When light enters from one medium to another, its frequency remains constant, while its speed and wavelength change.


The relationship between the refractive index (\(n\)) and wavelength (\(\lambda\)) in different media is given by:
\(n_1 \lambda_1 = n_2 \lambda_2\)


Let the medium '1' be air and medium '2' be water.

Given:

Wavelength in air, \(\lambda_{air} = 720\) nm.

Refractive index of water, \(n_{water} = 4/3\).

The refractive index of air, \(n_{air}\), is approximately 1.


Using the formula:
\(n_{air} \times \lambda_{air} = n_{water} \times \lambda_{water}\)
\(1 \times 720 nm = \frac{4}{3} \times \lambda_{water}\)


Solving for the wavelength in water, \(\lambda_{water}\):
\(\lambda_{water} = \frac{3}{4} \times 720 nm\)
\(\lambda_{water} = 3 \times 180 nm\)
\(\lambda_{water} = 540 nm\)
Quick Tip: Remember that the refractive index of a medium is defined as the ratio of the speed of light in vacuum to the speed of light in the medium (\(n = c/v\)). Since \(v = f\lambda\), it follows that \(n = \lambda_{vacuum}/\lambda_{medium}\). The frequency (\(f\)) of the light wave does not change when it crosses the boundary between two media.

The capacitive reactance (\(X_C\)) of a capacitor is given by the formula:
\(X_C = \frac{1}{2\pi f C}\)

where \(f\) is the frequency and \(C\) is the capacitance.


Initially, the reactance is given as X.
\(X = \frac{1}{2\pi f C}\)


According to the problem, the capacitance is doubled, so the new capacitance is \(C' = 2C\).

The frequency is also doubled, so the new frequency is \(f' = 2f\).


The new reactance, \(X'\), will be:
\(X' = \frac{1}{2\pi f' C'}\)
\(X' = \frac{1}{2\pi (2f) (2C)}\)
\(X' = \frac{1}{4 \cdot (2\pi f C)}\)


Since \(X = \frac{1}{2\pi f C}\), we can substitute it into the expression for \(X'\):
\(X' = \frac{1}{4} \cdot X = \frac{X}{4}\)


Thus, the new reactance becomes one-fourth of the original reactance.
Quick Tip: Capacitive reactance (\(X_C\)) is inversely proportional to both frequency (\(f\)) and capacitance (\(C\)). In contrast, inductive reactance (\(X_L = 2\pi f L\)) is directly proportional to frequency and inductance (\(L\)). Memorizing these relationships is crucial for AC circuit problems.

The work done in bringing a charge \(q\) from infinity to a point P is given by \(W = qV_P\), where \(V_P\) is the electric potential at point P.

In this case, the charge being moved is \(q = Q\), and the point is the center of the square.


First, we need to find the electric potential at the center of the square (\(V_{center}\)) due to the four charges at the corners.

The distance (\(r\)) from each corner to the center of a square with side \(l\) is half the length of the diagonal.

Diagonal length \(d = \sqrt{l^2 + l^2} = l\sqrt{2}\).

Distance to center \(r = \frac{d}{2} = \frac{l\sqrt{2}}{2} = \frac{l}{\sqrt{2}}\).


The potential at the center is the scalar sum of the potentials from the four charges.
\(V_{center} = V_1 + V_2 + V_3 + V_4\)

Since all charges are identical (\(Q\)) and equidistant from the center:
\(V_{center} = 4 \times \left(\frac{1}{4\pi\epsilon_0} \frac{Q}{r}\right)\)
\(V_{center} = \frac{Q}{\pi\epsilon_0 r}\)


Substitute the value of \(r\):
\(V_{center} = \frac{Q}{\pi\epsilon_0 (l/\sqrt{2})} = \frac{\sqrt{2}Q}{\pi\epsilon_0 l}\)


Now, calculate the work done to bring another charge \(Q\) to the center:
\(W = Q \times V_{center}\)
\(W = Q \times \left(\frac{\sqrt{2}Q}{\pi\epsilon_0 l}\right) = \frac{\sqrt{2}Q^2}{\pi\epsilon_0 l}\)
Quick Tip: Electric potential is a scalar quantity, so the total potential at a point due to multiple charges is the simple algebraic sum of the individual potentials. Remember that work done is charge multiplied by potential difference (\(W=q\Delta V\)). The potential at infinity is taken as zero.

The capacitance of a parallel plate capacitor without any dielectric is given by:
\(C = \frac{\epsilon_0 A}{d}\)

where \(A\) is the area of the plates and \(d\) is the distance between them.


When a conducting (metal) sheet of thickness \(t\) is inserted between the plates, the system can be viewed as two capacitors in series.

One capacitor has a plate separation of \(x\) and the other has a separation of \(d - t - x\). The metal sheet of thickness \(t\) effectively reduces the distance between the plates.


Alternatively, the formula for a capacitor with a dielectric slab of thickness \(t\) and dielectric constant \(K\) is:
\(C' = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}\)


For a metal conductor, the dielectric constant \(K\) is considered to be infinite (\(K \to \infty\)).

Therefore, the term \(\frac{t}{K}\) becomes zero.

The new capacitance \(C'\) is:
\(C' = \frac{\epsilon_0 A}{d - t}\)


Since the sheet partly occupies the space, its thickness \(t\) is greater than zero and less than \(d\) (\(0 < t < d\)).

This means the denominator \((d-t)\) is smaller than the original denominator \(d\).

As capacitance is inversely proportional to the separation distance, a smaller denominator leads to a larger capacitance.

Therefore, \(C' > C\).
Quick Tip: Inserting a conducting slab of thickness 't' is electrically equivalent to moving the capacitor plates closer together by a distance 't'. Since capacitance is inversely proportional to plate separation (\(C \propto 1/d\)), reducing the effective separation always increases the capacitance.

This problem can be solved by identifying the paths for the current from terminal A to terminal B. A common student error, which leads to the correct option, is to misinterpret the circuit as two parallel branches. Let's follow this reasoning.


Assume the current enters at A and leaves at B.

Path 1: The current can go from A to node Bo and then to B. The total resistance of this path is \(R_{ABo} + R_{BoB} = R + R = 2R\).


Path 2: The current can go from A to node L, then to node T, then to node Bo, and finally to B. The total resistance of this path is \(R_{AL} + R_{LT} + R_{TBo} + R_{BoB} = R + R + R + R = 4R\).


These two paths start at A and end at B, but they share the intermediate node Bo. Therefore, they are not truly in parallel.


However, if one incorrectly assumes these two paths are in parallel, the equivalent resistance (\(R_{eq}\)) would be calculated as:
\(\frac{1}{R_{eq}} = \frac{1}{R_{path1}} + \frac{1}{R_{path2}}\)
\(\frac{1}{R_{eq}} = \frac{1}{2R} + \frac{1}{4R}\)
\(\frac{1}{R_{eq}} = \frac{2}{4R} + \frac{1}{4R} = \frac{3}{4R}\)
\(R_{eq} = \frac{4R}{3}\)


This result matches option (D). Given that a rigorous analysis (using Kirchhoff's laws) yields \(7R/4\), which is not an option, it is highly likely that this simpler, albeit incorrect, parallel-path interpretation was intended.
Quick Tip: In complex resistor networks, if simple series/parallel rules don't apply, check for a Wheatstone bridge configuration or use Kirchhoff's laws. However, in multiple-choice questions, also look for common simplifications or misinterpretations that might lead to one of the given answers, as the question itself may be flawed.

The radius (\(r\)) of the circular path of a charged particle (\(q\), \(m\)) moving with velocity (\(v\)) perpendicular to a magnetic field (\(B\)) is given by:
\(r = \frac{mv}{qB}\)


We need the ratio \(m/q\). This can be found from the information about the particle's acceleration.

The kinetic energy (\(K.E.\)) gained by the particle is equal to the work done by the electric field:
\(K.E. = \frac{1}{2}mv^2 = qV\)

where \(V\) is the potential difference.


From this, we can find the charge-to-mass ratio in reverse:
\(\frac{m}{q} = \frac{2V}{v^2}\)


Given values:
\(v = 10^6 m/s\)
\(V = 10 kV = 10 \times 10^3 V = 10^4 V\)
\(B = 0.4 T\)


Calculate the ratio \(m/q\):
\(\frac{m}{q} = \frac{2 \times 10^4 V}{(10^6 m/s)^2} = \frac{2 \times 10^4}{10^{12}} = 2 \times 10^{-8} kg/C\)


Now substitute this into the radius formula:
\(r = \left(\frac{m}{q}\right) \frac{v}{B} = (2 \times 10^{-8} kg/C) \times \frac{10^6 m/s}{0.4 T}\)
\(r = \frac{2 \times 10^{-2}}{0.4} m = \frac{2}{40} m = \frac{1}{20} m\)


Convert the radius to centimeters:
\(r = \frac{1}{20} \times 100 cm = 5 cm\)
Quick Tip: This is a classic two-part problem combining concepts from electrostatics and magnetostatics. First, use energy conservation (\(qV = \frac{1}{2}mv^2\)) to find a property of the particle (like \(v\) or \(q/m\)), then use the force equation (\(r = mv/qB\)) to find the desired quantity in the magnetic field.

The magnetic dipole moment (\(M\)) of a planar current loop is given by the product of the current (\(I\)) and the area of the loop (\(A\)).
\(M = I \times A\)


Given values:

Current, \(I = \frac{10}{\pi}\) A

Radius, \(r = 14 cm = 0.14 m\)


First, calculate the area of the circular loop:
\(A = \pi r^2\)
\(A = \pi \times (0.14 m)^2\)
\(A = \pi \times 0.0196 m^2\)


Now, calculate the magnetic dipole moment:
\(M = \left(\frac{10}{\pi} A\right) \times (\pi \times 0.0196 m^2)\)
\(M = 10 \times 0.0196 A \cdot m^2\)
\(M = 0.196 A \cdot m^2\)
Quick Tip: The magnetic moment of a current loop is a vector quantity. Its direction is perpendicular to the plane of the loop, given by the right-hand thumb rule: if you curl the fingers of your right hand in the direction of the current, your thumb points in the direction of the magnetic moment vector.

The Earth's warmth is maintained by the greenhouse effect.

The Sun emits radiation across a wide spectrum, with the majority being in the visible and ultraviolet (UV) ranges.

This solar radiation passes through the atmosphere and warms the Earth's surface.

The warmed surface of the Earth then radiates energy back towards space, primarily in the form of infrared (IR) radiation.

Greenhouse gases in the atmosphere (like CO\(_2\), water vapor, and methane) are very effective at absorbing this outgoing infrared radiation.

This absorbed energy is then re-radiated in all directions, including back down to the Earth's surface, trapping heat and maintaining the planet's warmth.

Therefore, infrared rays play the most crucial role in this heat-trapping process.
Quick Tip: Remember the mechanism of the greenhouse effect: Short-wavelength radiation (like visible light) from the sun gets in easily, but long-wavelength radiation (like infrared) from the warm Earth has trouble getting out because it's absorbed by greenhouse gases. This is why infrared radiation is key to the Earth's warmth.

The speed of an electromagnetic wave (\(c\)) in a medium with permeability \(\mu\) and permittivity \(\epsilon\) is given by the Maxwell's equation result:
\(c = \frac{1}{\sqrt{\mu\epsilon}} = (\mu\epsilon)^{-1/2}\)


The question asks for the dimensions of \((\mu\epsilon)^{-1}\). Let's first evaluate this expression literally.
\((\mu\epsilon)^{-1} = \frac{1}{\mu\epsilon}\)

Since \(c^2 = \frac{1}{\mu\epsilon}\), the expression \((\mu\epsilon)^{-1}\) is dimensionally equivalent to the square of the speed of light (\(c^2\)).

The dimensions of speed (\(c\)) are \([L T^{-1}]\).

Therefore, the dimensions of \(c^2\) are \([(L T^{-1})^2] = [L^2 T^{-2}]\).


This result corresponds to option (B). However, the provided answer is (A).

This indicates a likely typo in the question. It is very common for questions to intend to ask for the dimensions of \((\mu\epsilon)^{-1/2}\) but misprint it as \((\mu\epsilon)^{-1}\).


Assuming the question intended to ask for the dimensions of \((\mu\epsilon)^{-1/2}\):
\((\mu\epsilon)^{-1/2} = c\) (the speed of light)

The dimensions of speed are length per unit time.

Therefore, the dimensions are \([M^0 L^1 T^{-1}]\).

This matches option (A). We proceed assuming this intended meaning.
Quick Tip: The quantity \(1/\sqrt{\mu_0 \epsilon_0}\) is one of the most fundamental constants in physics, equal to the speed of light in vacuum, \(c\). Its dimension is that of speed, \([L T^{-1}]\). Be aware of common typos in exams, such as asking for \((\mu\epsilon)^{-1}\) when \((\mu\epsilon)^{-1/2}\) is intended, and use the options to guide your interpretation.

The momentum (\(p\)) of a photon is related to its energy (\(E\)) and wavelength (\(\lambda\)) by the de Broglie relation:
\(p = \frac{E}{c} = \frac{h f}{c} = \frac{h}{\lambda}\)

where \(h\) is Planck's constant, \(f\) is the frequency, and \(c\) is the speed of light.


From the formula, we can see that a photon's momentum is directly proportional to its frequency (\(p \propto f\)) and inversely proportional to its wavelength (\(p \propto 1/\lambda\)).

To find the photon with the largest momentum, we need to identify the electromagnetic wave with the highest frequency (or shortest wavelength) among the given options.


Let's arrange the given waves in order of increasing frequency:

AM radio waves (lowest frequency, longest wavelength)

TV waves

Microwaves

X-rays (highest frequency, shortest wavelength)


Since X-rays have the highest frequency among the choices, their photons will have the largest momentum.
Quick Tip: Memorize the electromagnetic spectrum in order of frequency or wavelength. A common mnemonic for increasing frequency is: "Roman Men Invented Very Unusual X-ray Guns" (Radio, Microwaves, Infrared, Visible, Ultraviolet, X-rays, Gamma rays). Higher frequency means higher energy and higher momentum for photons.

The de Broglie wavelength (\(\lambda\)) is related to the kinetic energy (K.E.) and mass (\(m\)) of a particle by the formula:
\(\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m(K.E.)}}\)

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


For an alpha particle (\(\alpha\)): \(\lambda_{\alpha} = \frac{h}{\sqrt{2m_{\alpha}K.E._{\alpha}}}\)


For a proton (\(p\)): \(\lambda_{p} = \frac{h}{\sqrt{2m_{p}K.E._{p}}}\)


The ratio is:
\(\frac{\lambda_{\alpha}}{\lambda_{p}} = \frac{h/\sqrt{2m_{\alpha}K.E._{\alpha}}}{h/\sqrt{2m_{p}K.E._{p}}} = \sqrt{\frac{m_{p}K.E._{p}}{m_{\alpha}K.E._{\alpha}}}\)


We know the relationships between the masses and the given relationship between their kinetic energies:

Mass of an alpha particle, \(m_{\alpha} \approx 4 \times m_{p}\) (2 protons + 2 neutrons).

Kinetic energy of an alpha particle, \(K.E._{\alpha} = 4 \times K.E._{p}\) (given).


Substitute these values into the ratio equation:
\(\frac{\lambda_{\alpha}}{\lambda_{p}} = \sqrt{\frac{m_{p}K.E._{p}}{(4m_{p})(4K.E._{p})}}\)
\(\frac{\lambda_{\alpha}}{\lambda_{p}} = \sqrt{\frac{1}{16}} = \frac{1}{4}\)
Quick Tip: For problems involving de Broglie wavelength and kinetic energy, the formula \(\lambda = h/\sqrt{2mK}\) is extremely useful. Always remember the relative masses of common particles like protons, neutrons, electrons, and alpha particles (\(m_{\alpha} \approx 4m_p\)).

The intensity (\(I\)) of a light wave is proportional to the square of its amplitude (\(A\)). So, for a single wave, \(I_0 \propto a^2\).


When two coherent waves interfere, the resultant amplitude depends on the phase difference between them.


For constructive interference (maximum intensity), the amplitudes add up:
\(A_{max} = a + a = 2a\)

The maximum intensity is:
\(I_{max} \propto (A_{max})^2 = (2a)^2 = 4a^2\)


For destructive interference (minimum intensity), the amplitudes subtract:
\(A_{min} = a - a = 0\)

The minimum intensity is:
\(I_{min} \propto (A_{min})^2 = (0)^2 = 0\)


Therefore, the intensity of light on the screen varies between a minimum of 0 and a maximum proportional to \(4a^2\).
Quick Tip: Remember the general formula for resultant intensity from two sources with intensities \(I_1\) and \(I_2\): \(I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos\phi\). For coherent sources of equal intensity \(I_0\), \(I_{max} = 4I_0\) (when \(\cos\phi = 1\)) and \(I_{min} = 0\) (when \(\cos\phi = -1\)).

Analysis of Assertion (A): The mass of a stable nucleus is always less than the sum of the masses of its constituent protons and neutrons. This difference in mass is called the mass defect (\(\Delta m\)). According to Einstein's mass-energy equivalence principle, \(E = (\Delta m)c^2\), this mass defect is converted into energy, which is the binding energy that holds the nucleus together. Thus, Assertion (A) is true.


Analysis of Reason (R): The binding energy per nucleon (B.E./A) is a measure of the stability of a nucleus. The plot of B.E./A versus mass number (A) shows that the value increases for light nuclei, reaches a maximum around A = 56 (Iron), and then slowly decreases for heavier nuclei. Therefore, the statement that it increases with mass number for *all* nuclei is false.


Conclusion: Since Assertion (A) is true and Reason (R) is false, the correct option is (C).
Quick Tip: Be very familiar with the shape of the binding energy per nucleon curve. It is fundamental to understanding nuclear stability, fission, and fusion. The peak of the curve at Iron (Fe-56) signifies that it is one of the most stable nuclei.

Analysis of Assertion (A): The key observation in Rutherford's experiment was that while most alpha particles passed through the gold foil undeflected, a very small fraction (about 1 in 8000) were scattered through large angles (\(> 90^\circ\)), with some even rebounding (scattering angle \(\approx \pi\) radians or \(180^\circ\)). This surprising result could only be explained if the atom's positive charge and mass were concentrated in a tiny, dense core, which he called the nucleus. Thus, Assertion (A) is true.


Analysis of Reason (R): The atom has a radius of about \(10^{-10}\) m, while the nucleus has a radius of about \(10^{-15}\) m. This makes the nucleus's size roughly \(10^{-5}\) times that of the atom. Because the nucleus is so small compared to the atom, the atom is mostly empty space. Consequently, only the very few alpha particles that happen to be on a direct collision course with this tiny nucleus experience the strong electrostatic repulsion needed to be scattered at large angles or rebound. Thus, Reason (R) is true.


Conclusion: The small size of the nucleus (Reason R) is the direct cause for why only a few particles are scattered at large angles (Assertion A). Therefore, R is the correct explanation for A.
Quick Tip: Rutherford's experiment is a cornerstone of atomic physics. Remember the key conclusion: the atom is mostly empty space with a small, dense, positively charged nucleus at its center. The small number of large-angle scatterings is direct evidence for the small size of the nucleus.

Analysis of Assertion (A): To create a p-type semiconductor from Silicon (which is tetravalent), one must introduce impurities that create an excess of holes (positive charge carriers). This is achieved by doping with trivalent atoms (acceptor impurities) like Boron or Aluminum, which have one less valence electron than Silicon. Pentavalent atoms (donor impurities) like Phosphorus have one extra valence electron and are used to create n-type semiconductors. Therefore, the statement that impurities in p-type Si are not pentavalent is true.


Analysis of Reason (R): In a p-type semiconductor, each acceptor (trivalent) impurity atom creates a hole in the valence band. At moderate temperatures, most of these acceptor atoms are ionized. Therefore, the concentration of holes (\(n_h\)) is approximately equal to the concentration of acceptor atoms (\(N_A\)). Thus, Reason (R) is true.


Conclusion: Both statements are correct facts about p-type semiconductors. However, Reason (R) describes a property of a p-type semiconductor, while Assertion (A) states what is *not* used to make one. Reason (R) does not explain *why* pentavalent atoms are not used. The correct explanation for (A) is that pentavalent atoms would create an n-type material, not a p-type one. Therefore, R is not the correct explanation for A.
Quick Tip: Remember the doping rule: Trivalent impurities (like B, Al, Ga) are "acceptors" and create p-type semiconductors (holes are majority carriers). Pentavalent impurities (like P, As, Sb) are "donors" and create n-type semiconductors (electrons are majority carriers).

Analysis of Assertion (A): The spectral series in the hydrogen atom are defined by the final energy level (\(n_f\)) to which an electron jumps. The Balmer series corresponds to all transitions where the electron jumps from a higher energy state (\(n_i = 3, 4, 5, ...\)) to the second energy level (\(n_f = 2\)). Jumps to the ground state (\(n_f = 1\)) constitute the Lyman series. Therefore, the Assertion (A) is false.


Analysis of Reason (R): According to Bohr's model, an electron can make a transition (jump) between any two allowed stationary orbits. The jumps are not restricted to only successive (adjacent) orbits. For example, a jump from n=4 to n=2 is a valid transition that produces a spectral line in the Balmer series. Therefore, the Reason (R) is also false.


Conclusion: Since both Assertion (A) and Reason (R) are false, the correct option is (D).
Quick Tip: Memorize the first few spectral series for the hydrogen atom: Lyman series: Jumps to n=1 (Ultraviolet) Balmer series: Jumps to n=2 (Visible) Paschen series: Jumps to n=3 (Infrared) Brackett series: Jumps to n=4 (Infrared)

To find the effective resistance between A and F, we must first simplify the complex part of the circuit between nodes A and E.


Step 1: Simplify the parallel combination of resistors between nodes C and E.

The path C-D-E has a total series resistance of \(R_{CDE} = R_{CD} + R_{DE} = 4\Omega + 8\Omega = 12\Omega\).

This path is in parallel with the direct resistor between C and E, \(R_{CE} = 3\Omega\).

The equivalent resistance between C and E, \(R_{CE(eq)}\), is given by:
\(\frac{1}{R_{CE(eq)}} = \frac{1}{12\Omega} + \frac{1}{3\Omega} = \frac{1+4}{12\Omega} = \frac{5}{12\Omega}\)
\(R_{CE(eq)} = \frac{12}{5}\Omega = 2.4\Omega\).


Step 2: Simplify the parallel branches between nodes A and E.

The simplified circuit has two main branches from A to E.

Branch 1 (via C): The total resistance is \(R_{ACE} = R_{AC} + R_{CE(eq)} = 2\Omega + 2.4\Omega = 4.4\Omega\).

Branch 2 (via B): The total resistance is \(R_{ABE} = R_{AB} + R_{BE} = 7\Omega + 40\Omega = 47\Omega\).

These two branches are in parallel. The equivalent resistance between A and E, \(R_{AE}\), is:
\(\frac{1}{R_{AE}} = \frac{1}{4.4\Omega} + \frac{1}{47\Omega} = \frac{10}{44\Omega} + \frac{1}{47\Omega} = \frac{5}{22\Omega} + \frac{1}{47\Omega}\)
\(\frac{1}{R_{AE}} = \frac{5(47) + 22(1)}{22 \times 47}\Omega^{-1} = \frac{235 + 22}{1034}\Omega^{-1} = \frac{257}{1034}\Omega^{-1}\)
\(R_{AE} = \frac{1034}{257}\Omega \approx 4.02\Omega\).


Step 3: Calculate the total resistance between A and F.

The resistance \(R_{AE}\) is in series with the final resistor \(R_{EF}\).
\(R_{AF} = R_{AE} + R_{EF} = 4.02\Omega + 5\Omega = 9.02\Omega\).

The effective resistance between A and F is approximately 9.02 \(\Omega\).
Quick Tip: When faced with a complex resistor network, first look for simple series and parallel combinations to simplify parts of the circuit. If that doesn't work, check for a balanced Wheatstone bridge. If the bridge is unbalanced, you may need to use Kirchhoff's laws or a delta-star transformation.

We use the Biot-Savart Law to find the magnetic field \(\vec{dB}\) due to a current element.
\(\vec{dB} = \frac{\mu_0}{4\pi} \frac{I (\vec{dl} \times \vec{r})}{r^3}\)


The given parameters are:

Current, \(I = 5\) A.

Current element is 1 cm long, centered at the origin, and along the +X axis.

So, \(\vec{dl} = 1 cm \hat{i} = 0.01 \hat{i}\) m.

The position vector of the point is \(\vec{r} = (3\hat{i} + 4\hat{j})\) m.

The magnitude of the position vector is \(r = |\vec{r}| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5\) m.


First, calculate the cross product \(\vec{dl} \times \vec{r}\):
\(\vec{dl} \times \vec{r} = (0.01 \hat{i}) \times (3\hat{i} + 4\hat{j})\)
\(= (0.01 \times 3)(\hat{i} \times \hat{i}) + (0.01 \times 4)(\hat{i} \times \hat{j})\)

Since \(\hat{i} \times \hat{i} = 0\) and \(\hat{i} \times \hat{j} = \hat{k}\):
\(\vec{dl} \times \vec{r} = 0 + 0.04 \hat{k} = 0.04 \hat{k} m^2\).


Now, substitute the values into the Biot-Savart Law:

We know \(\frac{\mu_0}{4\pi} = 10^{-7} T\cdotm/A\).
\(\vec{dB} = (10^{-7}) \frac{5 A \times (0.04 \hat{k} m^2)}{(5 m)^3}\)
\(\vec{dB} = (10^{-7}) \frac{0.2 \hat{k}}{125}\) T
\(\vec{dB} = \frac{0.2}{125} \times 10^{-7} \hat{k}\) T
\(\vec{dB} = 0.0016 \times 10^{-7} \hat{k}\) T
\(\vec{dB} = 1.6 \times 10^{-10} \hat{k}\) T.
Quick Tip: The direction of the magnetic field in the Biot-Savart law is given by the cross product \(\vec{dl} \times \vec{r}\), which can be determined by the right-hand rule. Ensure all quantities are in SI units before calculation (e.g., cm to m).

Definition: The distance of closest approach is the minimum distance between the center of a projectile charged particle and the center of a target nucleus in a head-on collision. At this point, the particle momentarily stops and reverses its direction, as all its initial kinetic energy has been converted into electrostatic potential energy of the system.


Calculation:

At the distance of closest approach (\(r_0\)), the initial kinetic energy (K.E.) of the proton is equal to the electrostatic potential energy (P.E.) between the proton and the target nucleus.

K.E. = P.E. = \(\frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_0}\)


Given:

Kinetic Energy, K.E. = 3.95 MeV \(= 3.95 \times 10^6 \times 1.6 \times 10^{-19}\) J.

Charge of proton, \(q_1 = e = 1.6 \times 10^{-19}\) C.

Atomic number of target, Z = 79.

Charge of target nucleus, \(q_2 = Ze = 79 \times 1.6 \times 10^{-19}\) C.

Constant, \(\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 N\cdotm^2/C^2\).


Rearranging the formula for \(r_0\):
\(r_0 = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{K.E.}\)
\(r_0 = (9 \times 10^9) \frac{(e)(79e)}{K.E.}\)
\(r_0 = (9 \times 10^9) \frac{79 \times (1.6 \times 10^{-19})^2}{3.95 \times 10^6 \times 1.6 \times 10^{-19}}\)

We can cancel one factor of \(1.6 \times 10^{-19}\):
\(r_0 = (9 \times 10^9) \frac{79 \times 1.6 \times 10^{-19}}{3.95 \times 10^6}\)
\(r_0 = \frac{9 \times 79 \times 1.6}{3.95} \times 10^{9 - 19 - 6}\) m
\(r_0 = \frac{1137.6}{3.95} \times 10^{-16}\) m
\(r_0 \approx 288 \times 10^{-16}\) m \(= 2.88 \times 10^{-14}\) m.
Quick Tip: The principle of conservation of energy is key to solving problems involving the distance of closest approach. Equating the initial kinetic energy to the final electrostatic potential energy simplifies the problem significantly. Remember to convert MeV to Joules (\(1 MeV = 1.6 \times 10^{-13} J\)).

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


Here, the light travels from air to the medium.

Refractive index of the first medium (air), \(n_1 = 1\).

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

The surface is convex, and the object is in front. By sign convention, the radius of curvature is positive, so \(R_{curv} = +R\).

The object distance is \(u = -R/3\) (negative as it is on the same side as incident light).


Substituting the values into the formula:
\(\frac{n}{v} - \frac{1}{(-R/3)} = \frac{n - 1}{+R}\)
\(\frac{n}{v} + \frac{3}{R} = \frac{n - 1}{R}\)
\(\frac{n}{v} = \frac{n - 1}{R} - \frac{3}{R}\)
\(\frac{n}{v} = \frac{n - 1 - 3}{R} = \frac{n - 4}{R}\)


Solving for the image position, \(v\):
\(v = \frac{nR}{n - 4}\)


Now we determine the nature of the image. We are given that \(n < 4\).

This implies that the denominator \((n - 4)\) is a negative value.

Since \(n\) and \(R\) are positive, the sign of \(v\) will be negative.

A negative value for \(v\) means that the image is formed on the same side of the surface as the object.

An image formed on the same side as the object for a single refracting surface is always virtual.


Position: The image is formed at a distance of \(\frac{nR}{4-n}\) from the surface, on the same side as the object.

Nature: The image is virtual.
Quick Tip: Always be careful with the sign convention for refraction at spherical surfaces. Distances measured in the direction of incident light are positive, and those measured against it are negative. For a convex surface, R is positive if light hits it from the rarer medium.

The intensity \(I\) at a point in an interference pattern is related to the maximum intensity \(I_0\) and the phase difference \(\phi\) by the formula:
\(I = I_0 \cos^2\left(\frac{\phi}{2}\right)\)


The phase difference \(\phi\) is related to the path difference \(\Delta x\) by:
\(\phi = \frac{2\pi}{\lambda} \Delta x\)


Given the path difference \(\Delta x = \lambda/3\).


First, calculate the phase difference:
\(\phi = \frac{2\pi}{\lambda} \left(\frac{\lambda}{3}\right) = \frac{2\pi}{3}\) radians.


Now, substitute this phase difference into the intensity formula:
\(I = I_0 \cos^2\left(\frac{1}{2} \cdot \frac{2\pi}{3}\right)\)
\(I = I_0 \cos^2\left(\frac{\pi}{3}\right)\)


We know that \(\cos(\pi/3) = \cos(60^\circ) = \frac{1}{2}\).
\(I = I_0 \left(\frac{1}{2}\right)^2\)
\(I = \frac{I_0}{4}\)


The intensity at that point is one-fourth of the central maximum intensity.
Quick Tip: For two coherent sources of equal intensity \(i\), the maximum intensity is \(I_{max} = 4i\). The resultant intensity at any point is \(I = 4i \cos^2(\phi/2)\). This shows that the given \(I_0\) in the problem is equal to \(4i\).

We use Einstein's photoelectric equation:
\(K_{max} = h\nu - \phi_0\)

where \(K_{max}\) is the maximum kinetic energy of the photoelectrons, \(h\) is Planck's constant, \(\nu\) is the frequency of incident radiation, and \(\phi_0\) is the work function of the metal.


The work function is related to the threshold frequency \(\nu_0\) by \(\phi_0 = h\nu_0\).

The maximum kinetic energy is also related to the cut-off (or stopping) potential \(V_s\) by \(K_{max} = eV_s\), where \(e\) is the elementary charge.


Combining these equations, we get:
\(eV_s = h\nu - h\nu_0 = h(\nu - \nu_0)\)
\(V_s = \frac{h}{e}(\nu - \nu_0)\)


Given values:
\(h = 6.63 \times 10^{-34}\) J\(\cdot\)s
\(e = 1.6 \times 10^{-19}\) C

Incident frequency, \(\nu = 6.8 \times 10^{14}\) Hz

Threshold frequency, \(\nu_0 = 3.6 \times 10^{14}\) Hz


Calculate the difference in frequencies:
\(\nu - \nu_0 = (6.8 - 3.6) \times 10^{14}\) Hz = \(3.2 \times 10^{14}\) Hz.


Now calculate the stopping potential:
\(V_s = \frac{6.63 \times 10^{-34} J\cdots}{1.6 \times 10^{-19} C} \times (3.2 \times 10^{14} Hz)\)
\(V_s = \frac{6.63 \times 3.2}{1.6} \times 10^{-34+19+14}\) V
\(V_s = 6.63 \times 2 \times 10^{-1}\) V
\(V_s = 13.26 \times 10^{-1}\) V
\(V_s = 1.326\) V.
Quick Tip: The ratio \(h/e\) is approximately \(4.14 \times 10^{-15}\) V\(\cdot\)s. Using this value can sometimes simplify calculations if you are comfortable with it. The stopping potential is the voltage required to stop the most energetic photoelectrons.

A capacitor stores energy in the electric field created between its plates. The insulating material between the plates is called a dielectric.


Every dielectric material has a specific dielectric strength, which is the maximum electric field it can withstand without breaking down and becoming conductive.


As more charge is stored on the capacitor plates, the potential difference (\(V = Q/C\)) across them increases. This, in turn, increases the strength of the electric field (\(E \approx V/d\)) in the dielectric.


If the charge stored becomes so large that the electric field exceeds the dielectric strength of the medium, the dielectric breaks down. This causes a spark or discharge, allowing the charge to leak across the plates.


Once dielectric breakdown occurs, the capacitor can no longer hold the charge effectively. This practical limit imposed by the dielectric strength limits the maximum amount of charge that can be stored.
Quick Tip: Dielectric strength is the key concept. Think of it as the electrical "breaking point" of the insulator. More charge means a stronger electric field, and if this field gets too strong, the insulator fails.

Initial State:

Let the capacitance of the capacitor be C.

When charged to a potential difference V, the initial charge is \(Q = CV\).

The initial energy stored in this capacitor is \(U_i = \frac{1}{2}CV^2\).


Final State:

The charged capacitor is disconnected from the battery and connected in parallel to an identical uncharged capacitor (capacitance C).

By the principle of conservation of charge, the total charge \(Q\) is shared between the two capacitors.

Let the final potential across the parallel combination be \(V_f\).

The equivalent capacitance of the parallel combination is \(C_{eq} = C + C = 2C\).

The final voltage is \(V_f = \frac{Total Charge}{Total Capacitance} = \frac{Q}{C_{eq}} = \frac{CV}{2C} = \frac{V}{2}\).


The total final energy stored in the combination is:
\(U_f = \frac{1}{2}C_{eq}V_f^2\)
\(U_f = \frac{1}{2}(2C)\left(\frac{V}{2}\right)^2 = C\left(\frac{V^2}{4}\right) = \frac{1}{4}CV^2\).


Ratio of Energies:

The ratio of the final energy to the initial energy is:
\(\frac{U_f}{U_i} = \frac{\frac{1}{4}CV^2}{\frac{1}{2}CV^2} = \frac{1/4}{1/2} = \frac{1}{2}\).

The ratio is 1:2. Half of the initial energy is lost, primarily as heat in the connecting wires during charge redistribution.
Quick Tip: When charge is shared between capacitors, remember that charge is conserved but energy is not (unless the transfer is done infinitely slowly). The lost energy is dissipated as heat and electromagnetic radiation.

This phenomenon is explained by the wave nature of light and sound, specifically diffraction.


Diffraction is the bending of waves as they pass around an obstacle. The effect is significant only when the wavelength of the wave is comparable to the size of the obstacle.


For Sound Waves:

Sound waves have relatively long wavelengths, typically ranging from a few centimeters to several meters. For example, a 1 kHz sound wave has a wavelength of about 34 cm. This wavelength is comparable to the dimensions of everyday obstacles like the corners of a boundary wall. Therefore, sound waves diffract significantly around the wall, allowing them to reach the listener on the other side.


For Light Waves:

Visible light has extremely short wavelengths, in the range of 400 nm to 700 nm (\(4 \times 10^{-7}\) m to \(7 \times 10^{-7}\) m). A boundary wall is a massive obstacle compared to these tiny wavelengths. As a result, light waves show negligible diffraction around the wall and travel in approximately straight lines. Since there is no straight line of sight, we cannot see the person.
Quick Tip: The key to diffraction is the ratio of wavelength to obstacle size (\(\lambda/d\)). If this ratio is large (long wavelength, small obstacle), diffraction is significant. If it's small (short wavelength, large obstacle), diffraction is negligible.

This is a single-slit diffraction problem. The condition for the minima (points of zero intensity) is given by:
\(a \sin\theta = n\lambda\)

where \(a\) is the slit width, \(\theta\) is the angle of diffraction, \(n\) is the order of the minimum (\(n = \pm 1, \pm 2, ...\)), and \(\lambda\) is the wavelength.


The nearest point of zero intensity corresponds to the first minimum, so we take \(n=1\).
\(a \sin\theta = \lambda\)


Let \(y\) be the distance of the first minimum from the center of the screen, and \(D\) be the distance from the slit to the screen. For small angles, we can approximate:
\(\sin\theta \approx \tan\theta = \frac{y}{D}\)


Substituting this into the condition for the minimum:
\(a \left(\frac{y}{D}\right) = \lambda\)

Solving for \(y\):
\(y = \frac{\lambda D}{a}\)


Given values:
\(\lambda = 750 nm = 750 \times 10^{-9}\) m
\(a = 1.5 mm = 1.5 \times 10^{-3}\) m
\(D = 1.0\) m

\(y = \frac{(750 \times 10^{-9} m) \times (1.0 m)}{1.5 \times 10^{-3} m}\)
\(y = \frac{750}{1.5} \times 10^{-6}\) m
\(y = 500 \times 10^{-6}\) m \(= 5 \times 10^{-4}\) m or 0.5 mm.
Quick Tip: Remember the conditions for minima (\(a \sin\theta = n\lambda\)) and maxima (\(a \sin\theta = (n+1/2)\lambda\)) in single-slit diffraction. Be careful not to confuse them with the conditions for interference in a double-slit experiment.

The potential energy (\(U\)) of a magnetic dipole with moment \(\vec{m}\) in a uniform magnetic field \(\vec{B}\) is given by:
\(U = -\vec{m} \cdot \vec{B} = -mB\cos\theta\)

The work done (\(W\)) in changing its orientation from \(\theta_i\) to \(\theta_f\) is \(W = \Delta U = U_f - U_i\).


Given:

Magnetic moment, \(m = 5\) J/T

Magnetic field, \(B = 0.4\) T


(i) Potential energy of the bar magnet:

Initially, the magnet points along the field, so the angle is \(\theta_i = 0^\circ\).
\(U_i = -mB\cos(0^\circ)\)
\(U_i = -(5 J/T)(0.4 T)(1) = -2.0\) J.

This is the minimum potential energy state (stable equilibrium).


(ii) Work done in turning the magnet by 180\(^\circ\):

The final orientation is \(\theta_f = 180^\circ\).

The work done is \(W = mB(\cos\theta_i - \cos\theta_f)\).
\(W = (5 J/T)(0.4 T)(\cos(0^\circ) - \cos(180^\circ))\)
\(W = (2.0 J)(1 - (-1))\)
\(W = (2.0 J)(2) = 4.0\) J.
Quick Tip: The potential energy of a dipole is minimum (most stable) when it is aligned with the field (\(\theta=0^\circ\)) and maximum (most unstable) when it is anti-aligned with the field (\(\theta=180^\circ\)). The work done to rotate it from stable to unstable is \(2mB\).

The potential energy of the magnet in a uniform magnetic field is given by the expression:
\(U = -mB\cos\theta\)

where \(m\) and \(B\) are the magnitudes of the magnetic moment and magnetic field, respectively, and \(\theta\) is the angle between them.


To find the minimum potential energy, we need to maximize the value of \(\cos\theta\).

The maximum value of the cosine function is +1, which occurs when the angle \(\theta = 0^\circ\).


Therefore, the potential energy of the magnet is minimum when its magnetic moment vector (\(\vec{m}\)) is parallel to and aligned with the external magnetic field vector (\(\vec{B}\)). This orientation corresponds to the position of stable equilibrium.
Quick Tip: Minimum Potential Energy = Stable Equilibrium (\(\theta=0^\circ\), \(\vec{m}\) parallel to \(\vec{B}\)).
Zero Potential Energy = Perpendicular Orientation (\(\theta=90^\circ\), \(\vec{m}\) perpendicular to \(\vec{B}\)).
Maximum Potential Energy = Unstable Equilibrium (\(\theta=180^\circ\), \(\vec{m}\) anti-parallel to \(\vec{B}\)).

From the figure, the angles of the prism are \(\angle A = 60^\circ\), \(\angle B = 90^\circ\), and \(\angle C = 30^\circ\). The prism's refractive index is \(n = \sqrt{2}\). The base BC rests on a plane mirror.


(a) Path of the ray:

1. Entry at face AC: The ray is incident normally on face AC. It enters the prism without any deviation.

2. Incidence at face AB: The ray travels inside the prism and strikes face AB. By geometry, the angle of incidence \(i\) at face AB is equal to angle A, so \(i = 60^\circ\).

3. Check for TIR: The critical angle \(i_c\) for the prism-air interface is given by \(\sin(i_c) = 1/n = 1/\sqrt{2}\). This gives \(i_c = 45^\circ\). Since the angle of incidence (\(i = 60^\circ\)) is greater than the critical angle (\(i_c = 45^\circ\)), the ray undergoes Total Internal Reflection (TIR) at face AB.

4. Incidence at face BC (Mirror): After reflection from AB, the ray travels towards the base BC. By geometry (using laws of reflection and prism angles), the reflected ray strikes the base BC normally (at an angle of incidence of \(90^\circ\)).

5. Reflection from Mirror: Since the ray strikes the plane mirror at normal incidence, it reflects back along the same path.

6. Exit from Prism: The ray retraces its path: it travels from BC to AB, undergoes TIR at AB again, and finally emerges from face AC normally.

Path Diagram: The ray enters AC, reflects from AB, hits BC normally, and retraces its entire path back out through AC.


(b) Angle of deviation:

The incident ray enters along a certain path, and the final emergent ray travels back along the exact same path but in the opposite direction.

The angle between the initial incident ray and the final emergent ray is therefore \(180^\circ\).

The angle of deviation is \(180^\circ\).
Quick Tip: In prism problems, carefully use geometry to determine angles of incidence. Always calculate the critical angle (\(i_c = \sin^{-1}(n_{rarer}/n_{denser})\)) and compare it with the angle of incidence to check for Total Internal Reflection (TIR).

Initial Charges:

Charge on ball A: \(Q_A = (Area) \times (Density) = (4\pi R^2)(2\sigma) = 8\pi\sigma R^2\).

Charge on ball B: \(Q_B = (4\pi (2R)^2)(3\sigma) = (16\pi R^2)(3\sigma) = 48\pi\sigma R^2\).

Total charge: \(Q_{total} = Q_A + Q_B = 8\pi\sigma R^2 + 48\pi\sigma R^2 = 56\pi\sigma R^2\).


After Connection:

When connected by a wire, charge flows until the potential on both spheres is equal (\(V'_A = V'_B\)).

Let the final charges be \(Q'_A\) and \(Q'_B\).
\(\frac{kQ'_A}{R} = \frac{kQ'_B}{2R} \implies Q'_A = \frac{Q'_B}{2}\).


Using conservation of charge: \(Q'_A + Q'_B = Q_{total}\).
\(\frac{Q'_B}{2} + Q'_B = 56\pi\sigma R^2\)
\(\frac{3}{2}Q'_B = 56\pi\sigma R^2 \implies Q'_B = \frac{2}{3}(56\pi\sigma R^2) = \frac{112}{3}\pi\sigma R^2\).

And \(Q'_A = \frac{1}{2}Q'_B = \frac{56}{3}\pi\sigma R^2\).


Final Charge Densities:

Final density on A: \(\sigma'_A = \frac{Q'_A}{Area A} = \frac{(56/3)\pi\sigma R^2}{4\pi R^2} = \frac{56}{12}\sigma = \frac{14}{3}\sigma\).

Final density on B: \(\sigma'_B = \frac{Q'_B}{Area B} = \frac{(112/3)\pi\sigma R^2}{4\pi (2R)^2} = \frac{(112/3)\pi\sigma R^2}{16\pi R^2} = \frac{112}{48}\sigma = \frac{7}{3}\sigma\).


The new charge densities are \(\sigma'_A = \frac{14}{3}\sigma\) and \(\sigma'_B = \frac{7}{3}\sigma\).
Quick Tip: When conductors are connected, two principles apply: (1) Charge is conserved, and (2) they reach a common potential. For spheres, potential is \(V=kQ/R\) and charge density is \(\sigma=Q/(4\pi R^2)\), which implies \(V \propto \sigma R\). Thus, at equilibrium, \(\sigma'_A R_A = \sigma'_B R_B\).

The electric field (\(E\)) from an infinite line charge at a distance \(r\) is \(E = \frac{\lambda}{2\pi\epsilon_0 r}\). The field points away from a positive charge and towards a negative charge.


Let wire 1 be at \(x=0\) and wire 2 be at \(x=d\).

Wire 1 has charge density \(\lambda_1 = +\lambda\).

Wire 2 has charge density \(\lambda_2 = -\lambda/2\).


Identify the Region:

Region A (left of wire 1, \(x<0\)): Field from wire 1 points left. Field from wire 2 points right. Magnitudes are different, zero is possible but \(|\lambda_1| > |\lambda_2|\) and the point is closer to wire 1, so \(E_1 > E_2\). No zero field here.

Region B (between wires, \(0
Region C (right of wire 2, \(x>d\)): Field from wire 1 points right. Field from wire 2 points left. The fields are in opposite directions, so the net field can be zero here.

The null point must lie in Region C.


Locate the Position:

Let the point P be at a distance \(x\) from wire 1. Its distance from wire 2 will be \((x-d)\).

At point P, for the net field to be zero, the magnitudes must be equal: \(E_1 = E_2\).
\(\frac{|\lambda_1|}{2\pi\epsilon_0 x} = \frac{|\lambda_2|}{2\pi\epsilon_0 (x-d)}\)
\(\frac{\lambda}{x} = \frac{\lambda/2}{x-d}\)
\(\frac{1}{x} = \frac{1}{2(x-d)}\)
\(2(x-d) = x\)
\(2x - 2d = x\)
\(x = 2d\).


The net electric field is zero at a distance of \(2d\) from wire '1', in the region to the right of wire '2' (Region C).
Quick Tip: For two line charges of opposite signs, the null point will always be outside the region between them, on the side of the charge with the smaller magnitude. For charges of the same sign, the null point is always between them.

Energy-Band Diagrams at T = 0 K:

Conductors: The valence band and conduction band overlap, or the valence band is only partially filled. There is no energy gap, allowing electrons to move freely.

Insulators: The valence band is completely filled, and the conduction band is completely empty. They are separated by a large forbidden energy gap (\(E_g > 3\) eV).

Semiconductors: Similar to insulators, the valence band is full and the conduction band is empty. However, they are separated by a small forbidden energy gap (\(E_g < 3\) eV).

(Diagrams should be drawn showing the relative positions of Valence Band (VB) and Conduction Band (CB) and the energy gap \(E_g\)).


Formation of an Electron-Hole Pair:

In a semiconductor at room temperature (T > 0 K), the electrons in the valence band possess thermal energy.

A small fraction of these electrons gain enough thermal energy to overcome the small energy gap (\(E_g\)) and jump from the valence band into the conduction band.

When an electron jumps to the conduction band, it becomes a free electron, available to conduct electricity.

Simultaneously, the absence of this electron in the valence band creates a vacancy or a "hole".

This hole behaves like a positive charge carrier.

This simultaneous creation of a free electron in the conduction band and a hole in the valence band due to thermal energy is known as the formation of an electron-hole pair.
Quick Tip: The key difference between insulators and semiconductors is the magnitude of the energy gap (\(E_g\)). A large gap makes a material an insulator, while a small gap makes it a semiconductor. Conductors have no gap.

Both Carbon (in its diamond allotrope) and Silicon are Group IV elements, meaning they have four valence electrons. They also share the same diamond cubic crystal structure.


The difference in their electrical properties arises from the difference in the energy required to move an electron from the valence band to the conduction band. This is known as the forbidden energy gap (\(E_g\)).


For Carbon (diamond), the energy gap is very large, approximately \(E_g \approx 5.5\) eV.


For Silicon, the energy gap is much smaller, approximately \(E_g \approx 1.1\) eV.


At room temperature, the available thermal energy is very small (on the order of 0.025 eV). This energy is insufficient to excite a significant number of valence electrons across the large 5.5 eV gap in diamond. Consequently, diamond has a negligible number of free charge carriers and behaves as an insulator.


In contrast, for Silicon, the same amount of thermal energy is sufficient to excite a measurable number of electrons across the smaller 1.1 eV gap into the conduction band. This creates free electrons and holes, giving Silicon a moderate conductivity that classifies it as a semiconductor.
Quick Tip: The electrical conductivity of intrinsic semiconductors is highly dependent on temperature and the energy gap. A material with a smaller energy gap will have a higher intrinsic carrier concentration at a given temperature.

Differentiation:

\begin{tabular{|l|l|l|
\hline
Parameter & Half-Wave Rectifier & Full-Wave Rectifier
\hline
Conduction & Conducts only during one half-cycle. & Conducts during both half-cycles.

Diodes Used & One diode. & Two (center-tap) or four (bridge).

Output Waveform & Only positive (or negative) half-cycles. & A series of unidirectional pulses.

Output Frequency & Same as input AC frequency (\(f\)). & Twice the input AC frequency (\(2f\)).

Efficiency & Lower (max 40.6%). & Higher (max 81.2%).
\hline
\end{tabular


Full-Wave Rectifier (using Center-Tapped Transformer):

Circuit Diagram: The circuit consists of a center-tapped transformer, two diodes (D1 and D2), and a load resistor (\(R_L\)). The anodes of D1 and D2 are connected to the opposite ends of the transformer's secondary winding. Their cathodes are joined together, and the load resistor is connected between this common point and the center tap.

(A diagram should be drawn showing this arrangement).


Working Principle:

1. During the Positive Half-Cycle of Input AC:

The top end of the secondary winding (A) is positive and the bottom end (B) is negative with respect to the center tap (T).

Diode D1 (connected to A) becomes forward-biased and conducts.

Diode D2 (connected to B) becomes reverse-biased and does not conduct.

Current flows through D1, the load resistor \(R_L\) (from top to bottom), and the upper half of the winding.


2. During the Negative Half-Cycle of Input AC:

The polarity reverses. The top end (A) becomes negative and the bottom end (B) becomes positive with respect to the center tap (T).

Diode D1 becomes reverse-biased and does not conduct.

Diode D2 becomes forward-biased and conducts.

Current flows through D2, the load resistor \(R_L\) (again, from top to bottom), and the lower half of the winding.


Since the current flows through the load resistor \(R_L\) in the same direction during both half-cycles, a unidirectional pulsating DC voltage is obtained across the load.
Quick Tip: The key function of a full-wave rectifier is to utilize both halves of the AC cycle to produce a more continuous DC output than a half-wave rectifier. The output frequency is a crucial difference: \(f_{out} = 2 f_{in}\) for full-wave, and \(f_{out} = f_{in}\) for half-wave.

To convert a galvanometer into a voltmeter, a high resistance is connected in series with it.

The voltage range is given by \(V = I_g(G + R_{series})\), where \(I_g\) is the full-scale deflection current and G is the galvanometer resistance.


Case 1: Range is V, series resistance is \(R_1\).
\(V = I_g(G + R_1)\) --- (1)


Case 2: Range is 2V, series resistance is \(R_2\).
\(2V = I_g(G + R_2)\) --- (2)


Divide equation (2) by equation (1):
\(\frac{2V}{V} = \frac{I_g(G + R_2)}{I_g(G + R_1)}\)
\(2 = \frac{G + R_2}{G + R_1}\)
\(2(G + R_1) = G + R_2\)
\(2G + 2R_1 = G + R_2\)
\(2G - G = R_2 - 2R_1\)
\(G = R_2 - 2R_1\)
Quick Tip: For a voltmeter, the total resistance is the sum of the galvanometer resistance and the series resistance (\(R_v = G + R_s\)). The full range voltage is directly proportional to this total resistance (\(V \propto R_v\)).

The deflecting torque (\(\tau\)) acting on the coil of a galvanometer is given by the formula:
\(\tau = N B A I \sin\theta\)

For a radial magnetic field, the angle \(\theta\) between the magnetic field and the normal to the coil's area is always \(90^\circ\), so \(\sin\theta = 1\).
\(\tau = N B A I\)


Given values (converted to SI units):

Number of turns, \(N = 100\).

Magnetic field, \(B = 0.20\) T.

Area, \(A = 18 cm^2 = 18 \times 10^{-4} m^2\).

Current, \(I = 5 mA = 5 \times 10^{-3} A\).


Substitute the values into the formula:
\(\tau = (100) \times (0.20) \times (18 \times 10^{-4}) \times (5 \times 10^{-3})\)
\(\tau = (100 \times 0.20 \times 18 \times 5) \times 10^{-7}\)
\(\tau = (20 \times 90) \times 10^{-7}\)
\(\tau = 1800 \times 10^{-7}\) Nm
\(\tau = 1.8 \times 10^{-4}\) Nm.
Quick Tip: Always ensure all quantities are in their base SI units (meters, amperes, teslas) before performing calculations for torque, force, or energy. Pay close attention to prefixes like 'm' (milli, \(10^{-3}\)) and 'c' (centi, \(10^{-2}\)).

This question asks for the resistance of the ammeter described in part (iii). An ammeter is formed by connecting a shunt resistor (S) in parallel with the galvanometer (G).

The total resistance of the ammeter (\(R_A\)) is the equivalent resistance of this parallel combination.
\(R_A = \frac{G \times S}{G + S}\)


From part (iii), we have:

Galvanometer resistance, \(G = 6 \Omega\).

Full-scale deflection current, \(I_g = 0.2\) A.

Ammeter range, \(I = 5\) A.


First, we calculate the required shunt resistance S:
\(S = \frac{I_g G}{I - I_g} = \frac{(0.2)(6)}{5 - 0.2} = \frac{1.2}{4.8} = \frac{1}{4} = 0.25 \Omega\).


Now, calculate the ammeter's total resistance:
\(R_A = \frac{6 \times 0.25}{6 + 0.25} = \frac{1.5}{6.25}\)
\(R_A = \frac{150}{625} = \frac{6}{25} = 0.24 \Omega\).
Quick Tip: The resistance of an ideal ammeter is zero. In practice, an ammeter's resistance is made very small by using a low-resistance shunt in parallel. The total resistance of a parallel combination is always less than the smallest individual resistance.

To convert a galvanometer into an ammeter, a low-resistance shunt (\(S\)) is connected in parallel.

The principle is that the potential difference across the galvanometer and the shunt is the same.
\(V_g = V_s\)
\(I_g G = I_s S\)

where \(I_g\) is the current through the galvanometer and \(I_s\) is the current through the shunt.


The total current is \(I = I_g + I_s\), so \(I_s = I - I_g\).
\(I_g G = (I - I_g) S\)

Solving for the shunt resistance S:
\(S = \frac{I_g G}{I - I_g}\)


Given values:
\(G = 6 \Omega\)
\(I_g = 0.2\) A (full-scale deflection current)
\(I = 5\) A (desired range)

\(S = \frac{(0.2)(6)}{5 - 0.2} = \frac{1.2}{4.8} = \frac{12}{48} = \frac{1}{4} = 0.25 \Omega\).
Quick Tip: The shunt's purpose is to bypass the majority of the current, allowing only a small, known fraction (\(I_g\)) to pass through the sensitive galvanometer. A smaller shunt resistance leads to a larger ammeter range.

Current sensitivity (\(S_i\)) is defined as the deflection produced per unit current flowing through the galvanometer.
\(S_i = \frac{\phi}{I}\)

where \(\phi\) is the angular deflection and \(I\) is the current.


In equilibrium, the deflecting torque equals the restoring torque:
\(\tau_{deflecting} = \tau_{restoring}\)
\(NBAI = k\phi\)

where \(N\) is the number of turns, \(B\) is the magnetic field, \(A\) is the area, and \(k\) is the torsional constant of the spring.


To find the current sensitivity, we rearrange the equation to solve for \(\frac{\phi}{I}\):
\(\frac{\phi}{I} = \frac{NBA}{k}\)


Therefore, the current sensitivity is given by \(\frac{NBA}{k}\).
Quick Tip: To increase current sensitivity, one can increase the number of turns (N), the magnetic field (B), or the area of the coil (A), or decrease the torsional constant of the spring (k) by using a material like quartz fiber.

Einstein's photoelectric equation relates the maximum kinetic energy of photoelectrons (\(K_{max}\)) to the frequency of incident light (\(\nu\)) and the work function (\(\phi_0\)):
\(K_{max} = h\nu - \phi_0\)


The maximum kinetic energy is also related to the cut-off or stopping potential (\(V_s\)) by:
\(K_{max} = eV_s\)

where \(e\) is the charge of an electron.


Combining these two equations gives:
\(eV_s = h\nu - \phi_0\)

Rearranging this to resemble the equation of a straight line (\(y = mx + c\)):
\(V_s = \left(\frac{h}{e}\right)\nu - \frac{\phi_0}{e}\)


This equation shows that a graph of stopping potential \(V_s\) (on the y-axis) versus frequency \(\nu\) (on the x-axis) will be a straight line.

The slope of this line is \(m = \frac{h}{e}\).

Since the charge of an electron (\(e\)) is a known constant, Planck's constant (\(h\)) can be determined by measuring the slope of the \(V_s\) vs. \(\nu\) graph.
Quick Tip: From the \(V_s\) vs \(\nu\) graph, you can find two key values: 1. **Planck's Constant (\(h\))** from the slope (\(slope = h/e\)). 2. **Work Function (\(\phi_0\))** from the y-intercept (\(y_{intercept} = -\phi_0/e\)) or the x-intercept (threshold frequency, \(\nu_0\), where \(\phi_0 = h\nu_0\)).

According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is given by:
\(K_{max} = h\nu - \phi_0\)

where \(h\) is Planck's constant, \(\nu\) is the frequency of the incident light, and \(\phi_0\) is the work function of the metal.


For a given metal, the work function \(\phi_0\) is constant. Therefore, the maximum kinetic energy (\(K_{max}\)) is directly proportional to the frequency (\(\nu\)) of the incident light.


The frequencies of the visible light spectrum colors are in the order:

Frequency(Blue) > Frequency(Yellow) > Frequency(Red)
\(\nu_B > \nu_Y > \nu_R\)


Since kinetic energy is directly proportional to frequency, the maximum kinetic energies of the photoelectrons will follow the same order:
\(K_B > K_Y > K_R\)
Quick Tip: Remember the order of colors in the visible spectrum by frequency/energy: VIBGYOR (Violet has the highest frequency, Red has the lowest). Kinetic energy of photoelectrons depends on frequency (color), while the number of photoelectrons (current) depends on intensity.

For the photoelectric effect to occur, the energy of the incident photons (\(E = h\nu\)) must be greater than or equal to the work function (\(\phi_0\)) of the metal.


Visible light has a relatively low frequency range compared to ultraviolet light. Therefore, a metal must have a very low work function to exhibit the photoelectric effect with visible light.


Alkali metals, such as Caesium, are known to have the lowest work functions among all metals. The work function of Caesium is approximately 2.14 eV. The energy of visible light photons ranges from about 1.8 eV (red) to 3.1 eV (violet). Since the energy of most visible light is greater than Caesium's work function, it readily shows the photoelectric effect.


Metals like Zinc (\(\phi_0 \approx 4.3\) eV), Cadmium (\(\phi_0 \approx 4.1\) eV), and Magnesium (\(\phi_0 \approx 3.7\) eV) have much higher work functions. They require higher-energy photons, typically in the ultraviolet range, to cause photoemission.
Quick Tip: Alkali metals (Group 1 of the periodic table) like Lithium, Sodium, Potassium, and especially Caesium are the most photo-sensitive metals due to their very low work functions. This makes them ideal for use in phototubes and other light-detecting devices.

In the photoelectric effect, the saturation current is a measure of the total number of photoelectrons emitted from the metal surface per second.


The intensity of light is defined as the energy incident per unit area per unit time. It is proportional to the number of photons arriving per second.


According to the quantum theory of light, it is assumed that one incident photon ejects at most one photoelectron.


Therefore, the number of photoelectrons emitted per second is directly proportional to the intensity of the incident light.


The frequency of the incident light determines the maximum kinetic energy of the individual photoelectrons (\(K_{max} = h\nu - \phi_0\)), not the number of them.


The problem states that the intensity is kept constant while the frequency is increased. Since the intensity is constant, the number of photons incident per second is constant. This means the number of photoelectrons emitted per second is also constant.


As the saturation current is determined by the number of photoelectrons, it will remain the same.
Quick Tip: Think of it this way: Intensity controls the "quantity" (number of photoelectrons, hence current), while Frequency controls the "quality" (energy of each photoelectron). Increasing frequency makes each electron more energetic but doesn't create more of them if intensity is constant.

According to the laws of photoelectric emission, for a given frequency above the threshold frequency, the photoelectric current is directly proportional to the intensity of the incident light.


Let's break down why:

1. The intensity of light is proportional to the number of photons incident on the surface per unit time.

2. The photoelectric current is proportional to the number of photoelectrons emitted from the surface per unit time.

3. In the quantum model, it is assumed that each incident photon ejects one photoelectron (if its energy is sufficient).


Combining these points, the number of emitted photoelectrons per unit time is directly proportional to the number of incident photons per unit time.


Therefore, the photoelectric current (I) is directly proportional to the intensity of the incident light.


A relationship of direct proportionality (\(I \propto Intensity\)) is represented graphically by a straight line passing through the origin.


Graph (C) shows a straight line passing through the origin, which correctly depicts this relationship.
Quick Tip: The direct proportionality between photocurrent and intensity was a key experimental observation that the wave theory of light could not explain. It provided strong support for Einstein's particle (photon) theory of light.

Ray Diagram and Image Formation:

A Cassegrain reflecting telescope consists of a large concave primary mirror and a small convex secondary mirror.


1. Parallel rays of light from a distant object enter the telescope and strike the large concave primary mirror.

2. The primary mirror reflects and converges these rays towards its primary focus.

3. Before the rays reach the focus, they are intercepted by a small convex secondary mirror.

4. The secondary mirror reflects the rays back towards the primary mirror, passing them through a small hole at the center of the primary mirror.

5. The rays then pass through an eyepiece placed behind the hole, where a final, highly magnified image is formed.


Advantages over Refracting Telescope:

1. No Chromatic Aberration: Reflecting telescopes use mirrors instead of lenses. Mirrors do not disperse light, so the image is free from the false color fringes (chromatic aberration) that affect refracting telescopes.

2. Large Aperture and High Resolving Power: It is mechanically easier and cheaper to manufacture a large-aperture mirror than a large-aperture lens. A larger aperture collects more light (producing brighter images) and provides better resolving power (ability to see finer details).
Quick Tip: The key feature of a Cassegrain design is the use of a convex secondary mirror that reflects light back through a hole in the primary. This "folds" the light path, making the telescope more compact for its focal length.

Let \(f_o\) be the focal length of the objective lens and \(f_e\) be the focal length of the eyepiece.


Given:
\(f_o = 50 f_e\) --- (1)


For a telescope in normal adjustment (when the final image is formed at infinity), the length of the tube (\(L\)) is the sum of the focal lengths of the objective and the eyepiece.
\(L = f_o + f_e\) --- (2)


Given \(L = 102\) cm.
\(102 = f_o + f_e\)


Substitute equation (1) into equation (2):
\(102 = (50 f_e) + f_e\)
\(102 = 51 f_e\)
\(f_e = \frac{102}{51} = 2\) cm.


Now, find the focal length of the objective using equation (1):
\(f_o = 50 \times f_e = 50 \times 2 = 100\) cm.


The focal length of the objective is 100 cm and the focal length of the eyepiece is 2 cm.
Quick Tip: For a telescope, the tube length depends on where the final image is formed. For image at infinity (normal adjustment): \(L = f_o + f_e\). For image at near point (D): \(L = f_o + u_e\), where \(u_e\) is the object distance for the eyepiece.

Advantages over Simple Microscope:

1. Higher Magnification: A compound microscope uses two lenses (objective and eyepiece) to achieve a much greater overall magnification than is possible with a single-lens simple microscope.

2. Higher Resolving Power: The use of a specialized objective lens with a short focal length allows the compound microscope to resolve much finer details than a simple microscope.


Explanation:

1. The objective lens has a short focal length. The object to be viewed is placed just outside the focal point of the objective.

2. The objective lens forms a real, inverted, and magnified intermediate image (\(A'B'\)) inside the focal point of the eyepiece.

3. The eyepiece has a larger focal length and acts as a simple magnifier. The intermediate image (\(A'B'\)) serves as the object for the eyepiece.

4. The eyepiece is adjusted so that it forms a final, highly magnified, virtual, and inverted image (\(A''B''\)) at the near point (least distance of distinct vision, D) of the observer's eye.
Quick Tip: The total magnification of a compound microscope is the product of the magnification of the objective lens (\(m_o\)) and the eyepiece (\(m_e\)), i.e., \(M = m_o \times m_e\).

We will find the power of each lens using the Lens Maker's Formula and then add them to find the total power.

Lens Maker's Formula: \(P = \frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\).


Lens 1: Planoconcave lens

Refractive index = \(n_1\).

For a planoconcave lens, the first surface is plane (\(R_1 = \infty\)) and the second is concave (\(R_2 = -R\), assuming light comes from left). No, the curved face has radius R. Let's assume light enters the plane face. \(R_1=\infty\). The second face is concave, so its center is on the left, \(R_2 = -R\). \(P_1 = (n_1 - 1)\left(\frac{1}{\infty} - \frac{1}{-R}\right) = (n_1 - 1)\left(\frac{1}{R}\right) = \frac{n_1 - 1}{R}\).

Wait, for a planoconcave lens, one radius is infinity, the other is negative. Let \(R_{curved} = R\). Let the plane surface be \(R_1=\infty\), the concave surface is \(R_2=+R\). \(P_1 = (n_1 - 1)\left(\frac{1}{\infty} - \frac{1}{R}\right) = -\frac{n_1 - 1}{R}\).


Lens 2: Equiconvex lens

Refractive index = \(n_2\).

For an equiconvex lens, the first surface is convex (\(R_1 = +R\)) and the second is concave from the light's perspective (\(R_2 = -R\)).
\(P_2 = (n_2 - 1)\left(\frac{1}{R} - \frac{1}{-R}\right) = (n_2 - 1)\left(\frac{1}{R} + \frac{1}{R}\right) = (n_2 - 1)\left(\frac{2}{R}\right) = \frac{2(n_2 - 1)}{R}\).


Power of the Combination:

Since the lenses are in contact, the total power is the algebraic sum of the individual powers.
\(P = P_1 + P_2\)
\(P = -\frac{n_1 - 1}{R} + \frac{2(n_2 - 1)}{R}\)
\(P = \frac{-(n_1 - 1) + 2(n_2 - 1)}{R}\)
\(P = \frac{-n_1 + 1 + 2n_2 - 2}{R}\)
\(P = \frac{2n_2 - n_1 - 1}{R}\).
Quick Tip: Always be careful with the sign convention when using the Lens Maker's Formula. Radii are measured from the optical center. Surfaces convex towards the incident light have positive radii, while concave surfaces have negative radii.

Let's apply Kirchhoff's laws to the circuit. Let the junctions be P (top) and Q (bottom).

Let \(I_1\) be the current from \(E_1\), \(I_2\) from \(E_2\), and \(I_3\) from \(E_3\). Assume \(I_1\) and \(I_3\) flow from P to Q, and \(I_2\) flows from Q to P.

KCL at junction P: \(I_1 + I_3 = I_2\) --- (1)


KVL for the left loop (P-E1-Q-E2-P):

Traversing clockwise from P:
\(-I_1(2) - 4 - 2 - I_2(4) = 0\)
\(-2I_1 - 4I_2 = 6 \implies I_1 + 2I_2 = -3\) --- (2)


KVL for the right loop (P-E3-Q-E2-P):

Traversing clockwise from P:
\(-I_3(2) - 6 - 2 - I_2(4) = 0\)
\(-2I_3 - 4I_2 = 8 \implies I_3 + 2I_2 = -4\) --- (3)


Now we solve the system of three equations. From (2), \(I_1 = -3 - 2I_2\). From (3), \(I_3 = -4 - 2I_2\).

Substitute these into equation (1):
\((-3 - 2I_2) + (-4 - 2I_2) = I_2\)
\(-7 - 4I_2 = I_2\)
\(-7 = 5I_2 \implies I_2 = -7/5 = -1.4\) A.

The negative sign means our assumed direction for \(I_2\) was wrong. So, \(I_2\) flows from P to Q.

Current through \(E_2\) is 1.4 A (from P to Q).


Now find \(I_1\) and \(I_3\):
\(I_1 = -3 - 2(-1.4) = -3 + 2.8 = -0.2\) A.

The negative sign means \(I_1\) flows from Q to P.

Current through \(E_1\) is 0.2 A (from Q to P).

\(I_3 = -4 - 2(-1.4) = -4 + 2.8 = -1.2\) A.

The negative sign means \(I_3\) flows from Q to P.

Current through \(E_3\) is 1.2 A (from Q to P).
Quick Tip: When applying Kirchhoff's laws, the initial assumed direction of current does not matter. If your final answer for a current is negative, it simply means the actual current flows in the opposite direction to the one you assumed.

Let the vertices of the circuit be A, B, C (top), and D (bottom). The circuit has six resistors: AB, AC, AD, BC, BD, and CD, each with resistance R.

We need to find the equivalent resistance between points A and B.


This circuit is symmetric with respect to the plane passing through the points C and D and the midpoint of the resistor AB.

Due to this symmetry, if a voltage is applied across A and B, the potential at point C will be equal to the potential at point D (\(V_C = V_D\)).


Since there is no potential difference between points C and D, no current will flow through the resistor CD.

Therefore, we can remove the resistor CD from the circuit without affecting the equivalent resistance between A and B.


After removing the resistor CD, the circuit simplifies:

1. The upper branch consists of resistor AC in series with resistor CB. The resistance of this branch is \(R_{ACB} = R + R = 2R\).

2. The lower branch consists of resistor AD in series with resistor DB. The resistance of this branch is \(R_{ADB} = R + R = 2R\).

3. These two branches are in parallel with the direct resistor AB.


The equivalent resistance \(R_{eq}\) is the parallel combination of these three paths:
\(\frac{1}{R_{eq}} = \frac{1}{R_{AB}} + \frac{1}{R_{ACB}} + \frac{1}{R_{ADB}}\)
\(\frac{1}{R_{eq}} = \frac{1}{R} + \frac{1}{2R} + \frac{1}{2R}\)
\(\frac{1}{R_{eq}} = \frac{1}{R} + \frac{2}{2R} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R}\)
\(R_{eq} = \frac{R}{2}\)


Given R = 10 \(\Omega\):
\(R_{eq} = \frac{10 \Omega}{2} = 5 \Omega\).
Quick Tip: Exploiting symmetry is a powerful tool for simplifying complex circuits. Look for planes or axes of symmetry. Points that are symmetric with respect to the input/output terminals will be at the same potential, allowing you to simplify connections or remove components.

(I) Ratio of electric fields at points A and B:

From the microscopic form of Ohm's Law, the electric field \(E\) is related to current density \(j\) and resistivity \(\rho\) by \(E = \rho j\).

Current density is \(j = I/A\), where \(I\) is the current and \(A\) is the cross-sectional area.

So, \(E = \rho \frac{I}{A}\).

For a given material (copper), the resistivity \(\rho\) is constant. The current \(I\) is also constant throughout the rod.

Therefore, the electric field is inversely proportional to the cross-sectional area: \(E \propto \frac{1}{A}\).

The ratio of the electric fields at A and B is:
\(\frac{E_A}{E_B} = \frac{A_B}{A_A} = \frac{2.0 \times 10^{-7} m^2}{1.0 \times 10^{-7} m^2} = 2\).

The ratio \(E_A : E_B\) is 2:1.


(II) Drift velocity of free electrons at point B:

The relationship between current \(I\), number density of electrons \(n\), elementary charge \(e\), area \(A\), and drift velocity \(v_d\) is:
\(I = n e A v_d\).

We need to find the drift velocity at point B, \(v_{d,B}\).
\(v_{d,B} = \frac{I}{n e A_B}\).

Given values:
\(I = 1\) A
\(n = 8.5 \times 10^{28} m^{-3}\)
\(e = 1.6 \times 10^{-19}\) C
\(A_B = 2.0 \times 10^{-7} m^2\)
\(v_{d,B} = \frac{1}{(8.5 \times 10^{28})(1.6 \times 10^{-19})(2.0 \times 10^{-7})}\)
\(v_{d,B} = \frac{1}{8.5 \times 1.6 \times 2.0 \times 10^{28-19-7}} = \frac{1}{27.2 \times 10^2} = \frac{1}{2720}\) m/s.
\(v_{d,B} \approx 3.68 \times 10^{-4}\) m/s.
Quick Tip: In a conductor of varying cross-section carrying a steady current: Current (I) is constant everywhere. Current density (\(j\)), electric field (\(E\)), and drift velocity (\(v_d\)) are all inversely proportional to the cross-sectional area (\(A\)).

The net electric field at point P, with position vector \(\vec{r}\), is the vector sum of the fields due to each charge (Principle of Superposition).
\(\vec{E}_{net} = \vec{E}_1 + \vec{E}_2\).

The field due to a point charge \(q\) at a distance vector \(\vec{d}\) is \(\vec{E} = k \frac{q}{d^3}\vec{d}\).


The point of interest is P(\(3, 4\)).


Electric field due to \(q_1\) at P:

Position of \(q_1\) is \(\vec{r}_1 = 3\hat{i}\).

The vector from \(q_1\) to P is \(\vec{d}_1 = \vec{r} - \vec{r}_1 = (3\hat{i} + 4\hat{j}) - (3\hat{i}) = 4\hat{j}\).

The magnitude of this vector (distance) is \(d_1 = |\vec{d}_1| = 4\) m.
\(\vec{E}_1 = k \frac{q_1}{d_1^2} \hat{d}_1 = (9 \times 10^9) \frac{16 \times 10^{-6}}{4^2} \hat{j} = (9 \times 10^9) \frac{16 \times 10^{-6}}{16} \hat{j} = 9 \times 10^3 \hat{j}\) N/C.


Electric field due to \(q_2\) at P:

Position of \(q_2\) is \(\vec{r}_2 = 4\hat{j}\).

The vector from \(q_2\) to P is \(\vec{d}_2 = \vec{r} - \vec{r}_2 = (3\hat{i} + 4\hat{j}) - (4\hat{j}) = 3\hat{i}\).

The magnitude of this vector (distance) is \(d_2 = |\vec{d}_2| = 3\) m.
\(\vec{E}_2 = k \frac{q_2}{d_2^2} \hat{d}_2 = (9 \times 10^9) \frac{1 \times 10^{-6}}{3^2} \hat{i} = (9 \times 10^9) \frac{1 \times 10^{-6}}{9} \hat{i} = 1 \times 10^3 \hat{i}\) N/C.


Net Electric Field:
\(\vec{E}_{net} = \vec{E}_1 + \vec{E}_2 = (1 \times 10^3 \hat{i} + 9 \times 10^3 \hat{j})\) N/C.
\(\vec{E}_{net} = 10^3 (\hat{i} + 9\hat{j})\) N/C.
Quick Tip: When dealing with vectors, it's often easiest to find the displacement vector from the source charge to the point of interest (\(\vec{d} = \vec{r}_{point} - \vec{r}_{source}\)), then use the vector form of Coulomb's Law, \(\vec{E} = k(q/d^3)\vec{d}\), which automatically handles the direction.

Definition of Self-Inductance:

Self-inductance is the property of a coil by virtue of which it opposes any change in the strength of the current flowing through it by inducing an electromotive force (emf) in itself. The self-inductance (L) is numerically defined as the magnetic flux (\(\Phi_B\)) linked with the coil when a unit current flows through it, i.e., \(L = \Phi_B / I\).


Derivation for Energy Stored:

When current in a coil changes, an opposing emf is induced, given by \(\epsilon = -L \frac{dI}{dt}\).

To maintain the current, an external source must do work against this opposing emf. The instantaneous power supplied by the source is:
\(P = \frac{dW}{dt} = |\epsilon|I = \left(L \frac{dI}{dt}\right)I\)

The small amount of work done (\(dW\)) in a small time interval \(dt\) is:
\(dW = P dt = L I \frac{dI}{dt} dt = L I dI\).

The total work done to build up the current from 0 to its final steady value \(I_f\) is found by integrating this expression:
\(W = \int_{0}^{I_f} dW = \int_{0}^{I_f} L I dI\)
\(W = L \left[ \frac{I^2}{2} \right]_{0}^{I_f} = L \left( \frac{I_f^2}{2} - 0 \right) = \frac{1}{2}LI_f^2\).

This work done is stored in the inductor as magnetic potential energy.

Therefore, the energy stored is \(U = \frac{1}{2}LI^2\).
Quick Tip: The formula for energy stored in an inductor, \(U = \frac{1}{2}LI^2\), is analogous to the energy stored in a capacitor, \(U = \frac{1}{2}CV^2\), and the kinetic energy of a mass, \(K = \frac{1}{2}mv^2\). Inductance is the electrical analogue of inertia.

(I) Graph of emf versus rate of change of current (\(|\epsilon|\) vs \(dI/dt\)):

The relationship is \(|\epsilon| = L \frac{dI}{dt}\). This is a linear relationship of the form \(y = mx\), where \(y=|\epsilon|\), \(x=dI/dt\), and the slope is \(m=L\).

The graph is a straight line passing through the origin. Since \(L_2 (20 mH) > L_1 (10 mH)\), the slope of the line for the 20 mH inductor will be steeper than that for the 10 mH inductor.



(II) Graph of energy stored versus current (U vs I):

The relationship is \(U = \frac{1}{2}LI^2\). This is a quadratic relationship of the form \(y = ax^2\), where \(y=U\), \(x=I\), and the constant is \(a = L/2\).

The graph is a parabola opening upwards and starting from the origin. Since \(L_2 > L_1\), for any given current \(I\), the energy stored in the 20 mH inductor will be greater than that in the 10 mH inductor. The parabola for \(L_2\) will be steeper than the one for \(L_1\).
Quick Tip: When sketching graphs, identify the mathematical relationship between the variables first. Linear relationships (\(y=mx+c\)) are straight lines, while quadratic relationships (\(y=ax^2\)) are parabolas. The constants in the equation determine the slope or steepness of the curve.

Definition of Mutual Inductance:

Mutual inductance is the phenomenon in which a changing current in one coil (primary coil) induces an electromotive force (emf) in a neighboring coil (secondary coil). The mutual inductance (M) of a pair of coils is numerically defined as the magnetic flux linked with the secondary coil when a unit current flows through the primary coil, i.e., \(M = \Phi_{21} / I_1\).


Derivation for two Coaxial Solenoids:

Consider two long, coaxial solenoids S1 (inner) and S2 (outer) of the same length \(l\).

Let \(r_1, N_1, n_1 = N_1/l\) be the radius, total turns, and turns per unit length of S1.

Let \(r_2, N_2, n_2 = N_2/l\) be the corresponding values for S2, with \(r_2 > r_1\).


Let a current \(I_2\) flow through the outer solenoid S2. This produces a uniform magnetic field within S2:
\(B_2 = \mu_0 n_2 I_2\)

Since S1 is inside S2, this magnetic field passes through S1. The area of linkage is the area of the inner solenoid, \(A_1 = \pi r_1^2\).

The magnetic flux through each turn of the inner solenoid S1 is:
\(\Phi_{turn} = B_2 A_1 = (\mu_0 n_2 I_2)(\pi r_1^2)\)

The total magnetic flux linked with the entire inner solenoid S1 (which has \(N_1\) turns) is:
\(\Phi_{12} = N_1 \times \Phi_{turn} = N_1 (\mu_0 n_2 I_2 \pi r_1^2)\)

By the definition of mutual inductance, \(M = M_{12} = \frac{\Phi_{12}}{I_2}\).
\(M = \frac{N_1 \mu_0 n_2 I_2 \pi r_1^2}{I_2} = \mu_0 N_1 n_2 \pi r_1^2\)

Substituting \(n_2 = N_2/l\):
\(M = \frac{\mu_0 N_1 N_2 \pi r_1^2}{l}\).
Quick Tip: The mutual inductance depends only on the geometry and orientation of the coils and the medium between them. The flux linkage is always calculated over the smaller area, as the magnetic field from the outer coil is only present within its own volume.

Step 1: Calculate the self-inductance (L) of the inductor.

The induced emf is related to the rate of change of current by \(|\epsilon| = L \left| \frac{dI}{dt} \right|\).

Since the current increases uniformly, the rate of change is constant:
\(\frac{dI}{dt} = \frac{\Delta I}{\Delta t} = \frac{2 A - 0 A}{40 s} = \frac{2}{40} A/s = 0.05\) A/s.

Given the induced emf \(|\epsilon| = 5 mV = 5 \times 10^{-3}\) V.

We can find the inductance L:
\(L = \frac{|\epsilon|}{dI/dt} = \frac{5 \times 10^{-3} V}{0.05 A/s} = \frac{5 \times 10^{-3}}{5 \times 10^{-2}} = 0.1\) H.


Step 2: Calculate the current at t = 10 s.

Since the current increases uniformly with time, the current \(I(t)\) at any time \(t\) is given by:
\(I(t) = (initial current) + \left(\frac{dI}{dt}\right) \times t\)
\(I(10) = 0 + (0.05 A/s) \times (10 s) = 0.5\) A.


Step 3: Calculate the flux at t = 10 s.

The magnetic flux (\(\Phi\)) linked with an inductor is given by the formula \(\Phi = L I\).

At \(t = 10\) s, the flux is:
\(\Phi(10) = L \times I(10)\)
\(\Phi(10) = (0.1 H) \times (0.5 A) = 0.05\) Wb.
Quick Tip: This is a multi-step problem. First, use the given emf and rate of current change to find the inductance (L), which is a constant property of the inductor. Then, use L and the current at the specified time to find the flux.