Solution:

1. Kinetic Energy Relation:
The kinetic energy (KE) is given by:
KE = P² / 2m, where P is the linear momentum and m is the mass.

2. Proportionality of Momentum:
For equal kinetic energies:
P √m

3. Masses of the Bodies:
The masses of A, B, and C are:
m_A = 0.4 kg, m_B = 1.2 kg, m_C = 1.6 kg.

4. Calculate the Momentum Ratios:
Using P √m:
P_A : P_B : P_C = √0.4 : √1.2 : √1.6.
Simplify:
P_A : P_B : P_C = 1 : √3 : 2.

Solution:

1. Relation Between Pressure, Intensity, and Force:
The pressure exerted by light is given by:
Pressure = I / c = F / A.
Hence:
I / c = F / A.

2. Substitute the Known Values:
- Intensity: I = 360 W/cm² = 360 × 10⁴ W/m²
- Speed of light: c = 3 × 10⁸ m/s
- Force: F = 2.4 × 10^-4 N
Substituting:
(360 × 10⁴) / (3 × 10⁸) = (2.4 × 10^-4) / A.

3. Simplify the Expression:
(360 / 3) × 10^-4 = (2.4 × 10^-4) / A
120 × 10^-4 = (2.4 × 10^-4) / A.

4. Solve for A:
A = (2.4 × 10^-4) / (120 × 10^-4) = 2.4 / 120
Simplify:
A = 2 × 10^-2 m² = 0.02 m².

Solution:

1. Given Condition:
The de-Broglie wavelength (λ) is the same for both the proton and the electron.

2. de-Broglie Relation:
The momentum (P) is given by:
P = h / λ.
Since λ is the same for both particles, the momentum P is also the same.

3. Relation Between Momentum and Kinetic Energy:
The momentum (P) is related to the kinetic energy (K) as:
P = √(2mK).
Rearrange to find K:
K √(1/m).

4. Kinetic Energy Ratio:
Since K √(1/m), the ratio of the kinetic energies of the proton and electron is:
K_p / K_e = m_e / m_p.
Substituting m_p = 1836m_e:
K_p / K_e = 1 / 1836.

5. Final Answer:
The ratio of their kinetic energies is:
K_p : K_e = 1 : 1836.

Solution:

1. Specific Heat at Constant Volume (C_v):
For a monoatomic gas:
(C_v)_mono = 3/2 R
For a diatomic gas (rigid):
(C_v)_dia = 5/2 R

2. Ratio of Specific Heats:
The ratio of C_v for the monoatomic and diatomic gases is:
(C_v)_mono / (C_v)_dia = (3/2 R) / (5/2 R).
Simplify:
(C_v)_mono / (C_v)_dia = 3/5.

Solution:

1. Expression Analysis:
The expression a × 10^b is written in scientific notation. The value of b determines the order of magnitude.

2. Rules for Scientific Notation:
If a ≤ 5, the order of magnitude is b.
If a > 5, the order of magnitude increases by 1, i.e., b + 1.

3. Conclusion:
For a ≤ 5, b is the order of magnitude.

Solution:

1. Simplification of the Circuit:
The 4 Ω and 4 Ω resistors are in parallel: \[ R_{\text{parallel}} = \frac{1}{\frac{1}{4} + \frac{1}{4}} = 2 \, \Omega. \]

The simplified circuit becomes a 3 V cell with internal resistance 1 Ω, in series with an external resistance of 2 Ω.

2. Current in the Circuit:
Using Ohm's law: \[ i = \frac{E}{R_{\text{internal}} + R_{\text{external}}}. \] Substituting the values: \[ i = \frac{3}{1 + 2} = \frac{3}{3} = 1 \, \text{A}. \]

3. Terminal Potential Difference:
The terminal potential difference is given by: \[ v = E - i r, \] where \( E \) is the emf of the cell, \( i \) is the current, and \( r \) is the internal resistance. Substituting the values: \[ v = 3 - (1 \times 1) = 3 - 1 = 2 \, \text{V}. \]

Solution:

1. Relation Between Binding Energy and Mass Defect:
Using Einstein's mass-energy equivalence: \[ \Delta m c^2 = \text{Binding Energy}. \]

2. Substitute the Given Values:
- Binding energy: \( \Delta E = 18 \times 10^8 \, \text{J} \),
- Speed of light: \( c = 3 \times 10^8 \, \text{m/s}. \)

Rearrange to find \( \Delta m \):
\[ \Delta m = \frac{\Delta E}{c^2}. \] Substituting: \[ \Delta m = \frac{18 \times 10^8}{(3 \times 10^8)^2}. \]

3. Simplify the Expression:
\[ \Delta m = \frac{18 \times 10^8}{9 \times 10^{16}} = 2 \times 10^{-8} \, \text{kg}. \]

4. Convert to Micrograms (μg):
\[ \Delta m = 2 \times 10^{-8} \, \text{kg} = 20 \, \mu\text{g}. \]

Solution:

1. Properties of Paramagnetic Substances:
- Paramagnetic substances align themselves along the direction of the external magnetic field (A is correct).
- They are weakly attracted towards an external magnetic field, not strongly (B is incorrect).
- Their magnetic susceptibility (\( \chi \)) is small but positive, meaning it is slightly more than zero (C is correct).
- Paramagnetic substances move from a region of weak magnetic field to a strong magnetic field (D is incorrect).

2. Most Appropriate Answer:
The correct statements are A and C.

Solution:

1. Length of Minute and Second Hand:
- Length of the minute hand: \( r_{\text{min}} = 60 \, \text{cm} = \frac{60}{100} \, \text{m} = 0.6 \, \text{m}
- Length of the second hand: \( r_{\text{sec}} = 75 \, \text{cm} = \frac{75}{100} \, \text{m} = 0.75 \, \text{m}.

2. Distance Traveled by the Minute Hand:
In 30 minutes, the minute hand completes half a rotation. The distance traveled is: \( x_{\text{min}} = \pi \cdot r_{\text{min}}. \)
Substituting: \( x_{\text{min}} = 3.14 \cdot 0.6 = 1.884 \, \text{m}. \)

3. Distance Traveled by the Second Hand:
In 30 minutes, the second hand completes 30 full rotations. The distance traveled is: \( x_{\text{sec}} = 30 \cdot 2\pi \cdot r_{\text{sec}}. \)
Substituting: \( x_{\text{sec}} = 30 \cdot 2 \cdot 3.14 \cdot 0.75 = 141.3 \, \text{m}. \)

4. Difference in Distance Traveled:
The difference \( x \) is: \( x = x_{\text{sec}} - x_{\text{min}}. \)
Substituting: \( x = 141.3 - 1.884 = 139.416 \, \text{m}. \)

5. Final Answer:
\( x \approx 139.4 \, \text{m}. \)

Solution:

1. Formula for Percentage Error in Y:
The percentage error in Y is given by:
(ΔY / Y) = (ΔM / M) + (Δℓ / ℓ).

2. Errors in Measurement:
The smallest scale division for mass M is ΔM = 5 g.
The smallest scale division for extension ℓ is Δℓ = 0.02 cm.

3. Substitute the Given Values:
Mass M = 500 g, Extension ℓ = 2 cm.
The percentage error in M is:
(ΔM / M) × 100 = (5 / 500) × 100 = 1%.
The percentage error in ℓ is:
(Δℓ / ℓ) × 100 = (0.02 / 2) × 100 = 1%.

4. Total Percentage Error in Y:
Adding the percentage errors:
(ΔY / Y) × 100 = 1 + 1 = 2%.

Solution:

1. Adiabatic Process Equation:
For an adiabatic process: T • V^{γ - 1} = constant.

2. Relation for Points a and d:
Using the adiabatic process equation between points a and d: T_a • V_a^{γ - 1} = T_d • V_d^{γ - 1}.
Rearrange: ( V_a / V_d )^{γ - 1} = T_d / T_a.

3. Relation for Points b and c:
Similarly, for points b and c: ( V_b / V_c )^{γ - 1} = T_c / T_b.

4. Comparing Temperatures:
From the diagram, since T_d = T_c and T_a = T_b: T_d / T_a = T_c / T_b.

5. Final Relation:
Using the above equality: ( V_a / V_d )^{γ - 1} = ( V_b / V_c )^{γ - 1}.
Simplify: V_a / V_d = V_b / V_c.

Solution:

1. Relation Between Angular Momentum and Time Period:
For planet A: π r₁² • T_A = L / 2m₁.
For planet B: π r₂² • T_B = 3L / 2m₂.

2. Ratio of Time Periods:
Divide equations for T_A and T_B: T_A / T_B = ( L / 2m₁ ) / ( 3L / 2m₂ ) • ( r₂² / r₁² ).
Simplify: T_A / T_B = ( m₂ / 3m₁ ) • ( r₁ / r₂ )².

3. Final Expression:
Rearrange to express T_A / T_B: T_A / T_B = 1 / 27 • ( m₂ / m₁ )³.

Solution:

1. At Resonance in an LCR Circuit:
At resonance, the impedance Z is equal to the resistance R:
    Z = R.
The current amplitude is given by:
    I = V / Z = V / R.

2. Effect of Halving Resistance:
If R is halved (R → R / 2):
    I = V / R → V / (R / 2) = 2 × (V / R).
Therefore, the current amplitude I becomes double.

3. Conclusion:
When the resistance is halved, the current amplitude at resonance doubles.

Solution:

1. Understanding the Circuit:
The circuit involves a combination of NOT, AND, and OR gates. The inputs A and B are processed through the gates to produce the output Y.

2. Constructing the Truth Table:

3. Output Analysis:
For all combinations of A and B, the output Y is consistently 0.

4. Conclusion:
The output of the circuit is always 0, regardless of the input values.

Solution:

1. Concept of Potential on the Surface of Conductors:
When two conducting spheres are connected by a conducting wire, their potentials become equal. The potential V on the surface of a sphere is given by:
    V = Kq∕r,
where K is Coulomb's constant, q is the charge, and r is the radius of the sphere.

2. Equating the Potentials:
For the two spheres:
    Kq1∕a = Kq2∕b.
Cancel K and rearrange:
    q1∕q2 = a∕b.

3. Conclusion:
The ratio of charges of the two spheres is:
    q1∕q2 = a∕b.

Solution:

1. Bernoulli's Equation:
Bernoulli's principle for an ideal, incompressible, and non-viscous fluid is expressed as:
    P + ρgh + ½ρv² = constant.
- P: Pressure of the fluid.
- ρ: Density of the fluid.
- g: Acceleration due to gravity.
- h: Height above a reference level.
- v: Velocity of the fluid.

2. Analysis of Options:
- Option (1): Incorrect because it uses mass m instead of density ρ.
- Option (2): Correct as it matches the standard Bernoulli equation.
- Option (3): Incorrect because it lacks the ½ factor in the kinetic energy term.
- Option (4): Incorrect because the gravitational potential energy term is divided by 2, which is not standard.

3. Conclusion:
The correct equation is:
    P + ρgh + ½ρv² = constant.

Solution:

1. Impulse-Momentum Theorem:
The force exerted is calculated using:
    F = ∆P∕∆t,
where:
    - ∆P = Change in momentum = m•v - m•u,
    - ∆t = Time duration of the impact.

2. Substitute the Values:
- Mass m = 150 g = 150 × 10⁻³ kg,
- Initial velocity u = 20 m/s,
- Final velocity v = 0 m/s,
- Time ∆t = 0.1 s.
Change in momentum:
    ∆P = m•v - m•u = 150 × 10⁻³ × 20 - 0 = 3 kg m/s.
Force:
    F = ∆P∕∆t = 3∕0.1 = 30 N.

3. Conclusion:
The force exerted by the ball on the hand of the player is:
    F = 30 N.

Solution:

1. Initial Momentum Conservation:
Since the particle is stationary, the initial momentum is zero. After breaking, the total momentum is conserved:
P_A = P_B, or m_Av_A = m_Bv_B. (Equation 1)

2. Kinetic Energy Expressions:
Kinetic energy for each part is given by:
K_A = ½ m_Av_A², and K_B = ½ m_Bv_B².

3. Take the Ratio:
Divide the kinetic energies:
K_B / K_A = (½ m_Bv_B²) / (½ m_Av_A²).
Simplify:
K_B / K_A = (m_B / m_A) × (v_B / v_A).

4. Substitute Momentum Conservation:
From Equation 1, v_B / v_A = m_A / m_B.
Substitute this:
K_B / K_A = v_B / v_A.

5. Conclusion:
The ratio of kinetic energies is:
K_B : K_A = v_B : v_A.

Solution:

1. Critical Angle Formula:
The critical angle θ_c is related to the refractive indices μ₁ (denser medium) and μ₂ (rarer medium) as:
sin(θ_c) = μ₂ / μ₁.

2. Substitution:
Given θ_c = 45°:
sin(45°) = μ₂ / μ₁.
Using sin(45°) = 1/√2:
1/√2 = μ₂ / μ₁.

3. Refractive Index Ratio:
Rearrange:
μ₁ / μ₂ = √2 : 1.

4. Conclusion:
The refractive index ratio is:
μ₁ : μ₂ = √2 : 1.

Solution:

1. Least Count of Vernier Caliper:
- 9 MSD = 10 VSD,
- Length of 1 MSD = 1 mm = 0.1 cm,
- Least Count (LC) = 1 MSD - 1 VSD = 0.01 cm.

2. Diameter of the Sphere:
- Main Scale Reading (MSR) = 2 cm,
- Vernier Scale Reading (VSR) = 2,
- Diameter = MSR + (LC × VSR) = 2 + (0.01 × 2) = 2.02 cm.

3. Volume of the Sphere:
- Radius r = Diameter / 2 = 2.02 / 2 = 1.01 cm,
- Volume V = (4/3)πr³ = (4/3) × 3.1416 × (1.01)³ ≈ 4.32 cm³.

4. Density of the Sphere:
- Mass m = 8.635 g,
- Density ρ = m / V = 8.635 / 4.32 ≈ 2.00 g/cm³.

Solution:

1. Mass Distribution and Areas:
The plate is divided into smaller sections:
- Section 1 (main rectangle): Area = 3 × 2 = 6, Mass = 6 × 1 = 6 kg,
- Section 2 (removed square): Area = 1 × 1 = 1, Mass = 1 × 1 = 1 kg,
- Section 3: Remaining rectangle area = 4 × 2 = 8, Mass = 8 kg.

2. Calculate Center of Mass (COM):
x_COM = Σ (m_i x_i) / Σ m_i, y_COM = Σ (m_i y_i) / Σ m_i.

For x_COM:
Mass centers for each section:
- Section 1: x₁ = 1.5, m₁ = 6,
- Section 2: x₂ = 1, m₂ = 1,
- Section 3: x₃ = 3, m₃ = 8.
Compute:
x_COM = (6 × 1.5 + 1 × 1 + 8 × 3) / 15 = 2.1.

For y_COM:
Centers for each section:
y₁ = 1, y₂ = 0.5, y₃ = 1.5.
Compute:
y_COM = (6 × 1 + 1 × 0.5 + 8 × 1.5) / 15 = 1.4.

3. Ratio of x and y Coordinates:
x_COM : y_COM = 15 : 9, n = 15.

Solution:

1. Balanced Forces Condition:
For no deflection, qE = qvB or E = vB.

2. Relating Velocity to Kinetic Energy:
KE = (1/2) m v² → v = √(2 KE / m).

3. Substituting Values:
KE = 5 × 1.6 × 10^-19 J, m = 9 × 10^-31 kg, B = 3 × 10^-6 T.
Velocity:
v = √((2 × 5 × 1.6 × 10^-19) / (9 × 10^-31)) ≈ 1.3 × 10⁶ m/s.
Electric Field:
E = vB = (1.3 × 10⁶) × (3 × 10^-6) = 4 N/C.

Solution:

1. Work Formula:
W = (N² B² A²) / (R × t).

2. Substituting Values:
N = 10, B = 0.5 T, A = 3.6 × 10^-3 m², R = 100 Ω, t = 1 s.
W = ((10)² (0.5)² (3.6 × 10^-3)²) / (100 × 1) ≈ 3.24 × 10^-6 J.

Solution:

1. Temperature Dependence of Resistance:
The resistance R at a given temperature is related to the initial resistance R₀ as:
R = R0 (1 + α ΔT), where:
- ΔR = R - R₀,
- α = temperature coefficient of resistance,
- ΔT = change in temperature.

Rearranging to find α:
α = ΔR / (R0 ΔT).

2. Case-I: 0°C to 100°C:
Resistance values:
R₀ = 10 Ω, R = 10.2 Ω at 100°C.
Substituting:
α = (10.2 - 10) / (10 × 100) = 0.002 per °C.

3. Case-II: 0°C to t°C:
Resistance values:
R₀ = 10 Ω, R = 10.95 Ω.
Substituting:
10.95 - 10 = 0.002 × 10 × t.
Simplify:
t = 0.95 / 0.02 = 475°C.

4. Convert to Kelvin:
Temperature in Kelvin:
T = t + 273 = 475 + 273 = 748 K.

Solution:

1. Formula for Electric Flux:
Electric flux is given by:
Φ = &vec;E · &vec;A, where:
- &vec;A = A × &hat;n,
- A = 4 m²,
- &hat;n = (2î + ĵ + &kcirc;) / √6.

2. Calculate &vec;A:
Substituting:
&vec;A = 4 × (2î + ĵ + &kcirc;) / √6 = (8î + 4ĵ + 4&kcirc;) / √6.

3. Dot Product of &vec;E and &vec;A:
Substituting &vec;E and &vec;A:
Φ = ((2î + 6ĵ + 8&kcirc;) / √6) · ((8î + 4ĵ + 4&kcirc;) / √6).
Simplify:
Φ = (1 / 6) × (2 × 8 + 6 × 4 + 8 × 4).

4. Simplify the Terms:
Φ = (1 / 6) × (16 + 24 + 32) = (1 / 6) × 72 = 12 Vm.

Solution:

1. Excess Pressure in a Soap Bubble:
The excess pressure inside a soap bubble is:
ΔP = 4S / R, where:
- S = 0.28 N/m (surface tension),
- R = radius of the soap bubble.

2. Balancing Pressure with Liquid Column:
The pressure due to the liquid column is:
ΔP = ρ g h, where:
- ρ = 8 × 10³ kg/m³ (density of liquid),
- g = 10 m/s²,
- h = 0.04 cm = 4 × 10&sup4; m.
Equating pressures:
4S / R = ρ g h.

3. Solve for R:
Substituting:
4 × 0.28 / R = 8 × 10³ × 10 × 4 × 10&sup4;.
Simplify:
R = (4 × 0.28) / 32 = 0.035 m = 3.5 cm.

4. Diameter of the Soap Bubble:
D = 2R = 2 × 3.5 = 7 cm.

Solution:

1. Frequency of a Closed Organ Pipe:
The frequency of the n-th overtone of a closed organ pipe is:
fc = (2n + 1) * v / 4ℓ.
For the seventh overtone (n = 7):
fc = (2 * 7 + 1) * v / 4ℓ = 15v / 4ℓ.

2. Frequency of an Open Organ Pipe:
The frequency of the n-th overtone of an open organ pipe is:
fo = (n + 1) * v / 2ℓ.
For the seventh overtone (n = 7):
fo = (7 + 1) * v / 2ℓ = 8v / 2ℓ = 4v / ℓ.

3. Ratio of Frequencies:
The ratio of frequencies is given as:
fc / fo = (15v / 4ℓ) / (4v / ℓ) = 15 / 16.
According to the problem, this ratio is also equal to:
fc / fo = (a - 1) / a.

4. Equate the Ratios:
(a - 1) / a = 15 / 16.
Simplify:
16(a - 1) = 15a.
16a - 16 = 15a.
a = 16.

Solution:

1. Vectors and Geometry:
- ⟶OQ points vertically upward.
- ⟶OP makes an angle of 90° with ⟶OQ.
- ⟶OR makes an angle of 45° with ⟶OQ and lies in the same plane.

2. Resolve the Vectors into Components:
Components:
- ⟶OPx = A, ⟶ORx = A * cos 45° = A / √2.
- ⟶OQy = A, ⟶ORy = A * sin 45° = A / √2.

3. Resultant Components:
- Rx = ⟶OPx + ⟶ORx = A + A / √2.
- Ry = ⟶OQy + ⟶ORy = A + A / √2.

4. Magnitude of Resultant Vector:
R = √(Rx2 + Ry2).
Substituting:
R = √[(A + A / √2)2 + (A + A / √2)2].
Simplify:
R = √2 * (A + A / √2).
Factorize:
R = A√2 * (1 + 1 / √2).
Further simplify:
R = A√3.

Solution:

1. Condition for Minima:
The angular position of minima in single-slit diffraction is given by:
sin θ = (nλ)/b,
where:
n = 2 (order of minima),
λ = 600 nm = 600 × 10^-9 m (wavelength of light),
b = 0.4 mm = 4 × 10^-4 m (width of the slit).

2. Angular Position for Second Minima:
Substituting the values:
θ ≈ (2λ)/b.
θ = (2 × 600 × 10^-9) / (4 × 10^-4).
Simplify:
θ = (1200 × 10^-9) / (4 × 10^-4) = 3 × 10^-3 rad.

3. Total Divergence:
For the second-order minima on both sides of the central maximum:
Total divergence = 2 × θ = 2 × 3 × 10^-3 = 6 × 10^-3 rad.

4. Conclusion:
The total angular divergence for the second-order minima is:
6 × 10^-3 rad.

Solution:

1. Formula for Closest Approach:
The distance of closest approach (r_min) is related to the velocity (v) by:
r_min = (4KZe²) / (mv²),
where:
K = (1 / 4πε₀) = 9 × 10⁹ SI unit,
Z = 80 (atomic number of the nucleus),
e = 1.6 × 10^-19 C (charge of the electron),
m = 6.72 × 10^-27 kg (mass of the alpha particle),
r_min = 4.5 × 10^-14 m.

2. Rearrange for Velocity:
Rearranging the formula:
v = √[(4KZe²) / (mr_min)].

3. Substitute the Given Values:
v = √[(4 × 9 × 10⁹ × 80 × (1.6 × 10^-19)²) / (6.72 × 10^-27 × 4.5 × 10^-14)]

4. Simplify:
Numerator:
4 × 9 × 80 = 2880,
(1.6 × 10^-19)² = 2.56 × 10^-38,
Numerator = 2880 × 2.56 × 10^-38 = 7.3728 × 10^-35.
Denominator:
6.72 × 10^-27 × 4.5 × 10^-14 = 3.024 × 10^-40.

Final Calculation:
v = √[(7.3728 × 10^-35) / (3.024 × 10^-40)].
v = √[2.437 × 10⁵].
v ≈ 1.56 × 10⁵ m/s.

5. Conclusion:
The maximum velocity of the alpha particle is:
156 × 10⁵ m/s.