CBSE Class 12 Physics Set 1 - (55/4/1) Question Paper 2026 with Solution PDF

Concept:

The resistance of a conductor depends upon its resistivity, length, and area of cross-section. The relation is
\[ R=\rho\frac{L}{A} \]

where
\[ \rho = resistivity, \qquad L = length, \qquad A = area of cross-section. \]

When two resistors are connected in series, the equivalent resistance is equal to the sum of their individual resistances.

To determine the equivalent resistivity of a combination, we first calculate the equivalent resistance and then compare it with the standard relation
\[ R_{eq}=\rho_{eq}\frac{L_{eq}}{A}. \]



Step 1: Write the resistance of each wire.

Let each wire have length \(L\) and area of cross-section \(A\).

For the first wire,
\[ R_1=\rho_1\frac{L}{A}. \]

Similarly, for the second wire,
\[ R_2=\rho_2\frac{L}{A}. \]



Step 2: Determine the equivalent resistance of the series combination.

Since the wires are connected in series,
\[ R_{eq}=R_1+R_2. \]

Substituting the values,
\[ R_{eq} = \rho_1\frac{L}{A} + \rho_2\frac{L}{A}. \]

Taking \(\frac{L}{A}\) common,
\[ R_{eq} = \frac{L}{A}(\rho_1+\rho_2). \]



Step 3: Express the combination as a single wire.

The total length of the combination is
\[ L_{eq}=L+L=2L. \]

The area of cross-section remains
\[ A_{eq}=A. \]

Therefore,
\[ R_{eq} = \rho_{eq} \frac{2L}{A}. \]



Step 4: Compare both expressions of equivalent resistance.

We have
\[ \rho_{eq} \frac{2L}{A} = \frac{L}{A}(\rho_1+\rho_2). \]

Cancelling \(\frac{L}{A}\) from both sides,
\[ 2\rho_{eq} = \rho_1+\rho_2. \]

Hence,
\[ \rho_{eq} = \frac{\rho_1+\rho_2}{2}. \]



Step 5: Identify the correct option.

Thus the equivalent resistivity of the series combination is
\[ \boxed{\rho_{eq}=\frac{\rho_1+\rho_2}{2}}. \]

Therefore, the correct answer is
\[ \boxed{(A)} \] Quick Tip: For wires connected end-to-end (series combination), first find the equivalent resistance and then compare it with \(R=\rho L/A\). Remember that resistivity is a material property and does not simply add like resistances.

Concept:
When a charged particle moves in a uniform magnetic field with a velocity perpendicular to the field lines, it experiences a magnetic Lorentz force. This force always acts perpendicular to both the velocity vector and the magnetic field vector, providing the necessary centripetal force required for the particle to move in a stable circular trajectory.

The magnitude of the magnetic force (\(F_B\)) is given by: \[ F_B = q v B \sin(\theta) \]
Since the particles are projected perpendicularly to the magnetic field, \(\theta = 90^\circ\), which implies \(\sin(90^\circ) = 1\). Thus, the force is: \[ F_B = q v B \]
This magnetic force balances the centripetal force (\(F_C\)) required to maintain a circle of radius \(R\): \[ F_C = \frac{m v^2}{R} \]
Equating these two forces yields the fundamental formula for the radius of curvature of a charged particle in a magnetic field: \[ q v B = \frac{m v^2}{R} \quad \Rightarrow \quad R = \frac{m v}{q B} \]
By evaluating this relation under constraints of constant parameters, we can readily deduce the relationship between the physical properties of the two particles.


Step 1: Expressing the radius for both particles based on given constraints.

We are given that both particles are projected with the exact same velocity, let this common speed be \(v_1 = v_2 = v\).
Furthermore, they are moving within the same uniform magnetic field region, so the magnetic field intensity is identical for both, let it be \(B\).

Using the general radius formula derived from the balance of forces, we write the explicit radii expressions for particle 1 and particle 2:
For particle 1: \[ R_1 = \frac{m_1 v}{q_1 B} \quad \cdots (1) \]
For particle 2: \[ R_2 = \frac{m_2 v}{q_2 B} \quad \cdots (2) \]


Step 2: Comparing the trajectories from the given diagram.

By analyzing the circular paths provided in the graphic, we observe the curves described by each mass.
Path 1 corresponds to particle 1 with mass \(m_1\), and Path 2 corresponds to particle 2 with mass \(m_2\).

The curve for path 1 is sharper and tighter, meaning it turns more abruptly, which visually indicates a smaller radius of curvature. Conversely, path 2 is flatter and more extended, indicating a larger radius of curvature.
Therefore, by inspection of the geometric trajectories in the figure: \[ R_1 < R_2 \]


Step 3: Setting up the algebraic inequality and simplifying.

Substitute the expressions from equation (1) and equation (2) into the geometric radius inequality \(R_1 < R_2\): \[ \frac{m_1 v}{q_1 B} < \frac{m_2 v}{q_2 B} \]
Since the velocity \(v\) and the magnetic field magnitude \(B\) are positive scalar constants shared by both systems, we can divide both sides of the inequality by the common non-zero term \(\frac{v}{B}\): \[ \frac{m_1}{q_1} < \frac{m_2}{q_2} \]


Step 4: Rearranging into the final requested form.

The question presents the choices as comparisons between the mass ratio \(\frac{m_1}{m_2}\) and the charge ratio.
To isolate \(\frac{m_1}{m_2}\) on the left side of our inequality, we multiply both sides by \(q_1\) (which is positive as both undergo identical directional deflection trends indicating identical sign charges) and divide both sides by \(m_2\): \[ \frac{m_1}{m_2} < \frac{q_1}{q_2} \]
This mathematically derived relation perfectly corresponds to Option (C). Quick Tip: When tracking particles in magnetic field problems, always remember that the radius of a circular trajectory is directly proportional to the mass-to-charge ratio (\(m/q\)) when the velocity \(v\) and field \(B\) are held constant: \[ R \propto \frac{m}{q} \] A smaller radius directly implies a smaller value of \(\frac{m}{q}\). Therefore, seeing that curve 1 is tighter (\(R_1 < R_2\)), you can immediately write \(\frac{m_1}{q_1} < \frac{m_2}{q_2}\) and cross-multiply to reach the final answer in seconds.

Concept:

Electric current is defined as the rate of flow of charge.

Mathematically,
\[ I=\frac{Q}{t} \]

where
\[ Q=charge flowing, \]

and
\[ t=time taken. \]

An electron revolving around the nucleus constitutes a circulating charge. Such a moving charge behaves like a current loop and therefore produces a magnetic field.



Step 1: Determine the charge passing a point in one revolution.

The charge carried by one electron is
\[ e=1.6\times10^{-19}\,C. \]

Whenever the electron completes one revolution, the charge \(e\) effectively passes a given point once.

Thus,
\[ Q=e. \]



Step 2: Relate frequency to time period.

The electron makes \(n\) revolutions per second.

Therefore,
\[ n=\frac{1}{T} \]

where \(T\) is the time period.

Hence,
\[ T=\frac{1}{n}. \]



Step 3: Use the definition of current.

Current is
\[ I=\frac{Q}{T}. \]

Substituting
\[ Q=e \]

and
\[ T=\frac{1}{n}, \]

we get
\[ I=\frac{e}{1/n}. \]
\[ I=en. \]



Step 4: Interpret the result physically.

The current increases if the electron completes more revolutions per second because charge passes the observation point more frequently.

Thus the equivalent current associated with the orbiting electron is
\[ \boxed{I=ne}. \]



Step 5: Select the correct option.

Comparing with the given options,
\[ \boxed{(B) ne} \]

is the correct answer. Quick Tip: Whenever a charge \(q\) completes \(f\) revolutions per second, the equivalent current is \[ I=qf. \] Here \(q=e\) and \(f=n\), therefore \(I=en\).

Concept:

According to Faraday's law of electromagnetic induction,
\[ \mathcal{E} = -\frac{d\Phi}{dt}, \]

where
\[ \Phi = magnetic flux. \]

Thus, the rate of change of magnetic flux has the same dimensions as emf.

Therefore, instead of directly finding dimensions of magnetic flux and differentiating, we can use the dimensions of emf.



Step 1: Write the dimensions of emf.

Emf is defined as work done per unit charge.
\[ EMF = \frac{Work}{Charge}. \]



Step 2: Find dimensions of work.

Work
\[ = ForceX Distance. \]

Dimensions of force are
\[ [M\,L\,T^{-2}]. \]

Therefore,
\[ [Work] = [M\,L\,T^{-2}] \times [L] = [M\,L^{2}\,T^{-2}]. \]



Step 3: Find dimensions of charge.

Charge is
\[ Q=It. \]

Hence,
\[ [Q]=[A\,T]. \]



Step 4: Determine dimensions of emf.

Using
\[ EMF = \frac{Work}{Charge}, \]

we obtain
\[ [EMF] = \frac{[M\,L^{2}\,T^{-2}]}{[A\,T]}. \]

Therefore,
\[ [EMF] = [M\,L^{2}\,T^{-3}\,A^{-1}]. \]



Step 5: Apply Faraday's law.

Since
\[ \mathcal{E} = -\frac{d\Phi}{dt}, \]

the dimensions of
\[ \frac{d\Phi}{dt} \]

are exactly the same as those of emf.

Hence,
\[ \boxed{ \left[\frac{d\Phi}{dt}\right] = [M\,L^{2}\,T^{-3}\,A^{-1}] }. \]



Step 6: Choose the correct option.

Comparing with the given options,
\[ \boxed{(B)} \]

is the correct answer. Quick Tip: Remember Faraday's law: \[ \mathcal{E}=-\frac{d\Phi}{dt}. \] Therefore, the dimensions of the rate of change of magnetic flux are always the same as the dimensions of emf: \[ [M\,L^{2}\,T^{-3}\,A^{-1}]. \]

Concept:
This problem can be comprehensively analyzed using the principles of Electromagnetic Induction (Faraday's Law and Lenz's Law) and the Magnetic Lorentz Force acting on a current-carrying conductor.


Motional Electromotive Force (EMF): When a conducting rod of length \(L\) moves with a velocity \(\vec{v}\) perpendicular to a uniform magnetic field \(\vec{B}\), an EMF (\(\varepsilon\)) is induced across its ends, given by:
\[ \varepsilon = (\vec{v} \times \vec{B}) \cdot \vec{L} \]
The direction of the driving force on positive charge carriers inside the moving conductor is given by the vector cross product \(\vec{v} \times \vec{B}\).

Lenz's Law: The direction of an induced current is always such that it will oppose the change in magnetic flux that produced it.

Magnetic Force on a Straight Conductor: A wire carrying a current \(I\) with a length vector \(\vec{L}\) placed inside a uniform external magnetic field \(\vec{B}\) experiences a mechanical deflecting force expressed by:
\[ \vec{F} = I (\vec{L} \times \vec{B}) \]



Step 1: Define the coordinate system and given vectors.

From the reference coordinate axes provided in the diagram:

The positive \(x\)-axis is directed to the right, represented by the unit vector \(\hat{i}\).
The positive \(y\)-axis is directed vertically upwards within the plane of the paper, represented by the unit vector \(\hat{j}\).
The positive \(z\)-axis points perpendicularly outwards from the plane of the paper towards the reader, represented by the unit vector \(\hat{k}\).


The uniform magnetic field \(\vec{B}\) is directed vertically downward into the plane of the page, so it can be vectorially written as: \[ \vec{B} = -B\hat{k} \]
Conductor 1 is forced to move horizontally to the left with a constant velocity. Therefore, its velocity vector \(\vec{v}_1\) is written as: \[ \vec{v}_1 = -v\hat{i} \]


Step 2: Determine the direction of the induced current using motional EMF.

As conductor 1 moves through the uniform magnetic field, the free charges inside it experience a magnetic force. The direction of this force on positive charge carriers is given by the cross product: \[ \vec{v}_1 \times \vec{B} = (-v\hat{i}) \times (-B\hat{k}) \]
Using the scalar property of vector multiplication: \[ \vec{v}_1 \times \vec{B} = vB (\hat{i} \times \hat{k}) \]
According to the standard cross product rules for unit vectors (\(\hat{i} \times \hat{j} = \hat{k}\), \(\hat{j} \times \hat{k} = \hat{i}\), and \(\hat{k} \times \hat{i} = \hat{j}\)), we know that: \[ \hat{i} \times \hat{k} = -\hat{j} \]
Substituting this back into our expression: \[ \vec{v}_1 \times \vec{B} = vB (-\hat{j}) = -vB\hat{j} \]
Since the direction of the force on positive charges is along \(-\hat{j}\) (downwards), the induced current flows downwards within conductor 1.

Tracing this current through the closed loop formed by conductor 1, rail S, conductor 2, and rail R:

The current flows downwards in conductor 1 (\(-\hat{j}\) direction).
It moves to the right along the bottom conducting rail S (\(+\hat{i}\) direction).
It travels upwards through conductor 2 (\(+\hat{j}\) direction).
It travels back to the left along the top conducting rail R (\(-\hat{i}\) direction).

Thus, the induced current circles in a counter-clockwise direction around the loop. For conductor 2, the current vector points vertically upwards: \[ \vec{I}_2 = I\hat{j} \]


Step 3: Alternative Approach using Lenz's Law (Flux Change Analysis).

Let us cross-verify the current direction using Lenz's Law to ensure absolute certainty:

The initial magnetic field lines are pointing straight into the page (\(\times\)).
As conductor 1 is pulled towards the left, the enclosed surface area of the rectangular loop bounded by rails R, S and conductors 1, 2 is continuously increasing.
Because the area increases, the total inward magnetic flux (\(\Phi_B = B \cdot A\)) piercing through the loop increases.
According to Lenz's law, the system will establish an induced current to generate an opposing magnetic field pointing out of the page (\(\cdot\)) to mitigate this flux increase.
By applying the right-hand grip rule, a counter-clockwise current is required to produce a magnetic field pointing outwards.

In a counter-clockwise loop, the current must flow from bottom to top through the rightmost vertical branch, which is conductor 2. This perfectly confirms that the current flows in the \(+\hat{j}\) direction in conductor 2.


Step 4: Calculate the magnetic force acting on conductor 2.

Now that we have established that conductor 2 carries an upward current in the presence of an inward magnetic field, we can compute the mechanical force \(\vec{F}_2\) acting on it: \[ \vec{F}_2 = I (\vec{L}_2 \times \vec{B}) \]
Here, the length vector \(\vec{L}_2\) is in the direction of the current, so \(\vec{L}_2 = L\hat{j}\). The magnetic field is \(\vec{B} = -B\hat{k}\). Substituting these vectors into the formula: \[ \vec{F}_2 = I \left( L\hat{j} \times (-B\hat{k}) \right) \]
Factoring out the scalar constants: \[ \vec{F}_2 = -ILB (\hat{j} \times \hat{k}) \]
Using the cross product rule for the unit vectors where \(\hat{j} \times \hat{k} = \hat{i}\): \[ \vec{F}_2 = -ILB (\hat{i}) = -ILB\hat{i} \]
The negative sign attached to the unit vector \(\hat{i}\) indicates that the net magnetic force on conductor 2 is directed horizontally to the left. Therefore, the force acts along the \(-\hat{i}\) direction, matching Option (A). Quick Tip: Save time by analyzing this via energy conservation and Lenz's Law directly: Lenz's Law states that any induced effect always attempts to destroy the cause that created it. Here, the cause of the induction is the expansion of the loop's area} due to conductor 1 moving left. To counteract this growth and try to decrease the area, the free-to-move conductor 2 will experience an electromagnetic force pushing it inward to contract the loop. Since it is on the right side, it must move to the left (\(-\hat{i}\)) to shrink the area! No cross products are required.

Concept:

An alternating voltage varies sinusoidally with time and is generally represented as
\[ v = V_0 \sin \omega t \]

where
\[ V_0 = peak voltage (amplitude) \]

and
\[ \omega = angular frequency. \]

For a sinusoidal alternating voltage:
\[ V_{avg} = 0 \]

over one complete cycle because the positive half-cycle and negative half-cycle are equal in magnitude and opposite in sign.

The rms (root mean square) value or effective value is given by
\[ V_{rms}=\frac{V_0}{\sqrt{2}}. \]

The rms value represents the dc voltage that would produce the same heating effect in a resistor.



Step 1: Identify the peak voltage.

The given alternating voltage is
\[ v=14\sin(314t). \]

Comparing with
\[ v=V_0\sin\omega t, \]

we obtain
\[ V_0=14\,V. \]

Thus, the peak voltage is
\[ 14\,V. \]



Step 2: Find the average value over one complete cycle.

For a complete cycle of a sine wave,
\[ V_{avg}=0. \]

This is because the positive and negative halves cancel each other exactly.

Hence,
\[ V_{avg}=0\,V. \]



Step 3: Calculate the rms (effective) value.

Using
\[ V_{rms} = \frac{V_0}{\sqrt2}, \]

we get
\[ V_{rms} = \frac{14}{\sqrt2}. \]

Multiplying numerator and denominator by \(\sqrt2\),
\[ V_{rms} = \frac{14\sqrt2}{2} = 7\sqrt2. \]

Using
\[ \sqrt2 \approx 1.414, \]
\[ V_{rms} = 7\times1.414 = 9.898. \]

Therefore,
\[ V_{rms}\approx10\,V. \]



Step 4: Write the final answer.

Thus,
\[ V_{avg}=0\,V \]

and
\[ V_{rms}=10\,V. \]

Therefore,
\[ \boxed{(C) 0 and 10} \]

is the correct answer. Quick Tip: For any sinusoidal alternating voltage or current: \[ V_{avg}=0 \] (over a complete cycle) and \[ V_{rms}=\frac{V_0}{\sqrt2}. \] Always remember the factor \(\sqrt2\) while converting peak values into rms values.

Concept:

An electromagnetic wave consists of mutually perpendicular electric and magnetic fields.

For an electromagnetic wave propagating in a medium, the ratio of the electric field amplitude to the magnetic field amplitude is equal to the speed of the wave in that medium.

Thus,
\[ \frac{E_0}{B_0}=v. \]

The speed of light in a medium of refractive index \(n\) is
\[ v=\frac{c}{n}, \]

where
\[ c=3\times10^8\,ms^{-1} \]

is the speed of light in vacuum.



Step 1: Write the relation between refractive index and speed.

The refractive index is given by
\[ n=\frac{c}{v}. \]

Rearranging,
\[ v=\frac{c}{n}. \]



Step 2: Substitute the given values.

Given,
\[ n=1.5 \]

and
\[ c=3\times10^8\,ms^{-1}. \]

Therefore,
\[ v=\frac{3\times10^8}{1.5}. \]



Step 3: Perform the calculation.
\[ v=2\times10^8\,ms^{-1}. \]

Thus,
\[ \frac{E_0}{B_0} = 2\times10^8\,ms^{-1}. \]



Step 4: Identify the correct option.

Hence,
\[ \boxed{\frac{E_0}{B_0}=2\times10^8\,ms^{-1}} \]

and therefore
\[ \boxed{(B)} \]

is the correct answer. Quick Tip: Always remember: \[ \frac{E_0}{B_0}=c \] in vacuum and \[ \frac{E_0}{B_0}=v=\frac{c}{n} \] inside a medium of refractive index \(n\).

Concept:

According to Einstein's photoelectric equation,
\[ h\nu=\phi+K_{\max} \]

where
\[ h\nu=energy of incident photon, \]
\[ \phi=work function of the metal, \]

and
\[ K_{\max}=maximum kinetic energy of emitted electrons. \]

The stopping potential \(V_0\) is related to maximum kinetic energy by
\[ eV_0=K_{\max}. \]

Hence, a larger photon energy produces a larger stopping potential.

Also, saturation current mainly depends on the intensity of incident light and not on its frequency.



Step 1: Examine the change in wavelength.

The wavelength changes from
\[ 600\,nm \]

to
\[ 400\,nm. \]

Since
\[ \nu=\frac{c}{\lambda}, \]

a decrease in wavelength implies an increase in frequency.

Thus,
\[ \nu_{400}>\nu_{600}. \]



Step 2: Determine the effect on photon energy.

Photon energy is
\[ E=h\nu. \]

Since frequency increases,
\[ E_{400}>E_{600}. \]

Therefore, each photon now carries more energy.



Step 3: Determine the effect on maximum kinetic energy.

Using Einstein's equation,
\[ K_{\max}=h\nu-\phi. \]

Since \(h\nu\) increases while \(\phi\) remains constant,
\[ K_{\max} \]

increases.



Step 4: Determine the effect on stopping potential.

Since
\[ eV_0=K_{\max}, \]

an increase in \(K_{\max}\) leads to an increase in stopping potential.

Hence,
\[ V_0 \]

increases.



Step 5: Discuss saturation current.

The intensity of radiation is kept constant.

Saturation current depends primarily on the number of emitted photoelectrons per second, which in turn depends on intensity.

Since intensity remains unchanged,
\[ Saturation Current \]

remains approximately unchanged.

Thus options (C) and (D) are incorrect.



Step 6: Write the final conclusion.

Reducing the wavelength from \(600\,nm\) to \(400\,nm\) increases the frequency, increases the photon energy, increases the maximum kinetic energy of photoelectrons, and hence increases the stopping potential.

Therefore,
\[ \boxed{(B) cut-off potential will increase} \]

is the correct answer. Quick Tip: In photoelectric effect: \[ K_{\max}=h\nu-\phi \] and \[ eV_0=K_{\max}. \] Higher frequency (or lower wavelength) means higher stopping potential. Saturation current depends mainly on intensity.

Concept:

According to Einstein's photoelectric equation, photoelectric emission occurs only when the energy of the incident photon is greater than or equal to the work function of the metal.

The energy of a photon is given by
\[ E=\frac{hc}{\lambda}. \]

For practical calculations in electron volts,
\[ E(eV)=\frac{1240}{\lambda(nm)}. \]

If
\[ E \ge \phi, \]

photoelectric emission takes place.

If
\[ E < \phi, \]

photoelectric emission does not occur.



Step 1: Calculate the energy of the incident photon.

Given,
\[ \lambda=331\ nm. \]

Using
\[ E=\frac{1240}{331}, \]

we obtain
\[ E=3.746\ eV. \]

Thus, the energy of each incident photon is approximately
\[ E\approx3.75\ eV. \]



Step 2: Compare the photon energy with the work function of sodium.

For sodium,
\[ \phi_{Na}=1.92\ eV. \]

Since
\[ 3.75>1.92, \]

photoelectric emission occurs from sodium.



Step 3: Compare the photon energy with the work function of potassium.

For potassium,
\[ \phi_{K}=2.15\ eV. \]

Since
\[ 3.75>2.15, \]

photoelectric emission occurs from potassium.



Step 4: Compare the photon energy with the work function of calcium.

For calcium,
\[ \phi_{Ca}=3.20\ eV. \]

Since
\[ 3.75>3.20, \]

photoelectric emission occurs from calcium.



Step 5: Compare the photon energy with the work function of molybdenum.

For molybdenum,
\[ \phi_{Mo}=4.17\ eV. \]

Since
\[ 3.75<4.17, \]

photoelectric emission does not occur from molybdenum.



Step 6: Write the final conclusion.

Photoelectric emission occurs for
\[ Na,\ K,\ and\ Ca \]

but does not occur for
\[ Mo. \]

Therefore,
\[ \boxed{Only Mo will not show photoelectric emission.} \]

Hence,
\[ \boxed{(B)} \]

is the correct answer. Quick Tip: For photoelectric effect questions, quickly calculate the photon energy using \[ E(eV)=\frac{1240}{\lambda(nm)}. \] If the photon energy exceeds the work function, photoelectric emission occurs.

Concept:

According to de Broglie's hypothesis,
\[ \lambda=\frac{h}{p}, \]

where \(h\) is Planck's constant and \(p\) is the momentum of the particle.

For a non-relativistic particle,
\[ K=\frac{p^2}{2m}. \]

Therefore,
\[ p=\sqrt{2mK}. \]

Substituting into the de Broglie relation,
\[ \lambda=\frac{h}{\sqrt{2mK}}. \]

Hence,
\[ \lambda\propto\frac{1}{\sqrt{K}}. \]



Step 1: Write the relation between wavelength and kinetic energy.

Since
\[ \lambda\propto\frac{1}{\sqrt{K}}, \]

we can write
\[ \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{K_1}{K_2}}. \]



Step 2: Use the given information.

The final kinetic energy is four times the initial kinetic energy.

Therefore,
\[ K_2=4K_1. \]

Substituting,
\[ \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{K_1}{4K_1}}. \]
\[ \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{1}{4}}. \]
\[ \frac{\lambda_2}{\lambda_1} = \frac{1}{2}. \]

Thus,
\[ \lambda_2=\frac{\lambda_1}{2}. \]



Step 3: Determine the percentage decrease.

Initial wavelength
\[ =\lambda_1. \]

Final wavelength
\[ =\frac{\lambda_1}{2}. \]

Decrease
\[ =\lambda_1-\frac{\lambda_1}{2} =\frac{\lambda_1}{2}. \]

Percentage decrease
\[ = \frac{\frac{\lambda_1}{2}}{\lambda_1}\times100 = 50%. \]



Step 4: Write the final answer.

The de Broglie wavelength becomes half of its original value.

Hence, it decreases by
\[ 50%. \]

Therefore,
\[ \boxed{(D)} \]

is the correct answer. Quick Tip: Remember the important relation \[ \lambda\propto\frac{1}{\sqrt{K}}. \] If kinetic energy becomes \(n\) times, the wavelength becomes \(1/\sqrt{n}\) times.

Concept:

The spectral lines of hydrogen are classified into different series according to the final energy level to which the electron falls.

The major hydrogen series are:



Each series lies in a particular region of the electromagnetic spectrum.



Step 1: Recall the spectral regions of the hydrogen series.

The important hydrogen series are:
\[ Lyman Series \rightarrow Ultraviolet Region \]
\[ Balmer Series \rightarrow Visible Region \]
\[ Paschen Series \rightarrow Infrared Region \]
\[ Brackett Series \rightarrow Infrared Region \]
\[ Pfund Series \rightarrow Infrared Region \]



Step 2: Identify the Paschen series.

The Paschen series corresponds to transitions ending at
\[ n=3. \]

These transitions have relatively smaller energy differences than the Balmer and Lyman series.

Smaller energy differences correspond to longer wavelengths.

Longer wavelengths lie in the infrared region.



Step 3: Write the final conclusion.

Therefore, the Paschen series is found in the
\[ \boxed{Infrared Region}. \]

Hence,
\[ \boxed{(A)} \]

is the correct answer. Quick Tip: Remember the sequence: Lyman \(\rightarrow\) Ultraviolet Balmer \(\rightarrow\) Visible Paschen, Brackett, Pfund \(\rightarrow\) Infrared

Concept:

A p-n junction consists of:
\[ p-region \]

and
\[ n-region. \]

At the junction, diffusion of charge carriers creates a depletion region containing immobile ions.

This depletion region behaves as an insulating layer and possesses very high resistance compared to the p-region and n-region.



Step 1: Understand reverse biasing.

In reverse bias,
\[ p-side \]

is connected to the negative terminal and
\[ n-side \]

is connected to the positive terminal.

The external voltage increases the width of the depletion layer.



Step 2: Analyze the resistance of different regions.

The p-region and n-region contain majority charge carriers and therefore possess relatively low resistance.

The depletion region contains almost no free charge carriers and therefore has very high resistance.



Step 3: Apply the voltage division principle.

In any circuit, the larger voltage drop occurs across the region having the larger resistance.

Since the depletion region offers maximum resistance,
\[ most of the applied reverse voltage \]

appears across the depletion region.



Step 4: Write the final conclusion.

Hence, in a reverse-biased p-n junction diode, the applied voltage mostly drops across the
\[ \boxed{depletion region}. \]

Therefore,
\[ \boxed{(C)} \]

is the correct answer. Quick Tip: In reverse bias, the depletion layer widens and behaves like a high-resistance region. Therefore, almost the entire reverse voltage appears across the depletion region.

Concept:

According to Kirchhoff's loop rule, the algebraic sum of potential differences around a closed loop is zero.
\[ \sum \Delta V = 0 \]

This implies that the net work done per unit charge in moving around a complete closed loop is zero.



Step 1: Examine the Assertion.

The assertion states that the work done in taking a unit charge around a complete closed loop containing cells and resistors is zero.

According to Kirchhoff's voltage law,
\[ \sum \Delta V = 0 \]

for a complete closed loop.

Hence, the net work done per unit charge after completing one round of the circuit is zero.

Therefore, the Assertion is true.



Step 2: Examine the Reason.

The reason states that the potential at a point depends on the location of the point in the loop and after completing one round the charge comes back to the starting point.

The second statement that the charge returns to the starting point is correct.

However, the statement
\[ ``The potential at a point depends on the location of the point in the loop'' \]

is not the reason for the net work done becoming zero.

The actual reason is that potential is a state function and the algebraic sum of all potential changes around a closed loop is zero.

Thus the given Reason is considered false.



Step 3: Final conclusion.

Assertion is true but Reason is false.
\[ \boxed{(C)} \] Quick Tip: Always remember Kirchhoff's Loop Rule: \[ \sum \Delta V = 0 \] around any closed circuit loop. Therefore, net work done per unit charge in one complete round is zero.

Concept:

Ferromagnetic materials possess strong spontaneous magnetisation because magnetic dipoles align in domains.

When the temperature reaches the Curie temperature, thermal agitation destroys the ordered alignment of domains and the material becomes paramagnetic.



Step 1: Examine the Assertion.

Ferromagnetic materials such as iron, cobalt and nickel lose their ferromagnetic properties above the Curie temperature.

Beyond this temperature, thermal motion prevents permanent alignment of magnetic domains.

Therefore, they behave as paramagnetic substances.

Hence, the Assertion is true.



Step 2: Examine the Reason.

Near the Curie temperature, spontaneous magnetisation falls rapidly and disappears at the Curie point.

Thus, the disappearance of magnetisation is often described as abrupt rather than gradual.

Hence, the Reason is also true.



Step 3: Check whether the Reason explains the Assertion.

The actual reason for conversion into a paramagnetic state is thermal agitation destroying domain alignment.

The statement regarding abrupt disappearance of magnetisation does not explain why the substance becomes paramagnetic.

Therefore, both statements are true but the Reason is not the correct explanation.
\[ \boxed{(B)} \] Quick Tip: Above the Curie temperature, ferromagnetic materials lose domain alignment and become paramagnetic due to thermal agitation.

Concept:

The focal length of a lens in a medium is given by the lens maker's formula
\[ \frac{1}{f} = \left( \frac{n_{lens}}{n_{medium}} -1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right). \]

The focal length depends on both refractive indices and radii of curvature.



Step 1: Examine the Assertion.

When a glass lens is immersed in water,
\[ \frac{n_{lens}}{n_{medium}} \]

decreases because water has a refractive index greater than air.

Therefore,
\[ \frac{1}{f} \]

decreases and the focal length increases.

Since power
\[ P=\frac{1}{f}, \]

the power decreases.

Hence, the converging power does not increase.

The Assertion is false.



Step 2: Examine the Reason.

The focal length does not depend only on radii of curvature.

It also depends on the refractive index of the lens material and the surrounding medium.

Hence, the Reason is false.



Step 3: Final conclusion.

Both Assertion and Reason are false.
\[ \boxed{(D)} \] Quick Tip: When a convex lens is immersed in water, its power decreases because the refractive index contrast between lens and surrounding medium decreases.

Concept:

Conductivity of a semiconductor depends on both the concentration and mobility of charge carriers.
\[ \sigma = nq\mu \]

where
\[ n=carrier concentration, \]
\[ q=charge, \]

and
\[ \mu=mobility. \]



Step 1: Examine the Assertion.

In n-type semiconductors, the majority charge carriers are electrons.

In p-type semiconductors, the majority charge carriers are holes.

Electrons generally have greater mobility than holes.

Therefore, for comparable doping levels, n-type semiconductors exhibit higher conductivity.

Hence, the Assertion is true.



Step 2: Examine the Reason.

Electrons move in the conduction band and possess higher mobility than holes moving in the valence band.

Therefore, the Reason is also true.



Step 3: Check the explanation.

The higher mobility of electrons directly leads to greater conductivity in n-type semiconductors.

Thus, the Reason correctly explains the Assertion.
\[ \boxed{(A)} \] Quick Tip: Electrons are more mobile than holes. Hence, n-type semiconductors generally have higher conductivity than p-type semiconductors under similar conditions.

Concept:

An electric dipole placed in a uniform electric field possesses potential energy due to its orientation in the field.

The potential energy of an electric dipole in a uniform electric field is given by
\[ U=-pE\cos\theta \]

where
\[ p=q(2l) \]

is the dipole moment,
\[ E \]

is the magnitude of the electric field and
\[ \theta \]

is the angle between the dipole moment vector and the electric field.

The work done by an external agent in rotating the dipole slowly from one orientation to another is equal to the increase in its potential energy.
\[ W=\Delta U \]

The stable equilibrium position corresponds to minimum potential energy, while the unstable equilibrium position corresponds to maximum potential energy.



Step 1: Calculate the dipole moment.

Magnitude of each charge is
\[ q=1\,\mu C =1\times10^{-6}\,C \]

Distance between the charges is
\[ d=10\,cm =0.10\,m \]

Therefore,
\[ p=qd \]
\[ p=(1\times10^{-6})(0.10) \]
\[ p=1\times10^{-7}\,C\,m \]



Step 2: Determine the potential energy in stable equilibrium.

Stable equilibrium occurs when the dipole aligns parallel to the electric field.

Hence,
\[ \theta=0^\circ \]

Using
\[ U=-pE\cos\theta \]

we get
\[ U_s=-pE\cos0^\circ \]
\[ U_s=-pE \]

Substituting values,
\[ U_s=-(1\times10^{-7})(100) \]
\[ U_s=-10^{-5}\,J \]



Step 3: Determine the potential energy in unstable equilibrium.

Unstable equilibrium occurs when the dipole is antiparallel to the electric field.

Thus,
\[ \theta=180^\circ \]

Hence,
\[ U_u=-pE\cos180^\circ \]
\[ U_u=-pE(-1) \]
\[ U_u=+pE \]
\[ U_u=(1\times10^{-7})(100) \]
\[ U_u=10^{-5}\,J \]



Step 4: Calculate the work done.

Work done by the external agent equals the increase in potential energy.
\[ W=U_u-U_s \]
\[ W=(10^{-5})-(-10^{-5}) \]
\[ W=2\times10^{-5}\,J \]



Step 5: Write the final answer.

Therefore, the work required to rotate the dipole from stable equilibrium to unstable equilibrium is
\[ \boxed{2\times10^{-5}\,J} \] Quick Tip: For rotation of a dipole from stable equilibrium \((0^\circ)\) to unstable equilibrium \((180^\circ)\), \[ W=2pE \] This shortcut directly gives the required work done.

Concept:

The wave theory of light proposed by the Dutch physicist Huygens provides a geometrical method for understanding the propagation of wavefronts.

Using this principle, one can explain reflection, refraction and propagation of light waves.



Step 1: State Huygens' Principle.

Huygens' Principle consists of the following two statements:


Every point on a given wavefront acts as a source of secondary spherical wavelets.

The new wavefront at any later instant is the forward envelope (common tangent surface) of all these secondary wavelets.


Thus, a wavefront continuously advances through the formation of secondary wavelets.



Step 2: Explain the concept of secondary wavelets.

Consider a spherical wavefront emitted from a point source.

Each point on this wavefront behaves as a new source and emits secondary spherical wavelets in all directions.

The forward envelope of these wavelets gives the next position of the wavefront.



Step 3: Justification for the absence of backwave.

If every secondary wavelet propagated equally in all directions, one would expect a backward travelling wave called a backwave.

However, such backwaves are not observed experimentally.

To explain this, Fresnel modified Huygens' theory.

According to the Huygens-Fresnel principle:


Secondary wavelets interfere with one another.
In the backward direction, destructive interference occurs.
In the forward direction, constructive interference occurs.


As a result, the resultant intensity in the backward direction becomes negligible.

Therefore, no observable backwave is produced.



Final Answer:


Every point on a wavefront acts as a source of secondary spherical wavelets.
The forward envelope of these wavelets gives the new wavefront.
The absence of backwave is explained by destructive interference of secondary wavelets in the backward direction and constructive interference in the forward direction. Quick Tip: Remember: Huygens proposed secondary wavelets, while Fresnel explained the absence of backwave through interference of those wavelets.

Concept:
In single-slit diffraction, light passing through a narrow slit of width \(a\) undergoes diffraction due to the interference of secondary wavelets originating from different parts of the same wavefront.
Key mathematical relations:

Condition for Minima (Dark Fringes):
\[ a \sin\theta = \pm n\lambda \quad (n = 1, 2, 3, \ldots) \]
For small angles (\(\sin\theta \approx \theta\)):
\[ \theta = \pm \frac{\lambda}{a}, \pm \frac{2\lambda}{a}, \pm \frac{3\lambda}{a}, \ldots \]

Condition for Secondary Maxima (Bright Fringes):
\[ a \sin\theta \approx \pm \left(n + \frac{1}{2}\right)\lambda \quad (n = 1, 2, 3, \ldots) \]
For small angles:
\[ \theta \approx \pm \frac{3\lambda}{2a}, \pm \frac{5\lambda}{2a}, \pm \frac{7\lambda}{2a}, \ldots \]

Intensity Profile Formula:
\[ I(\theta) = I_0 \left( \frac{\sin\beta}{\beta} \right)^2 \quad where \beta = \frac{\pi a \sin\theta}{\lambda} \]



Step 1: Analyzing key features of the intensity distribution curve.


Central Maximum (\(\theta = 0\)):
At the center (\(\theta = 0\)), all secondary wavelets arrive in phase (\(\beta = 0\)). The intensity reaches its maximum peak, \(I = I_0\).

First Minima (\(\theta = \pm \frac{\lambda}{a}\)):
The path difference between extreme rays is \(\lambda\). The slit can be divided into two equal halves with wavelets cancelling out in pairs, resulting in zero intensity (\(I = 0\)).

Secondary Maxima:

First secondary maximum at \(\theta \approx \pm \frac{3\lambda}{2a}\) with intensity \(I_1 \approx \frac{I_0}{22} \approx 0.045 I_0\).
Second secondary maximum at \(\theta \approx \pm \frac{5\lambda}{2a}\) with intensity \(I_2 \approx \frac{I_0}{61} \approx 0.016 I_0\).

Unlike double-slit interference (where all bright fringes have equal intensity), single-slit secondary maxima decrease rapidly in intensity as \(|\theta|\) increases.




Step 2: Graphical Representation (Intensity vs Angular Position \(\theta\)).

Below is the mathematical plot depiction of the intensity distribution:



Key Characteristics to Note in Drawing:

The central maximum is twice as wide (\(2\lambda / a\)) compared to any secondary maximum (\(\lambda / a\)).
The height (intensity) of the central peak is substantially larger than the secondary peaks. Quick Tip: Remember key single-slit diffraction values: - Central Peak: Position \(\theta = 0\), Intensity \(= I_0\). - Minima Points: \(\theta = \pm \frac{\lambda}{a}, \pm \frac{2\lambda}{a}, \pm \frac{3\lambda}{a}\). - Angular Width of Central Peak: \(2\theta_0 = \frac{2\lambda}{a}\) (Double the width of secondary maxima!).

Concept:
The diffraction pattern formed on a screen placed at distance \(D\) from a single slit of width \(a\) consists of a central bright fringe bounded by the first minima on either side.
Key formulas involved:

Angular position of first minimum:
\[ \sin\theta = \frac{\lambda}{a} \approx \theta \quad (for small angles) \]

Angular width of central maximum (\(\theta_0\)):
\[ 2\theta = \frac{2\lambda}{a} \]

Linear width of central maximum (\(\beta_0\) or \(y_0\)):
\[ \beta_0 = 2 \cdot y_1 = 2 \cdot (D \cdot \theta) = \frac{2\lambda D}{a} \]
where:

\(\lambda = wavelength of light used\)
\(D = distance between the slit and the screen\)
\(a = width of the single slit\)




Step 1: Establishing the functional dependence on screen separation \(D\).

From the linear width formula derived above: \[ \beta_0 = \frac{2\lambda D}{a} \]
Since the wavelength of light (\(\lambda\)) and slit width (\(a\)) are kept constant, the linear width of the central maximum is directly proportional to the distance \(D\): \[ \beta_0 \propto D \]



Step 2: Analyzing the effect of decreasing distance \(D\).

When the separation distance \(D\) between the slit and the screen is decreased :

The factor \(D\) in the numerator decreases.
Consequently, the linear spread of light on the screen decreases proportionally.
Therefore, the linear width of the central maximum (\(\beta_0\)) decreases .


Note on Angular Width vs. Linear Width:
While the linear width (\(\beta_0 = \frac{2\lambda D}{a}\)) decreases with a reduction in \(D\), the angular width (\(2\theta = \frac{2\lambda}{a}\)) remains unchanged because it depends solely on \(\lambda\) and \(a\). Quick Tip: Distinguish between widths in Wave Optics: - Linear Width (\(\beta_0 = \frac{2\lambda D}{a}\)): Proportional to \(D \rightarrow\) Decreases when \(D\) decreases. - Angular Width (\(2\theta = \frac{2\lambda}{a}\)): Independent of \(D \rightarrow\) Remains constant when \(D\) changes!

Concept:

When a source of light is placed inside a denser medium, light rays incident at the liquid-air interface with angle of incidence less than or equal to the critical angle emerge into air.

Rays incident at angles greater than the critical angle suffer total internal reflection and do not emerge.

Therefore, only those rays lying inside a cone of semi-vertical angle equal to the critical angle can come out of the liquid surface.

The intersection of this cone with the liquid surface forms a circular region through which light emerges.

The required area is therefore the area of this circular patch.



Step 1: Determine the critical angle for the liquid-air interface.

For a denser medium of refractive index \(n\) and air of refractive index \(1\),
\[ \sin C=\frac{1}{n}. \]

Given,
\[ n=\sqrt{2}. \]

Therefore,
\[ \sin C=\frac{1}{\sqrt{2}}. \]

Since
\[ \sin 45^\circ=\frac{1}{\sqrt{2}}, \]

we obtain
\[ C=45^\circ. \]

Thus, the critical angle of the liquid-air interface is
\[ 45^\circ. \]



Step 2: Determine the radius of the circular region.

Let
\[ h=1\,m \]

be the depth of the bulb.

The extreme ray that can emerge from the liquid strikes the surface at the critical angle.

From the right triangle formed,
\[ \tan C=\frac{r}{h}, \]

where \(r\) is the radius of the circular patch.

Substituting values,
\[ \tan45^\circ=\frac{r}{1}. \]

Since
\[ \tan45^\circ=1, \]

we obtain
\[ r=1\,m. \]



Step 3: Calculate the area of the circular patch.

Area through which light emerges is
\[ A=\pi r^2. \]

Substituting \(r=1\,m\),
\[ A=\pi(1)^2. \]
\[ A=\pi\,m^2. \]

Hence,
\[ A\approx3.14\,m^2. \]



Step 4: Write the final answer.

Therefore, the area of the liquid surface through which light emerges is
\[ \boxed{\pi\,m^2} \]

or
\[ \boxed{3.14\,m^2}. \] Quick Tip: For such total internal reflection problems, \[ r=h\tan C \] and \[ A=\pi r^2. \] Always calculate the critical angle first and then determine the radius of the illuminated circular region.

Concept:

The Geiger-Marsden experiment, commonly known as Rutherford's alpha-particle scattering experiment, played a crucial role in the development of the nuclear model of the atom.

In this experiment, alpha particles were directed towards a thin gold foil and the number of scattered particles was measured at different scattering angles.

The observations obtained from the experiment provided direct information about the internal structure of atoms.



Step 1: Describe the variation of scattered particles with scattering angle.

The number of alpha particles detected decreases rapidly as the scattering angle increases.

A very large number of particles are detected at small scattering angles.

Only a few particles are detected at large scattering angles.

The graph is represented schematically as:



The curve starts with a large value of \(N\) for small \(\theta\) and decreases sharply as \(\theta\) increases.



Step 2: Interpret the observation of small-angle scattering.

Most alpha particles pass through the foil without any appreciable deflection.

This indicates that most of the volume of an atom contains no matter capable of significantly deflecting alpha particles.

Hence,
\[ \boxed{Most of the atom is empty space.} \]



Step 3: Interpret the observation of large-angle scattering.

A very small fraction of alpha particles are scattered through large angles.

Some are even reflected back.

Such large deflections can occur only if the positive charge and most of the mass of the atom are concentrated in a very small region.

Therefore,
\[ \boxed{Positive charge and mass are concentrated in a tiny nucleus.} \]



Step 4: State the conclusions.

The experiment led to the following important conclusions:


Most of the space inside an atom is empty.
Nearly the entire positive charge and almost all the mass of the atom are concentrated in a very small central nucleus.


These conclusions formed the basis of Rutherford's nuclear model of the atom. Quick Tip: Remember Rutherford's two most important conclusions: Atom is mostly empty space. Positive charge and mass are concentrated in a tiny nucleus.

Concept:

In an intrinsic semiconductor,
\[ n_i=p_i, \]

where
\[ n_i \]

is the intrinsic electron concentration and
\[ p_i \]

is the intrinsic hole concentration.

For any semiconductor in thermal equilibrium, the law of mass action states that
\[ np=n_i^2, \]

where
\[ n \]

is the electron concentration after doping and
\[ p \]

is the hole concentration after doping.

This relation remains valid even after doping.



Step 1: Determine the intrinsic carrier concentration.

Given intrinsic hole concentration,
\[ p_i=5\times10^8\,m^{-3}. \]

For an intrinsic semiconductor,
\[ n_i=p_i. \]

Therefore,
\[ n_i=5\times10^8\,m^{-3}. \]



Step 2: Apply the mass action law.

The law of mass action gives
\[ np=n_i^2. \]

Substituting the given values,
\[ (4\times10^{12})p = (5\times10^8)^2. \]



Step 3: Calculate the square of the intrinsic concentration.
\[ (5\times10^8)^2 = 25\times10^{16} = 2.5\times10^{17}. \]

Hence,
\[ (4\times10^{12})p = 2.5\times10^{17}. \]



Step 4: Calculate the new hole concentration.
\[ p = \frac{2.5\times10^{17}} {4\times10^{12}}. \]
\[ p = 0.625\times10^5. \]
\[ p = 6.25\times10^4\,m^{-3}. \]

Thus, the new hole concentration is
\[ \boxed{6.25\times10^4\,m^{-3}}. \]



Step 5: Identify the type of semiconductor.

After doping,
\[ n=4\times10^{12}\,m^{-3} \]

whereas
\[ p=6.25\times10^4\,m^{-3}. \]

Clearly,
\[ n \gg p. \]

Therefore, electrons are the majority charge carriers.

Hence, the doped semiconductor is an
\[ \boxed{n-type semiconductor}. \]



Step 6: Write the final answer.

The new hole concentration is
\[ \boxed{6.25\times10^4\,m^{-3}} \]

and the semiconductor formed is
\[ \boxed{n-type semiconductor}. \] Quick Tip: For semiconductor numericals involving carrier concentrations, always use \[ np=n_i^2. \] If \(n>p\), the semiconductor is n-type. If \(p>n\), the semiconductor is p-type.

Concept:

Kirchhoff's Rules are:


Kirchhoff's Current Law (KCL): The algebraic sum of currents at any junction is zero.
\[ \sum I_{in}=\sum I_{out} \]

Kirchhoff's Voltage Law (KVL): The algebraic sum of all potential differences around any closed loop is zero.
\[ \sum \Delta V=0 \]


Step 1: Assume mesh currents.

Let

\(I_1\) be the clockwise current in the upper loop.
\(I_2\) be the clockwise current in the lower loop.


The current through the common branch \(FC\) is
\[ I_{FC}=I_2-I_1. \]

Step 2: Apply KVL to the upper loop.


Traversing the upper loop clockwise,
\[ -5+3-I_1(1)-3I_1+(I_2-I_1)+2=0. \]

Hence,
\[ 5I_1-I_2=0 \]

or
\[ 5I_1=I_2. \]

Step 3: Apply KVL to the lower loop.


Traversing the lower loop clockwise,
\[ -2-(I_2-I_1)-4I_2=0. \]

Therefore,
\[ I_1-5I_2=2. \]

Substituting \(I_2=5I_1\),
\[ I_1-25I_1=2 \]
\[ -24I_1=2 \]
\[ I_1=-\frac{1}{12} A. \]

Hence,
\[ I_2=5I_1=-\frac{5}{12} A. \]

Step 4: Calculate the current through branch \(FC\).

\[ I_{FC}=I_2-I_1 \]
\[ =-\frac{5}{12}-\left(-\frac{1}{12}\right) =-\frac{4}{12} =-\frac{1}{3} A. \]

The negative sign indicates that the actual current flows from \(C\) to \(F\) with magnitude
\[ \boxed{\frac{1}{3} A}. \]

Thus, the current through branch \(FC\) is
\[ \boxed{\frac{1}{3} A (from C to F)}. \] Quick Tip: When applying Kirchhoff's Voltage Law (KVL), a powerful shortcut is the Nodal Potential Method. By setting one node (such as \(F\)) to \(0 V\), you can find all other node potentials quickly without solving messy multi-variable loop equations.

Concept:

Magnetic Field due to a Long Straight Wire: The magnitude of the magnetic field \(B\) at a perpendicular distance \(d\) from an infinitely long straight conductor carrying current \(I\) is given by Ampere's Law:
\[ B = \frac{\mu_0 I}{2\pi d} \]
The direction of this magnetic field is determined by the Right-Hand Thumb Rule.

Magnetic Force between Parallel Wires: The magnetic force per unit length (\(F/L\)) experienced by a conductor carrying current \(I_1\) due to another parallel conductor carrying current \(I_2\) separated by a distance \(d\) is:
\[ \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} \]
Parallel currents flowing in the same direction attract each other, whereas currents flowing in opposite directions repel each other.



Net magnetic field at a point on conductor 1.

Conductor 1 is located at the top. The magnetic field at any point on conductor 1 is the vector sum of the fields produced by conductor 2 and conductor 3.


Field due to Conductor 2 (\(B_2\)): Conductor 2 is at a distance \(d\) below conductor 1 and carries a current \(I\) in the \(+x\) direction. Applying the Right-Hand Thumb Rule, pointing the thumb along \(+x\) causes the fingers to curl out of the page (\(+z\) direction) at the position of conductor 1.
\[ \vec{B}_2 = \frac{\mu_0 I}{2\pi d} \hat{k} \]

Field due to Conductor 3 (\(B_3\)): Conductor 3 is at a total distance of \(2d\) below conductor 1 and carries a current \(3I\) in the \(-x\) direction. Pointing the right thumb along \(-x\) causes the fingers to curl into the page (\(-z\) direction) at the position of conductor 1.
\[ \vec{B}_3 = -\frac{\mu_0 (3I)}{2\pi (2d)} \hat{k} = -\frac{3\mu_0 I}{4\pi d} \hat{k} \]


The net magnetic field \(\vec{B}_{net}\) is the vector sum of these two components: \[ \vec{B}_{net} = \vec{B}_2 + \vec{B}_3 = \frac{\mu_0 I}{2\pi d} \hat{k} - \frac{3\mu_0 I}{4\pi d} \hat{k} \]
Taking a common denominator of \(4\pi d\): \[ \vec{B}_{net} = \left( \frac{2\mu_0 I - 3\mu_0 I}{4\pi d} \right) \hat{k} = -\frac{\mu_0 I}{4\pi d} \hat{k} \]
Thus, the magnitude of the net magnetic field is \(\frac{\mu_0 I}{4\pi d}\) and its direction is pointing into the plane of the page (\(-\hat{k}\)).


Part (ii): Net magnetic force acting on unit length of conductor 1.

We can calculate the force per unit length using the magnetic force equation \(\vec{F}/L = \vec{I} \times \vec{B}_{net}\):
The current vector for unit length of conductor 1 is: \[ \vec{I}_1 = 2I \hat{i} \]
Using the net magnetic field calculated in part (i): \[ \frac{\vec{F}}{L} = \vec{I}_1 \times \vec{B}_{net} = (2I \hat{i}) \times \left( -\frac{\mu_0 I}{4\pi d} \hat{k} \right) \]
Factoring out the scalar components: \[ \frac{\vec{F}}{L} = - \frac{2\mu_0 I^2}{4\pi d} (\hat{i} \times \hat{k}) \]
Since \(\hat{i} \times \hat{k} = -\hat{j}\): \[ \frac{\vec{F}}{L} = - \frac{\mu_0 I^2}{2\pi d} (-\hat{j}) = \frac{\mu_0 I^2}{2\pi d} \hat{j} \]
Thus, the net force per unit length has a magnitude of \(\frac{\mu_0 I^2}{2\pi d}\) and is directed vertically upwards (\(+\hat{j}\) direction). Quick Tip: For parallel wires, remember the simple rule: Like currents attract, unlike currents repel}. Conductor 1 and 2 carry currents in the same direction, so 2 attracts 1 upwards (\(+\hat{j}\)). Conductor 1 and 3 carry opposite currents, so 3 repels 1 upwards (\(+\hat{j}\)). Simply add their scalar forces together!

Concept:

Magnetic Field due to a Long Straight Wire: The magnitude of the magnetic field \(B\) at a perpendicular distance \(d\) from an infinitely long straight conductor carrying current \(I\) is given by Ampere's Law:
\[ B = \frac{\mu_0 I}{2\pi d} \]
The direction of this magnetic field is determined by the Right-Hand Thumb Rule.

Magnetic Force between Parallel Wires: The magnetic force per unit length (\(F/L\)) experienced by a conductor carrying current \(I_1\) due to another parallel conductor carrying current \(I_2\) separated by a distance \(d\) is:
\[ \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} \]
Parallel currents flowing in the same direction attract each other, whereas currents flowing in opposite directions repel each other.




Part (ii): Net magnetic force acting on unit length of conductor 1.

We can calculate the force per unit length using the magnetic force equation \(\vec{F}/L = \vec{I} \times \vec{B}_{net}\):
The current vector for unit length of conductor 1 is: \[ \vec{I}_1 = 2I \hat{i} \]
Using the net magnetic field calculated in part (i): \[ \frac{\vec{F}}{L} = \vec{I}_1 \times \vec{B}_{net} = (2I \hat{i}) \times \left( -\frac{\mu_0 I}{4\pi d} \hat{k} \right) \]
Factoring out the scalar components: \[ \frac{\vec{F}}{L} = - \frac{2\mu_0 I^2}{4\pi d} (\hat{i} \times \hat{k}) \]
Since \(\hat{i} \times \hat{k} = -\hat{j}\): \[ \frac{\vec{F}}{L} = - \frac{\mu_0 I^2}{2\pi d} (-\hat{j}) = \frac{\mu_0 I^2}{2\pi d} \hat{j} \]
Thus, the net force per unit length has a magnitude of \(\frac{\mu_0 I^2}{2\pi d}\) and is directed vertically upwards (\(+\hat{j}\) direction). Quick Tip: For parallel wires, remember the simple rule: Like currents attract, unlike currents repel}. Conductor 1 and 2 carry currents in the same direction, so 2 attracts 1 upwards (\(+\hat{j}\)). Conductor 1 and 3 carry opposite currents, so 3 repels 1 upwards (\(+\hat{j}\)). Simply add their scalar forces together!

Concept:
According to Faraday's Law of Electromagnetic Induction, whenever there is a change in magnetic flux linked with a closed circuit or loop, an electromotive force (emf) is induced in it.
Key physical laws and relations involved:

Magnetic Flux (\(\Phi_B\)): The total magnetic field passing normally through a surface of area \(A\):
\[ \Phi_B = \vec{B} \cdot \vec{A} = B A \cos\theta \]
where \(\theta\) is the angle between the magnetic field vector \(\vec{B}\) and the normal vector to the area \(\vec{A}\). Assuming the loop plane is perpendicular to the magnetic field (\(\theta = 0^\circ\)), \(\cos 0^\circ = 1\):
\[ \Phi_B = B \cdot A \]

Faraday's Law of Electromagnetic Induction:
\[ \varepsilon = -\frac{d\Phi_B}{dt} \]
The negative sign represents Lenz's Law, indicating that the induced emf opposes the change in magnetic flux that produces it.



Step 1: Calculating the area and magnetic flux linked with the rectangular loop.

Given:

Dimensions of the rectangular loop: length \(= l\), breadth \(= b\).
Area of the rectangular loop: \(A = l \times b\).
Magnetic field variation with time \(t\): \(B(t) = B_0 \sin \omega t\).


Assuming the magnetic field is uniform across the area of the loop and perpendicular to the plane of the loop (\(\theta = 0^\circ\)), the magnetic flux \(\Phi_B\) at any instant \(t\) is: \[ \Phi_B(t) = B(t) \cdot A = (B_0 \sin \omega t) \cdot (l \cdot b) \] \[ \Phi_B(t) = B_0 l b \sin \omega t \]



Step 2: Deriving the expression for induced electromotive force (emf).

Differentiating the magnetic flux \(\Phi_B(t)\) with respect to time \(t\) using Faraday's Law: \[ \varepsilon = -\frac{d\Phi_B}{dt} \]
Substitute \(\Phi_B(t) = B_0 l b \sin \omega t\): \[ \varepsilon = -\frac{d}{dt} \left( B_0 l b \sin \omega t \right) \]
Since \(B_0\), \(l\), and \(b\) are constants with respect to time \(t\): \[ \varepsilon = - B_0 l b \frac{d}{dt} (\sin \omega t) \]
Applying the chain rule for differentiation (\(\frac{d}{dt} \sin \omega t = \omega \cos \omega t\)): \[ \varepsilon = - B_0 l b \omega \cos \omega t \]

Alternatively, expressing the peak/maximum induced emf as \(\varepsilon_0 = B_0 l b \omega\): \[ \varepsilon = -\varepsilon_0 \cos \omega t \]
The magnitude of the induced emf is given by: \[ |\varepsilon| = B_0 l b \omega \cos \omega t \] Quick Tip: Remember the differentiation steps for induced EMF: - \(\Phi_B = B_0 A \sin \omega t\) - \(\varepsilon = -\frac{d\Phi_B}{dt} = -B_0 A \omega \cos \omega t\) - Peak EMF \(\varepsilon_0 = B_0 A \omega = B_0 l b \omega\).

Concept:
When a sinusoidally varying electromotive force is induced in a closed conductive circuit of resistance \(R\), an alternating current (AC) flows through the circuit.
Key principles involved:

Ohm's Law for Instantaneous Current:
\[ i(t) = \frac{\varepsilon(t)}{R} \]

Peak Current (\(I_0\)): The maximum amplitude of alternating current:
\[ I_0 = \frac{\varepsilon_0}{R} \]

Effective / Root Mean Square Current (\(I_{rms}\)): The equivalent direct current that produces the same heating effect in a given resistor over a complete cycle:
\[ I_{rms} = \frac{I_0}{\sqrt{2}} \]



Step 1: Determining the instantaneous and peak current in the loop.

From sub-part 23.(b)(i), the induced emf in the loop at any time \(t\) is: \[ \varepsilon = - B_0 l b \omega \cos \omega t \]
The maximum value (peak amplitude) of the induced emf \(\varepsilon_0\) is: \[ \varepsilon_0 = B_0 l b \omega \]
Using Ohm's Law, the peak current \(I_0\) flowing through the rectangular loop of resistance \(R\) is: \[ I_0 = \frac{\varepsilon_0}{R} = \frac{B_0 l b \omega}{R} \]



Step 2: Calculating the effective (rms) value of current.

The effective value of current (also known as the root mean square or rms current \(I_{rms}\)) for a sinusoidal AC waveform is related to peak current \(I_0\) by: \[ I_{rms} = \frac{I_0}{\sqrt{2}} \]
Substituting the value of peak current \(I_0 = \frac{B_0 l b \omega}{R}\): \[ I_{rms} = \frac{\left(\frac{B_0 l b \omega}{R}\right)}{\sqrt{2}} \] \[ I_{rms} = \frac{B_0 l b \omega}{\sqrt{2} R} \]

Hence, the effective value of current flowing in the loop is \(\frac{B_0 l b \omega}{\sqrt{2} R}\). Quick Tip: Formula relationship for AC quantities: - Peak Current \(I_0 = \frac{\varepsilon_0}{R} = \frac{B_0 l b \omega}{R}\) - Effective/RMS Current \(I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{B_0 l b \omega}{\sqrt{2} R}\).

Concept:
This problem involves Faraday's Law of Electromagnetic Induction and Lenz's Law, describing how changing magnetic flux produces an electromotive force (emf) and current in a closed loop.


Magnetic Flux (\(\Phi\)): The magnetic flux passing through a loop of area \(A\) completely inside a uniform magnetic field \(B\) directed perpendicular to its plane is given by:
\[ \Phi = B \cdot A \]
When the loop is partially outside, \(A\) represents the shaded region of the loop still submerged in the magnetic field.

Faraday's Law of Induction: The magnitude of the induced electromotive force (\(\varepsilon\)) is directly proportional to the time rate of change of magnetic flux through the circuit:
\[ \varepsilon = -\frac{d\Phi}{dt} \]

Lenz's Law: The direction of the induced current is always such that it sets up an opposing magnetic field to counteract the change in original magnetic flux that created it.



Step 1: Determine the direction of the induced current as the loop exits the field.

As the square loop \(MNOP\) is pulled horizontally to the right out of the uniform inward magnetic field (indicated by the \(\times\) symbols):

The effective surface area of the loop remaining inside the magnetic field decreases continuously over time.
Consequently, the inward magnetic flux passing through the interior area of the loop decreases.
According to Lenz's Law, the induced current must create an auxiliary magnetic field pointing into the plane of the page (\(\times\)) to reinforce the diminishing flux.
Applying the Right-Hand Grip Rule, a clockwise current produces a magnetic field directed into the page.

Therefore, the direction of the induced current around the perimeter of the loop is clockwise (along the path \(M \rightarrow N \rightarrow O \rightarrow P \rightarrow M\)).


Step 2: Calculate the specific time intervals for the motion.

Let us establish the exact timing of the loop's entry, transit, and exit phases. Let the side length of the square loop be \(l = 25 cm = 0.25 m\). The velocity at which it is pulled is \(v = 25 cm/s = 0.25 m/s\). The total horizontal extent of the uniform magnetic field region is \(L = 1 m\).


Initial Position (\(t = 0\)): The leading vertical edge \(NO\) of the loop is just aligned at the left boundary of the \(1 m\) wide field.

Completely inside the field (\(t_1\)): The loop becomes fully immersed when its trailing vertical edge \(MP\) travels a distance equal to its own width \(l = 0.25 m\):
\[ t_1 = \frac{l}{v} = \frac{25 cm}{25 cm/s} = 1 s \]
Between \(t = 0 s\) and \(t = 1 s\), the flux increases steadily until the loop is completely inside.

Reaching the right boundary (\(t_2\)): The leading vertical edge \(NO\) reaches the right edge of the magnetic field after traveling a full distance of \(L = 1 m\):
\[ t_2 = \frac{L}{v} = \frac{1 m}{0.25 m/s} = 4 s \]
Between \(t = 1 s\) and \(t = 4 s\), the loop remains fully submerged inside the uniform field. The flux stays constant, so the induced emf is zero during this phase.

Completely outside the field (\(t_3\)): The loop begins exiting at \(t = 4 s\). The trailing edge \(MP\) clears the right edge of the field after traveling an additional distance equal to the loop width \(l = 0.25 m\):
\[ t_3 = \frac{L + l}{v} = \frac{1 m + 0.25 m}{0.25 m/s} = 4 s + 1 s = 5 s \]



Step 3: Determine the duration for which the current persists.

An induced current only flows when there is a non-zero induced electromotive force, which requires a changing magnetic flux (\(\frac{d\Phi}{dt} \neq 0\)). As found in Step 2, the loop is exiting the field from \(t = 4 s\) to \(t = 5 s\).

The duration (\(\Delta t\)) for which the loop undergoes this exit phase is: \[ \Delta t = t_3 - t_2 = 5 s - 4 s = 1 s \]
Thus, the induced current during the exit phase persists for exactly \(1 second\).


Step 4: Formulating values for the graphs of Flux and Induced EMF.

Let us model the equations mathematically over the intervals:

Interval \(0 \le t \le 1 s\) (Entry phase):
The length of the loop inside the field increases as \(x(t) = v \cdot t\).
\[ \Phi(t) = B \cdot A(t) = B \cdot l \cdot (vt) = B l v t \]
The maximum flux occurs at \(t = 1 s\): \(\Phi_{max} = B l^2 = B \cdot (0.25)^2 = 0.0625 B\).
The induced emf magnitude is:
\[ |\varepsilon| = \left|\frac{d\Phi}{dt}\right| = Blv \quad (constant value) \]

Interval \(1 s < t \le 4 s\) (Fully immersed phase):
The area inside the field remains constant and equal to the full area of the square, \(l^2\).
\[ \Phi(t) = B \cdot l^2 = \Phi_{max} \quad (constant flat line) \]
Since the flux is steady:
\[ |\varepsilon| = \frac{d\Phi}{dt} = 0 \]

Interval \(4 s < t \le 5 s\) (Exit phase):
The length inside decreases linearly. The remaining length inside is \(l - v(t - 4)\).
\[ \Phi(t) = B \cdot l \cdot [l - v(t - 4)] \]
This is a linearly decreasing line from \(\Phi_{max}\) down to \(0\).
The induced emf magnitude during this exit window is:
\[ |\varepsilon| = \left|\frac{d\Phi}{dt}\right| = Blv \quad (constant value) \]



Step 5: Graphical representation of the results.

Based on our calculated functions, the plots look like this:


Magnetic Flux (\(\Phi\)) vs Time (\(t\)):

From \(t = 0\) to \(t = 1 s\): Straight line rising linearly from \(0\) to \(\Phi_{max}\).
From \(t = 1 s\) to \(t = 4 s\): Flat horizontal line staying at \(\Phi_{max}\).
From \(t = 4 s\) to \(t = 5 s\): Straight line falling linearly from \(\Phi_{max}\) to \(0\).


Magnitude of Induced EMF (\(|\varepsilon|\)) vs Time (\(t\)):

From \(t = 0\) to \(t = 1 s\): Constant positive horizontal step at a value of \(Blv\).
From \(t = 1 s\) to \(t = 4 s\): Drops down to zero and runs flat along the time axis.
From \(t = 4 s\) to \(t = 5 s\): Jumps back up to the constant positive horizontal step value of \(Blv\). Quick Tip: Whenever a closed loop moves through a localized uniform magnetic field of width \(L\), the time taken to enter or leave depends solely on the size of the loop itself (\(\Delta t = side/v\)), whereas the duration of zero induction inside depends on the gap width (\(t_{flat} = (L - side)/v\)). This structural intuition makes sketching verification values instantly straightforward!

Concept:
This problem involves Faraday's Law of Electromagnetic Induction and Lenz's Law, describing how changing magnetic flux produces an electromotive force (emf) and current in a closed loop.


Magnetic Flux (\(\Phi\)): The magnetic flux passing through a loop of area \(A\) completely inside a uniform magnetic field \(B\) directed perpendicular to its plane is given by:
\[ \Phi = B \cdot A \]
When the loop is partially outside, \(A\) represents the shaded region of the loop still submerged in the magnetic field.

Faraday's Law of Induction: The magnitude of the induced electromotive force (\(\varepsilon\)) is directly proportional to the time rate of change of magnetic flux through the circuit:
\[ \varepsilon = -\frac{d\Phi}{dt} \]

Lenz's Law: The direction of the induced current is always such that it sets up an opposing magnetic field to counteract the change in original magnetic flux that created it.



Step 1: Determine the direction of the induced current as the loop exits the field.

As the square loop \(MNOP\) is pulled horizontally to the right out of the uniform inward magnetic field (indicated by the \(\times\) symbols):

The effective surface area of the loop remaining inside the magnetic field decreases continuously over time.
Consequently, the inward magnetic flux passing through the interior area of the loop decreases.
According to Lenz's Law, the induced current must create an auxiliary magnetic field pointing into the plane of the page (\(\times\)) to reinforce the diminishing flux.
Applying the Right-Hand Grip Rule, a clockwise current produces a magnetic field directed into the page.

Therefore, the direction of the induced current around the perimeter of the loop is clockwise (along the path \(M \rightarrow N \rightarrow O \rightarrow P \rightarrow M\)).


Step 2: Calculate the specific time intervals for the motion.

Let us establish the exact timing of the loop's entry, transit, and exit phases. Let the side length of the square loop be \(l = 25 cm = 0.25 m\). The velocity at which it is pulled is \(v = 25 cm/s = 0.25 m/s\). The total horizontal extent of the uniform magnetic field region is \(L = 1 m\).


Initial Position (\(t = 0\)): The leading vertical edge \(NO\) of the loop is just aligned at the left boundary of the \(1 m\) wide field.

Completely inside the field (\(t_1\)): The loop becomes fully immersed when its trailing vertical edge \(MP\) travels a distance equal to its own width \(l = 0.25 m\):
\[ t_1 = \frac{l}{v} = \frac{25 cm}{25 cm/s} = 1 s \]
Between \(t = 0 s\) and \(t = 1 s\), the flux increases steadily until the loop is completely inside.

Reaching the right boundary (\(t_2\)): The leading vertical edge \(NO\) reaches the right edge of the magnetic field after traveling a full distance of \(L = 1 m\):
\[ t_2 = \frac{L}{v} = \frac{1 m}{0.25 m/s} = 4 s \]
Between \(t = 1 s\) and \(t = 4 s\), the loop remains fully submerged inside the uniform field. The flux stays constant, so the induced emf is zero during this phase.

Completely outside the field (\(t_3\)): The loop begins exiting at \(t = 4 s\). The trailing edge \(MP\) clears the right edge of the field after traveling an additional distance equal to the loop width \(l = 0.25 m\):
\[ t_3 = \frac{L + l}{v} = \frac{1 m + 0.25 m}{0.25 m/s} = 4 s + 1 s = 5 s \]



Step 3: Determine the duration for which the current persists.

An induced current only flows when there is a non-zero induced electromotive force, which requires a changing magnetic flux (\(\frac{d\Phi}{dt} \neq 0\)). As found in Step 2, the loop is exiting the field from \(t = 4 s\) to \(t = 5 s\).

The duration (\(\Delta t\)) for which the loop undergoes this exit phase is: \[ \Delta t = t_3 - t_2 = 5 s - 4 s = 1 s \]
Thus, the induced current during the exit phase persists for exactly \(1 second\).


Step 4: Formulating values for the graphs of Flux and Induced EMF.

Let us model the equations mathematically over the intervals:

Interval \(0 \le t \le 1 s\) (Entry phase):
The length of the loop inside the field increases as \(x(t) = v \cdot t\).
\[ \Phi(t) = B \cdot A(t) = B \cdot l \cdot (vt) = B l v t \]
The maximum flux occurs at \(t = 1 s\): \(\Phi_{max} = B l^2 = B \cdot (0.25)^2 = 0.0625 B\).
The induced emf magnitude is:
\[ |\varepsilon| = \left|\frac{d\Phi}{dt}\right| = Blv \quad (constant value) \]

Interval \(1 s < t \le 4 s\) (Fully immersed phase):
The area inside the field remains constant and equal to the full area of the square, \(l^2\).
\[ \Phi(t) = B \cdot l^2 = \Phi_{max} \quad (constant flat line) \]
Since the flux is steady:
\[ |\varepsilon| = \frac{d\Phi}{dt} = 0 \]

Interval \(4 s < t \le 5 s\) (Exit phase):
The length inside decreases linearly. The remaining length inside is \(l - v(t - 4)\).
\[ \Phi(t) = B \cdot l \cdot [l - v(t - 4)] \]
This is a linearly decreasing line from \(\Phi_{max}\) down to \(0\).
The induced emf magnitude during this exit window is:
\[ |\varepsilon| = \left|\frac{d\Phi}{dt}\right| = Blv \quad (constant value) \]



Step 5: Graphical representation of the results.

Based on our calculated functions, the plots look like this:


Magnetic Flux (\(\Phi\)) vs Time (\(t\)):

From \(t = 0\) to \(t = 1 s\): Straight line rising linearly from \(0\) to \(\Phi_{max}\).
From \(t = 1 s\) to \(t = 4 s\): Flat horizontal line staying at \(\Phi_{max}\).
From \(t = 4 s\) to \(t = 5 s\): Straight line falling linearly from \(\Phi_{max}\) to \(0\).


Magnitude of Induced EMF (\(|\varepsilon|\)) vs Time (\(t\)):

From \(t = 0\) to \(t = 1 s\): Constant positive horizontal step at a value of \(Blv\).
From \(t = 1 s\) to \(t = 4 s\): Drops down to zero and runs flat along the time axis.
From \(t = 4 s\) to \(t = 5 s\): Jumps back up to the constant positive horizontal step value of \(Blv\). Quick Tip: Whenever a closed loop moves through a localized uniform magnetic field of width \(L\), the time taken to enter or leave depends solely on the size of the loop itself (\(\Delta t = side/v\)), whereas the duration of zero induction inside depends on the gap width (\(t_{flat} = (L - side)/v\)). This structural intuition makes sketching verification values instantly straightforward!

Concept:

Infrared radiations are electromagnetic waves lying beyond the red end of the visible spectrum. These waves were discovered by William Herschel and form an important part of the electromagnetic spectrum.

Electromagnetic waves are classified according to their wavelength and frequency. Infrared radiations occupy the region between visible light and microwaves.



Step 1: Define infrared waves.

Infrared waves are electromagnetic radiations having wavelengths greater than those of visible red light and smaller than those of microwaves.

Their wavelength range is approximately
\[ 7\times10^{-7}\,m \]

to
\[ 10^{-3}\,m. \]

Thus, infrared radiations lie immediately beyond the red region of the visible spectrum.



Step 2: Explain why infrared waves are called heat waves.

Every hot object continuously emits infrared radiations.

When infrared radiations fall on a substance, they are readily absorbed by the molecules of the substance.

The absorbed energy increases the vibrational and rotational motion of molecules, resulting in an increase in temperature.

Thus, infrared radiations produce a strong heating effect.

Since these radiations are mainly responsible for the heating sensation received from the Sun, heaters and other hot bodies, they are commonly known as
\[ \boxed{Heat Waves}. \]



Step 3: Discuss the properties responsible for their heating effect.

Infrared radiations possess lower frequencies than visible light but carry sufficient energy to excite molecular vibrations.

Most materials absorb infrared radiations efficiently.

As a result, the absorbed radiant energy is converted into thermal energy.

This is the reason why infrared radiations are extensively used wherever controlled heating is required.



Step 4: State any two uses of infrared waves.

Use 1: Infrared Photography

Infrared photography is used for photographing objects in fog, haze, smoke and darkness where visible light photography becomes difficult.



Use 2: Remote Control Devices

Infrared radiations are used in television remotes, air-conditioner remotes and various wireless control systems for transmission of signals.



Other important applications include:


Physiotherapy and heat treatment.
Night vision devices.
Thermal imaging cameras.
Drying of paints and agricultural products.
Satellite and astronomical observations.




Final Answer:

Infrared waves are electromagnetic waves having wavelengths longer than visible red light and shorter than microwaves.

They are called heat waves because they produce a strong heating effect when absorbed by matter.

Any two uses are:


Infrared photography.
Remote control devices. Quick Tip: Remember: Visible Red \(\rightarrow\) Infrared \(\rightarrow\) Microwaves Infrared radiations are strongly absorbed by matter and therefore produce heating effects, which is why they are called heat waves.

Concept:

One of the most revolutionary developments in modern physics was the discovery that matter exhibits both particle nature and wave nature.

This idea was proposed by Louis de Broglie in 1924.

According to de Broglie, every moving material particle is associated with a wave known as the matter wave or de Broglie wave.



Step 1: State the de Broglie hypothesis.

De Broglie proposed that every moving particle possesses an associated wavelength called the de Broglie wavelength.

The de Broglie relation is
\[ \lambda=\frac{h}{p} \]

where
\[ \lambda \]

is the de Broglie wavelength,
\[ h \]

is Planck's constant and
\[ p \]

is the momentum of the particle.



Step 2: Identify the particle aspect in the relation.

The quantity
\[ p=mv \]

represents momentum.

Momentum is a characteristic property of particles.

A particle possesses mass, momentum and energy.

Therefore, the appearance of momentum in the de Broglie relation indicates the particle nature of matter.



Step 3: Identify the wave aspect in the relation.

The quantity
\[ \lambda \]

represents wavelength.

Wavelength is a characteristic property of waves.

Phenomena such as interference, diffraction and superposition are associated with waves.

Therefore, the appearance of wavelength in the relation indicates the wave nature of matter.



Step 4: Explain the dual aspect.

The de Broglie equation connects a wave quantity \((\lambda)\) with a particle quantity \((p)\).

Thus, a single relation simultaneously contains both wave and particle characteristics.

This demonstrates that matter cannot be described exclusively as a particle or exclusively as a wave.

Instead, matter possesses a dual nature.



Step 5: Experimental verification.

The wave nature predicted by de Broglie was later verified experimentally through electron diffraction experiments conducted by Davisson and Germer.

The observation of diffraction of electrons confirmed that particles can exhibit wave behaviour.



Final Answer:

The de Broglie relation
\[ \lambda=\frac{h}{p} \]

connects wavelength, which is a wave property, with momentum, which is a particle property.

Thus, it clearly establishes the dual nature of matter and shows that every moving particle possesses an associated wave character. Quick Tip: The de Broglie relation is the bridge between wave and particle concepts: \[ \lambda=\frac{h}{p} \] Wavelength \(\rightarrow\) Wave Nature Momentum \(\rightarrow\) Particle Nature

Concept:

When electromagnetic radiation falls on a photosensitive surface, electrons may be emitted through the photoelectric effect.

According to Einstein's photoelectric equation,
\[ h\nu=\phi+K_{\max}. \]

If the work function is negligible,
\[ \phi \approx 0. \]

In that case, the entire photon energy is converted into the kinetic energy of the emitted electron.

The de Broglie wavelength of the emitted electron can then be calculated using the kinetic energy obtained from the incident photon.



Step 1: Calculate the energy of the incident photon.

For radiation of wavelength
\[ \lambda, \]

the photon energy is
\[ E=\frac{hc}{\lambda}. \]

Since the work function is negligible,
\[ K=\frac{hc}{\lambda}. \]

Thus, the kinetic energy of the emitted electron is
\[ \boxed{K=\frac{hc}{\lambda}}. \]



Step 2: Relate kinetic energy and momentum of the electron.

For a non-relativistic electron,
\[ K=\frac{p^2}{2m}. \]

Substituting the value of kinetic energy,
\[ \frac{p^2}{2m} = \frac{hc}{\lambda}. \]

Multiplying both sides by \(2m\),
\[ p^2 = \frac{2mhc}{\lambda}. \]

Therefore,
\[ p = \sqrt{\frac{2mhc}{\lambda}}. \]



Step 3: Apply de Broglie relation.

The de Broglie wavelength of the emitted electron is
\[ \lambda_d=\frac{h}{p}. \]

Substituting the value of momentum,
\[ \lambda_d = \frac{h} {\sqrt{\dfrac{2mhc}{\lambda}}}. \]



Step 4: Simplify the expression.
\[ \lambda_d = \sqrt{ \frac{h^2\lambda} {2mhc} }. \]

Cancelling one factor of \(h\),
\[ \lambda_d = \sqrt{ \frac{h\lambda} {2mc} }. \]

Hence,
\[ \boxed{ \lambda_d = \sqrt{\frac{h\lambda}{2mc}} }. \]



Step 5: Write the final answer.

Therefore, the de Broglie wavelength of the emitted photoelectron is
\[ \boxed{ \lambda_d = \sqrt{\frac{h\lambda}{2mc}} }. \] Quick Tip: For photoelectric problems with negligible work function, \[ K=\frac{hc}{\lambda}. \] Then use \[ \lambda_d=\frac{h}{\sqrt{2mK}} \] to obtain the de Broglie wavelength of the emitted electron.

Concept:
Nuclear reactions are governed by fundamental conservation laws that dictate how mass, charge, and energy are distributed between reactants and products.


Law of Conservation of Nucleon Number (Mass Number): The total number of protons and neutrons (nucleons) must be equal before and after a nuclear reaction. In a standard balanced equation \({}_{Z_1}^{A_1}P + {}_{Z_2}^{A_2}Q \rightarrow {}_{Z_3}^{A_3}R + {}_{Z_4}^{A_4}S\), this requires:
\[ A_1 + A_2 = A_3 + A_4 \]

Law of Conservation of Atomic Number (Charge): The total electric charge of the reacting nuclei must equal the total electric charge of the resulting products:
\[ Z_1 + Z_2 = Z_3 + Z_4 \]

Mass-Energy Equivalence and Q-value: In nuclear reactions, the total rest mass of the products is typically less than the total rest mass of the initial reacting nuclei. This difference in mass, known as the mass defect (\(\Delta m\)), is converted directly into nuclear energy. The released energy (\(Q\)) is given by Albert Einstein's mass-energy relationship:
\[ Q = \Delta m \cdot c^2 \]
When the atomic mass units (u) are used, the energy equivalent of 1u is approximately \(931.5 MeV\). Thus, the energy released can be directly computed as:
\[ Q = \Delta m (in u) \times 931.5 MeV \]



Part (a): Find the value of A.

Let us analyze the given nuclear fusion equation involving two deuterium nuclei: \[ {}_{1}^{2}H + {}_{1}^{2}H \rightarrow {}_{2}^{A}X + {}_{0}^{1}n \]

To find the unknown mass number \(A\) of the product nucleus \(X\), we apply the Law of Conservation of Mass Number.
Summing up the total mass numbers on the left-hand side (reactants): \[ \sum A_{reactants} = 2 + 2 = 4 \]

Summing up the total mass numbers on the right-hand side (products): \[ \sum A_{products} = A + 1 \]

Equating the two sides according to the conservation principle: \[ 4 = A + 1 \]
Isolating \(A\) by subtracting 1 from both sides of the equation: \[ A = 4 - 1 \] \[ A = 3 \]

Note on Verification: We can also verify this by checking the conservation of atomic number (\(Z\)): sum Z_{reactants = 1 + 1 = 2 sum Z_{products} = 2 + 0 = 2
Since the atomic number matches perfectly, the product nucleus is an isotope of Helium, specifically Helium-3 (\({}_{2}^{3}He\)). Thus, the value of \(A\) is 3.


Part (b): Calculate the amount of energy released in the reaction.

Step 1: Calculate the total mass of the reactants.

The reactants consist of two identical Deuterium nuclei (\({}_{1}^{2}H\)).
The mass of one Deuterium atom is given as \(m\left({}_{1}^{2}H\right) = 2.014102 u\). \[ Total initial mass (m_{reactants}) = 2 \times m\left({}_{1}^{2}H\right) \] \[ m_{reactants} = 2 \times 2.014102 u \]
Performing the multiplication step-by-step: \[ m_{reactants} = 4.028204 u \]


Step 2: Calculate the total mass of the products.

The products consist of one nucleus \({}_{2}^{3}X\) and one neutron (\({}_{0}^{1}n\)).
We are given: \[ m\left({}_{2}^{3}X\right) = 3.016049 u \] \[ m\left({}_{0}^{1}n\right) = 1.008665 u \]
The total final mass of the products is: \[ Total final mass (m_{products}) = m\left({}_{2}^{3}X\right) + m\left({}_{0}^{1}n\right) \] \[ m_{products} = 3.016049 u + 1.008665 u \]
Adding the numbers carefully by aligning their decimal positions:


\[ m_{products} = 4.024714 u \]


Step 3: Calculate the mass defect (\(\Delta m\)).

The mass defect is the difference between the total mass of the reactants and the total mass of the products: \[ \Delta m = m_{reactants} - m_{products} \]
Substituting our calculated values: \[ \Delta m = 4.028204 u - 4.024714 u \]
Subtracting the decimals:

\[ \Delta m = 0.003490 u = 0.00349 u \]


Step 4: Convert the mass defect into released energy.

We are given that \(1 u = 931.5 MeV/c^2\). Therefore, the energy released (\(E\)) in units of Mega-electron Volts (MeV) is: E = m times 931.5text{  \[ E = 0.00349 \times 931.5 MeV \]
Let us carry out the long multiplication to ensure absolute accuracy: \[ 349 \times 9315 = 3250935 \]
Since there are 5 decimal places in total (\(3\) from \(0.00349\) and \(1\) from \(931.5\)), we position the decimal point: \[ E = 3.250935 MeV \]
Rounding the energy value to three decimal places yields: \[ E \approx 3.251 MeV \]
Thus, the total energy released during this nuclear fusion event is approximately \(3.251 MeV\). Quick Tip: When computing values for mass defects in nuclear reactions, never round off numbers prematurely in intermediate steps! The differences occur past the third or fourth decimal place, so maintaining all six decimal places provided in the given values is essential to avoid significant rounding errors in your final energy calculation.

Concept:
Nuclear reactions are governed by fundamental conservation laws that dictate how mass, charge, and energy are distributed between reactants and products.


Law of Conservation of Nucleon Number (Mass Number): The total number of protons and neutrons (nucleons) must be equal before and after a nuclear reaction. In a standard balanced equation \({}_{Z_1}^{A_1}P + {}_{Z_2}^{A_2}Q \rightarrow {}_{Z_3}^{A_3}R + {}_{Z_4}^{A_4}S\), this requires:
\[ A_1 + A_2 = A_3 + A_4 \]

Law of Conservation of Atomic Number (Charge): The total electric charge of the reacting nuclei must equal the total electric charge of the resulting products:
\[ Z_1 + Z_2 = Z_3 + Z_4 \]

Mass-Energy Equivalence and Q-value: In nuclear reactions, the total rest mass of the products is typically less than the total rest mass of the initial reacting nuclei. This difference in mass, known as the mass defect (\(\Delta m\)), is converted directly into nuclear energy. The released energy (\(Q\)) is given by Albert Einstein's mass-energy relationship:
\[ Q = \Delta m \cdot c^2 \]
When the atomic mass units (u) are used, the energy equivalent of 1u is approximately \(931.5 MeV\). Thus, the energy released can be directly computed as:
\[ Q = \Delta m (in u) \times 931.5 MeV \]



Part (a): Find the value of A.

Let us analyze the given nuclear fusion equation involving two deuterium nuclei: \[ {}_{1}^{2}H + {}_{1}^{2}H \rightarrow {}_{2}^{A}X + {}_{0}^{1}n \]

To find the unknown mass number \(A\) of the product nucleus \(X\), we apply the Law of Conservation of Mass Number.
Summing up the total mass numbers on the left-hand side (reactants): \[ \sum A_{reactants} = 2 + 2 = 4 \]

Summing up the total mass numbers on the right-hand side (products): \[ \sum A_{products} = A + 1 \]

Equating the two sides according to the conservation principle: \[ 4 = A + 1 \]
Isolating \(A\) by subtracting 1 from both sides of the equation: \[ A = 4 - 1 \] \[ A = 3 \]

Note on Verification: We can also verify this by checking the conservation of atomic number (\(Z\)):sum Z_{reactants = 1 + 1 = 2 sum Z_{products} = 2 + 0 = 2 
Since the atomic number matches perfectly, the product nucleus is an isotope of Helium, specifically Helium-3 (\({}_{2}^{3}He\)). Thus, the value of \(A\) is 3.


Part (b): Calculate the amount of energy released in the reaction.

Step 1: Calculate the total mass of the reactants.

The reactants consist of two identical Deuterium nuclei (\({}_{1}^{2}H\)).
The mass of one Deuterium atom is given as \(m\left({}_{1}^{2}H\right) = 2.014102 u\). \[ Total initial mass (m_{reactants}) = 2 \times m\left({}_{1}^{2}H\right) \] \[ m_{reactants} = 2 \times 2.014102 u \]
Performing the multiplication step-by-step: \[ m_{reactants} = 4.028204 u \]


Step 2: Calculate the total mass of the products.

The products consist of one nucleus \({}_{2}^{3}X\) and one neutron (\({}_{0}^{1}n\)).
We are given: \[ m\left({}_{2}^{3}X\right) = 3.016049 u \] \[ m\left({}_{0}^{1}n\right) = 1.008665 u \]
The total final mass of the products is: \[ Total final mass (m_{products}) = m\left({}_{2}^{3}X\right) + m\left({}_{0}^{1}n\right) \] \[ m_{products} = 3.016049 u + 1.008665 u \]
Adding the numbers carefully by aligning their decimal positions:


\[ m_{products} = 4.024714 u \]


Step 3: Calculate the mass defect (\(\Delta m\)).

The mass defect is the difference between the total mass of the reactants and the total mass of the products: \[ \Delta m = m_{reactants} - m_{products} \]
Substituting our calculated values: \[ \Delta m = 4.028204 u - 4.024714 u \]
Subtracting the decimals:

\[ \Delta m = 0.003490 u = 0.00349 u \]


Step 4: Convert the mass defect into released energy.

We are given that \(1 u = 931.5 MeV/c^2\). Therefore, the energy released (\(E\)) in units of Mega-electron Volts (MeV) 
Let us carry out the long multiplication to ensure absolute accuracy: \[ 349 \times 9315 = 3250935 \]
Since there are 5 decimal places in total (\(3\) from \(0.00349\) and \(1\) from \(931.5\)), we position the decimal point: \[ E = 3.250935 MeV \]
Rounding the energy value to three decimal places yields: \[ E \approx 3.251 MeV \]
Thus, the total energy released during this nuclear fusion event is approximately \(3.251 MeV\). Quick Tip: When computing values for mass defects in nuclear reactions, never round off numbers prematurely in intermediate steps! The differences occur past the third or fourth decimal place, so maintaining all six decimal places provided in the given values is essential to avoid significant rounding errors in your final energy calculation.

Concept:

A rectifier is an electronic device used to convert alternating current (AC) into direct current (DC).

In a half-wave rectifier, only one half cycle of the input AC signal is utilized. As a result, the efficiency is low and a large amount of input power is wasted.

To overcome this drawback, a full-wave rectifier is used. A full-wave rectifier utilizes both the positive and negative half cycles of the AC input signal and converts them into pulsating DC.

Consequently, the output current flows through the load resistor in the same direction during both half cycles of the input voltage.

Hence, a full-wave rectifier is more efficient than a half-wave rectifier and produces a higher average output voltage.



Step 1: Draw the circuit diagram of a full-wave rectifier.

A full-wave rectifier consists of:


A centre-tapped transformer.
Two p-n junction diodes \(D_1\) and \(D_2\).
A load resistance \(R_L\).


The circuit diagram is shown below:
\[ \begin{array}{c} Secondary Winding
[0.3cm] A \hspace{1cm} C \hspace{1cm} B \end{array} \]
\[ \begin{array}{c} A \longrightarrow |>| D_1 \longrightarrow +
[0.2cm] \hspace{2.6cm} R_L
[0.2cm] B \longrightarrow |>| D_2 \longrightarrow +
[0.2cm] \hspace{1.8cm} |
[0.1cm] \hspace{1.8cm} C \end{array} \]

where


\(A\) and \(B\) are the ends of the secondary winding.
\(C\) is the centre tap.
\(R_L\) is the load resistor.




Step 2: Explain the working during the positive half cycle.

During the positive half cycle of the AC input:


End \(A\) becomes positive with respect to the centre tap \(C\).
End \(B\) becomes negative with respect to \(C\).
Diode \(D_1\) becomes forward biased.
Diode \(D_2\) becomes reverse biased.


Therefore, only \(D_1\) conducts.

The current path is
\[ A \rightarrow D_1 \rightarrow R_L \rightarrow C. \]

Hence current flows through the load resistor in a particular direction.



Step 3: Explain the working during the negative half cycle.

During the negative half cycle:


End \(B\) becomes positive with respect to \(C\).
End \(A\) becomes negative with respect to \(C\).
Diode \(D_2\) becomes forward biased.
Diode \(D_1\) becomes reverse biased.


Therefore, only \(D_2\) conducts.

The current path is
\[ B \rightarrow D_2 \rightarrow R_L \rightarrow C. \]

Again the current through \(R_L\) flows in the same direction as in the previous half cycle.

Thus, both half cycles contribute to the output current.



Step 4: Explain why the output becomes pulsating DC.

Since current flows through the load resistor in the same direction during both half cycles:


The output voltage never changes polarity.
Both positive and negative half cycles are utilized.
The output consists of successive positive pulses.


Therefore, the output is a pulsating direct voltage.



Step 5: Draw the input and output waveforms.

Input AC waveform



The waveform contains both positive and negative half cycles.



Output waveform of full-wave rectifier



All portions of the waveform remain above the time axis, indicating pulsating DC.



Step 6: State the advantages of a full-wave rectifier.

Compared with a half-wave rectifier:


Both half cycles of AC are utilized.
Average output voltage is higher.
Rectification efficiency is greater.
Ripple content is lower.
Smoother DC output is obtained after filtering.




Final Answer:

A full-wave rectifier uses two diodes and a centre-tapped transformer to convert both half cycles of an AC signal into pulsating DC. During the positive half cycle diode \(D_1\) conducts, while during the negative half cycle diode \(D_2\) conducts. In both cases, current through the load resistor flows in the same direction. Therefore, the output obtained is a pulsating DC voltage, as shown in the output waveform. Quick Tip: In a full-wave rectifier: Both half cycles of AC are utilized. Current through the load always flows in the same direction. Output frequency is twice the input AC frequency. Efficiency is much higher than that of a half-wave rectifier.

Concept:
According to Ohm's Law, the electric current \(I\) flowing through a conductor is directly proportional to the potential difference \(V\) across its ends, provided the physical conditions such as temperature, tension, and material composition remain constant. Mathematically, this relationship is expressed as: \[ V = I \cdot R \quad \implies \quad I = \left(\frac{1}{R}\right)V \]
Here, \(R\) represents the electrical resistance of the material, which acts as the constant of proportionality. On an \(I\) versus \(V\) plot (where \(I\) is plotted on the vertical \(y\)-axis and \(V\) is plotted on the horizontal \(x\)-axis), the slope of the curve is equal to the reciprocal of the resistance (\(Slope = \frac{1}{R}\)).

When electric current passes through any resistor continuously without switching off the circuit, heat is generated inside the material due to Joule heating, which is given by the formula: \[ H = I^2 \cdot R \cdot t \]
This heat energy leads to a rise in the internal temperature of the resistor. For standard conductors, a change in temperature modifies the resistance according to the relationship: \[ R(T) = R_0 [1 + \alpha(T - R_0)] \]
where \(\alpha\) is the temperature coefficient of resistance.


Step 1: Analyzing the material characteristics of Manganin.

Manganin is a specialized alloy typically composed of approximately 84% copper, 12% manganese, and 4% nickel. It belongs to a unique category of materials known as precision resistance alloys. The defining characteristic of manganin is that it possesses an exceptionally low, nearly negligible temperature coefficient of resistance (\(\alpha \approx 0\)).

This implies that even if the internal temperature of a manganin wire increases significantly due to prolonged current flow or continuous operation without switching off the circuit, its electrical resistance \(R\) remains virtually unaltered and stays perfectly stable at its initial value.


Step 2: Determining the behavior of the \(I\)-\(V\) graph based on its material properties.

Since the resistance \(R\) of the manganin resistor stays constant throughout the experiment despite the continuous flow of current and subsequent Joule heating: \[ R = constant \]
The relationship between current \(I\) and voltage \(V\) remains strictly linear at all times: \[ I \propto V \]
Because the slope \(\frac{1}{R}\) is constant and does not change, the graph between the current \(I\) and the potential difference \(V\) must be a straight line passing through the origin, as depicted in option (A). There will be no deviation or bending towards either axis because the material does not exhibit significant non-ohmic heating characteristics under normal laboratory conditions. Quick Tip: Alloys like manganin and constantan are specifically utilized to manufacture standard resistance coils because their resistance values are highly immune to temperature variations. Consequently, their \(I\)-\(V\) characteristics always yield a perfect linear relationship (straight line passing through the origin), adhering strictly to Ohm's law.

Concept:
The unknown resistance \(X\) is determined experimentally by applying Ohm's Law to the recorded values of potential difference (\(V\)) measured by the voltmeter and electric current (\(I\)) measured by the ammeter. The governing formula is: \[ X = \frac{V}{I} \]
In any physical measurement, the quantities \(V\) and \(I\) are prone to experimental uncertainties or instrumental errors (\(\Delta V\) and \(\Delta I\)). When calculating a quantity derived via multiplication or division, the absolute errors do not simply add up or divide; instead, the maximum relative (or fractional) error of the result is equal to the sum of the relative errors of the individual measured quantities.


Step 1: Applying the theory of error propagation to the quotient formula.

To mathematically determine how errors combine, we take the natural logarithm (\(\ln\)) on both sides of the equation \(X = \frac{V}{I}\): \[ \ln(X) = \ln\left(\frac{V}{I}\right) = \ln(V) - \ln(I) \]
Differentiating both sides to find the relationship between small fractional increments yields: \[ \frac{dX}{X} = \frac{dV}{V} - \frac{dI}{I} \]
For estimating the maximum possible fractional error (worst-case uncertainty), the negative signs are replaced with positive signs because errors can accumulate destructively in the same direction: \[ \frac{\Delta X}{X} = \frac{\Delta V}{V} + \frac{\Delta I}{I} \]


Step 2: Evaluating the total error composition.

The expression clearly shows that the total relative error in finding \(X\) is bounded by the combined contributions of the fractional error from the voltmeter measurement (\(\frac{\Delta V}{V}\)) and the fractional error from the ammeter measurement (\(\frac{\Delta I}{I}\)). Thus, any variation or error observed in \(X\) from different sets of readings is due to both devices, and the mathematical framework for maximum limits dictates that it is equal to the combined sum of these fractional errors. Hence, option (C) is the most accurate choice among the given qualitative options. Quick Tip: Whenever a calculated quantity involves a multiplication or division of variables (such as \(R = V/I\) or \(\rho = R A / L\)), always add the relative or percentage errors of each component together to determine the total maximum error. Errors are never subtracted or divided.

Concept:
A rheostat acts as a variable resistor in a circuit. The resistance of a uniform conductor is directly proportional to its effective length connected in the path of the current: \[ R_{rheo} = \rho \frac{l}{A} \]
The total equivalent resistance of a series circuit consisting of a fixed resistance \(X\) and a variable rheostat resistance \(R_{rheo}\) connected to a source of constant electromotive force (\(V_{total}\)) is given by: \[ R_{total} = X + R_{rheo} \]
According to Ohm's Law for the entire circuit, the total circuit current (\(I\)), which is measured directly by the ammeter, is: \[ I = \frac{V_{total}}{R_{total}} = \frac{V_{total}}{X + R_{rheo}} \]


Step 1: Analyzing the variation in rheostat resistance when moving towards terminal P.

When the sliding contact (movable end) of the rheostat is shifted towards the terminal labeled P, the length (\(l\)) of the resistance wire actively included within the closed circuit is reduced. Since resistance is linearly proportional to the active length of the wire: \[ l \downarrow \quad \implies \quad R_{rheo} \downarrow \]
Therefore, moving the slider towards P decreases the electrical resistance contribution of the rheostat.


Step 2: Determining the effect on ammeter and voltmeter readings.

As the rheostat resistance \(R_{rheo}\) drops, the overall net resistance \(R_{total}\) of the series network diminishes: \[ R_{total} = X + R_{rheo} \downarrow \]
Since the supply voltage \(V_{total}\) remains fixed, a lower overall resistance causes the total current \(I\) flowing through the circuit to rise: \[ I = \frac{V_{total}}{R_{total}} \uparrow \]
Thus, the ammeter reading increases .

The voltmeter is connected in parallel across the unknown fixed resistance \(X\) and measures the localized potential drop \(V_X\) across it. By Ohm's Law, this localized potential drop is: \[ V_X = I \cdot X \]
Since the resistance value \(X\) is constant and the passing current \(I\) has increased, the product \(I \cdot X\) must increase: \[ V_X \uparrow \]
Consequently, the voltmeter reading also increases . Therefore, both instrument readings increase simultaneously, corresponding to option (B). Quick Tip: Reducing resistance in a single-loop series circuit always increases the global loop current. For any fixed resistor in that loop, a higher loop current translates directly to a larger individual potential drop (\(V = IR\)). Hence, both the ammeter and voltmeter readings go up.

Concept:
When multiple segments of a conductor are joined end-to-end (in a series arrangement), the principle of conservation of charge requires that the total electrical current (\(I\)) passing through each cross-section per unit time remains strictly identical. Therefore, the same current \(I\) flows through segment AC, segment CD, and segment DB: \[ I_{AC} = I_{CD} = I_{DB} = I \]
The relationship between macroscopic electric current \(I\) and microscopic electron drift velocity \(v_d\) inside a conductor is given by the formula: \[ I = n \cdot e \cdot A \cdot v_d \]
where:

\(n\) is the number density of free conduction electrons (dependent only on the type of metal).
\(e\) is the elementary charge of a single electron.
\(A\) is the cross-sectional area of the specific segment (\(A = \pi \cdot r_{wire}^2\)).
\(v_d\) is the average drift velocity of electrons.



Step 1: Expressing drift velocity in terms of the variable radius.

Since all three segments (AC, CD, and DB) are made from the exact same metal, the electron concentration \(n\) is identical across all parts. Isolating the drift velocity \(v_d\) from our fundamental equation gives: \[ v_d = \frac{I}{n \cdot e \cdot A} = \frac{I}{n \cdot e \cdot (\pi \cdot r_{wire}^2)} \]
Given that \(I\), \(n\), \(e\), and \(\pi\) are completely constant throughout this specific setting, we can see that the drift velocity is inversely proportional to the square of the segment's radius: \[ v_d \propto \frac{1}{r_{wire}^2} \]


Step 2: Comparing the drift velocities for the three sections.

Let us write down the explicit radii given for each of the three sequential parts:

For part AC: \( r_1 = r \)
For part CD: \( r_2 = \frac{r}{3} \)
For part DB: \( r_3 = \frac{r}{2} \)

Comparing the dimensions, we observe the following strict order for the radii: \[ r > \frac{r}{2} > \frac{r}{3} \quad \implies \quad r_1 > r_3 > r_2 \]
Squaring these terms maintains the same order inequality: \[ r_1^2 > r_3^2 > r_2^2 \]
Since the drift velocity is inversely proportional to the square of the radius (\(v_d \propto \frac{1}{r_{wire}^2}\)), reversing the inequality gives: \[ \frac{1}{r_2^2} > \frac{1}{r_3^2} > \frac{1}{r_1^2} \quad \implies \quad v_2 > v_3 > v_1 \]
Hence, the drift velocity is highest where the wire is narrowest (part CD) and lowest where the wire is widest (part AC). This matches option (B). Quick Tip: For conductors connected in series carrying a steady current, the drift velocity of charge carriers is inversely proportional to the cross-sectional area (\(v_d \propto 1/A \propto 1/r^2\)). Therefore, narrower sections always exhibit faster drift velocities to maintain a constant current flow.

Concept:
The localized electric field \(E\) inside a current-carrying cylindrical conductor of uniform cross-section and length \(L\) is related to the potential difference \(V\) across its boundaries by the electrostatic relationship: \[ E = \frac{V}{L} \]
According to Ohm's law, the potential difference across a specific segment is determined by the constant series current \(I\) and the segment's individual resistance \(R\): \[ V = I \cdot R \]
The resistance of a uniform conductor is defined by its dimensions and material properties as: \[ R = \rho \frac{L}{A} \]
where \(\rho\) is the electrical resistivity of the metal and \(A\) is the cross-sectional area (\(A = \pi \cdot r_{wire}^2\)).


Step 1: Deriving the electric field equation as a function of radius.

Let us substitute the expression for resistance \(R\) directly into Ohm's Law to find the voltage drop: \[ V = I \cdot \left(\rho \frac{L}{A}\right) \]
Now, substituting this expression for \(V\) back into our electric field relationship yields: \[ E = \frac{V}{L} = \frac{I \cdot \rho \cdot L}{A \cdot L} = \frac{I \cdot \rho}{A} \]
Replacing the cross-sectional area \(A\) with \(\pi \cdot r_{wire}^2\), we obtain: \[ E = \frac{I \cdot \rho}{\pi \cdot r_{wire}^2} \]
Because the wire segments are coupled in a series configuration, the current \(I\) is identical in all parts. Furthermore, since they are composed of the exact same metal, the resistivity \(\rho\) is also identical. This shows that the internal electric field is inversely proportional to the square of the radius: \[ E \propto \frac{1}{r_{wire}^2} \]


Step 2: Comparing the electric fields for the three regions.

The given dimensions for the radii of sections AC, CD, and DB are: \[ r_1 = r, \quad r_2 = \frac{r}{3}, \quad r_3 = \frac{r}{2} \]
Arranging these values in descending order gives: \[ r_1 > r_3 > r_2 \]
Squaring the radii yields: \[ r^2 > \frac{r^2}{4} > \frac{r^2}{9} \quad \implies \quad r_1^2 > r_3^2 > r_2^2 \]
Since the electric field magnitude shares an inverse square relationship with the radius (\(E \propto \frac{1}{r_{wire}^2}\)), taking the reciprocals reverses the inequalities: \[ \frac{1}{r_2^2} > \frac{1}{r_3^2} > \frac{1}{r_1^2} \quad \implies \quad E_2 > E_3 > E_1 \]
Thus, the electric field is strongest in the thinnest region (CD) and weakest in the thickest region (AC). This corresponds to option (C). Quick Tip: For a continuous series conductor of a given material, both the electron drift velocity and the internal electric field follow the exact same structural dependence: they are inversely proportional to the cross-sectional area (\(E \propto 1/A\) and \(v_d \propto 1/A\)). Consequently, their mathematical order profiles are identical.

Concept:

In an astronomical telescope, the objective lens receives nearly parallel rays coming from a very distant object.

The objective lens forms a real, inverted and diminished image at its focal plane.

This real image acts as the object for the eyepiece.

The eyepiece behaves like a simple microscope and produces a virtual, enlarged image for comfortable viewing.



Step 1: Nature of image formed by objective lens.

Since the object is at a very large distance, the rays reaching the objective lens are almost parallel.

The objective lens therefore forms the image at its principal focus.

This image is:


Real
Inverted
Diminished




Step 2: Nature of image formed by eyepiece.

The real image formed by the objective serves as an object for the eyepiece.

The eyepiece magnifies this image and produces a final image which is:


Virtual
Enlarged
Inverted with respect to the original object




Therefore,
\[ \boxed{Objective image = Real} \]

and
\[ \boxed{Eyepiece image = Virtual} \]

Hence,
\[ \boxed{Correct Option (B)} \] Quick Tip: In an astronomical telescope: Objective \(\rightarrow\) Real image Eyepiece \(\rightarrow\) Virtual image

Concept:

The magnifying power of an astronomical telescope in normal adjustment is
\[ M=\frac{f_o}{f_e} \]

where
\[ f_o \]

is the focal length of the objective lens and
\[ f_e \]

is the focal length of the eyepiece.



Step 1: Examine the expression for magnifying power.

From
\[ M=\frac{f_o}{f_e} \]

it is clear that magnifying power depends upon:


Focal length of objective lens
Focal length of eyepiece




Step 2: Role of aperture.

The aperture determines:


Brightness of image
Light gathering power
Resolving power


However, it does not appear in the expression for magnification.

Therefore, magnification is independent of aperture.



Hence,
\[ \boxed{Correct Option (D)} \] Quick Tip: Remember: \[ M=\frac{f_o}{f_e} \] Magnifying power depends only on focal lengths and not on aperture.

Step 1: Recall the condition for normal adjustment.

In normal adjustment,
\[ Distance between objective and eyepiece = f_o+f_e. \]

It is exactly equal to the sum of the focal lengths.



Step 2: Check each statement.

Statement (A):
\[ f_o>f_e \]

This is true because telescope objectives have large focal lengths.



Statement (B):
\[ M=\frac{f_o}{f_e} \]

Increasing \(f_o\) increases \(M\).

Hence true.



Statement (C):

It says distance between lenses is more than
\[ (f_o+f_e). \]

This is incorrect because in normal adjustment the separation equals
\[ f_o+f_e. \]



Statement (D):

Increasing \(f_e\) decreases
\[ M=\frac{f_o}{f_e}. \]

Hence true.



Therefore,
\[ \boxed{Correct Option (C)} \] Quick Tip: For normal adjustment: \[ L=f_o+f_e \] where \(L\) is the separation between objective and eyepiece.

Concept:

For normal adjustment of an astronomical telescope,
\[ L=f_o+f_e. \]



Step 1: Substitute the given values.

Given,
\[ f_o=80\,cm \]

and
\[ f_e=4\,cm. \]

Hence,
\[ L=f_o+f_e \]
\[ L=80+4 \]
\[ L=84\,cm. \]



Step 2: Write the final answer.

Therefore,
\[ \boxed{L=84\,cm} \]

Hence,
\[ \boxed{Correct Option (A)} \] Quick Tip: For normal adjustment: \[ L=f_o+f_e \] Always add the focal lengths of objective and eyepiece.

Concept:

The magnifying power of an astronomical telescope in normal adjustment is
\[ M=\frac{f_o}{f_e}. \]



Step 1: Substitute the given focal lengths.

Given,
\[ f_o=80\,cm \]

and
\[ f_e=4\,cm. \]

Therefore,
\[ M=\frac{80}{4}. \]
\[ M=20. \]



Step 2: Interpret the result.

The telescope makes the distant object appear twenty times larger in angular size than that seen by the unaided eye.



Thus,
\[ \boxed{M=20} \]

Hence,
\[ \boxed{Correct Option (D)} \] Quick Tip: For astronomical telescope in normal adjustment: \[ M=\frac{f_o}{f_e} \] Large objective focal length and small eyepiece focal length produce high magnification.

Concept:
The absolute refractive index (\(n\) or \(\mu\)) of an optical medium is a dimensionless fundamental property that quantifies how much the speed of light is reduced inside that medium relative to a vacuum. It acts as a measure of the optical density of the medium.

When an electromagnetic wave passes from an empty space (vacuum) into a material medium, its interactions with the electronic structure of the atoms or molecules cause a net reduction in its phase velocity. The relation is expressed mathematically as: mu = frac{c{v

Where:

\(c\) represents the speed of light in a vacuum (approximately \(3 \times 10^8 m/s\)).
\(v\) represents the speed of light in the specific material medium.


Since \(c\) is always greater than or equal to \(v\) in any physical material medium, the absolute refractive index \(\mu\) is always greater than or equal to \(1\) (\(\mu \ge 1\)). Being a ratio of two identical physical quantities (speeds), it has no units or dimensions.


Step 1: Formal Definition Expression.

By structural definition, the absolute refractive index (\(\mu\)) of a given medium is defined as the ratio of the velocity of light in a vacuum (or air as a close approximation) to the velocity of light in that particular medium.mu = frac{text{Speed of light in vacuum (c){text{Speed of light in medium (v)

Since this perfectly matches statement (A), the conceptual definition is established directly. Quick Tip: Refractive index is a pure number with no units. If a medium has a higher refractive index, it is optically denser, meaning light travels slower inside it (\(v = c/\mu\)).

Concept:
When a monochromatic ray of light travels through a triangular glass prism, it undergoes refraction twice—once at the first incident face and once at the second emerging face. The structural orientation of the two faces causes the emergent ray to bend away from its original path. This angular separation between the direction of the incident ray and the emergent ray is known as the angle of deviation (\(\delta\)).



Let us define the parameters involved in a standard triangular prism setup:

\(A\): Angle of the prism (refracting angle).
\(i\): Angle of incidence at the first face.
\(r_1\): Angle of refraction inside the first face.
\(r_2\): Angle of incidence inside the second face.
\(e\): Angle of emergence from the second face.
\(\delta\): Total angle of deviation.



Step 1: Geometrical relations inside the prism.

Consider a ray of light passing through a prism \(ABC\). Let the refraction at the first face \(AB\) have an angle of incidence \(i\) and angle of refraction \(r_1\). At the second face \(AC\), let the internal angle be \(r_2\) and the final angle of emergence be \(e\).

From the cyclic quadrilateral formed by the normals and the vertices of the prism, the sum of the internal angles satisfies: \(\)A + angle N = 180^circ\(\)
In the interior triangle formed by the light ray and the normal intersection: \(\)r_1 + r_2 + angle N = 180^circ\(\)
Equating these two geometric properties yields the fundamental structural relation of a prism: \(\)A = r_1 + r_2 quad cdots (1)\(\)

Now, considering the total angular deviation  suffered by the ray, it is the sum of deviations at both refracting surfaces: delta = (i - r_1) + (e - r_2)
Rearranging the terms gives: delta = (i + e) - (r_1 + r_2)Substituting equation (1) into this expression provides the second fundamental prism formula: delta = i + e - A quad cdots (2)


Step 2: Condition for minimum deviation.

Experimentally and theoretically, as the angle of incidence \(i\) is gradually increased, the angle of deviation \(\delta\) first decreases, reaches a certain minimum value (\(\delta_m\)), and then increases. At this unique condition of minimum deviation (\(\delta = \delta_m\)), the light ray passes completely symmetrically through the prism.

Under this perfect symmetric state, we have: \(\)i = e\(\) \(\)r_1 = r_2 = r\(\)

Substituting these symmetric conditions back into equations (1) and (2):
From equation (1):A = r + r = 2r quad Rightarrow quad r = frac{A{2 quad cdots (3)

From equation (2): delta_m = i + i - A = 2i - A 2i = A + delta_m quad Rightarrow quad i = frac{A + delta_m{2 quad cdots (4)


Step 3: Applying Snell's Law to determine the refractive index.

According to Snell's Law at the first refracting boundary interface (moving from air to glass): mu = frac{sin i{sin r

Substituting the explicit symmetric structural values of \(i\) and \(r\) derived in equations (3) and (4) directly into Snell's formula: mu = frac{sinleft(frac{A + delta_m{2right){sinleft(frac{A{2right)

This is the famous prism formula that determines the exact refractive index of the material, validating choice (A). Quick Tip: At the position of minimum deviation (\(\delta = \delta_m\)), the refracted ray passing through the interior of the prism becomes perfectly parallel to the base of the prism if the prism is isosceles or equilateral.

Concept:

When a ray is incident normally on a surface, it enters the medium without deviation. At a denser-to-rarer interface, the ray undergoes Total Internal Reflection (TIR) if the angle of incidence is greater than the critical angle.

For glass, \[ \sin i_c=\frac{1}{\mu} \]
where \(i_c\) is the critical angle and \(\mu\) is the refractive index of the glass.

Step 1: Refraction at face \(BC\).


The ray \(QP\) is incident normally on face \(BC\).

Hence,
\[ i=0^\circ. \]

Therefore, the ray enters the prism without any deviation and travels along the normal to face \(BC\).

Step 2: Determine the angle of incidence at face \(AC\).


Since the ray travels along the normal to face \(BC\), the angle made by the ray with the normal to face \(AC\) is equal to
\[ \angle B=60^\circ. \]

Thus,
\[ i=60^\circ. \]

Step 3: Calculate the critical angle.


Given,
\[ \mu=1.5. \]

Therefore,
\[ \sin i_c=\frac{1}{1.5}=\frac{2}{3}. \]

Hence,
\[ i_c=\sin^{-1}\!\left(\frac23\right)\approx41.8^\circ. \]

Step 4: Compare the angles.


Since
\[ i=60^\circ>i_c=41.8^\circ, \]

the ray undergoes Total Internal Reflection at face \(AC\).

After reflection, the reflected ray travels towards face \(AB\).

Since the reflected ray strikes face \(AB\) normally, it emerges from the prism without deviation.

Thus, the ray enters normally through BC, undergoes TIR at AC, and finally emerges normally through AB. Quick Tip: For a right-angled prism with angles \(30^\circ\), \(60^\circ\), and \(90^\circ\), a ray entering normally through the hypotenuse strikes one of the other faces at \(60^\circ\). Since the critical angle for glass \((\mu=1.5)\) is about \(41.8^\circ\), the ray undergoes Total Internal Reflection and then emerges normally from the remaining face.

Concept:

Light can be described using two complementary concepts in optics: rays and wavefronts. A ray represents the direction in which light energy propagates, whereas a wavefront represents the surface joining all points of a light wave that are in the same phase of vibration at a particular instant. These two concepts are closely related, as rays are always normal (perpendicular) to the corresponding wavefronts.



Difference between a Ray and a Wavefront:


Definition

Ray: An imaginary straight line that indicates the direction in which light travels.
Wavefront: An imaginary surface joining all points of a wave that are in the same phase of vibration.


Nature

Ray: One-dimensional and represented by a straight line.
Wavefront: A two-dimensional or three-dimensional surface depending on the nature of the light source.


Represents

Ray: Represents the direction of propagation of light energy.
Wavefront: Represents the position of the advancing light wave at a given instant.


Direction

Ray: Directly indicates the direction of propagation of light.
Wavefront: The direction of propagation is obtained by drawing a normal to the wavefront.


Relationship

Ray: Always perpendicular to the wavefront.
Wavefront: Always perpendicular to the rays passing through it.


Examples

Ray: Used in geometrical optics to study reflection, refraction and image formation.
Wavefront: May be plane, spherical or cylindrical depending on the nature of the light source.





Conclusion:

A ray gives the direction in which light travels.
A wavefront represents a surface of constant phase.
Rays are always drawn perpendicular to the corresponding wavefronts. Quick Tip: Remember:} Wavefront = Surface of constant phase.} Ray = Direction of propagation of light.} Plane wavefront \(\leftrightarrow\) Parallel rays. Spherical wavefront \(\leftrightarrow\) Diverging or converging rays. These concepts are fundamental in understanding Huygens' Principle, reflection, refraction and interference of light.

Concept:
Huygens' Principle states that every point on a primary wavefront acts as a source of secondary spherical wavelets, which spread out in all directions with the speed of the wave in that medium. The new envelope touching these secondary wavelets tangentially at any subsequent instant forms the new secondary wavefront.

To prove the law of reflection (\(\angle i = \angle r\)), we construct a geometric framework using a plane wavefront striking a smooth reflecting barrier.




Step 1: Constructing the geometry of incidence.

Let \(MN\) be a smooth plane reflecting surface. Consider a plane wavefront \(AB\) incident obliquely on this surface at an angle \(i\).
The wavefront first touches the reflecting surface at point \(A\) at time \(t = 0\). The other end of the wavefront, point \(B\), is still traveling through space and takes a finite time \(t\) to hit the surface at point \(C\).

If \(v\) is the velocity of the wave wavelets in the medium, the distance traveled by the wavefront from \(B\) to \(C\) in time \(t\) is: \(\)Distance BC = v cdot t quad cdots (1)\(\)


Step 2: Activating Huygens' wavelets at the surface.

According to Huygens' principle, as soon as the wavefront hits point \(A\), point \(A\) starts acting as a source of secondary wavelets. In the same time interval \(t\) during which the incident disturbance travels from \(B\) to \(C\), the secondary spherical wavelet originating from point \(A\) expands to a radius equal to: \(\)text{Radius AD = v cdot t quad cdots (2)\(\)

To find the new reflected wavefront, we draw a sphere of radius \(vt\) centered at \(A\), and then construct a tangential line or plane \(CD\) from point \(C\) to this spherical arc. This tangent plane \(CD\) represents the newly formed reflected wavefront .


Step 3: Proving congruence of the wave triangles.

Let us examine the two right-angled triangles formed on the base surface \(AC\): \(\triangle ABC\) and \(\triangle ADC\).

\(\angle ABC = \angle ADC = 90^\circ\) (Rays are always perpendicular to their respective wavefront surfaces \(AB\) and \(CD\)).
text{Hypotenuse AC = AC) (Common base side shared by both triangles).
\(Side BC = AD = v \cdot t\) (From equations 1 and 2, distances are equal).


By the Right angle-Hypotenuse-Side (RHS) congruence criterion: \(\)triangle ABC cong triangle ADC\(\)

Since the triangles are congruent, their corresponding matching angles must be exactly equal by CPCTC (Corresponding Parts of Congruent Triangles are Congruent): \(\)angle BAC = angle DCA quad cdots (3)\(\)

From the definition of angles in the diagram:

\(\angle BAC = i\) (Angle of incidence, which is the angle between the incident wavefront and the reflecting surface).
\(\angle DCA = r\) (Angle of reflection, which is the angle between the reflected wavefront and the reflecting surface).


Substituting these definitions into equation (3): \(\)i = r\(\)

This proves that the angle of incidence is equal to the angle of reflection, successfully verifying the fundamental law of reflection using wave theory. Quick Tip: When applying Huygens' principle for reflection, remember that because the medium does not change, the speed (\(v\)) and frequency (\(f\)) of the wave remain perfectly identical before and after reflection.

Concept:

According to Huygens' principle, every point on a wavefront acts as a source of secondary wavelets. When a plane wavefront passes through a convex lens, the central portion of the lens is thicker than its edges. Since light travels more slowly in glass than in air, different portions of the wavefront suffer different time delays. As a result, the plane wavefront is converted into a converging spherical wavefront.



Step 1: Incident Plane Wavefront


A plane wavefront is incident normally on the convex lens. The corresponding rays are parallel to the principal axis.



Step 2: Refraction through the Lens



The central part of the lens is thicker than the edges.
Therefore, the central portion of the wavefront travels through a greater thickness of glass and experiences a larger optical delay.
The edge portions pass through a thinner part of the lens and experience comparatively less delay.


Because of this unequal delay, the emerging wavefront is no longer plane.



Step 3: Formation of the Emergent Wavefront


The emergent wavefront becomes a converging spherical wavefront. The normals to this spherical wavefront represent the refracted rays, which converge towards the principal focus \(F\) of the convex lens.
\[ \boxed{Plane wavefront \;\longrightarrow\; Convex Lens \;\longrightarrow\; Converging Spherical Wavefront} \]





Conclusion:

Thus, a convex lens converts an incident plane wavefront into a converging spherical wavefront. The corresponding refracted rays converge at the principal focus \(F\). Quick Tip: Remember: Plane wavefront \(\rightarrow\) Parallel rays. Convex lens \(\rightarrow\) Converging spherical wavefront. Normals to the emergent wavefront meet at the principal focus \(F\).

Concept:
In a series \(LCR\) alternating current circuit, the electrical impedance \(Z\) is given by the expression: \[ Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2} \]
where \(R\) is resistance, \(X_L = \omega L\) is inductive reactance, and \(X_C = \frac{1}{\omega C}\) is capacitive reactance.


Resonant Frequency: It is the specific frequency \(\nu_r\) (or angular frequency \(\omega_r\)) of the applied AC voltage source at which the inductive reactance becomes equal to the capacitive reactance (\(X_L = X_C\)). At this frequency, the total impedance of the circuit drops to its minimum possible value (\(Z = R\)), resulting in maximum current amplitude.
Power Dissipation and Power Factor: Average power dissipated in a series \(LCR\) circuit is:
\[ P_{avg} = V_{rms} I_{rms} \cos \phi \]
where \(\cos \phi = \frac{R}{Z}\) is defined as the power factor, and \(\phi\) is the phase difference between current and voltage.


Step 1: Definition of Resonant Frequency.

Setting inductive reactance equal to capacitive reactance: \[ X_L = X_C \quad \Rightarrow \quad \omega_r L = \frac{1}{\omega_r C} \] \[ \omega_r^2 = \frac{1}{LC} \quad \Rightarrow \quad \omega_r = \frac{1}{\sqrt{LC}} \quad (in rad/s) \]
Since angular frequency \(\omega_r = 2\pi \nu_r\), the resonant frequency \(\nu_r\) in Hertz (Hz) is given by: \[ \nu_r = \frac{1}{2\pi \sqrt{LC}} \]

Step 2: Condition for Maximum Power Dissipation.

The expression for average power dissipated in an AC circuit is: \[ P_{avg} = V_{rms} I_{rms} \cos \phi \]
For a given voltage and current, the power dissipated depends directly on the power factor \(\cos \phi\).
Since the cosine function has a maximum value of \(1\) (when \(\phi = 0^\circ\)): \[ (P_{avg})_{max} = V_{rms} I_{rms} \times 1 \]
Therefore, the power dissipated in the circuit will be maximum when the power factor is \(\cos \phi = 1\) (purely resistive condition/resonance). Quick Tip: At resonance: - \(X_L = X_C\) and Impedance \(Z = R\) (minimum impedance). - Current amplitude is maximum (\(I_0 = \frac{V_0}{R}\)). - Phase angle \(\phi = 0^\circ\), which makes Power Factor \(\cos \phi = 1\) (maximum power dissipation).

Concept:
The electrical behavior of a series \(LCR\) circuit driven by an AC voltage \(v = v_0 \sin(\omega t)\) is governed by resistance \(R\), inductance \(L\), and capacitance \(C\).

Key properties used here:

Power Factor (\(\cos \phi\)): It is defined as the cosine of the phase angle (\(\phi\)) by which voltage leads or lags current in an AC circuit. Mathematically, it is the ratio of true power (resistance \(R\)) to apparent power (impedance \(Z\)):
\[ \cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + (X_L - X_C)^2}} \]
Average Power Dissipated: Given by \(P_{avg} = V_{rms} I_{rms} \cos \phi\).


Step 1: Definition and mathematical expression for Power Factor.

The power factor of a series \(LCR\) circuit represents the efficiency with which electrical energy is converted into real work (heat/mechanical energy). \[ Power Factor (\cos \phi) = \frac{Resistance (R)}{Total Impedance (Z)} \]
where \(Z = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2}\). The power factor ranges strictly between \(0\) and \(1\): \[ 0 \le \cos \phi \le 1 \]

Step 2: Determining maximum power dissipation condition.

Average power consumed in one full cycle is given by: \[ P_{avg} = V_{rms} I_{rms} \cos \phi \]
To maximize \(P_{avg}\) for given root-mean-square voltage \(V_{rms}\) and current \(I_{rms}\), the parameter \(\cos \phi\) must achieve its maximum possible mathematical value.
Since \(\max(\cos \phi) = 1\) (occurring when \(\phi = 0^\circ\)): \[ P_{max} = V_{rms} I_{rms} \]
Thus, the power dissipated is maximum when the power factor \(\cos \phi = 1\), which occurs in a purely resistive circuit or at resonance (\(X_L = X_C\)). Quick Tip: Power Factor Summary: - Purely Inductive or Capacitive circuit: \(\phi = 90^\circ \Rightarrow \cos \phi = 0\) (Wattless current). - Purely Resistive or Resonant LCR circuit: \(\phi = 0^\circ \Rightarrow \cos \phi = 1\) (Maximum power dissipation).

Concept:
For a series \(LCR\) circuit connected to an AC voltage source \(v = v_0 \sin(\omega t)\), the total opposition offered by the circuit to the flow of AC is called impedance (\(Z\)).
Key formulas:

Angular frequency: \(\omega\) from standard equation \(v = v_0 \sin(\omega t)\)
Inductive Reactance: \(X_L = \omega L\)
Capacitive Reactance: \(X_C = \frac{1}{\omega C}\)
Total Impedance: \(Z = \sqrt{R^2 + (X_L - X_C)^2}\)


Step 1: Extracting given values from the question statement.

Given parameters:

Inductance, \( L = \frac{5}{\pi} H \)
Capacitance, \( C = \frac{50}{\pi} \, muF = \frac{50}{\pi} \times 10^{-6} F \)
Resistance, \( R = 400 \, \Omega \)
Alternating voltage equation: \( v = 140 \sin(100\pi t) V \)

Comparing \( v = 140 \sin(100\pi t) \) with standard equation \( v = v_0 \sin(\omega t) \): \[ Peak Voltage v_0 = 140 V \] \[ Angular Frequency \omega = 100\pi rad/s \]

Step 2: Calculating Inductive Reactance (\(X_L\)).

Substituting \(\omega = 100\pi\) and \(L = \frac{5}{\pi}\): \[ X_L = \omega L = (100\pi) \times \left(\frac{5}{\pi}\right) = 100 \times 5 = 500 \, \Omega \]

Step 3: Calculating Capacitive Reactance (\(X_C\)).

Substituting \(\omega = 100\pi\) and \(C = \frac{50}{\pi} \times 10^{-6}\): \[ X_C = \frac{1}{\omega C} = \frac{1}{(100\pi) \times \left(\frac{50}{\pi} \times 10^{-6}\right)} \]
Simplify denominator: \[ (100\pi) \times \frac{50}{\pi} \times 10^{-6} = 5000 \times 10^{-6} = 5 \times 10^{-3} \]
Therefore: \[ X_C = \frac{1}{5 \times 10^{-3}} = \frac{1000}{5} = 200 \, \Omega \]

Step 4: Calculating Total Impedance (\(Z\)).

Using the impedance formula: \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Substitute \(R = 400 \, \Omega\), \(X_L = 500 \, \Omega\), and \(X_C = 200 \, \Omega\): \[ X_L - X_C = 500 - 200 = 300 \, \Omega \] \[ Z = \sqrt{(400)^2 + (300)^2} = \sqrt{160000 + 90000} = \sqrt{250000} = 500 \, \Omega \]
The impedance of the circuit is \(500 \, \Omega\). Quick Tip: When evaluating impedance, always calculate reactances \(X_L\) and \(X_C\) first. Notice that \((3, 4, 5)\) form a Pythagorean triple: \(\sqrt{400^2 + 300^2} = 500 \, \Omega\).

Concept:
The root-mean-square (rms) current \(I_{rms}\) flowing in an AC circuit is determined by the Ohm's law equivalent for AC circuits: \[ I_{rms} = \frac{V_{rms}}{Z} \]
where:

\(V_{rms} = \frac{v_0}{\sqrt{2}}\) (\(v_0\) is peak voltage)
\(Z = \sqrt{R^2 + (X_L - X_C)^2}\) (total circuit impedance)


Step 1: Determining the Impedance (\(Z\)) of the circuit.

From the given data:

\(L = \frac{5}{\pi} H\), \(C = \frac{50}{\pi} \times 10^{-6} F\), \(R = 400 \, \Omega\)
Applied voltage: \(v = 140 \sin(100\pi t)\) \(\Rightarrow v_0 = 140 V, \, \omega = 100\pi rad/s\)

Calculating reactances: \[ X_L = \omega L = 100\pi \times \frac{5}{\pi} = 500 \, \Omega \] \[ X_C = \frac{1}{\omega C} = \frac{1}{100\pi \times \frac{50}{\pi} \times 10^{-6}} = \frac{1}{5 \times 10^{-3}} = 200 \, \Omega \]
Calculating Impedance \(Z\): \[ Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{400^2 + (500 - 200)^2} = \sqrt{400^2 + 300^2} = 500 \, \Omega \]

Step 2: Calculating RMS Voltage (\(V_{rms}\)).

Given peak voltage \(v_0 = 140 V\) and taking \(\sqrt{2} = 1.4\): \[ V_{rms} = \frac{v_0}{\sqrt{2}} = \frac{140}{1.4} = 100 V \]

Step 3: Calculating RMS Current (\(I_{rms}\)).

Using AC Ohm's Law: \[ I_{rms} = \frac{V_{rms}}{Z} \]
Substitute \(V_{rms} = 100 V\) and \(Z = 500 \, \Omega\): \[ I_{rms} = \frac{100}{500} = \frac{1}{5} = 0.20 A \]
Thus, the rms current flowing through the series LCR circuit is \(0.20 A\). Quick Tip: Always convert peak voltage (\(v_0\)) to rms voltage (\(V_{rms} = \frac{v_0}{\sqrt{2}}\)) before finding the rms current. Standard AC ammeters measure \(I_{rms}\), not peak current \(I_0\).

Concept:
A transformer is an electrical device based on the principle of mutual induction. It is used to step up (increase) or step down (decrease) AC voltage.
Key principles:

Faraday's Law of Electromagnetic Induction: Induced EMF in a coil is proportional to the rate of change of magnetic flux linked with it.
Ideal Transformer Assumption: Zero flux leakage and negligible resistance in winding coils.


Step 1: Description and Labelling of Step-Up Transformer.

A step-up transformer consists of:

Laminated Soft Iron Core: Minimizes energy loss due to eddy currents.
Primary Coil (\(P\)): Consists of a smaller number of turns (\(N_p\)) made of thick copper wire, connected to an input AC source.
Secondary Coil (\(S\)): Consists of a larger number of turns (\(N_s\)) made of relatively thinner copper wire, from which output voltage is derived.

For a step-up transformer: \(N_s > N_p\).

Step 2: Derivation of Voltage Ratio in Terms of Turns Ratio.

Let \(\phi\) be the magnetic flux linked with each turn of both primary and secondary coils at any instant \(t\).

1. Induced EMF in Primary Coil (\(e_p\)):
According to Faraday's Law: \[ e_p = -N_p \frac{d\phi}{dt} \quad \cdots (1) \]
For an ideal transformer with zero resistance, input voltage \(V_p \approx e_p\): \[ V_p = -N_p \frac{d\phi}{dt} \quad \cdots (2) \]

2. Induced EMF in Secondary Coil (\(e_s\)):
Similarly, the output voltage \(V_s\) across the secondary coil is: \[ V_s = -N_s \frac{d\phi}{dt} \quad \cdots (3) \]

3. Taking the ratio of Equation (3) to Equation (2): \[ \frac{V_s}{V_p} = \frac{-N_s \frac{d\phi}{dt}}{-N_p \frac{d\phi}{dt}} = \frac{N_s}{N_p} \]
The ratio \(K = \frac{N_s}{N_p}\) is known as the transformation ratio. For a step-up transformer, \(K > 1\), hence \(V_s > V_p\). Quick Tip: Transformer Transformation Ratio Formula: \[ \frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s} = K \] For Step-Up: \(N_s > N_p \Rightarrow V_s > V_p\) and \(I_s < I_p\).

Concept:
For any electrical device, power supplied at the input is given by the product of primary voltage (\(V_p\)) and primary current (\(I_p\)): \[ P_{in} = V_p \times I_p \]
For an ideal transformer, there are no power losses (100% efficiency), meaning: \[ P_{in} = P_{out} = 3.3 kW \]

Step 1: Extracting given parameters and converting units.


Number of turns in primary coil, \( N_p = 100 \)
Number of turns in secondary coil, \( N_s = 5000 \)
Primary input voltage, \( V_p = 220 V \)
Input power supplied, \( P_{in} = 3.3 kW = 3.3 \times 10^3 W = 3300 W \)


Step 2: Calculating primary current (\(I_p\)).

Using the input power relation: \[ P_{in} = V_p \times I_p \]
Substitute \(P_{in} = 3300 W\) and \(V_p = 220 V\): \[ 3300 = 220 \times I_p \]
Rearranging to solve for \(I_p\): \[ I_p = \frac{3300}{220} = \frac{330}{22} = 15 A \]
Thus, the current flowing through the primary coil is \(15 A\). Quick Tip: To find primary current when input power is given, you do NOT need the turn ratio! Simply use \(I_p = \frac{P_{in}}{V_p}\). Remember to convert kW to Watts (\(1 kW = 1000 W\)).

Concept:
The voltage transformation in an ideal transformer follows the fundamental turn ratio equation: \[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \]
where:

\(V_p\) = Primary input voltage
\(V_s\) = Secondary output voltage
\(N_p\) = Number of primary turns
\(N_s\) = Number of secondary turns


Step 1: Extracting given numerical values.


Primary turns, \( N_p = 100 \)
Secondary turns, \( N_s = 5000 \)
Primary voltage, \( V_p = 220 V \)


Step 2: Calculating the Transformation Ratio (\(K\)).
\[ K = \frac{N_s}{N_p} = \frac{5000}{100} = 50 \]
This means the transformer steps up the input voltage by a factor of 50.

Step 3: Calculating Secondary Output Voltage (\(V_s\)).

Using the transformer equation: \[ V_s = V_p \times \left(\frac{N_s}{N_p}\right) \]
Substitute \(V_p = 220 V\) and \(\frac{N_s}{N_p} = 50\): \[ V_s = 220 \times 50 = 11,000 V = 11 kV \]
Therefore, the output voltage across the secondary coil is \(11,000 V\). Quick Tip: For ideal step-up transformers, output voltage increases in direct proportion to turns ratio: \[ V_s = V_p \times \frac{N_s}{N_p} \] Since \(N_s = 50 \times N_p\), output voltage must be 50 times the input voltage (\(220 \times 50 = 11,000 V\)).

Concept:
An equipotential surface is defined as a surface over which the electric potential \(V\) remains constant at all points.
The relationship between the electric field \(\mathbf{E}\) and electrostatic potential \(V\) is given by the differential potential difference equation: \[ dV = -\mathbf{E} \cdot d\mathbf{r} = -E \, dr \cos\theta \]
where \(d\mathbf{r}\) represents a small displacement vector along the surface, and \(\theta\) is the angle between the electric field vector \(\mathbf{E}\) and the displacement vector \(d\mathbf{r}\).


Step 1: Analyzing potential difference on an equipotential surface.

By definition, the electric potential \(V\) is identical at every point on an equipotential surface. Therefore, for any two points separated by a small displacement vector \(d\mathbf{r}\) lying entirely within the surface, the potential difference \(dV\) between them must be zero: \[ dV = V_B - V_A = 0 \]


Step 2: Relating potential difference to work done and electric field.

The small work done \(dW\) by an external agent in moving a test charge \(q_0\) through a displacement \(d\mathbf{r}\) on the equipotential surface is: \[ dW = q_0 dV \]
Since \(dV = 0\), the work done in moving a test charge along an equipotential surface is identically zero: \[ dW = 0 \]
We also know that work done in terms of the electric force \(\mathbf{F} = q_0 \mathbf{E}\) is: \[ dW = -\mathbf{F} \cdot d\mathbf{r} = -q_0 (\mathbf{E} \cdot d\mathbf{r}) = 0 \]
Assuming \(q_0 \neq 0\), this simplifies to: \[ \mathbf{E} \cdot d\mathbf{r} = 0 \]


Step 3: Applying the scalar product (dot product) condition.

Using the definition of the scalar product: \[ E \, |d\mathbf{r}| \cos\theta = 0 \]
For a non-zero electric field (\(E \neq 0\)) and a non-zero displacement along the surface (\(|d\mathbf{r}| \neq 0\)), we must have: \[ \cos\theta = 0 \quad \Rightarrow \quad \theta = 90^\circ = \frac{\pi}{2} radians \]


Step 4: Physical Interpretation.

Since \(\theta = 90^\circ\), the electric field vector \(\mathbf{E}\) at any given point is perpendicular (normal) to the displacement vector \(d\mathbf{r}\) lying on the equipotential surface.

If \(\mathbf{E}\) were not normal to the equipotential surface, it would have a non-zero tangential component along the surface. This component would exert a force on a test charge moving along the surface, performing work and causing a potential difference between points on the surface, which contradicts the definition of an equipotential surface.

Thus, the electric field line at any point must always be normal to the equipotential surface passing through that point. Quick Tip: Key Equipotential Surface Rules: - Work done moving a charge on an equipotential surface: \( W = q \Delta V = 0 \). - Electric field direction: Always perpendicular to the surface (\( \mathbf{E} \perp Surface \)). - Direction of field lines: Point from regions of higher potential to lower potential.

Concept:
A dielectric material is an electrical insulator that can be polarized by an applied electric field. When an electric field is applied to a dielectric, free charges do not flow through the material as they do in an electrical conductor, but bound charges shift slightly from their average equilibrium positions, giving rise to dielectric polarization.


Step 1: Understanding Polarization in an External Field.

Let an external uniform electric field \(\mathbf{E}_0\) be applied across a dielectric slab.
- In non-polar dielectrics, the external field induces electric dipole moments by stretching positive nuclei and negative electron clouds in opposite directions.
- In polar dielectrics, permanent dipoles that are initially randomly oriented due to thermal agitation tend to align themselves parallel to the direction of \(\mathbf{E}_0\).

This process is known as dielectric polarization .


Step 2: Formation of Bound Induced Surface Charges.

Due to the alignment of dipoles throughout the volume of the dielectric:
1. On the left surface (facing the positive plate producing \(\mathbf{E}_0\)), a net negative bound charge density \(-\sigma_p\) is induced.
2. On the right surface (facing the negative plate producing \(\mathbf{E}_0\)), a net positive bound charge density \(+\sigma_p\) is induced.
3. Inside the bulk volume of the slab, positive and negative charges cancel each other locally, so the net charge inside the volume remains zero.


Step 3: Creation of Induced Internal Electric Field.

These induced surface charges \(-\sigma_p\) and \(+\sigma_p\) create an internal induced electric field \(\mathbf{E}_p\) inside the dielectric slab.
By electrostatics, electric fields originate on positive charges and terminate on negative charges. Therefore, the direction of the induced electric field \(\mathbf{E}_p\) is directed from the positive induced surface to the negative induced surface, which is directly opposite to the external applied electric field \(\mathbf{E}_0\).

The magnitude of this induced field is given by: \[ E_p = \frac{\sigma_p}{\varepsilon_0} \]


Step 4: Calculating the Net Reduced Electric Field.

By the principle of superposition, the net effective electric field \(\mathbf{E}_{net}\) inside the dielectric medium is the vector sum of the external field \(\mathbf{E}_0\) and the induced field \(\mathbf{E}_p\): \[ \mathbf{E}_{net} = \mathbf{E}_0 + \mathbf{E}_p \]
Since \(\mathbf{E}_p\) is antiparallel to \(\mathbf{E}_0\), taking magnitudes yields: \[ E_{net} = E_0 - E_p \]
Since \(E_p > 0\), it follows directly that: \[ E_{net} < E_0 \]
The magnitude of the internal electric field is reduced by a factor equal to the dielectric constant \(K\) (\(K > 1\)) of the material: \[ E_{net} = \frac{E_0}{K} \]
Hence, the electric field inside the dielectric is strictly less than the external electric field outside it. Quick Tip: Dielectric Field Reduction Summary: - Reduced Electric Field: \( E = \frac{E_0}{K} \), where \( K > 1 \) is the dielectric constant (relative permittivity \(\varepsilon_r\)). - Induced Charge Density: \( \sigma_p = \sigma \left(1 - \frac{1}{K}\right) \). - In conductors, complete cancellation occurs (\(E = 0\)) because \(K = \infty\). In dielectrics, partial cancellation occurs (\(E_0 / K\)).

Concept:
For an isolated parallel plate capacitor charged to a charge \(Q\) and then disconnected from the charging battery, the total stored charge \(Q\) remains strictly constant due to the law of conservation of charge.

The fundamental relationship between charge \(Q\), capacitance \(C\), and potential difference \(V\) across the capacitor plates is: \[ Q = C \cdot V \quad \Rightarrow \quad V = \frac{Q}{C} \]


Step 1: Formula for the Capacitance of a Parallel Plate Capacitor.

The capacitance \(C\) of a parallel plate capacitor having plate area \(A\), separation distance \(d\), and permittivity of free space \(\varepsilon_0\) is given by: \[ C = \frac{\varepsilon_0 A}{d} \]
From this formula, capacitance \(C\) is inversely proportional to the plate separation distance \(d\): \[ C \propto \frac{1}{d} \]


Step 2: Analyzing the effect of reducing distance \(d\).

When the plates of the capacitor are brought closer together, the separation distance \(d\) decreases (\(d \to d'\) such that \(d' < d\)).
Since \(C\) is inversely proportional to \(d\), decreasing \(d\) causes the capacitance to increase : \[ C' = \frac{\varepsilon_0 A}{d'} > C \]


Step 3: Calculating the new Potential Difference \(V'\).

Since the capacitor is isolated, charge \(Q\) is conserved and remains constant (\(Q' = Q\)).
The potential difference across the plates is: \[ V' = \frac{Q}{C'} \]
Substituting \(C' = \frac{\varepsilon_0 A}{d'}\): \[ V' = \frac{Q}{\left(\frac{\varepsilon_0 A}{d'}\right)} = \frac{Q \cdot d'}{\varepsilon_0 A} \]
Alternatively, expressed in terms of the uniform electric field \(E = \frac{Q}{\varepsilon_0 A}\) between the plates: \[ V' = E \cdot d' \]


Step 4: Comparison between initial and final potential differences.

Because the electric field \(E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{A \varepsilon_0}\) depends only on charge \(Q\) and plate area \(A\), it remains constant as distance changes.
Thus, as the plate separation \(d\) is reduced (\(d' < d\)): \[ V' = E \cdot d' < E \cdot d = V \]
Therefore, the potential difference \(V\) between the plates decreases proportionately with the separation distance \(d\). Quick Tip: Isolated vs. Connected Capacitors (Distance Decreased \(d \downarrow\)): - Disconnected from Battery (Constant Charge \(Q\)): \(C \uparrow\), \(Q = constant\), \(V \downarrow\) (\(V = Ed\)), \(E = constant\), Energy \(U \downarrow\) (\(U = \frac{Q^2}{2C}\)). - Connected to Battery (Constant Voltage \(V\)): \(C \uparrow\), \(V = constant\), \(Q \uparrow\), \(E \uparrow\), Energy \(U \uparrow\) (\(U = \frac{1}{2}CV^2\)).

Concept:
1. Electrostatic Potential Energy (\(U\)): The electrostatic potential energy of a system of discrete point charges is defined as the work done by an external agent in assembling the charges from an initial state of infinite separation to their given positions without acceleration.
For a system of \(N\) point charges, total potential energy is: \[ U = \frac{1}{4\pi\varepsilon_0} \sum_{1 \le i < j \le N} \frac{q_i q_j}{r_{ij}} \]
2. Work Done to Dissociate a System (\(W_{dissociate}\)): The work required to completely dissociate (disassemble) a system of charges means moving all charges from their initial bound configuration to infinite separation (\(r \to \infty\)) where final potential energy \(U_f = 0\).
By the work-energy theorem: \[ W_{ext} = U_f - U_i = 0 - U_i = -U_i \]


Step 1: Identify charges and pairwise distances.

The charges positioned at the vertices of the equilateral triangle of side length \(a\) are:

Charge at vertex A: \(q_1 = q\)
Charge at vertex B: \(q_2 = -4q\)
Charge at vertex C: \(q_3 = 2q\)

Since ABC is an equilateral triangle with side length \(a\): \[ r_{AB} = r_{12} = a, \quad r_{BC} = r_{23} = a, \quad r_{AC} = r_{13} = a \]


Step 2: Calculate the initial electrostatic potential energy (\(U_i\)) of the system.

The initial electrostatic potential energy \(U_i\) is the sum of potential energies of all unique charge pairs: \[ U_i = U_{AB} + U_{BC} + U_{AC} \] \[ U_i = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q_1 q_2}{r_{12}} + \frac{q_2 q_3}{r_{23}} + \frac{q_1 q_3}{r_{13}} \right] \]

Substitute the given charge values and distances into the formula: \[ U_i = \frac{1}{4\pi\varepsilon_0} \left[ \frac{(q)(-4q)}{a} + \frac{(-4q)(2q)}{a} + \frac{(q)(2q)}{a} \right] \]


Step 3: Algebraic simplification of \(U_i\).

Factoring out \(\frac{q^2}{a}\): \[ U_i = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q^2}{a} \left[ -4 - 8 + 2 \right] \] \[ U_i = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q^2}{a} \left[ -12 + 2 \right] \] \[ U_i = \frac{1}{4\pi\varepsilon_0} \left( -\frac{10 q^2}{a} \right) = -\frac{10 q^2}{4\pi\varepsilon_0 a} \]


Step 4: Calculate work done to dissociate the system.

Dissociation implies separating all three charges to an infinite mutual distance apart.
Final state potential energy at infinite separation: \[ U_f = 0 \]
The work required by an external agent to dissociate the system is: \[ W_{dissociate} = U_f - U_i \] \[ W_{dissociate} = 0 - \left( -\frac{10 q^2}{4\pi\varepsilon_0 a} \right) \] \[ W_{dissociate} = +\frac{10 q^2}{4\pi\varepsilon_0 a} = \frac{1}{4\pi\varepsilon_0} \frac{10 q^2}{a} \]

Thus, the work required to dissociate the system is \(\frac{1}{4\pi\varepsilon_0} \frac{10 q^2}{a}\). Quick Tip: System Potential Energy vs. Dissociation Work: - Binding Energy / Assembly Work: \( W_{assemble} = U_{system} \). - Dissociation Work: \( W_{dissociate} = -U_{system} \). - Total pairs for \(N\) charges is given by \(\frac{N(N-1)}{2}\). For \(N=3\), there are 3 distinct interaction pairs.

Concept:
It is important to distinguish between two completely different velocities in electric circuits:
1. Drift Velocity (\(v_d\)): The slow average velocity with which free electrons drift towards the positive terminal under the influence of an internal electric field. It is typically very small (\(\approx 10^{-4} m/s = 0.1 mm/s\)).
2. Signal Propagation Speed (\(c\)): The speed at which the governing electromagnetic field propagates along the transmission path/circuit, which occurs at or near the speed of light in the medium (\(\approx 10^8 m/s\)).


Step 1: Understanding electron availability in conductors.

A metallic conductor contains an immense density of free conduction electrons. For copper, free electron density is: \[ n \approx 8.5 \times 10^{28} m^{-3} \]
Free electrons are present continuously throughout the entire length of the conducting wire before any voltage source is connected.


Step 2: Role of the electric field upon closing the switch.

When the circuit switch is closed, a potential difference is applied across the ends of the wire. This action sets up an electromagnetic wave that propagates along the surface and space surrounding the conducting wires at nearly the speed of light: \[ v_{signal} \approx 3 \times 10^8 m/s \]


Step 3: Simultaneous drift motion throughout the conductor.

As this electromagnetic wave sweeps through the circuit, it establishes a local electric field \(\mathbf{E}\) inside the conductor almost instantaneously at every point along the entire wire loop.

Every free electron located throughout the circuit experiences an electrostatic force \(\mathbf{F} = -e\mathbf{E}\) at virtually the same instant, initiating a localized drift motion \(v_d = \frac{e E \tau}{m_e}\) simultaneously everywhere.


Step 4: Conclusion.

Electric current is defined as the net rate of flow of charge across any cross-section: \[ I = n e A v_d \]
Current does not wait for an individual electron to travel all the way from the battery to the appliance (e.g., a lamp). Instead, because free electrons are already present everywhere and begin drifting simultaneously as soon as the electric field reaches them, current is established almost instantaneously throughout the circuit. Quick Tip: Analogy for Current Flow: - Think of a long pipe completely filled with water from end to end. As soon as you push water in at one end, water immediately pours out from the other end. The pressure wave travels fast, even if the individual water molecules move slowly.

Concept:
Every realistic voltage source (like a battery or DC power supply) possesses an internal resistance \(r\) due to the electrolyte/materials inside the cell.
When a load resistance \(R\) is connected across a cell of emf \(\mathcal{E}\) and internal resistance \(r\), the circuit current \(I\) is governed by Ohm's Law for a complete circuit: \[ I = \frac{\mathcal{E}}{R + r} \]
The terminal potential difference \(V\) across the load is: \[ V = \mathcal{E} - I r \]


Step 1: Determining maximum deliverable current.

To draw the maximum possible current \(I_{max}\) from a supply, the external load resistance \(R\) must be reduced towards zero (\(R \to 0\), short-circuit condition): \[ I_{max} = \lim_{R \to 0} \left( \frac{\mathcal{E}}{R + r} \right) = \frac{\mathcal{E}}{r} \]


Step 2: Analyzing the effect of a low emf \(\mathcal{E}\).

We are given that the voltage supply has a low emf \(\mathcal{E}\) (for instance, \(\mathcal{E} = 2 V\) or \(6 V\)).
From \(I_{max} = \frac{\mathcal{E}}{r}\), if \(\mathcal{E}\) is inherently small, the maximum current \(I_{max}\) that the supply can deliver is bounded by the ratio \(\frac{\mathcal{E}}{r}\).


Step 3: Analyzing the requirement for internal resistance \(r\).

If the internal resistance \(r\) were high (say \(r = 100 \ \Omega\)), then even under a short-circuit (\(R=0\)), the maximum possible current would be: \[ I_{max} = \frac{6 V}{100 \ \Omega} = 0.06 A = 60 mA \]
This is far below the required high current (e.g., several amperes needed for automobile starters or heavy heating elements).

To make \(I_{max}\) large when \(\mathcal{E}\) is small, the internal resistance \(r\) must be extremely small (\(r \ll 1 \ \Omega\)): \[ r must be very small so that I_{max} = \frac{\mathcal{E}}{r} is sufficiently large. \]
For example, if \(\mathcal{E} = 6 V\) and \(r = 0.005 \ \Omega\): \[ I_{max} = \frac{6}{0.005} = 1200 A \]


Step 4: Minimizing internal power loss.

Additionally, power lost internally as heat within the power supply is \(P_{loss} = I^2 r\). If high current \(I\) flows through a large internal resistance \(r\), almost all the energy supplied by the source will be converted into heat internally, causing severe thermal damage and dropping the terminal voltage \(V = \mathcal{E} - Ir\) close to zero.

Hence, a low voltage supply must have a extremely low internal resistance to deliver large currents safely and efficiently. Quick Tip: Internal Resistance & Maximum Current: - Maximum current drawn from a cell: \( I_{max} = \frac{\mathcal{E}}{r} \). - Terminal voltage under load: \( V = \mathcal{E} - Ir \). - For high output current from low voltage, \( r \) must be minimized.

Concept:
There is a fundamental difference between a definition equation and a physical law :
1. \(V = IR\) (or \(R = \frac{V}{I}\)) is simply the definition of resistance \(R\) for any circuit component, regardless of whether the component obeys Ohm's law or not.
2. Ohm's Law is a specific empirical statement about material behavior under constant physical conditions.


Step 1: Statement of Ohm's Law.

Ohm's Law states that:
"The electric current \(I\) flowing through a conductor is directly proportional to the potential difference \(V\) applied across its ends, provided physical conditions such as temperature, mechanical strain, and pressure remain strictly constant." \[ I \propto V \quad \Rightarrow \quad V \propto I \quad (at constant physical conditions) \]
Mathematically: frac{V{I} = Constant = R
The essence of Ohm's law is that the resistance \(R\) is independent of the applied voltage \(V\) and the resulting current \(I\).


Step 2: Microscopic form of Ohm's law.

At a fundamental materials level, Ohm's law is expressed as: \[ \mathbf{J} = \sigma \mathbf{E} \]
where \(\mathbf{J}\) is current density, \(\mathbf{E}\) is electric field, and \(\sigma\) is electrical conductivity. Ohm's law holds only if \(\sigma\) (or resistivity \(\rho = \frac{1}{\sigma}\)) is independent of the applied electric field \(\mathbf{E}\).


Step 3: Why \(V = IR\) alone is not Ohm's law.

The relation \(R = \frac{V}{I}\) defines resistance for any device.
For non-ohmic devices (e.g., p-n junction diodes, transistors, vacuum tubes, thermistors, gas discharge tubes):
- One can still calculate an instantaneous ratio \(\frac{V}{I}\) and call it resistance at that operating point.
- However, for these non-ohmic devices, the ratio \(\frac{V}{I}\) changes as \(V\) or \(I\) changes (the I-V characteristic graph is non-linear).


Step 4: Conclusion.

Therefore, \(V = IR\) is merely the defining formula for electric resistance \(R\).

The actual statement of Ohm's law is the physical assertion that \(R\) remains constant (linear relationship between \(V\) and \(I\)) over a wide range of voltages for Ohmic conductors. Quick Tip: Ohmic vs. Non-Ohmic Conductors: - Ohmic Devices: \(I \propto V\); I-V graph is a straight line through origin (Slope = \(\frac{1}{R} = constant\)). Examples: metals at constant temperature. - Non-Ohmic Devices: \(I \not\propto V\); I-V graph is non-linear. Examples: Diodes, LEDs, Transistors, Electrolytes. - Equation \(R = V/I\) holds for both, but Ohm's law holds ONLY for Ohmic devices.

Concept:
When two cells of emfs \(\mathcal{E}_1, \mathcal{E}_2\) and internal resistances \(r_1, r_2\) are connected in parallel across terminals with like polarities connected together:
1. The equivalent internal resistance \(r_{eq}\) is calculated as a standard parallel combination of individual internal resistances: \[ \frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} \quad \Rightarrow \quad r_{eq} = \frac{r_1 r_2}{r_1 + r_2} \]
2. The equivalent electromotive force (emf) \(\mathcal{E}_{eq}\) of the parallel combination is given by: \[ \frac{\mathcal{E}_{eq}}{r_{eq}} = \frac{\mathcal{E}_1}{r_1} + \frac{\mathcal{E}_2}{r_2} \quad \Rightarrow \quad \mathcal{E}_{eq} = \frac{\frac{\mathcal{E}_1}{r_1} + \frac{\mathcal{E}_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} = \frac{\mathcal{E}_1 r_2 + \mathcal{E}_2 r_1}{r_1 + r_2} \]


Step 1: Identify the given parameters from the problem statement.

From the problem description:

First cell emf: \(\mathcal{E}_1 = 12 V\)
First cell internal resistance: \(r_1 = 1\ \Omega\)
Second cell emf: \(\mathcal{E}_2 = 6 V\)
Second cell internal resistance: \(r_2 = 0.5\ \Omega = \frac{1}{2}\ \Omega\)



Step 2: Calculate the equivalent internal resistance \(r_{eq}\).

Using the formula for parallel combination of internal resistances: \[ \frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} \]
Substitute \(r_1 = 1\ \Omega\) and \(r_2 = 0.5\ \Omega\): \[ \frac{1}{r_{eq}} = \frac{1}{1} + \frac{1}{0.5} = 1 + 2 = 3\ \Omega^{-1} \]
Taking the reciprocal: \[ r_{eq} = \frac{1}{3}\ \Omega \approx 0.333\ \Omega \]


Step 3: Calculate the equivalent EMF \(\mathcal{E}_{eq}\).

Using the parallel combination formula for equivalent emf: \[ \mathcal{E}_{eq} = \frac{\frac{\mathcal{E}_1}{r_1} + \frac{\mathcal{E}_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} \]

Substitute the individual terms into the numerator: \[ \frac{\mathcal{E}_1}{r_1} = \frac{12}{1} = 12 A \] \[ \frac{\mathcal{E}_2}{r_2} = \frac{6}{0.5} = 12 A \]
Sum of ratios in numerator: \[ \frac{\mathcal{E}_1}{r_1} + \frac{\mathcal{E}_2}{r_2} = 12 + 12 = 24 A \]

Now, divide by \(\frac{1}{r_{eq}} = 3\ \Omega^{-1}\): \[ \mathcal{E}_{eq} = \frac{24}{3} = 8 V \]


Step 4: Verification using the alternative formula.

Using \(\mathcal{E}_{eq} = \frac{\mathcal{E}_1 r_2 + \mathcal{E}_2 r_1}{r_1 + r_2}\): \[ \mathcal{E}_{eq} = \frac{(12 \times 0.5) + (6 \times 1)}{1 + 0.5} = \frac{6 + 6}{1.5} = \frac{12}{1.5} = 8 V \]

Hence, the equivalent cell between terminals A and B has an equivalent emf of \(8 V\) and an equivalent internal resistance of \(\frac{1}{3}\ \Omega \approx 0.33\ \Omega\) . Quick Tip: Parallel Cells Formula Summary: - If cells assist each other: \( \mathcal{E}_{eq} = \frac{\mathcal{E}_1/r_1 + \mathcal{E}_2/r_2}{1/r_1 + 1/r_2} \). - If one cell is connected with opposite polarity: \( \mathcal{E}_{eq} = \frac{\mathcal{E}_1/r_1 - \mathcal{E}_2/r_2}{1/r_1 + 1/r_2} \). - Internal resistance is ALWAYS combined in parallel: \( \frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} \).