CBSE Class 12 Physics Set 3 - (55/3/3) Question Paper 2026 with Solution PDF

Concept:

A plane electromagnetic wave's electric and magnetic fields oscillate sinusoidally.

The standard mathematical representation for a wave traveling in the positive x-direction is \( B = B_0 \sin(\omega t - kx) \) or \( B = B_0 \sin(kx - \omega t) \).

Here, \( \omega \) represents the angular frequency in rad/s, and \( k \) represents the wave number in rad/m.

The propagation speed (velocity) \( v \) of the wave is related to these parameters by the formula \( v = \frac{\omega}{k} \).



Step 1: {\color{redExtract parameters from the given wave equation

The given equation for the magnetic field is:
\[ B = 5 \times 10^{-8} \sin (3 \times 10^{10} t - 150 x) T \]

By comparing this with the standard form \( B = B_0 \sin(\omega t - kx) \), we can identify:

The angular frequency \( \omega \) is the coefficient of \( t \), so \( \omega = 3 \times 10^{10} rad/s \).

The wave number \( k \) is the coefficient of \( x \), so \( k = 150 rad/m \).


Step 2: {\color{redCalculate the velocity of the wave

Substitute the extracted values of \( \omega \) and \( k \) into the wave velocity formula:
\[ v = \frac{\omega}{k} \]
\[ v = \frac{3 \times 10^{10}}{150} \]

To simplify, break down the numerator:
\[ v = \frac{300 \times 10^8}{150} \]
\[ v = 2.0 \times 10^8 m/s \]


Step 3: {\color{redConclusion

The calculated velocity of the electromagnetic wave is \( 2.0 \times 10^8 ms^{-1} \).

This result matches option (A). Quick Tip: Always remember that wave speed is simply the coefficient of \( t \) divided by the coefficient of \( x \), regardless of the order they appear inside the sine or cosine function.

Concept:

When light undergoes reflection, it stays in the same medium, so its speed and wavelength remain constant.

When light undergoes refraction, it enters a different medium with a different optical density (refractive index), causing its speed and wavelength to change.

Frequency is a fundamental characteristic of the light source and relates to the energy of the photon (\( E = h\nu \)).

Therefore, frequency does not change when light travels across different optical media or undergoes boundary phenomena like reflection and refraction.



Step 1: {\color{redAnalyze the changing properties (Speed, Wavelength, Intensity)

The reflected ray remains in air, so it travels at speed \( c \) with wavelength \( \lambda \).

The refracted ray enters the glass slab (an optically denser medium), so its speed decreases to \( v = \frac{c}{\mu} \).

Because \( v = f \lambda \) and speed decreases, its wavelength also decreases to \( \lambda' = \frac{\lambda}{\mu} \).

The incident energy is split between the reflected and refracted rays, so their individual intensities are lower than the original ray and generally unequal.


Step 2: {\color{redAnalyze the invariant property (Frequency)

The frequency of a wave corresponds to the number of oscillations per second generated by the source.

It is independent of the medium's properties.

Since both the reflected and refracted rays originate from the same incident monochromatic light ray, they must share the exact same frequency.


Step 3: {\color{redConclusion

Since wavelength, speed, and intensity differ between the two rays, but frequency remains constant, option (B) is the correct choice. Quick Tip: A useful mnemonic for optical transitions: "Frequency is Fundamental." It never changes during reflection, refraction, or interference, as it solely depends on the source oscillator.

Concept:

A semiconductor diode is a non-linear device, meaning it does not obey Ohm's Law in a simple linear fashion.

Instead of a static resistance, we define its resistance for small signal changes as "dynamic resistance" or "AC resistance".

Dynamic resistance \( r_d \) is defined as the ratio of a small change in applied voltage (\( \Delta V \)) to the corresponding small change in current (\( \Delta I \)).

The formula is mathematically expressed as \( r_d = \frac{\Delta V}{\Delta I} \).



Step 1: {\color{redIdentify the changes in voltage and current

The initial forward bias voltage is \( V_1 = 0.8 V \).

The final forward bias voltage is \( V_2 = 1.0 V \).

The change in voltage is \( \Delta V = V_2 - V_1 = 1.0 - 0.8 = 0.2 V \).

The corresponding change in forward current is given as \( \Delta I = 2.0 mA \).

Convert this current to Amperes for standard SI unit calculations: \( \Delta I = 2.0 \times 10^{-3} A \).


Step 2: {\color{redCalculate the dynamic resistance

Substitute the values of \( \Delta V \) and \( \Delta I \) into the dynamic resistance formula:
\[ r_d = \frac{\Delta V}{\Delta I} \]
\[ r_d = \frac{0.2}{2.0 \times 10^{-3}} \]

Multiply the numerator and denominator by 1000 to remove the power of 10:
\[ r_d = \frac{0.2 \times 1000}{2.0} \]
\[ r_d = \frac{200}{2.0} \]
\[ r_d = 100 \ \Omega \]


Step 3: {\color{redConclusion

The forward bias resistance (dynamic resistance) of the diode is \( 100 \ \Omega \).

This explicitly matches option (C). Quick Tip: Whenever "change in voltage" or "change in current" is mentioned in diode problems, immediately think of dynamic resistance (\( r = \Delta V / \Delta I \)).
Always ensure current is converted from milliamperes (mA) to Amperes (A) before dividing.

Concept:

Magnetic flux (\( \Phi \)) represents the total number of magnetic field lines passing perpendicular to a given surface area.

Mathematically, it is defined as the dot product of the uniform magnetic field vector (\( \vec{B} \)) and the area vector (\( \vec{A} \)) of the surface.

\( \Phi = \vec{B} \cdot \vec{A} \).

The area vector \( \vec{A} \) has a magnitude equal to the surface area and a direction strictly perpendicular (normal) to the surface plane.



Step 1: {\color{redDetermine the Area Vector of the square loop

The loop is a square with side length \( L \).

The magnitude of its area is \( A = L \times L = L^2 \).

The problem states that the loop lies in the x-y plane.

The vector strictly perpendicular to the x-y plane is the z-axis, represented by the unit vector \( \hat{k} \).

Therefore, the area vector is: \( \vec{A} = L^2 \hat{k} \).


Step 2: {\color{redCalculate the Magnetic Flux using the Dot Product

The given magnetic field vector is \( \vec{B} = B_0 (2\hat{i} + 3\hat{j} + 4\hat{k}) \).

Apply the dot product formula for flux:
\[ \Phi = \vec{B} \cdot \vec{A} \]
\[ \Phi = \left[ B_0 (2\hat{i} + 3\hat{j} + 4\hat{k}) \right] \cdot \left[ L^2 \hat{k} \right] \]

In a dot product, orthogonal unit vectors yield zero (\( \hat{i} \cdot \hat{k} = 0 \) and \( \hat{j} \cdot \hat{k} = 0 \)), and parallel unit vectors yield one (\( \hat{k} \cdot \hat{k} = 1 \)).
\[ \Phi = B_0 L^2 (2(\hat{i} \cdot \hat{k}) + 3(\hat{j} \cdot \hat{k}) + 4(\hat{k} \cdot \hat{k})) \]
\[ \Phi = B_0 L^2 (0 + 0 + 4(1)) \]
\[ \Phi = 4 B_0 L^2 \]


Step 3: {\color{redConclusion

Only the z-component of the magnetic field contributes to the flux through a loop in the x-y plane.

The resulting magnetic flux is \( 4 B_0 L^2 \), making option (B) the correct answer. Quick Tip: To find flux quickly, just multiply the area of the loop by the component of the magnetic field that is perpendicular to the loop's plane.
For an x-y loop, take the \( \hat{k} \) component of \( \vec{B} \). For a y-z loop, take the \( \hat{i} \) component of \( \vec{B} \).

Concept:

When an electric dipole with dipole moment \( \vec{P} \) is placed in a uniform electric field \( \vec{E} \), it possesses electrostatic potential energy.

This potential energy is given by the dot product: \( U = -\vec{P} \cdot \vec{E} \).

To slowly rotate a dipole from an initial orientation to a final orientation, an external agent must do work against the electric field.

The total work done by the external agent equals the change in the dipole's potential energy: \( W_{ext} = \Delta U = U_{final} - U_{initial} \).



Step 1: {\color{redExpress the given vectors and the work energy formula

Let the scalar constant be \( p_0 = 2.54 \times 10^{-28} C.m \).

The initial dipole moment is \( \vec{P}_i = p_0 (2.00\hat{i} + 3.00\hat{j}) \).

The final dipole moment is \( \vec{P}_f = p_0 (-3.00\hat{i} + 2.00\hat{j}) \).

The uniform electric field is \( \vec{E} = 1000\hat{i} N/C \).

The work done formula is \( W = (-\vec{P}_f \cdot \vec{E}) - (-\vec{P}_i \cdot \vec{E}) = (\vec{P}_i - \vec{P}_f) \cdot \vec{E} \).


Step 2: {\color{redPerform the vector subtraction and dot product

First, calculate the difference vector \( (\vec{P}_i - \vec{P}_f) \):
\[ \vec{P}_i - \vec{P}_f = p_0 [ (2.00\hat{i} + 3.00\hat{j}) - (-3.00\hat{i} + 2.00\hat{j}) ] \]
\[ \vec{P}_i - \vec{P}_f = p_0 [ (2.00 + 3.00)\hat{i} + (3.00 - 2.00)\hat{j} ] \]
\[ \vec{P}_i - \vec{P}_f = p_0 (5.00\hat{i} + 1.00\hat{j}) \]

Next, perform the dot product with the electric field \( \vec{E} = 1000\hat{i} \):
\[ W = [ p_0 (5.00\hat{i} + 1.00\hat{j}) ] \cdot [ 1000\hat{i} ] \]

Since \( \hat{i} \cdot \hat{i} = 1 \) and \( \hat{j} \cdot \hat{i} = 0 \), only the x-component survives.
\[ W = p_0 \times 5.00 \times 1000 \]
\[ W = 5000 \times (2.54 \times 10^{-28}) J \]


Step 3: {\color{redCalculate final numerical value
\[ W = 12700 \times 10^{-28} J \]

Rewrite this in proper scientific notation by shifting the decimal 4 places to the left:
\[ W = 1.27 \times 10^4 \times 10^{-28} J \]
\[ W = 1.27 \times 10^{-24} J \]


Step 4: {\color{redConclusion

The total work done by the external agent is mathematically calculated to be \( 1.27 \times 10^{-24} J \).

Therefore, option (B) is the correct answer. Quick Tip: When dealing with vectors in dipole problems, it is much faster and less error-prone to use the vector dot product directly \( W = \vec{E} \cdot (\vec{P}_i - \vec{P}_f) \) rather than calculating angles and using the \( pE(\cos\theta_1 - \cos\theta_2) \) formula.

Concept:

The focal length of an individual thin lens is determined by its physical shape (radii of curvature) and material (refractive index) using the Lens Maker's Formula: \( \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \).

When two thin lenses with focal lengths \( f_1 \) and \( f_2 \) are placed perfectly in contact, their optical powers add up.

The equivalent focal length \( F \) of the combination is given by: \( \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \).

Strict adherence to the Cartesian sign convention is crucial: surfaces bulging towards the incident light have positive \( R \), and surfaces cupping away have negative \( R \).



Step 1: {\color{redDetermine the focal length of the plano-convex lens (\( f_1 \))

Assume light travels from left to right.

For the plano-convex lens, the first surface is plane, so its radius of curvature \( R_1 = \infty \).

The second surface is convex, bulging outward to the right, so its center of curvature lies to the left (against incident light direction). Thus, \( R_2 = -R \).

Apply the Lens Maker's Formula:
\[ \frac{1}{f_1} = (\mu - 1) \left( \frac{1}{\infty} - \frac{1}{-R} \right) \]
\[ \frac{1}{f_1} = (\mu - 1) \left( 0 + \frac{1}{R} \right) = \frac{\mu - 1}{R} \]


Step 2: {\color{redDetermine the focal length of the equi-concave lens (\( f_2 \))

For the equi-concave lens, the first surface is concave, cupping inward. Its center of curvature lies to the left, so \( R_1 = -R \).

The second surface is also concave from the perspective of exiting light, its center lies to the right, so \( R_2 = +R \).

Apply the Lens Maker's Formula:
\[ \frac{1}{f_2} = (\mu - 1) \left( \frac{1}{-R} - \frac{1}{+R} \right) \]
\[ \frac{1}{f_2} = (\mu - 1) \left( -\frac{2}{R} \right) = \frac{-2(\mu - 1)}{R} \]


Step 3: {\color{redCalculate the equivalent focal length of the combination

Sum the inverse focal lengths:
\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \]
\[ \frac{1}{F} = \frac{\mu - 1}{R} + \left( \frac{-2(\mu - 1)}{R} \right) \]
\[ \frac{1}{F} = \frac{(\mu - 1) - 2(\mu - 1)}{R} \]
\[ \frac{1}{F} = \frac{-(\mu - 1)}{R} \]

Inverting both sides gives the equivalent focal length:
\[ F = -\frac{R}{\mu - 1} \]


Step 4: {\color{redConclusion

The effective focal length of the two-lens combination is \( -\frac{R}{\mu - 1} \), which corresponds to option (B).
Quick Tip: Sign convention is the only place students make mistakes here.
Remember: A plane surface has \( R = \infty \). A standard convex lens has \( R_1 > 0, R_2 < 0 \). A standard concave lens has \( R_1 < 0, R_2 > 0 \).

Concept:

In a p-n junction diode, a depletion region naturally forms at the junction, creating a built-in potential barrier \( V_0 \).

When a forward bias is applied (p-side connected to positive, n-side to negative), the external electric field strictly opposes the built-in field.

This opposition shrinks the depletion region and reduces the height of the potential barrier to \( V_0 - V_{applied} \).

Conversely, when a reverse bias is applied, the external electric field aligns with and supports the built-in field.

This alignment widens the depletion region and increases the height of the potential barrier to \( V_0 + V_{applied} \).



Step 1: {\color{redAnalyze the given graph for barrier heights

The graph plots electric potential on the y-axis against distance on the x-axis.

Curve 2 represents the reference potential barrier \( V_0 \) in the unbiased equilibrium state.

Curve 3 displays a potential barrier height that is visually lower than \( V_0 \).

Curve 1 displays a potential barrier height that is visually higher than \( V_0 \).


Step 2: {\color{redCorrelate barrier height to biasing conditions

Since a forward bias effectively lowers the barrier height to allow easier flow of majority charge carriers, Curve 3 represents the forward biased condition.

Since a reverse bias effectively raises the barrier height, impeding the flow of majority charge carriers, Curve 1 represents the reverse biased condition.


Step 3: {\color{redConclusion

Curve 3 correlates to forward bias, and Curve 1 correlates to reverse bias.

Scanning the given options, statement (B) perfectly describes this conclusion.
Quick Tip: Visual trick: "Forward" bias pushes charges "forward" over the hill, so it must lower the hill (barrier). "Reverse" bias makes it harder, so the hill (barrier) grows taller.

Concept:

Extensive scattering experiments have demonstrated that the volume of an atomic nucleus is directly proportional to its total number of nucleons, known as the mass number (\( A \)).

Assuming the nucleus is a perfect sphere (\( V = \frac{4}{3}\pi r^3 \)), this implies that \( r^3 \propto A \).

Taking the cube root yields the empirical relationship for nuclear radius: \( r = r_0 A^{1/3} \), where \( r_0 \) is a constant (\( \approx 1.2 \times 10^{-15} m \)).

When comparing two different nuclei, their radii ratio is determined purely by the cube roots of their respective mass numbers: \( \frac{r_1}{r_2} = \left( \frac{A_1}{A_2} \right)^{1/3} \).



Step 1: {\color{redIdentify the mass numbers and setup the ratio

From the question text, the mass number of the first nucleus is \( A_1 = 64 \).

The mass number of the second nucleus is \( A_2 = 27 \).

Using the derived ratio formula:
\[ \frac{r_1}{r_2} = \left( \frac{A_1}{A_2} \right)^{1/3} \]
\[ \frac{r_1}{r_2} = \left( \frac{64}{27} \right)^{1/3} \]


Step 2: {\color{redEvaluate the cube roots

The mathematical expression requires calculating the cube root of both the numerator and the denominator independently.

For the numerator: The cube root of 64 is 4, because \( 4 \times 4 \times 4 = 64 \).

For the denominator: The cube root of 27 is 3, because \( 3 \times 3 \times 3 = 27 \).

Substitute these roots back into the ratio:
\[ \frac{r_1}{r_2} = \frac{4}{3} \]


Step 3: {\color{redConclusion

The ratio of the radii \( r_1 / r_2 \) simplifies precisely to \( 4/3 \).

This corresponds directly to option (B).
Quick Tip: To solve these nuclear physics questions rapidly, commit to memory the cubes of integers up to 6:
\( 1^3 = 1 \), \( 2^3 = 8 \), \( 3^3 = 27 \), \( 4^3 = 64 \), \( 5^3 = 125 \), \( 6^3 = 216 \).

Concept:

Total Internal Reflection (TIR) occurs strictly when a light ray traveling in a denser medium strikes a rarer medium boundary at an angle of incidence \( i \) that is greater than the critical angle \( i_c \).

The critical angle \( i_c \) is mathematically related to the refractive index \( \mu \) of the denser medium by \( \sin(i_c) = \frac{1}{\mu} \).

According to Cauchy's dispersion formula (\( \mu = A + \frac{B}{\lambda^2} \)), the refractive index of a given material inversely depends on the wavelength \( \lambda \) of the light passing through it.

Consequently, light colors with smaller wavelengths experience higher refractive indices, which in turn correspond to smaller critical angles.



Step 1: {\color{redEstablish the order of wavelengths and critical angles

The visible light spectrum ordered by increasing wavelength is Violet, Indigo, Blue, Green, Yellow, Orange, Red (VIBGYOR).

From this, we extract the order for the given colors:
\[ \lambda_{blue} < \lambda_{green} < \lambda_{yellow} < \lambda_{red} \]

Because refractive index is inversely related to wavelength:
\[ \mu_{blue} > \mu_{green} > \mu_{yellow} > \mu_{red} \]

Since a larger refractive index yields a smaller critical angle (\( \sin i_c = 1/\mu \)):
\[ i_{c,blue} < i_{c,green} < i_{c,yellow} < i_{c,red} \]


Step 2: {\color{redApply the condition for Total Internal Reflection

The problem explicitly states that the yellow ray successfully undergoes TIR at a specific angle of incidence \( i \).

This strictly means that the angle of incidence \( i \) must be greater than the critical angle for yellow light:
\[ i > i_{c,yellow} \]

Looking at our inequality chain from Step 1, both the critical angles for blue and green light are strictly smaller than the critical angle for yellow light.

Therefore, by mathematical transitivity:
\[ i > i_{c,yellow} > i_{c,green} > i_{c,blue} \]

This proves that the fixed angle of incidence \( i \) is simultaneously greater than \( i_{c,green} \) and \( i_{c,blue} \).

Consequently, both green and blue rays will easily satisfy the condition for TIR.

However, \( i_{c,red} \) is larger than \( i_{c,yellow} \), so we cannot guarantee that \( i > i_{c,red} \).


Step 3: {\color{redConclusion

Both green and blue light rays will absolutely undergo total internal reflection under the identical incident conditions.

This logic points directly to option (C). Quick Tip: A quick rule of thumb for TIR color replacement: If a specific color undergoes TIR, all colors with a shorter wavelength (appearing to its left in the VIBGYOR sequence) will also undergo TIR at that same incident angle.

Concept:

The de Broglie wavelength \( \lambda \) of a moving particle is inversely proportional to its momentum \( p = mv \), given by the equation \( \lambda = \frac{h}{mv} \).

For an electron in a hydrogen atom, Niels Bohr postulated that its orbital angular momentum must be quantized: \( mvr = \frac{nh}{2\pi} \), where \( n \) is the principal quantum number.

By rearranging this quantization condition, we can relate the electron's momentum directly to its orbit radius and quantum number: \( mv = \frac{nh}{2\pi r} \).

Additionally, Bohr's model proves that the radius \( r \) of the \( n^{th} \) stable orbit is directly proportional to the square of the principal quantum number: \( r \propto n^2 \).



Step 1: {\color{redDerive the proportionality between wavelength and quantum number

Start with the de Broglie wavelength formula and substitute momentum from Bohr's condition:
\[ \lambda = \frac{h}{mv} \]

Substitute \( mv = \frac{nh}{2\pi r} \):
\[ \lambda = \frac{h}{\left( \frac{nh}{2\pi r} \right)} \]
\[ \lambda = \frac{2\pi r}{n} \]

Now, apply the known fact that orbit radius scales with the square of the quantum number (\( r \propto n^2 \)):
\[ \lambda \propto \frac{n^2}{n} \]

Simplify the expression to reveal a direct linear relationship:
\[ \lambda \propto n \]

This indicates that the de Broglie wavelength of an orbiting electron is simply directly proportional to the principal quantum number of its orbit.


Step 2: {\color{redCalculate the required ratio

We are asked to find the ratio of the wavelength in the first orbit (\( n_1 = 1 \)) to the wavelength in the third orbit (\( n_3 = 3 \)).

Using the direct proportionality \( \lambda \propto n \):
\[ \frac{\lambda_1}{\lambda_3} = \frac{n_1}{n_3} \]

Substitute the respective orbit numbers:
\[ \frac{\lambda_1}{\lambda_3} = \frac{1}{3} \]


Step 3: {\color{redConclusion

The calculated ratio of the de Broglie wavelengths is \( 1/3 \).

This corresponds exactly to option (D). Quick Tip: Memorizing proportionalities in Bohr's model saves massive amounts of exam time:
Radius: \( r \propto n^2 \)
Velocity: \( v \propto 1/n \)
Energy: \( E \propto 1/n^2 \)
Wavelength: \( \lambda \propto n \)

Concept:

The photoelectric effect is governed by Einstein's photoelectric equation, which is essentially an energy conservation statement.

The equation states that the maximum kinetic energy (\( K_{max} \)) of an emitted photoelectron equals the energy of the incident photon (\( E \)) minus the work function (\( \Phi \)) of the metal surface.

Mathematically: \( K_{max} = E - \Phi \).

The energy of an incident photon is inversely proportional to its wavelength, given by \( E = \frac{hc}{\lambda} \).



Step 1: {\color{redCalculate the energy of the incident photon

The wavelength of the incident radiation is provided as \( \lambda = 200 nm \).

To compute the photon energy directly in electron-volts (eV), we use the highly practical approximation formula:
\[ E (in eV) = \frac{1240}{\lambda (in nm)} \quad (or 1242 for slightly higher precision) \]

Substituting the given wavelength:
\[ E = \frac{1240}{200} \]
\[ E = 6.2 eV \]


Step 2: {\color{redCalculate the maximum kinetic energy

The work function of the photosensitive surface is provided as \( \Phi = 4.2 eV \).

Apply Einstein's photoelectric equation:
\[ K_{max} = E - \Phi \]

Substitute the known energy values:
\[ K_{max} = 6.2 eV - 4.2 eV \]
\[ K_{max} = 2.0 eV \]


Step 3: {\color{redConclusion

The kinetic energy of the fastest emitted photoelectrons is calculated to be 2.0 eV.

This perfectly matches option (D). Quick Tip: Using \( E = \frac{1240 eV\cdotnm}{\lambda (nm)} \) is an essential shortcut for modern physics problems.
It completely bypasses the tedious process of calculating \( \frac{(6.63 \times 10^{-34}) \times (3 \times 10^8)}{\lambda \times 10^{-9}} \) and then dividing by \( 1.6 \times 10^{-19} \) to convert from Joules to eV.

Concept:

In a Young's Double-Slit Experiment (YDSE), the pattern on the screen is formed by the superposition of two coherent light waves originating from two slits.

The resultant intensity \( I \) at any point on the screen depends heavily on the phase difference \( \Delta \phi \) between the two arriving waves.

The intensity formula is \( I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\Delta \phi) \).

A "bright spot" (or maxima) is produced by purely constructive interference, which happens when the intensity is maximized.



Step 1: {\color{redDetermine the mathematical condition for maximum intensity

Looking at the intensity equation \( I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\Delta \phi) \), the only variable factor is the cosine term.

To maximize the overall intensity \( I \), the cosine function must reach its maximum possible value.

The maximum value of the cosine function is \( +1 \).

Therefore, the mathematical condition for a bright spot is:
\[ \cos(\Delta \phi) = 1 \]


Step 2: {\color{redSolve for the phase difference

We need to find the angles (phase differences) for which the cosine equals 1.

From basic trigonometry, cosine equals 1 at \( 0, 2\pi, 4\pi, 6\pi, \dots \).

In other words, the phase difference must be an even integer multiple of \( \pi \).

This sequence can be concisely represented algebraically by using an integer variable \( n \) (where \( n = 0, \pm 1, \pm 2, \dots \)):
\[ \Delta \phi = 2n\pi \]

Conversely, a dark spot (destructive interference) requires minimum intensity, meaning \( \cos(\Delta \phi) = -1 \), which occurs at odd multiples of \( \pi \) or \( (2n+1)\pi \).

Since the question asks specifically for a bright spot, the condition is definitively \( 2n\pi \).


Step 3: {\color{redConclusion

The phase difference giving rise to a bright spot is universally \( 2n\pi \).

This corresponds directly to option (A). Quick Tip: Bright spot (Constructive Interference): Phase difference = \( 2n\pi \), Path difference = \( n\lambda \).
Dark spot (Destructive Interference): Phase difference = \( (2n \pm 1)\pi \), Path difference = \( (n \pm 0.5)\lambda \).

Concept:

The strong nuclear force is the fundamental interaction responsible for binding nucleons (protons and neutrons) tightly together to form atomic nuclei.

It is the strongest of the four fundamental forces in nature, vastly overpowering electrostatic repulsion between protons at short distances.

However, its behavior is highly distance-dependent. It is strongly attractive between distances of roughly 0.7 fm and 2.5 fm.

Crucially, at distances less than approximately 0.7 fm, the strong nuclear force becomes fiercely repulsive. This is known as the "repulsive core" and prevents the nucleus from collapsing in on itself.



Step 1: {\color{redEvaluate the Assertion (A)

Assertion (A) states: "Nuclear forces are always attractive."

Based on the physical characteristics described in the concepts, this statement is factually incorrect.

While it is primarily attractive to hold the nucleus together, it possesses a critical repulsive core at extremely short internucleon distances (less than 0.7 femtometers).

If the force were exclusively and always attractive, nucleons would crush together infinitely.

Therefore, Assertion (A) is definitively False.


Step 2: {\color{redEvaluate the Reason (R)

Reason (R) states: "The nuclear force between protons and neutrons in a nucleus is a weak force."

There are four distinct fundamental forces: Gravity, Electromagnetism, Weak Nuclear, and Strong Nuclear.

The force responsible for holding protons and neutrons together in a nucleus is specifically the \textit{strong nuclear force.

The "weak force" (or weak interaction) is an entirely different fundamental force responsible for phenomena like radioactive beta decay.

Describing the primary binding force of the nucleus as a "weak force" is completely scientifically inaccurate.

Therefore, Reason (R) is also definitively False.


Step 3: {\color{redConclusion

Since both the Assertion (A) and the Reason (R) contain factually incorrect statements regarding nuclear physics, the correct choice is (D). Quick Tip: Never confuse the "strong nuclear force" (which binds the nucleus together) with the "weak nuclear force" (which mediates radioactive decay).
Also, remember the strong force's repulsive core property; it is the reason nuclei maintain a relatively constant density.

Concept:

In the photoelectric effect, the "intensity" of incident light is directly related to the number of photons striking the photosensitive surface per unit area per unit time.
Provided the incident frequency is above the threshold frequency, each photon has a probability of ejecting exactly one electron.
Therefore, increasing the intensity increases the number of emitted photoelectrons per second, which proportionally increases the photoelectric current.
The "stopping potential" relates directly to the maximum kinetic energy of the emitted photoelectrons (\( eV_0 = K_{max} \)).
According to Einstein's photoelectric equation (\( K_{max} = h\nu - \Phi \)), this maximum kinetic energy depends solely on the frequency of the incident light and the nature of the material, making it completely independent of intensity.


Step 1: {\color{redAnalyze the Assertion (A)

Assertion (A) states that photoelectric current depends on the intensity of the incident radiation.
As established in the concepts, a higher intensity means a greater number of incident photons per second.
More photons result in more photoelectrons being ejected per second, thereby increasing the photoelectric current.
Thus, Assertion (A) is a true statement.

Step 2: {\color{redAnalyze the Reason (R) and its explanatory link

Reason (R) states that stopping potential is independent of the intensity of the incident radiation.
Since intensity only affects the \textit{number of photons and not their individual \textit{energies, the maximum kinetic energy of the ejected electrons remains unchanged when intensity is varied.
Consequently, the stopping potential required to halt these electrons also remains unchanged.
Thus, Reason (R) is also a true statement.
However, when determining if (R) explains (A), we see they describe two independent characteristics of the photoelectric effect. The independence of stopping potential from intensity does not serve as the logical cause or explanation for why photoelectric current relies on intensity.

Step 3: {\color{redConclusion

Both Assertion (A) and Reason (R) are experimentally true statements, but Reason (R) is not the correct explanation of Assertion (A).
Therefore, option (B) is the correct choice. Quick Tip: For photoelectric effect questions, strictly memorize these two relationships: 1. Intensity governs the number of photons \(\rightarrow\) affects Saturation Current. 2. Frequency governs photon energy \(\rightarrow\) affects Maximum Kinetic Energy and Stopping Potential.

Concept:

A Wheatstone bridge consists of four resistive arms (P, Q, R, S), a voltage source (cell), and a null detector (galvanometer).
The bridge is in a "balanced" state when the potential difference across the galvanometer is zero, resulting in no current flow through it.
For a standard configuration with the cell across one diagonal and the galvanometer across the other, the balance condition is \( \frac{P}{Q} = \frac{R}{S} \).
The Reciprocity Theorem for electrical networks dictates that interchanging the voltage source and the detector in a linear network will not alter the balanced state of the bridge.


Step 1: {\color{redAnalyze the Assertion (A)

Assertion (A) states that interchanging the cell and galvanometer leaves the balance condition unchanged.
Consider the standard bridge: arms AB=P, BC=Q, AD=R, CD=S. The cell is across AC, and the galvanometer is across BD. Balance requires equal potentials at B and D, yielding the condition \( \frac{P}{Q} = \frac{R}{S} \).
If we interchange the components (cell across BD, galvanometer across AC), the bridge configuration is simply rotated. By the reciprocity theorem, it will remain balanced.
Thus, Assertion (A) is True.

Step 2: {\color{redAnalyze the Reason (R) and its explanatory link

Reason (R) provides the mathematical justification: \( \frac{P}{Q} = \frac{R}{S} \Rightarrow \frac{Q}{S} = \frac{P}{R} \).

Let's analyze the new circuit with the cell connected across nodes B and D.

Current enters at B and splits into two paths: path B-A-D (resistances P and R) and path B-C-D (resistances Q and S).
For the galvanometer connected across A and C to show zero current, the voltage at A must equal the voltage at C.

This requires the ratio of the voltage drops across the first resistors in each branch to be equal: \( \frac{V_P}{V_R} = \frac{V_Q}{V_S} \), which simplifies to the resistance ratio \( \frac{P}{R} = \frac{Q}{S} \).

Does the original condition mathematically guarantee this new requirement? Yes.

Taking the original condition \( \frac{P}{Q} = \frac{R}{S} \) and cross-multiplying gives \( PS = QR \). Rearranging this equation yields exactly \( \frac{P}{R} = \frac{Q}{S} \).

Because the mathematical requirement for the interchanged setup is inherently satisfied by the algebraic equivalence of the original condition, the bridge remains balanced.

Thus, Reason (R) is True and perfectly explains the physical phenomenon stated in Assertion (A).


Step 3: {\color{redConclusion

Both statements are true, and the algebraic manipulation shown in (R) represents the exact reason why the interchanged circuit stays balanced.
Therefore, option (A) is the correct choice. Quick Tip: The balance condition of a Wheatstone bridge depends on the product of opposite arms being equal (\( P \times S = Q \times R \)). Because multiplication is commutative, swapping the input (battery) and output (galvanometer) terminals never breaks this fundamental equality.

Concept:

A Moving Coil Galvanometer (MCG) operates on the principle that a current-carrying coil placed in a magnetic field experiences a deflecting torque.
The magnitude of this torque is given by \( \tau = NIAB \sin(\theta) \), where \( \theta \) is the angle between the magnetic field vector and the area vector (the normal to the coil's plane).
To ensure linear operation and maximum sensitivity, the torque must always be maximum. This requires \( \sin(\theta) = 1 \), meaning the area vector must always be perpendicular to the field lines (\( \theta = 90^\circ \)).
Consequently, the \textit{plane of the coil itself must remain \textit{parallel to the magnetic field lines at all times.
A radial magnetic field is used to achieve this parallel alignment regardless of the coil's rotation.
A soft iron core is placed centrally because it is a highly permeable ferromagnetic material. It concentrates the magnetic flux lines, which significantly increases the overall magnetic field strength \( B \) in the air gap.


Step 1: {\color{redAnalyze the Assertion (A)

Assertion (A) claims that the soft iron core "only makes the magnetic field radial and does not affect the strength".
While its cylindrical shape (along with concave pole pieces) helps create the radial field, the second part of the statement is entirely false.
Soft iron is chosen specifically for its extremely high magnetic permeability (\( \mu_r \gg 1 \)). This property allows it to pull in and concentrate the magnetic flux lines, dramatically increasing the magnetic field strength (\( B \)) in the gap where the coil rotates.
Because it heavily affects and increases the field strength, Assertion (A) is False.

Step 2: {\color{redAnalyze the Reason (R)

Reason (R) states that the plane of the coil is always perpendicular to the magnetic field.

As outlined in the concepts, for the torque to be maximum, the angle between the normal (area vector) and the field must be \( 90^\circ \).

If the normal is perpendicular to the field, it logically means the plane of the coil is parallel to the magnetic field lines.
The entire engineering purpose of the radial magnetic field is to ensure that the plane of the coil is always parallel to the field lines as it rotates, not perpendicular.

Therefore, Reason (R) is fundamentally incorrect and False.


Step 3: {\color{redConclusion

Both the Assertion (A) and the Reason (R) contain clear factual errors regarding the construction and operating principles of a moving coil galvanometer.
This leads us to select option (D). Quick Tip: Read statements about angles in electromagnetism extremely carefully. Maximum torque occurs when the Area Vector is PERPENDICULAR to the field, which means the Coil Plane is PARALLEL to the field. Also, inserting a ferromagnetic material like soft iron always INCREASES the magnetic field strength due to high permeability.

Concept:

Semiconductors are materials with electrical conductivity between that of a conductor and an insulator.
They are broadly classified into two categories based on their purity: intrinsic (pure) and extrinsic (doped).


Step 1: {\color{redPoint of Difference 1: Purity and Composition

Intrinsic Semiconductors: These are pure semiconductor materials (like pure Silicon or Germanium) without any significant dopant atoms present.
Extrinsic Semiconductors: These are impure semiconductors formed by deliberately adding (doping) a small, controlled amount of trivalent or pentavalent impurity atoms to a pure semiconductor.

Step 2: {\color{redPoint of Difference 2: Charge Carrier Concentration

Intrinsic Semiconductors: The number of thermally generated electrons (\(n_e\)) in the conduction band is exactly equal to the number of holes (\(n_h\)) in the valence band (\(n_e = n_h = n_i\)).
Extrinsic Semiconductors: The number of electrons and holes are entirely unequal. In n-type, electrons are majority carriers (\(n_e \gg n_h\)), whereas in p-type, holes are majority carriers (\(n_h \gg n_e\)).

Step 3: {\color{redConclusion

Intrinsic semiconductors have low conductivity that depends solely on temperature, whereas extrinsic semiconductors have high conductivity controlled by the doping concentration. Quick Tip: To easily remember: "Intrinsic" means "In its pure state." "Extrinsic" means something "Extra" (impurity) has been added to increase conductivity.

Concept:

The nucleus of an atom is composed of protons and neutrons (nucleons).
Precise measurements show that the actual mass of a stable nucleus is always less than the sum of the individual masses of its constituent nucleons.
This missing mass is converted into energy that holds the nucleus together.


Step 1: {\color{redExplain Mass Defect (\(\Delta m\))

Mass defect is defined as the difference between the sum of the resting masses of the individual nucleons (protons and neutrons) making up a nucleus and the actual rest mass of that nucleus.
Mathematically, for a nucleus with \(Z\) protons and \((A-Z)\) neutrons: \[ \Delta m = [Z \cdot m_p + (A - Z) \cdot m_n] - M \]

where \(m_p\) is the mass of a proton, \(m_n\) is the mass of a neutron, and \(M\) is the actual mass of the nucleus.

Step 2: {\color{redExplain Binding Energy (\(E_b\))

Binding energy is the minimum amount of energy required to completely separate a nucleus into its constituent protons and neutrons to infinite distance.
Alternatively, it is the energy released when individual nucleons bind together to form a stable nucleus.

Step 3: {\color{redRelation between Mass Defect and Binding Energy

The mass defect and binding energy are directly related by Albert Einstein's mass-energy equivalence principle. The "missing" mass (\(\Delta m\)) is converted entirely into the binding energy (\(E_b\)).
The relationship is given by the formula: \[ E_b = \Delta m \cdot c^2 \]

where \(c\) is the speed of light in vacuum.

Step 4: {\color{redConclusion

Mass defect is the missing mass upon nucleus formation, and binding energy is the energy equivalent of this missing mass, governing the stability of the nucleus. Quick Tip: A higher binding energy per nucleon indicates a more tightly bound and stable nucleus. When calculating in atomic mass units (u), remember the convenient conversion factor: \(1 u \approx 931.5 MeV\).

Concept:

The photoelectric effect is governed by Einstein's photoelectric equation: \(K_{max} = h\nu - h\nu_0\).
Here, \(K_{max}\) is the maximum kinetic energy of the emitted photoelectrons, \(h\nu\) is the energy of the incident light, and \(h\nu_0\) is the work function (minimum energy required to eject an electron).
\(\nu\) is the incident frequency and \(\nu_0\) is the threshold frequency.


Step 1: {\color{redIdentify the given parameters

Incident frequency, \(\nu = 5.0 \times 10^{14} Hz\).
Maximum speed of photoelectrons, \(v_{max} = 6.63 \times 10^5 m/s\).
Mass of an electron, \(m = 9.1 \times 10^{-31} kg\) (standard constant).
Planck's constant, \(h = 6.63 \times 10^{-34} J s\) (standard constant).

Step 2: {\color{redCalculate the maximum kinetic energy (\(K_{max}\))

Using the kinetic energy formula: \[ K_{max} = \frac{1}{2} m v_{max}^2 \]
\[ K_{max} = \frac{1}{2} \times (9.1 \times 10^{-31} kg) \times (6.63 \times 10^5 m/s)^2 \]
\[ K_{max} = 0.5 \times 9.1 \times 10^{-31} \times 43.9569 \times 10^{10} J \]
\[ K_{max} = 199.998 \times 10^{-21} J \approx 2.0 \times 10^{-19} J \]


Step 3: {\color{redCalculate the energy of the incident photon (\(E\))
\[ E = h\nu \]
\[ E = (6.63 \times 10^{-34} J s) \times (5.0 \times 10^{14} Hz) \]
\[ E = 33.15 \times 10^{-20} J = 3.315 \times 10^{-19} J \]


Step 4: {\color{redCalculate the threshold frequency (\(\nu_0\))

From Einstein's photoelectric equation: \[ h\nu_0 = h\nu - K_{max} \]
\[ h\nu_0 = 3.315 \times 10^{-19} J - 2.0 \times 10^{-19} J \]
\[ h\nu_0 = 1.315 \times 10^{-19} J \]

Now, solve for \(\nu_0\): \[ \nu_0 = \frac{1.315 \times 10^{-19}}{6.63 \times 10^{-34}} Hz \]
\[ \nu_0 = 0.19834 \times 10^{15} Hz \]
\[ \nu_0 = 1.98 \times 10^{14} Hz \]


Step 5: {\color{redConclusion

The threshold frequency for the metal surface is approximately \(1.98 \times 10^{14} Hz\). Quick Tip: When dealing with \(6.63\) in numerical problems, remember that squaring it gives roughly \(44\) (\(43.95\)). Using standard constants thoughtfully can help avoid rounding errors early in the calculation.

Concept:

When points A and C are connected by a wire of zero resistance, the potential at A becomes equal to the potential at C (\(V_A = V_C\)).
The circuit forms a single closed loop containing two batteries and two resistors.
We can apply Kirchhoff's Voltage Law (KVL) around this loop to find the steady current, and then use Ohm's law to find the potential difference across specific points.


Step 1: {\color{redEstablish the loop equation

The circuit branch contains two cells. Looking at the standard battery symbols (long line is positive, short thick line is negative):
Between A and B: \(E_1 = 6V\), \(R_1 = 1\ \Omega\). The positive terminal faces A.
Between B and C: \(E_2 = 4V\), \(R_2 = 3\ \Omega\). The positive terminal faces B.
Let us assume a steady current \(I\) flows from left to right (from A to C inside the branch).
Applying KVL from A to C: \[ V_A - E_1 - I \cdot R_1 - E_2 - I \cdot R_2 = V_C \]
\[ V_A - 6 - I(1) - 4 - I(3) = V_C \]
\[ V_A - V_C = 10 + 4I \]


Step 2: {\color{redCalculate the current in the circuit

Since A and C are connected by a wire of negligible resistance, they are at the same potential. \[ V_A = V_C \implies V_A - V_C = 0 \]

Substitute this into the equation: \[ 0 = 10 + 4I \]
\[ 4I = -10 \]
\[ I = -2.5 A \]

The negative sign indicates that the actual current flows in the opposite direction, i.e., from C to A through the branch components.
So, actual current \(I_{actual} = 2.5 A\) flowing from right to left (C \(\rightarrow\) B \(\rightarrow\) A).

Step 3: {\color{redCalculate the potential difference between B and C

We need to find \(V_B - V_C\). We will trace the path from C to B.
When moving from C to B, we are moving with the direction of the actual current (\(2.5 A\)).
As we cross the \(3\ \Omega\) resistor in the direction of current, potential drops.
As we cross the \(4V\) battery from the negative terminal to the positive terminal, potential rises. \[ V_C - I_{actual \cdot R_2 + E_2 = V_B \]
\[ V_C - (2.5)(3) + 4 = V_B \]
\[ V_C - 7.5 + 4 = V_B \]
\[ V_C - 3.5 = V_B \]

Rearranging for \(V_B - V_C\): \[ V_B - V_C = -3.5 V \]

The magnitude of the potential difference between points B and C is \(3.5 V\).

Step 4: {\color{redConclusion

The magnitude of the potential difference between points B and C is \(3.5 V\). Quick Tip: Always clearly define your assumed current direction. A negative result just means the current physically flows opposite to your assumption. Keep the sign consistent when calculating voltage drops.

Concept:

When a battery is connected to an external resistor, the total resistance of the circuit is the sum of the external resistance (\(R\)) and the internal resistance (\(r\)) of the battery.
Ohm's law for a complete circuit relates the electromotive force (emf, \(E\)) to the total current (\(I\)) and total resistance (\(R + r\)).


Step 1: {\color{redIdentify given values

Electromotive force (emf) of the battery, \(E = 21 V\).
Internal resistance of the battery, \(r = 3\ \Omega\).
Current flowing in the circuit, \(I = 3 A\).
Let the external resistance be \(R\).

Step 2: {\color{redApply Ohm's law for the complete circuit

The formula relating these quantities is: \[ I = \frac{E}{R + r} \]

Substitute the known values into the formula: \[ 3 = \frac{21}{R + 3} \]


Step 3: {\color{redSolve for the external resistance \(R\)

Rearrange the equation to isolate \(R + 3\): \[ R + 3 = \frac{21}{3} \]
\[ R + 3 = 7 \]

Subtract 3 from both sides: \[ R = 7 - 3 \]
\[ R = 4\ \Omega \]


Step 4: {\color{redConclusion

The resistance of the external resistor is \(4\ \Omega\). Quick Tip: Remember that the internal resistance acts as a resistor strictly in series with the ideal voltage source inside the battery.

Concept:

The terminal voltage (\(V\)) is the actual potential difference available across the terminals of the battery when a current is drawn from it.
Due to the voltage drop across the internal resistance, the terminal voltage is always less than the emf when the battery is discharging.
It can be calculated using \(V = E - Ir\) or by calculating the voltage drop across the external resistor \(V = IR\).


Step 1: {\color{redIdentify given values

Electromotive force, \(E = 21 V\).
Internal resistance, \(r = 3\ \Omega\).
Current in the circuit, \(I = 3 A\).
External resistance calculated previously, \(R = 4\ \Omega\).

Step 2: {\color{redCalculate the terminal voltage using the battery parameters

The formula for terminal voltage during discharge is: \[ V = E - I \cdot r \]

Substitute the values: \[ V = 21 - (3 \times 3) \]
\[ V = 21 - 9 \]
\[ V = 12 V \]


Step 3: {\color{redAlternative calculation using external resistance

The terminal voltage is also equal to the potential drop entirely across the external circuit: \[ V = I \cdot R \]

Substitute the values: \[ V = 3 \times 4 \]
\[ V = 12 V \]

Both methods yield the exact same result.

Step 4: {\color{redConclusion

The terminal voltage of the battery is \(12 V\). Quick Tip: Always cross-verify your answer in exam scenarios. If \(E - Ir\) matches \(I \times R\), you can be 100% confident your calculated resistance and terminal voltage are correct.

Concept:

In a standard Young's Double Slit Experiment (YDSE), the sources are exactly in phase (initial phase difference = \(0\)). At the center of the screen, the path lengths from both slits are equal (\(\Delta x = 0\)).
This zero path difference normally results in a net phase difference of zero, leading to constructive interference (a central bright maximum).
The total phase difference \(\Delta \phi\) at any point on the screen is the sum of the initial phase difference \(\phi_0\) and the phase difference due to the path difference (\(\frac{2\pi}{\lambda} \Delta x\)).


Step 1: {\color{redAnalyze the new phase condition at the center

The problem states that the waves emitted from slits \(S_1\) and \(S_2\) now have an inherent initial phase difference, \(\phi_0 = \pi\) radians.
At the exact geometric center of the screen, the distance travelled by the light from both slits is perfectly identical.
Therefore, the path difference at the center is still zero: \(\Delta x = 0\).
The phase difference created by the path length difference is: \[ \phi_{path} = \frac{2\pi}{\lambda} \times \Delta x = \frac{2\pi}{\lambda} \times 0 = 0 radians \]


Step 2: {\color{redCalculate the total effective phase difference

The total phase difference \(\Delta \phi\) at the central point is: \[ \Delta \phi = \phi_0 + \phi_{path} \]
\[ \Delta \phi = \pi + 0 = \pi radians \]


Step 3: {\color{redDetermine the nature of interference

When the total phase difference between two superimposing waves is an odd multiple of \(\pi\) (e.g., \(\pi, 3\pi, 5\pi\)), they are completely out of phase.
This condition strictly leads to destructive interference.
Because destructive interference minimizes intensity, a dark spot will be formed.

Step 4: {\color{redConclusion

Instead of a central bright fringe, a central dark fringe (minimum) will be obtained on the screen. The entire interference pattern will effectively shift, swapping the positions of all maxima and minima. Quick Tip: When sources are completely out of phase by \(\pi\), the conditions for maxima and minima completely swap. Constructive interference becomes destructive and vice-versa everywhere on the screen.

Concept:

The phenomenon of interference explicitly requires the superposition of two or more coherent wave trains.
In YDSE, these two coherent waves originate from the two separate slits, \(S_1\) and \(S_2\).
Diffraction, on the other hand, is the bending and spreading of light as it passes through a single narrow aperture.


Step 1: {\color{redAnalyze the physical change to the setup

When one of the slits is completely closed, light can only pass through the single remaining open slit.
Because only one source of light is present, there is no longer a second coherent wave available to superimpose with the first.

Step 2: {\color{redDetermine the resulting phenomenon

Without two superimposing waves, the sharp, evenly spaced interference fringes (alternating bright and dark bands of equal width) completely disappear.
Instead, the light passing through the single remaining narrow slit undergoes diffraction.
This results in a single-slit diffraction pattern being projected on the screen.

Step 3: {\color{redDescribe the visual outcome on the screen

The single-slit diffraction pattern is characterized by a very broad and extremely bright central maximum.
This central maximum is flanked on both sides by alternating dark and bright secondary bands (maxima and minima).
Crucially, these secondary diffraction maxima are much wider and their intensity decreases very rapidly as you move away from the center, completely unlike the uniform intensity fringes of interference.

Step 4: {\color{redConclusion

The sharp interference pattern will completely vanish, and it will be replaced by a single-slit diffraction pattern characterized by a broad, bright central maximum with rapidly fading secondary maxima. Quick Tip: Interference = Multiple slits (superposition of multiple waves). Diffraction = Single slit (spreading of a single wavefront). Closing a slit transitions the experiment fundamentally from the former to the latter.

Concept:

An electromagnetic (EM) wave consists of time-varying electric and magnetic fields propagating through space.
These fields behave according to Maxwell's equations and exhibit specific directional and phase relationships.


Step 1: {\color{redDepict the variation (Diagram placeholder)




Description of the required diagram:
Draw a 3-dimensional Cartesian coordinate system (x, y, and z axes).

Assume the wave propagates along the positive x-axis.

Draw a sine wave oscillating strictly in the x-y plane to represent the Electric Field (\(\vec{E\)).

Draw a second sine wave oscillating strictly in the x-z plane to represent the Magnetic Field (\(\vec{B}\)).

Ensure that both sine waves cross the x-axis (zero amplitude) at the exact same points and reach their peaks at the exact same points along the x-axis.


Step 2: {\color{redState Two Important Characteristics

1. Transverse Nature: The electric field vector (\(\vec{E}\)) and the magnetic field vector (\(\vec{B}\)) are mutually perpendicular to each other, and both are completely perpendicular to the direction of wave propagation (e.g., if propagation is along \(\hat{i}\), \(\vec{E}\) could be along \(\hat{j}\) and \(\vec{B}\) along \(\hat{k}\)).

2. Phase Synchronization: The oscillating electric and magnetic fields are always in identical phase. This means they reach their maximum (peak) values and their minimum (zero) values at the exact same position and instant in time.


Step 3: {\color{redConclusion

The diagram must show orthogonal oscillations, and the core characteristics are their mutually perpendicular transverse nature and in-phase oscillation. Quick Tip: Remember the right-hand rule for EM waves: The direction of wave propagation is strictly given by the cross product vector \(\vec{E} \times \vec{B}\).

Concept:

According to Maxwell's electromagnetic theory, the speed of light (an electromagnetic wave) in a vacuum is dictated by the fundamental electric and magnetic properties of free space.
These properties are the permittivity of free space (\(\varepsilon_0\)) and the permeability of free space (\(\mu_0\)).


Step 1: {\color{redIdentify the fundamental constants

The permeability of free space, representing the magnetic capability of vacuum, has a standard value: \[ \mu_0 = 4\pi \times 10^{-7} T m A^{-1} \]

The permittivity of free space, representing the electrostatic capability of vacuum, is derived from Coulomb's constant (\(\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 N m^2 C^{-2}\)): \[ \varepsilon_0 = \frac{1}{4\pi \times (9 \times 10^9)} C^2 N^{-1} m^{-2} = \frac{1}{36\pi \times 10^9} F m^{-1} \]


Step 2: {\color{redSubstitute values into the expression

We need to evaluate the expression \(\frac{1}{\sqrt{\varepsilon_0 \mu_0}}\). Let's substitute the known values: \[ \varepsilon_0 \mu_0 = \left( \frac{1}{36\pi \times 10^9} \right) \times (4\pi \times 10^{-7}) \]


Step 3: {\color{redSimplify the product

Cancel out the \(\pi\) terms and simplify the fraction: \[ \varepsilon_0 \mu_0 = \frac{4\pi \times 10^{-7}}{36\pi \times 10^9} \]
\[ \varepsilon_0 \mu_0 = \frac{4}{36} \times \frac{10^{-7}}{10^9} \]
\[ \varepsilon_0 \mu_0 = \frac{1}{9} \times 10^{-16} \]


Step 4: {\color{redCalculate the final velocity

Now, take the square root of this product and invert it: \[ \sqrt{\varepsilon_0 \mu_0} = \sqrt{\frac{1}{9} \times 10^{-16}} \]
\[ \sqrt{\varepsilon_0 \mu_0} = \frac{1}{3} \times 10^{-8} \]

Finally, find the reciprocal: \[ v = \frac{1}{\sqrt{\varepsilon_0 \mu_0}} = \frac{1}{\frac{1}{3} \times 10^{-8}} \]
\[ v = 3 \times 10^8 m/s \]


Step 5: {\color{redConclusion

The resulting value, \(3 \times 10^8 m/s\), perfectly matches the known velocity of an electromagnetic wave (light) in free space, conventionally denoted as \(c\). Thus, it is proved that \(c = \frac{1}{\sqrt{\varepsilon_0 \mu_0}}\). Quick Tip: Using the fractional form of \(\varepsilon_0\) (via \(\frac{1}{4\pi\varepsilon_0}\)) makes this calculation incredibly fast and avoids dealing with the messy decimal \(8.854 \times 10^{-12}\).

Concept:

When a ray of light passes "symmetrically" through a prism, the angle of incidence (\(i\)) is exactly equal to the angle of emergence (\(e\)).
This symmetric path occurs only when the prism is in the state of minimum deviation.
The general prism formula relating incidence, emergence, prism angle (\(A\)), and deviation (\(\delta\)) is: \(i + e = A + \delta\).


Step 1: {\color{redIdentify the given parameters

The prism has an equilateral triangular base, which means all its internal angles are equal.
Therefore, the angle of the prism, \(A = 60^\circ\).
The angle of incidence is given as \(i = 45^\circ\).
Because the light passes symmetrically, we know the deviation is at its minimum (\(\delta_m\)), and the angle of emergence equals the angle of incidence: \(e = i = 45^\circ\).

Step 2: {\color{redApply the minimum deviation formula

Use the standard prism equation: \[ i + e = A + \delta_m \]

Substitute the known values (\(e = i\)): \[ i + i = A + \delta_m \]
\[ 2i = A + \delta_m \]
\[ 2(45^\circ) = 60^\circ + \delta_m \]
\[ 90^\circ = 60^\circ + \delta_m \]


Step 3: {\color{redSolve for \(\delta_m\)

Rearrange to find the angle of minimum deviation: \[ \delta_m = 90^\circ - 60^\circ \]
\[ \delta_m = 30^\circ \]


Step 4: {\color{redConclusion

The angle of minimum deviation for the equilateral prism is \(30^\circ\). Quick Tip: The keyword "symmetrically" in prism problems is a massive hint. It instantly tells you two things: \(i = e\) and the deviation is at its absolute minimum.

Concept:

The refractive index (\(\mu\)) of the material of a prism can be determined if the angle of the prism (\(A\)) and the angle of minimum deviation (\(\delta_m\)) are known.
The relationship is given by the Prism Formula, which is derived from Snell's law applied at the symmetric position.
The formula is: \(\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}\).


Step 1: {\color{redIdentify the required angles

From the problem description and the previous part, we have:
Angle of the equilateral prism, \(A = 60^\circ\).
Angle of minimum deviation, \(\delta_m = 30^\circ\).

Step 2: {\color{redApply the Prism Formula

Substitute these angles into the refractive index formula: \[ \mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]
\[ \mu = \frac{\sin\left(\frac{60^\circ + 30^\circ}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)} \]
\[ \mu = \frac{\sin\left(\frac{90^\circ}{2}\right)}{\sin(30^\circ)} \]
\[ \mu = \frac{\sin(45^\circ)}{\sin(30^\circ)} \]


Step 3: {\color{redEvaluate the trigonometric functions

Recall standard trigonometric values: \(\sin(45^\circ) = \frac{1}{\sqrt{2}}\) \(\sin(30^\circ) = \frac{1}{2}\)
Substitute these into the equation: \[ \mu = \frac{\left( \frac{1}{\sqrt{2}} \right)}{\left( \frac{1}{2} \right)} \]
\[ \mu = \frac{1}{\sqrt{2}} \times 2 \]
\[ \mu = \frac{2}{\sqrt{2}} \]

Rationalize the fraction: \[ \mu = \sqrt{2} \]
\[ \mu \approx 1.414 \]


Step 4: {\color{redConclusion

The refractive index of the material of the prism is \(\sqrt{2}\) or approximately \(1.414\). Quick Tip: Always memorize standard trigonometric values like \(\sin 30^\circ\), \(\sin 45^\circ\), and \(\sin 60^\circ\). Leaving the answer in surd form (\(\sqrt{2}\)) is perfectly acceptable and often preferred in physics board exams.

Concept:

A p-n junction diode is a two-terminal semiconductor device that allows electric current to flow predominantly in one direction.
Its behavior changes dramatically depending on the polarity of the external voltage applied across it, a process known as biasing.


Step 1: {\color{redExplain Forward Biasing

When the positive terminal of an external battery is connected to the p-type semiconductor and the negative terminal to the n-type semiconductor, the junction is said to be forward biased.



\textit{Circuit Diagram: A battery connected with its positive terminal to the P-side (triangle base) of a diode symbol and negative terminal to the N-side (straight line).

In this configuration, the applied external voltage creates an electric field that opposes the built-in potential barrier of the depletion region.

Because the applied voltage opposes the barrier, the effective barrier height is reduced, and the width of the depletion layer decreases.

Majority charge carriers (holes from the p-side and electrons from the n-side) are pushed towards the junction, easily crossing it.

This results in a significant forward current flowing through the diode due to the continuous recombination of majority carriers at the junction.

Step 2: {\color{redExplain Reverse Biasing

When the positive terminal of an external battery is connected to the n-type semiconductor and the negative terminal to the p-type semiconductor, the junction is reverse biased.



\textit{Circuit Diagram: A battery connected with its positive terminal to the N-side (straight line) of a diode symbol and negative terminal to the P-side (triangle base).

In this configuration, the applied external voltage creates an electric field that perfectly aligns with the built-in potential barrier.

This alignment increases the effective barrier height and significantly widens the depletion layer.

Majority charge carriers are pulled away from the junction by the external battery, effectively stopping the flow of majority current.

Only a very tiny leakage current (reverse saturation current) flows due to the minority charge carriers generated by thermal agitation.

Step 3: {\color{redConclusion

Forward biasing reduces the depletion region and allows heavy current flow, acting like a closed switch. Reverse biasing widens the depletion region and blocks current flow, acting like an open switch. Quick Tip: A simple memory trick: "Forward" bias connects Positive to P-type and Negative to N-type (P-P, N-N). "Reverse" bias connects them oppositely (P-N, N-P).

Concept:

When an electric field is applied across a conductor, the free electrons drift towards the positive terminal with an average velocity known as the drift velocity (\(v_d\)).
The macroscopic electric current (\(I\)) flowing through the conductor is fundamentally derived from this microscopic drifting motion of the electrons.


Step 1: {\color{redDefine the parameters of the conductor

Consider a cylindrical conductor of length \(L\) and uniform cross-sectional area \(A\).

Let \(n\) be the number density of free electrons, which is the number of free electrons per unit volume.

The total volume of the considered conductor segment is \(V_{vol} = A \times L\).

Therefore, the total number of free electrons in this segment is \(N = n \times A \times L\).

Step 2: {\color{redCalculate the total mobile charge

Let \(e\) be the magnitude of the charge of a single electron.

The total charge \(Q\) contained within this section of the conductor is:
\[ Q = N \times e \]
\[ Q = (n A L) e = n e A L \]


Step 3: {\color{redRelate charge to current using drift velocity

When a potential difference is applied, all these free electrons drift with an average velocity \(v_d\).

The time \(t\) required for an electron to travel the entire length \(L\) of the conductor is given by:
\[ t = \frac{L}{v_d} \]

Electric current \(I\) is defined as the rate of flow of electric charge across a cross-section:
\[ I = \frac{Q}{t} \]

Substitute the expressions for \(Q\) and \(t\) into the current equation:
\[ I = \frac{n e A L}{\left(\frac{L}{v_d}\right)} \]


Step 4: {\color{redSimplify the final expression

The length parameter \(L\) cancels out from the numerator and denominator:
\[ I = n e A v_d \]


Step 5: {\color{redConclusion

The relationship between electric current and drift velocity is established as \(I = n e A v_d\). This shows that current is directly proportional to the drift velocity. Quick Tip: You can remember this formula using the mnemonic "I = venA" (current equals velocity, charge, number density, Area) or just "\(neAv_d\)". This relation is crucial for deriving Ohm's law from microscopic principles.

Concept:

Drift velocity (\(v_d\)) is directly driven by the electric field (\(E\)) established inside the conductor.
The relationship is \(v_d = \frac{e E}{m} \tau\), where \(\tau\) is the average relaxation time.
The electric field is related to the applied voltage (\(V\)) and the length of the conductor (\(L\)) by the equation \(E = \frac{V}{L}\).


Step 1: {\color{redExpress drift velocity in terms of voltage and length

Start with the fundamental equation for drift velocity:
\[ v_d = \frac{e E \tau}{m} \]

Substitute the expression for the uniform electric field \(E = \frac{V}{L}\):
\[ v_d = \frac{e \left(\frac{V}{L}\right) \tau}{m} \]
\[ v_d = \frac{e V \tau}{m L} \]

This equation shows that for a constant applied voltage \(V\) and a given material at a constant temperature (so \(\tau\) is constant), the drift velocity is inversely proportional to the length of the conductor.
\[ v_d \propto \frac{1}{L} \]


Step 2: {\color{redAnalyze the effect of doubling the length

Let the initial length be \(L\) and the initial drift velocity be \(v_{d1}\).

When the length is doubled, the new length is \(L' = 2L\).

The applied voltage \(V\) remains constant.

The new drift velocity \(v_{d2}\) will be:
\[ v_{d2} = \frac{e V \tau}{m (2L)} \]
\[ v_{d2} = \frac{1}{2} \left( \frac{e V \tau}{m L} \right) \]
\[ v_{d2} = \frac{v_{d1}}{2} \]


Step 3: {\color{redConclusion

When the length of the conductor is doubled while keeping the applied voltage constant, the internal electric field is halved. Consequently, the drift velocity of the electrons is also halved. Quick Tip: Always check what is kept constant in the problem. If the problem had said "keeping the current constant", then since \(I = neAv_d\), changing length wouldn't affect \(v_d\) as long as \(A\) is unchanged! But here, Voltage is constant, which changes the Electric Field.

Concept:

A current-carrying coil placed in a magnetic field experiences a magnetic torque.
The magnitude of this torque is given by \(\tau = N I A B \sin(\theta)\).
Here, \(N\) is the number of turns, \(I\) is the current, \(A\) is the cross-sectional area, \(B\) is the magnetic field strength, and crucially, \(\theta\) is the angle between the magnetic field vector and the area vector (which is the normal to the plane of the coil).
To prevent the coil from turning, an external torque of exactly equal magnitude must be applied in the opposite direction.


Step 1: {\color{redExtract the given parameters

Number of turns, \(N = 30\).

Radius of the coil, \(r = 8.0 cm = 0.08 m\).

Current, \(I = 6 A\).

Magnetic field, \(B = 1.0 T\).

The angle the magnetic field makes with the \textit{plane of the coil is given as \(30^\circ\).

Therefore, the angle \(\theta\) between the normal to the coil (area vector) and the magnetic field is \(\theta = 90^\circ - 30^\circ = 60^\circ\).

Step 2: {\color{redCalculate the area of the circular coil

The area \(A\) of a circle is \(\pi r^2\).
\[ A = \pi \times (0.08 m)^2 \]
\[ A = \pi \times 0.0064 m^2 \]


Step 3: {\color{redCalculate the magnitude of the torque

Apply the magnetic torque formula:
\[ \tau = N I A B \sin(\theta) \]
\[ \tau = 30 \times 6 \times (\pi \times 0.0064) \times 1.0 \times \sin(60^\circ) \]
\[ \tau = 180 \times 0.0064\pi \times \frac{\sqrt{3}{2} \]
\[ \tau = 1.152\pi \times 0.866 \]
\[ \tau = 3.133 N m \]

The magnitude of the external torque required to balance this is exactly \(3.13 N m\).

Step 4: {\color{redAnalyze the replacement with an irregular coil

The formula for magnetic torque (\(\tau = N I A B \sin\theta\)) solely depends on the total enclosed area \(A\) of the coil, not on its geometric shape.

The problem explicitly states that the new irregular planar coil encloses the \textit{same area and all other parameters (\(N, I, B, \theta\)) remain unchanged.

Therefore, the magnetic moment (\(M = N I A\)) remains identical.


Step 5: {\color{redConclusion

The required external torque is \(3.13 N m\). If replaced by an irregular coil of the same area, the torque experienced by the coil would remain exactly the same. Quick Tip: The most common mistake here is using the angle given directly (\(30^\circ\)). Always remember that the formula uses \(\sin(\theta)\) where \(\theta\) is the angle with the NORMAL to the plane. If the angle with the plane is \(\alpha\), use \(\theta = 90^\circ - \alpha\).

Concept:

When a charged particle enters a uniform magnetic field perpendicularly, it experiences a maximum Lorentz force that acts entirely as a centripetal force, causing it to move in a perfectly circular path.
The radius of this circular path is derived by equating the magnetic force to the centripetal force: \(q v B = \frac{m v^2}{r} \implies r = \frac{m v}{q B} = \frac{p}{q B}\).
The momentum \(p\) can be related to the kinetic energy \(K\) via the relation \(p = \sqrt{2 m K}\).


Step 1: {\color{redIdentify and convert the given parameters

Mass of alpha particle, \(m = 6.4 \times 10^{-27} kg\).

Charge of alpha particle, \(q = 3.2 \times 10^{-19} C\).

Magnetic field strength, \(B = 0.5 T\).

Kinetic Energy, \(K = 8.0 MeV\).

First, convert the kinetic energy from MeV to Joules (Standard SI unit):
\[ K = 8.0 \times 10^6 eV \]
\[ K = 8.0 \times 10^6 \times 1.6 \times 10^{-19} J \]
\[ K = 12.8 \times 10^{-13} J \]


Step 2: {\color{redCalculate the momentum of the alpha particle

Use the energy-momentum relation:
\[ p = \sqrt{2 m K} \]
\[ p = \sqrt{2 \times (6.4 \times 10^{-27}) \times (12.8 \times 10^{-13})} \]
\[ p = \sqrt{12.8 \times 10^{-27} \times 12.8 \times 10^{-13}} \]
\[ p = \sqrt{(12.8)^2 \times 10^{-40}} \]
\[ p = 12.8 \times 10^{-20} kg m/s \]


Step 3: {\color{redCalculate the radius of the circular path

Substitute momentum into the radius formula:
\[ r = \frac{p}{q B} \]
\[ r = \frac{12.8 \times 10^{-20}}{(3.2 \times 10^{-19}) \times 0.5} \]
\[ r = \frac{12.8 \times 10^{-20}}{1.6 \times 10^{-19}} \]
\[ r = 8 \times 10^{-1} m \]
\[ r = 0.8 m \]


Step 4: {\color{redProvide conditions for helical and undeviated paths

(i) Helical Path: The particle will describe a helical path if its initial velocity vector makes an oblique angle \(\theta\) (where \(0^\circ < \theta < 90^\circ\) or \(90^\circ < \theta < 180^\circ\)) with the direction of the uniform magnetic field. The perpendicular velocity component provides the circular motion, while the parallel component provides the linear translation.

(ii) Straight Undeviated Path: The particle will pass straight through undeviated if it is injected exactly parallel or exactly anti-parallel to the magnetic field lines (\(\theta = 0^\circ\) or \(\theta = 180^\circ\)). Under these conditions, the magnetic force \(F_m = q v B \sin\theta\) becomes entirely zero.

Step 5: {\color{redConclusion

The radius of the circular path is \(0.8 m\). A helical path requires an oblique entry angle, and an undeviated straight path requires parallel or anti-parallel entry relative to the magnetic field. Quick Tip: Notice how \(2 \times 6.4 = 12.8\), which perfectly matched the \(12.8\) from the kinetic energy. Examiners often design the numbers in these square roots to form perfect squares. Always group terms carefully before multiplying blindly!

Concept:

Electromagnetic induction is the phenomenon where a changing magnetic environment induces an electromotive force (emf) within a conductor.
This fundamental principle governs the operation of electrical generators and transformers.
The total magnetic flux \(\Phi_B\) through a surface area \(A\) is given by \(\Phi_B = B A \cos\theta\). Therefore, an emf can be induced by altering any of these three variables (\(B\), \(A\), or \(\theta\)).


Step 1: {\color{redState Faraday's Law of Electromagnetic Induction

Faraday's Law states that the magnitude of the induced electromotive force (emf) in a closed circuit is directly proportional to the time rate of change of the magnetic flux linked with that circuit.

Mathematically, it is expressed as: \[ |\varepsilon| = \left| \frac{d\Phi_B}{dt} \right| \]

where \(\varepsilon\) is the induced emf and \(\frac{d\Phi_B}{dt}\) is the rate of change of magnetic flux. For a coil with \(N\) turns, it is \(\varepsilon = -N \frac{d\Phi_B}{dt}\) (the negative sign indicates Lenz's Law).

Step 2: {\color{redDescribe two methods for producing induced emf

Based on the flux formula \(\Phi_B = B \cdot A \cdot \cos\theta\), we can induce an emf by changing the magnetic field strength (\(B\)), the area of the coil (\(A\)), or the orientation angle (\(\theta\)).

Method 1: By changing the magnetic field strength (\(B\)).

An induced emf can be generated by varying the magnitude of the magnetic field passing through a stationary coil. This is practically achieved by moving a permanent bar magnet towards or away from the coil, or by placing the coil near another circuit carrying a continuously varying alternating current (mutual induction).

Method 2: By changing the orientation angle (\(\theta\)) of the coil.

An induced emf can be produced by continuously rotating a closed coil within a uniform, constant magnetic field. As the coil rotates, the angle \(\theta\) between the area vector and the magnetic field lines constantly changes, which alters the magnetic flux continuously. This specific method is the foundational working principle of all commercial AC generators.

Step 3: {\color{redConclusion

Faraday's law links changing magnetic flux to induced voltage. The emf can be practically generated either by manipulating the external magnetic field itself or by mechanically rotating the coil within a steady field. Quick Tip: A third valid method you can mention is changing the Area (\(A\)) of the coil inside the magnetic field, often demonstrated as "motional emf" by pulling a rectangular loop out of a uniform magnetic field region.

Concept:

When a positively charged particle is fired directly at a target nucleus, it experiences strong electrostatic repulsion.
As it approaches, its initial kinetic energy is gradually converted into electrostatic potential energy.
At the "distance of closest approach" (\(r_0\)), the particle momentarily stops before rebounding. Here, its entire initial kinetic energy is completely transformed into electrostatic potential energy.
By conservation of energy: \(\frac{1}{2} m v^2 = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r_0}\).


Step 1: {\color{redEstablish the general formula for distance of closest approach

Let the target nucleus have a charge \(Z e\).

Let the approaching particle have mass \(m\), charge \(q\), and initial velocity \(v\).

Equating initial kinetic energy to potential energy at closest approach:
\[ \frac{1}{2} m v^2 = \frac{1}{4\pi\varepsilon_0} \frac{(Z e) (q)}{r_0} \]

Rearranging to solve for the distance of closest approach, \(r_0\):
\[ r_0 = \frac{1}{4\pi\varepsilon_0} \frac{2 Z e \cdot q}{m v^2} \]

This indicates that for a given target (\(Z\)) and a constant velocity (\(v\)), the distance of closest approach is directly proportional to the charge-to-mass ratio of the incident particle:
\[ r_0 \propto \frac{q}{m} \]


Step 2: {\color{redIdentify properties of the interacting particles

For a deuterium nucleus (Deuteron, d): It consists of one proton and one neutron.
Its charge is \(q_d = +e\).
Its mass is approximately \(m_d = 2m_p\) (where \(m_p\) is the proton mass).

For an alpha particle (\(\alpha\)): It consists of two protons and two neutrons.
Its charge is \(q_\alpha = +2e\).
Its mass is approximately \(m_\alpha = 4m_p\).

Step 3: {\color{redCalculate the ratio for the case of same velocity

Since both particles have the identical initial velocity \(v\), we utilize the proportionality \(r_0 \propto \frac{q}{m}\).

The ratio of their closest approach distances will be:
\[ \frac{r_{0d}}{r_{0\alpha}} = \frac{\left( \frac{q_d}{m_d} \right)}{\left( \frac{q_\alpha}{m_\alpha} \right)} \]

Substitute the known charges and masses:
\[ \frac{r_{0d}}{r_{0\alpha}} = \frac{\left( \frac{e}{2m_p} \right)}{\left( \frac{2e}{4m_p} \right)} \]

Simplify the denominator fraction (\(\frac{2e}{4m_p} = \frac{e}{2m_p}\)):
\[ \frac{r_{0d}}{r_{0\alpha}} = \frac{\left( \frac{e}{2m_p} \right)}{\left( \frac{e}{2m_p} \right)} \]
\[ \frac{r_{0d}}{r_{0\alpha}} = 1 \]


Step 4: {\color{redConclusion

When approaching with the same initial velocity, the ratio of their distances of closest approach is exactly \(1 : 1\). Quick Tip: An alpha particle is essentially twice as heavy and has twice the charge of a deuteron. Because kinetic energy depends on mass linearly, and potential energy depends on charge linearly, the doubling effects cancel each other out entirely when velocity is constant.

Concept:

As established, the distance of closest approach occurs when initial kinetic energy entirely equals electrostatic potential energy.
Kinetic energy \(K\) can be expressed in terms of momentum \(p\) using the relation \(K = \frac{p^2}{2m}\).
Substituting this into the energy conservation equation provides a different proportional relationship suitable for constant momentum conditions.


Step 1: {\color{redDerive \(r_0\) in terms of momentum

Equate kinetic energy (in terms of momentum) to potential energy at closest approach:
\[ K = U \]
\[ \frac{p^2}{2m} = \frac{1}{4\pi\varepsilon_0} \frac{(Z e) (q)}{r_0} \]

Rearranging to solve for the distance of closest approach, \(r_0\):
\[ r_0 = \frac{1}{4\pi\varepsilon_0} \frac{2 m Z e \cdot q}{p^2} \]

This indicates that for a given target (\(Z\)) and a strictly constant momentum (\(p\)), the distance of closest approach is directly proportional to the product of the particle's mass and its charge:
\[ r_0 \propto m \cdot q \]


Step 2: {\color{redIdentify properties of the interacting particles

As detailed previously:

For a deuterium nucleus (d): Charge \(q_d = e\), Mass \(m_d = 2m_p\).

For an alpha particle (\(\alpha\)): Charge \(q_\alpha = 2e\), Mass \(m_\alpha = 4m_p\).

Step 3: {\color{redCalculate the ratio for the case of same momentum

Since both particles share identical initial momentum \(p\), we utilize the proportionality \(r_0 \propto m \cdot q\).

The ratio of their closest approach distances will be:
\[ \frac{r_{0d}}{r_{0\alpha}} = \frac{m_d \cdot q_d}{m_\alpha \cdot q_\alpha} \]

Substitute the known masses and charges into the ratio:
\[ \frac{r_{0d}}{r_{0\alpha}} = \frac{(2m_p) \cdot (e)}{(4m_p) \cdot (2e)} \]

Simplify the numerator and denominator:
\[ \frac{r_{0d}}{r_{0\alpha}} = \frac{2 m_p e}{8 m_p e} \]

Cancel the common terms (\(m_p e\)) and reduce the fraction:
\[ \frac{r_{0d}}{r_{0\alpha}} = \frac{2}{8} = \frac{1}{4} \]


Step 4: {\color{redConclusion

When approaching with the same initial momentum, the ratio of their distances of closest approach is \(1 : 4\). Quick Tip: Whenever a question presents a "same [quantity]" scenario, always immediately manipulate the core formula to explicitly include that constant quantity (like converting \(v\) to \(p\) here) before taking any ratios. It prevents critical substitution errors.

Concept:

When two or more capacitors are connected head-to-tail in a single branch, they are said to be in a series combination.
The reciprocal of the equivalent (total) capacitance of capacitors in series is equal to the sum of the reciprocals of their individual capacitances.
For two capacitors in series, the formula simplifies conveniently to the "product over sum" rule: \(C_{eq} = \frac{C_1 \times C_2}{C_1 + C_2}\).


Step 1: {\color{redIdentify the capacitor configuration

By carefully examining the provided circuit diagram, we see a specific branch located between node A and node B.
This specific branch exclusively contains the \(6 \ \mu\)F capacitor and the \(3 \ \mu\)F capacitor connected end-to-end sequentially.
Therefore, these two capacitors are strictly connected in a series configuration relative to each other.

Step 2: {\color{redCalculate the total equivalent capacitance

Let \(C_1 = 6 \ \mu\)F and \(C_2 = 3 \ \mu\)F.

Apply the series equivalent capacitance formula:
\[ C_{eq} = \frac{C_1 \times C_2}{C_1 + C_2} \]

Substitute the given values:
\[ C_{eq} = \frac{6 \times 3}{6 + 3} \]
\[ C_{eq} = \frac{18}{9} \]
\[ C_{eq} = 2 \ \muF \]


Step 3: {\color{redConclusion

The total equivalent capacitance of the capacitor branch in the circuit is exactly \(2 \ \mu\)F. This directly matches option (D). Quick Tip: A quick check for series capacitors: The equivalent capacitance will ALWAYS be mathematically smaller than the smallest individual capacitor in that series chain. Since 2 is less than 3, our answer makes physical sense.

Concept:

A fully charged capacitor acts essentially as an open circuit (infinite resistance) to direct current (DC).
The phrase "connected... for a long time" implies the circuit has completely reached its DC steady state.
In this steady state, absolutely zero current flows through any branch containing a capacitor.
Current will exclusively flow through the purely resistive branches, governed by Ohm's Law (\(I = \frac{V}{R_{eq}}\)).


Step 1: {\color{redAnalyze the steady-state circuit

Because the circuit has been running "for a long time" with a DC battery, the \(6 \ \mu\)F and \(3 \ \mu\)F capacitors are fully charged.
Thus, the middle branch between A and B containing these capacitors blocks all DC current. The current in this specific branch is \(0\) A.
Therefore, the total steady current from the 3V battery flows out, passes entirely through the top \(5 \ \Omega\) resistor, and then must route completely through the \(10 \ \Omega\) resistor to return to the battery.

Step 2: {\color{redCalculate total resistance and total current

Since the current paths through the \(5 \ \Omega\) and \(10 \ \Omega\) resistors are sequential, they act as a simple series combination.
Total resistance of the active circuit, \(R_{eq} = 5 \ \Omega + 10 \ \Omega = 15 \ \Omega\).
The applied battery voltage is \(V = 3 V\).
Applying Ohm's law to find the main circuit current:
\[ I = \frac{V}{R_{eq}} \]
\[ I = \frac{3 V}{15 \ \Omega} \]
\[ I = 0.2 A \]


Step 3: {\color{redConclusion

Because all the active current flows through the \(10 \ \Omega\) resistor, the current passing through it is precisely \(0.2 A\). This matches option (C). Quick Tip: Whenever a DC circuit problem states "after a long time" or "steady state", simply physically erase or ignore any branches containing capacitors. Analyze what remains purely as a resistor circuit.

Concept:

The potential difference between two specific nodes in a circuit can be found by calculating the voltage drop across any single branch connecting those two nodes.
Nodes A and B are physically connected by the \(10 \ \Omega\) resistor branch.
According to Ohm's Law, the potential difference across a resistor is exactly the product of the current flowing through it and its resistance (\(V = I \times R\)).


Step 1: {\color{redIdentify the active component between A and B

From our previous steady-state analysis, we know that nodes A and B have two parallel branches between them.
One branch contains the fully charged capacitors (which blocks DC current).
The other branch contains the \(10 \ \Omega\) resistor.
The potential difference \(V_{AB}\) must strictly be the voltage drop appearing across this \(10 \ \Omega\) resistor.

Step 2: {\color{redCalculate the voltage drop

We already successfully calculated that the steady current \(I\) flowing entirely through the \(10 \ \Omega\) resistor is \(0.2 A\).
Apply Ohm's law specifically across this resistor:
\[ V_{AB} = I_{resistor} \times R \]
\[ V_{AB} = 0.2 A \times 10 \ \Omega \]
\[ V_{AB} = 2 V \]


Step 3: {\color{redConclusion

The potential difference heavily maintained between points A and B is exactly \(2 V\). This definitively matches option (A). Quick Tip: Alternatively, you can view the circuit as a voltage divider. The 3V battery is split across the \(5 \ \Omega\) and \(10 \ \Omega\) series resistors. Voltage across the \(10 \ \Omega\) part is \(V_{10} = 3V \times \frac{10}{10 + 5} = 3 \times \frac{10}{15} = 2V\). Both methods are equally valid and fast.

Concept:

Capacitors connected in a series configuration inherently store the exact same magnitude of charge (\(Q\)) on their plates, regardless of their individual capacitance values.
The total charge stored in this series combination is completely determined by the equivalent capacitance (\(C_{eq}\)) of the branch and the total potential difference applied directly across that entire branch.
The relationship is given by the fundamental formula: \(Q = C_{eq} \times V_{branch}\).


Step 1: {\color{redIdentify the parameters for the capacitor branch

The capacitor branch is connected directly between node A and node B.
From a previous calculation, we established that the potential difference across nodes A and B is exactly \(V_{AB} = 2 V\). This is the voltage forcefully applied across the entire series capacitor combination.
We also previously calculated the equivalent capacitance of this series branch to be \(C_{eq} = 2 \ \mu\)F.

Step 2: {\color{redCalculate the total charge on the series branch

Use the standard capacitance charge formula:
\[ Q = C_{eq} \times V_{AB} \]

Substitute the known values:
\[ Q = (2 \ \muF) \times (2 V) \]
\[ Q = 4 \ \muC \]

Because the \(6 \ \mu\)F capacitor and the \(3 \ \mu\)F capacitor are connected in strict series, they must forcefully share this exact identical charge.

Step 3: {\color{redConclusion

The charge residing strictly on the plates of the \(6 \ \mu\)F capacitor is \(4 \ \mu\)C. This corresponds directly to option (B). Quick Tip: Remember: In series, Charge (\(Q\)) is the exact same for all components, but Voltage (\(V\)) divides. In parallel, Voltage (\(V\)) is the exact same, but Charge (\(Q\)) divides based on capacity.

Concept:

The question asks about the overall current in the resistive portion of the circuit after a physical modification is made.
It's crucial to evaluate what role the modified branch was actively playing in the circuit \textit{just before it was altered.
In a fully settled DC steady state, a branch containing a capacitor possesses infinite resistance and actively passes completely zero current.


Step 1: {\color{redAnalyze the initial steady state

Before any wires are cut, the circuit has been connected "for a long time," meaning it has achieved a perfect steady state.
In this state, the capacitors are fully charged. As established previously, the entire capacitor branch between node A and node B acts as an open circuit.
The steady-state current flowing through this specific capacitor branch is \(0 A\).
The only active current in the entire system is the \(0.2 A\) flowing through the \(5 \ \Omega\) and \(10 \ \Omega\) resistors.

Step 2: {\color{redAnalyze the effect of cutting the wire

The wire is physically cut at point P, which is located directly between the two series capacitors.
Cutting this wire physically creates an open circuit in that specific branch.
However, electrically, that branch was \textit{already acting identically to an open circuit regarding the DC current flow.
Since the branch was previously carrying zero current, physically severing it completely fails to alter the resistance profile or the current distribution of the remaining active resistive loops in the circuit.

Step 3: {\color{redConclusion

Because the capacitor branch was entirely inactive in terms of continuous DC current flow, severing it changes nothing. The current in the main circuit will undeniably remain the same at \(0.2 A\). This makes option (C) correct. Quick Tip: This is a classic conceptual trick question. Examiners try to make you think a physical change always implies an electrical change. Always evaluate the pre-change electrical state first!

Concept:

When a particle with charge \(q\) is accelerated from rest across a potential difference \(V\), the work done entirely transforms into the particle's kinetic energy.
The kinetic energy \(K\) acquired is governed strictly by the formula: \(K = |q|V\), where \(|q|\) is the magnitude of the charge.
This relationship is notably completely independent of the mass of the accelerated particle.


Step 1: {\color{redCalculate kinetic energy for Particle 1

Particle 1 has a charge of \(-q\). Since it is being accelerated, we utilize the magnitude of the charge to calculate the positive kinetic energy gained.

The accelerating potential difference is \(V\).
\[ K_1 = |-q| \times V = qV \]


Step 2: {\color{redCalculate kinetic energy for Particle 2

Particle 2 has a charge of \(+2q\).

The accelerating potential difference is exactly the same, \(V\).
\[ K_2 = |+2q| \times V = 2qV \]


Step 3: {\color{redDetermine the ratio

We are asked to find the specific ratio \(\frac{K_1}{K_2}\).
\[ \frac{K_1}{K_2} = \frac{qV}{2qV} \]

Cancel the common terms \(q\) and \(V\) from the numerator and denominator:
\[ \frac{K_1}{K_2} = \frac{1}{2} \]


Step 4: {\color{redConclusion

The ratio of their acquired kinetic energies is strictly \(1/2\). This elegantly corresponds to option (A). Quick Tip: A common trap is trying to involve the masses (\(m\) and \(m/2\)) or calculating velocities first. Remember, \(K = qV\) relies only on charge and potential. Ignore extraneous information designed to confuse you!

Concept:

When a charged particle enters a uniform magnetic field perpendicularly, it traces a circular trajectory.
The radius \(r\) of this circular path is heavily dependent on the particle's momentum \(p\), charge \(q\), and the magnetic field \(B\): \(r = \frac{p}{qB}\).
Momentum can be smoothly linked back to kinetic energy \(K\) via the relation \(p = \sqrt{2mK}\).
Combining these provides a comprehensive radius formula: \(r = \frac{\sqrt{2mK}}{qB}\).


Step 1: {\color{redDerive expression for the radius of Particle 1

For Particle 1:

Mass = \(m\)

Charge magnitude = \(q\)

Kinetic energy = \(K_1 = qV\)

Apply the comprehensive radius formula:
\[ r_1 = \frac{\sqrt{2 m K_1}}{q B} \]
\[ r_1 = \frac{\sqrt{2 m (qV)}}{q B} = \frac{\sqrt{2 m q V}}{q B} \]


Step 2: {\color{redDerive expression for the radius of Particle 2

For Particle 2:

Mass = \(\frac{m}{2}\)

Charge magnitude = \(2q\)

Kinetic energy = \(K_2 = 2qV\)

Apply the comprehensive radius formula:
\[ r_2 = \frac{\sqrt{2 \left(\frac{m}{2}\right) K_2}}{(2q) B} \]
\[ r_2 = \frac{\sqrt{2 \left(\frac{m}{2}\right) (2qV)}}{2q B} \]
\[ r_2 = \frac{\sqrt{m (2qV)}}{2q B} = \frac{\sqrt{2 m q V}}{2q B} \]


Step 3: {\color{redCalculate the ratio of the radii

Now, divide the expression for \(r_1\) by the expression for \(r_2\):
\[ \frac{r_1}{r_2} = \frac{ \left( \frac{\sqrt{2 m q V}}{q B} \right) }{ \left( \frac{\sqrt{2 m q V}}{2q B} \right) } \]

The massive square root terms (\(\sqrt{2 m q V}\)) and the \(B\) terms cancel out entirely:
\[ \frac{r_1}{r_2} = \frac{ \left( \frac{1}{q} \right) }{ \left( \frac{1}{2q} \right) } \]
\[ \frac{r_1}{r_2} = \frac{1}{q} \times \frac{2q}{1} = 2 \]


Step 4: {\color{redConclusion

The ratio of their circular radii \(\left(\frac{r_1}{r_2}\right)\) is exactly 2. This perfectly matches option (D). Quick Tip: Notice that the numerators (which represent momentum, \(\sqrt{2mK}\)) were mathematically identical for both particles. \(p_1 = \sqrt{2m(qV)}\) and \(p_2 = \sqrt{2(m/2)(2qV)} = \sqrt{mqV \cdot 2} = \sqrt{2mqV}\). Since momentum is equal, the ratio of radii simply becomes the inverse ratio of their charges.

Concept:

The direction of the magnetic force \(\vec{F}\) acting on a moving charged particle is strictly determined by the Lorentz force vector cross product: \(\vec{F} = q(\vec{v} \times \vec{B})\).
The resultant direction of this force dictates the initial bending direction of the particle's path, and consequently, its direction of circular revolution (clockwise or anticlockwise).
We evaluate this from the standard perspective, typically looking down directly from the \(+z\) axis towards the xy-plane.


Step 1: {\color{redDetermine force and direction for Particle 1

Particle 1 has a strictly negative charge: \(q_1 = -q\).

Its initial velocity vector is entirely along the +x axis: \(\vec{v}_1 = v_1 \hat{i}\).

The uniform magnetic field vector is entirely along the +z axis: \(\vec{B} = B_0 \hat{k}\).

Apply the Lorentz force equation:
\[ \vec{F}_1 = (-q) (\vec{v}_1 \times \vec{B}) \]
\[ \vec{F}_1 = (-q) [(v_1 \hat{i}) \times (B_0 \hat{k})] \]

Using the right-hand rule for unit vectors, \(\hat{i} \times \hat{k} = -\hat{j}\).
\[ \vec{F}_1 = -q \cdot v_1 B_0 (-\hat{j}) \]
\[ \vec{F}_1 = +q v_1 B_0 \hat{j} \]

The force points strongly in the \(+y\) direction. An initial velocity in the \(+x\) direction being violently pulled toward the \(+y\) direction causes the particle to curve leftwards. Looking from the \(+z\) axis, this curving path forms an anticlockwise (counter-clockwise) revolution in the xy-plane.

Step 2: {\color{redDetermine force and direction for Particle 2

Particle 2 has a strictly positive charge: \(q_2 = +2q\).

Its initial velocity vector is identical in direction: \(\vec{v}_2 = v_2 \hat{i}\).

The uniform magnetic field remains: \(\vec{B} = B_0 \hat{k}\).

Apply the Lorentz force equation:
\[ \vec{F}_2 = (+2q) (\vec{v}_2 \times \vec{B}) \]
\[ \vec{F}_2 = (+2q) [(v_2 \hat{i}) \times (B_0 \hat{k})] \]

Again, \(\hat{i} \times \hat{k} = -\hat{j}\).
\[ \vec{F}_2 = +2q \cdot v_2 B_0 (-\hat{j}) \]
\[ \vec{F}_2 = -2q v_2 B_0 \hat{j} \]

The force points strongly in the \(-y\) direction. An initial velocity in the \(+x\) direction being violently pulled toward the \(-y\) direction causes the particle to curve rightwards. Looking from the \(+z\) axis, this curving path forms a clockwise revolution in the xy-plane.

Step 3: {\color{redConclusion

Particle 1 revolves securely in an anticlockwise direction, and particle 2 revolves securely in a clockwise direction. This precisely matches option (D). Quick Tip: You can use Fleming's Left-Hand Rule as a physical shortcut. For a positive charge moving right (\(+x\)) in a field pointing out/up (\(+z\)), the thumb (force) points down (\(-y\)), meaning it curves into a clockwise circle. A negative charge always experiences the exact opposite force, thereby curving upwards (\(+y\)) into an anticlockwise circle.

Concept:

The time period (\(T\)) of a charged particle revolving in a uniform magnetic field is the time taken to complete one full circular orbit.
It depends exclusively on the particle's mass \(m\), charge magnitude \(q\), and the magnetic field \(B\).
The standard derived formula is \(T = \frac{2\pi m}{qB}\).
Notably, the time period is entirely independent of the particle's velocity or kinetic energy.


Step 1: {\color{redEstablish the time period for Particle 1

For Particle 1, mass is \(m\) and charge magnitude is \(q\).

The given time period is \(T_1 = 4 s\).

According to the standard formula:
\[ T_1 = \frac{2\pi m}{q B} = 4 s \]


Step 2: {\color{redDerive the time period formula for Particle 2

For Particle 2, the mass is given as \(\frac{m}{2}\) and the charge magnitude is \(2q\).

Apply these specific values into the standard time period formula to find \(T_2\):
\[ T_2 = \frac{2\pi \left(\frac{m}{2}\right)}{(2q) B} \]

Simplify the complex fraction:
\[ T_2 = \frac{\pi m}{2q B} \]


Step 3: {\color{redRelate \(T_2\) to \(T_1\) and calculate

We can rewrite \(T_2\) to expose the structure of \(T_1\) within it:
\[ T_2 = \frac{1}{4} \times \left( \frac{2\pi m}{q B} \right) \]

Since we know that the parenthetical term is exactly \(T_1\):
\[ T_2 = \frac{1}{4} \times T_1 \]

Substitute the known value of \(T_1 = 4 s\):
\[ T_2 = \frac{1}{4} \times 4 s \]
\[ T_2 = 1 s \]


Step 4: {\color{redConclusion

The period of revolution for particle 2 will be firmly established as \(1 s\). This matches option (A). Quick Tip: Always check the dependencies first! Because \(T \propto \frac{m}{q}\), if mass is halved (factor of \(1/2\)) and charge is doubled (factor of \(1/2\) in denominator), the overall effect is multiplying by \(\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\). Thus, \(4 s / 4 = 1 s\).

Concept:

When a charged particle is accelerated from rest by an electric potential difference \(V\), it acquires kinetic energy \(K = qV\).
This acquired kinetic energy can be smoothly translated into the particle's momentum \(p\).
The foundational relationship between momentum, mass, and kinetic energy is \(p = \sqrt{2mK}\).


Step 1: {\color{redCalculate momentum for Particle 1

Particle 1 has mass \(m\) and acquired kinetic energy \(K_1 = qV\).

Utilize the momentum-energy relation:
\[ p_1 = \sqrt{2 \cdot m \cdot K_1} \]

Substitute the kinetic energy:
\[ p_1 = \sqrt{2 m (qV)} = \sqrt{2mqV} \]


Step 2: {\color{redCalculate momentum for Particle 2

Particle 2 has a distinctly different mass \(\frac{m}{2}\) and acquired kinetic energy \(K_2 = 2qV\).

Utilize the momentum-energy relation:
\[ p_2 = \sqrt{2 \cdot \left(\frac{m}{2}\right) \cdot K_2} \]

Substitute the kinetic energy:
\[ p_2 = \sqrt{2 \cdot \left(\frac{m}{2}\right) \cdot (2qV)} \]

Simplify the terms inside the massive square root:

The \(2\) and the \(\frac{1}{2}\) conveniently cancel each other out, leaving:
\[ p_2 = \sqrt{m \cdot (2qV)} = \sqrt{2mqV} \]


Step 3: {\color{redCompare the two momentums

Comparing the completely simplified expressions from Step 1 and Step 2:
\(p_1 = \sqrt{2mqV}\)
\(p_2 = \sqrt{2mqV}\)

It is mathematically obvious that \(p_1\) is exactly equal to \(p_2\).

Step 4: {\color{redConclusion

Despite having vastly different masses and charges, their resulting momentums are entirely identical due to how the mass and charge ratios perfectly canceled out. Thus, \(p_1 = p_2\), corresponding to option (B). Quick Tip: This equality of momentum (\(p_1 = p_2\)) is precisely why, in question 30(ii), the ratio of their radii was dictated strictly by the inverse ratio of their charges (\(r \propto 1/q\)), as radius equals momentum divided by \(qB\).

Concept:

Refraction at a single spherical surface forms the foundational basis of image formation by lenses.

The fundamental equation relates object distance, image distance, and the refractive indices of the respective media.

We utilize the paraxial approximation, meaning the rays make exceedingly small angles with the principal axis.



Step 1: {\color{redRay Diagram Formulation

Let a point object \(O\) be placed on the principal axis in a rarer medium of refractive index \(n_1\).

The light ray strikes a convex spherical refracting surface separating the rarer medium from a denser medium of refractive index \(n_2\).





A paraxial ray from \(O\) is incident at point \(A\) on the surface and bends towards the normal \(CN\), forming a real image at \(I\).


Step 2: {\color{redGeometrical Analysis

Let the incident ray make an angle \(\alpha\) with the principal axis, the refracted ray make an angle \(\beta\), and the normal make an angle \(\gamma\).

From the basic geometry of the triangles formed, the exterior angle is always equal to the sum of interior opposite angles.

In triangle \(OAC\), the angle of incidence \(i\) is given by \(i = \alpha + \gamma\).

In triangle \(AIC\), the angle of refraction \(r\) is given by \(\gamma = r + \beta\), which implies \(r = \gamma - \beta\).


Step 3: {\color{redApplying Paraxial Approximations

For paraxial rays, the point \(A\) is assumed very close to the pole \(P\), making the angles extremely small.

We can approximate the angles using their tangents:
\(\alpha \approx \tan \alpha \approx \frac{AM}{PO} \approx \frac{AM}{-u}\).
\(\beta \approx \tan \beta \approx \frac{AM}{PI} \approx \frac{AM}{v}\).
\(\gamma \approx \tan \gamma \approx \frac{AM}{PC} \approx \frac{AM}{R}\).


Step 4: {\color{redSnell's Law Application

According to Snell's law, for small angles, \(n_1 i = n_2 r\).

Substituting the expressions for \(i\) and \(r\):
\(n_1 (\alpha + \gamma) = n_2 (\gamma - \beta)\).

Substituting the tangent approximations alongside the Cartesian sign convention gives:
\(n_1 \left(\frac{AM}{-u} + \frac{AM}{R}\right) = n_2 \left(\frac{AM}{R} - \frac{AM}{v}\right)\).

Canceling out the common altitude \(AM\) from both sides leaves us with:
\(\frac{-n_1}{u} + \frac{n_1}{R} = \frac{n_2}{R} - \frac{n_2}{v}\).

Rearranging the terms yields the final comprehensive relation:
\(\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}\).

If the first medium is vacuum or air (\(n_1 = 1\)) and the second medium has a refractive index \(n\) (\(n_2 = n\)), the formula simplifies to:
\(\frac{n}{v} - \frac{1}{u} = \frac{n - 1}{R}\).
Quick Tip: Always define the variables and state the Cartesian sign convention explicitly when deriving optical formulas in board exams.
Remember that the paraxial approximation is the critical assumption that logically permits substituting angles with their tangents.

Concept:

The thin lens formula precisely relates the object distance \(u\), image distance \(v\), and focal length \(f\).

The linear magnification relates the height of the image to the height of the object and reveals the image's nature.

Correct application of the Cartesian sign convention is mandatory for accurate results.



Step 1: {\color{redIdentify the Given Parameters

The lens is a convex lens, which inherently possesses a positive focal length.

Focal length, \(f = +20 cm\).

The object is placed in front of the lens, so the object distance is taken as negative.

Object distance, \(u = -30 cm\).


Step 2: {\color{redApply the Thin Lens Formula

The thin lens formula is mathematically defined as:
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \]

Substitute the known values into the equation:
\[ \frac{1}{v} - \frac{1}{-30} = \frac{1}{20} \]
\[ \frac{1}{v} + \frac{1}{30} = \frac{1}{20} \]

Rearrange the terms to isolate the image distance variable:
\[ \frac{1}{v} = \frac{1}{20} - \frac{1}{30} \]


Step 3: {\color{redCalculate the Image Position

Find a common denominator to subtract the fractions smoothly:
\[ \frac{1}{v} = \frac{3 - 2}{60} \]
\[ \frac{1}{v} = \frac{1}{60} \]

Taking the reciprocal of both sides gives the final position:
\[ v = +60 cm \]

The positive sign firmly indicates that the image is formed on the other side of the lens.


Step 4: {\color{redDetermine the Nature of the Image

We calculate the linear magnification \(m\) to confirm the image traits:
\[ m = \frac{v}{u} \]
\[ m = \frac{+60}{-30} = -2 \]

The negative sign of magnification proves that the image is strictly inverted and real.

The magnitude being greater than 1 signifies that the image is significantly magnified.
Quick Tip: For a convex lens, if the object is placed beyond the principal focus, the resulting image will always be real, inverted, and formed on the opposite side.
Always double-check your sign conventions before performing reciprocal math.

Concept:

When two thin lenses are placed perfectly in contact, they act as a single equivalent lens.

The image intelligently formed by the first lens serves as a virtual object for the second lens.

The total power of the optical combination is simply the algebraic sum of the individual powers.



Step 1: {\color{redAnalyze the First Lens

Consider two thin converging lenses \(L_1\) and \(L_2\) of focal lengths \(f_1\) and \(f_2\) respectively, placed coaxially in contact.

Let a point object \(O\) be placed strictly on the principal axis at a distance \(u\) from the optical center of the combination.

The first lens \(L_1\) attempts to form an image at a point \(I_1\) at a distance \(v_1\).

Using the thin lens formula for the first lens alone:
\[ \frac{1}{v_1} - \frac{1}{u} = \frac{1}{f_1} \]


Step 2: {\color{redAnalyze the Second Lens

This intermediate image \(I_1\) behaves precisely as a virtual object for the second lens \(L_2\).

The second lens intercepts the converging rays and forms the final real image \(I\) at a distance \(v\).

Applying the thin lens formula for the second lens, treating \(v_1\) as the object distance:
\[ \frac{1}{v} - \frac{1}{v_1} = \frac{1}{f_2} \]


Step 3: {\color{redCombine the Mathematical Equations

We algebraically add the two equations obtained from Step 1 and Step 2:
\[ \left( \frac{1}{v_1} - \frac{1}{u} \right) + \left( \frac{1}{v} - \frac{1}{v_1} \right) = \frac{1}{f_1} + \frac{1}{f_2} \]

The term containing the intermediate image distance cleanly cancels out:
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f_1} + \frac{1}{f_2} \]


Step 4: {\color{redDetermine the Equivalent Focal Length

If this entire two-lens system is replaced by a single equivalent lens of focal length \(F\) that forms the image \(I\) at distance \(v\) for an object at distance \(u\), its formula would be:
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{F} \]

Comparing this generalized equation with our derived combination equation yields:
\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \]

This comprehensively proves the relation for lenses in contact.
Quick Tip: This specific derivation heavily relies on the lenses being thin enough so that their optical centers effectively coincide.
If the lenses are slightly separated by a distance \(d\), the formula uniquely changes to \(1/F = 1/f_1 + 1/f_2 - d/(f_1 f_2)\).

Concept:

Young's Double Slit Experiment (YDSE) produces interference fringes whose width is directly proportional to wavelength and screen distance, and inversely proportional to slit separation.

Dark fringes dynamically occur at positions where destructive interference takes place, governed by odd multiples of half-wavelengths.

When an optical setup is submerged in a denser medium, the wavelength of light proportionally decreases, directly affecting the fringe width.



Step 1: {\color{redList the parameters and calculate Fringe Width (I)

Given wavelength, \(\lambda = 550 nm = 5.5 \times 10^{-7} m\).

Slit separation distance, \(d = 1.1 mm = 1.1 \times 10^{-3} m\).

Screen distance, \(D = 2.2 m\).

The expression for fringe width \(\beta\) is formally given by:
\[ \beta = \frac{\lambda D}{d} \]

Substituting the listed numerical values:
\[ \beta = \frac{5.5 \times 10^{-7} \times 2.2}{1.1 \times 10^{-3}} \]
\[ \beta = 5.5 \times 10^{-7} \times 2 \times 10^3 = 11 \times 10^{-4} m \]

Converting to a readable metric unit gives:
\[ \beta = 1.1 mm \]


Step 2: {\color{redCalculate Distance of Second Dark Fringe (II)

The generalized position for the \(n\)-th dark fringe from the central maximum is given by:
\[ y_n = \left(n - \frac{1}{2}\right) \beta \]

For the second completely dark fringe, we specifically set \(n = 2\):
\[ y_2 = \left(2 - \frac{1}{2}\right) \beta = 1.5 \beta \]

Substitute the previously calculated fringe width:
\[ y_2 = 1.5 \times 1.1 mm = 1.65 mm \]


Step 3: {\color{redAnalyze Immersion in Water (III)

When the entire YDSE apparatus is completely immersed in water, the refractive index of the surrounding medium physically changes.

The new wavelength \(\lambda'\) in water is inherently shorter than in air, expressed as \(\lambda' = \lambda / \mu_{water}\).

Because the fringe width \(\beta\) is strictly proportional to the wavelength, the new fringe width \(\beta'\) will similarly decrease:
\[ \beta' = \frac{\beta}{\mu_{water}} \]

Consequently, the entire interference fringe pattern visually shrinks, causing the fringes to become noticeably narrower and closer together.
Quick Tip: Always clearly distinguish between slit width (often denoted by \(a\), dictating diffraction) and slit separation (\(d\), dictating interference).
The position formula for dark fringes uses \((n - 0.5)\) when counting starts naturally from \(n=1\).

Concept:

The scalar electric potential \(V\) uniquely created by an isolated point charge \(Q\) at a radial distance \(r\) is explicitly given by the formula \(V = \frac{1}{4\pi\epsilon_0}\frac{Q}{r}\).

Consequently, if a mathematical graph is meticulously plotted between the potential \(V\) on the y-axis and the inverse distance \(1/r\) on the x-axis, the resulting curve is structurally a straight line passing perfectly through the origin.

The mathematical slope of this specific straight line is strictly proportional to both the magnitude and the fundamental sign of the source charge \(Q\).



Step 1: {\color{redAnalyze the mathematical relationship

The governing electrostatic equation is:
\[ V = \left(\frac{Q}{4\pi\epsilon_0}\right) \cdot \frac{1}{r} \]

By comparing this directly to the standard linear equation form \(y = mx\), we can easily map the variables:

Here, \(y\) strictly corresponds to \(V\), and \(x\) strictly corresponds to \((1/r)\).

The resulting slope \(m\) of the graphed line is therefore definitively equal to the constant term:
\[ m = \frac{Q}{4\pi\epsilon_0} \]

Because the factor \(1/(4\pi\epsilon_0)\) is a strictly positive universal constant, the sign of the slope \(m\) is completely determined by the physical sign of the charge \(Q\).


Step 2: {\color{redEvaluate Line OA to determine \(Q_1\)


Looking closely at the graphical lines provided in the figure, we mathematically assess their respective plotted slopes.

The plotted line OA visibly lies entirely within the first quadrant, making an angle of \(+60^\circ\) with the positive x-axis.

Because the tangent of an acute positive angle is positive, the mathematical slope of line OA is demonstrably positive.

Since a positive slope implies a strictly positive generating charge according to our derived relation, \(Q_1\) must definitively be a positive point charge.


Step 3: {\color{redEvaluate Line OB to determine \(Q_2\)

Conversely, the plotted line OB inherently ventures downward into the fourth quadrant, generating negative potential values for positive distances.

It geometrically makes an angle of \(-30^\circ\) (or \(330^\circ\)) with the primary positive x-axis.

The tangent of this specific angle is mathematically negative, firmly meaning the slope of line OB is inherently negative.

Therefore, driven by the same logic, \(Q_2\) must definitively be a negative point charge.
Quick Tip: Graphical interpretations are a major recurring staple in CBSE physics; always carefully double-check which specific physical variables are rigidly assigned to the respective coordinate axes.
A line projecting into the first quadrant denotes a positive proportionality constant, while the fourth quadrant strictly denotes a negative one.

Concept:

As established, the slope of the \(V\) versus \(1/r\) graph serves as a direct mathematical proxy for the magnitude and sign of the source charge.

The precise slope of any straight line graphed on a Cartesian plane can be easily calculated by taking the trigonometric tangent of the angle it makes with the positive x-axis.

By mathematically finding the exact ratio of the two calculated slopes, we can systematically discover the exact ratio of the two corresponding point charges.



Step 1: {\color{redEstablish the Slope-Charge Proportionality

From the standard electrostatic potential formula, the linear slope \(m\) of the \(V\) versus \(1/r\) graph is formally given by:
\[ m = \frac{V}{(1/r)} = \frac{Q}{4\pi\epsilon_0} \]

This fundamentally means that the charge \(Q\) is strictly and directly proportional to the calculated slope \(m\):
\[ Q = m \cdot (4\pi\epsilon_0) \]


Step 2: {\color{redCalculate the Slope for Charge \(Q_1\)

For the first point charge \(Q_1\), the corresponding line is denoted as OA.

The graph explicitly indicates that line OA makes an angle of \(\theta_1 = 60^\circ\) strictly with the positive x-axis.

The mathematical slope \(m_1\) is computed using the standard tangent function:
\[ m_1 = \tan(60^\circ) = \sqrt{3} \]


Step 3: {\color{redCalculate the Slope for Charge \(Q_2\)

For the second point charge \(Q_2\), the corresponding line is denoted as OB.

The graph visually shows that line OB makes an angle of \(30^\circ\) below the positive x-axis.

In standard mathematical convention, an angle measured clockwise from the positive x-axis is rigorously taken as negative, so \(\theta_2 = -30^\circ\).

The mathematical slope \(m_2\) is computed similarly:
\[ m_2 = \tan(-30^\circ) = -\tan(30^\circ) = -\frac{1}{\sqrt{3}} \]


Step 4: {\color{redDetermine the Final Charge Ratio

We algebraically divide the derived charge equations to meticulously find the explicit ratio of the point charges.

Because the \(4\pi\epsilon_0\) constant perfectly cancels out during division, the charge ratio is strictly identical to the slope ratio:
\[ \frac{Q_1}{Q_2} = \frac{m_1 \cdot 4\pi\epsilon_0}{m_2 \cdot 4\pi\epsilon_0} = \frac{m_1}{m_2} \]

Substitute the exact trigonometrically calculated slope values securely into the fraction:
\[ \frac{Q_1}{Q_2} = \frac{\sqrt{3}}{-1/\sqrt{3}} \]

Simplifying this fractional mathematical expression mechanically yields the final answer:
\[ \frac{Q_1}{Q_2} = \sqrt{3} \times (-\sqrt{3}) = -3 \]

This robustly concludes both the exact magnitude ratio and firmly confirms the opposing signs of the charges.
Quick Tip: A negative angle physically denotes a clockwise angular measurement strictly from the primary positive x-axis.
Always remember basic trigonometric values like \(\tan(60^\circ)\) and \(\tan(30^\circ)\) to quickly execute slope calculations without relying on external tables.

Concept:

The total electrostatic potential energy of a multi-charge system situated within a pre-existing external electric field is a scalar sum of multiple independent energy components.

First, it fundamentally includes the individual physical work done in securely bringing each charge from infinity to its spatial position within the external field (\(U = qV\)).

Secondly, it includes the mutual interaction potential energy exclusively existing between the localized charges themselves, computed using Coulomb's electrostatic potential formula.



Step 1: {\color{redDetermine the External Electric Potential Function

The external electric field is mathematically given strictly as \(\vec{E} = \frac{A}{x^2} \hat{i}\).

The scalar electric potential \(V(x)\) intricately and universally relates to the electric field via the line integral equation:
\[ V(x) = -\int E dx \]

Integrating the provided field function perfectly from infinity yields the exact potential function:
\[ V(x) = -\int \frac{A}{x^2} dx = -\left( -\frac{A}{x} \right) = \frac{A}{x} \]


Step 2: {\color{redCalculate Individual Energies of Charges in the External Field

We logically compute the exact potential strictly at the spatial position of the first charge, \(x_1 = -30 cm = -0.3 m\):
\[ V_1 = \frac{9 \times 10^5}{-0.3} = -3 \times 10^6 V \]

The energy of the first charge \(q_1 = -2 \mu C = -2 \times 10^{-6} C\) inside this localized field is:
\[ U_1 = q_1 V_1 = (-2 \times 10^{-6}) \times (-3 \times 10^6) = 6 J \]

Similarly, systematically compute the potential exactly at the position of the second charge, \(x_2 = 30 cm = 0.3 m\):
\[ V_2 = \frac{9 \times 10^5}{0.3} = 3 \times 10^6 V \]

The energy of the second charge \(q_2 = 5 \mu C = 5 \times 10^{-6} C\) securely inside this localized field is:
\[ U_2 = q_2 V_2 = (5 \times 10^{-6}) \times (3 \times 10^6) = 15 J \]


Step 3: {\color{redCalculate Mutual Interaction Energy Between the Charges

The absolute spatial distance separating the two fixed point charges is \(r = |0.3 - (-0.3)| = 0.6 m\).

The mutual electrostatic potential energy is computed explicitly by Coulomb's two-charge potential formula:
\[ U_{12} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r} \]

Substitute the universally known standard constant and the exact charge values carefully:
\[ U_{12} = 9 \times 10^9 \frac{(-2 \times 10^{-6}) \times (5 \times 10^{-6})}{0.6} \]
\[ U_{12} = \frac{-90 \times 10^{-3}}{0.6} = -150 \times 10^{-3} J = -0.15 J \]


Step 4: {\color{redCalculate Total Electrostatic Potential Energy

The absolute total energy \(U_{total}\) of the entire configuration is the direct scalar sum of all three previously calculated independent energy terms:
\[ U_{total} = U_1 + U_2 + U_{12} \]
\[ U_{total} = 6 + 15 - 0.15 = 20.85 J \]

The final computed potential energy of the comprehensive system is strictly 20.85 Joules.
Quick Tip: Never casually ignore the critical algebraic signs of charges when calculating potential energy, as they directly dictate whether the energy actively contributes positively or formally acts as a binding negative factor.
Ensure all spatial distances are appropriately converted strictly to standard meters before invoking SI constant calculations.

Concept:

An infinitely long, perfectly straight charged wire inherently creates a radially outward or inward electric field in its completely surrounding three-dimensional space.

A second charged wire placed gracefully within this already established electric field will consequently and inevitably experience a continuous electrostatic force.

The total experienced force per unit length is elegantly and systematically derived using the mathematical definition of linear charge density tightly coupled with Gauss's law for electrostatics.



Step 1: {\color{redDetermine the Electric Field of the First Wire

Let the primary infinite wire possess a strictly negative linear charge density mathematically defined as \(\lambda_1 = -\lambda\).

Using the standard application of Gauss's law for a cylindrical geometry, the magnitude of the electric field perfectly generated by this infinite wire at a perpendicular radial distance \(r\) is:
\[ E_1 = \frac{|\lambda_1|}{2\pi\epsilon_0 r} = \frac{\lambda}{2\pi\epsilon_0 r} \]

Because the primary wire's charge density is firmly negative, this resulting electric field vector is directed radially strictly inwards, pointing directly toward the primary wire.


Step 2: {\color{redCalculate the Force Exerted on the Second Wire

The secondary parallel wire possesses a distinctly positive linear charge density logically given as \(\lambda_2 = 3\lambda\).

We carefully and meticulously consider a microscopically small arbitrary length segment \(dl\) of this secondary wire.

The accumulated static electric charge meticulously contained entirely within this tiny segment is strictly \(dq = \lambda_2 dl = 3\lambda dl\).

The infinitesimal electrostatic force \(dF\) experienced physically by this charged segment residing strictly in the presence of the external field \(E_1\) is:
\[ dF = dq \cdot E_1 \]

Substituting our securely established mathematical expressions into the foundational force equation forcefully gives:
\[ dF = (3\lambda dl) \left( \frac{\lambda}{2\pi\epsilon_0 r} \right) \]


Step 3: {\color{redDetermine Force Per Unit Length and its Fundamental Nature

To firmly and absolutely find the exact force per unit length acting continuously along the wires, we systematically divide the derived infinitesimal force by the segment length \(dl\):
\[ \frac{F}{l} = \frac{dF}{dl} = \frac{3\lambda^2}{2\pi\epsilon_0 r} \]

This mathematically represents the exact, uncompromising magnitude of the continuous electrostatic force per unit length strictly acting mutually on the wires.

To decisively determine the physical nature of this specific force, we must deeply examine the opposing algebraic signs of the respective charge densities.

Since one wire is heavily negatively charged (\(-\lambda\)) and the other is heavily positively charged (\(+3\lambda\)), they actively and constantly experience mutual electrostatic attraction.

Thus, the resulting continuous force is decisively attractive in its fundamental physical nature.
Quick Tip: Do not mistakenly confuse this pure electrostatic geometrical setup with the widely known Ampere's magnetic force existing between two parallel current-carrying wires; linear charge density \(\lambda\) refers exclusively to static, non-moving charges.
Opposite charges attract, identical charges repel—a universal fundamental rule spanning all physics.

Concept:

Gauss's law universally and strictly dictates that the total electric flux piercing entirely through any closed mathematical surface is precisely equal to the dynamically enclosed net charge divided directly by the permittivity of free space \(\epsilon_0\).

To successfully apply this law, one must carefully systematically define the exact imaginary Gaussian surface and meticulously sum all internal charges existing strictly within that spatial boundary.

The spherical symmetry of concentric conducting shells makes applying a spherical Gaussian surface the optimal geometric choice.



Step 1: {\color{redEvaluate Electric Flux for region (1) \(x < r_1\)

We systematically apply Gauss's integral law mathematically stated as \(\Phi = \frac{Q_{enc}}{\epsilon_0}\) for distinct radial geometric regions.

For this innermost region strictly defined by radial distance \(x < r_1\), the arbitrarily chosen spherical Gaussian surface perfectly encloses only the central space inside the first hollow sphere.

The only electrical entity existing in this deep central space is the centrally placed point charge.

Thus, the definitively enclosed net charge is mathematically calculated as exactly \(Q_{enc} = 2q\).

Applying Gauss's law directly, the total electric flux piercing outward through this inner surface is firmly computed as:
\[ \Phi_{(x < r_1)} = \frac{2q}{\epsilon_0} \]


Step 2: {\color{redEvaluate Electric Flux for region (2) \(r_1 < x < r_2\)

We logically shift our analytical focus to the intermediate spatial region located squarely between the inner conducting sphere and the outer conducting shell.

A spherical Gaussian surface meticulously drawn at any radius \(x\) strictly bounded by \(r_1 < x < r_2\) physically encompasses everything inside the inner sphere's radius.

This overarching Gaussian surface successfully surrounds both the exact central point charge (\(2q\)) and the entire primary inner conducting sphere, which inherently carries a net total charge of \(Q\).

The newly enclosed net total charge strictly aggregates by scalar addition to:
\[ Q_{enc} = 2q + Q \]

Invoking Gauss's law once again using this updated enclosed charge value, the corresponding electric flux physically passing entirely through this intermediate surface is:
\[ \Phi_{(r_1 < x < r_2)} = \frac{2q + Q}{\epsilon_0} \]

This comprehensively concludes the explicit flux calculations for both highly specific interior regions.
Quick Tip: When painstakingly calculating enclosed charge for Gauss's law, you must absolutely include every single charge residing physically inside the drawn boundary—point charges, surface charges, and volume charges alike.
Flux strictly cares exclusively about the quantity of enclosed charge, not its detailed spatial arrangement, provided the surface is completely closed.

Concept:

Gauss's law serves as a powerful mathematical tool to decisively determine the electric field generated by highly symmetric continuous charge distributions.

For universally symmetric spherical systems, the electric field vector perfectly aligns radially with the area vector everywhere precisely on a concentric spherical Gaussian surface.

This elegant geometric alignment simplifies the complex flux integral strictly into the algebraic product of the constant electric field magnitude and the entire surface area: \(E \cdot 4\pi x^2 = \frac{Q_{enc}}{\epsilon_0}\).



Step 1: {\color{redDetermine Electric Field for region (1) \(x > r_3\)

For this completely outermost region, the mathematical Gaussian surface is drawn broadly to safely encompass the entire physical apparatus.

The overarching Gaussian surface universally encloses all localized point and shell charges contained anywhere in the entire physical system.

We must meticulously sum every single charge to firmly find the absolute total enclosed charge:

The central point charge is \(+2q\).

The net charge purposefully given to the inner hollow sphere is \(+Q\).

The net charge purposefully given to the massive outer conducting shell is \(-3q\).

The absolute total enclosed charge is mathematically computed by combining these algebraically:
\[ Q_{enc} = 2q + Q + (-3q) = Q - q \]

Using the standard robust spherical symmetry relation, we easily derive the external radial electric field:
\[ E \cdot 4\pi x^2 = \frac{Q - q}{\epsilon_0} \]
\[ E_{(x > r_3)} = \frac{Q - q}{4\pi\epsilon_0 x^2} \]


Step 2: {\color{redDetermine Electric Field for region (2) \(r_1 < x < r_2\)

For this specific intermediate region cleanly located between the nested conductors, the Gaussian surface is rigidly restricted in radius.

As previously proven definitively in the prior flux calculation, the comprehensively enclosed total charge here consists strictly of the central charge and the inner sphere's net charge.
\[ Q_{enc} = 2q + Q \]

Applying identically the exact same geometric symmetry arguments directly yields the electric field strictly residing within this empty gap:
\[ E \cdot 4\pi x^2 = \frac{2q + Q}{\epsilon_0} \]
\[ E_{(r_1 < x < r_2)} = \frac{2q + Q}{4\pi\epsilon_0 x^2} \]

This accurately establishes the precise magnitude of the electric fields in the requested distinct spatial domains.
Quick Tip: For any point strictly located outside a perfectly spherically symmetric charge distribution, the entire complex system behaves mathematically exactly as if all its combined charge were tightly concentrated at the central origin.
Always remember the surface area of a sphere is strictly \(4\pi x^2\), completely avoiding common confusion with the circle's area \(\pi x^2\).

Concept:

A fundamental cornerstone property of solid conductors in stable electrostatic equilibrium is that the electric field deeply robustly inside the actual solid metallic material is strictly zero.

This mandatory zero internal field phenomenon inherently forces mobile induced charges to perfectly and dynamically arrange themselves entirely on the available inner and outer surfaces of the conductors.

By strategically placing a mathematical Gaussian surface entirely within the conductive thickness, we mathematically mandate that the net enclosed charge must total exactly zero.



Step 1: {\color{redAnalyze the Inner Surface of the Primary Sphere (1)

Consider the innermost hollow conducting sphere which possesses an exact radius \(r_1\).

To absolutely guarantee that the internal electric field deeply inside the solid shell of this conducting sphere is completely zero, a Gaussian surface drawn precisely inside its metallic body must logically enclose a net zero total charge (\(Q_{enc} = 0\)).

The spatial region enclosed essentially contains the central point charge (\(+2q\)) and whatever new charge (\(q_{inner1}\)) aggressively induces on the exact inner surface at radius \(r_1\).

Therefore, the sum must perfectly equate to zero:
\[ 2q + q_{inner1} = 0 \implies q_{inner1} = -2q \]

This physically requires that an induced charge of precisely \(-2q\) must form perfectly on its inner surface to actively counteract the central charge.

The resulting surface charge density \(\sigma\) is calculated by dividing this induced charge by the precise spherical inner surface area:
\[ \sigma_{sphere, inner} = \frac{q_{inner1}}{A_1} = \frac{-2q}{4\pi r_1^2} \]


Step 2: {\color{redAnalyze the Inner Surface of the Secondary Shell (2)

Now consider the massive outer conducting spherical shell which strictly possesses an inner radius \(r_2\).

Similarly, a Gaussian surface flawlessly drawn completely inside the outer conducting shell's solid metallic thickness must absolutely also enclose exactly zero net charge.

The total enclosed charge inside this vast hollow cavity space is clearly the algebraic sum of the central point charge and the total net charge safely residing on the entire primary inner sphere (both its inner and outer surfaces combined).

The combined inner charge total is exactly \((+2q) + (+Q)\).

Let the desperately needed induced charge on the shell's inner surface be \(q_{inner2}\).

To maintain zero net enclosed charge, the sum must precisely equal zero:
\[ (+2q + Q) + q_{inner2} = 0 \implies q_{inner2} = -(2q + Q) \]

Therefore, a massive compensating induced charge of \(-(2q + Q)\) must systematically form entirely on the inner surface of the outer shell.

The corresponding surface charge density decisively becomes the induced charge divided by that specific surface area:
\[ \sigma_{shell, inner} = \frac{q_{inner2}}{A_2} = \frac{-(2q + Q)}{4\pi r_2^2} \]
Quick Tip: Conductors invariably manipulate their vast supply of free charges to effectively eliminate internal electric fields; analyzing nested shells is purely a rigorous logical exercise in successfully balancing inner enclosed charges step-by-step from the inside out.
Surface charge density rigorously requires mathematically dividing the specific bounded charge by the exact corresponding geometrical spherical surface area, not the volume.

Concept:

The visible brightness or visual glow of a standard resistive light bulb is strictly determined by the real power seamlessly dissipated across its internal filament, heavily dependent on the RMS current (\(P = I_{rms}^2 R\)).

In a practical series AC circuit containing resistance and inductance, the driving operating current is meticulously governed by the total circuit impedance.

The total impedance mathematically adjusts based on the inductive reactance, which is fiercely dependent on the physical inductance of the inserted coil.



Step 1: {\color{redUnderstand the Initial State

The total electrical impedance \(Z\) of the operating series circuit is robustly formulated as \(Z = \sqrt{R_{bulb}^2 + X_L^2}\), where \(X_L = \omega L\) physically represents the inductive reactance.

Initially, the inductor is an open air-core coil, meaning it has a relatively low baseline self-inductance \(L_0\), dictated largely by the magnetic permeability of free space \(\mu_0\).

The circuit draws a specific steady RMS current strictly based on this baseline impedance, causing the bulb to visibly glow with a certain stable brightness.


Step 2: {\color{redAnalyze the Effect of Inserting the Iron Bar

When a highly permeable soft iron bar is deliberately and deeply inserted inside the open coil, it forcibly completely replaces the air core with an iron core.

Soft iron intrinsically possesses a magnetic permeability (\(\mu\)) that is thousands of times significantly higher than that of empty air.

Because the physical self-inductance \(L\) of a coil is mathematically strictly proportional to the permeability of its core material (\(L \propto \mu\)), introducing the iron bar drastically increases the fundamental self-inductance \(L\) of the inductor coil.

As the self-inductance \(L\) surges heavily upward, the inductive reactance \(X_L\) proportionately and aggressively spikes up.


Step 3: {\color{redConclude the Effect on Bulb Brightness

Consequently, due to the massively inflated inductive reactance, the overall total impedance \(Z\) of the entire series circuit substantially and noticeably increases.

Because the applied driving AC voltage source is held securely constant by definition of the problem, a much higher total impedance actively chokes the flow of electrons, strictly decreasing the operational RMS current \(I_{rms}\) flowing continuously through the interconnected bulb.

With a noticeably reduced driving current, the actual thermal power actively dissipated by the glowing bulb (\(P = I_{rms}^2 R_{bulb}\)) heavily drops.

Since power strictly dictates brightness, the visual glow of the light bulb will dramatically and visibly decrease.
Quick Tip: Iron heavily amplifies localized magnetic flux linkage, acting as a massive multiplier for physical inductance; always strongly associate soft iron core insertion directly with significantly increased inductive reactance.
Think of the inductor exactly as a frequency-dependent resistor; increasing its "resistance" strictly throttles the overall current.

Concept:

The total current strictly flowing through a series RL (resistor-inductor) circuit is intricately controlled by the overarching impedance, which beautifully combines both fixed resistance and variable reactance.

Inductive reactance is not a static constant; it is fundamentally a dynamic quantity directly and linearly proportional to the physical operating frequency of the applied alternating current.

Manipulating the source frequency inevitably alters the reactance, which heavily manipulates the current, ultimately dictating the power entirely dissipated by the light bulb.



Step 1: {\color{redRelate Frequency to Inductive Reactance

We strictly analyze the specific scenario where the AC voltage source's operating frequency \(f\) is intentionally and manually dialed down.

The inductive reactance \(X_L\) of a coil is mathematically defined precisely by the foundational AC equation:
\[ X_L = \omega L = 2\pi f L \]

This fundamental mathematical relation unequivocally shows that the inductive reactance is directly and strictly proportional to the applied source frequency (\(X_L \propto f\)).

Therefore, deliberately decreasing the source frequency actively causes the corresponding inductive reactance \(X_L\) to cleanly and linearly decrease.


Step 2: {\color{redRelate Reactance to Circuit Impedance

The total working impedance \(Z\) for this series configuration is given exactly by the standard magnitude formula:
\[ Z = \sqrt{R_{bulb}^2 + X_L^2} \]

Since the physical resistance \(R_{bulb}\) of the light bulb essentially remains solidly constant, a considerably lower inductive reactance \(X_L\) logically leads directly to a markedly reduced total electrical impedance \(Z\) for the entire series combination.


Step 3: {\color{redConclude the Effect on Bulb Brightness

According to Ohm's law strictly adapted for AC circuits, the RMS current strictly depends inversely on the impedance:
\[ I_{rms} = \frac{V_{rms}}{Z} \]

Driven by the assumed constant supply voltage, this newly lowered total impedance graciously permits a significantly higher operational RMS current \(I_{rms}\) to successfully course freely through the circuit.

Since the current has vigorously increased, the thermal power seamlessly dissipated within the bulb's resistive tungsten filament (\(P = I_{rms}^2 R\)) aggressively surges higher.

Consequently, emitting much more thermal and radiant energy, the visual brightness and overall glow of the light bulb will undeniably and visibly increase.
Quick Tip: An inductor inherently acts to aggressively block high frequencies while letting low frequencies pass easily; radically reducing the AC frequency always allows the electrical current to aggressively pass much more freely.
Ensure you always logically step through the chain reaction: Frequency \(\rightarrow\) Reactance \(\rightarrow\) Impedance \(\rightarrow\) Current \(\rightarrow\) Power \(\rightarrow\) Glow.

Concept:

A classic series LCR circuit possesses a unified total impedance constructed physically from three key components: ohmic resistance, inductive reactance, and capacitive reactance.

The driving AC voltage elegantly follows a standard sinusoidal mathematical waveform, where the coefficient of time \(t\) uniquely represents the angular frequency.

These distinct reactances strictly operate entirely out of phase with the standard resistance, necessitating a careful vector-like Pythagorean addition to solidly find total impedance.



Step 1: {\color{redExtract Fundamental AC Parameters

From the rigorously standard mathematical voltage form \(V = V_m \sin(\omega t)\), we intelligently compare it directly with the heavily given equation \(V = 280 \sin(100\pi t)\).

This comparison decisively reveals two critical physical parameters:

The absolute peak driving voltage is \(V_m = 280 V\).

The operating angular frequency is rigidly \(\omega = 100\pi rad/s\).


Step 2: {\color{redCalculate Individual Component Reactances

We logically compute the explicit inductive reactance \(X_L\) exactly using its defining formula:
\[ X_L = \omega L = (100\pi) \times \left(\frac{5}{\pi}\right) \]

The \(\pi\) cleanly cancels out, yielding:
\[ X_L = 500 \Omega \]

Next, we strictly compute the explicit capacitive reactance \(X_C\), carefully incorporating the crucial microfarad (\(10^{-6}\)) conversion:
\[ X_C = \frac{1}{\omega C} = \frac{1}{(100\pi) \times \left(\frac{50}{\pi} \times 10^{-6}\right)} \]

The \(\pi\) again cleanly cancels out from the denominator:
\[ X_C = \frac{1}{5000 \times 10^{-6}} = \frac{10^6}{5000} \]

Simplify the fraction completely:
\[ X_C = \frac{1000}{5} = 200 \Omega \]


Step 3: {\color{redCalculate the Total Circuit Impedance

The overarching comprehensive impedance \(Z\) for a fully functioning series LCR circuit is mathematically formalized rigorously as:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]

We systematically substitute the confirmed physical resistance and calculated reactance values directly into the square root:
\[ Z = \sqrt{400^2 + (500 - 200)^2} \]

Perform the strict internal subtraction first:
\[ Z = \sqrt{400^2 + 300^2} \]

Recognizing the ubiquitous 3-4-5 mathematical Pythagorean triplet drastically and beautifully simplifies the heavy arithmetic square root calculation without pain:
\[ Z = \sqrt{160000 + 90000} = \sqrt{250000} \]
\[ Z = 500 \Omega \]

The total effective impedance strictly opposing current flow in this complex circuit is 500 Ohms.
Quick Tip: Continuously watch actively for familiar Pythagorean triplets like 3-4-5, 5-12-13, or 8-15-17 during complex impedance calculations to cleanly and rapidly bypass tedious arithmetic.
Always meticulously convert microfarads (\(\muF\)) entirely into standard farads by aggressively applying the critical \(10^{-6}\) multiplier before inserting them deeply into denominator fractions.

Concept:

In alternating current (AC) circuit analysis, peak values exclusively describe the absolute maximum instantaneous amplitude reached strictly by the oscillating waveform.

Root Mean Square (RMS) values are fundamentally far more practical, accurately representing the equivalent constant DC value that would unequivocally deliver the exact same average thermal power.

Ohm's law cleanly applies to AC circuits by meticulously relating peak current strictly to peak voltage, and RMS current strictly to RMS voltage, universally bridged by the total impedance.



Step 1: {\color{redExtract Parameters and Previous Results

From the rigorously standard mathematical voltage equation \(V = 280 \sin(100\pi t)\), we readily identify the absolute peak driving voltage:
\[ V_m = 280 V \]

From our meticulous, detailed calculation in the prior sub-question (I), we robustly established the total operational circuit impedance:
\[ Z = 500 \Omega \]


Step 2: {\color{redCalculate the Peak Current

To systematically find the robust RMS current, we generally must first accurately locate the absolute peak current \(I_m\) physically circulating heavily in the closed circuit.

Using Ohm's law rigidly adapted for peak AC variables, we divide the peak voltage by the total opposing impedance:
\[ I_m = \frac{V_m}{Z} \]

Substitute the known exact numerical values safely into the ratio:
\[ I_m = \frac{280}{500} \]

Simplify the numerical fraction completely to acquire a workable decimal:
\[ I_m = \frac{28}{50} = \frac{56}{100} = 0.56 A \]

This signifies that at the exact peak of its sinusoidal cycle, the current reaches exactly 0.56 Amperes.


Step 3: {\color{redConvert to RMS Current

The operating RMS (Root Mean Square) current is tightly and universally tied mathematically to the peak current strictly by the classic root-two divisor for pure sine waves:
\[ I_{rms} = \frac{I_m}{\sqrt{2}} \]

The question explicitly and kindly instructs us to strictly use the specific numerical approximation \(\sqrt{2} = 1.4\).

Substitute the calculated peak current and the provided approximation perfectly into the equation:
\[ I_{rms} = \frac{0.56}{1.4} \]

To execute this division smoothly, aggressively multiply both numerator and denominator exactly by 100:
\[ I_{rms} = \frac{56}{140} \]

Simplify the resulting fraction efficiently by recognizing common multiples:
\[ I_{rms} = \frac{4}{10} = 0.4 A \]

The effective, power-producing RMS current reliably flowing through this entire setup is exactly 0.4 Amperes.
Quick Tip: Always strictly ensure you do not inadvertently mix peak values with RMS values within a single Ohm's Law equation (\(V = IZ\)); either use peak entirely for both or use RMS entirely for both.
When specific approximations like \(\sqrt{2} = 1.4\) or \(\pi = 3.14\) are heavily explicitly provided in the question stem, strictly using them is mandatory to exactly match the marking scheme.

Concept:

In any complex alternating current setup, power is not merely the simple product of total voltage and total current; it is heavily dictated by relative phase.

The power factor fundamentally acts as a strict efficiency multiplier, geometrically quantifying exactly what fraction of the total apparent power is actively converted into highly useful true real power.

Mathematically, it is elegantly defined precisely as the exact cosine of the phase angle separating the total voltage waveform and the operational current waveform.



Step 1: {\color{redUnderstand the Power Factor Formula

From the standard foundational impedance triangle specifically constructed for series AC circuits, the base strictly represents the true Resistance \(R\), while the long hypotenuse rigidly represents the total Impedance \(Z\).

The angle securely nestled exactly between them is the fundamental phase angle \(\phi\).

Therefore, the power factor, denoted officially as \(\cos\phi\), can be extracted geometrically straight from the lengths of the triangle's rigid sides:
\[ Power Factor (\cos\phi) = \frac{Adjacent Side}{Hypotenuse} = \frac{R}{Z} \]

This elegant fractional formula effectively serves as the absolute fastest and most reliable method to precisely determine the required factor.


Step 2: {\color{redExtract Required Values

From the original problem statement text, the pure ohmic resistance heavily installed in the circuit is rigidly given as:
\[ R = 400 \Omega \]

From our own meticulous, detailed calculation securely performed back in sub-question (I), we robustly established the total operational circuit impedance:
\[ Z = 500 \Omega \]

We also previously calculated \(X_L = 500 \Omega\) and \(X_C = 200 \Omega\), establishing firmly that \(X_L > X_C\), which fundamentally means the circuit is definitively inductive overall.


Step 3: {\color{redCalculate the Power Factor

We systematically substitute our definitively verified resistance and impedance values directly into the established geometric power factor ratio:
\[ \cos\phi = \frac{R}{Z} \]
\[ \cos\phi = \frac{400}{500} \]

Simplify the numerical fraction completely by canceling the clearly obvious trailing hundreds:
\[ \cos\phi = \frac{4}{5} \]

Convert the remaining simple fraction smoothly into a final readable decimal format:
\[ \cos\phi = 0.8 \]

Because the overall inductive reactance heavily outweighs the overall capacitive reactance in this specific setup, this circuit operates definitively with a "lagging" power factor of precisely 0.8, meaning the current waveform trails behind the voltage waveform.
Quick Tip: The absolute value of any physically valid power factor inherently must always rigidly fall strictly between 0 and 1 inclusive; getting a number larger than 1 means a fraction was flipped upside down.
While merely providing the numerical value usually earns full marks, additionally specifying whether the factor is 'leading' or 'lagging' vividly demonstrates a supreme mastery of the core AC concepts.

Concept:

Electromagnetic induction is a profound natural phenomenon firmly governed by Faraday's laws for predicting magnitude and Lenz's law for predicting strict polarity.

The universal principle of energy conservation acts as an impenetrable physical barrier, fiercely preventing the spontaneous creation of limitless electrical energy from absolutely nothing.

Lenz's law physically manifests this conservation requirement perfectly by strictly dictating an oppositional reactive response to any changing magnetic flux.



Step 1: {\color{redFormal Statement of Lenz's Law

Lenz's Law formally and unequivocally states that the specific polarity of the induced electromotive force (emf), and heavily consequently the exact direction of the newly induced current circulating in a completely closed electrical circuit, is rigorously always such that it actively attempts to vehemently oppose the physical change in the magnetic flux that primarily produced it.

In much simpler, conceptual terms, the induced electrical effect fiercely and stubbornly combats the fundamental physical cause of its own induction.


Step 2: {\color{redJustification via Conservation of Energy

To brilliantly explain precisely why this specific law inherently and perfectly obeys the universal law of conservation of energy, let us logically conduct a rigorous physical thought experiment.

Imagine deliberately pushing the highly magnetic North pole of a solid, heavy bar magnet directly towards a conductive closed stationary coil.

According to Lenz's law, the new induced current operating in the coil will stubbornly circulate in a specific direction that cleverly establishes a repelling North magnetic pole directly on the very coil face confronting the advancing magnet.

Because similar magnetic poles notoriously and fiercely repel each other, an external agent (like a human hand or a moving machine) must firmly and continuously perform tangible mechanical work to forcibly overcome this aggressive magnetic repulsion and continue pushing the magnet relentlessly forward.

This exerted physical mechanical work definitely does not simply vanish into thin air; it is seamlessly and meticulously converted directly into the useful electrical energy actively powering the induced current running aggressively through the coil, which eventually perfectly dissipates as thermal heat due to internal Joule heating.


Step 3: {\color{redAnalyzing the Impossible Alternative

Now, hypothesize a physically impossible, broken universe scenario where Lenz's law is entirely reversed in nature.

If the induced current instead enthusiastically created an attracting South magnetic pole directly facing the approaching magnet, it would forcefully and irresistibly pull the heavy magnet inward entirely on its own.

The magnet would furiously accelerate completely by itself, effortlessly generating massive kinetic energy while simultaneously pumping out boundless, free electrical energy in the closed circuit.

This wild, unchecked scenario implies actively creating infinite, limitless energy completely out of nothing, violently and fatally violating the supreme foundational law of conservation of energy.

Therefore, the strict oppositional nature stubbornly championed by Lenz's law is definitely not just a quirky arbitrary rule; it is an absolute, non-negotiable physical necessity fundamentally required to tightly maintain global energy conservation.
Quick Tip: When tackling strictly theoretical, descriptive questions extensively about Lenz's law, always emphatically connect the concept of "magnetic opposition" directly to the indispensable requirement of doing tangible "mechanical work" to generate electrical power.
Nature structurally and fundamentally abhors a free lunch; raw electrical energy can absolutely never be cleanly harvested without forcefully expending a completely equivalent amount of physical effort.

Concept:

Self-inductance firmly and mathematically measures a conductive coil's inherent natural ability to dynamically oppose any rapid, sudden changes in its internal flowing electrical current.

Dimensional formulas elegantly and systematically break down complex, derived electromagnetic quantities strictly into their fundamental baseline mechanical and electrical base units.

The induced electromotive force strictly links a coil's self-inductance mathematically with the precise time-rate of temporal current change occurring within it.



Step 1: {\color{redDerive the Dimensional Formula for Self-Inductance

We start strongly and strategically from the well-known, foundational energy formula involving an energized inductor:
\[ U = \frac{1}{2} L I^2 \]

Here, \(U\) physically represents the stored magnetic potential energy, \(L\) critically represents the self-inductance, and \(I\) rigidly represents the operating current.

We strategically and algebraically rearrange this specific formula to clearly isolate the self-inductance variable on one side:
\[ L = \frac{2U}{I^2} \]

Since the numerical mathematical constant \(2\) is completely dimensionless, the final physical dimension of \(L\) entirely depends strictly on the fundamental physical dimensions of energy and current.

The universally accepted, standard dimensional formula for standard energy (equivalent to Work) is \([ML^2T^{-2}]\).

The firmly established standard dimensional formula for foundational electrical current is inherently \([A]\).

Substituting these specific reliable base dimensions smoothly and carefully into the algebraically rearranged equation yields:
\[ [L] = \frac{[ML^2T^{-2}]}{[A^2]} \]

Moving the current dimension cleanly into the upper numerator gives the final expression:
\[ [L] = [ML^2T^{-2}A^{-2}] \]


Step 2: {\color{redIdentify Parameters for Numerical Calculation

The initial electrical current strongly flowing in the coil is strictly recorded as \(I_1 = 8.0 A\).

The drastically and rapidly reduced final current is firmly recorded as \(I_2 = 2.0 A\).

The exact time interval over which this rapid, dynamic change thoroughly occurs is given as \(\Delta t = 0.6 s\).

The absolute average magnitude of the opposing induced electromotive force securely registered during this event is \(|e| = 50 V\).


Step 3: {\color{redCalculate the Exact Self-Inductance

The primary governing mathematical formula firmly connecting all these dynamic variables is famously given by Faraday's and Lenz's laws combined together:
\[ |e| = L \left| \frac{\Delta I}{\Delta t} \right| \]

We mathematically and carefully determine the absolute magnitude of the rapid change in the operating current:
\[ |\Delta I| = |I_2 - I_1| = |2.0 - 8.0| = |-6.0| = 6.0 A \]

Now, we meticulously substitute all the carefully collected and verified numerical values squarely back into the primary algebraic formula:
\[ 50 = L \times \left( \frac{6.0}{0.6} \right) \]

Calculating the simple internal mathematical fraction cleanly provides a very neat factor of exactly ten:
\[ 50 = L \times 10 \]

Isolating the self-inductance variable firmly by performing basic division elegantly yields the final answer:
\[ L = \frac{50}{10} = 5 H \]

The total self-inductance of the rigorously tested induction coil is precisely evaluated to be strictly 5 Henrys.
Quick Tip: Using the standard magnetic energy formula \(U = \frac{1}{2}LI^2\) is frequently the absolute fastest, most reliably error-free algebraic pathway to reliably deriving the often-forgotten dimensional formula for inductance.
When meticulously calculating absolute physical properties strictly like inductance or resistance, strictly using the absolute scalar magnitude of the dynamic current change safely and effectively prevents horribly confusing negative results.