UP Board Class 12 Physics Question Paper 2024 (Code 347 FX) Available- Download Solution PDF with Answer Key

Step 1: The lens maker's formula is given as:
\[ \frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

Step 2: Here, \( R_1 \) and \( R_2 \) are the radii of curvature of the two lens surfaces. The sign convention is used to determine which radius corresponds to the first and second surfaces.

Step 3: Generally, the formula considers the second focal length in practical applications as it corresponds to the image formation side.
\[ \therefore The correct answer is Always second. \]
\[ \boxed{Always second} \] Quick Tip: Always consider the lens maker's formula with proper sign convention and determine the focal length based on whether the lens is convex or concave.

Step 1: An n-type semiconductor is created by doping a pure semiconductor with elements having more electrons.

Step 2: Despite having free electrons, the overall charge of the semiconductor remains neutral because the number of positive protons in the nuclei balances the charge.
\[ \therefore The correct answer is Neutral. \]
\[ \boxed{Neutral} \] Quick Tip: N-type semiconductors are neutral overall, even though they contain free electrons as charge carriers.

Step 1: Analyze the given nuclear reaction:
\[ _{7}^{15}N + _{0}^{1}n \rightarrow _{6}^{14}C + P \]

Step 2: By balancing atomic and mass numbers:
\[ 7 + 0 = 6 + X \quad \Rightarrow \quad X = 1 \quad (Proton) \]
\[ 15 + 1 = 14 + Y \quad \Rightarrow \quad Y = 2 \quad (Mass of proton) \]
\[ \therefore The correct answer is Proton. \]
\[ \boxed{Proton} \] Quick Tip: In nuclear reactions, balance both atomic and mass numbers to identify unknown particles.

Step 1: Equivalent resistance for resistors in series:
\[ R_1 = 5R \]

Step 2: Equivalent resistance for resistors in parallel:
\[ \frac{1}{R_2} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} + \frac{1}{R} + \frac{1}{R} \]
\[ R_2 = \frac{R}{5} \]

Step 3: Given \( R_1 = nR_2 \):
\[ 5R = n \cdot \frac{R}{5} \]
\[ n = 25 \]
\[ \therefore The correct answer is 25. \]
\[ \boxed{25} \] Quick Tip: For resistors in series, simply add their resistances; for parallel resistors, sum their reciprocals and take the inverse.

Step 1: Electric flux is given by the formula:
\[ \Phi_E = \mathbf{E} \cdot \mathbf{A} \]

Step 2: The unit of electric field \( E \) is volt per metre (\( V/m \)), and the unit of area \( A \) is metre squared (\( m^2 \)).
\[ Unit of flux = \left( \frac{volt}{metre} \right) \times metre^2 = volt \times metre \]
\[ \therefore The correct answer is volt × metre. \]
\[ \boxed{volt × metre} \] Quick Tip: Remember that electric flux is the product of electric field strength and surface area, leading to the unit \( volt \times metre \).

Step 1: An ammeter is designed to measure larger currents with lower resistance, whereas a milli-ammeter measures smaller currents and therefore requires higher sensitivity.

Step 2: Since a milli-ammeter is more sensitive, it has higher internal resistance compared to an ammeter.
\[ \therefore x_1 < x_2 \]
\[ \boxed{x_1 < x_2} \] Quick Tip: Milli-ammeters have higher resistance than regular ammeters because they are designed for lower current measurements.

Step 1: In the SI system, electric power is measured in watts (W), where:
\[ 1 watt = 1 joule/second \]

Step 2: In the MKS system, power is also expressed in terms of mechanical energy per unit time, which leads to the same value.
\[ \therefore The relation is 1 watt = 1 joule per second. \]
\[ \boxed{1 watt = 1 joule/second} \] Quick Tip: Power in the SI and MKS systems is numerically the same, with both units related to energy conversion per second.

Step 1: When current flows through a conductor, it generates a magnetic field around it according to Ampere's circuital law.

Step 2: The right-hand rule helps determine the direction of the magnetic field, which forms concentric circles around the conductor.
\[ \therefore The field produced is a magnetic field. \]
\[ \boxed{Magnetic field} \] Quick Tip: Use the right-hand thumb rule to determine the direction of the magnetic field around a current-carrying conductor.

Step 1: The speed of electromagnetic waves in free space is given by the formula:
\[ c = \frac{1}{\sqrt{\varepsilon_0 \mu_0}} \]

Step 2: The values of permittivity \( \varepsilon_0 \) and permeability \( \mu_0 \) in free space determine the propagation speed of light.
\[ \therefore The correct answer is the speed of light. \]
\[ \boxed{Speed of light} \] Quick Tip: The relation \( c = \frac{1}{\sqrt{\varepsilon_0 \mu_0}} \) is fundamental in electromagnetism and defines the speed of light in vacuum.

Step 1: A wavefront represents the locus of points having the same phase in a wave.

Step 2: An optical ray indicates the direction of propagation of the wave.

Step 3: The optical ray is always perpendicular to the wavefront at any given point.
\[ \therefore The correct answer is: An optical ray is perpendicular to the wavefront. \]
\[ \boxed{An optical ray is perpendicular to the wavefront} \] Quick Tip: To understand wave behavior, remember that optical rays are always normal to wavefronts, following Huygens' principle.

Step 1: In solids, atomic energy levels overlap due to the close proximity of atoms, forming continuous ranges called energy bands.

Step 2: The two main types of bands in a solid are the valence band (occupied by electrons) and the conduction band (where free electrons exist).

Step 3: The gap between these bands determines the electrical properties of the material.
\[ \therefore Energy bands define the allowed energy levels in a solid. \]
\[ \boxed{Energy bands are ranges of electron energy levels in solids.} \] Quick Tip: In semiconductors and insulators, the energy gap between valence and conduction bands determines their conductivity.

Step 1: The photoelectric effect is the phenomenon in which light incident on a metal surface ejects electrons.

Step 2: Albert Einstein explained this effect by proposing the quantization of light energy in the form of photons.

Step 3: This discovery confirmed the particle nature of light and won Einstein the Nobel Prize in Physics in 1921.
\[ \therefore The correct answer is: Albert Einstein. \]
\[ \boxed{Albert Einstein} \] Quick Tip: Remember that the photoelectric effect supports the quantum theory of light, leading to the concept of photons.

Step 1: The total energy supplied by the source is calculated using:
\[ Energy = CV^2 \]
\[ = (10 \times 10^{-6}) (2)^2 \]
\[ = 4 \times 10^{-5} \, J \]

Step 2: The energy stored in the capacitor is:
\[ Stored Energy = \frac{1}{2} CV^2 \]
\[ = \frac{1}{2} (10 \times 10^{-6}) (2)^2 \]
\[ = 2 \times 10^{-5} \, J \]
\[ \therefore The correct answers are 4 \times 10^{-5} \, J and 2 \times 10^{-5} \, J. \]
\[ \boxed{4 \times 10^{-5} \, J, \, 2 \times 10^{-5} \, J} \] Quick Tip: In capacitors, only half of the energy supplied by the source is stored; the rest is dissipated in the circuit.

Step 1: Using Ohm's Law, total resistance is:
\[ R_{total} = 0.2 + 2.8 = 3 \, \Omega \]

Step 2: The current obtained from the cell:
\[ I = \frac{E}{R_{total}} = \frac{1.5}{3} = 0.5 \, A \]

Step 3: Potential difference across the terminals:
\[ V = E - Ir = 1.5 - (0.5 \times 0.2) \]
\[ = 1.4 \, V \]
\[ \therefore The correct answers are 1.4 \, V and 0.5 \, A. \]
\[ \boxed{1.4 \, V, \, 0.5 \, A} \] Quick Tip: Always include internal resistance when calculating the total circuit resistance in practical cells.

Step 1: The expression \( \lambda = \frac{h}{p} \) represents the de Broglie wavelength, which describes the wave nature of particles.

Step 2: The expression \( p = \frac{h}{\lambda} \) represents the particle's momentum derived from its wavelength.
\[ \therefore The correct answers are de Broglie wavelength and momentum. \]
\[ \boxed{de Broglie wavelength, momentum} \] Quick Tip: De Broglie's equation shows the wave-particle duality, linking wavelength with momentum.

Step 1: The binding energy is calculated using Einstein’s equation:
\[ E = mc^2 \]
\[ = (2 \times 10^{-6}) (3 \times 10^8)^2 \]
\[ = 1.8 \times 10^{11} \, J \]

Step 2: Converting joules to electron-volts:
\[ 1 J = 6.24 \times 10^{18} eV \]
\[ E = 1.8 \times 10^{11} \times 6.24 \times 10^{18} \]
\[ = 1.12 \times 10^{30} \, eV \]
\[ \therefore The correct answers are 1.8 \times 10^{11} J and 1.12 \times 10^{30} eV. \]
\[ \boxed{1.8 \times 10^{11} J, \, 1.12 \times 10^{30} eV} \] Quick Tip: Use \( E = mc^2 \) to calculate binding energy, and remember the conversion factor \( 1 \, J = 6.24 \times 10^{18} eV \).

i. Internal resistance of cell

Step 1: The internal resistance of a cell depends on several factors:
\[ r \propto \frac{distance between electrodes}{area of electrodes} \]

- It increases if the distance between electrodes increases.

- It decreases if the cross-sectional area of electrodes increases.

- It depends on the electrolyte's concentration and nature.
\[ Thus, the internal resistance can be minimized by using a highly conductive electrolyte and optimizing electrode placement. \]
\[ \boxed{Factors: Distance, area, electrolyte nature, and temperature} \]


\hrule


ii. Resistance of conductor

Step 1: The resistance of a conductor is given by:
\[ R = \rho \frac{L}{A} \]

where:
- \( R \) is resistance,
- \( \rho \) is resistivity,
- \( L \) is length,
- \( A \) is cross-sectional area.

Step 2: Factors affecting resistance:

- It increases with an increase in length (\( L \)).

- It decreases with an increase in cross-sectional area (\( A \)).

- Different materials have different resistivities.

- Resistance increases with temperature due to an increase in resistivity.
\[ Therefore, resistance can be controlled by choosing appropriate materials and dimensions. \]
\[ \boxed{Factors: Length, cross-sectional area, material, and temperature} \] Quick Tip: To reduce conductor resistance, choose materials with low resistivity and increase the cross-sectional area.

i. The refractive index of the medium

Step 1: The given equation of the wave is:
\[ E = 100 \cos(6 \times 10^8 t + 4x) \]

Step 2: Comparing with the general wave equation \( E = E_0 \cos(\omega t + kx) \), we get:
\[ \omega = 6 \times 10^8, \quad k = 4 \]

Step 3: Using the relation \( v = \frac{\omega}{k} \):
\[ v = \frac{6 \times 10^8}{4} = 1.5 \times 10^8 \, m/s \]

Step 4: The refractive index is:
\[ n = \frac{c}{v} = \frac{3 \times 10^8}{1.5 \times 10^8} = 2 \]
\[ \boxed{n = 2} \]





ii. The velocity of electromagnetic wave in the medium

Step 1: From the previous step, we obtained the velocity as:
\[ v = \frac{\omega}{k} = 1.5 \times 10^8 \, m/s \]
\[ \boxed{v = 1.5 \times 10^8 \, m/s} \]





iii. The expression for magnetic field

Step 1: In an electromagnetic wave, the magnetic field \( B \) is related to the electric field \( E \) by:
\[ B = \frac{E}{c} \]

Step 2: Substituting the given electric field expression:
\[ B = \frac{100}{3 \times 10^8} \cos(6 \times 10^8 t + 4x) \]
\[ B = \frac{E_0}{c} \cos(6 \times 10^8 t + 4x) \]
\[ \boxed{B = \frac{E_0}{c} \cos(6 \times 10^8 t + 4x)} \] Quick Tip: For plane electromagnetic waves, the velocity can be calculated using \( v = \frac{\omega}{k} \) and the magnetic field using \( B = \frac{E}{c} \).

i. Paramagnetic Substances

Step 1: Paramagnetic materials have unpaired electrons in their atomic structure, which cause weak attraction to external magnetic fields.

Step 2: They have a small and positive magnetic susceptibility and align weakly with the applied magnetic field.

Examples: Aluminum, Platinum, Oxygen.
\[ \boxed{Weak attraction, e.g., Aluminum} \]

ii. Diamagnetic Substances

Step 1: Diamagnetic materials have all paired electrons, causing them to create an induced magnetic field in the opposite direction to the applied field.

Step 2: They have a small and negative magnetic susceptibility and are weakly repelled by the applied magnetic field.

Examples: Copper, Gold, Bismuth.
\[ \boxed{Weak repulsion, e.g., Copper} \]





iii. Ferromagnetic Substances

Step 1: Ferromagnetic materials have domains of atoms that align strongly with an external magnetic field, creating a strong attraction.

Step 2: They have high positive susceptibility and retain magnetization even after removing the external field.

Examples: Iron, Nickel, Cobalt.
\[ \boxed{Strong attraction, e.g., Iron} \] Quick Tip: Ferromagnetic materials retain magnetism, while paramagnetic and diamagnetic materials do not.

i. The angle of incidence

Step 1: At minimum deviation condition, the angle of incidence \( i \) is related to the prism angle \( A \) and minimum deviation angle \( \delta_m \) by:
\[ i = A + \frac{\delta_m}{2} \]

Step 2: Given \( A = \delta_m \), we substitute:
\[ i = A + \frac{A}{2} = \frac{3A}{2} \]
\[ \boxed{i = \frac{3A}{2}} \]




ii. The angle of refraction

Step 1: At minimum deviation, the refracted ray passes symmetrically through the prism, meaning:
\[ r = \frac{A}{2} \]
\[ \boxed{r = \frac{A}{2}} \]





iii. The refractive index of the material of the prism

Step 1: The refractive index \( n \) of the prism material is given by:
\[ n = \frac{\sin \left( \frac{A + \delta_m}{2} \right)}{\sin \left( \frac{A}{2} \right)} \]

Step 2: Since \( A = \delta_m \):
\[ n = \frac{\sin \left( \frac{A + A}{2} \right)}{\sin \left( \frac{A}{2} \right)} \]
\[ n = \frac{\sin A}{\sin \left( \frac{A}{2} \right)} \]
\[ \boxed{n = \frac{\sin A}{\sin \left( \frac{A}{2} \right)}} \] Quick Tip: In minimum deviation condition, the refracted ray is symmetric inside the prism, and refractive index can be calculated using the formula \( n = \frac{\sin \left( \frac{A + \delta_m}{2} \right)}{\sin \left( \frac{A}{2} \right)} \).

Step 1: Forward Biasing

In forward biasing, the external voltage applied reduces the potential barrier of the depletion region, allowing current to flow easily through the junction.

- The positive terminal of the battery is connected to the \( p \)-side, and the negative terminal to the \( n \)-side.
- As the barrier potential decreases, more charge carriers (electrons and holes) cross the junction.
- The current increases exponentially with applied voltage.

Example: Forward biasing is used in rectifiers and LEDs.
\[ Current in forward bias I_F \propto e^{\frac{V}{kT}} \]
\[ \boxed{Forward bias allows current flow.} \]



Step 2: Reverse Biasing

In reverse biasing, the external voltage increases the potential barrier, preventing the flow of majority carriers and widening the depletion region.

- The positive terminal is connected to the \( n \)-side, and the negative terminal to the \( p \)-side.
- Only a small leakage current flows due to minority carriers.
- Beyond a certain voltage, breakdown occurs, causing a large current to flow.

Example: Reverse biasing is used in Zener diodes and photodiodes.
\[ Reverse current I_R \approx 0 \]
\[ \boxed{Reverse bias restricts current flow.} \]



Step 3: Difference Between Forward and Reverse Current

Step 1: The time period of a charged particle in a magnetic field is given by the formula:
\[ T = \frac{2\pi m}{q B} \]

Step 2: Given that the velocity and magnetic field are the same for all particles, the ratio of time periods depends on the ratio \( \frac{m}{q} \).

Step 3: The mass and charge values of the particles:
\[ Proton: m_p, \quad q_p = e \]
\[ Deuteron: m_d = 2m_p, \quad q_d = e \]
\[ Alpha particle: m_{\alpha} = 4m_p, \quad q_{\alpha} = 2e \]

Step 4: Using the formula \( T \propto \frac{m}{q} \):
\[ \frac{T_{proton}}{T_{deuteron}} = \frac{m_p/e}{2m_p/e} = \frac{1}{2} \]
\[ \frac{T_{proton}}{T_{\alpha}} = \frac{m_p/e}{4m_p/2e} = \frac{1}{4} \]

Thus, the ratio of time periods is:
\[ 1:2:4 \]
\[ \boxed{1:2:4} \] Quick Tip: The time period of a charged particle in a magnetic field is proportional to the mass-to-charge ratio \( \frac{m}{q} \).

Given:

Resistance \( R = 400 \, \Omega \), Inductive voltage \( V_L = 120 \, V \), Capacitive voltage \( V_C = 120 \, V \), and source voltage:
\[ V = 200 \cos(50\pi t) \, volt \]

(i) Current in the circuit


Step 1: The impedance of the circuit:

Since the inductor and capacitor voltages are equal, they cancel each other out, leaving only the resistance.

Step 2: Applying Ohm's law:
\[ I = \frac{V}{R} \]
\[ = \frac{200}{400} \]
\[ = 0.5 \, A \]
\[ \boxed{I = 0.5 \, A} \]

(ii) The potential across resistance





Step 1: Using Ohm's law to find the potential across the resistor:
\[ V_R = IR \]
\[ = (0.5)(400) \]
\[ = 200 \, V \]
\[ \boxed{V_R = 200 \, V} \]



(iii) The phase difference between the potentials across inductor and capacitor




Step 1: In an LC circuit, the voltage across the inductor leads the current by \(90^\circ\), while the voltage across the capacitor lags the current by \(90^\circ\).

Step 2: Therefore, the phase difference between the inductor and capacitor voltages is:
\[ \theta = 90^\circ + 90^\circ = 180^\circ \]
\[ \boxed{180^\circ} \] Quick Tip: In an LC circuit, inductor voltage leads the current by \( 90^\circ \), and capacitor voltage lags by \( 90^\circ \), resulting in a \( 180^\circ \) phase difference.

Step 1: The focal length \( f \) of a spherical mirror is the distance between the pole and the focal point.

Step 2: Given that the distance between the focus and centre of curvature is \( 20 \, cm \), we know that:
\[ f = 20 \, cm \]
\[ \boxed{20 \, cm (Focal length)} \]



(ii) The distance of pole of mirror from its centre of curvature

\( 40 \, cm \) (Radius of curvature)


Step 1: The radius of curvature \( R \) is the distance between the pole and the centre of curvature.

Step 2: Since the focal length is given as \( f = 20 \, cm \), we use the relation:
\[ R = 2f \]

Step 3: Substituting the given value:
\[ R = 2 \times 20 = 40 \, cm \]
\[ \boxed{40 \, cm (Radius of curvature)} \] Quick Tip: In spherical mirrors, the focal length is half of the radius of curvature, given by \( f = \frac{R}{2} \). The radius of curvature is always twice the focal length in spherical mirrors.

Step 1: Threshold Wavelength and Threshold Frequency

Threshold Wavelength:

Threshold wavelength \( \lambda_0 \) is the longest wavelength of incident light that can cause photoemission. It is related to the work function \( \phi \) of the material by the equation:
\[ \lambda_0 = \frac{hc}{\phi} \]

If the wavelength \( \lambda \) of the incident light is greater than \( \lambda_0 \), photoemission does not occur.

Threshold Frequency:

Threshold frequency \( \nu_0 \) is the minimum frequency of the incident light required to eject electrons from the metal surface. It is given by:
\[ \nu_0 = \frac{\phi}{h} \]

If the frequency \( \nu \) of the incident light is less than \( \nu_0 \), no photoelectric emission occurs.
\[ \boxed{\lambda_0 = \frac{hc}{\phi}, \quad \nu_0 = \frac{\phi}{h}} \]




Step 2: Factors affecting Saturation Current and Cut-off Voltage

Saturation Current:

Saturation current is the maximum current obtained when all photoelectrons emitted from the surface reach the anode. It depends on:

- Intensity of Incident Light: As the intensity increases, more photons strike the surface, resulting in the emission of more electrons.
\[ Saturation current \propto Intensity of light \]

Cut-off Voltage:

Cut-off voltage (stopping potential) is the minimum negative potential required to stop the photoelectrons from reaching the anode. It depends on:

- Frequency of Incident Light: Higher frequency photons carry more energy, requiring a higher cut-off voltage to stop the electrons.

Using Einstein's photoelectric equation:
\[ eV_{cut-off} = h\nu - \phi \]
\[ \boxed{Saturation current \propto intensity, \quad V_{cut-off} \propto frequency.} \] Quick Tip: Remember: Saturation current depends on light intensity, whereas cut-off voltage is determined by light frequency and the material's work function.

(i) Half-Wave Rectifier (HWR)

Step 1: In a half-wave rectifier, a single diode is used to allow the flow of current only during the positive half of the AC input.

Step 2: During the negative half-cycle, the diode blocks the current flow.
\[ \boxed{HWR allows current in one half-cycle.} \]


(ii) Full-Wave Rectifier (FWR)

Step 1: A full-wave rectifier uses two diodes (center-tap transformer) or four diodes (bridge rectifier) to allow current in both positive and negative cycles.

Step 2: This results in a pulsating DC output.
\[ \boxed{FWR allows current in both half-cycles.} \] Quick Tip: A half-wave rectifier uses one diode, while a full-wave rectifier uses two or four diodes for continuous current flow.

(i) The nature of doping material




- In \( n \)-type semiconductors, donor atoms with 5 valence electrons provide extra electrons.
- In \( p \)-type semiconductors, acceptor atoms with 3 valence electrons create holes.
\[ \boxed{Pentavalent for n-type, Trivalent for p-type} \]



(ii) Majority and minority charge carriers




- In \( n \)-type semiconductors, electrons contribute to conduction.
- In \( p \)-type semiconductors, holes dominate conduction.
\[ \boxed{Electrons in n-type, Holes in p-type} \]



(iii) The relation between conductivity and mobility




Step 1: The conductivity \( \sigma \) is given by:
\[ \sigma = nq\mu \]

where:
- \( n \) is the number of charge carriers,
- \( q \) is the charge of the carriers,
- \( \mu \) is the mobility of the carriers.

Step 2: Since electrons have higher mobility than holes:
\[ \mu_n > \mu_p \Rightarrow \sigma_n > \sigma_p \]
\[ \boxed{\sigma_n > \sigma_p} \] Quick Tip: Conductivity increases with higher charge carrier concentration and mobility.

(i) Capacity

Step 1: The capacitance of a parallel plate capacitor with dielectric is given by:
\[ C' = kC \]
\[ \boxed{C' = kC} \]

(ii) Potential Difference





Step 1: Since the capacitor is disconnected, the charge remains constant. Using the formula:
\[ V' = \frac{q}{C'} = \frac{q}{kC} \]
\[ V' = \frac{V}{k} \]
\[ \boxed{V' = \frac{V}{k}} \]



(iii) Charge

\( q' = q \) (remains constant)



Step 1: As the capacitor is disconnected, the charge remains the same:
\[ q' = q \]
\[ \boxed{q' = q} \]



(iv) Electric Field

\( E' = \frac{E}{k} \)



Step 1: The electric field is given by:
\[ E = \frac{V}{d} \]

Step 2: Substituting the new potential difference:
\[ E' = \frac{V'}{d} = \frac{V}{kd} \]
\[ E' = \frac{E}{k} \]
\[ \boxed{E' = \frac{E}{k}} \]

(v) Electrostatic Potential Energy

\( U' = \frac{U}{k} \)



Step 1: Electrostatic potential energy is given by:
\[ U = \frac{1}{2} C V^2 \]

Step 2: Substituting new values:
\[ U' = \frac{1}{2} kC \left( \frac{V}{k} \right)^2 \]
\[ U' = \frac{U}{k} \]
\[ \boxed{U' = \frac{U}{k}} \] Quick Tip: When a dielectric is inserted, the capacitance increases by a factor of \( k \), while the potential and electric field decrease proportionally.

Step 1: A cell generates electrical energy from a chemical reaction within a single unit.

Step 2: A battery consists of multiple interconnected cells to provide higher output for more demanding applications.
\[ \boxed{Cell: Single unit, Battery: Multiple cells} \]

(i) Series Combination



In a series combination, the voltages add up while the current remains the same.


Step 1: Cells are connected end-to-end, with the positive terminal of one connected to the negative terminal of the next.

Step 2: The total voltage is:
\[ V_{total} = V_1 + V_2 + V_3 + \dots \]

Step 3: This configuration is useful when high voltage is required, such as in electronic circuits.
\[ \boxed{Used for high voltage applications.} \]


\hrule


(ii) Parallel Combination





Step 1: Cells are connected with all positive terminals together and all negative terminals together.

Step 2: The total current is:
\[ I_{total} = I_1 + I_2 + I_3 + \dots \]

Step 3: This configuration is useful when a higher current capacity is needed, such as in power backup systems.
\[ \boxed{Used for high current applications.} \]



(iii) Mixed Combination




Step 1: In mixed combination, cells are connected in both series and parallel configurations.

Step 2: This provides an increase in both voltage and current, depending on the number of cells.

Step 3: It is useful in applications where both high voltage and current are needed, such as in electric vehicles.
\[ \boxed{Used for balanced voltage and current applications.} \] Quick Tip: Series combination increases voltage, parallel combination increases current, and mixed combination balances both.

(i) Condition: \( f_1 > f_2 \)


Step 1: The effective focal length of the lens combination is given by:
\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \]

Step 2: Since \( f_1 > f_2 \) and \( f_2 \) is negative for a diverging lens:
\[ \frac{1}{F} = \frac{1}{f_1} - \frac{1}{|f_2|} \]

Step 3: If \( f_1 \) dominates, the net effect is converging.
\[ \boxed{Combination acts as a converging lens.} \]



(ii) Condition: \( f_1 < f_2 \)



Step 1: Using the same formula:
\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \]

Step 2: If \( f_1 < f_2 \), the negative focal length of the diverging lens dominates.
\[ \frac{1}{F} < 0 \]

Step 3: The net effect is a diverging lens.
\[ \boxed{Combination acts as a diverging lens.} \]



(iii) Condition: \( f_1 = f_2 \)


Step 1: Using the lens formula:
\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \]

Step 2: If \( f_1 \) and \( f_2 \) are equal in magnitude but opposite in sign:
\[ \frac{1}{F} = 0 \Rightarrow F = \infty \]

Step 3: The combination acts as a plane parallel glass.
\[ \boxed{Combination acts as a plane glass.} \] Quick Tip: The focal length of a lens combination is determined by the sum of reciprocals of individual focal lengths.

(i) Components

An astronomical telescope consists of an objective lens and an eyepiece with a longer focal length, while a compound microscope consists of an objective lens and an eyepiece with a shorter focal length.


Astronomical Telescope:

- Used for viewing distant objects.
- Components:
- Objective lens: Large aperture, long focal length.
- Eyepiece lens: Shorter focal length for magnification.

Compound Microscope:

- Used for viewing tiny nearby objects.
- Components:
- Objective lens: Short focal length, higher magnification.
- Eyepiece lens: Further magnifies the intermediate image.
\[ \boxed{Telescope: Long focal length, Microscope: Short focal length.} \]



(ii) Magnifying Power




The magnifying power of an astronomical telescope is the ratio of focal lengths, while for a compound microscope, it is the product of magnifications of the objective and eyepiece lenses.


Astronomical Telescope:
\[ Magnifying Power = \frac{f_{objective}}{f_{eyepiece}} \]

Compound Microscope:
\[ Magnifying Power = m_{objective} \times m_{eyepiece} \]

Conclusion:

- Telescope magnifies distant objects by collecting more light.
- Microscope magnifies small objects by successive magnification.
\[ \boxed{Telescope: Focal length ratio, Microscope: Successive magnification.} \]

Can one device be used as the other?



A telescope is designed for distant objects with long focal length, while a microscope is designed for close objects with short focal length.
The image formation principles and optical configurations are different, making them unsuitable for interchange.
\[ Thus, one cannot be used as the other. \]
\[ \boxed{Telescope and microscope serve distinct purposes.} \] Quick Tip: Remember: Telescopes are for distant objects with long focal lengths, while microscopes magnify tiny objects with short focal lengths.

Step 1: The binding energy per nucleon varies with mass number.

Step 2: The curve shows that lighter nuclei and very heavy nuclei have lower binding energy, while medium-sized nuclei (e.g., iron) have the highest binding energy, making them more stable.
\[ \boxed{Most stable nucleus: Fe (A = 56)} \]

(i) Nuclear Fission



Nuclear fission is the process where a heavy nucleus splits into smaller nuclei, releasing energy.


Step 1: Heavy elements such as uranium-235 split into lighter elements when bombarded with neutrons.

Step 2: This process releases energy due to the increase in binding energy per nucleon.
\[ \boxed{Heavy nucleus \rightarrow Two lighter nuclei + Energy} \]

(ii) Nuclear Fusion


Nuclear fusion is the process where lighter nuclei combine to form a heavier nucleus, releasing energy.


Step 1: Light elements such as hydrogen nuclei fuse together under high temperature and pressure to form helium.

Step 2: This results in a release of energy due to an increase in binding energy per nucleon.
\[ \boxed{Light nuclei \rightarrow Heavier nucleus + Energy} \]

(iii) Nuclear Energy


Nuclear energy is the energy released during fission or fusion due to changes in binding energy.


Step 1: The energy released during fission and fusion reactions is calculated using Einstein's equation:
\[ E = mc^2 \]

Step 2: This energy is harnessed in nuclear reactors and stars.
\[ \boxed{Energy is released due to mass defect.} \] Quick Tip: Nuclear fusion occurs in stars, while nuclear fission is used in power plants.

(i) Binding energy

Binding energy is the energy required to remove the electron from the given state.
\[ Binding energy = 3.4 eV \]
\[ \boxed{3.4 eV} \]

(ii) Ionization Potential

Ionization potential is the energy required to ionize the atom by removing an electron from the current state.
\[ Ionization potential = 3.4 eV \]
\[ \boxed{3.4 eV} \]

(iii) The Number of Energy States


The number of possible states for an electron in the nth energy level is given by:
\[ Number of states = n^2 \]

For \( n = 2 \):
\[ 2^2 = 4 \]
\[ \boxed{4} \]

(iv) The Angular Momentum


The angular momentum of an electron in the nth orbit is:
\[ L = n \hbar \]

For \( n = 2 \):
\[ L = 2\hbar \]
\[ \boxed{2\hbar} \]

(v) Kinetic Energy

In hydrogen atoms, kinetic energy in an energy state is numerically equal to the binding energy.
\[ Kinetic Energy = 3.4 eV \]
\[ \boxed{3.4 eV} \] Quick Tip: For hydrogen atoms, ionization energy equals the binding energy in a given state.

Step 1: Magnetic flux (\( \Phi \)) linked with a coil depends on:
\[ \Phi = B A \cos \theta \]

where:
- \( B \) = Magnetic field strength
- \( A \) = Area of the coil
- \( \theta \) = Angle between the field and the normal to the coil
\[ \boxed{Flux depends on B, A, and \theta.} \]

(i) Cause
The changing magnetic field causes an induced electromotive force (EMF).

Step 1: According to Faraday's law, EMF is induced when the magnetic flux through a coil changes with time.
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
\[ \boxed{Changing flux induces EMF.} \]

(ii) Magnitude

The magnitude of induced EMF is proportional to the rate of change of flux.

Step 1: Using Faraday's law, the magnitude of the induced EMF is:
\[ |\mathcal{E}| = \left| \frac{d\Phi}{dt} \right| \]

Step 2: Faster changes in flux result in higher EMF.
\[ \boxed{\mathcal{E} \propto \frac{d\Phi}{dt}} \]



(iii) Direction


The direction of induced EMF is given by Lenz's Law.

Step 1: According to Lenz's Law, the induced EMF opposes the change in magnetic flux.

Step 2: The negative sign in Faraday's law signifies this opposition.
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
\[ \boxed{Opposes the flux change.} \] Quick Tip: Remember, a change in flux induces an EMF that opposes the cause producing it.

A transformer works on the principle of electromagnetic induction.

Step 1: An alternating current in the primary coil creates a changing magnetic field.

Step 2: This varying magnetic flux induces an EMF in the secondary coil according to Faraday's law.
\[ V_s = \frac{N_s}{N_p} V_p \]
\[ \boxed{Transformer works on mutual induction.} \]



Conditions for an ideal transformer

(i) Leakage of magnetic flux


There should be no flux leakage; all flux should link both coils.

Step 1: Ideally, all magnetic flux produced by the primary coil should pass through the secondary coil.
\[ \boxed{Zero flux leakage for ideal transformer.} \]

(ii) Resistance of primary coil



The primary coil should have zero resistance.

Step 1: An ideal transformer assumes no resistive losses in the primary coil to prevent energy dissipation.
\[ \boxed{Zero resistance in the primary coil.} \]

(iii) Resistance of secondary coil




The secondary coil should also have zero resistance.

Step 1: In an ideal transformer, the secondary coil is assumed to be perfect, with no resistance.
\[ \boxed{Zero resistance in the secondary coil.} \]

(iv) Dissipation of power


There should be no power loss in an ideal transformer.


Step 1: In an ideal transformer, the entire power input is transferred to the output.

Step 2: Efficiency is 100%, meaning no loss due to resistance, flux leakage, or core losses.
\[ \boxed{No power dissipation in ideal transformer.} \] Quick Tip: An ideal transformer assumes perfect energy transfer with no losses.