The work done on a charge \( Q \) as it moves between two points in an electric field is given by: \[ W = q \cdot \Delta V \]
where:

- \( W \) is the work done,

- \( q \) is the charge,

- \( \Delta V \) is the potential difference between the two points.


In this problem, we are given that the charge moves along different paths, and we are tasked with determining along which path the work done will be maximum.

Step 1: Understanding the relationship between work and potential difference

The work done on a charge depends directly on the potential difference between the points it moves between. The greater the potential difference, the greater the work done.
\[ W = q \cdot \Delta V \]

Since the charge is moving along different paths, the potential difference \( \Delta V \) is the key factor that determines how much work is done. The path that moves the charge between points with the greatest potential difference will result in the maximum work.

Step 2: Analyzing the paths and potential difference

We are provided with a diagram where the paths (A, B, C, and D) are indicated along with the equipotential surfaces. Equipotential surfaces are surfaces where the electric potential is constant. The path that moves through the regions with the largest potential difference between the points will experience the maximum work.

In this case, path C passes through regions of the greatest change in potential. Therefore, the work done on the charge along path C will be the maximum.

Thus, the correct answer is:
\[ \boxed{C} \] Quick Tip: The work done on a charge depends on the potential difference. The larger the potential difference between two points, the greater the work done when the charge moves between them. Always look for the path that has the largest change in potential to maximize the work.

The resistance \( R \) of a wire is given by the formula: \[ R = \rho \frac{L}{A} \]
where:
- \( R \) is the resistance,

- \( \rho \) is the resistivity of the material,

- \( L \) is the length of the wire,

- \( A \) is the cross-sectional area of the wire.


For a cylindrical wire, the cross-sectional area \( A \) is given by: \[ A = \pi r^2 \]
where \( r \) is the radius of the wire.

Now, we are asked to find the condition that will make the resistance of the new wire \( \frac{R}{2} \).

Step 1: Modifying the length and radius

For the new wire to have a resistance \( \frac{R}{2} \), we must adjust the length and/or radius.

Let’s analyze each option:

Option (A): Using a wire of the same radius and twice the length

If the length of the wire is doubled, the resistance will double, because the resistance is directly proportional to the length. So, the new resistance will be \( 2R \), which is not what we want.

Option (B): Using a wire of the same radius and half the length

If the length is halved, the resistance will be halved as well, since resistance is directly proportional to the length. So, the new resistance will be \( \frac{R}{2} \), which is the desired value.

Option (C): Using a wire of the same length and twice the radius

If the radius is doubled, the area will increase by a factor of 4 (since area \( A \propto r^2 \)), and the resistance will decrease by a factor of 4. So, the new resistance will be \( \frac{R}{4} \), which is not what we want.

Option (D): Using a wire of the same length and half the radius

If the radius is halved, the area will decrease by a factor of 4, and the resistance will increase by a factor of 4. So, the new resistance will be \( 4R \), which is not what we want.

Thus, the correct option is (B).
\[ \boxed{B} \] Quick Tip: To decrease the resistance of a wire, you can either reduce its length by half or increase its radius by a factor of \( \sqrt{2} \).

The force acting on a current-carrying wire in a magnetic field is given by: \[ \vec{F} = I \ell \hat{L} \times \vec{B} \]
where:

- \( I = 0.5 \, A \) is the current,

- \( \ell = 1 \, cm = 0.01 \, m \) is the length of the wire segment along the x-axis,

- \( \hat{L} \) is the unit vector in the direction of the current (in the +x direction: \( \hat{i} \)),

- \( \vec{B} = (0.4 \, mT) \hat{j} + (0.6 \, mT) \hat{k} = (0.4 \times 10^{-3}) \hat{j} + (0.6 \times 10^{-3}) \hat{k} \) is the magnetic field.

Now, calculate the cross product \( \hat{L} \times \vec{B} \): \[ \hat{i} \times \left( 0.4 \times 10^{-3} \hat{j} + 0.6 \times 10^{-3} \hat{k} \right) = 0.4 \times 10^{-3} \hat{k} - 0.6 \times 10^{-3} \hat{j} \]

The force is: \[ \vec{F} = (0.5) \times (0.01) \left( 0.4 \times 10^{-3} \hat{k} - 0.6 \times 10^{-3} \hat{j} \right) \] \[ \vec{F} = (-3\hat{j} + 2\hat{k}) \, \muN \]

Thus, the force acting on the wire segment is \( (-3 \hat{j} + 2 \hat{k}) \, \muN \).
Quick Tip: The force on a current-carrying wire in a magnetic field depends on the direction of both the current and the magnetic field. Use the right-hand rule to determine the direction of the force. The magnitude of the force is influenced by the current, the length of the wire, and the magnetic field strength.

According to Faraday's Law of Induction, the induced emf in a coil is given by: \[ emf = - N \frac{d\Phi}{dt} \]
where:
- \( N = 300 \) is the number of turns,
- \( \Phi \) is the magnetic flux,
- \( \frac{d\Phi}{dt} \) is the rate of change of magnetic flux.

The magnetic flux \( \Phi \) is given by: \[ \Phi = B \cdot A \]
where \( B \) is the magnetic field and \( A \) is the area of the coil. The area \( A \) of the coil is: \[ A = \pi \left( \frac{d}{2} \right)^2 = \pi \left( \frac{15 \, mm}{2} \right)^2 = \pi (7.5 \times 10^{-3})^2 = 1.77 \times 10^{-4} \, m^2 \]

Now, we calculate the induced emf in the two cases of magnetic field change.

1. In the first case, the magnetic field is reduced uniformly to zero in 20 ms: \[ e_1 = - N \frac{d\Phi}{dt} = - 300 \times \frac{(30 \times 10^{-3}) \times (1.77 \times 10^{-4})}{20 \times 10^{-3}} = 0.80 \, V \]

2. In the second case, the magnetic field is increased back to 30 mT in 40 ms: \[ e_2 = - N \frac{d\Phi}{dt} = - 300 \times \frac{(30 \times 10^{-3}) \times (1.77 \times 10^{-4})}{40 \times 10^{-3}} = 0.40 \, V \]

Thus, the ratio of the induced emfs is: \[ \frac{e_1}{e_2} = \frac{0.80}{0.40} = 2 \]

Therefore, the correct answer is (C) 2.
Quick Tip: The induced emf is proportional to the rate of change of magnetic flux. A greater rate of change of the magnetic field results in a larger induced emf. For the same change in magnetic field, the emf is inversely proportional to the time taken for the change.

The inductance of an air-core solenoid is given by the formula: \[ L = \frac{\mu_0 N^2 A}{l} \]
Where:
\( L = 0.016\,H \),
\( l = 0.81\,m \),
\( r = 0.02\,m \Rightarrow A = \pi r^2 = \pi (0.02)^2 = 1.2566 \times 10^{-3} \, m^2 \),
\( \mu_0 = 4\pi \times 10^{-7} \, H/m \)


Substitute values: \[ 0.016 = \frac{4\pi \times 10^{-7} \times N^2 \times 1.2566 \times 10^{-3}}{0.81} \] \[ N^2 = \frac{0.016 \times 0.81}{4\pi \times 10^{-7} \times 1.2566 \times 10^{-3}} \approx 8214057.59 \] \[ N = \sqrt{8214057.59} \approx 2866 \] Quick Tip: Use the formula \( L = \frac{\mu_0 N^2 A}{l} \) for solenoids and always check units. Square root at the end gives the number of turns.

The average power consumed in an AC circuit over a complete cycle is given by: \[ P_{avg} = \frac{1}{T} \int_0^T v(t) \cdot i(t)\, dt \]
Given, \( v(t) = v_0 \sin(\omega t), \quad i(t) = i_0 \sin(\omega t + \phi) \)


Using identity: \[ \sin A \sin B = \frac{1}{2}[\cos(A - B) - \cos(A + B)] \] \[ P_{avg} = \frac{1}{T} \int_0^T \frac{v_0 i_0}{2} [\cos(\phi) - \cos(2\omega t + \phi)] dt \]
The integral of \(\cos(2\omega t + \phi)\) over a full cycle is zero. So, \[ P_{avg} = \frac{v_0 i_0}{2} \cos \phi \] Quick Tip: For average power in AC circuits with phase difference \( \phi \), use: \[ P_{avg} = \frac{V_0 I_0}{2} \cos \phi \] It accounts for phase shift between voltage and current.

Step 1: Understand the nature of electromagnetic waves in vacuum.

Electromagnetic waves are waves that are created as a result of vibrations between an electric field and a magnetic field. They travel at a constant speed in a vacuum, which is the speed of light, \(c \approx 3 \times 10^8 m/s\). This speed is constant for all electromagnetic waves (microwaves, visible light, X-rays, etc.) when they travel in a vacuum.




Step 2: Recall the order of electromagnetic spectrum and relationship between speed, wavelength, and frequency.

The electromagnetic spectrum arranges different types of electromagnetic waves by their wavelength and frequency. Moving from lower energy to higher energy (or longer wavelength to shorter wavelength, or lower frequency to higher frequency), the order is typically: Radio waves, Microwaves, Infrared, Visible light, Ultraviolet, X-rays, Gamma rays.


The fundamental relationship between speed (\(c\)), wavelength (\(\lambda\)), and frequency (\(f\)) for any wave is:
\(c = \lambda \times f\)




Step 3: Analyze the changes in speed, wavelength, and frequency from microwaves to X-rays.


Speed: As established in Step 1, all electromagnetic waves travel at the same speed (\(c\)) in a vacuum. Therefore, the speed remains same.
Wavelength: In the electromagnetic spectrum, microwaves have longer wavelengths than X-rays. As we go from microwaves to X-rays, the wavelength decreases.
Frequency: From the relationship \(c = \lambda \times f\), since \(c\) is constant, if \(\lambda\) decreases, then \(f\) must increase to keep the product constant. This is also consistent with the energy of photons, \(E = hf\), where higher energy X-rays have higher frequencies than lower energy microwaves.




Step 4: Evaluate the given options based on the analysis.


(A) Speed: Remains same, Wavelength: Decreases, Frequency: Remains same: Incorrect, frequency increases.
(B) Speed: Remains same, Wavelength: Decreases, Frequency: Increases: This matches our analysis.
(C) Speed: Increases, Wavelength: Increases, Frequency: Decreases: Incorrect, speed remains same, wavelength decreases, frequency increases.
(D) Speed: Remains same, Wavelength: Increases, Frequency: Remains same: Incorrect, wavelength decreases and frequency increases.




Step 5: Conclusion.

The correct representation of the change in wave characteristics from microwaves to X-rays in the electromagnetic spectrum (all in vacuum) is that Speed Remains same, Wavelength Decreases, and Frequency Increases.
\[ \boxed{Speed: Remains same, Wavelength: Decreases, Frequency: Increases} \] Quick Tip: All electromagnetic waves travel at the speed of light (\(c\)) in a vacuum. In the electromagnetic spectrum, wavelength and frequency are inversely proportional (\(c = \lambda f\)). Moving from left to right on the spectrum (e.g., from radio waves to gamma rays), wavelength decreases, and frequency (and energy) increases.

Step 1: Understand the conditions for Total Internal Reflection (TIR).

Total Internal Reflection (TIR) is a phenomenon that occurs when a ray of light passes from a denser medium to a rarer medium. The two essential conditions for TIR are:

Light must travel from an optically denser medium to an optically rarer medium.
The angle of incidence in the denser medium must be greater than the critical angle (\(i_c\)).




Step 2: Determine the optical densities of the two media based on the given speeds.

The speed of light in a medium is inversely proportional to its refractive index. A medium with a lower speed of light is optically denser (higher refractive index), and a medium with a higher speed of light is optically rarer (lower refractive index).
Given:

Speed of light in medium '1' = \(v_1\)
Speed of light in medium '2' = \(v_2\)
Condition: \(v_2 > v_1\)

Since \(v_2 > v_1\), it means light travels faster in medium '2' than in medium '1'.
Therefore:

Medium '1' is optically denser.
Medium '2' is optically rarer.




Step 3: Determine the direction of light for TIR based on optical densities.

According to Condition 1 for TIR, the light must travel from the optically denser medium to the optically rarer medium.

Thus, the light must be incident from medium '1' (the denser medium) to medium '2' (the rarer medium).



Step 4: Calculate the critical angle (\(i_c\)).

The refractive index of a medium (\(\mu\)) is given by \(\mu = \frac{c}{v}\), where \(c\) is the speed of light in a vacuum.

So, \(\mu_1 = \frac{c}{v_1}\) and \(\mu_2 = \frac{c}{v_2}\).
The critical angle \(i_c\) for light going from a denser medium (\(\mu_1\)) to a rarer medium (\(\mu_2\)) is given by:
\(\sin i_c = \frac{\mu_{rarer}}{\mu_{denser}} = \frac{\mu_2}{\mu_1}\)

Substitute the expressions for refractive indices in terms of speeds: \(\sin i_c = \frac{c/v_2}{c/v_1} = \frac{v_1}{v_2}\)

Therefore, the critical angle is \(i_c = \sin^{-1}\left(\frac{v_1}{v_2}\right)\).



Step 5: Combine all conditions to find the correct option.

For total internal reflection to occur, the ray of light must be incident from medium '1' (denser) and the angle of incidence must be greater than the critical angle, \(\sin^{-1}\left(\frac{v_1}{v_2}\right)\).



Step 6: Evaluate the options.


(A) medium '1' and at an angle greater than \(\sin^{-1}\left(\frac{v_1}{v_2}\right)\): This perfectly matches our derived conditions.
(B) medium '1' and at an angle greater than \(\cos^{-1}\left(\frac{v_1}{v_2}\right)\): Incorrect critical angle expression.
(C) medium '2' and at an angle greater than \(\sin^{-1}\left(\frac{v_1}{v_2}\right)\): Incorrect medium of incidence. Light must be incident from the denser medium ('1').
(D) medium '2' and at an angle greater than \(\cos^{-1}\left(\frac{v_1}{v_2}\right)\): Incorrect medium and critical angle expression.

Based on the physics principles and the given condition \(v_2 > v_1\), option (A) is the correct choice.
\[ \boxed{medium '1' and at an angle greater than \sin^{-1}\left(\frac{v_1}{v_2}\right)} \] Quick Tip: Remember that Total Internal Reflection (TIR) occurs only when light passes from an optically denser medium to an optically rarer medium. The critical angle for TIR is given by \(\sin i_c = \frac{\mu_{rarer}}{\mu_{denser}}\). If speeds are given, remember that the medium with lower speed is denser (\(v_{denser} < v_{rarer}\)), and thus \(\mu_{denser} > \mu_{rarer}\).

The energy of a photon is given by the equation: \[ E = h f \]
where \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, J \cdot s \)), and \( f \) is the frequency of the light. Substituting the given values: \[ E = 6.626 \times 10^{-34} \times 5.0 \times 10^{14} = 3.313 \times 10^{-19} \, J \]

Now, the number of photons \( n \) emitted per second can be calculated using the power \( P \) of the source and the energy \( E \) of one photon: \[ n = \frac{P}{E} \]
Given that the power emitted is 3.31 mW, or \( 3.31 \times 10^{-3} \) W, we have: \[ n = \frac{3.31 \times 10^{-3}}{3.313 \times 10^{-19}} = 10^{16} \, photons per second \]

Thus, the correct answer is (A). Quick Tip: To find the number of photons emitted by a light source, use the relationship between power, photon energy, and the number of photons per second: \( n = \frac{P}{E} \).

The binding energy per nucleon increases rapidly for light nuclei, reaches a maximum around iron (mass number 56), and then gradually decreases for heavier nuclei. This curve explains nuclear stability and why energy is released during both fusion (of light nuclei) and fission (of heavy nuclei).

Option (A) correctly depicts this: a steep rise followed by a gentle decline, with a peak at A = 56.
Quick Tip: Binding energy per nucleon peaks near iron (A = 56). Use this to identify the correct curve in such questions.

In a forward-biased p-n junction diode, an external voltage is applied such that it reduces the built-in potential barrier. This causes more charge carriers to move across the junction, reducing the width of the depletion layer.

Hence, both the potential barrier and depletion width decrease.
Quick Tip: In forward bias, think “easier flow” — the depletion region shrinks and the barrier becomes lower to allow current.

Step 1: Use de Broglie wavelength formula: \[ \lambda = \frac{h}{mv} \]
Where:

\( \lambda \) = de Broglie wavelength

\( h \) = Planck's constant

\( m \) = mass of particle

\( v \) = speed of particle


Step 2: Given that all particles have the same speed, \[ \lambda \propto \frac{1}{m} \]
So, the smaller the mass, the greater the wavelength.

Step 3: Compare masses:

Electron mass \( m_e \) is smallest

Proton mass \( m_p \approx 1836 \times m_e \)

Deuteron mass \( m_d \approx 2 \times m_p \)


Step 4: Therefore, \[ (B) \quad \lambda_e > \lambda_p > \lambda_d \] Quick Tip: For particles at equal speed, lighter particles have larger de Broglie wavelength. So compare masses to deduce the correct order.

Step 1: Understanding Assertion (A):

In Bohr’s model, the potential energy of an electron in an orbit is negative and equal to \(-2\) times the kinetic energy. Hence, Assertion (A) is false.


Step 2: Understanding Reason (R):

The total energy (kinetic + potential) of a bound system like an electron in an atom is negative, not always positive. So Reason (R) is also false.


Step 3: Final evaluation:

Both statements are incorrect, so option (D) is the right choice.
Quick Tip: In bound systems, total energy is negative — indicating the electron is trapped in the atom.

Step 1: Evaluate Assertion (A):

You cannot create a working p-n junction diode just by physically joining p-type and n-type slabs because there will be no depletion region or electric field formed at the junction. Therefore, Assertion (A) is true.


Step 2: Evaluate Reason (R):

The Reason (R) is incorrect because the carrier concentrations are reversed: in a p-type semiconductor, hole concentration \(\eta_h \gg \eta_e\), and in an n-type semiconductor, electron concentration \(\eta_e \gg \eta_h\). So Reason (R) is false.


Step 3: Relationship between A and R:

Assertion is true, but Reason is false — hence, option (C) is correct.
Quick Tip: Always remember: p-n junctions must be chemically fabricated to allow diffusion and depletion region formation.

Step 1: Verifying Assertion (A):

Galvanometers are designed such that their deflection is proportional to the current. The torque on the coil causes angular deflection proportional to current. Hence, Assertion (A) is true.


Step 2: Verifying Reason (R):

In a galvanometer, a radial magnetic field is used so that the plane of the coil is always perpendicular to the magnetic field. This ensures torque is always proportional to current. So, Reason (R) is true.


Step 3: Explanation linkage:

Since the radial magnetic field ensures consistent torque-current relation, Reason (R) correctly explains Assertion (A).
Quick Tip: Radial magnetic fields ensure uniform behavior in measuring instruments like galvanometers.

Step 1: Understanding Assertion (A):

When a magnet is moved into a coil with many turns and the circuit is closed, a large amount of current is induced due to electromagnetic induction. This current opposes the motion of the magnet (Lenz's Law), making it harder to move the magnet. So, Assertion (A) is true.


Step 2: Understanding Reason (R):

Lenz’s Law states that the direction of induced current is such that it opposes the cause producing it — in this case, the motion of the magnet. Hence, Reason (R) is also true.


Step 3: Checking the relationship between A and R:

The opposition to motion described in the assertion is explained by Lenz’s Law stated in the reason. Therefore, Reason (R) correctly explains Assertion (A).
Quick Tip: Lenz’s Law is key to explaining resistance to magnet motion in closed loops.

We start with the equation for the electric field and the current density in a material: \[ \vec{E} = \rho \vec{j} \]
where:
\( \vec{E} \) is the electric field (measured in volts per meter),

\( \rho \) is the resistivity of the material (measured in ohm meters),

\( \vec{j} \) is the current density (measured in amperes per square meter).

This equation describes the relationship between the electric field and the current density in a material. The resistivity \( \rho \) is a proportional constant that characterizes how strongly the material opposes the flow of current.

Now, let’s use Ohm's law, which states that: \[ \vec{E} = \frac{V}{L} \]
where:

\( V \) is the potential difference (in volts),

\( L \) is the length of the conductor (in meters).

At the same time, the current density \( \vec{j} \) is given by: \[ \vec{j} = \frac{I}{A} \]
where:

\( I \) is the current (in amperes),

\( A \) is the cross-sectional area of the conductor (in square meters).

Substituting \( \vec{j} \) into the first equation, we get: \[ \vec{E} = \rho \frac{I}{A} \]
This is the form of Ohm's law, where \( \vec{E} \) is proportional to \( I \), and the proportionality constant is \( \frac{\rho}{A} \), which can also be interpreted as the resistance \( R \) of the material: \[ R = \frac{\rho L}{A} \]
Thus, we have: \[ \vec{E} = I R \]
This is the familiar form of Ohm's law, \( \vec{E} = I R \), where \( R \) is the resistance of the material.

However, Ohm's law does not hold for materials where the resistivity \( \rho \) is not constant. For example:

Non-linear materials such as semiconductors, where the current density \( \vec{j} \) and electric field \( \vec{E} \) do not follow a linear relationship due to the material’s intrinsic properties.

Temperature-dependent resistivity, where the resistivity changes with temperature, can cause deviations from Ohm’s law. As the current increases, the material heats up, which in turn changes its resistivity.

Therefore, Ohm's law is not valid when the resistivity \( \rho \) depends on factors such as temperature, current, or the electric field. Quick Tip: Ohm's law applies to materials with constant resistivity. If the resistivity depends on factors like temperature or current, the material is non-ohmic, and Ohm’s law does not hold.

Step 1: Use the formula for the first minimum in single-slit diffraction.

The condition for the first minimum is: \[ a \sin \theta = \lambda \]
where:

\( a \) is the width of the slit,

\( \lambda = 600 \, nm = 600 \times 10^{-9} \, m \),

\( \theta = 30^\circ \Rightarrow \sin(30^\circ) = 0.5 \)

Step 2: Substitute the known values into the formula. \[ a = \frac{\lambda}{\sin \theta} = \frac{600 \times 10^{-9}}{0.5} = 1.2 \times 10^{-6} \, m \]

Step 3: Express the answer in micrometres. \[ a = 1.2 \, \mum \] Quick Tip: In single-slit diffraction, the position of the first minimum is given by \( a \sin \theta = m \lambda \), with \( m = 1 \) for the first minimum.

Step 1: Use the formula for phase difference in terms of path difference. \[ \Delta \phi = \frac{2\pi}{\lambda} \times path difference \]
Given: path difference \( \Delta x = \frac{\lambda}{8} \) \[ \Rightarrow \Delta \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{8} = \frac{\pi}{4} \]

Step 2: Use the intensity formula in interference.

When two waves of intensity \( I_0 \) each interfere with phase difference \( \Delta \phi \), the resultant intensity is: \[ I = I_0 + I_0 + 2\sqrt{I_0 \cdot I_0} \cos(\Delta \phi) = 2I_0 + 2I_0 \cos\left(\frac{\pi}{4}\right) \] \[ \Rightarrow I = 2I_0 + 2I_0 \cdot \frac{1}{\sqrt{2}} = 2I_0\left(1 + \frac{1}{\sqrt{2}}\right) = I_0(2 + \sqrt{2}) \approx 3.414 \, I_0 \] Quick Tip: To calculate intensity in interference, convert path difference to phase using \( \phi = \frac{2\pi}{\lambda} \Delta x \), then apply \( I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi \).

We can use the refraction formula for a spherical surface: \[ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \]
where:

\( \mu_2 = 1.5 \) (refractive index of glass)

\( \mu_1 = 1 \) (refractive index of air)

\( u = -\frac{R}{2} \) (object distance, negative because the object is in air)

\( R \) is the radius of curvature of the surface

\( v \) is the image distance we need to calculate.


Substituting the given values: \[ \frac{1.5}{v} - \frac{1}{-\frac{R}{2}} = \frac{1.5 - 1}{R} \]
Simplifying: \[ \frac{1.5}{v} + \frac{2}{R} = \frac{0.5}{R} \] \[ \frac{1.5}{v} = \frac{0.5}{R} - \frac{2}{R} = -\frac{1.5}{R} \] \[ v = -\frac{R}{1.5} = -R \]

Thus, the image is formed at a distance \( -R \) from the surface, which means the image is real and on the opposite side of the surface compared to the object. Quick Tip: When solving problems involving spherical surfaces, remember to use the refraction formula and pay attention to the sign conventions. A negative image distance indicates a real image formed on the opposite side of the surface.

In the Bohr model of the hydrogen atom, the angular momentum \( L \) of an electron in the \( n \)-th orbit is quantized and is given by: \[ L = n \hbar \]
where \( n \) is the principal quantum number, and \( \hbar \) is the reduced Planck's constant (\( \hbar = 1.055 \times 10^{-34} \, J \cdot s \)).

The energy of the electron in the \( n \)-th orbit is given by: \[ E_n = - \frac{13.6 \, eV}{n^2} \]
where \( 13.6 \, eV \) is the energy of the electron in the first orbit (for \( n = 1 \)).

Given that the energy of the electron in this problem is \( -3.4 \, eV \), we can use the formula to find \( n \): \[ E_n = - \frac{13.6 \, eV}{n^2} = -3.4 \, eV \]
Solving for \( n \): \[ n^2 = \frac{13.6}{3.4} = 4 \quad \Rightarrow \quad n = 2 \]

Now, using the formula for angular momentum, since \( n = 2 \): \[ L = n \hbar = 2 \times 1.055 \times 10^{-34} \, J \cdot s = 2.11 \times 10^{-34} \, J \cdot s \]

Thus, the angular momentum of the electron is \( L = 2.11 \times 10^{-34} \, J \cdot s \). Quick Tip: In the Bohr model, the angular momentum is quantized and given by \( L = n\hbar \), where \( n \) is the principal quantum number. The energy of the electron in the \( n \)-th orbit is given by \( E_n = - \frac{13.6}{n^2} \, eV \).

Step 1: Understand the doping ratio

1 dopant atom is added for every \( 5 \times 10^7 \) silicon atoms.


Step 2: Use the given silicon atom density

Number density of silicon atoms is: \[ 5 \times 10^{28} \, atoms/m^3 \]

Step 3: Calculate the number of dopant atoms per m\(^3\)
\[ No. of holes created per m^3 = \frac{5 \times 10^{28}}{5 \times 10^7} = 10^{21} \]

Step 4: Convert to holes per cm\(^3\)
\[ No. of holes per cm^3 = \frac{10^{21}}{10^6} = 10^{15} \]

Step 5: Example of a dopant

One example of such a trivalent dopant is: Aluminium (Al), Indium (In), or Gallium (Ga).
Quick Tip: To find charge carriers from doping, divide atom density by the number of host atoms per dopant, then convert units properly.

Let:

\( E_1 = 3\,V, \quad r_1 = 0.2\,\Omega \)

\( E_2 = 6\,V, \quad r_2 = 0.4\,\Omega \)


Step 1: Equivalent EMF of the parallel combination: \[ E_{eq} = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2} = \frac{\frac{3}{0.2} + \frac{6}{0.4}}{\frac{1}{0.2} + \frac{1}{0.4}} = \frac{15 + 15}{5 + 2.5} = \frac{30}{7.5} = 4\,V \]

Step 2: Equivalent internal resistance of the combination: \[ r_{eq} = \frac{r_1 r_2}{r_1 + r_2} = \frac{0.2 \times 0.4}{0.2 + 0.4} = \frac{0.08}{0.6} = \frac{2}{15}\,\Omega \approx 0.133\,\Omega \]

Step 3: Total resistance in the circuit: \[ R_{total} = r_{eq} + R = 0.133 + 4 = 4.133\,\Omega \]

Step 4: Current drawn from the combination: \[ I = \frac{E_{eq}}{R_{total}} = \frac{4}{4.133} \approx 0.968\,A \] Quick Tip: To combine batteries in parallel, use the formula: \[ E_{eq} = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2}, \quad r_{eq} = \frac{r_1 r_2}{r_1 + r_2} \] Then apply Ohm’s law with total external resistance.

Part (i): Relation between \( R' \) and \( R \)

We know resistance of a wire: \[ R = \rho \frac{l}{A} \]
When length becomes \( 2l \), assuming volume is constant: \[ A \propto \frac{1}{l} \Rightarrow A' = \frac{A}{2} \] \[ R' = \rho \frac{2l}{A/2} = 4 \cdot \frac{\rho l}{A} = 4R \]
So, \( R' = 4R \)

Part (ii): Relation between \( v_d' \) and \( v_d \)

Drift velocity is given by: \[ v_d = \frac{eE\tau}{m} \]
For a given potential \( V \), \[ E = \frac{V}{l} \Rightarrow v_d \propto \frac{1}{l} \]
When length becomes \( 2l \): \[ v_d' = \frac{v_d}{2} \] Quick Tip: When a wire is stretched to double its length, its resistance becomes 4 times and drift velocity becomes half due to the inverse dependence on length.

Step 1: Understanding electron motion in a conductor

In a conductor, free electrons are always in random thermal motion, even without an electric field. These motions are in all directions and are very fast (random velocities).

Step 2: Effect of applying an electric field

When a potential difference is applied, an electric field is established. This field causes a small net velocity (called drift velocity) superimposed on the random motion.

Step 3: Direction of drift velocity

Though the electrons still move randomly, the average motion of all electrons is in a direction opposite to the electric field (from lower to higher potential).

Step 4: Conclusion

Therefore, not all electrons move in the same direction. They still move randomly, but with a small average (drift) velocity in a particular direction. Quick Tip: Drift velocity is a small net movement superimposed on random thermal motion. Electrons do not all move in a straight line; instead, they exhibit a slow drift on top of rapid, random motion.

The magnetic moment (\( \mu \)) of a current-carrying coil is defined as the product of the current \( I \) and the area \( A \) of the coil, with the direction given by the right-hand rule. The magnetic moment is a vector quantity that points perpendicular to the plane of the coil, in the direction of the normal to the coil's surface.

The formula for the magnetic moment is given by: \[ \mu = I \cdot A \]
where:
\( I \) is the current flowing through the coil (in Amperes),

\( A \) is the area of the coil (in square meters).


SI Unit of Magnetic Moment:
The SI unit of magnetic moment is Ampere square meter (\( A \cdot m^2 \)). Quick Tip: The magnetic moment of a coil is a vector quantity, which depends on the current and the area of the coil. Its SI unit is \( A \cdot m^2 \).

The torque \( \tau \) experienced by a current-carrying coil in a magnetic field is given by the formula: \[ \tau = N I A B \sin \theta \]
where:
- \( \tau \) is the torque,

- \( N \) is the number of turns of the coil,

- \( I \) is the current flowing through the coil,

- \( A \) is the area of the coil,

- \( B \) is the magnetic field strength,

- \( \theta \) is the angle between the normal to the coil's plane and the magnetic field.


Step 1: Use the initial torque to find the magnetic field.

Given:

Number of turns: \( N = 60 \),

Area of the coil: \( A = 1.5 \times 10^{-3} \, m^2 \),

Current: \( I = 2 \, A \),

Initial torque \( \tau_1 = 0.12 \, Nm \) (when the angle between the normal to the coil's plane and the magnetic field is \( \theta = 0^\circ \)).

When \( \theta = 0^\circ \), \( \sin 0^\circ = 0 \), so the formula becomes: \[ \tau_1 = N I A B \]
Solving for \( B \): \[ B = \frac{\tau_1}{N I A} = \frac{0.12}{60 \times 2 \times 1.5 \times 10^{-3}} = \frac{0.12}{0.18} = 0.6667 \, T \]

Thus, the magnetic field strength is \( B = 0.6667 \, T \).

Step 2: Use the final torque to find the magnetic field.

Now, the coil is rotated by \( 90^\circ \), so the angle \( \theta = 90^\circ \), and the torque changes to \( \tau_2 = 0.05 \, Nm \). When \( \theta = 90^\circ \), \( \sin 90^\circ = 1 \), so the torque formula becomes: \[ \tau_2 = N I A B \sin 90^\circ = N I A B \]
Using the values: \[ B = \frac{\tau_2}{N I A} = \frac{0.05}{60 \times 2 \times 1.5 \times 10^{-3}} = \frac{0.05}{0.18} = 0.2778 \, T \]

Step 3: Calculate the resultant magnetic field strength.

Now, we need to use the relationship between the two torques, \( \tau_1 \) and \( \tau_2 \), to find the magnitude of the magnetic field. Given: \[ \tau_1 = N I A B \sin \theta_1 \quad and \quad \tau_2 = N I A B \sin \theta_2 \]
For \( \tau_1 = 0.12 \, Nm \), \( \theta_1 = 90^\circ \) (maximum torque), and for \( \tau_2 = 0.05 \, Nm \), the coil is rotated by 90°.

The total magnetic field is given by: \[ B = \sqrt{B_1^2 \sin^2 \theta + B_2^2 \cos^2 \theta} \]

Substituting the values: \[ B = \sqrt{\left(\frac{2}{3}\right)^2 + \left(\frac{5}{18}\right)^2} = \frac{13}{18} \, T \]

Conclusion:
- The resultant magnetic field strength is \( B = \frac{13}{18} \, T \). Quick Tip: The torque on a coil in a magnetic field depends on the number of turns, current, area of the coil, magnetic field strength, and the angle between the normal to the coil's plane and the magnetic field. For a coil rotated by 90°, the torque equation simplifies to \( \tau = N I A B \).

The mutual inductance \( M \) between two solenoids depends on the flux linkage and the induced EMF. For two solenoids with magnetic fields interacting, the mutual inductance is given by:
\[ M_{12} = \frac{\mu_0 n_1 n_2 A l}{l} \]
Where:
\( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, T \cdot m/A \)),

\( n_1 \) and \( n_2 \) are the number of turns per unit length of solenoid 1 and 2 respectively,

\( A \) is the cross-sectional area of the solenoid,

\( l \) is the length of the solenoid.

The magnetic flux through solenoid \( S_2 \) due to solenoid \( S_1 \) is given by:
\[ \Phi_{B2} = B_1 \cdot A = \left( \frac{\mu_0 n_1 I_1}{l} \right) \cdot A \]

For the induced EMF in solenoid \( S_2 \), the formula for mutual inductance becomes:
\[ M_{12} = \frac{\mu_0 n_1 n_2 A l}{l} \]

Similarly, the mutual inductance of solenoid \( S_1 \) with respect to solenoid \( S_2 \) is:
\[ M_{21} = \frac{\mu_0 n_1 n_2 A l}{l} \]

Thus, we can see that:
\[ M_{21} = M_{12} \]

This shows that the mutual inductance is symmetric. Quick Tip: The mutual inductance between two solenoids depends on their turns per unit length, cross-sectional area, and the permeability of free space. The mutual inductance is symmetric, meaning \( M_{12} = M_{21} \).

Step 1: Understanding the two types of current

\( I_c \): The conduction current in the wires and resistor parts of the circuit.

\( I_d \): The displacement current between the plates of the capacitor (where no conduction current flows).

Step 2: Maxwell’s correction to Ampere’s Law

To maintain continuity of current in the entire AC circuit, Maxwell introduced displacement current: \[ I_d = \epsilon_0 \frac{d\Phi_E}{dt} \]
This makes the modified Ampere’s law: \[ \oint \vec{B} \cdot d\vec{l} = \mu_0 (I_c + I_d) \]

Step 3: Applying to the capacitor circuit

As the current \( I_c \) flows into one plate of the capacitor, it changes the electric field between the plates. This changing electric field gives rise to \( I_d \).

Step 4: Conclusion

The net current (conduction + displacement) is the same throughout the circuit at every point: \[ I_{net} = I_c + I_d = constant \] Quick Tip: In an AC circuit with a capacitor, conduction current in the wires is matched by displacement current in the gap between plates. This keeps total current continuous.

Step 1: Kirchhoff’s first rule (junction rule):

It states that the algebraic sum of currents meeting at a junction is zero: \[ \sum I = 0 \]

Step 2: Problem at capacitor plates

At the capacitor plates, conduction current flows into the plate but no conduction current flows through the gap. This seems to violate the junction rule.

Step 3: Inclusion of displacement current

Maxwell resolved this by defining displacement current \( I_d \), which flows through the dielectric gap and acts like a current to maintain continuity.

Step 4: Conclusion

With the displacement current considered, the current entering a plate equals the current (displacement) leaving the plate. So Kirchhoff’s junction rule still holds true. Quick Tip: Junction rule remains valid at the capacitor plates if we include displacement current as part of the current flow. This is crucial in time-varying electric fields.

The plot of frequency \( \nu \) of incident radiation versus the stopping potential \( V_0 \) is a linear relationship. According to Einstein's photoelectric equation: \[ E_k = h\nu - \phi \]
where:

\( E_k \) is the kinetic energy of the emitted electrons,

\( h \) is Planck's constant,

\( \nu \) is the frequency of the incident radiation,

\( \phi \) is the work function of the material.


The stopping potential \( V_0 \) is related to the kinetic energy of the emitted electrons by: \[ E_k = eV_0 \]
where \( e \) is the charge of the electron.

Therefore, combining these two equations: \[ eV_0 = h\nu - \phi \]
This gives the equation of a straight line: \[ V_0 = \frac{h}{e} \nu - \frac{\phi}{e} \]
Thus, the plot of \( V_0 \) versus \( \nu \) will be a straight line with slope \( \frac{h}{e} \) and intercept \( -\frac{\phi}{e} \) on the \( V_0 \)-axis. The intercept on the stopping potential axis gives information about the work function \( \phi \) of the photo-emissive material. Quick Tip: The slope of the plot \( V_0 \) vs \( \nu \) gives \( \frac{h}{e} \), where \( h \) is Planck's constant and \( e \) is the charge of the electron. The intercept on the \( V_0 \)-axis corresponds to the work function \( \phi \) of the material.

(i) Momentum:

The momentum of an electron is given by: \[ p = \sqrt{2mK} \]
where:

\( m = 9.1 \times 10^{-31} \, kg \) is the mass of the electron,

\( K = 80 \, eV = 80 \times 1.6 \times 10^{-19} \, J = 1.28 \times 10^{-17} \, J \) is the kinetic energy of the electron.

Now, substituting the known values: \[ p = \sqrt{2 \times 9.1 \times 10^{-31} \times 1.28 \times 10^{-17}} \, kg \cdot m/s \] \[ p = 4.8 \times 10^{-24} \, kg \cdot m/s \]

Thus, the momentum \( p \approx 4.8 \times 10^{-24} \, kg \cdot m/s \).


(ii) de Broglie Wavelength:

The de Broglie wavelength \( \lambda \) is given by the formula: \[ \lambda = \frac{h}{p} \]
where:

\( h = 6.63 \times 10^{-34} \, J \cdot s \) is Planck’s constant,

\( p = 4.8 \times 10^{-24} \, kg \cdot m/s \) is the momentum.

Substituting the known values: \[ \lambda = \frac{6.63 \times 10^{-34}}{4.8 \times 10^{-24}} \, m \] \[ \lambda \approx 1.38 \times 10^{-10} \, m \]

Thus, the de Broglie wavelength \( \lambda \approx 1.38 \times 10^{-10} \, m \). Quick Tip: To find the momentum, use \( p = \sqrt{2mK} \) and to find the de Broglie wavelength, use \( \lambda = \frac{h}{p} \).

Step 1: Components of the setup:

Power supply (variable DC source)

Diode under test

Resistor (current limiting)

Voltmeter across diode

Ammeter in series to measure current


Step 2: Forward bias arrangement:

Positive terminal of battery connected to p-side and negative terminal to n-side through resistor. Voltmeter is parallel to diode and ammeter in series.




Step 3: Reverse bias arrangement:

Battery polarity reversed — positive to n-side and negative to p-side. Rest of the arrangement remains the same.



Quick Tip: Forward bias: p to +, n to –. Reverse bias: p to –, n to +. Use voltmeter and ammeter correctly.

Step 1: Forward bias region:

Very small current until threshold (knee) voltage.

After this, current rises exponentially with small increase in voltage.

Step 2: Reverse bias region:

Very small leakage current (reverse saturation current).

Sharp increase only if breakdown occurs.


Graph shape:

Forward bias: Exponential rise after knee voltage.

Reverse bias: Almost flat line near zero (reverse current), till breakdown.


Quick Tip: Remember: Diode conducts significantly only after crossing threshold voltage in forward bias.

Step 1: Mass Defect

The mass defect of a nucleus is the difference between the sum of the masses of its individual protons and neutrons and the actual mass of the nucleus. It arises due to the conversion of mass into binding energy.
\[ \Delta m = Z m_p + (A - Z) m_n - M_nucleus \]

Step 2: Binding Energy

Binding energy is the energy required to disassemble a nucleus into its constituent protons and neutrons. It is equivalent to the mass defect by Einstein’s relation: \[ E_b = \Delta m \cdot c^2 \]

Step 3: Fission Process and Binding Energy Per Nucleon

In fission, a heavy nucleus (like uranium) splits into smaller nuclei with higher binding energy per nucleon. Since products have higher binding energy per nucleon, the total binding energy increases, and the difference is released as energy. This explains the huge energy release in nuclear fission.
Quick Tip: Higher binding energy per nucleon means greater stability. Fission releases energy because smaller fragments are more stable.

Step 1: Use known values
\[ m_p = 1.007277\, u, \quad m_n = 1.008665\, u, \quad m_d = 2.013553\, u \]

Step 2: Calculate expected mass without binding
\[ m_p + m_n = 1.007277 + 1.008665 = 2.015942\, u \]

Step 3: Find mass defect
\[ \Delta m = 2.015942 - 2.013553 = 0.002389\, u \]

Step 4: Convert mass defect into energy using \( 1\, u = 931.5\, MeV/c^2 \)
\[ E_b = 0.002389 \times 931.5 = 2.224\, MeV \] Quick Tip: Use accurate atomic mass values and multiply the mass defect by 931.5 to get binding energy in MeV.

Step 1: Understanding the setup:

The lens is convex and the screen is placed on the opposite side of the object. A real image is formed when object is beyond the focal length.


Step 2: Nature of real image:

A real image formed by a convex lens is always inverted and appears on the side opposite to the object (on the screen).


Step 3: Image matching:

Option (C) shows an inverted image on the correct side — the screen. Hence, it is correct.
Quick Tip: Convex lenses form real, inverted images on the opposite side of the screen when the object is beyond focus.

Step 1: Understanding convex mirrors:

Convex mirrors always form virtual, erect, and diminished images — so their magnification is always positive, not negative.


Step 2: Review other statements:

(B) is correct — virtual images have positive magnification.

(C) is correct — concave lenses also always form virtual, erect, diminished images (positive magnification).

(D) is correct — real and inverted implies negative magnification.


Conclusion:

Only (A) is incorrect.
Quick Tip: Virtual images always have positive magnification; convex mirrors form virtual images only.

Step 1: Understanding the type of cut

Here, the lens is cut parallel to the principal axis, i.e., it is sliced horizontally into two thinner lenses.


Step 2: Impact on power and focal length

The power of a lens is given by: \[ P = \frac{1}{f} \]
Cutting along the principal axis results in lenses with half the power: \[ P' = \frac{P}{2} \Rightarrow f' = \frac{1}{P'} = \frac{1}{P/2} = 2f \]

Step 3: Conclusion

Each half has a focal length of \( 2f \), hence the correct answer is (B).
Quick Tip: Cutting a lens \textbf{parallel} to the principal axis reduces its power by half, doubling the focal length.

Step 1: Optical Geometry

In biconvex lens geometry, using Newton’s form of the lens formula: \[ X_1 \cdot X_2 = f^2 \]

Step 2: Solve for \( f \)
\[ f = \sqrt{X_1 X_2} \]

Step 3: Final answer

So, focal length is the geometric mean of object and image distances from their respective focal points.
Quick Tip: Use Newton’s lens formula: \( X_1 X_2 = f^2 \) to find focal length in such cases.

Step 1: Understand the circuit configuration and the condition for final charge.

The problem describes an RC series circuit. When key \(S_1\) is closed and \(S_2\) is open, the capacitor C is connected in series with the resistor R and the battery of emf V. The capacitor will charge through the resistor until it is fully charged.



Step 2: Identify the given numerical values.

From the last line of the preceding paragraph, we are given:


Resistance, \(R = 20 k\Omega = 20 \times 10^3 \Omega\)
Capacitance, \(C = 500 \textmu F = 500 \times 10^{-6} F\)
Voltage of the battery, \(V = 10 V\)




Step 3: Calculate the final charge (Q) on the capacitor.

When the capacitor is fully charged, the potential difference across it equals the battery voltage. The formula for the final charge (Q) on a capacitor is given by:
\(Q = C \times V\)


Substitute the given values into the formula:
\(Q = (500 \times 10^{-6} F) \times (10 V)\)
\(Q = 5000 \times 10^{-6} C\)
\(Q = 5 \times 10^3 \times 10^{-6} C\)
\(Q = 5 \times 10^{-3} C\)




Step 4: Convert the final charge to appropriate units and evaluate the options.

The calculated final charge is \(5 \times 10^{-3} C\).

We know that 1 millicoulomb (mC) = \(10^{-3}\) C.

So, \(Q = 5 mC\).


Now, let's compare this with the given options:


(A) 5 \(\muC\) = \(5 \times 10^{-6}\) C (Incorrect)
(B) 5 mC = \(5 \times 10^{-3}\) C (Correct)
(C) 25 mC = \(25 \times 10^{-3}\) C (Incorrect)
(D) 0.1 C (Incorrect)




Step 5: Conclusion.

The final charge on the capacitor when key \(S_1\) is closed and \(S_2\) is open is 5 mC.
\[ \boxed{5 mC} \] Quick Tip: Always use \( Q = CV \) to find the maximum charge stored on a capacitor in a DC circuit.

Step 1: Initial condition — key \( S_1 \) closed, \( S_2 \) open:

In this condition, the capacitor gets charged fully by the battery. \[ Q = C \cdot V = 500 \times 10^{-6} \cdot 10 = 5 \times 10^{-3} = 5 mC \]

Step 2: Now \( S_2 \) is closed and \( S_1 \) is opened:

The capacitor is now connected across just the resistor (no battery). This leads to discharging of the capacitor through the resistor.

Step 3: Final condition after long time:

As time \( t \to \infty \), the charge on the capacitor: \[ q(t) = Q \cdot e^{-t/RC} \to 0 \]

So, the final charge on the capacitor is: \[ \boxed{0} \] Quick Tip: When a charged capacitor is connected across a resistor without any battery, it discharges completely over time. Final charge becomes zero.

Resistance: \( [R] = [M L^2 T^{-3} A^{-2}] \)

Capacitance: \( [C] = [M^{-1} L^{-2} T^4 A^2] \)

\[ [RC] = [R] \cdot [C] = [T] \] Quick Tip: Time constant \( RC \) always has the dimension of time \([T]\). Multiply units directly to confirm.

\[ I(t) = \frac{V}{R} e^{-t/RC} \] \[ RC = 10\,s,\quad t = 5\,s \Rightarrow \frac{t}{RC} = \frac{1}{2} \] \[ I = \frac{10}{20000} \cdot e^{-1/2} = \frac{1}{2000} \cdot \frac{1}{\sqrt{e}} = \frac{1}{2\sqrt{e}} mA \] Quick Tip: Use \( I(t) = \frac{V}{R} e^{-t/RC} \) to find current at any time \( t \).

Step 1: Use formula for initial charging current: \[ I(0) = \frac{V}{R} \]

Step 2: Given values:

\( V = 10\,V \)

\( R = 20\,k\Omega = 2 \times 10^4\,\Omega \)

\[ I(0) = \frac{10}{2 \times 10^4} = \frac{1}{2000} = 0.0005\,A = 0.5\,mA \] Quick Tip: The initial charging current in an RC circuit is \( I = \frac{V}{R} \). Be sure to convert resistance to ohms and express current in milliamperes.

Coherent sources are sources of light that emit waves with a constant phase relationship. In other words, the phase difference between the waves from coherent sources remains constant over time. For sustained interference patterns, the waves need to maintain a fixed phase difference, which results in a stable constructive or destructive interference.

This constant phase relationship is necessary to observe sustained and stable interference patterns. Without coherence, the interference pattern would fade or change over time, making it impossible to observe clear and stable fringes.

(2) Lights from two independent sources are not coherent. Explain.


% Solution
Solution:

When lights come from two independent sources, the phase difference between the waves from each source is random. These sources do not have a fixed phase relationship and therefore are not coherent. As a result, the interference pattern formed by two such sources would be irregular and would not sustain itself, as the random phase differences lead to fluctuating constructive and destructive interference. Quick Tip: For sustained interference, coherence is key, ensuring that the phase difference between the waves remains constant.

The distance between adjacent bright fringes (the fringe separation \( \Delta y \)) on the screen is given by the formula: \[ \Delta y = \frac{\lambda \cdot L}{d} \]
where:

\( \lambda \) is the wavelength of the light,

\( L \) is the distance from the slits to the screen,

\( d \) is the distance between the two slits.


Given:

\( \lambda = 600\,nm = 600 \times 10^{-9}\,m \)

\( L = 1.20\,m \)

\( d = 0.1\,mm = 0.1 \times 10^{-3}\,m \)


Substituting the values: \[ \Delta y = \frac{600 \times 10^{-9} \cdot 1.20}{0.1 \times 10^{-3}} = 7.2 \times 10^{-3}\,m = 7.2\,mm \]

Thus, the adjacent bright fringes are \( 7.2\,mm \) apart.

(2) Find the angular width (in degrees) of the first bright fringe.


% Solution
Solution:

The angular width of the first bright fringe \( \theta \) is given by: \[ \theta = \frac{\lambda}{d} \]
where:
\( \lambda \) is the wavelength of the light,

\( d \) is the distance between the slits.


Substituting the values: \[ \theta = \frac{600 \times 10^{-9}}{0.1 \times 10^{-3}} = 6^\circ \]

Thus, the angular width of the first bright fringe is \( 6^\circ \). Quick Tip: The fringe separation is dependent on the wavelength of the light, the distance between the slits, and the distance to the screen. The angular width is related to the slit separation and wavelength.

Step 1: Definition of a wavefront:

A wavefront is the locus of points in a medium that are all in phase, i.e., points where the waves have the same phase. For a plane wave, all points on the wavefront are at the same distance from the source and oscillate in unison.


Step 2: Incident and refracted wavefronts:

When a plane wave hits a convex lens, the wavefront bends as it passes through the lens.
The part of the wavefront passing through the lens is refracted and converges towards the focal point of the lens, changing the curvature.

Step 3: Diagram for incident and refracted wavefronts:



Quick Tip: For a converging lens, the incident plane wavefront gets refracted and converges at the focal point, changing the curvature.

Ray Diagram:

A parallel beam of light strikes the spherical glass ball.

The rays get refracted as they pass through the spherical surface.

The rays converge to form an image inside the sphere, depending on the radius and refractive index.





Step 1: Refraction at the first surface (from rarer to denser medium)

For refraction from the first surface, we use the formula: \[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \]
Where:

\( n_1 = 1 \) (refractive index of air),

\( n_2 = 1.5 \) (refractive index of the glass),

\( R = 15 \, cm \) (radius of the sphere),

\( u = \infty \) (object at infinity).


Substituting the values into the equation: \[ \frac{1.5}{v} - \frac{1}{\infty} = \frac{1.5 - 1}{15} \] \[ \frac{1.5}{v} = \frac{0.5}{15} \] \[ v = 45 \, cm \]

Thus, the image formed by the first surface is at a distance of \( v = 45 \, cm \) inside the sphere.

Step 2: Refraction at the second surface (from denser to rarer medium)

Now, for the refraction at the second surface, we use the formula: \[ \frac{n_1}{v'} - \frac{n_2}{u'} = \frac{n_1 - n_2}{R} \]
Where:

\( n_1 = 1 \) (refractive index of air),

\( n_2 = 1.5 \) (refractive index of the glass),

\( R = -15 \, cm \) (radius of curvature of the second surface),

\( u' = 15 \, cm \) (object distance from the second surface).

Substituting the values into the equation: \[ \frac{1}{v'} - \frac{1.5}{15} = \frac{1 - 1.5}{-15} \] \[ \frac{1}{v'} - \frac{0.1}{1} = \frac{-0.5}{-15} \] \[ \frac{1}{v'} - 0.1 = 0.0333 \] \[ \frac{1}{v'} = 0.1333 \] \[ v' = 7.5 \, cm \]

Thus, the final image position after refraction through the second surface is at \( 7.5 \, cm \) inside the sphere.

Step 3: Final Image:

The final image is formed at a distance of \( 7.5 \, cm \) from the second surface.
The image is real and formed inside the spherical glass ball. Quick Tip: For spherical surfaces, the image formed depends on the object distance and the refractive index of the material.

Step 1: Electrostatic Potential Energy Without External Field
\[ U_{initial} = \frac{1}{4\pi\epsilon_0} \cdot \frac{q_1 q_2}{r} \]
where \( r = 6\,cm = 0.06\,m \). Substituting: \[ U_{initial} = \frac{1}{4\pi\epsilon_0} \cdot \frac{(5 \times 10^{-6})(-1 \times 10^{-6})}{0.06} = -\frac{5 \times 10^{-12}}{4\pi\epsilon_0 \times 0.06} \] \[ = -\frac{5 \times 10^{-12} \cdot 9 \times 10^9}{0.06} = -\frac{45 \times 10^{-3}}{0.06} = -0.75\,J \]

Step 2: Electrostatic Potential Energy Due to External Field

The electric potential due to the field is: \[ V(r) = -\int_\infty^r \vec{E} \cdot d\vec{r} = -\int_\infty^r \frac{3 \times 10^5}{r^2} \, dr = \frac{3 \times 10^5}{r} \]
\[ U_{ext} = q_1 V(r_1) + q_2 V(r_2) \]
with \( r_1 = r_2 = 3\,cm = 0.03\,m \):
\[ U_{ext} = \left( 5 \times 10^{-6} \cdot \frac{3 \times 10^5}{0.03} \right) + \left( -1 \times 10^{-6} \cdot \frac{3 \times 10^5}{0.03} \right) \]
\[ = (5 - 1) \times 10^{-6} \cdot 10^7 = 4 \times 10^{-6} \cdot 10^7 = 40\,J \]

Step 3: Total Potential Energy With External Field
\[ U_{total} = U_{initial} + U_{ext} = -0.75 + 40 = 39.25\,J \] Quick Tip: To calculate the change in electrostatic energy due to an external field, use: \[ \Delta U = \sum q_i V(\vec{r}_i) \] especially when mutual interaction remains constant.

\[ C = \frac{Q}{V} = \frac{80 \times 10^{-6}}{16} = 5 \times 10^{-6}\,F \] Quick Tip: Capacitance is defined by \( C = \frac{Q}{V} \), where \( Q \) is the magnitude of charge on either conductor.

Capacitance increases by factor of \( k \): \[ C' = kC = 3 \cdot C \Rightarrow V' = \frac{Q}{C'} = \frac{Q}{3C} = \frac{V}{3} \] \[ V' = \frac{16}{3} \approx 5.33\,V \] Quick Tip: Inserting a dielectric increases capacitance \( C \to kC \), so for same charge, potential reduces: \( V \to \frac{V}{k} \).

Capacitance depends only on geometry and dielectric of the system, not on the amount of charge stored. Even if the charges are doubled, capacitance remains: \[ C = \frac{Q}{V} \quad but both Q and V double \] Quick Tip: Capacitance is a geometric property and independent of charge. Changing the charge will change potential, but not capacitance.

For a spherical shell with a charge distributed over its surface, the electric field at a point outside the shell (at distance \( r \) from the center, where \( r > R \)) is given by Coulomb's Law: \[ E = \frac{KQ}{r^2} \]
where \( Q \) is the total charge on the shell, and \( r \) is the distance from the center.

At a distance \( 3R \) (which is outside all the shells), the electric field depends only on the total charge on the shell, and not on the distribution of charge inside. Therefore, for each shell at a distance \( 3R \), the electric field magnitude is given by: \[ E = \frac{KQ}{(3R)^2} = \frac{KQ}{9R^2} \]

So, for each shell at \( 3R \), the electric field magnitudes are:

For shell A: \( E_A = \frac{K \cdot 6q}{9R^2} \)

For shell B: \( E_B = \frac{K \cdot (-4q)}{9R^2} \)

For shell C: \( E_C = \frac{K \cdot 14q}{9R^2} \)


Since the magnitude depends only on the charge, the electric field magnitudes are proportional to the absolute values of the charges: \[ |E_A| = |E_B| = |E_C| \]
Thus, the electric field magnitudes at a distance of \( 3R \) are the same for all shells. Quick Tip: For a spherical shell with charge \( Q \), the electric field outside the shell depends only on the total charge and not on its internal charge distribution.

We are calculating the work done on the charge \( +5\,\mu C \) as it is moved from point \( C \) to point \( A \) along the semicircular path. The work done is given by: \[ W = q \cdot [V_A - V_C] \]

Step 1: Calculate the potential at point \( C \) (due to both charges at \( B \) and \( D \))
\[ V_C = \left[ \frac{k \cdot 6 \times 10^{-6}}{5 \times 10^{-2}} \right] - \left[ \frac{k \cdot 6 \times 10^{-6}}{5 \times 10^{-2}} \right] \] \[ V_C = 0 \]

Step 2: Calculate the potential at point \( A \):
\[ V_A = \left[ \frac{k \cdot 6 \times 10^{-6}}{15 \times 10^{-2}} \right] - \left[ \frac{k \cdot 6 \times 10^{-6}}{5 \times 10^{-2}} \right] \] \[ V_A = \frac{k \cdot 6 \times 10^{-6}}{15 \times 10^{-2}} \left[ 1 - \frac{3}{15} \right] \] \[ V_A = \frac{9 \times 10^9 \cdot 6 \times 10^{-6} \cdot 2}{15 \times 10^{-2}} \] \[ V_A = -7.2 \times 10^5\, V \]

Step 3: Calculate the work done: \[ W = q \cdot [V_A - V_C] \] \[ W = 5 \times 10^{-6} \cdot \left[ -7.2 \times 10^5 - 0 \right] \] \[ W = -3.6\, J \]

Thus, the work done is \( W = -3.6\, J \). Quick Tip: When moving a charge in an electric field, the work done is the charge times the potential difference between the initial and final positions. The potential at a point due to multiple charges is the sum of the potentials due to each charge.

Step 1: Understanding proton motion in a magnetic field:

The proton is positively charged, and the force acting on it is given by the Lorentz force: \[ \vec{F} = q (\vec{v} \times \vec{B}) \]
where \( q \) is the charge of the proton and \( \vec{B} \) is the magnetic field.


Step 2: Determining the direction of magnetic field:

The proton moves in the plane of the paper, and the force is perpendicular to the velocity, which means the magnetic field is directed perpendicular to the paper.

At point P, the proton is moving along the curved path, and the force must be directed towards the center of the curve. Using the right-hand rule, the direction of \( \vec{B} \) at point P is out of the paper (towards the viewer).

At point Q, since the proton is turning in the opposite direction, the magnetic field direction is into the paper.

At point R, as the proton is turning in the same direction again, the magnetic field direction is again out of the paper.


Step 3: Relative magnitude of magnetic fields:

The magnitude of the magnetic field depends on the curvature of the proton's path. Since the proton moves in a non-uniform magnetic field, the magnetic field strength is greater where the curvature is higher.
Therefore, the magnetic field magnitude at point P (with sharp curvature) is greater than at point R (less curvature). The field at point Q, where the proton changes direction, could be comparable to point P.
Quick Tip: For charged particles in a magnetic field, the direction of the field is determined using the right-hand rule, and the field strength correlates with the curvature of the path.

Step 1: Magnetic field at the center of a current-carrying loop:

The magnetic field \( B \) at the center of a circular loop of radius \( r \) carrying current \( I \) is given by: \[ B = \frac{\mu_0 I}{2r} \]
where \( \mu_0 \) is the permeability of free space.

Step 2: Relating current to magnetic moment:

The magnetic moment \( \vec{\mu} \) of a loop is given by: \[ \vec{\mu} = I A \hat{n} \]
where \( A \) is the area of the loop and \( \hat{n} \) is the unit vector perpendicular to the plane of the loop.

The magnetic moment is also related to the magnetic field at the center by: \[ B = \frac{\mu_0 \mu}{2\pi r^3} \]

Step 3: Substituting the expression for \( B \):

We substitute \( r = \sqrt{\frac{A}{\pi}} \) (from the area of the loop) into the equation for magnetic field: \[ B = \frac{2\mu_0 \mu}{\mu_0 r} \] \[ \Rightarrow \mu = \frac{B A}{2 \mu_0} \]

Thus, the magnetic moment is: \[ \mu = \frac{2BA}{\mu_0} \sqrt{\frac{A}{\pi}} \] Quick Tip: Magnetic moment of a current-carrying loop is proportional to both the current and the area of the loop. The magnetic field at the center of the loop is inversely proportional to the radius.

Let the current loop be a rectangle with dimensions \( l \) and \( w \) (length and width) carrying a current \( I \). The loop is placed in a uniform magnetic field \( \vec{B} \) with the plane of the loop perpendicular to the field.

The torque \( \tau \) acting on the loop is given by: \[ \tau = \vec{\mu} \times \vec{B} \]
where \( \mu \) is the magnetic dipole moment of the current loop, and \( \vec{B} \) is the external magnetic field.

The magnetic dipole moment is defined as: \[ \mu = I \cdot A \]
where \( A \) is the area of the loop. For a rectangular loop, the area is: \[ A = l \cdot w \]

Thus, the magnetic moment becomes: \[ \mu = I \cdot l \cdot w \]

Now, the torque is given by: \[ \tau = \mu B \sin \theta \]
where \( \theta \) is the angle between the magnetic moment and the magnetic field.

For the case where the magnetic moment is perpendicular to the field (\( \theta = 90^\circ \)), the torque simplifies to: \[ \tau = I \cdot l \cdot w \cdot B \]

Thus, the expression for the torque acting on a rectangular current loop suspended in a uniform magnetic field is: \[ \tau = I \cdot A \cdot B \cdot \sin \theta \] Quick Tip: The torque on a current loop in a magnetic field depends on the magnetic moment of the loop and the external field. When the loop is perpendicular to the field, the torque is maximized.

Let the mass of the particle be \( m \), the charge be \( q \), and its velocity be \( v \). The radius \( r \) of the circular path is related to the velocity by: \[ r = \frac{mv}{qB} \]
where \( B \) is the magnetic field strength.

Part (1) Change in Radius:

The particle loses half of its kinetic energy. The initial kinetic energy is: \[ K_i = \frac{1}{2} m v^2 \]
After passing through the sheet of lead, the final kinetic energy is: \[ K_f = \frac{1}{2} K_i = \frac{1}{4} m v^2 \]

The new velocity \( v' \) after the energy loss can be found by equating the final kinetic energy: \[ K_f = \frac{1}{2} m v'^2 \quad \Rightarrow \quad \frac{1}{4} m v^2 = \frac{1}{2} m v'^2 \] \[ v' = \frac{v}{\sqrt{2}} \]

The new radius \( r' \) is given by: \[ r' = \frac{m v'}{qB} = \frac{m \cdot \frac{v}{\sqrt{2}}}{qB} = \frac{r}{\sqrt{2}} \]

Thus, the radius decreases by a factor of \( \sqrt{2} \).

Part (2) Change in Time Period:

The time period \( T \) of revolution is given by: \[ T = \frac{2 \pi r}{v} \]
Substituting the expression for \( r' \), the new time period \( T' \) is: \[ T' = \frac{2 \pi r'}{v'} = \frac{2 \pi \cdot \frac{r}{\sqrt{2}}}{\frac{v}{\sqrt{2}}} = T \]

Thus, the time period remains unchanged. Quick Tip: When a charged particle moves through a material that reduces its kinetic energy, the radius of the path decreases because of the reduced velocity. However, the time period of revolution remains unchanged if only the velocity magnitude changes without altering the charge or magnetic field.