CBSE Class 12 Physics Question Paper 2024 (Set 3 - 55/1/3) Available: Download Solution PDF with Answer Key

Step 1: Understand the symmetry and net effect of the electric field due to a circular charge distribution.

In a uniformly charged ring, each infinitesimal charge element produces an electric field at the center. Due to the circular symmetry, the horizontal components of the electric fields due to opposite charge elements cancel out. Therefore, the net electric field at the center of a uniformly charged ring is zero. Quick Tip: For problems involving symmetry in charge distributions, it is often useful to consider the components of the electric fields and their directions. Symmetrical arrangements like rings, spheres, or cylinders often lead to cancellations of electric field components at certain points.

Step 1: Analyzing the forces acting on sphere X.

Since sphere X carries a positive charge \( +q \) and is placed between sphere Y with charge \( -q \) and sphere Z with charge \( +q \), it will experience an attractive force towards sphere Y and a repulsive force away from sphere Z. Given the equidistance and symmetry, the resultant force vector on X will be directed along the line connecting X and Y, pointing towards Y.

Step 2: Predicting the movement of X.

Due to the attractive force exerted by Y, sphere X will move directly towards Y, following path B in the diagram. Path B represents the direct line of action due to the unbalanced attractive force towards the negatively charged sphere Y. Quick Tip: In problems involving multiple charges, always consider the net force acting on each charge by vector addition of the individual forces exerted by all other charges in the system. The direction of the net force determines the direction of acceleration and movement.

Step 1: Understanding the direction of current and electron motion.

In metallic conductors, electric current is due to the flow of electrons. Since electrons have a negative charge, they move in the direction opposite to the electric field. Thus, if the electrons move in the \( +x \) direction, the electric field \( \vec{E} \) must be in the \( -x \) direction. The current density \( \vec{j} \), which is defined in the direction of positive charge flow, is thus in the \( +x \) direction. Quick Tip: Remember that in the context of current and electric fields in conductors, the direction of the electric field is opposite to the motion of electrons because they are negatively charged.

Step 1: Applying the formula for the radius of a particle's circular path in a magnetic field.

The radius \( r \) of the circular path of a charged particle in a magnetic field is given by the equation: \[ r = \frac{mv}{qB} \]
where \( m \) is the mass of the particle, \( v \) is its velocity, \( q \) is its charge, and \( B \) is the magnetic field strength. Given that both particles are charged the same (\( q \)) and accelerated through the same potential difference (\( V \)), they will have the same kinetic energy and thus the same velocity \( v \). Therefore, the ratio of their masses is inversely proportional to the square of the ratio of their radii: \[ \frac{m_1}{m_2} = \left(\frac{r_1}{r_2}\right)^2 = \left(\frac{a}{b}\right)^2 \] Quick Tip: When dealing with circular motion of charged particles in a magnetic field, remember that the radius of the path is directly proportional to the particle's mass and inversely proportional to the charge and magnetic field strength.

Step 1: Calculating the shunt resistance needed to convert the galvanometer.

To convert a galvanometer to an ammeter, a shunt resistor \( S \) is placed in parallel to allow the full range of current to pass. If the galvanometer reads 0.1% of the full scale (1 A), the galvanometer current \( I_g \) is 0.001 A. Using the formula: \[ \frac{1}{R_a} = \frac{1}{G} + \frac{1}{S} \]
where \( R_a \) is the desired ammeter resistance. Given \( I = 1 \, A \) and \( I_g = 0.001 \, A \), \( S \) can be derived to be approximately \( \frac{G}{999} \). Thus, the total resistance \( R_a \approx \frac{G}{1000} \). Quick Tip: When converting a galvanometer to an ammeter, the shunt resistor drastically reduces the overall resistance to allow more current through without damaging the sensitive galvanometer mechanism.

Step 1: Applying the right-hand rule to find the direction of the force.

Using the Lorentz force law: \[ \vec{F} = I \vec{L} \times \vec{B} \]
For \( \vec{L} = 0.1 \, m \hat{j} \) and \( \vec{B} = (5 \, mT) \hat{j} - (8 \, mT) \hat{k} \), the cross product yields: \[ \vec{F} = (1 \, A) (0.1 \, m) (0 \hat{i} + 0 \hat{j} + 8 \, mT \hat{i}) = 0.8 \, mN \hat{i} \]
The direction is negative \( \hat{i} \) due to the negative component in the magnetic field. Quick Tip: In calculating magnetic forces, ensure vector directions and units are correctly aligned, and apply the right-hand rule for determining force direction.

Step 1: Calculating the output voltage using the transformer equation.

The transformer equation relates the primary voltage (\( V_p \)), secondary voltage (\( V_s \)), primary turns (\( N_p \)), and secondary turns (\( N_s \)): \[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \]
Given \( V_p = 220 \, V \), \( N_p = 500 \), and \( N_s = 5000 \), solving for \( V_s \) yields: \[ V_s = \frac{5000}{500} \times 220 \, V = 2200 \, V \]
The frequency remains unchanged at 50 Hz. Quick Tip: In transformers, the voltage transformation ratio is directly proportional to the turns ratio, but the frequency of the AC supply remains unchanged.

Step 1: Identifying the scientist responsible for electromagnetic waves.

Heinrich Rudolf Hertz, a German physicist, was the first person to generate and observe electromagnetic waves in the laboratory. He performed experiments between 1886 and 1889, proving that light and other electromagnetic radiation travel in waves. Quick Tip: Hertz’s experiments confirmed the theoretical predictions made by James Clerk Maxwell about electromagnetic waves, thus providing solid experimental evidence of Maxwell's theory.

Step 1: Analyzing the relationship between wavelength and momentum.

The de Broglie wavelength \( \lambda \) of a particle is given by: \[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum. If the wavelength is the same for both particles (electron and proton), it implies that their momenta are equal. Quick Tip: The de Broglie relation connects the momentum of a particle with its wavelength, establishing that particles with the same wavelength must have the same momentum.

Step 1: Using the photoelectric effect equation.

The kinetic energy of the emitted electron is given by: \[ K.E. = E_{photon} - \phi \]
where \( E_{photon} \) is the photon energy and \( \phi \) is the work function. The speed of the emitted electron is related to its kinetic energy by: \[ K.E. = \frac{1}{2} m v^2 \]
For beam A and beam B, the respective kinetic energies are \( E_A - \phi \) and \( E_B - \phi \). Since the ratio of the speeds \( v_A \) and \( v_B \) depends on the square root of the ratio of their kinetic energies, the ratio of maximum speeds is: \[ \frac{v_A}{v_B} = \sqrt{\frac{E_A - \phi}{E_B - \phi}} = \sqrt{\frac{3.3 - 2.3}{11.3 - 2.3}} = \sqrt{\frac{1}{9}} = \frac{1}{3} \] Quick Tip: In the photoelectric effect, the kinetic energy of the emitted electron is determined by the energy of the incoming photon minus the work function of the material. The maximum speed of the electron is related to this kinetic energy.

Step 1: Understanding the Balmer series.

In the Balmer series, the electron transitions occur between \( n_i \) and \( n_f = 2 \). The second spectral line of the Balmer series corresponds to the transition from \( n_i = 3 \) to \( n_f = 2 \), which gives the second line in the visible spectrum. Quick Tip: The Balmer series corresponds to transitions where the final orbit is \( n_f = 2 \), and the transitions involve higher energy levels \( n_i \). The second spectral line comes from \( n_i = 3 \) to \( n_f = 2 \).

Step 1: Understanding the effect of doping with As.

When germanium (Ge) is doped with arsenic (As), which is a donor impurity, it donates an extra electron to the conduction band. This results in an increase in the number of conduction electrons in the material. Quick Tip: Doping with donor impurities like As introduces extra electrons into the conduction band, thus increasing the number of conduction electrons.

Step 1: Analyzing the situation.

The assertion is true because two parallel wires carrying current in the same direction will attract each other, while those carrying current in opposite directions will repel each other. This is a result of the magnetic fields produced by the currents. Therefore, the assertion is correct. The reason is also correct, and it explains why the wires move apart. Quick Tip: The magnetic force between two current-carrying conductors follows the rule: currents in the same direction attract and currents in opposite directions repel.

Step 1: Understanding the nature of images formed by plane and convex mirrors.

The assertion is correct because plane mirrors and convex mirrors only form virtual images, which cannot be projected onto a screen. However, the reason is incorrect because a virtual image can indeed act as the object for another mirror to produce a real image (such as in the case of compound optical systems). Quick Tip: Plane and convex mirrors only form virtual images. The fact that a virtual image cannot directly serve as an object for producing a real image is incorrect in some contexts, especially when using multiple mirrors or lenses.

Step 1: Understanding mutual inductance and flux linkage.

When two coils are wound on each other, the magnetic flux through one coil is maximized due to the close physical configuration, resulting in maximum flux linkage. This maximization of flux linkage leads to the highest mutual inductance, as the change in current in one coil will induce the greatest change in the other coil. Quick Tip: The mutual inductance between two coils is directly proportional to the amount of flux linkage, which is maximized when the coils are wound on each other.

Step 1: Understanding the photoelectric effect.

The assertion is true because in the photoelectric effect, the kinetic energy of emitted photoelectrons depends on the frequency (or energy) of the incident light, not its intensity. As intensity increases, the number of emitted electrons increases, but the kinetic energy of each electron remains constant.

Step 2: Correcting the reason.

The reason is false because the photoelectric current depends on the intensity of the light (which is proportional to the number of photoelectrons emitted), not on the wavelength. Quick Tip: In the photoelectric effect, the kinetic energy of the emitted electrons is determined by the frequency of the incident light, not its intensity. Intensity affects the number of emitted electrons, while frequency affects their kinetic energy.

Step 1: Understanding the relationship between resistance, length, and area.
The resistance \( R \) of a wire is given by the formula: \[ R = \rho \frac{L}{A} \]
where \( \rho \) is the resistivity of the material, \( L \) is the length, and \( A \) is the cross-sectional area.

Step 2: Applying the changes due to stretching.

When the wire is stretched, its length increases by 25%, so the new length is: \[ L' = L + 0.25L = 1.25L \]
Since the volume of the wire is conserved during stretching, the cross-sectional area decreases in such a way that the volume remains constant: \[ A' = \frac{A}{1.25} \]
The new resistance \( R' \) is given by: \[ R' = \rho \frac{L'}{A'} = \rho \frac{1.25L}{\frac{A}{1.25}} = \rho \frac{1.25^2L}{A} \]
Thus, the new resistance is: \[ R' = 1.5625R \]
The percentage increase in resistance is: \[ \frac{R' - R}{R} \times 100 = \frac{1.5625R - R}{R} \times 100 = 56.25% \] Quick Tip: When a wire is stretched, its length increases and its area decreases. The resistance increases more than the length due to the inverse relationship between resistance and cross-sectional area.

Given:

Object distance, \( u = -30 \, cm \) (since the object is in front of the concave mirror, it is negative).
Radius of curvature, \( R = 40 \, cm \).
Focal length, \( f = \frac{R}{2} = \frac{40}{2} = 20 \, cm \).


Step 1: To find the position of the image, we use the mirror equation: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Substitute the known values: \[ \frac{1}{20} = \frac{1}{v} + \frac{1}{-30} \]
Simplifying: \[ \frac{1}{v} = \frac{1}{20} + \frac{1}{-30} = \frac{3}{60} + \frac{-2}{60} = \frac{5}{60} \]
Therefore: \[ v = \frac{60}{5} = 12 \, cm \]
So, the image is formed at a distance of \( 12 \, cm \) in front of the mirror.


Step 2: To find the magnification, we use the magnification formula: \[ M = \frac{-v}{u} \]
Substitute the values: \[ M = \frac{-12}{-30} = \frac{12}{30} = 0.4 \]

Final Answers:

(i) The position of the image is \( 12 \, cm \) in front of the mirror.
(ii) The magnification of the image is \( 0.4 \). Quick Tip: For concave mirrors, if the object is placed outside the focal length, the image formed will be real, inverted, and reduced in size. The magnification gives the ratio of image size to object size.

Given:

• Kinetic energy of the neutron \( E \),
• The de Broglie wavelength of the neutron is \( \lambda_n \),
• The wavelength of the photon is \( \lambda_p \).

Step 1: Expression for the de Broglie wavelength of the neutron (\( \lambda_n \)):

The de Broglie wavelength of the neutron is given by: \[ \lambda_n = \frac{h}{p_n} \] where \( p_n \) is the momentum of the neutron.

The momentum of the neutron is related to its kinetic energy \( E \) as: \[ E = \frac{p_n^2}{2m} \] Rearranging for \( p_n \): \[ p_n = \sqrt{2mE} \] Thus, the de Broglie wavelength of the neutron becomes: \[ \lambda_n = \frac{h}{\sqrt{2mE}} \]

Step 2: Expression for the wavelength of the photon (\( \lambda_p \)):

The wavelength of the photon is related to its energy \( E \) by: \[ E = \frac{hc}{\lambda_p} \] Rearranging for \( \lambda_p \): \[ \lambda_p = \frac{hc}{E} \]

Step 3: Ratio of the wavelengths \( \frac{\lambda_n}{\lambda_p} \):

Now, we can find the ratio of the wavelengths: \[ \frac{\lambda_n}{\lambda_p} = \frac{\frac{h}{\sqrt{2mE}}}{\frac{hc}{E}} \] Simplifying: \[ \frac{\lambda_n}{\lambda_p} = \frac{h}{\sqrt{2mE}} \times \frac{E}{hc} = \frac{E}{\sqrt{2mE} \, c} \] Further simplifying: \[ \frac{\lambda_n}{\lambda_p} = \frac{\sqrt{E}}{\sqrt{2mc}} \]

Thus, the expression for the ratio of the de Broglie wavelength of the neutron to the wavelength of the photon is: \[ \frac{\lambda_n}{\lambda_p} = \frac{\sqrt{E}}{\sqrt{2mc}} \]

Step 1: Wavelength of the reflected light.
The wavelength of the reflected light remains the same as in air, because reflection does not affect the wavelength of light. Therefore, the wavelength of the reflected light is the same as the initial wavelength in air.

Step 2: Wavelength of the refracted light.
The wavelength of light in a medium is related to the wavelength in air by the refractive index \( n \): \[ \lambda_{refracted} = \frac{\lambda_{air}}{n} \]
where \( n = 1.5 \). The wavelength in air \( \lambda_{air} \) is calculated from the speed of light and the frequency: \[ \lambda_{air} = \frac{c}{f} = \frac{3.0 \times 10^8}{5.0 \times 10^{14}} = 6.0 \times 10^{-7} \, m \]
Now, the wavelength of the refracted light is: \[ \lambda_{refracted} = \frac{6.0 \times 10^{-7}}{1.5} = 4.0 \times 10^{-7} \, m \] Quick Tip: The refractive index affects the wavelength of light when it enters a new medium, but it does not affect the frequency of the light. The reflected light retains its original wavelength, as it does not enter the new medium.

Given:

• Focal length of the lens, \( f = 16 \, \text{cm} \)
• Refractive index of the lens material, \( n = 1.4 \)

For a plano-convex lens, the lens maker's formula is given by: \[ \frac{1}{f} = (n-1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \] where \( R_1 \) is the radius of curvature of the convex surface and \( R_2 = \infty \) for the flat surface.

This simplifies the formula to: \[ \frac{1}{f} = \frac{n-1}{R} \]

Rearranging to solve for \( R \): \[ R = \frac{n-1}{\frac{1}{f}} \]

Substituting the given values: \[ R = \frac{1.4 - 1}{\frac{1}{16}} = \frac{0.4}{\frac{1}{16}} = 0.4 \times 16 = 6.4 \, \text{cm} \]

Conclusion: The radius of the curved surface of the plano-convex lens is \( R = 6.4 \, \text{cm} \).

Step 1: Diffusion current.
Diffusion current arises when charge carriers (electrons or holes) move from regions of high concentration to low concentration. In a p-n junction, electrons move from the n-region to the p-region, and holes move from the p-region to the n-region, generating a diffusion current.

Step 2: Drift current.
Drift current arises due to the movement of charge carriers under the influence of an external electric field. In a p-n junction, once the electric field is established due to the potential difference across the junction, the electrons in the n-region and the holes in the p-region drift towards the opposite sides. Quick Tip: In a p-n junction, the diffusion current dominates when the junction is forward biased, while drift current plays a major role under reverse bias due to the internal electric field.

Given:

Initial capacitance \( C_1 = 20 \, pF = 20 \times 10^{-12} \, F \)
Charge on the capacitor \( Q = 4.0 \, µC = 4.0 \times 10^{-6} \, C \)
Initial plate separation \( d_1 = 1 \, mm = 1 \times 10^{-3} \, m \)
Final plate separation \( d_2 = 5 \, mm = 5 \times 10^{-3} \, m \)


\subsection*{Calculating the New Capacitance

The capacitance of a parallel plate capacitor is inversely proportional to the distance between the plates. The new capacitance \( C_2 \) when the distance is increased can be calculated by: \[ C_2 = C_1 \frac{d_1}{d_2} = (20 \times 10^{-12} \, F) \frac{1 \times 10^{-3} \, m}{5 \times 10^{-3} \, m} = 4 \times 10^{-12} \, F \]

\subsection*{Calculating the Energy Stored in the Capacitor

The energy stored in a capacitor is given by: \[ U = \frac{Q^2}{2C} \]

Initial Energy: \[ U_1 = \frac{(4.0 \times 10^{-6} \, C)^2}{2 \times 20 \times 10^{-12} \, F} = 0.4 \, J \]

Final Energy: \[ U_2 = \frac{(4.0 \times 10^{-6} \, C)^2}{2 \times 4 \times 10^{-12} \, F} = 2.0 \, J \]

\subsection*{Calculating the Work Done

The work done to pull the plates apart is equal to the change in stored energy: \[ W = U_2 - U_1 = 2.0 \, J - 0.4 \, J = 1.6 \, J \]

Conclusion: The amount of work done to increase the plate separation to 5 mm while keeping the charge constant is \( 1.6 \, J \). Quick Tip: When the plate separation in a capacitor is increased while maintaining the charge, the capacitance decreases and the potential energy stored in the capacitor increases, resulting in positive work done.

Step 1: Understanding current density.
Current density \( \vec{j} \) is defined as the amount of charge flowing per unit area per unit time, and it is a vector quantity that points in the direction of flow of positive charges (opposite to the flow of electrons).

Step 2: Deriving the expression for current density.
According to Drude's model: \[ \vec{j} = n e \vec{v_d} \]
where \( \vec{v_d} \) is the drift velocity of electrons, and it is related to the electric field by: \[ \vec{v_d} = -\frac{e \vec{E} \tau}{m} \]
Thus, substituting for \( \vec{v_d} \) we get: \[ \vec{j} = n e \left(-\frac{e \vec{E} \tau}{m}\right) = -\frac{ne^2 \tau}{m} \vec{E} \]
Defining \( \alpha = \frac{ne^2}{m} \tau \), we have: \[ \vec{j} = \alpha \vec{E} \] Quick Tip: Current density links the motion of charges to the applied electric field, providing a measure of how effectively a material conducts electricity.

Step 1: Defining a Wheatstone bridge.
A Wheatstone bridge is an electrical circuit used to measure an unknown electrical resistance by balancing two legs of a bridge circuit. It consists of four resistors \( R_1, R_2, R_3, \) and \( R_4 \) arranged in a quadrilateral with a voltage source connected across one diagonal and a galvanometer across the other.

Step 2: Conditions for balance.
For the Wheatstone bridge to be balanced (i.e., no current flows through the galvanometer), the following condition must be met: \[ \frac{R_1}{R_2} = \frac{R_3}{R_4} \]
This condition ensures that the potential drop across \( R_1 \) is proportional to the potential drop across \( R_2 \), and similarly for \( R_3 \) and \( R_4 \), leading to no potential difference across the galvanometer.

Step 3: Deriving the balance condition.
Consider the voltage at the nodes of the Wheatstone bridge. At balance, the voltage at the junction of \( R_1 \) and \( R_3 \) equals the voltage at the junction of \( R_2 \) and \( R_4 \), ensuring no current through the galvanometer. Applying Kirchhoff’s Voltage Law: \[ \frac{V_a}{R_1} = \frac{V_b}{R_2} \quad and \quad \frac{V_a}{R_3} = \frac{V_b}{R_4} \]
Simplifying, we get: \[ \frac{R_1}{R_2} = \frac{R_3}{R_4} \] Quick Tip: In practical applications, adjusting two of the resistors allows the Wheatstone bridge to measure the third resistor precisely, assuming the fourth is known.

(a) Net Force on the Coil
The net force \(\vec{F}\) on a current-carrying loop in a uniform magnetic field, which is perpendicular to the loop, is zero. This is because the magnetic forces on opposite segments of the loop cancel each other out.

Answer: The net force on the coil is 0 N.

(b) Torque on the Coil
The torque \(\vec{\tau}\) on a current-carrying coil in a magnetic field can be calculated using: \[ \vec{\tau} = \vec{\mu} \times \vec{B} \]
where \(\vec{\mu}\) is the magnetic moment of the coil given by: \[ \vec{\mu} = I \vec{A} \]
For a single loop: \[ \mu = I A = 4 \, A \times 0.2 \times 10^{-4} \, m^2 = 8 \times 10^{-4} \, Am^2 \]
The magnitude of the torque is then: \[ |\tau| = \mu B = 8 \times 10^{-4} \, Am^2 \times 0.5 \, T = 4 \times 10^{-4} \, Nm \]

Answer: The torque on the coil is \(4 \times 10^{-4} \, Nm\).

c) Average Force on Each Electron
The Lorentz force \(\vec{F}\) on an electron moving in a magnetic field is given by: \[ \vec{F} = q \vec{v} \times \vec{B} \]
where \( q = -e \), and \( \vec{v} \) is the drift velocity calculated as: \[ I = n e v A \Rightarrow v = \frac{I}{n e A} \]
Substituting given values: \[ v = \frac{4}{10^{28} \times 1.6 \times 10^{-19} \times 0.2 \times 10^{-4}} \approx 1.25 \times 10^6 \, m/s \]
Thus, the force on each electron is: \[ |F| = e v B = 1.6 \times 10^{-19} \times 1.25 \times 10^6 \times 0.5 \approx 1 \times 10^{-19} \, N \]

Answer: The average force on each electron due to the magnetic field is approximately \(1 \times 10^{-19} \, N\). Quick Tip: Remember that in uniform magnetic fields, loops of wire experience no net force but can experience significant torque if the plane of the loop is not parallel to the magnetic field lines.

Step 1: Understanding the graph behavior.

(i) Resistance \( R \) is independent of the frequency for a pure resistor, hence it remains constant regardless of frequency changes.

(ii) Capacitive reactance \( X_C \) decreases inversely with frequency, illustrating that as frequency increases, the impedance offered by a capacitor decreases.

(iii) Inductive reactance \( X_L \) increases linearly with frequency, showing that as frequency increases, the impedance offered by an inductor increases.


Step 2: Voltage drops in a series LCR circuit.

In a series LCR circuit at resonance, the inductive and capacitive reactances can cancel each other out, leaving only the resistive effect. However, individually, the voltage across the inductor \( V_L = I X_L \) and the capacitor \( V_C = I X_C \) can be quite high, especially at or near resonance conditions. At resonance, the impedance is minimal, maximizing the current, which can lead to voltage drops across the inductor or capacitor that exceed the applied source voltage, due to the energy stored in the magnetic field of the inductor or the electric field of the capacitor. Quick Tip: When analyzing circuits, especially at varying frequencies, consider how each component interacts with changes in frequency. At resonance, capacitive and inductive effects can amplify voltages beyond the source voltage, often leading to surprising circuit behavior.

Step 1: Properties of the nucleus.

The nucleus is extremely small compared to the atom and contains protons and neutrons (nucleons), which hold nearly all the atom's mass. It is positively charged due to the presence of protons.

Step 2: Explaining the high density of the nucleus.

The nucleus is very dense because it packs a significant amount of mass into a very small volume. While the electrons orbiting the nucleus occupy most of the atom's volume, they contribute very little to its mass.

Step 3: Uniformity of nuclear density.

The nuclear density \( \rho \) can be estimated using the formula: \[ \rho = \frac{mass}{volume} = \frac{A \times m_{nuc}}{\frac{4}{3} \pi R^3} \]
where \( A \) is the mass number (protons + neutrons), \( m_{nuc} \) is the mass of a nucleon, and \( R \approx r_0 A^{1/3} \) is the radius of the nucleus, with \( r_0 \) being a constant approximately \( 1.2 \times 10^{-15} \) m.
Substituting \( R \) and simplifying: \[ \rho = \frac{A \times m_{nuc}}{\frac{4}{3} \pi (r_0 A^{1/3})^3} = \frac{m_{nuc}}{\frac{4}{3} \pi r_0^3} \]
This shows that the density is independent of \( A \), indicating that the nuclear density is roughly constant across different nuclei. Quick Tip: Nuclear density being constant is a consequence of the nuclear force being short-ranged and equally effective across all nucleons within a nucleus, resulting in a similar compactness regardless of the size of the nucleus.

Step 1: Formula for energy levels in the Bohr model.
The energy \( E_n \) of an electron in the \( n \)-th orbit of hydrogen is given by: \[ E_n = -\frac{13.6 \, eV}{n^2} \]
where \( 13.6 \, eV \) is the ground state energy.

Step 2: Energy difference for a transition from \( n \) to \( n - 1 \). \[ \Delta E = E_{n-1} - E_n = -\frac{13.6 \, eV}{(n-1)^2} + \frac{13.6 \, eV}{n^2} \]
For large \( n \), approximate \( (n-1)^2 \) as \( n^2 \) (considering \( n \) large, the terms of higher order become negligible), the difference simplifies to: \[ \Delta E \approx \frac{13.6 \, eV}{n^3} \]

Step 3: Calculating the frequency of the emitted radiation.
Using the relationship \( \Delta E = h \nu \) where \( h \) is Planck's constant: \[ \nu = \frac{\Delta E}{h} \approx \frac{\frac{13.6 \, eV}{n^3}}{h} = \frac{\alpha}{n^3} \]
where \( \alpha = \frac{13.6 \, eV}{h} \) is a constant. Quick Tip: When solving problems involving the Bohr model, remember that approximations for large \( n \) values can simplify calculations significantly, revealing underlying patterns and relationships such as the \( 1/n^3 \) dependence here.

Step 1: Correlation between system size and wavelength.

The wavelength of the waves emitted by any system depends on the size of the system because the physical dimensions of the system determine the resonant frequencies at which it can effectively emit or absorb electromagnetic radiation.

Step 2: Justification for radio broadcast and astronomical observations.

(i) The ionosphere reflects shorter wavelengths (short-wave radio bands), facilitating long-distance communication without the need for satellite or cable transmission.

(ii) X-rays have very short wavelengths that are absorbed by the Earth's atmosphere, preventing them from reaching ground-based detectors. Thus, detecting them requires positioning the detectors in space, free from atmospheric interference. Quick Tip: Understanding the interaction between electromagnetic waves and the medium through which they travel or from which they originate is crucial in applications ranging from communications to astronomy.

The lens maker’s formula is given by: \[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
For a double-convex lens, both surfaces have the same radius of curvature \( R \), and: \[ \frac{1}{f} = (n - 1) \left( \frac{1}{R} - \frac{-1}{R} \right) \] \[ \frac{1}{f} = (n - 1) \cdot \frac{2}{R} \]
The power \( P \) of the lens is: \[ P = \frac{1}{f} = \frac{2(n - 1)}{R} \] Quick Tip: For a double-convex lens, use the lens maker’s formula with the radii of curvature appropriately signed. The total power is proportional to the refractive index and inversely proportional to the radius.

When a lens is cut into two equal parts perpendicular to its principal axis, the focal length of each part remains unchanged. Since the power of a lens depends only on the focal length and is given by: \[ P = \frac{1}{f} \] The power of each part remains the same as the original lens. Thus, the power of one part is \( P \).

Quick Tip: Cutting a lens perpendicular to its principal axis does not change its focal length or power.

When two lenses are kept in contact, the total power of the combination is the algebraic sum of their individual powers. Since both parts have power \( P \), the total power is: \[ P_{\text{total}} = P + P = 2P \]

Quick Tip: For lenses in contact, the combined power is the sum of their individual powers.