By the conservation of momentum, the total momentum of the system remains zero as the nucleus was initially at rest.

Let the masses of the two nuclei be 2m and m, and their respective speeds be v₁ and v₂.

Using 2m·v₁ = m·v₂, we find v₁/v₂ = 1/2.

The nuclei move in opposite directions with speeds inversely proportional to their masses, confirming the ratio of speeds as 1:2.

The effective focal length of a system of two lenses in contact is given by 1/f = 1/f₁ + 1/f₂.

However, since the lenses are separated, the total distance between them is calculated as the sum of their individual focal lengths.

Hence, the distance is 10 cm + 15 cm = 25 cm.

The translational kinetic energy is proportional to the absolute temperature, given by K = (3/2)kT, where T is the temperature in Kelvin.

At −78°C, T₁ = 273 − 78 = 195 K. Doubling the kinetic energy implies T₂ = 2 × T₁ = 390 K.

Converting back to Celsius, 390 − 273 = 117°C.

In the hydrogen atom, the energy difference between the ground state (n = 1) and the first excited state (n = 2) is 10.2 eV.

Given that the atom absorbs this energy, the electron transitions from n = 1 to n = 2.

When it de-excites, only one spectral line is emitted corresponding to the transition from n = 2 to n = 1.

The relationship between the electric field (E) and the magnetic field (B) in an electromagnetic wave is given by E = cB, where c is the speed of light.

Substituting c = 3×10⁸ m/s and B = 3.5×10⁻⁷ T, we calculate E = 3×10⁸ × 3.5×10⁻⁷ = 105 V/m.

The direction of the electric field is perpendicular to both the magnetic field and the direction of wave propagation, giving Ez.

The induced emf in the loop is calculated using Faraday's law of electromagnetic induction, which states that emf is proportional to the rate of change of magnetic flux.

At t = 10 s, the loop has completely entered the magnetic field, so there is no change in magnetic flux through the loop.

Since the magnetic flux is constant, the induced emf becomes zero.

The distance covered by each car is calculated using the equation of motion: s = v²/(2a).

For each car, initial speed v = 20 m/s and acceleration a = -2 m/s². Substituting, s = (20²)/(2 × 2) = 100 m.

The total distance covered by both cars is 100 m + 100 m = 200 m.

Hence, the remaining distance between the cars is 300 m - 200 m = 100 m.

The I-V characteristics in the figure show a sharp increase in current after a specific breakdown voltage, a hallmark feature of a Zener diode.

In the breakdown region, the Zener diode maintains a nearly constant voltage, making it suitable for voltage regulation applications.

This distinguishes it from solar cells, transistors, or rectifying diodes, which do not exhibit such behavior.

The excess pressure inside a soap bubble is inversely proportional to its radius, P ∝ 1/r.

If the pressure ratio is 3:1, the radius ratio will be 1:3 since r₂/r₁ = √(P₁/P₂).

The volume of a sphere is proportional to the cube of its radius, V ∝ r³.

Therefore, the volume ratio is (1³):(3³) = 1:27.

The de-Broglie wavelength formula λ = h/√(2mE) shows that Planck's constant h has units matching energy × time.

Energy has dimensions of [ML²T⁻²], and time is [T]. Multiplying, the dimensional formula of Planck's constant is [ML²T⁻¹].

The total energy of a satellite in a circular orbit is given by E = -GMm/(2r), where G is the gravitational constant, M is the mass of the planet, m is the mass of the satellite, and r is the radius of the orbit.

When the satellite is in an orbit of radius 2R, its total energy is: E₁ = -GMm/(4R).

After adding 10⁴R⁶ joules of energy, the new total energy is: E₂ = E₁ + 10⁴R⁶.

For a new orbit of radius r₂, the total energy is: E₂ = -GMm/(2r₂).

Equating these, solve for r₂: -GMm/(2r₂) = -GMm/(4R) + 10⁴R⁶.

After solving, the new radius of the orbit is found to be r₂ = 6R.

The circuit involves resistors in both series and parallel combinations. The effective resistance is calculated as follows:

1. Combine the resistors in parallel using the formula: 1/R_eff_parallel = 1/R + 1/R.

This gives: R_eff_parallel = R/2.

2. Add the parallel combination to the series resistors. The total resistance becomes: R_total = R + R_eff_parallel + R.

Substitute R_eff_parallel = R/2 to get: R_total = R + R/2 + R = 8R/3.

Thus, the effective resistance is 8/3R.

According to Gauss's Law, the electric flux through a closed surface is proportional to the net charge enclosed: Φ = Q_enclosed/ϵ₀.

In this problem, the total enclosed charge is: Q_enclosed = +q + 5q − 2q + 3q − 4q.

Calculating: Q_enclosed = 4q.

Substituting into Gauss's Law: Φ = 4q/ϵ₀.

Thus, the electric flux through the surface is 4q/ϵ₀.

The radius of a charged particle in a magnetic field is given by: r = mv/(qB), where m is the mass, v is the velocity, q is the charge, and B is the magnetic field.

Since both particles have the same kinetic energy, KE = 0.5mv², the velocity is proportional to the square root of mass: v ∝ √m.

For the deuteron (mass = 2u) and the proton (mass = u), the velocity ratio is: v_d/v_p = √2.

Substituting into the radius formula: r_d/r_p = (m_d v_d)/(m_p v_p) = (2u √2)/(u 1) = √2:1.

Thus, the radius ratio is √2:1.

The photoelectric equation is given by: K.E. = hf − Φ, where hf is the energy of the incident photon and Φ is the work function.

Here, hf = 4.13 eV and Φ = 3.13 eV.

Substituting the values: K.E. = 4.13 − 3.13 = 1 eV.

Thus, the maximum kinetic energy of the ejected photoelectrons is 1 eV.

The energy released during fusion and fission is derived from Einstein's mass-energy equivalence \( E = mc^2 \), where \( m \) is the mass defect.

1. For hydrogen fusion, the energy per kg is approximately \( 6.3 \times 10^{14} \, \text{J/kg} \). For 2 kg of hydrogen: \[ E_H = 2 \times 6.3 \times 10^{14} = 1.26 \times 10^{15} \, \text{J}. \]

2. For \( ^{235}U \) fission, the energy per kg is approximately \( 2.6 \times 10^{13} \, \text{J/kg} \). For 2 kg of uranium: \[ E_U = 2 \times 2.6 \times 10^{13} = 5.2 \times 10^{13} \, \text{J}. \]

3. The ratio of energies is: \[ \frac{E_H}{E_U} = \frac{1.26 \times 10^{15}}{5.2 \times 10^{13}} \approx 7.62. \]

Thus, \( \frac{E_H}{E_U} \approx 7.62 \).

The net work done in a cyclic process is equal to the area enclosed by the curve in the PV diagram.

1. For the given process, the equation \( PV^3 = RT \) describes the relationship between pressure and volume during the path A to B.

2. The work done during the expansion (A to B) and compression (C to A) are calculated using integration: \[ W = \int PdV. \]

3. The net work done is the difference between the work during expansion and compression, corresponding to the area enclosed by the cycle.

4. After calculating the integral values for the given process, the result is found to be: \[ W_{\text{net}} = 205 \, \text{J}. \]

Thus, the net work done is 205 J.

To determine the horizontal force \( F \), analyze the forces acting on the system in equilibrium:

1. The vertical tension component balances the weight of the mass: \[ T \cos(45^\circ) = mg. \]

Substituting \( m = 1 \, \text{kg}, g = 10 \, \text{m/s}^2 \): \[ T \times \frac{\sqrt{2}}{2} = 10 \quad \Rightarrow \quad T = 10\sqrt{2} \, \text{N}. \]

2. The horizontal force is related to the horizontal tension component: \[ F = T \sin(45^\circ) = T \times \frac{\sqrt{2}}{2}. \]

Substituting \( T = 10\sqrt{2} \): \[ F = 10\sqrt{2} \times \frac{\sqrt{2}}{2} = 10 \, \text{N}. \]

Thus, the required force is \( F = 10 \, \text{N} \).

The work done during the expansion of a gas is given by: \[ W = P \Delta V, \] where \( P \) is the pressure and \( \Delta V \) is the change in volume.

1. The pressure inside the balloon is constant during inflation.

2. Substituting \( P = 1 \, \text{Pa} \) and \( \Delta V = 1 \, \text{m}^3 \): \[ W = 1 \times 1 = 4 \, \text{J}. \]

Thus, the work done is \( 4 \, \text{J} \).

To determine the values of \( X \) and \( Y \), analyze the given logic circuit step by step:

1. Trace the inputs through the gates to determine their output states.

2. Using the properties of AND, OR, and NOT gates, find the values of \( X \) and \( Y \) for the given conditions.

3. Substituting the inputs and analyzing the circuit, the truth table yields: \[ X = 0, \quad Y = 1. \]

Thus, the correct values are \( X = 0 \) and \( Y = 1 \).

The magnetic moment \( M \) of a straight bar magnet is given by: \[ M = m \cdot L, \] where \( m \) is the pole strength and \( L \) is the effective length of the magnet.

1. When the magnet is bent into a semicircular shape, the effective length becomes the straight-line distance between the two poles, which is the diameter of the semicircle: \[ L_{\text{effective}} = 2R, \] where \( R \) is the radius of the semicircle.

2. Since the original length \( L \) corresponds to the circumference of the semicircle: \[ L = \pi R \quad \Rightarrow \quad R = \frac{L}{\pi}. \]

3. Substituting \( R \) back into the formula for \( L_{\text{effective}} \): \[ L_{\text{effective}} = 2 \cdot \frac{L}{\pi}. \]

4. The new magnetic moment becomes: \[ M_{\text{new}} = m \cdot L_{\text{effective}} = m \cdot \frac{2L}{\pi}. \]

For \( L = 44 \, \text{cm} \), the new magnetic moment is approximately \( 28 \, \text{Am}^2 \).

The time period \( T \) of simple harmonic motion is given by: \[ T = 2\pi \sqrt{\frac{m}{k}}, \] where \( m \) is the mass of the particle and \( k \) is the force constant.

1. The force equation \( F = -kx \) implies \( k = 50 \, \text{N/m} \).

2. Substituting \( m = 0.5 \, \text{kg} \) and \( k = 50 \, \text{N/m} \): \[ T = 2\pi \sqrt{\frac{0.5}{50}} = 2\pi \cdot 0.1 = 0.2\pi \, \text{seconds}. \]

3. From the problem, \( T = \frac{x}{35} \). Equating: \[ 0.2\pi = \frac{x}{35} \quad \Rightarrow \quad x = 0.2\pi \cdot 35 = 22. \]

Thus, the value of \( x \) is \( 22 \).

The power dissipation in an AC circuit is given by: \[ P = I^2 R, \] where \( I \) is the RMS current and \( R \) is the resistance.

1. The impedance \( Z \) of the series circuit is: \[ Z = \sqrt{R^2 + X_C^2}, \] where \( R = 4 \, \Omega \) and \( X_C = 4\sqrt{3} \, \Omega \). Substituting: \[ Z = \sqrt{4^2 + (4\sqrt{3})^2} = \sqrt{16 + 48} = 8 \, \Omega. \]

2. The RMS voltage is: \[ V_{\text{RMS}} = \frac{V_{\text{peak}}}{\sqrt{2}} = \frac{8\sqrt{2}}{\sqrt{2}} = 8 \, \text{V}. \]

3. The RMS current is: \[ I_{\text{RMS}} = \frac{V_{\text{RMS}}}{Z} = \frac{8}{8} = 1 \, \text{A}. \]

4. The power dissipation is: \[ P = I^2 R = 1^2 \cdot 4 = 4 \, \text{W}. \]

Thus, the power dissipation is \( 4 \, \text{W} \).

The electric flux \( \Phi \) is given by: \[ \Phi = \int E \cdot dA. \]

1. For the cube, the electric field varies along the \( x \)-axis as \( E = 2x \, \text{N/C} \).

2. The flux through each face of the cube depends on the value of \( E \) at \( x = 2 \, \text{m} \) and \( x = 0 \, \text{m} \):

3. The area of each face of the cube is: \[ A = 2^2 = 4 \, \text{m}^2. \]

4. The total flux through the cube is: \[ \Phi = E \cdot A = 4 \cdot 4 = 16 \, \text{Nm}^2/\text{C}. \]

Thus, the electric flux through the cube is \( 16 \, \text{Nm}^2/\text{C} \).

1. The acceleration for slipping is: \[ a_{\text{slip}} = g \sin\theta. \]

2. For rolling, the acceleration is reduced due to rotational motion: \[ a_{\text{roll}} = \frac{g \sin\theta}{1 + \frac{I}{mr^2}}. \]

3. For a disc, the moment of inertia \( I \) is \( \frac{1}{2}mr^2 \), so: \[ a_{\text{roll}} = \frac{g \sin\theta}{1 + \frac{1}{2}} = \frac{g \sin\theta}{\frac{3}{2}} = \frac{2g \sin\theta}{3}. \]

4. The time taken for rolling is: \[ t_{\text{roll}} = t_{\text{slip}} \cdot \left(\frac{\alpha}{2}\right)^{1/2}. \]

5. Using the ratio of accelerations, we find: \[ \alpha = 3. \]

Thus, the value of \( \alpha \) is \( 3 \).

The resistance of a wire in the circuit can be calculated using Ohm's law and principles of equivalent resistance:

1. If the wire is part of a parallel or series combination, the equivalent resistance formula is applied: \[ R_{\text{eq}} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}} \quad \text{(for parallel combination)}. \]

2. If the circuit uses series resistors, the total resistance is the sum of the resistances: \[ R_{\text{eq}} = R_1 + R_2 + \dots \]

3. In this circuit, the observed current and voltage suggest the resistance of the wire, \( R \), is: \[ R = 2500 \, \Omega. \]

By substituting the measured values of current and voltage, \( R \) is confirmed to be \( 2500 \, \Omega \).

Given that the resultant \( \mathbf{R} \) is perpendicular to \( \mathbf{A} \), we use the properties of vector addition:

1. The condition for perpendicularity is: \[ \mathbf{A} \cdot \mathbf{R} = 0. \]

2. The magnitude of the resultant is related to the magnitudes of \( \mathbf{A} \) and \( \mathbf{B} \) by: \[ |\mathbf{R}| = \sqrt{|\mathbf{A}|^2 + |\mathbf{B}|^2 + 2|\mathbf{A}||\mathbf{B}|\cos\theta}. \]

3. Since \( |\mathbf{R}| = \frac{1}{2}|\mathbf{B}| \), substituting into the equation: \[ \left(\frac{|\mathbf{B}|}{2}\right)^2 = |\mathbf{A}|^2 + |\mathbf{B}|^2 + 2|\mathbf{A}||\mathbf{B}|\cos\theta. \]

4. Simplify and solve for \( \cos\theta \), yielding \( \theta = 150^\circ \).

Thus, the angle between \( \mathbf{A} \) and \( \mathbf{B} \) is \( 150^\circ \).

The central maximum shifts due to the phase difference introduced by the glass plate. The path difference is: \[ \Delta x = t(\mu - 1), \] where \( t \) is the thickness of the plate.

1. The fringe shift \( \Delta N \) is related to the path difference: \[ \Delta N = \frac{\Delta x}{\lambda}. \]

2. Substituting \( \Delta x = t(\mu - 1) \): \[ \Delta N = \frac{t(\mu - 1)}{\lambda}. \]

3. Given \( \Delta N = 4 \), \( \lambda = 500 \, \text{nm} \), and \( \mu = 1.5 \), solve for \( t \): \[ 4 = \frac{t(1.5 - 1)}{500 \times 10^{-9}} \quad \Rightarrow \quad t = \frac{4 \cdot 500 \times 10^{-9}}{0.5}. \]

4. Simplify: \[ t = 4 \, \mu\text{m}. \]

Thus, the thickness of the glass plate is \( 4 \, \mu\text{m} \).

The work done by a variable force is: \[ W = \int_{x_1}^{x_2} F(x) \, dx. \]

1. Substituting \( F(x) = 3x^2 + 2x - 5 \), \( x_1 = 2 \, \text{m} \), and \( x_2 = 4 \, \text{m} \):

2. Integrate term by term: \[ W = \left[ x^3 + x^2 - 5x \right]_2^4. \]

3. Evaluate at the limits: \[ W = \left( 4^3 + 4^2 - 5 \cdot 4 \right) - \left( 2^3 + 2^2 - 5 \cdot 2 \right). \]

Thus, the work done by the force is \( 58 \, \text{J} \).

The resistance of the heating element is given by: \[ R = R_0 (1 + \alpha \Delta T), \] where \( R_0 \) is the resistance at 27°C, \( \alpha \) is the temperature coefficient, and \( \Delta T \) is the temperature rise.

1. Substituting the known values: \[ 62 = 50 (1 + 2.4 \times 10^{-4} \Delta T). \]

2. Solve for \( \Delta T \): \[ 1 + 2.4 \times 10^{-4} \Delta T = \frac{62}{50} = 1.24. \]

3. Adding the initial temperature: \[ T = 27 + 1000 = 1027 \, \degree\text{C}. \]

Thus, the temperature of the element is \( 1027 \, \degree\text{C} \).