From Ohm's law, we have the relation V = IR, where R is the resistance.


The resistance R can be expressed as R = V/I.


The given graph plots current (I) on the y-axis and voltage (V) on the x-axis.


The slope of the I-V graph is given by Slope = \(\frac{\Delta I}{\Delta V}\).


From the relation R = V/I, we can see that the slope is the reciprocal of the resistance, i.e., Slope = \(\frac{1}{R}\).


From the graph, it is clear that the slope of the line corresponding to temperature T₁ is greater than the slope of the line for temperature T₂.


Slope(T₁) > Slope(T₂)


This implies that \(\frac{1}{R_1} > \frac{1}{R_2}\), which leads to R₁ < R₂.


For metallic conductors, the resistance is directly proportional to the temperature. An increase in temperature causes an increase in resistance.


Since R₁ < R₂, it follows that the corresponding temperature T₁ must be less than T₂.


Therefore, we can conclude that T₁ < T₂.
Quick Tip: Always carefully check the axes of a graph. In a standard V-I graph, the slope represents resistance (R). In an I-V graph, as shown in this question, the slope represents the reciprocal of resistance (1/R). For metals, resistance increases with temperature.

First, let's find the equivalent resistance for the series combination (Rs).


When n identical resistors, each of resistance R, are connected in series, the total equivalent resistance is the sum of their individual resistances.

\(R_s = R + R + ... (n times) = nR\).


Next, let's find the equivalent resistance for the parallel combination (Rp).


When n identical resistors are connected in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of the individual resistances.

\(\frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} + ... (n times) = \frac{n}{R}\).


Inverting this gives the equivalent parallel resistance: \(R_p = \frac{R}{n}\).


Now, we need to calculate the difference (Rs - Rp).

\(R_s - R_p = nR - \frac{R}{n}\).


Factoring out R from the expression:

\(R_s - R_p = R\left(n - \frac{1}{n}\right)\).


To simplify the expression in the parenthesis, we find a common denominator:

\(R_s - R_p = R\left(\frac{n^2 - 1}{n}\right)\).


This result matches option (C).
Quick Tip: Memorize the simple formulas for 'n' identical resistors: In series, Rs = nR. In parallel, Rp = R/n. The ratio Rs/Rp is n², a commonly tested concept. The difference calculation is a straightforward application of these two formulas.

The wire consists of two straight segments and one semicircular segment. The total magnetic field at point O is the vector sum of the fields from these three parts.


For the two straight wire segments, the point O lies on the line extending from the wires (i.e., on their axis).


According to the Biot-Savart law, the magnetic field \(d\vec{B} = \frac{\mu_0}{4\pi} \frac{I(d\vec{l} \times \vec{r})}{r^3}\). For any point on the axis of a straight current element, the vector \(d\vec{l}\) and the position vector \(\vec{r}\) are collinear. Thus, their cross product \(d\vec{l} \times \vec{r}\) is zero.


Therefore, the magnetic field at point O due to both straight segments is zero.


Now, consider the semicircular arc. The magnetic field at the center of a full circular loop of radius R is given by \(B_{circle} = \frac{\mu_0 I}{2R}\).


A semicircle is exactly half of a full circle. Thus, the magnetic field it produces at its center O is half of the field from a full circle.

\(B_{semicircle} = \frac{1}{2} \times B_{circle} = \frac{1}{2} \left(\frac{\mu_0 I}{2R}\right) = \frac{\mu_0 I}{4R}\).


The total magnetic field at O is the sum of the fields from all segments: \(B_{total} = B_{straight1} + B_{straight2} + B_{semicircle} = 0 + 0 + \frac{\mu_0 I}{4R}\).


So, the net magnetic field at O is \(\frac{\mu_0 I}{4R}\).
Quick Tip: Remember two key results from the Biot-Savart law: 1) The magnetic field on the axis of a straight current-carrying wire is zero. 2) The magnetic field at the center of a circular arc subtending an angle \(\theta\) (in radians) is \(B = \frac{\mu_0 I \theta}{4\pi R}\). For a semicircle, \(\theta = \pi\), which simplifies to \(B = \frac{\mu_0 I}{4R}\).

Diamagnetic materials are substances that have a tendency to be repelled by magnetic fields.


When a diamagnetic substance is placed in an external magnetic field, magnetic dipoles are induced within the material.


According to Lenz's law, these induced dipoles align themselves in a direction that opposes the external magnetic field.


This opposition results in a net repulsive force on the material, pushing it away from the magnet or the source of the field.


Consequently, the diamagnetic material will tend to move from a region of stronger magnetic field to a region of weaker magnetic field.


This movement is described as "moving opposite to the field" because it is repelled by the field.


In an ideal, perfectly uniform field, the net force would be zero. However, any real magnetic field has edges with gradients, and the material will be pushed out from the field region, thus the description "moves opposite to the field" is the correct physical characterization.
Quick Tip: Remember the response of different magnetic materials to a non-uniform field: \textbf{Diamagnetic:} Repelled, move from stronger to weaker field region. \textbf{Paramagnetic:} Weakly attracted, move from weaker to stronger field region. \textbf{Ferromagnetic:} Strongly attracted, move from weaker to stronger field region.

An ammeter is a device used to measure the current flowing through a circuit element. It must be connected in series with that element.


To avoid altering the current it is meant to measure, an ideal ammeter should have a very low, preferably zero, resistance.


A galvanometer is a sensitive instrument that gives a full-scale deflection for a very small current, let's call it \(I_g\).


To measure a large current \(I\) (where \(I > I_g\)), most of the current must be diverted away from the galvanometer coil to prevent damage.


This is achieved by providing an alternative, low-resistance path for the excess current.


This path is created by connecting a very small resistance, known as a shunt resistor (S), in parallel with the galvanometer.


The majority of the current, \(I - I_g\), passes through the low-resistance shunt, while only the small current \(I_g\) flows through the galvanometer.


Therefore, a galvanometer is converted into an ammeter by connecting a small resistance in parallel.
Quick Tip: A simple mnemonic: To make an \textbf{A}mmeter, connect a shunt resistor in p\textbf{A}r\textbf{A}llel. To make a \textbf{V}oltmeter, connect a high resistance in se\textbf{R}ies (\textbf{V}e\textbf{R}y large series resistance). Ammeters have low resistance; voltmeters have high resistance.

When a charged particle with charge q and mass m enters a uniform magnetic field B with velocity v perpendicular to B, it experiences a magnetic force \(F_m = qvB\).


This force acts as the centripetal force, causing the particle to move in a circular path. The centripetal force required is \(F_c = \frac{mv^2}{r}\), where r is the radius of the path.


Equating the magnetic force and the centripetal force: \(qvB = \frac{mv^2}{r}\).


Solving for the radius r, we get: \(r = \frac{mv}{qB}\).


Let's define the properties for a proton (p) and an alpha particle (\(\alpha\)).


For a proton: mass \(m_p\), charge \(q_p = e\).


For an alpha particle: mass \(m_\alpha \approx 4m_p\), charge \(q_\alpha = 2e\).


Both particles have the same velocity v and enter the same magnetic field B.


The radius of the proton's path is \(r_p = \frac{m_p v}{eB}\).


The radius of the alpha particle's path is \(r_\alpha = \frac{m_\alpha v}{q_\alpha B} = \frac{(4m_p)v}{(2e)B} = 2 \frac{m_p v}{eB}\).


We can see that \(r_\alpha = 2r_p\).


The question asks for the ratio of the radii of their paths (proton to alpha particle).

\(\frac{r_p}{r_\alpha} = \frac{r_p}{2r_p} = \frac{1}{2}\).
Quick Tip: The radius of a charged particle's path in a magnetic field is directly proportional to its momentum (mv) and inversely proportional to its charge (q). Remember that an \(\alpha\)-particle is a Helium nucleus (\(^4_2He\)), so it has 4 times the mass (approx.) and 2 times the charge of a proton.

As the bar magnet falls through the ring, the magnetic flux linked with the ring changes.


According to Faraday's law of electromagnetic induction, this change in magnetic flux will induce an electromotive force (EMF) in the copper ring.


If the ring were a complete, closed loop, this induced EMF would drive a current through the ring.


According to Lenz's law, this induced current would create its own magnetic field that opposes the change in flux that produced it. This would result in a repulsive force on the magnet as it approaches and an attractive force as it leaves, slowing its fall. The acceleration would be less than g.


However, the copper ring in the question has a cut, meaning it is an open circuit.


Because the circuit is open, no continuous induced current can flow in the ring, despite the presence of an induced EMF.


Since there is no induced current, there is no opposing magnetic field and no magnetic force to oppose the motion of the falling magnet.


Therefore, the magnet falls solely under the influence of gravity, and its acceleration is equal to the acceleration due to gravity, g.
Quick Tip: Lenz's law requires an induced current to create an opposing magnetic field. If the conductive loop is incomplete (has a cut or a gap), no sustained current can flow. Without an induced current, there is no magnetic braking force, and the object falls freely under gravity.

In a series L-R AC circuit, the voltage across the resistor (\(V_R\)) and the voltage across the inductor (\(V_L\)) are not in phase.


The voltage across the resistor, \(V_R\), is in phase with the current.


The voltage across the inductor, \(V_L\), leads the current by 90 degrees (\(\pi/2\) radians).


Therefore, \(V_R\) and \(V_L\) are out of phase by 90 degrees with each other.


The total voltage of the source (\(V_S\)) is the phasor sum of the individual voltages, not their algebraic sum.


Using the Pythagorean theorem for the phasor diagram, the magnitude of the source voltage is given by:

\(V_S = \sqrt{V_R^2 + V_L^2}\).


We are given \(V_R = 8\) V and \(V_L = 6\) V.


Substituting these values into the equation:

\(V_S = \sqrt{(8 V)^2 + (6 V)^2}\).

\(V_S = \sqrt{64 V^2 + 36 V^2}\).

\(V_S = \sqrt{100 V^2}\).

\(V_S = 10 V\).
Quick Tip: In series AC circuits, voltages across different components (R, L, C) add as phasors (vectors), not as simple numbers. For a resistor and inductor, the voltages are perpendicular, so you use the Pythagorean theorem: \(V_S = \sqrt{V_R^2 + V_L^2}\). For a resistor and capacitor, the same formula applies: \(V_S = \sqrt{V_R^2 + V_C^2}\).

When two coherent waves interfere, the resultant intensity at a point is given by \(I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi)\), where \(\phi\) is the phase difference between the waves.


In this case, the intensity of each wave is the same, so \(I_1 = I_2 = I_0\).


The resultant intensity formula becomes \(I_R = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos(\phi)\).

\(I_R = 2I_0 + 2I_0 \cos(\phi) = 2I_0 (1 + \cos(\phi))\).


This formula shows that the intensity varies across the screen depending on the phase difference \(\phi\). The intensity oscillates between a minimum value, \(I_{min} = 2I_0 (1 - 1) = 0\) (for destructive interference), and a maximum value, \(I_{max} = 2I_0 (1 + 1) = 4I_0\) (for constructive interference).


The question asks for the average intensity over the entire screen. According to the principle of conservation of energy, the interference pattern merely redistributes the energy from the two sources. The total energy remains constant.


The average intensity is the total intensity spread over the area. Since the two waves are independent before interference, their total intensity is simply the sum of their individual intensities.


Average Intensity, \(I_{avg} = I_1 + I_2 = I_0 + I_0 = 2I_0\).


Alternatively, we can find the average value of the function \(I_R(\phi)\). The average value of \(\cos(\phi)\) over many cycles or a large area is zero.


Therefore, \(I_{avg} = \langle I_R \rangle = \langle 2I_0 + 2I_0 \cos(\phi) \rangle = 2I_0 + 2I_0 \langle \cos(\phi) \rangle = 2I_0 + 2I_0(0) = 2I_0\).
Quick Tip: In interference problems, the maximum possible intensity is \(I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2\) and the minimum is \(I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2\). For identical sources (\(I_1 = I_2 = I_0\)), \(I_{max} = 4I_0\) and \(I_{min} = 0\). The average intensity over the pattern, due to energy conservation, is simply the sum of the initial intensities: \(I_{avg} = I_1 + I_2 = 2I_0\).

The photoelectric effect occurs when the energy of the incident photon (E) is greater than or equal to the work function (\(\Phi\)) of the material.


The condition for photoemission is \(E \ge \Phi\).


We are given the work function \(\Phi = 2.21\) eV.


The energy of a photon is related to its wavelength (\(\lambda\)) by the formula \(E = \frac{hc}{\lambda}\). A useful constant is \(hc \approx 1240\) eV·nm.


For photoemission to be possible, the energy of the incident light must be at least 2.21 eV. Light that cannot produce photoelectrons will have an energy less than 2.21 eV.


Let's examine the approximate energy ranges for the colors in the options:

- Violet light (\(\lambda \approx 400\) nm): \(E \approx \frac{1240}{400} = 3.1\) eV. (E > \(\Phi\))

- Blue light (\(\lambda \approx 475\) nm): \(E \approx \frac{1240}{475} \approx 2.6\) eV. (E > \(\Phi\))

- Green light (\(\lambda \approx 540\) nm): \(E \approx \frac{1240}{540} \approx 2.3\) eV. (E > \(\Phi\))

- Red light (\(\lambda \approx 650\) nm): \(E \approx \frac{1240}{650} \approx 1.9\) eV. (E < \(\Phi\))


Comparing the photon energies with the work function:

- Violet, Blue, and Green light have enough energy to cause photoemission.

- The energy of red light (approx. 1.9 eV) is less than the work function (2.21 eV).


Therefore, red light cannot produce photoelectrons from this material.


Alternatively, we can calculate the threshold wavelength (\(\lambda_0\)) corresponding to the work function:
\(\lambda_0 = \frac{hc}{\Phi} = \frac{1240 eV·nm}{2.21 eV} \approx 561\) nm.

Any light with a wavelength longer than 561 nm cannot cause photoemission. Red light has a wavelength range of approximately 620-750 nm, which is longer than the threshold wavelength.
Quick Tip: Remember the relationship between color, wavelength, and energy for visible light (VIBGYOR). Violet light has the shortest wavelength and highest energy. Red light has the longest wavelength and lowest energy. For photoemission, the photon energy must exceed the work function. If low-energy red light can't do it, no other color with longer wavelength (infrared) can either.

The energy of a photon (E) is given by the Planck-Einstein relation, \(E = hf\), where h is Planck's constant and f is the frequency.


The energy of a particle is also related to its momentum (p) and mass by \(E = \sqrt{(pc)^2 + (m_0c^2)^2}\).


For a photon, the rest mass \(m_0\) is zero, so its energy is \(E = pc\).


Combining the two energy expressions, we get \(pc = hf\).


Solving for momentum p, we get \(p = \frac{hf}{c}\).


We are given the following values:

Frequency, \(f = 6.0 \times 10^{14}\) Hz.

Planck's constant, \(h = 6.63 \times 10^{-34}\) J·s.

Speed of light, \(c = 3 \times 10^8\) m/s.


Substituting these values into the momentum formula:
\(p = \frac{(6.63 \times 10^{-34} J·s) \times (6.0 \times 10^{14} Hz)}{3 \times 10^8 m/s}\).

\(p = \frac{6.63 \times 6.0}{3} \times 10^{-34 + 14 - 8}\) kg·m/s.

\(p = (6.63 \times 2) \times 10^{-28}\) kg·m/s.

\(p = 13.26 \times 10^{-28}\) kg·m/s.


Writing this in standard scientific notation:
\(p = 1.326 \times 10^{-27}\) kg·m/s.
Quick Tip: Remember the two key formulas for a photon's momentum: \(p = \frac{h}{\lambda}\) and \(p = \frac{hf}{c}\). Use the first formula if wavelength (\(\lambda\)) is given and the second if frequency (f) is given. They are equivalent since \(c = f\lambda\).

The strong nuclear force is the fundamental force responsible for holding the nucleons (protons and neutrons) together in the atomic nucleus.


A key property of the strong nuclear force is that it is 'charge-independent'.


This means that the force of attraction is the same between any two nucleons, regardless of whether they are protons or neutrons, provided other factors like distance and spin orientation are the same.


Therefore, the attractive nuclear force between two protons (\(F_{pp, nuclear}\)) is equal to the attractive force between a proton and a neutron (\(F_{pn}\)), which is also equal to the attractive force between two neutrons (\(F_{nn}\)).

\(F_{pp, nuclear} \approx F_{pn} \approx F_{nn}\).


It is important to note that the net force between two protons, \(F_{pp}\), also includes the electrostatic repulsive force. However, the question refers to the "nuclear forces", which are conventionally taken to be the same.


Within the standard model taught at this level, the principle of charge independence is paramount, leading to the conclusion that the nuclear forces are equal.


Hence, we consider \(F_{pp} = F_{pn} = F_{nn}\).


(Note: In a more advanced context, considering spin dependence and electrostatic repulsion, one might find \(F_{pn} > F_{nn} > F_{pp, net}\), but option D represents the fundamental principle of charge independence).
Quick Tip: A fundamental property of the strong nuclear force is its charge independence: it acts equally between proton-proton, proton-neutron, and neutron-neutron pairs. For exam purposes, unless specified otherwise, assume these forces are equal.

Let's analyze the Assertion (A).

A reflecting telescope uses a large concave mirror as its primary objective to collect and focus light. The law of reflection (angle of incidence equals angle of reflection) is independent of the wavelength (or color) of light. All colors are reflected at the same angle and focus at the same point. Therefore, reflecting telescopes do not suffer from chromatic aberration. Assertion (A) is true.


Now let's analyze the Reason (R).

Chromatic aberration is a defect found in lenses. It occurs because the refractive index of the lens material (like glass) is slightly different for different wavelengths of light. This phenomenon is called dispersion. As a result, when white light passes through a lens, different colors are refracted at slightly different angles and are brought to focus at different points. This is exactly what the reason states. Reason (R) is true.


Finally, let's see if the Reason explains the Assertion.

The reason states that chromatic aberration is caused by refraction. The assertion states that reflecting telescopes (which use mirrors, i.e., reflection) are free from this defect. Since reflecting telescopes use reflection and not refraction to focus light, they are free from the defect caused by refraction. Thus, the Reason correctly explains the Assertion.


Therefore, both A and R are true, and R is the correct explanation of A.
Quick Tip: Remember the key difference: Lenses work by \textbf{refraction}, which is wavelength-dependent (causes chromatic aberration). Mirrors work by \textbf{reflection}, which is wavelength-independent (no chromatic aberration). This is a primary advantage of reflecting telescopes over refracting ones.

Let's analyze the Assertion (A).

In semiconductor physics, when an electron from the valence band is excited to the conduction band, it leaves behind an empty energy state. This vacancy is called a 'hole'. An electron from a neighboring atom can move into this hole, effectively making the hole appear to move in the opposite direction. Since the region with the hole has a net local positive charge (one less electron), the hole is treated as a mobile charge carrier with an effective positive charge equal in magnitude to the electronic charge. So, Assertion (A) is true.


Now let's analyze the Reason (R).

The reason states that a hole is not necessarily a vacancy left behind by an electron in the valence band. This is incorrect. By its very definition in semiconductor physics, a hole is precisely the vacancy created in the valence band when an electron is removed or excited. There is no other definition for a hole in this context. Therefore, Reason (R) is false.


Since Assertion (A) is true and Reason (R) is false, the correct option is (C).
Quick Tip: Think of a hole as a bubble in water. When water molecules move to fill the bubble's space, the bubble itself appears to move in the opposite direction. Similarly, a hole "moves" when electrons in the valence band shift positions. It is defined as a vacancy.

Let's analyze the Assertion (A).

X-rays are produced when electrons that have been accelerated to very high kinetic energies strike a metal target. The sudden deceleration of these fast-moving electrons causes them to lose energy, which is emitted in the form of high-energy photons called X-rays (Bremsstrahlung radiation). The assertion states that "slow moving" electrons are used, which is incorrect. A high accelerating voltage is required to produce energetic electrons. Therefore, Assertion (A) is false.


Now let's analyze the Reason (R).

The electromagnetic spectrum is ordered by energy (or frequency/wavelength). X-rays are located on the high-energy side of the spectrum, beyond ultraviolet light and before gamma rays. They are characterized by their high energy, high frequency, and short wavelength. The reason claims that X-rays consist of "low-energy" photons, which is incorrect. Therefore, Reason (R) is false.


Since both Assertion (A) and Reason (R) are false, the correct option is (D).
Quick Tip: Remember the production of X-rays: Fast electrons hit a target. The process is like throwing a fast ball into a dense forest; it stops suddenly, releasing its energy. For X-rays, the kinetic energy of the electrons is high, and the emitted photons (X-rays) are also high-energy.

Let's analyze the Assertion (A).

The binding energy curve is a graph of binding energy per nucleon (BE/A) versus mass number (A). A key feature of this curve is that it rises sharply for light nuclei, reaches a broad maximum, and then gradually decreases for heavy nuclei. In the range of mass numbers from approximately A = 30 to A = 170, the curve is relatively flat, meaning the binding energy per nucleon is nearly constant at about 8.5 MeV. So, Assertion (A) is true.


Now let's analyze the Reason (R).

The reason states that the nuclear forces in this range are not short-range. This is incorrect. The strong nuclear force is a fundamental force of nature and it is always short-ranged, typically acting over distances of only a few interferometers (10 m). This property is independent of the size of the nucleus. The constancy of the BE/A is, in fact, a consequence of the short-range nature of the nuclear force (a property called saturation), as each nucleon only interacts with its immediate neighbors. Therefore, Reason (R) is false.


Since Assertion (A) is true and Reason (R) is false, the correct option is (C).
Quick Tip: The flatness of the binding energy curve for medium-sized nuclei (30 < A < 170) is due to the saturation of the nuclear force. Saturation means that a nucleon only interacts with a limited number of its neighbors, which is a direct consequence of the force being short-range. So, the constant BE/A is proof that the nuclear force IS short-range.

Let's analyze the circuit diagram. Let the node after the 1 \(\Omega\) resistor be C, and the node before the 20 \(\Omega\) resistor be D.


The circuit between nodes C and D consists of a Wheatstone bridge in parallel with a 25 \(\Omega\) resistor.


First, consider the Wheatstone bridge part. The arms are \(R_1 = 5 \Omega\), \(R_2 = 10 \Omega\), \(R_3 = 15 \Omega\), and \(R_4 = 30 \Omega\). The galvanometer arm has a resistance of 5 \(\Omega\).


We check the condition for a balanced bridge: \(\frac{R_1}{R_3} = \frac{R_2}{R_4}\).

\(\frac{5}{15} = \frac{1}{3}\) and \(\frac{10}{30} = \frac{1}{3}\).


Since the ratio is equal, the bridge is balanced. No current flows through the central 5 \(\Omega\) resistor, so it can be removed from the circuit for calculation.


The upper branch of the bridge now has resistors 5 \(\Omega\) and 10 \(\Omega\) in series. Their equivalent resistance is \(R_{upper} = 5 + 10 = 15 \Omega\).


The lower branch of the bridge has resistors 15 \(\Omega\) and 30 \(\Omega\) in series. Their equivalent resistance is \(R_{lower} = 15 + 30 = 45 \Omega\).


These two branches are in parallel between nodes C and D. The equivalent resistance of the bridge is \(R_{bridge}\).
\(\frac{1}{R_{bridge}} = \frac{1}{R_{upper}} + \frac{1}{R_{lower}} = \frac{1}{15} + \frac{1}{45} = \frac{3+1}{45} = \frac{4}{45}\).

So, \(R_{bridge} = \frac{45}{4} = 11.25 \Omega\).


This bridge combination is in parallel with the 25 \(\Omega\) resistor, also connected between C and D. The total equivalent resistance between C and D, \(R_{CD}\), is:
\(\frac{1}{R_{CD}} = \frac{1}{R_{bridge}} + \frac{1}{25} = \frac{4}{45} + \frac{1}{25} = \frac{20+9}{225} = \frac{29}{225}\).

So, \(R_{CD} = \frac{225}{29} \Omega\).


Finally, the total resistance between A and B, \(R_{AB}\), is the series combination of the 1 \(\Omega\) resistor, \(R_{CD}\), and the 20 \(\Omega\) resistor.
\(R_{AB} = 1 + R_{CD} + 20 = 21 + \frac{225}{29}\).
\(R_{AB} = \frac{21 \times 29 + 225}{29} = \frac{609 + 225}{29} = \frac{834}{29} \Omega\).
Quick Tip: When faced with a complex resistor network, first look for simple series and parallel combinations. If that's not possible, check for a Wheatstone bridge structure. Always test the balance condition (\(R_1/R_3 = R_2/R_4\)). If balanced, the middle resistor can be ignored, greatly simplifying the circuit.

The resultant intensity \(I_R\) of two interfering waves with individual intensities \(I_1\) and \(I_2\) is given by:
\(I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi)\), where \(\phi\) is the phase difference.


Given \(I_1 = I_2 = I_0\), the formula simplifies to \(I_R = 2I_0(1 + \cos(\phi))\).


The relationship between phase difference \(\phi\) and path difference \(\Delta x\) is \(\phi = \frac{2\pi}{\lambda} \Delta x\).


(i) Path difference \(\Delta x = \frac{\lambda}{3}\).

The phase difference is \(\phi = \frac{2\pi}{\lambda} \left(\frac{\lambda}{3}\right) = \frac{2\pi}{3}\) radians.
\(\cos\left(\frac{2\pi}{3}\right) = -0.5 = -\frac{1}{2}\).

Substituting this into the intensity formula:
\(I_R = 2I_0\left(1 - \frac{1}{2}\right) = 2I_0\left(\frac{1}{2}\right) = I_0\).


(ii) Path difference \(\Delta x = \frac{\lambda}{2}\).

The phase difference is \(\phi = \frac{2\pi}{\lambda} \left(\frac{\lambda}{2}\right) = \pi\) radians.
\(\cos(\pi) = -1\).

Substituting this into the intensity formula:
\(I_R = 2I_0(1 - 1) = 0\).

This corresponds to a condition of perfect destructive interference.
Quick Tip: Memorize the key relationships for interference: \(\phi = \frac{2\pi}{\lambda} \Delta x\) and \(I_R = I_{max} \cos^2(\phi/2)\) where \(I_{max}=4I_0\). For \(\Delta x = \lambda/3\), \(\phi=2\pi/3\), so \(\phi/2=\pi/3\). \(I_R = 4I_0 \cos^2(\pi/3) = 4I_0 (1/2)^2 = I_0\). For \(\Delta x = \lambda/2\), \(\phi=\pi\), so \(\phi/2=\pi/2\). \(I_R = 4I_0 \cos^2(\pi/2) = 0\).

We use the formula for refraction at a single spherical surface:
\(\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}\).


Here, light travels from air to glass.
Refractive index of the first medium (air), \(n_1 = 1\).

Refractive index of the second medium (glass), \(n_2 = 1.5\).


According to the sign convention:
The object is placed 12 cm in front of the surface, so the object distance \(u = -12\) cm.

The surface is convex, so its center of curvature is on the side where light goes. Thus, the radius of curvature \(R = +30\) cm.


Substituting these values into the formula:
\(\frac{1.5}{v} - \frac{1}{-12} = \frac{1.5 - 1}{+30}\).

\(\frac{1.5}{v} + \frac{1}{12} = \frac{0.5}{30}\).

\(\frac{1.5}{v} + \frac{1}{12} = \frac{1}{60}\).

\(\frac{1.5}{v} = \frac{1}{60} - \frac{1}{12} = \frac{1 - 5}{60} = \frac{-4}{60} = -\frac{1}{15}\).

\(1.5 \times 15 = -v\).
\(v = -22.5\) cm.


Position: The image is formed at 22.5 cm from the spherical surface.

Nature: The negative sign for \(v\) indicates that the image is formed on the same side of the surface as the object. Such an image is virtual.
Quick Tip: Always be careful with the sign convention for the lens maker's formula and spherical surface refraction. A standard convention is: light travels from left to right, the pole is the origin, distances to the right are positive, and distances to the left are negative. A convex surface has a positive radius of curvature (R > 0).

(i) To find the energy of a single photon (E), we use the Planck-Einstein relation:
\(E = hf\), where \(h\) is Planck's constant and \(f\) is the frequency.

Given: \(f = 3.0 \times 10^{14}\) Hz and \(h \approx 6.63 \times 10^{-34}\) J·s.
\(E = (6.63 \times 10^{-34} J·s) \times (3.0 \times 10^{14} s^{-1})\).
\(E = 19.89 \times 10^{-20}\) J.
\(E = 1.989 \times 10^{-19}\) J.


(ii) To find the number of photons emitted per second (n), we use the definition of power.

Power (P) is the total energy emitted per unit time. If n photons are emitted per second, then:
\(P = n \times E\).

Therefore, \(n = \frac{P}{E}\).

Given: Power \(P = 9 mW = 9 \times 10^{-3} W = 9 \times 10^{-3}\) J/s.

Using the energy E calculated in part (i):
\(n = \frac{9 \times 10^{-3} J/s}{1.989 \times 10^{-19} J}\).
\(n \approx 4.5248 \times 10^{16}\) photons/second.

Rounding to three significant figures, \(n = 4.52 \times 10^{16}\) photons/second.
Quick Tip: Remember the fundamental relationships: Photon energy \(E=hf\), and Power \(P = (Number of photons per second) \times (Energy per photon)\). Always ensure your units are consistent (e.g., convert mW to W) before calculating.

1. Entry into Prism ABC: The ray MN is incident normally on face AB, so it enters the prism without any deviation.


2. At Face AC: The ray travels horizontally inside prism ABC and strikes the face AC. Since prism ABC is a right-angled isosceles prism (\(\angle A = 45^\circ\)), the angle of incidence at face AC is \(i_1 = 45^\circ\).

The critical angle (\(i_c\)) for the glass-air interface is given by \(\sin(i_c) = 1/n = 1/1.6 = 0.625\).

Since \(\sin(45^\circ) \approx 0.707\), we have \(\sin(i_1) > \sin(i_c)\). Therefore, Total Internal Reflection (TIR) occurs.

The ray reflects from AC at an angle of 45°, turning through 90°. The ray now travels vertically downwards, perpendicular to the base BC.


3. At Interface BC: The ray strikes the interface BC between the two prisms at an angle of 90° (normal incidence). Since both prisms are made of the same material (\(n=1.6\)), the ray passes undeviated from prism ABC into prism DBC.


4. At Face DC: The ray continues vertically downwards and strikes the face DC. Prism DBC is equilateral, so \(\angle BCD = 60^\circ\). The angle of incidence at face DC is \(i_2 = 60^\circ\).

Since \(i_2 = 60^\circ > i_c\), TIR occurs again at face DC. The ray reflects at an angle of 60°.


5. Emergence from Face DB: The ray reflected from DC now travels towards face DB. From the geometry of the equilateral prism, this ray strikes the face DB at normal incidence (90°). Therefore, the ray emerges from face DB without any deviation. The final emergent ray is parallel to the base BC but travels in the opposite direction.
Quick Tip: For prism problems, carefully apply geometry to find the angles of incidence at each face. Always calculate the critical angle (\(\sin(i_c) = n_{rarer}/n_{denser}\)) and compare it with the angle of incidence to check for Total Internal Reflection (TIR) whenever light tries to exit into a rarer medium.

1. An n-type semiconductor is created by doping an intrinsic semiconductor (like silicon) with pentavalent (donor) impurity atoms (like phosphorus).


2. At room temperature, there are two processes that generate free charge carriers:

(a) Thermal Generation: Thermal energy breaks some covalent bonds, creating electron-hole pairs. This process creates an equal number of free electrons and holes.

(b) Impurity Donation: The fifth valence electron of each donor atom is very loosely bound and requires very little energy to become a free electron in the conduction band. This process creates a large number of free electrons without creating any holes.


3. The total concentration of electrons (\(n_e\)) is the sum of electrons from thermal generation and electrons from donor atoms.

The total concentration of holes (\(p_h\)) is due only to thermal generation.


4. While electron-hole recombination is a continuous process, at thermal equilibrium, the rate of generation of charge carriers equals the rate of recombination.


5. Since a vast majority of the electrons come from the donor atoms, which do not create holes, the total number of electrons is significantly greater than the number of holes. Therefore, in an n-type semiconductor, electrons are the majority charge carriers and holes are the minority charge carriers (\(n_e \gg p_h\)).
Quick Tip: Remember the source of carriers: \textbf{Intrinsic:} Thermal energy breaks bonds \(\rightarrow\) 1 electron + 1 hole. \textbf{n-type doping:} Donor atom gives \(\rightarrow\) 1 electron + 0 holes. \textbf{p-type doping:} Acceptor atom takes \(\rightarrow\) 0 electrons + 1 hole. The imbalance in generation is the reason for having majority and minority carriers.

Difference between EMF and Terminal Voltage:

\begin{tabular{p{0.45\textwidth | p{0.45\textwidth
\hline
EMF (Electromotive Force) & Terminal Voltage (V)

\hline
It is the maximum potential difference between the two terminals of a cell when no current is drawn from it (open circuit). & It is the potential difference between the two terminals of a cell when current is being drawn from it (closed circuit).

It is independent of the external resistance of the circuit. & It depends on the current flowing and the internal resistance (\(V = E - Ir\)).

It is the cause of the current flow in a circuit. & It is the effect of the current flow in a circuit.

\hline
\end{tabular


Derivation for Parallel Combination:

Consider two cells connected in parallel between points A and B.

Let the potential at point A be \(V_A\) and at point B be \(V_B\). The terminal voltage across the combination is \(V = V_A - V_B\).

Let the total current from the combination be \(I\), which splits into \(I_1\) through the first cell and \(I_2\) through the second cell. So, \(I = I_1 + I_2\).


For the first cell, the terminal voltage is \(V = E_1 - I_1 r_1\).

From this, we can write \(I_1 = \frac{E_1 - V}{r_1}\). \quad (1)


For the second cell, the terminal voltage is \(V = E_2 - I_2 r_2\).

From this, we can write \(I_2 = \frac{E_2 - V}{r_2}\). \quad (2)


Substitute equations (1) and (2) into \(I = I_1 + I_2\):
\(I = \frac{E_1 - V}{r_1} + \frac{E_2 - V}{r_2}\).
\(I = \frac{E_1}{r_1} - \frac{V}{r_1} + \frac{E_2}{r_2} - \frac{V}{r_2}\).
\(I = \left(\frac{E_1}{r_1} + \frac{E_2}{r_2}\right) - V\left(\frac{1}{r_1} + \frac{1}{r_2}\right)\).


Rearranging the terms to solve for V:
\(V\left(\frac{1}{r_1} + \frac{1}{r_2}\right) = \left(\frac{E_1}{r_1} + \frac{E_2}{r_2}\right) - I\).
\(V\left(\frac{r_1+r_2}{r_1r_2}\right) = \left(\frac{E_1r_2 + E_2r_1}{r_1r_2}\right) - I\).


Multiply the entire equation by \(\frac{r_1r_2}{r_1+r_2}\) to isolate V:
\(V = \frac{E_1r_2 + E_2r_1}{r_1+r_2} - I\left(\frac{r_1r_2}{r_1+r_2}\right)\).


This equation is of the form \(V = E_{eq} - I r_{eq}\).

By comparing the two equations, we get:

Equivalent EMF: \(E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}\).

Equivalent internal resistance: \(r_{eq} = \frac{r_1 r_2}{r_1 + r_2}\).
Quick Tip: For parallel combination of cells, think of the equivalent resistance \(r_{eq}\) as two resistors in parallel. The equivalent EMF \(E_{eq}\) can be remembered from the expression \( \frac{E_{eq}}{r_{eq}} = \frac{E_1}{r_1} + \frac{E_2}{r_2} \). This form is easier to generalize for more than two cells.

(i) Torque on the loop:


The magnetic field (\(\vec{B}\)) produced by the long straight wire is directed perpendicular to the plane of the loop (into the page, by the right-hand thumb rule).

The magnetic dipole moment of the loop is given by \(\vec{m} = I_{loop}\vec{A}\). The area vector \(\vec{A}\) is also perpendicular to the plane of the loop.

Therefore, the magnetic field vector \(\vec{B}\) and the magnetic moment vector \(\vec{m}\) are parallel (or anti-parallel) to each other. The angle \(\theta\) between them is 0° or 180°.

The torque on a current loop in a magnetic field is given by \(\vec{\tau} = \vec{m} \times \vec{B}\).

The magnitude of the torque is \(\tau = mB\sin\theta\).

Since \(\theta\) is 0° or 180°, \(\sin\theta = 0\).

Thus, the torque acting on the loop is zero.


(ii) Net force on the loop:

Let the sides of the loop be named AB (nearer to the wire), CD (farther), BC, and DA. Length of AB and CD is \(l = 5 cm = 0.05 m\).

The magnetic field is non-uniform; it is stronger near the wire.

The forces on the sides BC and DA are equal in magnitude and opposite in direction, so they cancel each other out.


Force on side AB (at distance \(d_1 = 1 cm = 0.01 m\)):

Magnetic field at AB is \(B_1 = \frac{\mu_0 I_{wire}}{2\pi d_1}\).

Force \(F_{AB} = I_{loop} l B_1 = \frac{\mu_0 I_{wire} I_{loop} l}{2\pi d_1}\). By Fleming's left-hand rule, this force is attractive (towards the wire).


Force on side CD (at distance \(d_2 = 1+1 = 2 cm = 0.02 m\)):

Magnetic field at CD is \(B_2 = \frac{\mu_0 I_{wire}}{2\pi d_2}\).

Force \(F_{CD} = I_{loop} l B_2 = \frac{\mu_0 I_{wire} I_{loop} l}{2\pi d_2}\). This force is repulsive (away from the wire).


Since \(d_1 < d_2\), \(B_1 > B_2\), and therefore \(F_{AB} > F_{CD}\).

The net force is \(F_{net} = F_{AB} - F_{CD}\) (in the direction of \(F_{AB}\)).
\(F_{net} = \frac{\mu_0 I_{wire} I_{loop} l}{2\pi} \left(\frac{1}{d_1} - \frac{1}{d_2}\right)\).

Substituting the values: \(I_{wire}=2\) A, \(I_{loop}=1\) A, \(l=0.05\) m, \(d_1=0.01\) m, \(d_2=0.02\) m.
\(F_{net} = \frac{4\pi \times 10^{-7} \times 2 \times 1 \times 0.05}{2\pi} \left(\frac{1}{0.01} - \frac{1}{0.02}\right)\).
\(F_{net} = (2 \times 10^{-7} \times 2 \times 0.05) \left(100 - 50\right)\).
\(F_{net} = (2 \times 10^{-8}) \times (50)\).
\(F_{net} = 100 \times 10^{-8} = 1 \times 10^{-6}\) N.

The direction is towards the long straight wire.
Quick Tip: For a planar loop in a magnetic field that is everywhere perpendicular to its plane, the torque is always zero. The net force is zero only if the magnetic field is uniform. If the field is non-uniform, there will be a net force directed towards the region of the stronger field.

Lenz's Law:

Lenz's law gives the direction of the induced electromotive force (emf) and current resulting from electromagnetic induction. The law states: The polarity of the induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it. It is a consequence of the conservation of energy.


Derivation of Induced EMF:

Consider a small element of length \(dx\) of the rod MN at a distance \(x\) from the end M.

This element is moving with a linear velocity \(v = x\omega\).

Since the rod is rotating in a magnetic field \(\vec{B}\) that is perpendicular to the plane of rotation, the velocity \(\vec{v}\) is perpendicular to \(\vec{B}\).

The motional emf induced across this small element \(dx\) is given by:
\(d\mathcal{E} = Bv\,dx\).

Substituting \(v = x\omega\), we get:
\(d\mathcal{E} = B(x\omega)dx\).


To find the total emf induced across the entire length of the rod (from M to N), we integrate this expression from \(x=0\) (at end M) to \(x=L\) (at end N).
\(\mathcal{E} = \int_{0}^{L} d\mathcal{E} = \int_{0}^{L} B\omega x\,dx\).


Since B and \(\omega\) are constant, we can take them out of the integral:
\(\mathcal{E} = B\omega \int_{0}^{L} x\,dx\).

\(\mathcal{E} = B\omega \left[ \frac{x^2}{2} \right]_{0}^{L}\).

\(\mathcal{E} = B\omega \left( \frac{L^2}{2} - \frac{0^2}{2} \right)\).

\(\mathcal{E} = \frac{1}{2}B\omega L^2\).

This is the expression for the emf induced between the ends of the rotating rod.
Quick Tip: The motional EMF in a rotating rod can also be found by considering the rate of change of area swept. In one revolution (time \(T=2\pi/\omega\)), the area swept is \(\pi L^2\). The flux change is \(\Delta\Phi = B(\pi L^2)\). The average EMF is \(\Delta\Phi/\Delta t = B\pi L^2 / (2\pi/\omega) = \frac{1}{2}B\omega L^2\). The instantaneous EMF is the same in this case.

Definition of Self-Inductance:

Self-inductance is the property of an electrical circuit (like a coil) by virtue of which it opposes any change in the electric current flowing through it. This opposition is in the form of an induced electromotive force (emf). Numerically, the self-inductance (L) of a coil is defined as the magnetic flux linkage (\(N\Phi_B\)) with the coil per unit current (I) flowing through it.
\(L = \frac{N\Phi_B}{I}\). The SI unit of self-inductance is the Henry (H).


Derivation for a Long Solenoid:

Consider a long solenoid of length \(l\), cross-sectional area A, and n turns per unit length. The total number of turns is \(N = n l\).


When a current I flows through the solenoid, it produces a nearly uniform magnetic field inside it. The magnitude of this magnetic field is given by:
\(B = \mu_0 n I\).


The magnetic flux (\(\Phi_B\)) through a single turn of the solenoid is the product of the magnetic field and the area of the turn:
\(\Phi_B = B \cdot A = (\mu_0 n I) A\).


The total magnetic flux linkage for the entire solenoid is the flux through one turn multiplied by the total number of turns (N):

Total Flux Linkage = \(N \Phi_B = (n l) \times (\mu_0 n I A)\).

Total Flux Linkage = \(\mu_0 n^2 A l I\).


By the definition of self-inductance, \(L = \frac{Total Flux Linkage}{I}\).
\(L = \frac{\mu_0 n^2 A l I}{I}\).

\(L = \mu_0 n^2 A l\).

This is the expression for the self-inductance of a long solenoid.
Quick Tip: The self-inductance \(L\) depends only on the geometry of the coil (number of turns, area, length) and the magnetic properties of the core material (\(\mu_0\)). It is independent of the current flowing through the coil.

(i) In radar systems:

- EM Wave: Microwaves.

- Reason: Microwaves have short wavelengths, allowing them to be transmitted as a narrow beam. They can travel long distances without significant loss of energy and can reflect off objects like airplanes, ships, and weather formations.

- Wavelength Range: Approximately \(10^{-3}\) m (1 mm) to 1 m.


(ii) In eye surgery (e.g., LASIK):

- EM Wave: Ultraviolet (UV) rays.

- Reason: High-energy UV photons from an excimer laser can be precisely focused to break molecular bonds in the cornea's tissue, allowing for its reshaping with minimal thermal damage to surrounding areas. This process is called photoablation.

- Wavelength Range: Approximately \(10^{-8}\) m (10 nm) to \(4 \times 10^{-7}\) m (400 nm).


(iii) As a diagnostic tool in medicine:

- EM Wave: X-rays.

- Reason: X-rays have high penetrating power. They can pass through soft tissues but are absorbed more by denser materials like bones. This differential absorption allows for the creation of images of the internal structures of the body.

- Wavelength Range: Approximately \(10^{-11}\) m (0.01 nm) to \(10^{-8}\) m (10 nm).
Quick Tip: Memorize the electromagnetic spectrum in order of increasing wavelength (decreasing energy): Gamma rays, X-rays, UV, Visible, Infrared, Microwaves, Radio waves. "Good Xylophones Use Very Interesting Musical Rhythms" can be a helpful mnemonic.

Ray Diagram:

The object AB is placed between the center of curvature (C) and the principal focus (F) of a concave mirror.

1. A ray from A parallel to the principal axis strikes the mirror at M and is reflected through the focus F.

2. A ray from A passing through the center of curvature C strikes the mirror normally and is reflected back along the same path.

The two reflected rays intersect at point A' beyond C. A perpendicular A'B' drawn to the principal axis is the real, inverted, and magnified image of the object AB.


Derivation of Mirror Formula:

In the ray diagram, consider the triangle \(\triangle ABC\) and \(\triangle A'B'C\).
\(\angle BAC = \angle B'A'C\) (Both are 90°).
\(\angle BCA = \angle B'CA'\) (Vertically opposite angles).

Therefore, \(\triangle ABC \sim \triangle A'B'C\) by AA similarity.

This gives us the relation: \(\frac{A'B'}{AB} = \frac{CB'}{CB}\) \quad (1)


Now consider the triangles \(\triangle MPF\) and \(\triangle A'B'F\). (Assuming the aperture of the mirror is small, the point M is very close to the pole P, so MP can be considered a straight line perpendicular to the principal axis).
\(\angle MPF \approx \angle A'B'F = 90^\circ\).
\(\angle MFP = \angle A'FB'\) (Vertically opposite angles).

Therefore, \(\triangle MPF \sim \triangle A'B'F\).

This gives us the relation: \(\frac{A'B'}{MP} = \frac{B'F}{PF}\).

Since the incident ray is parallel to the principal axis, \(MP = AB\).

So, \(\frac{A'B'}{AB} = \frac{B'F}{PF}\) \quad (2)


From equations (1) and (2), we have:
\(\frac{CB'}{CB} = \frac{B'F}{PF}\).


Now, express these distances in terms of object distance (u), image distance (v), and focal length (f) from the pole P.
\(CB = PB - PC\)
\(CB' = PC - PB'\)
\(B'F = PB' - PF\)


Using the sign convention: Object distance \(PB = -u\), Image distance \(PB' = -v\), Focal length \(PF = -f\), and Radius of curvature \(PC = -2f\).
\(\frac{-2f - (-v)}{-u - (-2f)} = \frac{-v - (-f)}{-f}\).
\(\frac{v - 2f}{2f - u} = \frac{f - v}{-f}\).
\(-f(v - 2f) = (f - v)(2f - u)\).
\(-vf + 2f^2 = 2f^2 - uf - 2vf + uv\).
\(-vf = -uf - 2vf + uv\).
\(vf + uf = uv\).


Divide the entire equation by \(uvf\):
\(\frac{vf}{uvf} + \frac{uf}{uvf} = \frac{uv}{uvf}\).
\(\frac{1}{u} + \frac{1}{v} = \frac{1}{f}\).

This is the required mirror formula.
Quick Tip: For deriving the mirror or lens formula, the key is to identify two pairs of similar triangles. One pair usually involves the center of curvature or the pole, and the other pair involves the principal focus. Equating the ratios of corresponding sides leads to the formula. Always be meticulous with the sign convention.

Necessary Force:

According to the Bohr model, the electron revolves around the nucleus in a fixed circular orbit. The necessary centripetal force required to maintain this circular motion is provided by the electrostatic force of attraction (Coulomb's force) between the positively charged nucleus and the negatively charged electron.

Centripetal Force = Electrostatic Force
\(\frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{(+Ze)(-e)}{r^2}\).

For a hydrogen atom, the atomic number Z = 1.
\(\frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}\).


Derivation of Total Energy:

The total energy (E) of the electron is the sum of its kinetic energy (KE) and potential energy (PE).

1. Kinetic Energy (KE):

From the force equation above, we can get an expression for \(mv^2\):
\(mv^2 = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r}\).

The kinetic energy is \(KE = \frac{1}{2}mv^2\).
\(KE = \frac{1}{2} \left( \frac{1}{4\pi\epsilon_0} \frac{e^2}{r} \right) = \frac{1}{8\pi\epsilon_0} \frac{e^2}{r}\).


2. Potential Energy (PE):

The electrostatic potential energy of a system of two charges \(q_1\) and \(q_2\) separated by a distance r is \(PE = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r}\).

Here, \(q_1 = +e\) (charge of nucleus) and \(q_2 = -e\) (charge of electron).
\(PE = \frac{1}{4\pi\epsilon_0} \frac{(+e)(-e)}{r} = -\frac{1}{4\pi\epsilon_0} \frac{e^2}{r}\).


3. Total Energy (E):
\(E = KE + PE\).
\(E = \left( \frac{1}{8\pi\epsilon_0} \frac{e^2}{r} \right) + \left( -\frac{1}{4\pi\epsilon_0} \frac{e^2}{r} \right)\).
\(E = \frac{e^2}{4\pi\epsilon_0 r} \left( \frac{1}{2} - 1 \right)\).
\(E = -\frac{1}{8\pi\epsilon_0} \frac{e^2}{r}\).


Significance of the Negative Sign:

The negative sign of the total energy indicates that the electron-nucleus system is a bound system. This means the electron is trapped in the electrostatic field of the nucleus and cannot escape. Energy must be supplied to the electron to overcome this binding and make its total energy zero (or positive), at which point it becomes a free electron. The magnitude of this negative energy is the binding energy of the electron in that orbit.
Quick Tip: A useful relationship to remember in the Bohr model is that Total Energy (E) = - Kinetic Energy (KE) and Potential Energy (PE) = 2 \(\times\) Total Energy (E). This can be a quick check during derivations.

The energy released (Q) in a nuclear reaction is due to the conversion of mass into energy, according to Einstein's mass-energy equivalence principle. It is calculated from the mass defect (\(\Delta m\)).


The nuclear reaction is: \({}^2_1H + {}^3_1H \rightarrow {}^4_2He + {}^1_0n + Q\).


Step 1: Calculate the total mass of the reactants.

Mass of reactants = \(m({}^2_1H) + m({}^3_1H)\).

Mass of reactants = \(2.014102 u + 3.016049 u = 5.030151 u\).


Step 2: Calculate the total mass of the products.

Mass of products = \(m({}^4_2He) + m({}^1_0n)\).

Mass of products = \(4.002603 u + 1.008665 u = 5.011268 u\).


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

Mass defect \(\Delta m\) = (Mass of reactants) - (Mass of products).
\(\Delta m = 5.030151 u - 5.011268 u\).
\(\Delta m = 0.018883 u\).


Step 4: Calculate the energy released (Q).

The energy released is given by \(Q = \Delta m \times c^2\).

We are given that \(1 u = 931 MeV/c^2\).

So, \(Q = (0.018883 u) \times (931 MeV/u)\).
\(Q = 17.580073 MeV\).

Rounding to two decimal places, the energy released is \(Q \approx 17.58 MeV\).
Quick Tip: For nuclear reactions, remember the simple rule: \(Q = (Mass_{initial} - Mass_{final}) \times c^2\). If \(Q\) is positive, energy is released (exothermic reaction, like fusion or fission). If \(Q\) is negative, energy is absorbed (endothermic reaction).

Step 1: Express the radius of a nucleus.

The radius (R) of a nucleus is experimentally found to be proportional to the cube root of its mass number (A).
\(R = R_0 A^{1/3}\), where \(R_0\) is an empirical constant approximately equal to \(1.2 \times 10^{-15}\) m (1.2 fm).


Step 2: Express the volume of a nucleus.

Assuming the nucleus to be spherical, its volume (V) is given by:
\(V = \frac{4}{3}\pi R^3\).

Substituting the expression for R from Step 1:
\(V = \frac{4}{3}\pi (R_0 A^{1/3})^3\).
\(V = \frac{4}{3}\pi R_0^3 (A^{1/3})^3 = \frac{4}{3}\pi R_0^3 A\).


Step 3: Express the mass of a nucleus.

The mass (M) of a nucleus is approximately equal to the mass number (A) multiplied by the average mass of a single nucleon (m). (Let m be the mass of a proton or neutron, \(\approx 1.67 \times 10^{-27}\) kg).
\(M \approx A \cdot m\).


Step 4: Calculate the nuclear density (\(\rho\)).

Density is defined as mass per unit volume.
\(\rho = \frac{M}{V}\).

Substituting the expressions for M and V:
\(\rho = \frac{A \cdot m}{\frac{4}{3}\pi R_0^3 A}\).


The mass number A in the numerator and the denominator cancels out.
\(\rho = \frac{m}{\frac{4}{3}\pi R_0^3}\).


Conclusion:

The final expression for nuclear density (\(\rho\)) contains only constants (m, \(\pi\), \(R_0\)). It does not contain the mass number A. Therefore, the nuclear density is constant for all nuclei and is independent of the mass number.
Quick Tip: The key to this proof is the relationship \(R \propto A^{1/3}\). This implies that Volume (\(\propto R^3\)) is directly proportional to A. Since Mass is also directly proportional to A, their ratio (Density) becomes independent of A.

The capacitance of the original parallel plate capacitor is given by the formula:
\(C = \frac{\epsilon_0 A}{d}\).


For the new configuration shown in Fig. (b):

The new distance between the plates is \(d' = 2d\).

The diagram shows that the plates are shifted such that the effective overlapping area is halved.

So, the new area of overlap is \(A' = A/2\).


The capacitance of the new system, \(C'\), is calculated using the same formula with the new parameters:
\(C' = \frac{\epsilon_0 A'}{d'}\).


Substituting the new values for area and distance:
\(C' = \frac{\epsilon_0 (A/2)}{2d} = \frac{1}{4} \frac{\epsilon_0 A}{d}\).


Since \(C = \frac{\epsilon_0 A}{d}\), we can substitute C into the expression for C':
\(C' = \frac{1}{4} C = \frac{C}{4}\).
Quick Tip: The capacitance of a parallel plate capacitor is directly proportional to the overlapping area of the plates and inversely proportional to the distance between them. Any change in these geometric factors will alter the capacitance accordingly.

Germanium (Ge) is a tetravalent semiconductor, belonging to Group 14 of the periodic table.

To create an n-type semiconductor, we need to introduce donor impurities.

Donor impurities are elements from Group 15 (pentavalent atoms), which have five valence electrons.

When a pentavalent atom replaces a Ge atom in the crystal lattice, four of its valence electrons form covalent bonds with neighboring Ge atoms, and the fifth electron is loosely bound and can be easily donated to the conduction band.

Let's check the options:

- Boron (B), Aluminium (Al), and Indium (In) are all Group 13 elements (trivalent). They are acceptor impurities used to create p-type semiconductors.

- Antimony (Sb) is a Group 15 element. It is a pentavalent atom and acts as a donor impurity.

Therefore, Antimony is a donor impurity atom for Ge.
Quick Tip: A simple mnemonic to remember dopants: "B-Al-Ga-In" (Group 13) are acceptor impurities for p-type. "P-As-Sb-Bi" (Group 15) are donor impurities for n-type.

Principle:

An AC generator works on the principle of electromagnetic induction.


It states that whenever the magnetic flux linked with a coil changes, an electromotive force (EMF) is induced in the coil.


Diagram:

A labelled diagram of an AC generator includes:

1. A strong magnetic field (provided by permanent magnets with N and S poles).

2. An armature coil (a rectangular coil ABCD with many turns).

3. Slip Rings (R₁ and R₂).

4. Brushes (B₁ and B₂).

5. A load resistor (R) in the external circuit.


Working:

1. The armature coil is rotated in the uniform magnetic field about an axis perpendicular to the field.


2. As the coil rotates, the angle \(\theta\) between the magnetic field vector and the area vector of the coil changes continuously.


3. This change in angle causes the magnetic flux linked with the coil, \(\Phi_B = NBA\cos(\theta) = NBA\cos(\omega t)\), to change with time.


4. According to Faraday's law of induction, an EMF is induced in the coil, given by \(\mathcal{E} = -\frac{d\Phi_B}{dt} = -NBA \frac{d}{dt}(\cos(\omega t)) = NBA\omega\sin(\omega t)\).


5. The ends of the coil are connected to two slip rings, R₁ and R₂, which rotate with the coil.


6. Two stationary carbon brushes, B₁ and B₂, are kept in contact with the rotating slip rings. The external circuit is connected across these brushes.


7. The sinusoidal nature of the EMF (\(\mathcal{E} = \mathcal{E}_0 \sin(\omega t)\)) means the direction of the current in the coil and the external circuit reverses periodically, thus producing an alternating current (AC).
Quick Tip: The key difference between an AC generator and a DC generator lies in the connection to the external circuit. AC generators use slip rings for a continuous connection, while DC generators use a split-ring commutator to reverse the current direction every half rotation.

Given the source voltage \(v = 140 \sin(100\pi t)\).


Comparing this with \(v = v_0 \sin(\omega t)\), we get:

Peak voltage \(v_0 = 140\) V.

Angular frequency \(\omega = 100\pi\) rad/s.


First, we calculate the reactances and impedance:

Inductive reactance: \(X_L = \omega L = (100\pi) \left(\frac{5}{\pi}\right) = 500 \, \Omega\).


Capacitive reactance: \(X_C = \frac{1}{\omega C} = \frac{1}{(100\pi) \left(\frac{50}{\pi} \times 10^{-6}\right)} = \frac{1}{5000 \times 10^{-6}} = \frac{10^6}{5000} = 200 \, \Omega\).


Impedance of the circuit: \(Z = \sqrt{R^2 + (X_L - X_C)^2}\).
\(Z = \sqrt{(400)^2 + (500 - 200)^2} = \sqrt{160000 + (300)^2} = \sqrt{160000 + 90000} = \sqrt{250000} = 500 \, \Omega\).


Now, calculate the RMS values:

RMS voltage of the source: \(V_{rms} = \frac{v_0}{\sqrt{2}} = \frac{140}{\sqrt{2}}\) V.


RMS current in the circuit: \(I_{rms} = \frac{V_{rms}}{Z} = \frac{140/\sqrt{2}}{500} = \frac{140}{500\sqrt{2}} = \frac{1.4}{5\sqrt{2}} = \frac{0.28}{\sqrt{2}}\) A \(\approx 0.198\) A.


RMS voltages across each element:

Across the resistor: \(V_{R,rms} = I_{rms} \times R = \frac{0.28}{\sqrt{2}} \times 400 = \frac{112}{\sqrt{2}} \approx 79.2\) V.


Across the inductor: \(V_{L,rms} = I_{rms} \times X_L = \frac{0.28}{\sqrt{2}} \times 500 = \frac{140}{\sqrt{2}} \approx 99.0\) V.


Across the capacitor: \(V_{C,rms} = I_{rms} \times X_C = \frac{0.28}{\sqrt{2}} \times 200 = \frac{56}{\sqrt{2}} \approx 39.6\) V.


Explanation for the sum of voltages:

Algebraic sum = \(V_{R,rms} + V_{L,rms} + V_{C,rms} \approx 79.2 + 99.0 + 39.6 = 217.8\) V.


Source RMS voltage \(V_{rms} \approx 99.0\) V.


The algebraic sum (217.8 V) is greater than the source voltage (99.0 V).


This is because the voltages across R, L, and C are not in phase with each other. They are phasors.

- \(V_R\) is in phase with the current \(I\).

- \(V_L\) leads the current \(I\) by 90°.

- \(V_C\) lags the current \(I\) by 90°.

The source voltage is the phasor sum, not the algebraic sum:
\(V_{rms} = \sqrt{V_{R,rms}^2 + (V_{L,rms} - V_{C,rms})^2}\).
\(V_{rms} = \sqrt{(79.2)^2 + (99.0 - 39.6)^2} = \sqrt{6272.64 + (59.4)^2} \approx \sqrt{6273 + 3528} \approx \sqrt{9801} \approx 99.0\) V.

This matches the source voltage, confirming that phasor addition must be used.
Quick Tip: In a series LCR circuit, voltages add like vectors (phasors) due to phase differences. The total voltage is found using the Pythagorean theorem on a phasor diagram, not by simple addition.

Principle:

A transformer works on the principle of mutual induction.


It states that when the electric current flowing through a coil changes, the magnetic flux linked with a nearby coil also changes, which induces an EMF in the second coil.


Diagram (Step-up Transformer):

A labelled diagram includes:

1. A soft iron laminated core.

2. A primary coil (P) with a fewer number of turns (\(N_p\)).

3. A secondary coil (S) with a greater number of turns (\(N_s\)).

4. An input AC voltage source connected to the primary.

5. A load connected to the secondary.


Working:

1. An alternating voltage is applied across the primary coil. This causes an alternating current to flow through it.


2. This alternating primary current produces a continuously changing magnetic flux in the soft iron core.


3. The changing magnetic flux gets linked with the secondary coil.


4. According to Faraday's law of induction, this changing flux induces an alternating EMF in the secondary coil.


5. Let \(\phi\) be the flux per turn. For an ideal transformer, the EMF induced in the primary is \(E_p = -N_p \frac{d\phi}{dt}\) and in the secondary is \(E_s = -N_s \frac{d\phi}{dt}\).


6. The ratio of the EMFs is \(\frac{E_s}{E_p} = \frac{N_s}{N_p}\).


7. In a step-up transformer, the number of turns in the secondary is greater than in the primary (\(N_s > N_p\)).


8. Therefore, the secondary voltage is greater than the primary voltage (\(V_s > V_p\)), and the voltage is stepped up.
Quick Tip: Remember the transformer equations: \(\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s}\). For a step-up transformer, voltage increases, so current must decrease (assuming ideal transformer where power is conserved).

The design specification is to convert 50 V to 250 V (RMS values).

The turns ratio is \(k = \frac{N_s}{N_p} = \frac{V_s}{V_p} = \frac{250}{50} = 5\).


The actual input voltage is given as \(v_i = 20 \sin(100\pi t)\) V.

From this, the peak input voltage is \(v_{p,peak} = 20\) V.

The RMS input voltage is \(V_{p,rms} = \frac{v_{p,peak}}{\sqrt{2}} = \frac{20}{\sqrt{2}}\) V.


The transformer draws power \(P_{in} = 200\) W.


(I) RMS value of input current (\(I_{p,rms}\)):

Power \(P_{in} = V_{p,rms} \times I_{p,rms}\).
\(I_{p,rms} = \frac{P_{in}}{V_{p,rms}} = \frac{200}{20/\sqrt{2}} = \frac{200\sqrt{2}}{20} = 10\sqrt{2}\) A.
\(I_{p,rms} \approx 14.14\) A.


(II) Expression for instantaneous output voltage (\(v_o(t)\)):

The peak output voltage is stepped up by the turns ratio \(k\). However, the design specification (50V to 250V) implies a turns ratio of 5, while the given instantaneous input voltage has a peak of 20V. Let's assume the turns ratio \(k=5\) is the fixed property of this transformer.

Peak output voltage \(v_{o,peak} = k \times v_{p,peak} = 5 \times 20 = 100\) V.

The output voltage will be in phase with the input voltage and has the same angular frequency.
\(v_o(t) = v_{o,peak} \sin(\omega t) = 100 \sin(100\pi t)\) V.


(III) Expression for instantaneous output current (\(i_o(t)\)):

For an ideal transformer, input power equals output power, so \(V_p I_p = V_s I_s\).

This gives the current ratio \(\frac{I_s}{I_p} = \frac{N_p}{N_s} = \frac{1}{k} = \frac{1}{5}\).

First find the peak input current: \(I_{p,peak} = I_{p,rms} \times \sqrt{2} = (10\sqrt{2}) \times \sqrt{2} = 20\) A.

The peak output current is \(I_{o,peak} = \frac{I_{p,peak}}{k} = \frac{20}{5} = 4\) A.

The instantaneous output current is:
\(i_o(t) = I_{o,peak} \sin(\omega t) = 4 \sin(100\pi t)\) A.
Quick Tip: Be careful to distinguish between RMS and peak values in AC problems. The instantaneous expression \(v(t) = v_0 \sin(\omega t)\) uses the peak value \(v_0\), while power calculations often use RMS values (\(P = V_{rms}I_{rms}\)). Remember \(V_{rms} = v_0/\sqrt{2}\).

Ray Diagram (Final Image at Infinity):

1. An objective lens (short focal length \(f_o\)) is placed near the object AB.

2. The objective forms a real, inverted, and magnified image A'B' of the object.

3. For the final image to be at infinity, this intermediate image A'B' must be formed exactly at the principal focus of the eyepiece lens.

4. The eyepiece lens (larger focal length \(f_e\)) acts as a simple magnifier for the image A'B'.

5. Since A'B' is at its focus, the rays emerging from the eyepiece are parallel, forming a final virtual and highly magnified image at infinity.


Magnification Expression:

The total magnifying power (M) of a compound microscope is the product of the linear magnification of the objective (\(m_o\)) and the angular magnification of the eyepiece (\(M_e\)).
\(M = m_o \times M_e\).


Magnification by Objective (\(m_o\)):

The objective forms image A'B' at a distance \(v_o\) from its optical center. The object distance is \(u_o\).
\(m_o = \frac{v_o}{u_o}\).

For a microscope, the object is placed just outside \(f_o\), so \(u_o \approx -f_o\). The image A'B' is formed near the eyepiece, so \(v_o \approx L\), where L is the tube length (distance between the objective and eyepiece).

So, \(m_o \approx -\frac{L}{f_o}\). (The negative sign indicates a real, inverted image).


Magnification by Eyepiece (\(M_e\)):

The eyepiece acts as a simple microscope. When the final image is formed at infinity, its angular magnification is given by:
\(M_e = \frac{D}{f_e}\), where D is the least distance of distinct vision (typically 25 cm).


Total Magnification (M):

Combining the two expressions:
\(M = m_o \times M_e = \left(-\frac{L}{f_o}\right) \left(\frac{D}{f_e}\right)\).
\(M = -\frac{L D}{f_o f_e}\).
Quick Tip: For a compound microscope, remember the two adjustment cases for the final image: \textbf{Image at Infinity (Normal Adjustment):} Intermediate image is at \(f_e\). \(M = m_o \times (D/f_e)\). \textbf{Image at Near Point (D):} Intermediate image is within \(f_e\). \(M = m_o \times (1 + D/f_e)\).

Given values:

Object distance for the objective, \(u_o = -1.5\) cm.

Focal length of the objective, \(f_o = +1.25\) cm.

Focal length of the eyepiece, \(f_e = +5\) cm.


Step 1: Find the image position for the objective (\(v_o\)).

We use the lens formula for the objective lens: \(\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}\).
\(\frac{1}{v_o} - \frac{1}{-1.5} = \frac{1}{1.25}\).
\(\frac{1}{v_o} + \frac{1}{1.5} = \frac{1}{1.25}\).
\(\frac{1}{v_o} = \frac{1}{1.25} - \frac{1}{1.5} = \frac{4}{5} - \frac{2}{3}\).
\(\frac{1}{v_o} = \frac{12 - 10}{15} = \frac{2}{15}\).

So, the image distance for the objective is \(v_o = \frac{15}{2} = 7.5\) cm.


Step 2: Determine the position of the intermediate image relative to the eyepiece.

The problem states that the final image is formed at infinity.

For the eyepiece to form an image at infinity, the object for it (which is the intermediate image formed by the objective) must be located at its principal focus.

Therefore, the distance of the intermediate image from the eyepiece is equal to the eyepiece's focal length.

Object distance for the eyepiece, \(u_e = f_e = 5\) cm.


Step 3: Calculate the distance between the lenses.

The distance between the objective and the eyepiece (L) is the sum of the image distance of the objective and the object distance of the eyepiece.
\(L = |v_o| + |u_e|\).
\(L = 7.5 cm + 5 cm\).
\(L = 12.5\) cm.
Quick Tip: The distance between the lenses in a microscope or telescope is the sum of the image distance for the first lens (\(v_o\)) and the object distance for the second lens (\(u_e\)). The condition "final image at infinity" immediately tells you that \(u_e = f_e\).

Consider a plane wavefront AB incident on a plane interface XY separating two media, air (rarer, speed \(v_1\)) and glass (denser, speed \(v_2\)).


1. Let the wavefront be incident at an angle i. At time \(t=0\), the wavefront touches the interface at point A.


2. According to Huygens' principle, every point on the wavefront acts as a source of secondary wavelets.


3. While the point B on the wavefront travels to point C on the interface, let the time taken be \(\tau\). The distance covered is \(BC = v_1 \tau\).


4. In the same time \(\tau\), the secondary wavelet from point A will travel a distance \(AE = v_2 \tau\) into the denser medium. We draw a hemispherical wavelet of radius AE centered at A.


5. The new refracted wavefront is the common tangent CE drawn from point C to this wavelet. The line AE is the refracted ray.


Verification of Snell's Law:

In the right-angled triangle \(\triangle ABC\):
\(\sin i = \frac{BC}{AC} = \frac{v_1 \tau}{AC}\).


In the right-angled triangle \(\triangle AEC\):
\(\sin r = \frac{AE}{AC} = \frac{v_2 \tau}{AC}\).


Now, we take the ratio of these two equations:
\(\frac{\sin i}{\sin r} = \frac{v_1 \tau / AC}{v_2 \tau / AC} = \frac{v_1}{v_2}\).


The refractive index of medium 2 with respect to medium 1 is defined as \(n_{21} = \frac{v_1}{v_2}\).

Therefore, \(\frac{\sin i}{\sin r} = n_{21}\).

This is Snell's law of refraction, which is thus verified using Huygens' principle.
Quick Tip: The key to deriving Snell's law from Huygens' principle is to relate the distances traveled by the wavelets in the two media (\(v_1 t\) and \(v_2 t\)) to the sines of the angles of incidence and refraction using a common hypotenuse (AC).

The mirror formula relates the object distance (u), image distance (v), and focal length (f):
\(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\).


We rearrange the formula to solve for the image distance, v:
\(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\).


Now, we apply the sign convention for a convex mirror and a real object:

1. For a convex mirror, the principal focus (F) is behind the mirror. Therefore, its focal length is positive: \(f > 0\).


2. For any real object placed in front of the mirror, the object distance is negative: \(u < 0\).


Let's analyze the expression for \(\frac{1}{v}\) using these conditions:

Since \(f > 0\), the term \(\frac{1}{f}\) is positive.


Since \(u < 0\), the term \(-\frac{1}{u}\) is also positive.


So, \(\frac{1}{v} = (a positive number) + (a positive number)\).


This means that the sum, \(\frac{1}{v}\), must always be positive.


If \(\frac{1}{v} > 0\), then the image distance \(v\) must also be positive.


According to the sign convention for mirrors, a positive image distance means the image is formed behind the mirror.


An image formed behind the mirror is always a virtual image.


Thus, a convex mirror always produces a virtual image for any real object placed in front of it.
Quick Tip: The nature of a mirror is defined by its focal length: \(f < 0\) for concave and \(f > 0\) for convex. The nature of an object/image is defined by its position: \(u, v < 0\) for real (in front) and \(u, v > 0\) for virtual (behind). Using these sign conventions in the mirror formula allows you to deduce the image properties.

The work done by an external agent (\(W_{ext}\)) in moving a charge q from an initial point A to a final point B in an electric field \(\vec{E}\) is given by the change in potential energy, which is \(W_{ext} = q(V_B - V_A)\).


First, we need to find the potential difference between the points A(0, 3m) and B(5m, 0).

The potential difference is related to the electric field by \(V_B - V_A = -\int_{A}^{B} \vec{E} \cdot d\vec{l}\).


Here, \(\vec{E} = 40x \hat{i}\) and the differential displacement vector is \(d\vec{l} = dx \hat{i} + dy \hat{j} + dz \hat{k}\).

So, \(\vec{E} \cdot d\vec{l} = (40x \hat{i}) \cdot (dx \hat{i} + dy \hat{j} + dz \hat{k}) = 40x \, dx\).


Now we integrate from point A to point B:
\(V_B - V_A = -\int_{A}^{B} 40x \, dx\).

The integral only depends on the x-coordinate. The initial x-coordinate is \(x_A = 0\) and the final x-coordinate is \(x_B = 5\).
\(V_B - V_A = -\int_{0}^{5} 40x \, dx = -40 \left[\frac{x^2}{2}\right]_{0}^{5}\).
\(V_B - V_A = -20 [x^2]_{0}^{5} = -20 (5^2 - 0^2) = -20(25) = -500\) V.


The work done in taking a unit positive charge (\(q = +1\) C) is:
\(W_{ext} = q(V_B - V_A) = (1 C) \times (-500 V)\).
\(W_{ext} = -500\) J.
Quick Tip: Remember the distinction: Work done by the field is \(W_{field} = q(V_A - V_B)\). Work done by an external agent (to move the charge) is \(W_{ext} = q(V_B - V_A)\). The external work is the negative of the field work.

Let the charge on the inner sphere be \(Q_1\) and on the outer sphere be \(Q_2\).

Given: \(Q_1 + Q_2 = Q\). (1)

Also, their surface charge densities (\(\sigma\)) are equal.
\(\sigma_1 = \sigma_2 \implies \frac{Q_1}{Area_1} = \frac{Q_2}{Area_2}\).
\(\frac{Q_1}{4\pi r^2} = \frac{Q_2}{4\pi R^2} \implies Q_1 R^2 = Q_2 r^2\).

From this, \(Q_2 = Q_1 \frac{R^2}{r^2}\). (2)


Substitute (2) into (1):
\(Q_1 + Q_1 \frac{R^2}{r^2} = Q \implies Q_1 \left(1 + \frac{R^2}{r^2}\right) = Q \implies Q_1 \left(\frac{r^2+R^2}{r^2}\right) = Q\).

This gives \(Q_1 = \frac{Q r^2}{r^2 + R^2}\).

And \(Q_2 = Q - Q_1 = Q - \frac{Q r^2}{r^2 + R^2} = Q \left(\frac{r^2+R^2-r^2}{r^2+R^2}\right) = \frac{Q R^2}{r^2 + R^2}\).


(I) Electric field at the common centre:

The electric field inside a uniformly charged hollow spherical shell is zero.

The common centre is inside both hollow spheres.

Therefore, the electric field at the center due to the inner sphere is zero, and the field due to the outer sphere is also zero.
\(E_{centre} = E_1 + E_2 = 0 + 0 = 0\).


(II) Potential at the common centre:

The potential inside a hollow spherical shell is constant and equal to the potential on its surface, which is \(V = \frac{1}{4\pi\epsilon_0}\frac{Charge}{Radius}\).

The potential at the center is the sum of the potentials due to both spheres.
\(V_{centre} = V_1 + V_2 = \frac{1}{4\pi\epsilon_0} \frac{Q_1}{r} + \frac{1}{4\pi\epsilon_0} \frac{Q_2}{R}\).

Substitute the expressions for \(Q_1\) and \(Q_2\):
\(V_{centre} = \frac{1}{4\pi\epsilon_0} \left[ \frac{1}{r}\left(\frac{Q r^2}{r^2 + R^2}\right) + \frac{1}{R}\left(\frac{Q R^2}{r^2 + R^2}\right) \right]\).
\(V_{centre} = \frac{Q}{4\pi\epsilon_0(r^2 + R^2)} \left[ \frac{r^2}{r} + \frac{R^2}{R} \right]\).
\(V_{centre} = \frac{Q}{4\pi\epsilon_0(r^2 + R^2)} [r + R]\).
\(V_{centre} = \frac{Q(r+R)}{4\pi\epsilon_0(r^2 + R^2)}\).
Quick Tip: Key results for a spherical shell of charge q and radius R: Field inside (\(r

Consider an electric dipole with charges +q and -q separated by a distance 2a. Let P be a point on the equatorial plane at a distance r from the center of the dipole.

The distance of P from both +q and -q is \(\sqrt{r^2 + a^2}\).


Electric field at P due to +q is \(E_{+q}\), directed away from +q.
\(|\vec{E}_{+q}| = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2 + a^2}\).


Electric field at P due to -q is \(E_{-q}\), directed towards -q.
\(|\vec{E}_{-q}| = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2 + a^2}\).


The vertical components of the fields (\(E_{+q}\sin\theta\) and \(E_{-q}\sin\theta\)) are equal and opposite, so they cancel out.

The horizontal components (\(E_{+q}\cos\theta\) and \(E_{-q}\cos\theta\)) add up.

The net electric field \(E_{net} = 2 E_{+q} \cos\theta\).

From the geometry, \(\cos\theta = \frac{a}{\sqrt{r^2 + a^2}}\).

\(E_{net} = 2 \left(\frac{1}{4\pi\epsilon_0} \frac{q}{r^2 + a^2}\right) \left(\frac{a}{\sqrt{r^2 + a^2}}\right) = \frac{1}{4\pi\epsilon_0} \frac{2qa}{(r^2 + a^2)^{3/2}}\).

The dipole moment magnitude is \(p = q(2a)\).
\(E_{net} = \frac{1}{4\pi\epsilon_0} \frac{p}{(r^2 + a^2)^{3/2}}\).


Direction: The net field is directed from P towards the left, which is opposite to the direction of the dipole moment vector \(\vec{p}\) (from -q to +q).

In vector form: \(\vec{E}_{eq} = -\frac{1}{4\pi\epsilon_0}\frac{\vec{p}}{(r^2+a^2)^{3/2}}\).


(I) At the centre of the dipole (r = 0):
\(E_{centre} = \frac{1}{4\pi\epsilon_0} \frac{p}{(0^2 + a^2)^{3/2}} = \frac{1}{4\pi\epsilon_0} \frac{p}{a^3}\).


(II) At a point r >> a:

The term \(a^2\) can be neglected in comparison to \(r^2\). So, \(r^2 + a^2 \approx r^2\).
\(E_{net} \approx \frac{1}{4\pi\epsilon_0} \frac{p}{(r^2)^{3/2}} = \frac{1}{4\pi\epsilon_0} \frac{p}{r^3}\).
Quick Tip: For a short dipole (\(r \gg a\)), remember the key results: Axial field: \(E \propto 1/r^3\), parallel to \(\vec{p}\). Equatorial field: \(E \propto 1/r^3\), anti-parallel to \(\vec{p}\). The magnitude of the axial field is twice the magnitude of the equatorial field for the same distance r.

The net electric flux through a closed surface is given by Gauss's law, \(\Phi_{net} = \oint \vec{E} \cdot d\vec{A}\).


The electric field \(\vec{E}\) is directed only along the x-axis. Therefore, the flux will be non-zero only through the faces of the cube that are perpendicular to the x-axis.

These are the left face (at \(x=0\)) and the right face (at \(x=L\)).

The flux through the other four faces is zero because their area vectors are perpendicular to \(\vec{E}\).


Flux through the left face (at x=0):

The electric field at this face is \(\vec{E}_{left} = (10(0) + 5)\hat{i} = 5\hat{i}\) N/C.

The area of the face is \(A = L^2\). The area vector points outwards, so \(\vec{A}_{left} = -L^2 \hat{i}\) m².
\(\Phi_{left} = \vec{E}_{left} \cdot \vec{A}_{left} = (5\hat{i}) \cdot (-L^2 \hat{i}) = -5L^2\) Nm²/C.


Flux through the right face (at x=L):

The electric field at this face is \(\vec{E}_{right} = (10L + 5)\hat{i}\) N/C.

The area vector points outwards, so \(\vec{A}_{right} = +L^2 \hat{i}\) m².
\(\Phi_{right} = \vec{E}_{right} \cdot \vec{A}_{right} = ((10L + 5)\hat{i}) \cdot (L^2 \hat{i}) = (10L + 5)L^2 = 10L^3 + 5L^2\) Nm²/C.


Net Flux through the cube:

The net flux is the sum of the fluxes through all six faces.
\(\Phi_{net} = \Phi_{left} + \Phi_{right} + 0 + 0 + 0 + 0\).
\(\Phi_{net} = (-5L^2) + (10L^3 + 5L^2)\).
\(\Phi_{net} = 10L^3\) Nm²/C.


Alternative Method (using differential form of Gauss's Law):
\(\Phi_{net} = \int_{V} (\nabla \cdot \vec{E}) \, dV\).

The divergence of the electric field is \(\nabla \cdot \vec{E} = \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} + \frac{\partial E_z}{\partial z}\).
\(\nabla \cdot \vec{E} = \frac{\partial}{\partial x}(10x + 5) = 10\).
\(\Phi_{net} = \int_{V} (10) \, dV = 10 \int_{V} dV = 10 \times (Volume of the cube)\).

The volume of the cube is \(L^3\).
\(\Phi_{net} = 10L^3\) Nm²/C.
Quick Tip: For non-uniform electric fields, you must calculate the flux through each face separately and then sum them up. If the field depends only on one coordinate (like x here), flux is non-zero only for the faces perpendicular to that axis. Using the divergence theorem (\(\Phi = \int (\nabla \cdot E) dV\)) can be a much faster method if you are comfortable with vector calculus.