CBSE Class 12 Physics Question Paper 2026 PDF Set-1 Code (55-4-1) with Solutions - Available

We are given two wires of same length \( l \) and same area of cross-section \( A \), but different resistivities \( \rho_1 \) and \( \rho_2 \). They are connected in series.


Step 1: Resistance of each wire

The resistance of a wire is given by: \[ R = \frac{\rho l}{A} \]

For the two wires: \[ R_1 = \frac{\rho_1 l}{A}, \quad R_2 = \frac{\rho_2 l}{A} \]


Step 2: Equivalent resistance in series

For series combination: \[ R_{eq} = R_1 + R_2 = \frac{\rho_1 l}{A} + \frac{\rho_2 l}{A} = \frac{(\rho_1 + \rho_2)l}{A} \]


Step 3: Equivalent resistivity

For the combination, total length becomes \( 2l \) and cross-section area remains \( A \).

If we consider the combination as a single wire of length \( 2l \), area \( A \), and equivalent resistivity \( \rho_{eq} \), then: \[ R_{eq} = \frac{\rho_{eq} (2l)}{A} \]

Equating with the expression from Step 2: \[ \frac{(\rho_1 + \rho_2)l}{A} = \frac{\rho_{eq} (2l)}{A} \]

Cancelling \( \frac{l}{A} \) from both sides: \[ \rho_1 + \rho_2 = 2\rho_{eq} \]
\[ \rho_{eq} = \frac{1}{2}(\rho_1 + \rho_2) \]


Conclusion: The equivalent resistivity is the average of the two resistivities.


Final Answer: (A) \( \frac{1}{2}(\rho_1 + \rho_2) \) Quick Tip: When wires of same length and cross-section but different resistivities are connected in series, equivalent resistivity = average of individual resistivities: \( \frac{\rho_1 + \rho_2}{2} \). For parallel combination, it would be harmonic mean.

When a charged particle enters a uniform magnetic field perpendicularly, it experiences a magnetic force that acts as centripetal force, causing it to move in a circular path.


Step 1: Formula for radius of circular path

The magnetic force provides the centripetal force: \[ qvB = \frac{mv^2}{r} \]

Solving for radius \( r \): \[ r = \frac{mv}{qB} \]

Given that both particles have the same velocity \( v \) and are in the same magnetic field \( B \), we have: \[ r \propto \frac{m}{q} \]


Step 2: Interpreting the figure

From the diagram (as described in the question), we can observe the relative sizes of the circular paths. Let's denote the radii as \( r_1 \) for \( m_1 \) and \( r_2 \) for \( m_2 \).

Based on the figure:
- If \( r_1 > r_2 \), then \( \frac{m_1}{q_1} > \frac{m_2}{q_2} \)
- If \( r_1 < r_2 \), then \( \frac{m_1}{q_1} < \frac{m_2}{q_2} \)


Step 3: Rearranging the inequality

From \( \frac{m_1}{q_1} > \frac{m_2}{q_2} \), we cross-multiply (all quantities positive): \[ m_1 q_2 > m_2 q_1 \]
\[ \frac{m_1}{m_2} > \frac{q_1}{q_2} \]


Conclusion: The ratio of masses is greater than the ratio of charges when the radius of the path for \( m_1 \) is larger than that for \( m_2 \).


Final Answer: (A) \( \frac{m_1}{m_2} > \frac{q_1}{q_2} \) Quick Tip: Radius of circular path in magnetic field: \( r = \frac{mv}{qB} \). For same \( v \) and \( B \), \( r \propto \frac{m}{q} \). Larger radius means larger \( m/q \) ratio.

When a charged particle moves in a circular orbit, it constitutes an electric current. The current is defined as the rate of flow of charge.


Step 1: Understanding current in terms of moving charge

Current \( I \) is given by: \[ I = \frac{q}{T} \]
where:

\( q \) = charge passing through a point
\( T \) = time taken for one complete revolution



Step 2: Applying to the electron

Given:

Charge of electron = \( e \) (magnitude)
Frequency of revolution = \( n \) revolutions per second
Time period \( T = \frac{1}{n} \) seconds


In each revolution, the electron passes any point on the circle once, carrying charge \( e \).


Step 3: Calculating current
\[ I = \frac{charge per revolution}{time per revolution} = \frac{e}{T} = \frac{e}{1/n} = e \times n \]
\[ I = ne \]


Conclusion: The equivalent current is \( ne \), which depends only on the charge and frequency, not on the radius \( r \). Quick Tip: Current due to orbiting charged particle = frequency × charge. For electron: \( I = ne \). Radius does not affect current magnitude—only frequency matters.

Step 1: Understanding the Concept:

According to Faraday's Law of Electromagnetic Induction, the induced EMF (\(\varepsilon\)) is equal to the negative rate of change of magnetic flux (\(d\Phi/dt\)).

Therefore, the dimensions of the rate of change of magnetic flux are identical to the dimensions of EMF (Potential difference).


Step 2: Key Formula or Approach:

EMF (\(V\)) is defined as work done per unit charge: \[ V = \frac{W}{q} \]
Dimensions of Work (\(W\)) = \([M L^2 T^{-2}]\)

Dimensions of Charge (\(q\)) = \([A T]\)


Step 3: Detailed Explanation:

Substitute the dimensions into the formula for \(V\): \[ [V] = \frac{[M L^2 T^{-2}]}{[A T]} \] \[ [V] = [M L^2 T^{-3} A^{-1}] \]
Since \(\frac{d\Phi}{dt} = \varepsilon\), the dimensions of the rate of change of magnetic flux are \([M L^2 T^{-3} A^{-1}]\).


Step 4: Final Answer:

The dimensions are \([M L^2 T^{-3} A^{-1}]\).
Quick Tip: Whenever asked for the dimensions of a "rate of change" of a quantity, check if there is a fundamental law (like Faraday's Law) that equates it to a simpler physical quantity.

Step 1: Understanding the Concept:

Moving conductor 1 in a magnetic field induces an EMF (Motional EMF).

This EMF creates a current in the closed loop formed by the rails and conductor 2.

The induced current in conductor 2, which is in a magnetic field, experiences a magnetic force (Lorentz force).


Step 2: Key Formula or Approach:

1. Direction of induced current in conductor 1: \(\vec{v} \times \vec{B}\).

2. Force on conductor 2: \(\vec{F} = I(\vec{l} \times \vec{B})\).

3. Lenz's Law: The system will try to oppose the change in flux.


Step 3: Detailed Explanation:

Let \(\hat{i}\) be right, \(\hat{j}\) be up, and \(\hat{k}\) be out of the page.

The magnetic field \(\vec{B}\) is into the page, so \(\vec{B} = -B \hat{k}\).

Conductor 1 moves right, so \(\vec{v} = v \hat{i}\).

Induced current direction in 1 is \(\vec{v} \times \vec{B} = (v \hat{i}) \times (-B \hat{k}) = vB (\hat{i} \times -\hat{k}) = vB (\hat{j})\).

So current flows "up" in conductor 1. In the closed loop, this means current flows "down" through conductor 2.

For conductor 2, the length vector \(\vec{l}\) is along \(-\hat{j}\).

Magnetic force on conductor 2: \(\vec{F} = I\vec{l} \times \vec{B} = I(-L\hat{j}) \times (-B\hat{k})\)
\[ \vec{F} = ILB (\hat{j} \times \hat{k}) = ILB \hat{i} \]
Alternatively, according to Lenz's law, to oppose the increase in flux caused by moving 1 to the right, the loop will try to expand or contract. Here, the repulsive force between opposing currents in the parallel wires 1 and 2 will push conductor 2 to the right.


Step 4: Final Answer:

The force on conductor 2 is along the \(+\hat{i}\) direction.
Quick Tip: Using Lenz's Law: The induced current always opposes the motion. If conductor 1 moves right, it feels a force left. By conservation of momentum or repulsion of anti-parallel currents, conductor 2 will feel a force to the right.

Step 1: Understanding the Concept:

For a purely sinusoidal alternating voltage, the average value over a complete cycle is zero because the positive half-cycle exactly cancels the negative half-cycle.

The effective value is the Root Mean Square (RMS) value.


Step 2: Key Formula or Approach:

1. \(V_{avg} (full cycle) = 0\)

2. \(V_{rms} = \frac{V_0}{\sqrt{2}}\)

Where \(V_0\) is the peak voltage.


Step 3: Detailed Explanation:

The given equation is \(v = 14 \sin(314t)\).

By comparison with \(v = V_0 \sin(\omega t)\), we find \(V_0 = 14\) V.

The average value over a full cycle is \(V_{avg} = 0\).

The effective (RMS) value is: \[ V_{rms} = \frac{14}{\sqrt{2}} \approx \frac{14}{1.414} \approx 9.9 \approx 10 V \]

Step 4: Final Answer:

The values are 0 V and 10 V respectively.
Quick Tip: Always remember: Average over a \textbf{full cycle} is 0. Average over a \textbf{half cycle} is \(2V_0/\pi \approx 0.637 V_0\). RMS is \(V_0/\sqrt{2} \approx 0.707 V_0\).

Step 1: Understanding the Concept:

For an electromagnetic wave, the ratio of the amplitude of the electric field (\(E_0\)) to the amplitude of the magnetic field (\(B_0\)) is equal to the speed of the wave in that medium (\(v\)).


Step 2: Key Formula or Approach:
\[ \frac{E_0}{B_0} = v = \frac{c}{n} \]
Where \(c\) is the speed of light in vacuum (\(3 \times 10^8\) m/s) and \(n\) is the refractive index of the medium.


Step 3: Detailed Explanation:

Given: \(n = 1.5\) and \(c = 3 \times 10^8\) m/s.

The speed of light in glass is: \[ v = \frac{3 \times 10^8}{1.5} = 2 \times 10^8 ms^{-1} \]
Thus, the ratio \(\frac{E_0}{B_0} = 2 \times 10^8 ms^{-1}\).


Step 4: Final Answer:

The ratio is \(2 \times 10^8 ms^{-1}\).
Quick Tip: In vacuum, the ratio is exactly \(c = 3 \times 10^8\) m/s. In any other medium, it will always be less than \(c\).

Step 1: Understanding the Concept:

The cut-off potential (stopping potential) depends on the maximum kinetic energy of the photoelectrons, which in turn depends on the energy of the incident photons.

Saturation current depends on the number of photoelectrons emitted per second, which is proportional to the intensity (number of photons) if frequency is above threshold.


Step 2: Key Formula or Approach:

Einstein's photoelectric equation: \[ eV_s = h\nu - \phi = \frac{hc}{\lambda} - \phi \]
Where \(V_s\) is the stopping potential, \(\lambda\) is the wavelength, and \(\phi\) is the work function.


Step 3: Detailed Explanation:

The wavelength is decreased from 600 nm to 400 nm.

As \(\lambda\) decreases, the energy of each photon \(E = \frac{hc}{\lambda}\) increases.

From the equation \(eV_s = \frac{hc}{\lambda} - \phi\), an increase in \(\frac{hc}{\lambda}\) results in an increase in \(V_s\), provided the intensity is kept constant.

The saturation current depends on intensity (energy per unit area per unit time). If intensity is constant while wavelength decreases, the number of photons per second actually decreases slightly, but certainly doesn't increase. Thus, (D) is wrong. The most prominent effect is the change in stopping potential.


Step 4: Final Answer:

The cut-off potential will increase.
Quick Tip: Remember: Frequency/Wavelength \(\rightarrow\) Stopping Potential. Intensity \(\rightarrow\) Saturation Current.

Step 1: Understanding the Concept:

Photoelectric emission occurs only if the energy of the incident photon (\(E\)) is greater than or equal to the work function (\(\phi\)) of the metal.


Step 2: Key Formula or Approach:

Energy of a photon: \[ E = \frac{hc}{\lambda} \]
In eV, using \(\lambda\) in nm: \[ E (eV) \approx \frac{1240}{\lambda (nm)} \]

Step 3: Detailed Explanation:

Incident wavelength \(\lambda = 331\) nm.

Energy of incident photons: \[ E = \frac{1240}{331} \approx 3.746 eV \]
Comparing this energy with the given work functions:

1. For Na (\(\phi = 1.92\) eV): \(E > \phi\), so emission occurs.

2. For K (\(\phi = 2.15\) eV): \(E > \phi\), so emission occurs.

3. For Ca (\(\phi = 3.20\) eV): \(E > \phi\), so emission occurs.

4. For Mo (\(\phi = 4.17\) eV): \(E < \phi\), so emission does NOT occur.


Step 4: Final Answer:

Na, K, and Ca show emission, while Mo does not. Thus, only Mo will not show photoelectric emission.
Quick Tip: Using \(1240 / \lambda\) is a fast way to get energy in eV. If the answer is close to the work function, use \(1242\) for better precision.

Step 1: Understanding the Concept:

The de Broglie wavelength is inversely proportional to the square root of the kinetic energy of the particle.


Step 2: Key Formula or Approach:
\[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} \]
Where \(K\) is the kinetic energy.


Step 3: Detailed Explanation:

Let initial wavelength be \(\lambda_1 = \frac{h}{\sqrt{2mK_1}}\).

The kinetic energy is increased to four times, so \(K_2 = 4K_1\).

The new wavelength \(\lambda_2\) is: \[ \lambda_2 = \frac{h}{\sqrt{2m(4K_1)}} = \frac{1}{\sqrt{4}} \left( \frac{h}{\sqrt{2mK_1}} \right) = \frac{\lambda_1}{2} \]
The change in wavelength is: \[ \Delta \lambda = \lambda_2 - \lambda_1 = \frac{\lambda_1}{2} - \lambda_1 = -\frac{\lambda_1}{2} \]
Percentage change: \[ \frac{\Delta \lambda}{\lambda_1} \times 100 = -50% \]
The negative sign indicates a decrease.


Step 4: Final Answer:

The de Broglie wavelength decreases by 50% of its initial value.
Quick Tip: If \(X \propto \frac{1}{\sqrt{Y}}\), and \(Y\) becomes \(n\) times, then \(X\) becomes \(\frac{1}{\sqrt{n}}\) times. Here \(\sqrt{4} = 2\), so wavelength becomes half.

Step 1: Understanding the Concept:

The spectral series of the hydrogen atom correspond to electronic transitions between different energy levels. Each series falls within a specific region of the electromagnetic spectrum.


Step 2: Detailed Explanation:

The regions for different hydrogen series are:

1. Lyman series (\(n_1 = 1\)): Ultraviolet (UV) region.

2. Balmer series (\(n_1 = 2\)): Visible region.

3. Paschen series (\(n_1 = 3\)): Near Infrared (IR) region.

4. Brackett series (\(n_1 = 4\)): Infrared (IR) region.

5. Pfund series (\(n_1 = 5\)): Far Infrared (IR) region.


Step 3: Final Answer:

The Paschen series lies in the infrared region.
Quick Tip: Mnemonics: \textbf{L}ike \textbf{B}oys \textbf{P}lay \textbf{B}all \textbf{P}roperly \(\rightarrow\) UV, Visible, IR, IR, IR. Only Balmer is visible.

Step 1: Understanding the Concept:

In reverse bias, the positive terminal of the battery is connected to the n-side and the negative terminal to the p-side. This widens the depletion region.


Step 2: Detailed Explanation:

The depletion region is devoid of mobile charge carriers (electrons and holes).

As a result, the resistance of the depletion region is extremely high compared to the p and n regions, which are doped semiconductors.

According to Ohm's law and potential divider logic, in a series arrangement, the largest potential drop occurs across the highest resistance.

Therefore, almost the entire applied external voltage drops across the depletion region.


Step 3: Final Answer:

The applied voltage drops mostly across the depletion region.
Quick Tip: The depletion region acts like an insulator in reverse bias. Most of the electric field and potential drop are concentrated in this narrow high-resistance zone.

Step 1: Understanding the Concept:

This question relates to Kirchhoff's Loop Rule (KVL), which is a statement of the conservation of energy in an electric circuit.


Step 2: Detailed Explanation:

Assertion (A) is True: Kirchhoff's loop law states that the algebraic sum of changes in potential around any closed loop is zero. This implies that the net work done by the electrostatic and non-electrostatic forces on a unit charge moving around a closed loop is zero.

Reason (R) is True: Electric potential is a state function. It depends only on the position. If a charge starts at a point with potential \(V\) and returns to the same point, the final potential is still \(V\). The change in potential \(\Delta V = V_{final} - V_{initial} = V - V = 0\).

Since potential difference is defined as work done per unit charge, \(\Delta V = 0\) directly explains why the work done around a closed loop is zero.


Step 3: Final Answer:

Both A and R are true and R is the correct explanation of A.
Quick Tip: KVL is the circuit equivalent of stating that the electrostatic field is conservative. Even with cells (non-conservative sources), the \textbf{net} energy change of a charge in a full cycle is zero.

Step 1: Understanding the Concept:

Ferromagnetic materials have domains that are aligned. Heating increases thermal agitation, which disrupts this order.


Step 2: Detailed Explanation:

Assertion (A) is True: At a specific temperature called the Curie temperature (\(T_C\)), the thermal energy becomes strong enough to overcome the exchange coupling between atomic magnets. Above \(T_C\), the material behaves as a paramagnet.

Reason (R) is True: The transition from ferromagnetism to paramagnetism at the Curie point is a phase transition where the spontaneous magnetization vanishes. In many contexts, this transition at \(T_C\) is considered "abrupt" because the long-range order disappears at that specific point.

Explanation: While both are true, the reason for the transition to paramagnetism is thermal agitation overcoming exchange forces, not simply the fact that the transition is abrupt. The "abruptness" describes the nature of the transition but doesn't explain "why" it becomes paramagnetic.


Step 3: Final Answer:

Both A and R are true, but R is not the correct explanation of A.
Quick Tip: Curie-Weiss Law describes the susceptibility above \(T_C\): \(\chi = \frac{C}{T - T_C}\). The change in properties at \(T_C\) marks the end of ferromagnetic domain structure.

Step 1: Understanding the Concept:

The focal length and power of a lens depend on both the geometry of the lens and the refractive indices of the lens material and the surrounding medium.


Step 2: Key Formula or Approach:

Lens Maker's Formula: \[ \frac{1}{f} = \left( \frac{n_{lens}}{n_{medium}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Power \(P = \frac{1}{f}\).


Step 3: Detailed Explanation:

Assertion (A) is False: When a glass lens (\(n \approx 1.5\)) is immersed in water (\(n \approx 1.33\)) instead of air (\(n=1\)), the relative refractive index \(\frac{n_{lens}}{n_{medium}}\) decreases. Consequently, the term \((\frac{n_l}{n_m} - 1)\) decreases, making \(\frac{1}{f}\) smaller. Thus, focal length increases and power decreases.

Reason (R) is False: As seen from the formula, focal length depends on the radii of curvature and the refractive indices of the lens and the medium. The word "only" makes the statement false.


Step 4: Final Answer:

Both Assertion and Reason are false.
Quick Tip: Immersing a lens in any medium denser than air always increases its focal length and decreases its power. If \(n_{med} = n_{lens}\), the lens becomes invisible and behaves like a plane sheet (power = 0).

Step 1: Understanding the Concept:

Conductivity in semiconductors depends on the concentration of charge carriers and their mobility.


Step 2: Key Formula or Approach:

Conductivity \(\sigma = e(n_e \mu_e + n_h \mu_h)\)

Where \(n\) is concentration and \(\mu\) is mobility.


Step 3: Detailed Explanation:

Assertion (A) is True: For similar doping levels and the same temperature, n-type semiconductors (where electrons are majority carriers) generally exhibit higher conductivity than p-type semiconductors (where holes are majority carriers).

Reason (R) is True: Mobility (\(\mu\)) is the drift velocity per unit electric field. Electrons move in the conduction band where they are relatively free. Holes move in the valence band by a process of bound electrons jumping into vacant sites, which is a slower process. Therefore, \(\mu_e > \mu_h\).

Explanation: Since \(\mu_e > \mu_h\), for the same carrier concentration, the conductivity \(\sigma \approx n e \mu\) will be higher for n-type than for p-type.


Step 4: Final Answer:

Both A and R are true and R is the correct explanation of A.
Quick Tip: Electrons are lighter and move through a "clearer" band (conduction band) compared to the "crowded" valence band where holes move. This makes \(\mu_e\) significantly higher than \(\mu_h\).

Step 1: Understanding the Concept:

The work done in rotating an electric dipole in a uniform electric field is equal to the change in its potential energy.

Stable equilibrium occurs when the dipole moment \(\vec{p}\) is parallel to the electric field \(\vec{E}\) (\(\theta = 0^\circ\)).

Unstable equilibrium occurs when \(\vec{p}\) is anti-parallel to \(\vec{E}\) (\(\theta = 180^\circ\)).


Step 2: Key Formula or Approach:

Work done (\(W\)) = \(\Delta U = U_2 - U_1\)

Potential energy of a dipole: \(U = -pE \cos\theta\)

Work done to rotate from \(\theta_1\) to \(\theta_2\): \[ W = pE(\cos\theta_1 - \cos\theta_2) \]
Dipole moment \(p = q \times 2l\)


Step 3: Detailed Explanation:

Given:

Charge \(q = 1 \muC = 10^{-6} C\)

Separation \(2l = 10 cm = 0.1 m\)

Electric field \(E = 100 N/C\)

Initial angle \(\theta_1 = 0^\circ\) (Stable)

Final angle \(\theta_2 = 180^\circ\) (Unstable)


First, calculate the dipole moment (\(p\)):
\[ p = q \times 2l = 10^{-6} C \times 0.1 m = 10^{-7} C\cdotm \]

Now, calculate the work done (\(W\)):
\[ W = pE(\cos 0^\circ - \cos 180^\circ) \] \[ W = pE(1 - (-1)) \] \[ W = 2pE \] \[ W = 2 \times 10^{-7} \times 100 \] \[ W = 2 \times 10^{-5} J \]

Step 4: Final Answer:

The amount of work done is \(2 \times 10^{-5} J\).
Quick Tip: Remember that work done to rotate a dipole through \(180^\circ\) from stable equilibrium is always \(2pE\). If rotated through \(90^\circ\), it is simply \(pE\).

Step 1: Understanding the Concept:

Huygens' Principle is a geometrical construction used to determine how a wavefront propagates through a medium.


Step 2: Detailed Explanation:

Huygens' Principle states:

1. Every point on a given wavefront (called the primary wavefront) acts as a source of secondary disturbance, sending out small spherical waves called wavelets.

2. These secondary wavelets spread out in all directions with the speed of light in that medium.

3. The new wavefront at any later time is the forward envelope (the common tangent in the forward direction) of these secondary wavelets.


Justification for the absence of the backwave:

Huygens argued that the amplitude of the secondary wavelets is not the same in all directions.

According to him, the amplitude of a secondary wavelet is proportional to \(\frac{1}{2}(1 + \cos \theta)\), where \(\theta\) is the angle between the direction of propagation and the normal to the wavelet.

1. In the forward direction, \(\theta = 0^\circ\), so \(\cos 0^\circ = 1\). The factor is \(\frac{1}{2}(1 + 1) = 1\) (Maximum).

2. In the backward direction, \(\theta = 180^\circ\), so \(\cos 180^\circ = -1\). The factor is \(\frac{1}{2}(1 - 1) = 0\) (Zero).

Hence, there is no backwave.


Step 3: Final Answer:

Huygens' principle describes wavefront propagation via secondary wavelets, and the backwave is absent because the intensity/amplitude factor \((1 + \cos \theta)\) vanishes at \(180^\circ\).
Quick Tip: While Huygens' \((1 + \cos \theta)\) factor was an assumption, it was later rigorously proven by Kirchhoff's integral theorem in wave optics.

Step 1: Understanding the Concept:

Single slit diffraction results in a broad central maximum surrounded by narrower and less intense secondary maxima and minima.


Step 2: Detailed Explanation:

(i) Intensity Variation Graph:

The graph of Intensity (\(I\)) versus angular position (\(\theta\)) or \(\sin \theta\) shows:

1. A central peak at \(\theta = 0\) with maximum intensity \(I_0\).

2. First minima occurring at \(\sin \theta = \pm \frac{\lambda}{a}\).

3. Secondary maxima occurring roughly at \(\sin \theta = \pm \frac{1.5\lambda}{a}, \pm \frac{2.5\lambda}{a}\), etc., with rapidly decreasing intensity.

The intensity of the central maximum is much higher than the secondary ones (roughly in the ratio \(1 : \frac{1}{22} : \frac{1}{61}\)).


(ii) Linear Width of Central Maximum:

The linear width of the central maximum (\(W\)) is defined as the distance between the first minima on either side of the center.

The angular position of the first minimum is \(\theta \approx \frac{\lambda}{a}\).

The linear distance from the center to the first minimum is \(y = D \tan \theta \approx D \frac{\lambda}{a}\).

Total linear width \(W = 2y = \frac{2D\lambda}{a}\).


From the formula \(W = \frac{2D\lambda}{a}\), we see that \(W \propto D\).

Therefore, if the separation between the slit and the screen (\(D\)) is decreased, the linear width of the central maximum will decrease.


Step 3: Final Answer:

(i) The intensity graph shows a central maximum with intensity fading as we move to higher orders.

(ii) The linear width decreases when \(D\) is decreased.
Quick Tip: Differentiate between \textbf{angular width} (\(2\lambda/a\)) and \textbf{linear width} (\(2D\lambda/a\)). Angular width depends only on the slit and light, whereas linear width depends on screen distance.

Step 1: Understanding the Concept:

Light from a source in a denser medium can only escape into a rarer medium if the angle of incidence is less than or equal to the critical angle (\(C\)).

This creates a circular "window" of light on the surface of the liquid.


Step 2: Key Formula or Approach:

1. Critical angle: \(\sin C = \frac{1}{\mu}\)

2. Radius of the circular area: \(R = h \tan C\)

3. Area: \(A = \pi R^2\)


Step 3: Detailed Explanation:

Given:

Depth \(h = 1 m\)

Refractive index \(\mu = \sqrt{2}\)


Calculate the critical angle \(C\):
\[ \sin C = \frac{1}{\sqrt{2}} \implies C = 45^\circ \]

Calculate the radius \(R\) of the circle:
\[ R = h \tan C = 1 m \times \tan 45^\circ = 1 m \times 1 = 1 m \]

Calculate the area \(A\):
\[ A = \pi R^2 = \pi \times (1)^2 = \pi m^2 \approx 3.14159 m^2 \]

Step 4: Final Answer:

The area of the surface through which light emerges is \(\pi m^2\).
Quick Tip: A useful direct formula for radius is \(R = \frac{h}{\sqrt{\mu^2 - 1}}\). This saves time if the angle isn't a standard value like \(45^\circ\).

Step 1: Understanding the Concept:

The Geiger-Marsden (\(\alpha\)-particle scattering) experiment was used to probe the internal structure of the atom.


Step 2: Detailed Explanation:

Graph Description:

The graph plots the number of scattered \(\alpha\)-particles (\(N\)) on the Y-axis (often on a log scale) against the scattering angle (\(\theta\)) on the X-axis.

The graph shows a very high count for small angles (\(\theta < 1^\circ\)) and drops exponentially as \(\theta\) increases, with very few particles scattered at angles near \(180^\circ\).


Conclusions inferred from the graph:

1. Most of the atom is empty space: Since the vast majority of \(\alpha\)-particles passed through the gold foil with little or no deflection (small \(\theta\)), it implies that most of the volume of the atom is empty.

2. The nucleus is small and positively charged: Only a tiny fraction of \(\alpha\)-particles (about 1 in 8000) were deflected by more than \(90^\circ\). This indicates that the positive charge and nearly all the mass of the atom are concentrated in an extremely small, dense region called the nucleus.


Step 3: Final Answer:

The graph shows a steep decrease in scattered particles as the angle increases. Conclusions: 1. Most of the atom is empty. 2. The positive charge is concentrated in a tiny nucleus.
Quick Tip: The number of particles scattered at angle \(\theta\) is proportional to \(1/\sin^4(\theta/2)\). This inverse fourth power dependence explains why the count drops so dramatically at larger angles.

Step 1: Understanding the Concept:

In an intrinsic semiconductor, hole concentration (\(n_h\)) equals electron concentration (\(n_e\)), both being equal to the intrinsic carrier concentration (\(n_i\)).

The Mass Action Law states that for any semiconductor (intrinsic or extrinsic) in thermal equilibrium, the product of electron and hole concentrations is constant at a given temperature.


Step 2: Key Formula or Approach:

Mass Action Law: \[ n_e \cdot n_h = n_i^2 \]

Step 3: Detailed Explanation:

Given:

Intrinsic carrier concentration \(n_i = 5 \times 10^8 m^{-3}\) (since intrinsic \(n_h = n_i\))

New electron concentration after doping \(n_e = 4 \times 10^{12} m^{-3}\)


Calculate the new hole concentration (\(n_h\)):
\[ n_h = \frac{n_i^2}{n_e} \] \[ n_h = \frac{(5 \times 10^8)^2}{4 \times 10^{12}} \] \[ n_h = \frac{25 \times 10^{16}}{4 \times 10^{12}} \] \[ n_h = 6.25 \times 10^4 m^{-3} \]

Identification of semiconductor type:

Comparing the concentrations:
\(n_e = 4 \times 10^{12} m^{-3}\)
\(n_h = 6.25 \times 10^4 m^{-3}\)

Since \(n_e > n_h\), electrons are the majority charge carriers. Therefore, the semiconductor is an n-type semiconductor.


Step 4: Final Answer:

The new hole concentration is \(6.25 \times 10^4 m^{-3}\) and the semiconductor formed is n-type.
Quick Tip: Doping with pentavalent impurities (Donor atoms) creates n-type semiconductors (\(n_e > n_h\)). Doping with trivalent impurities (Acceptor atoms) creates p-type semiconductors (\(n_h > n_e\)).

Step 1: Understanding the Concept:

Kirchhoff's rules are fundamental for circuit analysis:

1. Kirchhoff's Junction Rule (Current Law): The algebraic sum of currents at any junction is zero (\(\sum I = 0\)). This is based on the conservation of charge.

2. Kirchhoff's Loop Rule (Voltage Law): The algebraic sum of changes in potential around any closed loop is zero (\(\sum \Delta V = 0\)). This is based on the conservation of energy.


Step 2: Key Formula or Approach:

We use the loop rule for independent loops or nodal analysis (derived from Kirchhoff's laws).

Let the potential at junctions A, F, and E be 0 V (ground).

Let the potential at node B be \(V_B\) and at node C be \(V_C\).


Step 3: Detailed Explanation:

From the diagram, the branch AB has a 5V battery (positive left), a 1\(\Omega\) resistor, and a 3V battery (positive right).

Applying the Junction Rule at node B:
\[ \frac{V_B - (0 + 5 - 3)}{1} + \frac{V_B - V_C}{3} = 0 \] \[ V_B - 2 + \frac{V_B - V_C}{3} = 0 \implies 3V_B - 6 + V_B - V_C = 0 \implies 4V_B - V_C = 6 \quad ---(i) \]
Applying the Junction Rule at node C:

The branch FC has a 2V battery (positive left) and a 1\(\Omega\) resistor. The branch CD/ED has a 4\(\Omega\) resistor.
\[ \frac{V_C - V_B}{3} + \frac{V_C - (0 + 2)}{1} + \frac{V_C - 0}{4} = 0 \]
Multiply the entire equation by 12:
\[ 4(V_C - V_B) + 12(V_C - 2) + 3V_C = 0 \] \[ 4V_C - 4V_B + 12V_C - 24 + 3V_C = 0 \implies -4V_B + 19V_C = 24 \quad ---(ii) \]
Adding equations (i) and (ii):
\[ (4V_B - V_C) + (-4V_B + 19V_C) = 6 + 24 \] \[ 18V_C = 30 \implies V_C = \frac{30}{18} = \frac{5}{3} V \]
Current flowing through branch FC (\(I_{FC}\)) from F to C:
\[ I_{FC} = \frac{V_F + 2 - V_C}{1} = \frac{0 + 2 - 5/3}{1} = 2 - \frac{5}{3} = \frac{1}{3} A \]

Step 4: Final Answer:

The current flowing through branch FC is \(\frac{1}{3} A\) directed from F towards C.
Quick Tip: When applying Kirchhoff's Loop rule, be extremely careful with battery polarities. If you go from the negative terminal to the positive terminal, the potential change is \(+V\). If you move in the direction of current through a resistor, the potential change is \(-IR\).

Step 1: Understanding the Concept:

The magnetic field produced by a long straight wire at a distance \(r\) is \(B = \frac{\mu_0 I}{2\pi r}\). The direction is determined by the Right-Hand Thumb Rule.

The magnetic force per unit length between two parallel wires is \(F/L = \frac{\mu_0 I_1 I_2}{2\pi r}\). Parallel currents attract; anti-parallel currents repel.


Step 2: Key Formula or Approach:

Let the x-axis be along the wires and the y-axis be perpendicular to them. Conductor 1 is at \(y=d\), 2 at \(y=0\), and 3 at \(y=-d\).


Step 3: Detailed Explanation:

(i) Net Magnetic Field at Conductor 1:

Field due to conductor 2 (\(I\) along \(+\hat{i}\)) at distance \(d\):

Using the right-hand thumb rule, \(\vec{B}_2 = \frac{\mu_0 I}{2\pi d} \hat{k}\) (Out of page).

Field due to conductor 3 (\(3I\) along \(-\hat{i}\)) at distance \(2d\):

Using the right-hand thumb rule, \(\vec{B}_3 = \frac{\mu_0 (3I)}{2\pi (2d)} (-\hat{k}) = \frac{3\mu_0 I}{4\pi d} (-\hat{k})\) (Into page).

Net Field \(\vec{B}_{net} = \vec{B}_2 + \vec{B}_3 = \left( \frac{2\mu_0 I}{4\pi d} - \frac{3\mu_0 I}{4\pi d} \right) \hat{k} = -\frac{\mu_0 I}{4\pi d} \hat{k}\).

Magnitude is \(\frac{\mu_0 I}{4\pi d}\) and direction is into the page (\(-\hat{k}\)).


(ii) Net Magnetic Force per Unit Length on Conductor 1:

Force due to 2 (\(F_{12}/L\)): Currents are parallel (\(2I\) and \(I\)), so they attract.
\(\vec{F}_{12}/L = \frac{\mu_0 (2I)(I)}{2\pi d} (-\hat{j})\).

Force due to 3 (\(F_{13}/L\)): Currents are anti-parallel (\(2I\) and \(3I\)), so they repel.
\(\vec{F}_{13}/L = \frac{\mu_0 (2I)(3I)}{2\pi (2d)} (+\hat{j}) = \frac{3\mu_0 I^2}{2\pi d} (+\hat{j})\).

Net Force \(\vec{F}/L = \frac{\mu_0 I^2}{2\pi d} (-2\hat{j} + 3\hat{j}) = \frac{\mu_0 I^2}{2\pi d} \hat{j}\).


Step 4: Final Answer:

(i) Net field magnitude is \(\frac{\mu_0 I}{4\pi d}\) acting into the page.

(ii) Net force per unit length is \(\frac{\mu_0 I^2}{2\pi d}\) acting vertically upwards (\(+\hat{j}\)).
Quick Tip: Remember the mnemonic: "Like currents attract, unlike currents repel." This is the opposite of electrostatic charges.

Step 1: Understanding the Concept:

Faraday's Law states that the induced emf is the negative rate of change of magnetic flux: \(\varepsilon = -d\Phi/dt\).


Step 2: Key Formula or Approach:

Flux \(\Phi = \vec{B} \cdot \vec{A} = B A \cos \theta\). Assuming the loop is perpendicular to the field, \(\theta = 0\).

Effective (RMS) current \(I_{eff} = I_{peak} / \sqrt{2}\).


Step 3: Detailed Explanation:

(i) Expression for Induced emf:

Area of the loop \(A = l \times b\).

Magnetic Flux \(\Phi = B \times A = (B_0 \sin \omega t) \cdot (l b)\).

According to Faraday's law:
\[ \varepsilon = -\frac{d\Phi}{dt} = -\frac{d}{dt} (l b B_0 \sin \omega t) \] \[ \varepsilon = -l b B_0 \omega \cos \omega t \]

(ii) Effective value of Current:

Instantaneous current \(i = \frac{\varepsilon}{R} = -\frac{l b B_0 \omega}{R} \cos \omega t\).

The peak value of current is \(I_0 = \frac{l b B_0 \omega}{R}\).

The effective (RMS) value of current is:
\[ I_{eff} = \frac{I_0}{\sqrt{2}} = \frac{l b B_0 \omega}{\sqrt{2} R} \]

Step 4: Final Answer:

(i) Induced emf \(\varepsilon = -l b B_0 \omega \cos \omega t\).

(ii) Effective current \(I_{eff} = \frac{l b B_0 \omega}{\sqrt{2} R}\).
Quick Tip: Note that since the field varies with time, this is a case of transformer induction (stationary loop, varying field) rather than motional emf.

Step 1: Understanding the Concept:

Lenz's Law states that the induced current will flow in such a direction as to oppose the change in magnetic flux.


Step 2: Key Formula or Approach:

Motional emf \(\varepsilon = B l v\).

Time duration \(t = distance / velocity\).


Step 3: Detailed Explanation:

(a) Direction and Duration:

As the loop moves out to the right, the area inside the magnetic field decreases. Thus, the magnetic flux (directed downwards) decreases.

To oppose this decrease, the induced current must produce a magnetic field directed downwards (inside the loop). By the right-hand rule, the current must flow in the clockwise direction (M \(\rightarrow\) N \(\rightarrow\) O \(\rightarrow\) P).

The side of the square is \(a = 25 cm\). The current persists as long as the loop is exiting the field.

Time taken for the loop to completely exit:
\[ t = \frac{Length of side}{Velocity} = \frac{25 cm}{25 cm/s} = 1 s \]

(b) Graphs:

1. Magnetic Flux (\(\Phi\)) vs Time: Initially, flux is maximum (\(\Phi = B a^2\)). As it exits linearly at constant velocity, flux decreases linearly from \(t = 0\) to \(t = 1 s\) until it reaches zero.

2. Induced emf (\(|\varepsilon|\)) vs Time: Since \(d\Phi/dt\) is constant during the exit, the magnitude of induced emf is constant (\(|\varepsilon| = B a v\)) for the interval \(0 \le t \le 1 s\), and zero otherwise.


Step 4: Final Answer:

(a) Current is clockwise and persists for 1 second.

(b) The flux graph is a line with a negative slope, and the emf graph is a constant horizontal pulse.
Quick Tip: The induced emf only exists when the flux is changing. Once the loop is entirely out of the field, the flux is zero and constant, so the emf vanishes.

Step 1: Understanding the Concept:

Infrared (IR) waves are electromagnetic waves with wavelengths longer than visible light but shorter than microwaves.


Step 2: Detailed Explanation:

Definition: Infrared waves have wavelengths typically between 700 nm and 1 mm. They are produced by hot bodies and molecules.

Why 'Heat Waves'?: Infrared waves are referred to as 'heat waves' because they are readily absorbed by water molecules and other polar molecules present in most materials. This absorption increases the thermal motion (internal energy) of the molecules, which in turn increases the temperature of the material.

Uses:

1. TV Remote Controls: IR LEDs are used to transmit encoded signals to electronic devices.

2. Physical Therapy: IR lamps are used to treat muscular strains and provide warmth to tissues.

3. Night Vision: IR sensors are used in goggles and cameras to detect heat signatures in the dark.


Step 3: Final Answer:

Infrared waves are thermal EM waves. They are called heat waves because they increase molecular motion upon absorption. Uses include remote controls and physical therapy.
Quick Tip: The greenhouse effect is primarily caused by atmospheric gases (like \(CO_2\)) trapping outgoing infrared radiation from the Earth's surface.

Step 1: Understanding the Concept:
The de Broglie relation proposed that matter, like radiation, exhibits both particle-like and wave-like properties.


Step 2: Detailed Explanation:

The de Broglie relation is given by:
\[ \lambda = \frac{h}{p} = \frac{h}{mv} \]
Here:

1. \(\lambda\) (wavelength) is a characteristic property of waves (e.g., interference, diffraction).

2. \(p\) (momentum) is a characteristic property of particles (mass and velocity).

3. \(h\) (Planck's constant) acts as the bridge connecting these two aspects.

The relation shows that a moving particle of momentum \(p\) has an associated wave of wavelength \(\lambda\). This directly links the wave and particle nature of matter in a single equation.


Step 3: Final Answer:

The dual nature is evident because the equation relates a wave parameter (\(\lambda\)) to a particle parameter (\(p\)).
Quick Tip: Heisenberg's Uncertainty Principle is a direct consequence of this wave-particle duality. Because matter is a "wave packet," we cannot pinpoint its position and momentum simultaneously.

Step 1: Understanding the Concept:

The incident photon transfers its energy to the electron. The kinetic energy of the emitted photoelectron determines its de Broglie wavelength.


Step 2: Key Formula or Approach:

1. Photon energy: \(E = \frac{hc}{\lambda}\)

2. Einstein's photoelectric equation: \(K_{max} = E - \phi\)

3. de Broglie wavelength: \(\lambda_e = \frac{h}{\sqrt{2mK}}\)


Step 3: Detailed Explanation:

Since the work function \(\phi\) is negligible (\(\phi \approx 0\)):

The maximum kinetic energy of the emitted electron is equal to the energy of the incident photon:
\[ K = \frac{hc}{\lambda} \]
Now, substituting this value of kinetic energy into the de Broglie wavelength formula:
\[ \lambda_e = \frac{h}{\sqrt{2m \left( \frac{hc}{\lambda} \right)}} \] \[ \lambda_e = \frac{h \cdot \sqrt{\lambda}}{\sqrt{2 m h c}} \] \[ \lambda_e = \sqrt{\frac{h^2 \lambda}{2 m h c}} = \sqrt{\frac{h \lambda}{2 m c}} \]

Step 4: Final Answer:

The de Broglie wavelength of the emitted electrons is \(\sqrt{\frac{h \lambda}{2 m c}}\).
Quick Tip: Notice that \(\lambda_e\) depends on the square root of the incident photon's wavelength \(\lambda\).

Step 1: Understanding the Concept:

In a nuclear reaction, mass number (A) and atomic number (Z) are conserved. The energy released (Q-value) is due to the mass defect (\(\Delta m\)) converted into energy using \(E = \Delta m \cdot c^2\).


Step 2: Key Formula or Approach:

1. Conservation of A: \(\sum A_{reactants} = \sum A_{products}\).

2. Mass defect \(\Delta m = \sum m_{reactants} - \sum m_{products}\).

3. \(Q = \Delta m (in u) \times 931.5 MeV/u\).


Step 3: Detailed Explanation:

(a) Value of A:

Conservation of mass number: \(2 + 2 = A + 1 \implies A = 3\).

Conservation of atomic number: \(1 + 1 = Z + 0 \implies Z = 2\).

Thus, the product \(X\) is Helium-3 (\({}^3_2He\)).


(b) Energy Released:

Total mass of reactants: \(M_R = 2 \times 2.014102 = 4.028204 u\).

Total mass of products: \(M_P = 3.016049 + 1.008665 = 4.024714 u\).

Mass defect \(\Delta m = M_R - M_P\):
\[ \Delta m = 4.028204 - 4.024714 = 0.00349 u \]
Energy released \(Q\):
\[ Q = 0.00349 \times 931.5 \approx 3.2509 MeV \]

Step 4: Final Answer:

(a) \(A = 3\).

(b) The energy released is approximately \(3.25 MeV\).
Quick Tip: This is a fusion reaction (Deuterium-Deuterium fusion). Always double-check mass calculations to the 5th or 6th decimal place as nuclear energy changes are small mass fractions.

Step 1: Understanding the Concept:

A full-wave rectifier is a device that converts both halves of an AC cycle into a pulsating DC output.


Step 2: Detailed Explanation:

Circuit Diagram: The circuit consists of a center-tapped transformer, two p-n junction diodes (\(D_1\) and \(D_2\)), and a load resistor (\(R_L\)). The input AC is applied to the primary of the transformer. The cathodes of both diodes are connected together at one end of the load resistor.


Working:

1. During Positive Half Cycle: The center-tap is at a neutral potential. End A becomes positive and end B becomes negative. Diode \(D_1\) is forward-biased and conducts current, while \(D_2\) is reverse-biased and does not conduct. Current flows through \(R_L\) from top to bottom.

2. During Negative Half Cycle: End A becomes negative and end B becomes positive. Diode \(D_1\) is now reverse-biased and \(D_2\) is forward-biased. \(D_2\) conducts, and current again flows through \(R_L\) in the same direction (top to bottom).


Input and Output Waveforms:

1. Input: A continuous sinusoidal wave.

2. Output: A series of positive pulses. Unlike the half-wave rectifier, there are no gaps between the pulses, as power is delivered during both halves of the input cycle.


Step 3: Final Answer:

A full-wave rectifier uses two diodes and a center-tapped transformer to ensure current flows through the load in the same direction for both half-cycles of AC.
Quick Tip: The output frequency of a full-wave rectifier is twice the input AC frequency (\(f_{out} = 2f_{in}\)). This makes it easier to filter out ripples compared to a half-wave rectifier.

Step 1: Understanding the Concept:

The relationship between Voltage (\(V\)) and Current (\(I\)) for a conductor is governed by Ohm's Law (\(V = IR\)).

Manganin is an alloy specifically used for making standard resistors because it has a very high resistivity and an extremely low temperature coefficient of resistance.


Step 2: Key Formula or Approach:

According to Ohm's Law:
\[ V = IR \implies I = \left(\frac{1}{R}\right)V \]
This is the equation of a straight line (\(y = mx\)) passing through the origin.


Step 3: Detailed Explanation:

Since Manganin has a negligible temperature coefficient of resistance, its resistance \(R\) remains constant even if the circuit is not switched off (i.e., even if there is slight heating due to Joule effect).

Because \(R\) is constant, the ratio \(\frac{V}{I}\) remains constant.

A constant ratio results in a linear graph (straight line) passing through the origin in the \(V\)-\(I\) plane.


Step 4: Final Answer:

The graph will be a straight line passing through the origin.
Quick Tip: Manganin and Constantan are preferred for standard resistances because their resistance doesn't change significantly with temperature, unlike pure metals which would show a curved graph at high currents.

Step 1: Understanding the Concept:

This question pertains to the propagation of errors in physical measurements involving division.


Step 2: Key Formula or Approach:

The unknown resistance is calculated as:
\[ X = \frac{V}{I} \]
In the measurement of a quantity involving division (\(Z = A/B\)), the maximum relative error is the sum of the relative errors of the individual quantities.
\[ \frac{\Delta X}{X} = \frac{\Delta V}{V} + \frac{\Delta I}{I} \]

Step 3: Detailed Explanation:

The calculated value of \(X\) depends on both the voltage \(V\) measured by the voltmeter and the current \(I\) measured by the ammeter.

Any inaccuracy (systematic or random error) in either instrument will contribute to the total error in \(X\).

Based on the rules of error propagation, when quantities are divided, their fractional or percentage errors add up to give the total error in the result.


Step 4: Final Answer:

The total error is the sum of the errors in both measurements.
Quick Tip: Remember: For \(Z = AB\) or \(Z = A/B\), relative errors add: \(\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}\). For \(Z = A \pm B\), absolute errors add: \(\Delta Z = \Delta A + \Delta B\).

Step 1: Understanding the Concept:

A rheostat is a variable resistor. Moving its slider changes the total resistance of the circuit.


Step 2: Key Formula or Approach:

Total Resistance \(R_{total} = X + R_{rheostat}\).

Current \(I = \frac{E}{R_{total}}\).

Voltmeter reading \(V = IX\).


Step 3: Detailed Explanation:

In the given circuit, moving the slider towards P decreases the effective length of the rheostat wire included in the circuit.

Since Resistance \(R \propto Length\), the rheostat resistance \(R_{rheostat}\) decreases.

As \(R_{total}\) decreases, the current \(I\) in the circuit increases (\(I = \frac{E}{R_{total}}\)). Thus, the ammeter reading increases.

The voltmeter is connected across the fixed resistor \(X\). The potential drop across it is \(V = IX\).

Since \(I\) has increased and \(X\) is constant, the potential drop \(V\) also increases. Thus, the voltmeter reading increases.


Step 4: Final Answer:

Readings in both voltmeter and ammeter increase.
Quick Tip: Decreasing resistance in a series circuit increases current. For a fixed load resistor, increased current always results in a higher voltage drop across it.

Step 1: Understanding the Concept:

Drift velocity (\(v_d\)) is related to the current flowing through a conductor and its cross-sectional area.


Step 2: Key Formula or Approach:
\[ I = n e A v_d \implies v_d = \frac{I}{n e A} \]
Where \(A = \pi R^2\).


Step 3: Detailed Explanation:

The three parts of the wire are connected in series, so the same current \(I\) flows through each part.

The material is the same, so the number density of free electrons (\(n\)) is the same.

Thus, \(v_d \propto \frac{1}{A} \propto \frac{1}{R^2}\).

Let's calculate the ratios:

Part AC: \(R_1 = r \implies v_1 \propto \frac{1}{r^2}\).

Part CD: \(R_2 = \frac{r}{3} \implies v_2 \propto \frac{1}{(r/3)^2} = \frac{9}{r^2}\).

Part DB: \(R_3 = \frac{r}{2} \implies v_3 \propto \frac{1}{(r/2)^2} = \frac{4}{r^2}\).

Comparing the values: \(v_2 = 9v_1\) and \(v_3 = 4v_1\).

Therefore, \(v_2 > v_3 > v_1\).


Step 4: Final Answer:

The ranking of drift velocities is \(v_2 > v_3 > v_1\).
Quick Tip: In a series circuit, current is constant. Drift velocity is highest where the wire is thinnest (smallest cross-sectional area).

Step 1: Understanding the Concept:

The electric field (\(E\)) inside a current-carrying conductor is related to the current density and resistivity.


Step 2: Key Formula or Approach:
\[ E = \rho J = \rho \left(\frac{I}{A}\right) \]
Where \(J\) is current density and \(A = \pi R^2\).


Step 3: Detailed Explanation:

Since the wires are in series, the current \(I\) is constant throughout.

The material is the same, so the resistivity \(\rho\) is constant.

Thus, \(E \propto \frac{1}{A} \propto \frac{1}{R^2}\).

For part AC: \(E_1 \propto \frac{1}{r^2}\).

For part CD: \(E_2 \propto \frac{1}{(r/3)^2} = \frac{9}{r^2}\).

For part DB: \(E_3 \propto \frac{1}{(r/2)^2} = \frac{4}{r^2}\).

Comparing the magnitudes: \(E_2 > E_3 > E_1\).


Step 4: Final Answer:

The ranking of electric field magnitudes is \(E_2 > E_3 > E_1\).
Quick Tip: Similar to drift velocity, the electric field in a series wire is inversely proportional to the area. Thinner sections require a stronger electric field to push the same current through.

Step 1: Understanding the Concept:

In an astronomical telescope, the objective lens collects light from a distant source to form an intermediate image. The eyepiece then acts as a simple microscope to magnify this intermediate image.


Step 2: Detailed Explanation:

1. Objective Lens: Light from an object at infinity enters the large objective lens. It converges the light to form a diminished, inverted, and real image at its second focal point.

2. Eyepiece Lens: This intermediate real image is positioned within the focal length of the eyepiece. The eyepiece then forms a highly magnified, inverted (with respect to the original object), and virtual image.


Step 3: Final Answer:

The objective forms a real image, and the eyepiece forms a virtual image.
Quick Tip: Remember: The objective in a telescope always forms a real image. The eyepiece always acts as a magnifying glass, which by definition produces a virtual image.

Step 1: Understanding the Concept:

Magnification refers to how much larger the final image appears compared to the object when seen with the naked eye.


Step 2: Key Formula or Approach:

Magnifying power (\(M\)) for normal adjustment is:
\[ M = \frac{f_o}{f_e} \]

Step 3: Detailed Explanation:

- Magnification clearly depends on the focal lengths of the objective (\(f_o\)) and the eyepiece (\(f_e\)).

- It also depends on the color of light because the refractive index of glass (and thus the focal length) varies with wavelength (Chromatic Aberration).

- Aperture (diameter) of the lenses determines the light-gathering power and the resolving power (how sharp or bright the image is), but it does not enter the mathematical formula for magnification.


Step 4: Final Answer:

Magnification is independent of the apertures of the lenses.
Quick Tip: Large apertures are used in telescopes not for more magnification, but to collect more light from faint stars and to increase resolution.

Step 1: Understanding the Concept:

The construction and performance of a telescope depend on the relative sizes of focal lengths and the separation of the lenses (tube length).


Step 2: Key Formula or Approach:

Magnifying power \(M = f_o/f_e\).

Tube length (separation) for normal adjustment: \(L = f_o + f_e\).


Step 3: Detailed Explanation:

(A) Correct: In astronomical telescopes, \(f_o\) must be much larger than \(f_e\) to obtain high magnification.

(B) Correct: Since \(M = f_o/f_e\), increasing \(f_o\) directly increases \(M\).

(C) Incorrect: In \textit{normal adjustment (final image at infinity), the distance is exactly \(f_o + f_e\). If the final image is formed at the least distance of distinct vision (\(D\)), the eyepiece is moved closer to the objective, making the separation \(L = f_o + u_e\), where \(u_e < f_e\). Thus, \(L < f_o + f_e\). The separation is never \textit{more than \(f_o + f_e\) for standard operation.

(D) Correct: Since \(f_e\) is in the denominator, increasing it will decrease \(M\).


Step 4: Final Answer:

Statement (C) is incorrect.
Quick Tip: For a telescope, the length of the tube in normal adjustment (\(f_o + f_e\)) is the maximum possible separation between the lenses.

Step 1: Understanding the Concept:

"Normal adjustment" refers to the case where the final image is formed at infinity. This happens when the intermediate image formed by the objective coincides with the focal point of the eyepiece.


Step 2: Key Formula or Approach:

Distance between lenses (Tube length) \(L = f_o + f_e\).


Step 3: Detailed Explanation:

Given:

Focal length of objective \(f_o = 80 cm\).

Focal length of eyepiece \(f_e = 4 cm\).

In normal adjustment, the lenses are separated by the sum of their focal lengths:
\[ L = f_o + f_e = 80 cm + 4 cm = 84 cm \]

Step 4: Final Answer:

The separation distance is 84 cm.
Quick Tip: Always look for the keyword "normal adjustment" as it tells you both the final image location (infinity) and the tube length formula (\(f_o + f_e\)).

Step 1: Understanding the Concept:

Magnifying power (or angular magnification) of a telescope is the ratio of the focal length of the objective to the focal length of the eyepiece.


Step 2: Key Formula or Approach:
\[ M = \frac{f_o}{f_e} \]

Step 3: Detailed Explanation:

From the previous part (iv)(a), we have:

Focal length of objective \(f_o = 80 cm\).

Focal length of eyepiece \(f_e = 4 cm\).

Calculating magnifying power:
\[ M = \frac{80}{4} = 20 \]
This means the distant object will appear 20 times larger in angular size when viewed through the telescope.


Step 4: Final Answer:

The magnifying power is 20.
Quick Tip: Magnifying power is a dimensionless quantity. To increase magnification, either increase the objective's focal length or use an eyepiece with a shorter focal length.

Step 1: Understanding the Concept:

Refractive index is a dimensionless number that describes how light propagates through a medium. It characterizes the optical density of the medium.


Step 2: Key Formula or Approach:

The absolute refractive index (\(n\)) of a medium is defined as the ratio of the speed of light in vacuum (\(c\)) to the speed of light in that medium (\(v\)).
\[ n = \frac{c}{v} \]

Step 3: Detailed Explanation:

1. Definition: The absolute refractive index of a medium is defined as the ratio of the speed of light in vacuum (approximately \(3 \times 10^{8}\) m/s) to the speed of light in the given medium.

2. Since light travels fastest in vacuum, the value of the absolute refractive index is always greater than or equal to 1.

3. It indicates the factor by which the speed of light decreases when it enters the medium from vacuum.


Step 4: Final Answer:

Refractive index \(n = c/v\), where \(c\) is the speed of light in vacuum and \(v\) is the speed of light in the medium.
Quick Tip: Refractive index is a unitless and dimensionless quantity. A higher refractive index means light travels slower in that medium.

Step 1: Understanding the Concept:

At the condition of minimum deviation, the ray passes symmetrically through the prism. The angle of incidence equals the angle of emergence (\(i = e\)), and the internal refractive angles are equal (\(r_1 = r_2\)).


Step 2: Key Formula or Approach:

1. Prism relation: \(A = r_1 + r_2\)

2. Deviation formula: \(\delta = i + e - A\)

3. Snell's Law: \(\mu = \frac{\sin i}{\sin r}\)


Step 3: Detailed Explanation:

At minimum deviation (\(\delta = \delta_m\)):

1. \(i = e\) and \(r_1 = r_2 = r\).

2. From \(A = r_1 + r_2 \implies A = 2r \implies r = \frac{A}{2}\).

3. From \(\delta_m = i + e - A \implies \delta_m = 2i - A \implies i = \frac{A + \delta_m}{2}\).

4. Applying Snell's Law at the first surface: \[ \mu = \frac{\sin i}{\sin r_1} = \frac{\sin i}{\sin r} \]
Substituting the values of \(i\) and \(r\): \[ \mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]

Step 4: Final Answer:

The refractive index of the prism is \(\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}\).
Quick Tip: For a small-angled prism, the formula simplifies to \(\delta_m = (\mu - 1)A\).

Step 1: Understanding the Concept:

When a ray is incident normally (\(i = 0^\circ\)), it passes undeviated into the medium. It then hits another surface where its behavior depends on whether the angle of incidence exceeds the critical angle (\(i_c\)).


Step 2: Key Formula or Approach:

1. Critical angle \(i_c = \sin^{-1}(1/\mu)\).

2. If \(i > i_c\), Total Internal Reflection (TIR) occurs.


Step 3: Detailed Explanation:

1. Incident on BC: The ray QP is incident normally on BC, so it enters the prism without bending and travels straight.

2. Incident on face AC/AB: Assuming the prism is equilateral (\(A=60^\circ\)), the ray hitting the side AC will have an angle of incidence \(i = 60^\circ\).

3. Calculation of Critical Angle: \[ \mu = 1.5 \implies \sin i_c = \frac{1}{1.5} = \frac{2}{3} \approx 0.66 \] \[ i_c \approx 41.8^\circ \]
4. Comparison: Since the angle of incidence \(i = 60^\circ\) is greater than the critical angle \(i_c \approx 41.8^\circ\), the ray will undergo Total Internal Reflection (TIR) at the face. It will reflect back into the prism according to the laws of reflection.


Step 4: Final Answer:

The ray enters undeviated, hits the slanted face at \(60^\circ\), and undergoes total internal reflection because the angle of incidence exceeds the critical angle.
Quick Tip: Always check the critical angle \(\sin i_c = 1/\mu\) when a ray is traveling from a denser to a rarer medium to determine if it will refract or reflect.

Step 1: Understanding the Concept:

Rays and wavefronts are two different ways of representing the propagation of light.


Step 2: Detailed Explanation:

1. Ray: A ray is a straight line along which energy of light travels. It represents the direction of propagation of light.

2. Wavefront: A wavefront is defined as the locus of all points in a medium that are in the same phase of vibration at a given instant.

3. Geometrical Relation: Rays are always perpendicular to the wavefronts at every point.

4. Physical Meaning: Wavefront is a wave-concept (Huygens' Principle), while a ray is a geometric optics concept used to trace path.


Step 3: Final Answer:

A ray represents direction of light propagation, whereas a wavefront is the locus of points in the same phase. They are mutually perpendicular.
Quick Tip: For a point source, wavefronts are spherical and rays are radial. For a source at infinity, wavefronts are plane and rays are parallel.

Step 1: Understanding the Concept:

Huygens' Principle states that every point on a wavefront acts as a source of secondary wavelets which spread out in the forward direction.


Step 2: Detailed Explanation:

1. Let \(AB\) be a plane wavefront incident at an angle \(i\) on a reflecting surface \(MN\). Point \(A\) touches the surface first.

2. In the time \(t\) that the wavelet from \(B\) takes to reach \(C\) on the surface, the secondary wavelet from \(A\) would have traveled a distance \(v \times t\) and formed a hemisphere of radius \(v t\).

3. We have \(BC = v t\). To find the reflected wavefront, we draw a tangent \(CD\) from point \(C\) to the wavelet from \(A\). Thus \(AD = v t\).

4. Now, consider triangles \(\triangle ABC\) and \(\triangle ADC\):

- \(AC = AC\) (Common side)

- \(BC = AD = v t\) (Distances traveled by wavelets in time \(t\))

- \(\angle ABC = \angle ADC = 90^\circ\) (Tangents to wavelets/wavefronts)

5. By RHS congruence, \(\triangle ABC \cong \triangle ADC\).

6. Therefore, \(\angle BAC = \angle DCA\).

7. Since \(\angle BAC = i\) and \(\angle DCA = r\), we get \(i = r\).


Step 3: Final Answer:

Using geometric properties of wavelets and triangles, we prove the angle of incidence equals the angle of reflection (\(i = r\)).
Quick Tip: The key to these derivations is setting the distance traveled by wavelets in the same medium (\(BC = AD = vt\)).

Step 1: Understanding the Concept:

Refraction through a lens changes the shape of the wavefront because light travels slower in the denser lens material compared to air.


Step 2: Detailed Explanation:

1. A plane wavefront is incident on a convex lens.

2. The central part of the wavefront travels through the thickest part of the lens and is delayed most.

3. The marginal parts travel through thinner sections (or air) and are delayed less.

4. As a result, the emerging wavefront is curved and "sinks" at the center relative to the edges.

5. This resulting wavefront is a converging spherical wavefront that focuses at the focal point of the lens.


Step 3: Final Answer:

When a plane wavefront passes through a convex lens, it emerges as a converging spherical wavefront.
Quick Tip: Remember: Thicker glass \(\implies\) More delay \(\implies\) Wavefront lags behind.

Step 1: Understanding the Concept:

Resonance and power factor are critical parameters describing the behavior of AC circuits containing inductive and capacitive elements.


Step 2: Detailed Explanation:

1. Resonant Frequency (\(f_r\)): It is the frequency of the AC source at which the inductive reactance becomes equal to the capacitive reactance (\(X_L = X_C\)). At this frequency, the impedance of the series LCR circuit is minimum and equal to the resistance (\(Z = R\)), resulting in maximum current. \[ f_r = \frac{1}{2\pi\sqrt{LC}} \]
2. Power Factor (\(\cos \phi\)): It is defined as the ratio of real power to apparent power in an AC circuit. Mathematically, it is the cosine of the phase angle (\(\phi\)) between voltage and current. \[ \cos \phi = \frac{R}{Z} \]
3. Maximum Power Dissipation: The average power dissipated is \(P_{avg} = V_{rms} I_{rms} \cos \phi\). Power is maximum when \(\cos \phi = 1\). This happens when the circuit is purely resistive or at resonance.


Step 3: Final Answer:

Resonant frequency is where \(X_L = X_C\). Power factor is \(\cos\phi = R/Z\). Maximum power dissipation occurs when power factor = 1.
Quick Tip: At resonance, the circuit behaves as a purely resistive circuit, and the phase difference \(\phi = 0\), so \(\cos 0 = 1\).

Step 1: Understanding the Concept:

This is a standard calculation for a series LCR circuit. We need to find reactances first, then the total impedance, and finally the current.


Step 2: Key Formula or Approach:

1. Inductive reactance \(X_L = \omega L\).

2. Capacitive reactance \(X_C = \frac{1}{\omega C}\).

3. Impedance \(Z = \sqrt{R^2 + (X_L - X_C)^2}\).

4. Peak current \(I_0 = \frac{V_0}{Z}\); RMS current \(I_{rms} = \frac{I_0}{\sqrt{2}}\).


Step 3: Detailed Explanation:

Given: \(L = \frac{5}{\pi}\) H, \(C = \frac{50}{\pi} \muF = \frac{50}{\pi} \times 10^{-6}\) F, \(R = 400 \Omega\), \(V = 140 \sin(100\pi t)\).

From the voltage equation: \(V_0 = 140\) V and \(\omega = 100\pi\) rad/s.

1. Inductive Reactance: \[ X_L = \omega L = 100\pi \times \frac{5}{\pi} = 500 \Omega \]
2. Capacitive Reactance: \[ X_C = \frac{1}{\omega C} = \frac{1}{100\pi \times \frac{50}{\pi} \times 10^{-6}} = \frac{1}{5000 \times 10^{-6}} = \frac{10^6}{5000} = 200 \Omega \]
3. Impedance (Z): \[ Z = \sqrt{400^2 + (500 - 200)^2} = \sqrt{160000 + 90000} = \sqrt{250000} = 500 \Omega \]
4. RMS Current (\(I_{rms}\)):
Peak current \(I_0 = \frac{V_0}{Z} = \frac{140}{500} = 0.28\) A.
\[ I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{0.28}{1.4} = 0.2 A \]

Step 4: Final Answer:

(I) Impedance \(Z = 500 \Omega\).

(II) RMS current \(I_{rms} = 0.2 A\).
Quick Tip: Note that \(\omega = 2\pi f\). Always look at the \(\pi\) in the denominator of \(L\) or \(C\) values; it often cancels with \(\omega\) to simplify calculations.

Step 1: Understanding the Concept:

A transformer works on the principle of mutual induction. It changes the voltage levels without changing the frequency.


Step 2: Detailed Explanation:

1. Step-up Transformer: It increases the output voltage (\(V_s > V_p\)). This is achieved by having more turns in the secondary coil than in the primary coil (\(N_s > N_p\)).

2. Derivation of Voltage Ratio:

According to Faraday's law of induction, the emf induced in each turn is the same (assuming perfect flux linkage).

The induced emf in the primary coil is \(\varepsilon_p = -N_p \frac{d\Phi}{dt}\).

The induced emf in the secondary coil is \(\varepsilon_s = -N_s \frac{d\Phi}{dt}\).

For an ideal transformer, the terminal voltages \(V_p\) and \(V_s\) are approximately equal to the emfs. Dividing the two equations: \[ \frac{V_s}{V_p} = \frac{N_s}{N_p} = k \]
Where \(k\) is the transformation ratio. For a step-up transformer, \(k > 1\).


Step 3: Final Answer:

The voltage ratio is directly proportional to the turns ratio: \(V_s/V_p = N_s/N_p\).
Quick Tip: Remember: A transformer cannot step up DC. It only works with time-varying (AC) signals. In a step-up transformer, voltage increases but current decreases to conserve power.

Step 1: Understanding the Concept:

For an ideal transformer, there are no power losses, so power input equals power output. We use the turns ratio for voltage and current relations.


Step 2: Key Formula or Approach:

1. Power \(P = V_p I_p\).

2. Turns ratio relation: \(\frac{V_s}{V_p} = \frac{N_s}{N_p}\).


Step 3: Detailed Explanation:

Given: \(N_p = 100\), \(N_s = 5000\), \(P = 3.3 kW = 3300 W\), \(V_p = 220 V\).

1. Current in the Primary Coil (\(I_p\)): \[ P = V_p I_p \implies I_p = \frac{P}{V_p} = \frac{3300}{220} = \frac{330}{22} = 15 A \]
2. Output Voltage (\(V_s\)):
Using the transformer formula: \[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \] \[ V_s = V_p \times \frac{N_s}{N_p} = 220 \times \frac{5000}{100} \] \[ V_s = 220 \times 50 = 11,000 V = 11 kV \]

Step 4: Final Answer:

(I) Primary current \(I_p = 15 A\).

(II) Output voltage \(V_s = 11,000 V\).
Quick Tip: Check current in secondary: \(I_s = P / V_s = 3300 / 11000 = 0.3 A\). As expected, voltage stepped up 50 times, so current stepped down 50 times (\(15/0.3 = 50\)).

Step 1: Understanding the Concept:

These statements describe fundamental properties of electrostatic fields, dielectrics, and capacitors.


Step 2: Detailed Explanation:

1. (I) Equipotential Surface: Work done to move a charge \(q\) on an equipotential surface is zero (\(W = q \Delta V = 0\)). Since \(W = \int \vec{E} \cdot d\vec{l} = E dl \cos \theta\), for \(W\) to be zero with non-zero field and displacement, \(\cos \theta\) must be zero. Thus \(\theta = 90^\circ\), meaning the field is perpendicular to the surface.

2. (II) Dielectrics in Field: When a dielectric is placed in an external field (\(E_0\)), it undergoes polarization. Induced charges on the surfaces of the dielectric create an internal "induced" electric field (\(E_i\)) in the opposite direction. The net field inside is \(E = E_0 - E_i\), which is less than \(E_0\).

3. (III) Parallel Plate Capacitor: For a charged (and isolated) capacitor, the charge \(Q\) and the field \(E\) (which is \(Q/A\varepsilon_0\)) between the plates remain constant. The potential difference is \(V = E \times d\). When distance \(d\) decreases, \(V\) also decreases proportionally.


Step 3: Final Answer:

(I) Because no work is done along the surface. (II) Because of induced internal field due to polarization. (III) Because \(V = Ed\), and \(E\) is constant.
Quick Tip: In (III), if the capacitor is connected to a battery, \(V\) remains constant and \(Q\) increases. Always check if the battery is connected or disconnected.

Step 1: Understanding the Concept:

The work done to dissociate a system of charges is equal to the negative of the electrostatic potential energy of the system. Potential energy is the work done to assemble the charges from infinity.


Step 2: Key Formula or Approach:

Potential energy of a system of charges: \[ U = k \sum \frac{q_i q_j}{r_{ij}} \]
Work to dissociate \(W = -U_{system}\).


Step 3: Detailed Explanation:

Charges are \(q_1 = q\), \(q_2 = -4q\), and \(q_3 = 2q\). The distance between any two charges is \(r = a\).

Electrostatic Potential Energy (\(U\)): \[ U = k \left[ \frac{q_1 q_2}{a} + \frac{q_2 q_3}{a} + \frac{q_3 q_1}{a} \right] \] \[ U = \frac{k}{a} [ (q)(-4q) + (-4q)(2q) + (2q)(q) ] \] \[ U = \frac{k}{a} [ -4q^2 - 8q^2 + 2q^2 ] \] \[ U = \frac{k}{a} [ -10q^2 ] = -\frac{10 k q^2}{a} \]
The work required to dissociate (separate the charges to infinity) is: \[ W = 0 - U = - \left( -\frac{10 k q^2}{a} \right) = \frac{10 k q^2}{a} \]

Step 4: Final Answer:

The work done is \(\frac{10 k q^2}{a}\).
Quick Tip: "Work to dissociate" is the same as the binding energy. If \(U\) is negative, the system is stable and energy must be supplied to break it.

Step 1: Understanding the Concept:

These conceptual questions deal with the microscopic and macroscopic aspects of electric circuits.


Step 2: Detailed Explanation:

1. (I) Quick Establishment of Current: Although the individual electrons drift slowly, the current is established at the speed of an electromagnetic wave (nearly speed of light). When the circuit is closed, an electric field is set up throughout the wire almost instantly. This field pushes all the free electrons already present in the wire simultaneously.

2. (II) Low Internal Resistance: The maximum current that can be drawn from a source of emf \(E\) and internal resistance \(r\) is \(I_{max} = E/r\). If \(E\) is low, \(r\) must be extremely small to allow a high value of \(I_{max}\).

3. (III) V = IR vs Ohm's Law: \(V = IR\) is simply a definition of resistance. Ohm's Law specifically states that for many materials, the ratio \(V/I\) remains constant (i.e., \(R\) is independent of \(V\) and \(I\)) provided physical conditions like temperature remain unchanged. Not all components follow this (e.g., diodes, transistors).


Step 3: Final Answer:

(I) Electric field propagates at the speed of light. (II) Because \(I_{max} = E/r\). (III) Because Ohm's law requires \(R\) to be constant, whereas \(V = IR\) just defines \(R\).
Quick Tip: Ohmic conductors show a linear V-I graph. Non-ohmic conductors can still have a defined resistance \(R = V/I\) at any point, but they don't follow Ohm's Law.

Step 1: Understanding the Concept:

When cells are connected in parallel, we use the equivalent cell formula derived from Kirchhoff's laws.


Step 2: Key Formula or Approach:

1. Equivalent internal resistance: \(\frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2}\).

2. Equivalent emf: \(\varepsilon_{eq} = \frac{\frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}}\) (Assuming same polarity orientation).


Step 3: Detailed Explanation:

Given: \(\varepsilon_1 = 12\) V, \(r_1 = 1 \Omega\), \(\varepsilon_2 = 6\) V, \(r_2 = 0.5 \Omega\).

1. Internal Resistance (\(r_{eq}\)): \[ \frac{1}{r_{eq}} = \frac{1}{1} + \frac{1}{0.5} = 1 + 2 = 3 \] \[ r_{eq} = \frac{1}{3} \Omega \approx 0.33 \Omega \]
2. Equivalent EMF (\(\varepsilon_{eq}\)): \[ \varepsilon_{eq} = r_{eq} \left[ \frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2} \right] \] \[ \varepsilon_{eq} = \frac{1}{3} \left[ \frac{12}{1} + \frac{6}{0.5} \right] \] \[ \varepsilon_{eq} = \frac{1}{3} [ 12 + 12 ] = \frac{24}{3} = 8 V \]

Step 4: Final Answer:

The equivalent emf is 8 V and the internal resistance is \(\frac{1}{3} \Omega\).
Quick Tip: For parallel cells with different emfs, the equivalent emf always lies between the individual emfs (\(6 V < 8 V < 12 V\)).