 V.

Tertiary alkyl halides predominantly undergo elimination (E2 mechanism) due to steric hindrance, forming alkenes instead of ethers. For example:

(CH₃)₃C-Br + CH₃O^- → (CH₃)₂C=CH₂ + CH₃OH + NaBr.

We calculate powers of A:

A² = [[1, 0], [1/2, 1]] × [[1, 0], [1/2, 1]] = [[1, 0], [1, 1]].

A⁴ = A² × A² = [[1, 0], [1, 1]] × [[1, 0], [1, 1]] = [[1, 0], [2, 1]].

From this pattern, A⁵⁰ = [[1, 0], [50, 1]].

Thus, the given options do not match the result.

The function is f(x) = sin^-1(x - √x).

Step 1: Analyze g(x) = x - √x. For x ∈ [0, 1], the range of g(x) is [-1/4, 1].

Step 2: Apply sin^-1. The range of f(x) is [-sin^-1(1/4), π/2].

Compute a - b = [-i + 5j + 2k]. Magnitude: |a - b| = √30. For |a| = √14 and |b| = √38, verify that |a - b| > |b| - |a|.

Simplify the given equation: y² + 16x² = 8xy. Differentiate both sides and solve for dy/dx. Result: dy/dx = 4.

Differentiate f(x): f'(x) = √2 cos(x + π/4)/(1 + (sin x + cos x)²). f'(x) > 0 in (-π/2, π/4).

The function f(x) = ||x| - 1| has sharp edges at x = -1, 0, and 1. These points are where the function is non-differentiable.

Using substitution x → π/2 - x, the integral simplifies. Adding the original and substituted integrals results in the evaluation of limits and symmetry properties, yielding 7π/44.

A vector in the plane of a and b is u = a + λb. Solving for the projection condition on c using projection formula gives the vector 4i - j + 4k.

The statement "neither brave nor climb Mount Everest" corresponds to "not brave and not climbing," represented symbolically as ~p ∧ ~q.

Simplifying each trigonometric term using identities like sin(π + x) = -sin(x) and cos(π/2 + x) = -sin(x), and substituting into the given expression, results in a final value of 1.

The powers of i cycle every 4 terms: i, -1, -i, 1. Grouping terms into 1000 complete cycles and adding the remaining terms gives a result of -i.

Fix the first digit as 4 and the last digit as either 0 or 5. The second and third digits can take any value from 0 to 9, resulting in 200 valid numbers.

To distribute 500 dissimilar boxes equally among 50 people, each receiving 10 boxes, the total arrangements are 500! divided by (10!)⁵⁰ for the internal arrangement of each group.

Using the definitions of f(x + y) and f(x - y) and simplifying with exponential rules, the result is 2f(x)f(y).

Using the condition of equidistance and substituting into the line equation, the coordinates of the point are calculated as (4, 14).

The circle's center is at (5, 4), and the radius is the distance from the center to the Y-axis, which equals 5. The equation is derived as \( (x - 5)² + (y - 4)² = 5² \). Simplified, it becomes \( x² + y² - 10x - 8y + 16 = 0 \).

The given limit is:

Limit as theta approaches -pi/4 of (cos(theta) + sin(theta)) / (theta + pi/4).

Step 1: Simplify the denominator.
Let h = theta + pi/4. Then, as theta approaches -pi/4, h approaches 0. The expression becomes:

Limit as h approaches 0 of (cos(pi/4 - h) + sin(pi/4 - h)) / h.

Step 2: Simplify the numerator using trigonometric identities.
Expand cos(pi/4 - h) and sin(pi/4 - h):

cos(pi/4 - h) = cos(pi/4) * cos(h) + sin(pi/4) * sin(h),

sin(pi/4 - h) = sin(pi/4) * cos(h) - cos(pi/4) * sin(h).

Adding these:

cos(pi/4 - h) + sin(pi/4 - h) = (cos(pi/4) + sin(pi/4)) * cos(h) + (sin(pi/4) - cos(pi/4)) * sin(h).

Since cos(pi/4) = sin(pi/4) = 1/sqrt(2), this simplifies to:

cos(pi/4 - h) + sin(pi/4 - h) = sqrt(2) * cos(h).

Step 3: Evaluate the limit.
Substitute back into the limit:

Limit as h approaches 0 of (sqrt(2) * cos(h)) / h.

As h approaches 0, cos(h) approaches 1. The limit becomes:

sqrt(2).

Calculate total possible outcomes as 16. Identify 7 favorable outcomes where \( p² ≥ 4q \). Probability is \( 7/16 \).

The inverse of a rotation matrix is its transpose. Hence, \( F(α)⁻¹ = F(-α) \), which satisfies the rotation matrix properties.

From the given equation, we use the scalar triple product formula. After solving, we find that cos θ = -1/3. Using the identity sin²θ + cos²θ = 1, we find sin θ = 2√2 / 3.

Using parametric equations of the line, we find the coordinates of the foot of the perpendicular. Then, using the formula for the distance from a point to a plane, the calculated distance is 5.

Substitute y = -(3/4)x into the given equation and solve for c. Simplification leads to c = -3.

Compute dx/dθ and dy/dθ, then find dy/dx. Finally, differentiate dy/dx with respect to θ and simplify. The result is sec³θ / (aθ).

Find the critical points by solving f'(x) = 0. Evaluate f(x) at the critical points and endpoints. The maximum value is 53.

At x = 1, continuity gives 1 + b + c = 1, so b + c = 0. Differentiability gives 2x + b = 1 at x = 1, which means b = -1. From b + c = 0, we find c = 1. Therefore, b - c = -1 - 1 = -2.

Simplify the Boolean expression using De Morgan's laws and distributive properties. The result is equivalent to p → (p ∨ q).

Using the parametric relations x = cos(2θ) and y = sin(2θ), the derivative dy/dx simplifies to -x / y.

Find the tangent equation and verify the given curves. The point (x₁, y₁) does not satisfy (x / 3) - y² = 2.

Using substitution sin⁻¹(x) = t, the integral simplifies to x - √(1 - x²) sin⁻¹(x) + C.

Using the substitution xe^x = t, we simplify the integral to integrate (1/t - 1/(1 + t)) with respect to t. After integration, substituting back t = xe^x gives the final result: log | (xe^x) / (1 + xe^x) | + C.

Calculate the cross product of the diagonals, then divide by 2 to find the parallelogram's area. The magnitude of the result gives the final answer: 4√6 square units.

The mean is calculated as the sum of all possible values of X multiplied by their probabilities. Here, the sum simplifies to 0.

Using trigonometric identities, simplify the integrand into a sum of two simpler terms. Solve each integral to obtain e^x tan(x) + C as the final answer.

Divide the equation by y², substitute x = vy, and simplify. Separate the variables and integrate both sides, resulting in arctan(x/y) + ln(y) + C = 0.

Perpendicular lines have direction vectors with a dot product equal to zero. Compute the dot product of the direction vectors (2i + pj + 5k) and (3i - pj + pk), set it to zero, and solve for p. The solution gives p = 6.

Simplify the expression using the formula for adding and subtracting arctangents. Combine terms to get sin(π/2), which equals 1.

To ensure continuity, find the limit of f(x) as x approaches 0. Apply L’Hôpital’s Rule to resolve the indeterminate form and determine f(0) = -1/8.

Substitute tan(x) = t, transform the integral, and simplify. Compute partial fractions and integrate to get the result: -π/12 + (2/3) arctan(2).

The region is bounded by the circle x² + y² = 1 and the parabola y² = 1 - x. The total area is computed as A = 2∫[0 to 1] √(1 - x²) dx + ∫[0 to 1] √(1 - x) dx. Solving the integrals gives the final result as π/2 + 4/3.

The probability of throwing both a head and a 6 is 1/12. Using geometric progression for alternating events, the probability that B wins is calculated as 11/23.

The determinant of the coefficient matrix is calculated as k³ - 3k + 2. For the system to have no solution, the determinant must be zero, leading to k = 2 or -2. Thus, |k| = 2.

Using the definition of a derivative, the given limit simplifies to f'(1) = 5.

The given region is divided into two parts based on boundaries. For x between 0 and 1, the top boundary is y = x² + 1. For x between 1 and 2, the top boundary is y = x + 1. The areas are computed as ∫[0 to 1] (x² + 1) dx + ∫[1 to 2] (x + 1) dx, resulting in a total area of 23/6.

The given equation is rearranged as √(1 - y²) dx = (sin⁻¹(y) - x) dy. Dividing by √(1 - y²), the equation becomes dx = (sin⁻¹(y) - x)/√(1 - y²) dy. Using an integrating factor e^(sin⁻¹(y)), the solution is derived as x = sin⁻¹(y) - 1 + ce^(-sin⁻¹(y)).

Total probability is 1. Solve for k: 0.1 + 0.15 + 0.3 + 0.25 + 2k = 1, which gives k = 0.1. Calculate the mean μ = Σ(X * P(X)) = 3.4. Calculate E(X²) = Σ(X² * P(X)) = 13.5. Variance is E(X²) - μ² = 13.5 - 11.56 = 1.93.

The line passes through the point (2, 5/2, -1) with direction ratios (2, -3/2, 0). The z-component of the direction vector must be 0, so p = 0.

If A is symmetric, then (B'AB)' = B'A'B = B'AB. Therefore, B'AB is symmetric.

Volume rate of change is dV/dt = 50. Volume of the sphere is (4/3)πr³. Differentiate to find dr/dt: 4πr²(dr/dt) = 50. Substituting r = 15 gives dr/dt = 1/(18π).