1. Let A and B be two vectors with an angle θ between them.
2. The term B - B sin(θ)a_perp can be interpreted as the component of B along the direction of A. This is because a_perp is the unit vector perpendicular to A, and B sin(θ)a_perp represents the perpendicular component of B to A.
3. The remaining vector B - B sin(θ)a_perp is thus aligned with the direction of A. Therefore, the correct direction is along A.

1. In dimensional analysis, we match the dimensions of both sides of the equation. For power P, we consider its fundamental dimensions: [P] = [M L² T⁻³].
2. Dimensions of variables:
- [q] = [M⁰ L⁰ T⁰] (dimensionless)
- [a] = [L T⁻²] (acceleration)
- [c] = [L T⁻¹] (speed of light)
3. Equating the dimensions of both sides, we solve for m and n, giving m = 2 and n = 3.

1. For the first lens (L1), the object is placed at 4f from L1. Using the lens formula: 1/f = 1/v1 - 1/u1, where u1 = -4f.
Solving, v1 = 4f/3.

2. This image acts as the object for the second lens (L2). The distance between the lenses is f/2. The object distance for L2 is:
u2 = (4f/3) - (f/2) = 5f/6.
3. Using the lens formula for L2:
1/f = 1/v2 - 1/u2.
Solving, v2 = 5f/11.

The final image is located 5f/11 to the right of L2.

1. Recall dimensions of variables:
- h: [M L² T⁻¹]
- m: [M]
- c: [L T⁻¹].
2. Analyze each option:
- hc/m = [L² T⁻²], not length.
- h/mc² = [T], time, not length.
- h²/m²c² = [L²], area, not length.
- h/mc = [L], dimension of length.

Hence, the correct answer is h/mc.

1. The speed distribution function P(v) describes the number of particles as a function of speed. From the graph, P(v) is nonzero only for speeds between 0 and 2v₀.
2. The area under the curve between 1.2v₀ and 1.8v₀ represents the fraction of particles in that range.
3. From the graph, this fraction is 0.4. Therefore, the number of particles in this range is 0.4N.

1. Using the Euler relation for extensive properties: U = T s - P v.
2. Given U = a s^(4/3)v^α:
T = (∂U/∂s) = (4/3) a s^(1/3)v^α,
P = -(∂U/∂v) = -α a s^(4/3)v^(α-1).
3. Substituting into the Euler relation:
a s^(4/3)v^α = (4/3)a s^(4/3)v^α + α a s^(4/3)v^α.
4. Solving: 1 = (4/3) + α.
α = -1/3.

1. The potential energy U(x) is given as U(x) = (αx)/(x² + β²). The force F(x) is:
F(x) = -dU/dx.
2. For small oscillations, the force can be approximated as F(x) ≈ -kx, where k is the effective spring constant. For small x:
F(x) = (α(β² - x²))/(x² + β²)² ≈ -k x.
3. Using dimensional analysis, the spring constant k is proportional to α/β³.
4. The angular frequency ω is given by ω = √(k/m), so:
ω ∝ √(α/mβ³).

1. Longitudinal waves involve oscillations in the medium parallel to the direction of wave propagation. Examples include sound waves.
2. Polarization occurs only in transverse waves, where oscillations are perpendicular to the direction of wave propagation.
3. Longitudinal waves lack perpendicular oscillations and hence cannot exhibit polarization.
Therefore, longitudinal waves cannot be polarized.

1. In this circuit, forward-biased diodes behave as short circuits, and reverse-biased diodes act as open circuits.
2. The total resistance in the circuit is the sum of the resistors in series:
R_total = 10Ω + 20Ω = 30Ω.
3. Using Ohm's law, V = IR:
I = V/R = 2V/30Ω = 0.0667 A ≈ 0.2 A.

1. The total electric flux through a closed surface is given by Gauss's Law:
Φ = ∮ E · dA = Q_enclosed/ε₀.
2. Here, Q_enclosed = Q, and the cube is a closed surface.
3. By symmetry, the flux is distributed evenly over the six faces of the cube, but the total flux through all six surfaces is:
Φ = Q/ε₀.

The elastic potential energy stored in a strained body is the energy per unit volume stored due to deformation. It is given by the formula:

U = (1/2) σ ε,

where:
- U is the elastic potential energy per unit volume,
- σ is the stress,
- ε is the strain.

For the total elastic potential energy stored in the body, the energy per unit volume is multiplied by the total volume of the body V. Thus, the total elastic potential energy U_total is:

U_total = (1/2) σ ε V.

Hence, the elastic potential energy of a strained body is:

(1/2) Stress × Strain × Volume.

1. Statement (i): The mass energy of a nucleus is smaller than the sum of the mass energies of individual protons and neutrons due to the binding energy. Therefore, this statement is false.
2. Statement (ii): When a nucleus is separated into its nucleons, energy equal to the binding energy must be supplied to overcome the nuclear forces holding the nucleons together. This statement is true.
3. Statement (iii): The binding energy measures how strongly the nucleons (protons and neutrons) are held together within the nucleus. This statement is true.
4. Statement (iv): Nuclear fission involves the splitting of a heavy nucleus into lighter nuclei, releasing binding energy and leading to a higher binding energy per nucleon in the fragments. This statement is true.

Thus, the correct answer is that statements (ii), (iii), and (iv) are true.

1. Step 1: The angular momentum J is related to the kinetic energy K and the radius R by the formula:
J = m R v,
where v is the velocity of the satellite.

2. Step 2: The total kinetic energy is given by:
K = (1/2) m v².
From the relationship J = m R v, solve for v and substitute it into the equation for kinetic energy:
v = J / (m R), K = (1/2) m (J / m R)² = J² / 2mR².

3. Step 3: The total energy E of the satellite is the sum of its kinetic and potential energy. The potential energy is given by:
U = -GMm / R,
where G is the gravitational constant and M is the mass of the earth. The total energy is:
E = K + U = -J² / 2mR².

Thus, the correct answer is:
K = J² / 2mR², E = -J² / 2mR².

1. Step 1: The force of static friction f_s is given by:
f_s = µ_s N,
where µ_s is the coefficient of static friction and N is the normal force.

2. Step 2: The normal force is altered by the applied force F at an angle of 60°. The normal force N is given by:
N = mg - F sin 60°,
where m = 10 kg is the mass of the block, and g = 9.8 m/s² is the acceleration due to gravity.

3. Step 3: The applied force F must overcome the force of static friction, so:
F = f_s / µ_s = (10 × 9.8 × 0.6) / cos 60° = 24.97 N.

For destructive interference (dark fringe) to occur in reflected light, the condition is:
2 µ t cos r = (m + 1/2) λ,
where:
- µ = 1.5 is the refractive index of the glass plate,
- t is the thickness of the plate,
- r = 60° is the angle of refraction,
- λ = 6000 Å = 6 × 10⁻⁷ m is the wavelength of light in air,
- m = 0 for the smallest thickness (first-order dark fringe).

Substituting m = 0 into the formula:
2 µ t cos r = λ / 2.
Rearranging for t:
t = λ / (4 µ cos r).
Substituting the given values:
t = (6 × 10⁻⁷) / (4 × 1.5 × cos 60°).
Since cos 60° = 1/2:
t = (6 × 10⁻⁷) / (3) = 2 × 10⁻⁷ m.

The smallest thickness of the plate is:
t = 4 × 10⁻⁷ m.

1. Step 1: The total resistance in the circuit is the sum of the internal resistance r and the external resistance R:
R_total = R + r.

2. Step 2: The power generated in the circuit is given by:
P = E₀² / R_total² × R = E₀² / (R + r)² × R.

3. Step 3: To find the value of R that maximizes the power, take the derivative of P with respect to R and set it to zero:
dP/dR = 0.
Solving, we find:
R = r.

Thus, the value of R for which the power generated in the circuit is maximum is R = r.

1. Step 1: The given diagram involves capacitors in series and parallel. For capacitors in series, the equivalent capacitance C_eq is given by:
1 / C_eq = 1 / C₁ + 1 / C₂ + ...
For capacitors in parallel, the equivalent capacitance is the sum of the individual capacitances:
C_eq = C₁ + C₂ + ....

2. Step 2: By applying these formulas to the combination of capacitors in the given circuit, the equivalent capacitance between points P and N is:
2C / 3.

1. Step 1: In single-slit diffraction, the condition for the m-th diffraction minimum is given by:
a sin θ_m = m λ,
where a is the width of the slit, λ is the wavelength of light, and m is the diffraction order.

2. Step 2: For the 2nd order minimum of λ₁ and the 3rd order minimum of λ₂, the angles for both minima coincide:
2 λ₁ = 3 λ₂.

3. Step 3: Rearrange to find the ratio:
λ₁ / λ₂ = 3 / 2.

1. Step 1: From the given acceleration-time graph, observe that the acceleration is constant.
2. Step 2: The velocity-time graph is the integral of the acceleration-time graph with respect to time. Since acceleration is constant, the velocity increases linearly with time.
3. Step 3: With initial velocity zero, the velocity-time graph is a straight line starting at the origin and increasing with time.

1. Step 1: The angular momentum L of a particle about the origin is given by the cross product:
L = r × m u,
where r is the position vector and u is the velocity vector.

2. Step 2: The magnitude of the angular momentum is:
L = |r| · m |u| · sin θ,
where θ is the angle between r and u.

3. Step 3: Since b is constant and the particle moves in a straight line, the angular momentum varies with sin θ, and the correct graph is as shown.

1. The center of mass of a uniform rectangular plate lies at its geometric center.
2. For a symmetric plate with origin shifted, the center of mass is shifted accordingly relative to the origin.
3. Using calculations, the coordinates of the center of mass relative to the origin are:
(-a/6, -b/6).

1. The applied voltage rms is 50 V.
2. Using V_applied² = V_R² + V_C² + V_L²:
Substituting known values, V_L = 50 V.
3. Peak voltage across the inductor:
V_peak,L = 50√2 V.

1. The ball is moving in a horizontal circle. The forces acting are the gravitational force (mg) and the tension in the string (T).
2. Since the ball is in circular motion, the horizontal component of tension provides the centripetal force, and the vertical component balances the weight.
3. The torque about the center O is due to the force perpendicular to the radius. However, the tension acts along the string, and no torque is created about O.
4. Therefore, the torque about the center is zero.

1. The incident ray, reflected ray, and the normal to the surface lie in the same plane.
2. By the law of reflection, the angle of incidence equals the angle of reflection.
3. Using vector representation for reflection:
n₂ = n₁ + 2 (n₁ ⋅ t) t.

1. The energy of emitted electrons is related to the wavelength of incident light by the photoelectric equation:
E = hc/λ - φ,
where E is the kinetic energy of emitted electrons, h is Planck’s constant, c is the speed of light, and φ is the work function.

2. The kinetic energy is related to the radius R and the magnetic field B as:
E = eB²R² / 2m,
where e is the electron charge and m is the mass of the electron.

3. Substituting for E:
eB²R² / 2m = hc/λ - φ.

4. Rearranging for B²:
B² ∝ 1/λ.

5. Hence, the graph of B² vs 1/λ is a straight line.