The Earth's surface is warmed by absorbing radiation from the Sun.


The warmed Earth then radiates energy back into space, primarily in the form of infrared radiation due to its lower temperature compared to the Sun.


Greenhouse gases in the atmosphere, such as CO\(_2\) and H\(_2\)O, are effective at absorbing this outgoing infrared radiation.


This trapped energy warms the atmosphere and the Earth's surface, a phenomenon known as the greenhouse effect, which maintains the planet's warmth.


Therefore, infrared rays play the most crucial role in this heat-trapping process.
Quick Tip: Remember the electromagnetic spectrum in order of increasing wavelength (decreasing energy): Gamma rays, X-rays, UV, Visible, Infrared, Microwaves, Radio waves. Infrared radiation is often associated with heat.

The given circuit is a Wheatstone bridge arrangement for capacitors.


Let the capacitors be \(C_1 = 3 \, \mu F\), \(C_2 = 6 \, \mu F\), \(C_3 = 6 \, \mu F\), and \(C_4 = 12 \, \mu F\). The central capacitor is \(C_5 = 1 \, \mu F\).


The condition for a balanced Wheatstone bridge with capacitors is \(\frac{C_1}{C_3} = \frac{C_2}{C_4}\).


Checking the condition: \(\frac{3}{6} = \frac{1}{2}\) and \(\frac{6}{12} = \frac{1}{2}\).


Since the condition \(\frac{C_1}{C_3} = \frac{C_2}{C_4}\) is satisfied, the bridge is balanced.


This means no charge flows through the central capacitor (\(1 \, \mu F\)), and it can be removed from the circuit.


The upper branch consists of \(3 \, \mu F\) and \(6 \, \mu F\) capacitors in series. Their equivalent capacitance is \(C_{upper} = \left(\frac{1}{3} + \frac{1}{6}\right)^{-1} = \left(\frac{3}{6}\right)^{-1} = 2 \, \mu F\).


The lower branch consists of \(6 \, \mu F\) and \(12 \, \mu F\) capacitors in series. Their equivalent capacitance is \(C_{lower} = \left(\frac{1}{6} + \frac{1}{12}\right)^{-1} = \left(\frac{3}{12}\right)^{-1} = 4 \, \mu F\).


The total effective capacitance between A and B is the sum of the two parallel branches.

\(C_{eff} = C_{upper} + C_{lower} = 2 \, \mu F + 4 \, \mu F = 6 \, \mu F\).
Quick Tip: For capacitor circuits, always check for a balanced Wheatstone bridge (\(C_1/C_3 = C_2/C_4\)). If balanced, ignore the central component. Remember: capacitors in series add like resistors in parallel, and vice-versa.

The relationship between electric field \(\vec{E}\) and electric potential \(V\) is \(V = -\int \vec{E} \cdot d\vec{r}\).


The magnitude of the given electric field is \(E = |\vec{E}| = \left| \alpha \frac{\vec{r}}{|\vec{r}|^3} \right| = \alpha \frac{r}{r^3} = \frac{\alpha}{r^2}\).


The electric field is radial, so we can integrate along the radial direction. We find the potential difference by integrating from a reference point (usually infinity, where V=0) to the point r.

\(V(r) - V(\infty) = -\int_{\infty}^{r} E \, dr\).

\(V(r) - 0 = -\int_{\infty}^{r} \frac{\alpha}{r^2} \, dr\).

\(V(r) = -\alpha \left[ -\frac{1}{r} \right]_{\infty}^{r} = \alpha \left[ \frac{1}{r} \right]_{\infty}^{r}\).

\(V(r) = \alpha \left( \frac{1}{r} - \frac{1}{\infty} \right) = \alpha \left( \frac{1}{r} - 0 \right)\).

\(V(r) = \frac{\alpha}{r}\).


The magnitude of the potential is \(\frac{\alpha}{r}\).
Quick Tip: For radially symmetric fields like this one (\(E \propto 1/r^2\)), the relation \(E = -dV/dr\) is very useful. An electric field that follows an inverse square law, like that of a point charge, will always correspond to a potential that varies as \(1/r\).

The intensity (\(I\)) of a wave is directly proportional to the square of its amplitude (\(A\)), i.e., \(I \propto A^2\).


For constructive interference, the waves are in phase, and their amplitudes add up.


The maximum amplitude is \(A_{max} = a + a = 2a\).


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


For destructive interference, the waves are completely out of phase, and their amplitudes subtract.


The minimum amplitude is \(A_{min} = a - a = 0\).


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


Thus, the intensity on the screen varies between a minimum of 0 and a maximum proportional to \(4a^2\).
Quick Tip: For interference of two identical sources with amplitude 'a' and intensity \(I_0 \propto a^2\), the maximum intensity is always \(4I_0\) (from amplitude \(2a\)) and the minimum intensity is 0 (from amplitude 0).

The magnetic dipole moment (\(M\)) of a current-carrying loop is given by the formula \(M = NIA\), where N is the number of turns, I is the current, and A is the area of the loop.


For this problem, we have a single circular loop, so \(N=1\).


The given current is \(I = \frac{10}{\pi}\) A.


The given radius is \(r = 14\) cm, which must be converted to meters: \(r = 0.14\) m.


First, we calculate the area of the loop: \(A = \pi r^2\).

\(A = \pi (0.14)^2 = \pi (0.0196)\) m\(^2\).


Now, we calculate the magnetic dipole moment:

\(M = I \times A = \left(\frac{10}{\pi}\right) \times (\pi \times 0.0196)\).


The \(\pi\) terms cancel out.

\(M = 10 \times 0.0196 = 0.196\) Am\(^2\).
Quick Tip: Always ensure all units are in the SI system before calculation. For magnetic moment problems, this means converting cm to m. The formula \(M = NIA\) is fundamental for current loops.

A transformer works on the principle of mutual induction to change AC voltage levels.


The fundamental principle for an ideal transformer is the conservation of power: Power in = Power out.


Power is given by \(P = VI\) (Voltage \(\times\) Current).


Therefore, \(V_{primary} \times I_{primary} = V_{secondary} \times I_{secondary}\).


This relationship implies that if the voltage is increased (stepped up), the current must decrease, and if the voltage is decreased (stepped down), the current must increase.


Let's analyze the options:

(A) Violates power conservation (\(V_{high}I_{large} > V_{low}I_{small}\)).

(B) This describes a step-down transformer (\(V_{high} \rightarrow V_{low}\), \(I_{small} \rightarrow I_{large}\)). This is a valid function.

(C) This describes a step-up transformer (\(V_{low} \rightarrow V_{high}\), \(I_{large} \rightarrow I_{small}\)). This is also a valid function.

(D) Violates power conservation (\(V_{low}I_{small} < V_{high}I_{large}\)).


Both (B) and (C) describe functions of a transformer. However, option (C) is given as the correct choice, representing the function of a step-up transformer.
Quick Tip: For any ideal transformer problem, remember the power conservation rule: \(P_{in} = P_{out}\) or \(V_p I_p = V_s I_s\). This means voltage and current are inversely proportional. A step-up transformer increases voltage but decreases current.

This problem requires interpreting a non-standard circuit diagram. One interpretation that leads to the correct answer is to view the circuit as a combination of series and parallel resistors.


Let's assume the leftmost resistor is in series with the rest of the active circuit when the key K is open.


When the key K is open, the path containing the rightmost resistor is broken and can be ignored.


The circuit then consists of the leftmost resistor connected to a junction. At this junction, the current splits into three parallel paths: the top resistor (R), the vertical resistor (R), and the bottom resistor (R).


The equivalent resistance of these three parallel resistors is:

\(\frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{3}{R}\).


So, \(R_p = \frac{R}{3}\).


This parallel combination is in series with the first (leftmost) resistor.


The total equivalent resistance between the terminals A and B is:

\(R_{eq} = R_{left} + R_p = R + \frac{R}{3} = \frac{3R + R}{3} = \frac{4R}{3}\).


This interpretation, while not immediately obvious from a standard Wheatstone bridge view, logically leads to the provided correct answer.
Quick Tip: When faced with a complex or ambiguously drawn circuit, try to redraw it in a more standard form. If that fails, consider alternative interpretations of the connectivity, especially if a standard analysis does not yield any of the given options.

The momentum (\(p\)) of a photon is related to its energy (\(E\)) and wavelength (\(\lambda\)) by the following equations:

\(p = \frac{E}{c}\) and \(E = \frac{hc}{\lambda}\), where \(h\) is Planck's constant and \(c\) is the speed of light.


Combining these, we get the momentum of a photon as \(p = \frac{h}{\lambda}\).


This shows that a photon's momentum is inversely proportional to its wavelength.


To have the largest momentum, the photon must have the shortest wavelength.


Let's compare the wavelengths of the given electromagnetic waves:

- AM radio waves: Longest wavelength (hundreds of meters).

- TV waves: Long wavelength (meters).

- Microwaves: Shorter wavelength (centimeters to millimeters).

- X-rays: Shortest wavelength (angstroms to picometers).


Since X-rays have the shortest wavelength among the options, their photons have the largest momentum.
Quick Tip: Remember the relationship: High Energy = High Frequency = High Momentum = Short Wavelength. Knowing the order of the electromagnetic spectrum is key to solving such problems quickly.

The de Broglie wavelength (\(\lambda\)) of a particle is given by \(\lambda = \frac{h}{p}\), where \(p\) is the momentum.


Momentum can be expressed in terms of kinetic energy (\(K\)) and mass (\(m\)) as \(p = \sqrt{2mK}\).


So, the de Broglie wavelength formula becomes \(\lambda = \frac{h}{\sqrt{2mK}}\).


We are given the following information:

Kinetic energy of alpha particle: \(K_{\alpha} = 4 K_{p}\).

Mass of an alpha particle (\(m_{\alpha}\)) is approximately 4 times the mass of a proton (\(m_{p}\)), so \(m_{\alpha} = 4 m_{p}\).


Now, let's find the ratio of their wavelengths:

\(\frac{\lambda_{\alpha}}{\lambda_{p}} = \frac{h/\sqrt{2m_{\alpha}K_{\alpha}}}{h/\sqrt{2m_{p}K_{p}}} = \sqrt{\frac{m_{p}K_{p}}{m_{\alpha}K_{\alpha}}}\).


Substitute the known relationships for mass and kinetic energy:

\(\frac{\lambda_{\alpha}}{\lambda_{p}} = \sqrt{\frac{m_{p}K_{p}}{(4m_{p})(4K_{p})}} = \sqrt{\frac{1}{16}}\).

\(\frac{\lambda_{\alpha}}{\lambda_{p}} = \frac{1}{4}\).
Quick Tip: The formula \(\lambda = h/\sqrt{2mK}\) is crucial for problems relating de Broglie wavelength to kinetic energy. Remember the basic properties of subatomic particles, like \(m_{\alpha} \approx 4 m_{p}\) and \(q_{\alpha} = 2 q_{p}\).

When an object is viewed through a denser medium, its apparent depth is less than its real depth.


The real thickness of the glass slab is the real depth, \(t_{real} = 6\) cm.


The refractive index of the glass slab is \(\mu = 1.5\).


The apparent depth (\(t_{apparent}\)) is given by the formula:

\(t_{apparent} = \frac{t_{real}}{\mu}\).

\(t_{apparent} = \frac{6 cm}{1.5} = 4\) cm.


The shift in the position of the letters is the difference between the real depth and the apparent depth.


Shift = \(t_{real} - t_{apparent}\).


Shift = \(6 cm - 4 cm = 2\) cm.


Alternatively, the formula for shift can be directly used:


Shift = \(t_{real} \left(1 - \frac{1}{\mu}\right) = 6 \left(1 - \frac{1}{1.5}\right) = 6 \left(1 - \frac{2}{3}\right) = 6 \left(\frac{1}{3}\right) = 2\) cm.
Quick Tip: Remember the direct formula for the normal shift caused by a glass slab: Shift = thickness \(\times (1 - 1/\mu)\). This can save time in calculations. The shift is always towards the observer.

The speed of light (\(c\)) in a vacuum is related to the permeability (\(\mu_0\)) and permittivity (\(\epsilon_0\)) of free space by the equation:

\(c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} = (\mu_0 \epsilon_0)^{-1/2}\).


Similarly, the speed of light (\(v\)) in any medium is given by \(v = \frac{1}{\sqrt{\mu \epsilon}} = (\mu \epsilon)^{-1/2}\).


The question asks for the dimensions of \((\mu\epsilon)^{-1}\). There appears to be a typo in the question, as the correct option (A) corresponds to the dimensions of speed, which is \((\mu\epsilon)^{-1/2}\).


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


This quantity is the speed of light in the medium, \(v\).


The dimensions of speed are distance divided by time.


Therefore, the dimensions are \([L]/[T] = [M^0 L^1 T^{-1}]\).
Quick Tip: The expression \(\frac{1}{\sqrt{\mu_0 \epsilon_0}}\) is a very important relation for the speed of light in a vacuum. Be aware that questions may have typos, and sometimes you must infer the intended question from the given options, as in this case where the options point to speed (\(v\)) rather than speed squared (\(v^2\)).

When a charged particle moves perpendicular to a magnetic field, the magnetic force provides the necessary centripetal force for circular motion.

\(qvB = \frac{mv^2}{r}\)


This gives the radius of the circular path as \(r = \frac{mv}{qB}\).


We can write this as \(r = \frac{v}{(q/m)B}\). We need to find the charge-to-mass ratio (\(q/m\)).


The kinetic energy gained by the particle when accelerated through a potential difference \(V\) is given by:

\(K = qV = \frac{1}{2}mv^2\).


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


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

\(\frac{q}{m} = \frac{(10^6)^2}{2 \times 10^4} = \frac{10^{12}}{2 \times 10^4} = 0.5 \times 10^8\) C/kg.


Now, substitute this value into the radius formula:

\(r = \frac{v}{(q/m)B} = \frac{10^6}{(0.5 \times 10^8) \times 0.4} = \frac{10^6}{0.2 \times 10^8} = \frac{10^6}{2 \times 10^7}\).

\(r = 0.5 \times 10^{-1}\) m \(= 0.05\) m.


Converting to centimeters: \(r = 0.05 \times 100\) cm \(= 5\) cm.
Quick Tip: This is a classic two-step problem. First, use the kinetic energy equation (\(qV = \frac{1}{2}mv^2\)) to find the charge-to-mass ratio (\(q/m\)). Second, use this ratio in the radius formula for motion in a magnetic field (\(r = mv/qB\)).

Assertion (A) is true. The most surprising result of Rutherford's experiment was that a very small fraction of alpha particles (about 1 in 8000) were deflected by large angles (\(> 90^{\circ}\)), with some even rebounding (scattering angle \(\approx \pi\) or \(180^{\circ}\)). This could only be explained if the positive charge of the atom was concentrated in a very small, dense core, which he named the nucleus.


Reason (R) is true. The radius of an atom is of the order of \(10^{-10}\) m, while the radius of a nucleus is of the order of \(10^{-15}\) m. The ratio of their sizes is about \(\frac{10^{-15}}{10^{-10}} = 10^{-5}\).


The reason correctly explains the assertion. Because the nucleus is an extremely small target compared to the size of the atom, the vast majority of alpha particles pass through the empty space of the atom undeflected. Only the very few 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 and rebound.


Therefore, both statements are true, and the reason is the correct explanation for the assertion.
Quick Tip: For Assertion-Reason questions, follow a three-step process: 1. Check if the Assertion is true. 2. Check if the Reason is true. 3. If both are true, check if the Reason correctly explains the Assertion by asking "Why?" or "Because".

Assertion (A) is false. The Balmer series corresponds to electronic transitions from higher energy levels (\(n_i = 3, 4, 5, ...\)) to the second energy level (\(n_f = 2\)). Transitions to the ground state (\(n_f = 1\)) constitute the Lyman series.


Reason (R) is false. According to Bohr's postulates, an electron can jump from any higher energy orbit to any lower energy orbit, not just between successive (adjacent) orbits. For example, a jump from \(n=4\) to \(n=2\) is allowed and contributes a spectral line to the Balmer series.


Since both the Assertion and the Reason are false statements, the correct option is (D).
Quick Tip: Memorize the first few spectral series for the hydrogen atom: Lyman series (transitions to n=1, UV region), Balmer series (transitions to n=2, visible region), and Paschen series (transitions to n=3, infrared region).

Assertion (A) is true. 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, \(E = mc^2\), this mass defect is converted into energy, which is the binding energy that holds the nucleus together. So, \(BE = (\Delta m)c^2\).


Reason (R) is false. The binding energy per nucleon (BE/A) is not a constantly increasing function of the mass number (A). The BE/A curve shows that it increases rapidly for light nuclei, reaches a peak value of about 8.8 MeV for nuclei with mass number around A=56 (Iron), and then gradually decreases for heavier nuclei.


Therefore, the Assertion is true, but the Reason is false.
Quick Tip: Familiarize yourself with the shape of the binding energy per nucleon curve. It is a key concept in nuclear physics that explains both nuclear fission (splitting of heavy nuclei) and nuclear fusion (combining of light nuclei) as processes that release energy by moving towards the more stable, higher BE/A region.

Assertion (A) is true. A p-type semiconductor is formed by doping a pure semiconductor like Silicon (which is tetravalent) with trivalent impurities (e.g., Boron, Aluminium, Indium). These impurities have one less valence electron and create a "hole" (an electron vacancy). Pentavalent impurities (e.g., Phosphorus, Arsenic) have one extra valence electron and are used to create n-type semiconductors. Therefore, impurities in p-type Si are not pentavalent.


Reason (R) is true. Trivalent impurity atoms are called "acceptor" atoms because they can accept an electron from the valence band to complete their covalent bonds, thereby creating a mobile hole in the valence band. At normal operating temperatures, almost all acceptor atoms are ionized, so the concentration of holes (\(n_h\)) is approximately equal to the concentration of acceptor atoms (\(N_A\)).


However, the reason does not explain the assertion. The reason explains a resulting property of a p-type semiconductor. The explanation for why pentavalent atoms are not used for p-type Si is that they would donate electrons and create an n-type semiconductor, which is the opposite of what is desired. Since both statements are correct but the reason does not explain the assertion, option (B) is the correct choice.
Quick Tip: To remember dopants, think: P-type for Positive holes, which are created by impurities that are "short" of an electron (trivalent). N-type for Negative electrons, which are created by impurities that have an "extra" electron (pentavalent).

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, the light travels from air to the medium.

\(n_1\) (air) \(= 1\).

\(n_2\) (medium) \(= n\).


Object distance, \(u = -R/3\) (by sign convention).


Radius of curvature, \(R_{surface} = +R\) (for a convex surface).


Substituting the values into the formula:

\(\frac{n}{v} - \frac{1}{-R/3} = \frac{n - 1}{R}\)

\(\frac{n}{v} + \frac{3}{R} = \frac{n - 1}{R}\)

\(\frac{n}{v} = \frac{n - 1}{R} - \frac{3}{R} = \frac{n - 1 - 3}{R}\)

\(\frac{n}{v} = \frac{n - 4}{R}\)

\(v = \frac{nR}{n - 4}\)


Given that \(n < 4\), the denominator \((n - 4)\) is negative.


Since \(n\) and \(R\) are positive, the image distance \(v\) is negative.


A negative image distance means the image is formed on the same side as the object (in the air).


Therefore, the image is virtual and is formed at a distance of \(\frac{nR}{4-n}\) in front of the convex surface.
Quick Tip: Always be careful with the sign convention for object distance (\(u\)), image distance (\(v\)), and radius of curvature (\(R\)). For a single spherical surface, distances measured in the direction of incident light are positive, and those measured against it are negative.

The relationship between phase difference (\(\phi\)) and path difference (\(\Delta x\)) is given by:

\(\phi = \frac{2\pi}{\lambda} \Delta x\)


Given path difference \(\Delta x = \frac{\lambda}{3}\).


So, the phase difference is \(\phi = \frac{2\pi}{\lambda} \left(\frac{\lambda}{3}\right) = \frac{2\pi}{3}\).


The intensity (\(I\)) at any point in an interference pattern is given by:

\(I = I_{max} \cos^2\left(\frac{\phi}{2}\right)\)


Here, the intensity of the central maximum is given as \(I_0\), so \(I_{max} = I_0\).

\(I = I_0 \cos^2\left(\frac{2\pi/3}{2}\right) = I_0 \cos^2\left(\frac{\pi}{3}\right)\).


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


Therefore, \(I = I_0 \left(\frac{1}{2}\right)^2 = \frac{I_0}{4}\).


The intensity at the given point is one-fourth of the maximum intensity.
Quick Tip: The formula \(I = I_{max} \cos^2(\phi/2)\) is extremely useful for YDSE problems. First, convert the given path difference to a phase difference, then substitute into this formula to find the resultant intensity.

Let the initial length be \(L_1\), initial area of cross-section be \(A_1\), and initial resistance be \(R_1 = X\).


The formula for resistance is \(R = \rho \frac{L}{A}\), where \(\rho\) is the resistivity.


So, \(X = \rho \frac{L_1}{A_1}\).


When the wire is stretched, its volume (\(V = L \times A\)) remains constant, but its length and area change.


The new length is \(L_2 = 2L_1\).


Since volume is constant, \(L_1 A_1 = L_2 A_2\).

\(L_1 A_1 = (2L_1) A_2 \implies A_2 = \frac{A_1}{2}\).


The new resistance \(R_2\) is given by:

\(R_2 = \rho \frac{L_2}{A_2} = \rho \frac{2L_1}{A_1/2} = 4 \left(\rho \frac{L_1}{A_1}\right)\).


Substituting \(R_1 = X\), we get \(R_2 = 4R_1 = 4X\).


We are given that the new resistance \(R_2 = 40 \, \Omega\).

\(40 = 4X\).

\(X = \frac{40}{4} = 10 \, \Omega\).


The original resistance was 10 \(\Omega\).
Quick Tip: For problems involving stretching a wire, remember that resistance is proportional to the square of the length (\(R \propto L^2\)) because volume (\(AL\)) is constant. If length is doubled, resistance becomes \(2^2=4\) times.

The magnitude of the magnetic field at the center of a circular coil with N turns is given by the formula:

\(B = \frac{\mu_0 N I}{2r}\)


Given values are:


Number of turns, \(N = 200\).


Current, \(I = 0.40\) A.


Radius, \(r = 4.0\) cm \(= 0.04\) m.


Permeability of free space, \(\mu_0 = 4\pi \times 10^{-7}\) T m/A.


Substituting these values into the formula:

\(B = \frac{(4\pi \times 10^{-7}) \times 200 \times 0.40}{2 \times 0.04} = \frac{4\pi \times 10^{-7} \times 80}{0.08}\).

\(B = \frac{320\pi \times 10^{-7}}{8 \times 10^{-2}} = 40\pi \times 10^{-5} = 4\pi \times 10^{-4}\) T.

\(B \approx 4 \times 3.14159 \times 10^{-4} \approx 12.57 \times 10^{-4}\) T, or \(1.26 \times 10^{-3}\) T.


For the direction, we use the Right-Hand Thumb Rule.


The coil is in a horizontal plane with the current flowing in a clockwise direction (when viewed from above).


If we curl the fingers of our right hand in the direction of the current (clockwise), our thumb points downwards.


Therefore, the direction of the magnetic field at the center is vertically downwards (into the plane of the coil).
Quick Tip: The Right-Hand Thumb Rule is essential for finding the direction of the magnetic field. For a current loop, curl your fingers in the direction of the current; your thumb points in the direction of the magnetic field inside the loop.

When free protons and neutrons (nucleons) come together to form a stable nucleus, they are bound by the strong nuclear force.


This binding process releases a certain amount of energy, which is known as the binding energy of the nucleus.


According to Albert Einstein's mass-energy equivalence principle, \(E = mc^2\), energy and mass are interconvertible.


The release of binding energy corresponds to a decrease in the total mass of the system.


This difference between the sum of the masses of the individual free nucleons and the mass of the resulting nucleus is called the mass defect (\(\Delta m\)).


Therefore, the mass of a nucleus is always less than the sum of the masses of its constituents because some of their mass has been converted into binding energy to hold the nucleus together.
Quick Tip: Think of binding energy as "negative energy". To break a nucleus apart into its constituent nucleons, you must supply energy equal to the binding energy. This supplied energy manifests as an increase in mass, making the free nucleons heavier than the nucleus.

The Coulomb repulsion between the positively charged protons within a nucleus is overcome by the strong nuclear force.


Explanation:


1. Strength: The strong nuclear force is the most powerful of the four fundamental forces of nature, but it acts only over very short distances. At the typical separation of nucleons inside a nucleus (around \(10^{-15}\) m), it is much stronger than the electrostatic repulsive force between protons.


2. Short Range: This force is effective only within the confines of the nucleus. Its strength drops off very rapidly at distances greater than a few femtometers, which is why its effects are not felt outside the nucleus.


3. Charge Independence: The strong nuclear force acts equally between proton-proton, neutron-neutron, and proton-neutron pairs. The presence of neutrons adds to the total attractive nuclear force without contributing to the electric repulsion, thus helping to stabilize the nucleus.


In summary, the powerful, short-range, attractive strong nuclear force binds all nucleons together, overwhelming the long-range Coulomb repulsion between protons.
Quick Tip: A simple analogy: The strong nuclear force is like a very strong but short-range glue, while the Coulomb force is like a weaker but long-range repulsion. Within the nucleus, the "glue" wins, holding everything together.

According to Einstein's photoelectric equation, the maximum kinetic energy (\(K_{max}\)) of an emitted photoelectron is given by:

\(K_{max} = h\nu - \phi_0\)


where \(h\) 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 \(\phi_0 = h\nu_0\).


So, the equation becomes \(K_{max} = h\nu - h\nu_0 = h(\nu - \nu_0)\).


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 charge of an electron.


Combining these equations, we get: \(eV_s = h(\nu - \nu_0)\).

\(V_s = \frac{h(\nu - \nu_0)}{e}\).


Given values:

\(\nu = 6.8 \times 10^{14}\) Hz.

\(\nu_0 = 3.6 \times 10^{14}\) Hz.

\(h = 6.63 \times 10^{-34}\) J s.

\(e = 1.6 \times 10^{-19}\) C.

\(V_s = \frac{6.63 \times 10^{-34} \times (6.8 \times 10^{14} - 3.6 \times 10^{14})}{1.6 \times 10^{-19}}\).

\(V_s = \frac{6.63 \times 10^{-34} \times (3.2 \times 10^{14})}{1.6 \times 10^{-19}}\).

\(V_s = \frac{6.63 \times 3.2}{1.6} \times 10^{-34+14+19} = (6.63 \times 2) \times 10^{-1}\).

\(V_s = 13.26 \times 10^{-1} = 1.326\) V.


The cut-off potential is approximately 1.33 V.
Quick Tip: In photoelectric effect calculations, it's often easier to first calculate the energy of the incident photon (\(E = h\nu\)) and the work function (\(\phi_0 = h\nu_0\)) in electron-volts (eV) if possible. Then, \(K_{max} (eV) = E (eV) - \phi_0 (eV)\), and the stopping potential in volts is numerically equal to \(K_{max}\) in eV.

Let the initial charges on spheres A and B be \(Q_A\) and \(Q_B\).


Charge is given by density times surface area (\(Q = \sigma \times 4\pi r^2\)).

\(Q_A = (2\sigma) \times 4\pi R^2 = 8\pi\sigma R^2\).

\(Q_B = (3\sigma) \times 4\pi (2R)^2 = 3\sigma \times 16\pi R^2 = 48\pi\sigma R^2\).


When connected by a wire, charge flows until they reach a common potential, V.


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


Let the final charges be \(Q'_A\) and \(Q'_B\). So, \(Q'_A + Q'_B = 56\pi\sigma R^2\).


At common potential: \(V_A = V_B \implies \frac{kQ'_A}{R} = \frac{kQ'_B}{2R} \implies Q'_A = \frac{Q'_B}{2}\) or \(Q'_B = 2Q'_A\).


Substitute this into the conservation of charge equation:

\(Q'_A + 2Q'_A = 56\pi\sigma R^2 \implies 3Q'_A = 56\pi\sigma R^2 \implies Q'_A = \frac{56}{3}\pi\sigma R^2\).


Then, \(Q'_B = 2Q'_A = \frac{112}{3}\pi\sigma R^2\).


Now, find the new charge densities, \(\sigma'_A\) and \(\sigma'_B\).

\(\sigma'_A = \frac{Q'_A}{4\pi R^2} = \frac{(56/3)\pi\sigma R^2}{4\pi R^2} = \frac{14}{3}\sigma\).

\(\sigma'_B = \frac{Q'_B}{4\pi (2R)^2} = \frac{(112/3)\pi\sigma R^2}{16\pi R^2} = \frac{7}{3}\sigma\).
Quick Tip: When two conductors are connected, they reach a common potential. For spheres, this means \(Q_1/r_1 = Q_2/r_2\). Also remember that charge density is inversely proportional to radius for connected spheres (\(\sigma \propto 1/r\)).

The electric field due to an infinitely long straight wire is \(E = \frac{\lambda}{2\pi\epsilon_0 r}\).


Let the point P where the net field is zero be at a distance \(x\) from wire '1'.


Region A (to the left of wire 1): Let \(x < 0\). The distance is \(|x|\). Fields are in opposite directions. For the net field to be zero, \(|E_1| = |E_2|\).
\(\frac{\lambda}{2\pi\epsilon_0 |x|} = \frac{\lambda/2}{2\pi\epsilon_0 (d+|x|)} \implies \frac{1}{|x|} = \frac{1}{2(d+|x|)} \implies 2d + 2|x| = |x| \implies |x| = -2d\). This is not possible as distance cannot be negative.


Region B (between the wires): Let \(0 < x < d\). The field from wire 1 points right. The field from wire 2 (negative charge) also points right. The fields add up and can never be zero.


Region C (to the right of wire 2): Let \(x > d\). The field from wire 1 points right. The field from wire 2 points left. For the net field to be zero, \(|E_1| = |E_2|\).
\(\frac{\lambda}{2\pi\epsilon_0 x} = \frac{\lambda/2}{2\pi\epsilon_0 (x-d)}\).

\(\frac{1}{x} = \frac{1}{2(x-d)}\).

\(2x - 2d = x \implies x = 2d\).


This position is valid as it lies in Region C (\(x=2d > d\)).


Therefore, the point of zero electric field lies in Region C, at a distance of \(2d\) from wire '1'.
Quick Tip: When finding a point of zero electric field for two line charges of opposite signs, the point will always be outside the region between them and closer to the charge with the smaller magnitude.

Energy-band diagrams at T = 0 K:


Conductors: The valence band (VB) and the conduction band (CB) overlap. There is no forbidden energy gap, allowing electrons to move freely.

[Diagram showing overlapping VB and CB]


Insulators: There is a large forbidden energy gap (\(E_g > 3\) eV) between the fully occupied valence band and the empty conduction band.

[Diagram showing a full VB, empty CB, and a large \(E_g\)]


Semiconductors: There is a small forbidden energy gap (\(E_g < 3\) eV) between the filled valence band and the empty conduction band.

[Diagram showing a full VB, empty CB, and a small \(E_g\)]


Formation of an electron-hole pair:


At room temperature (T \(>\) 0 K), some electrons in the valence band of a semiconductor gain sufficient thermal energy to overcome the small forbidden energy gap.


These electrons jump from the valence band to the conduction band.


When an electron leaves the valence band, it creates a vacancy or a "hole" in its place.


This combination of a free electron in the conduction band and a hole in the valence band is called an electron-hole pair.
Quick Tip: The key difference between insulators and semiconductors is the size of the energy gap (\(E_g\)). For insulators, \(E_g\) is too large for thermal energy at room temperature to create charge carriers, while for semiconductors, it is small enough.

The electrical behavior of a material is primarily determined by its forbidden energy gap (\(E_g\)).


Both Carbon (in its diamond form) and Silicon are group IV elements and have a crystal structure where each atom forms four covalent bonds.


The energy gap for Carbon (diamond) is approximately 5.5 eV.


The energy gap for Silicon is approximately 1.1 eV.


At room temperature, the available thermal energy (of the order of 0.025 eV) is insufficient to excite electrons from the valence band to the conduction band across the large 5.5 eV gap in Carbon. Therefore, Carbon acts as an insulator.


In Silicon, the energy gap of 1.1 eV is much smaller. Thermal energy at room temperature is sufficient to excite a significant number of electrons from the valence band to the conduction band, creating electron-hole pairs and allowing for electrical conduction. Therefore, Silicon behaves as a semiconductor.
Quick Tip: Remember that the conductivity of semiconductors increases with temperature, while for conductors it decreases. This is because higher temperature creates more charge carriers (electron-hole pairs) in semiconductors.

Initial state: Capacitance \(C = \frac{\epsilon_0 A}{d}\), Voltage = V, Charge \(Q = CV = \frac{\epsilon_0 A V}{d}\).


When the battery is disconnected, the charge Q on the plates remains constant.


Final state: Separation = \(2d\). New capacitance \(C' = \frac{\epsilon_0 A}{2d} = \frac{C}{2}\).


(a) Potential difference between the plates (\(V'\)):
\(Q = C'V' \implies V' = \frac{Q}{C'} = \frac{CV}{C/2} = 2V\).

The new potential difference is twice the original.


(b) Electric field between the plates (\(E'\)):

The electric field between the plates is given by \(E = \frac{\sigma}{\epsilon_0} = \frac{Q}{A\epsilon_0}\).

Since Q, A, and \(\epsilon_0\) are constant, the electric field remains unchanged.
\(E' = E = \frac{V}{d}\).

Alternatively, \(E' = \frac{V'}{2d} = \frac{2V}{2d} = \frac{V}{d}\).


(c) Work done in pulling the plates apart (W):

Work done is equal to the change in the potential energy stored in the capacitor.
\(W = U_{final} - U_{initial} = \frac{Q^2}{2C'} - \frac{Q^2}{2C}\).
\(W = \frac{Q^2}{2(C/2)} - \frac{Q^2}{2C} = \frac{Q^2}{C} - \frac{Q^2}{2C} = \frac{Q^2}{2C}\).

Substituting \(Q=CV\), we get \(W = \frac{(CV)^2}{2C} = \frac{1}{2}CV^2\).
Quick Tip: The key to this type of problem is to identify what stays constant. If the battery is disconnected, charge (Q) is constant. If the battery remains connected, voltage (V) is constant. Use the energy formula involving the constant quantity (\(U=Q^2/2C\) for constant Q, \(U=\frac{1}{2}CV^2\) for constant V).

Description:

Let XY be a plane reflecting surface and AB be a plane wavefront incident at an angle \(i\).

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

Let the time taken for the disturbance to travel from B to C on the surface be \(t\). Thus, \(BC = vt\), where \(v\) is the speed of the wave.

In this time \(t\), the secondary wavelet from point A will travel a distance \(AD = vt\) in the same medium. We draw a sphere of radius \(AD\) with A as the center.

The new reflected wavefront is the common tangent CD to all such spheres.

[A diagram showing incident wavefront AB, reflecting surface XY, reflected wavefront CD, and relevant angles and distances is required here.]


Proof of Laws of Reflection:

In triangles \(\triangle ABC\) and \(\triangle ADC\):

1. \(AD = BC = vt\) (distance traveled by wavelets in same time t).

2. \(AC\) is the common side.

3. \(\angle ABC = \angle ADC = 90^\circ\) (wavefront is perpendicular to the direction of propagation).


Therefore, by RHS congruence, \(\triangle ABC \cong \triangle ADC\).

By CPCTC (Corresponding Parts of Congruent Triangles are Congruent):
\(\angle BAC = \angle DCA\).

From the diagram, the angle of incidence \(i = \angle BAC\).

And the angle of reflection \(r = \angle DCA\).

Hence, \(\angle i = \angle r\). This is the first law of reflection.


Also, the incident wavefront (AB), the reflected wavefront (CD), and the reflecting surface (XY) are all perpendicular to the plane of the paper. This implies that the incident ray, the reflected ray, and the normal all lie in the same plane. This is the second law of reflection.
Quick Tip: The key to the geometrical proof using Huygens' principle is to show the congruence of the two right-angled triangles formed by the incident and reflected wavefronts with the reflecting surface.

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


Light travels from the glass (where the object is) to the air (where the observer is).

So, \(n_1 = 1.5\) (glass) and \(n_2 = 1\) (air).


The object is the air bubble at P. The distance is measured from the surface B.

Object distance, \(u = -(Radius - CP) = -(7 - 1.75) = -5.25\) cm.


The surface is convex towards the object, so its center of curvature is on the left. Measured from B, the radius of curvature is negative.

Radius of curvature, \(R = -7\) cm.


Substitute the values into the formula:
\(\frac{1}{v} - \frac{1.5}{-5.25} = \frac{1 - 1.5}{-7}\).

\(\frac{1}{v} + \frac{1.5}{5.25} = \frac{-0.5}{-7} = \frac{1}{14}\).


Since \(5.25 = 1.5 \times 3.5\), the fraction becomes \(\frac{1}{3.5} = \frac{2}{7}\).

\(\frac{1}{v} + \frac{2}{7} = \frac{1}{14}\).

\(\frac{1}{v} = \frac{1}{14} - \frac{2}{7} = \frac{1 - 4}{14} = \frac{-3}{14}\).

\(v = -\frac{14}{3} \approx -4.67\) cm.


Nature: The negative sign for \(v\) indicates that the image is formed on the same side as the object (inside the glass sphere). Therefore, the image is virtual.

Position: The image is formed at a distance of 4.67 cm from the surface B, inside the glass ball.


Ray Diagram:

[A ray diagram must be drawn showing the spherical glass ball, the center C, the bubble P. Two rays diverging from P travel towards surface B. They refract away from the normal at surface B. When extended backward, they appear to meet at the virtual image point P' between C and P.]




% QuickTipBox
\begin{quicktipbox
For refraction at a spherical surface, correctly identifying \(n_1\) (medium of object) and \(n_2\) (medium of observer) is the first crucial step. The second is applying the sign convention correctly for \(u\) and \(R\) with respect to the point of incidence.
\end{quicktipbox Quick Tip: For refraction at a spherical surface, correctly identifying \(n_1\) (medium of object) and \(n_2\) (medium of observer) is the first crucial step. The second is applying the sign convention correctly for \(u\) and \(R\) with respect to the point of incidence.

Consider a long straight wire carrying a current \(I\). We want to find the magnetic field \(\vec{B}\) at a point P, at a perpendicular distance \(r\) from the wire.


Due to the symmetry of the wire, the magnetic field lines are concentric circles around the wire.


We choose a circular Amperian loop of radius \(r\), centered on the wire and passing through P.


According to Ampere's Circuital Law: \(\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}\).


Here, \(I_{enc}\) is the net current enclosed by the loop, which is \(I\).


At every point on the Amperian loop, the magnetic field vector \(\vec{B}\) is tangential to the circle, and so is the length element vector \(d\vec{l}\).


Therefore, \(\vec{B}\) and \(d\vec{l}\) are parallel, and the angle between them is \(0^\circ\).

\(\vec{B} \cdot d\vec{l} = B \, dl \cos(0^\circ) = B \, dl\).


Also, by symmetry, the magnitude of the magnetic field \(B\) is constant at all points on the circular loop.


The line integral becomes: \(\oint B \, dl = B \oint dl\).


The integral \(\oint dl\) is the total length of the loop, which is its circumference, \(2\pi r\).


So, \(B (2\pi r) = \mu_0 I\).

\(B = \frac{\mu_0 I}{2\pi r}\).


This is the expression for the magnetic field due to a long straight current-carrying wire.




% QuickTipBox
\begin{quicktipbox
Ampere's law is most useful for calculating magnetic fields in situations with high symmetry (infinite wire, infinite solenoid, toroid), where the line integral \(\oint \vec{B} \cdot d\vec{l}\) can be easily evaluated.
\end{quicktipbox Quick Tip: Ampere's law is most useful for calculating magnetic fields in situations with high symmetry (infinite wire, infinite solenoid, toroid), where the line integral \(\oint \vec{B} \cdot d\vec{l}\) can be easily evaluated.

Ampere's law is used for the derivation due to the high degree of cylindrical symmetry of the long straight wire.


Explanation:


1. Symmetry Advantage: Ampere's law, \(\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}\), is an integral law. For highly symmetric current distributions, we can choose an Amperian loop where the magnitude of \(\vec{B}\) is constant and its direction is parallel (or perpendicular) to \(d\vec{l}\). This makes the line integral very easy to calculate, turning it into a simple multiplication (\(B \times length\)).


2. Complexity of Biot-Savart's Law: The Biot-Savart law, \(d\vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{l} \times \hat{r}}{r^2}\), is a differential law and is more fundamental. However, using it requires integrating a vector cross product over the entire length of the wire (from \(-\infty\) to \(+\infty\)), which is a much more complex mathematical procedure than the algebraic simplification offered by Ampere's law in this symmetric case.


In summary, while both laws can derive the result, Ampere's law provides a much simpler and more elegant solution for a long straight wire because its symmetry perfectly matches the conditions where the law is most powerful.




% QuickTipBox
\begin{quicktipbox
Think of Ampere's law as the magnetic equivalent of Gauss's law for electric fields. Both are powerful tools that simplify calculations immensely in situations of high symmetry (spherical, cylindrical, or planar).
\end{quicktipbox Quick Tip: Think of Ampere's law as the magnetic equivalent of Gauss's law for electric fields. Both are powerful tools that simplify calculations immensely in situations of high symmetry (spherical, cylindrical, or planar).

Differentiation:

\begin{tabular{|l|l|l|
\hline
Parameter & Half-Wave Rectifier & Full-Wave Rectifier

\hline
Conduction & Conducts only during one half of the AC input cycle. & Conducts during both halves of the AC input cycle.

\hline
Components & Uses a single diode. & Uses two diodes (center-tapped) or four diodes (bridge).

\hline
Output & Pulsating DC with high ripple factor. & Smoother pulsating DC with lower ripple factor.

\hline
Efficiency & Low (max 40.6%). & High (max 81.2%).

\hline
\end{tabular


Working of a Full-Wave Rectifier (Center-Tapped):


Circuit Diagram:

[A diagram showing an AC source, a transformer with a center-tapped secondary coil. Diode D1 is connected to the top of the secondary, D2 to the bottom. A load resistor R_L is connected between the common point of the diodes and the center tap.]


Working Principle:


1. During the positive half-cycle of AC input: The top end (A) of the secondary coil is positive with respect to the center tap (C), and the bottom end (B) is negative. Diode D1 becomes forward-biased and conducts. Diode D2 is reverse-biased and does not conduct. Current flows through D1, the load resistor R_L (from P to C), and back to the source. A positive voltage appears across the load.


2. During the negative half-cycle of AC input: The top end (A) becomes negative, and the bottom end (B) becomes positive with respect to the center tap (C). Diode D2 is now forward-biased and conducts, while D1 is reverse-biased. Current flows through D2, the load resistor R_L (from P to C), and back to the source.


In both half-cycles, the current flows through the load resistor R_L in the same direction. This results in a unidirectional, pulsating DC voltage across the load for both halves of the input AC cycle.


[Input and Output Waveforms should be drawn, showing a sine wave input and a pulsating DC output with only positive peaks.]




% QuickTipBox
\begin{quicktipbox
The key to a full-wave rectifier is that it inverts the negative half-cycles of the AC input to produce a continuous series of positive pulses. This makes the output DC smoother and the rectification process more efficient.
\end{quicktipbox Quick Tip: The key to a full-wave rectifier is that it inverts the negative half-cycles of the AC input to produce a continuous series of positive pulses. This makes the output DC smoother and the rectification process more efficient.

The intensity of incident light is directly proportional to the number of photons striking the metal surface per unit time.


In the photoelectric effect, one incident photon (with energy greater than the work function) ejects one electron.


Therefore, the number of photoelectrons ejected per unit time is directly proportional to the number of incident photons per unit time.


The photoelectric current is the rate of flow of these ejected photoelectrons.


Thus, the photoelectric current is directly proportional to the intensity of the incident light.


A direct proportionality relationship is represented by a straight line passing through the origin. This corresponds to graph (C).
Quick Tip: Remember the core concepts of the photoelectric effect: Intensity determines the number of photoelectrons (current), while Frequency determines the kinetic energy of each photoelectron (stopping potential).

Saturation current is achieved when all the photoelectrons emitted from the cathode reach the anode.


The magnitude of the saturation current depends on the number of photoelectrons emitted per second.


The number of photoelectrons emitted per second is determined by the number of photons incident per second.


The number of incident photons per second is a measure of the intensity of the light.


The question states that the intensity is kept constant.


Therefore, the number of photons incident per second does not change, and the number of emitted photoelectrons per second also does not change.


Consequently, the saturation current remains the same.
Quick Tip: Increasing the frequency of light (while keeping intensity constant) increases the energy of each photon. This leads to a higher maximum kinetic energy for the photoelectrons and a larger stopping potential, but it does not change the number of electrons emitted per second.

Einstein's photoelectric equation is \(K_{max} = h\nu - \phi_0\).


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


Substituting this into the equation, we get: \(eV_s = h\nu - \phi_0\).


Rearranging for \(V_s\): \(V_s = \left(\frac{h}{e}\right)\nu - \frac{\phi_0}{e}\).


This equation is in the form of a straight line, \(y = mx + c\), where:
\(y = V_s\) (cut-off potential)
\(x = \nu\) (frequency)

The slope \(m = \frac{h}{e}\).


By plotting a graph of Cut-off potential (\(V_s\)) versus Frequency (\(\nu\)), we get a straight line.


The slope of this line can be measured experimentally. Since the charge of an electron, \(e\), is a known constant, Planck's constant, \(h\), can be calculated from the slope (\(h = e \times slope\)).
Quick Tip: The slope of the \(V_s\) vs \(\nu\) graph is \(h/e\) and is constant for all metals. The y-intercept is \(-\phi_0/e\) and the x-intercept is the threshold frequency \(\nu_0\). These features of the graph are frequently tested.

From Einstein's photoelectric equation, the maximum kinetic energy of photoelectrons is \(K_{max} = h\nu - \phi_0\).


For a given metal surface, 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 order of frequencies for the colors of the visible spectrum is:

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


Since \(K_{max}\) increases with frequency, the order of the maximum kinetic energies will be the same as the order of frequencies.
\(K_B > K_Y > K_R\).
Quick Tip: Remember the visible spectrum order (VIBGYOR) from violet to red. Frequency and energy decrease as you go from violet to red, while wavelength increases.

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


Visible light has a wavelength range of approximately 400 nm to 750 nm, which corresponds to an energy range of about 3.1 eV to 1.65 eV.


A metal will exhibit the photoelectric effect with visible light if its work function is less than the maximum energy of a visible light photon (approx. 3.1 eV).


Let's compare the work functions of the given metals:

- Caesium: \(\phi_0 \approx 2.14\) eV.

- Zinc: \(\phi_0 \approx 4.3\) eV.

- Cadmium: \(\phi_0 \approx 4.08\) eV.

- Magnesium: \(\phi_0 \approx 3.66\) eV.


Only Caesium has a work function (2.14 eV) that is low enough to be overcome by the energy of photons in the visible light spectrum. The work functions of Zinc, Cadmium, and Magnesium are too high; they require ultraviolet light to exhibit the photoelectric effect.
Quick Tip: Alkali metals (like Sodium, Potassium, Caesium) are known for their very low work functions, which is why they are often used in photodetectors designed for visible light.

Current sensitivity (\(S_i\)) is defined as the deflection produced per unit current flowing through the galvanometer.

\(S_i = \frac{\phi}{I}\).


From the equilibrium equation given in the passage: \(NBAI = k\phi\).


To find the expression for \(\frac{\phi}{I}\), we rearrange this equation.

\(\frac{\phi}{I} = \frac{NBA}{k}\).


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

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


Given:

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

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

Desired range of ammeter, \(I = 5\) A.


The current must split between the galvanometer and the shunt. The current through the shunt is \(I_s = I - I_g\).
\(I_s = 5 - 0.2 = 4.8\) A.


Since the galvanometer and shunt are in parallel, the potential difference across them must be equal.
\(V_g = V_s \implies I_g G = I_s S\).


Solving for the shunt resistance S:
\(S = \frac{I_g G}{I_s} = \frac{0.2 \times 6}{4.8}\).
\(S = \frac{1.2}{4.8} = \frac{1}{4} = 0.25 \, \Omega\).
Quick Tip: The shunt formula is \(S = \frac{I_g G}{I - I_g}\). A shunt is always a very small resistance connected in parallel to divert most of the current away from the sensitive galvanometer.

The ammeter consists of the galvanometer (G) and the shunt resistor (S) connected in parallel.


The total resistance of the ammeter (\(R_A\)) is the equivalent resistance of this parallel combination.


From the previous part, we have:

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

Shunt resistance, \(S = 0.25 \, \Omega\).


The formula for equivalent resistance in parallel is \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\).
\(\frac{1}{R_A} = \frac{1}{G} + \frac{1}{S} = \frac{1}{6} + \frac{1}{0.25}\).

\(\frac{1}{R_A} = \frac{1}{6} + \frac{1}{1/4} = \frac{1}{6} + 4\).

\(\frac{1}{R_A} = \frac{1 + 24}{6} = \frac{25}{6}\).

\(R_A = \frac{6}{25} = 0.24 \, \Omega\).
Quick Tip: The resistance of an ideal ammeter is zero. In practice, the resistance of a real ammeter is very low, as it's the parallel combination of the galvanometer and an even smaller shunt resistance. The equivalent resistance will always be smaller than the smallest individual resistance.

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

Let the galvanometer resistance be G and the full-scale deflection current be \(I_g\).


Case 1: Range is V, series resistance is \(R_1\).

According to Ohm's law, \(V = I_g (G + R_1)\). (Equation 1)


Case 2: Range is 2V, series resistance is \(R_2\).

Similarly, \(2V = I_g (G + R_2)\). (Equation 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: The formula for a voltmeter is \(V = I_g(G+R_{series})\). The key idea is that the series resistor drops most of the voltage, allowing only a small, safe current (\(I_g\)) to flow through the galvanometer.

The deflecting torque (\(\tau\)) on a coil in a magnetic field is given by \(\tau = NIAB \sin\theta\).


For a galvanometer with a radial magnetic field, the plane of the coil is always parallel to the magnetic field, which means the normal to the coil's area is always perpendicular to the field. So, \(\theta = 90^\circ\) and \(\sin\theta = 1\).


The formula simplifies to \(\tau = NIAB\).


Given values (converted to SI units):
\(N = 100\) turns.
\(I = 5\) mA \(= 5 \times 10^{-3}\) A.
\(A = 18\) cm\(^2 = 18 \times (10^{-2} m)^2 = 18 \times 10^{-4}\) m\(^2\).
\(B = 0.20\) T.


Now, calculate the torque:
\(\tau = (100) \times (5 \times 10^{-3}) \times (18 \times 10^{-4}) \times (0.20)\).

\(\tau = (100 \times 5 \times 18 \times 0.20) \times 10^{-3} \times 10^{-4}\).

\(\tau = (500 \times 18 \times 0.20) \times 10^{-7}\).

\(\tau = (9000 \times 0.20) \times 10^{-7} = 1800 \times 10^{-7}\).

\(\tau = 1.8 \times 10^3 \times 10^{-7} = 1.8 \times 10^{-4}\) Nm.
Quick Tip: A radial magnetic field is a clever design in galvanometers to make the torque directly proportional to the current (\(\tau \propto I\)) because \(\sin\theta\) is always 1. This results in a linear scale for the instrument.

Definition of Self-Inductance:

Self-inductance is the property of a coil by virtue of which it opposes any change in the strength of current flowing through it by inducing an electromotive force (emf) in itself.

The magnetic flux (\(\Phi_B\)) linked with a coil is directly proportional to the current (I) flowing through it, i.e., \(\Phi_B \propto I\) or \(\Phi_B = LI\).

The self-inductance (L) is numerically equal to the induced emf (\(e\)) in the coil when the rate of change of current (\(\frac{dI}{dt}\)) through the coil is unity (\(e = -L\frac{dI}{dt}\)).


Derivation for Energy Stored:

To build up a current in an inductor, work must be done against the back emf induced in it.

The instantaneous induced emf is \(e = -L\frac{dI}{dt}\).

The work done (\(dW\)) in moving a small charge \(dq\) against this emf is \(dW = -e \, dq\).

Since \(I = \frac{dq}{dt}\), we have \(dq = I \, dt\).
\(dW = -(-L\frac{dI}{dt}) (I \, dt) = LI \, dI\).

The total work done in establishing the current from 0 to a final value I is found by integrating \(dW\).
\(W = \int dW = \int_{0}^{I} LI \, dI\).
\(W = L \int_{0}^{I} I \, dI = L \left[ \frac{I^2}{2} \right]_{0}^{I} = L \left( \frac{I^2}{2} - 0 \right)\).
\(W = \frac{1}{2}LI^2\).

This work done is stored in the inductor as magnetic potential energy (\(U_B\)).

Thus, \(U_B = \frac{1}{2}LI^2\).
Quick Tip: The formula for energy stored in an inductor, \(U_B = \frac{1}{2}LI^2\), is analogous to the energy stored in a capacitor, \(U_E = \frac{1}{2}CV^2 = \frac{Q^2}{2C}\), and the kinetic energy of a mass, \(K.E. = \frac{1}{2}mv^2\).

Let \(L_1 = 10\) mH and \(L_2 = 20\) mH.


(I) Magnitude of emf vs. rate of change of current:

The magnitude of the induced emf is given by \(|e| = L \frac{dI}{dt}\).

This is an equation of a straight line of the form \(y = mx\), where \(y = |e|\), \(x = \frac{dI}{dt}\), and the slope \(m = L\).

Since \(L_2 > L_1\), the slope of the graph for the 20 mH inductor will be steeper than that for the 10 mH inductor.

[Graph showing two straight lines passing through the origin. The y-axis is labeled \(|e|\) and the x-axis is labeled \(dI/dt\). The line for \(L_2=20\) mH has a greater slope than the line for \(L_1=10\) mH.]


(II) Energy stored vs. current:

The energy stored in an inductor is given by \(U = \frac{1}{2}LI^2\).

This is an equation of a parabola of the form \(y = ax^2\), where \(y = U\), \(x = I\), and the constant \(a = \frac{L}{2}\).

The graph is a parabola opening upwards, starting from the origin.

Since \(L_2 > L_1\), the parabola for the 20 mH inductor will be steeper (rise more quickly) than the one for the 10 mH inductor.

[Graph showing two parabolas starting from the origin and opening upwards. The y-axis is labeled U and the x-axis is labeled I. The parabola for \(L_2=20\) mH is steeper/narrower than the parabola for \(L_1=10\) mH.]
Quick Tip: When comparing graphs related to inductor properties, remember that both induced emf (vs rate of change of current) and stored energy (vs current squared) are directly proportional to the inductance L. A larger L will always result in a steeper graph.

Definition of Mutual Inductance:

Mutual inductance is the phenomenon in which a changing current in one coil induces an electromotive force (emf) in a neighboring coil.

The mutual inductance (M) of a pair of coils is numerically equal to the magnetic flux linked with one coil when a unit current flows through the other coil (\(\Phi_2 = MI_1\)).


Derivation for two long coaxial solenoids:

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

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

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

Assume a current \(I_2\) flows through the outer solenoid S2.

The magnetic field produced inside S2 is uniform and given by \(B_2 = \mu_0 n_2 I_2 = \mu_0 \frac{N_2}{l} I_2\).

This magnetic field is confined within S2. Therefore, the magnetic flux is linked with the inner solenoid S1.

The area of the inner solenoid is \(A_1 = \pi r_1^2\).

The magnetic flux through a single turn of the inner solenoid S1 is \(\Phi_{turn, 1} = B_2 A_1 = \left(\mu_0 \frac{N_2}{l} I_2\right) (\pi r_1^2)\).

The total magnetic flux linked with the inner solenoid S1 (which has \(N_1\) turns) is \(\Phi_1 = N_1 \Phi_{turn, 1}\).
\(\Phi_1 = N_1 \left(\mu_0 \frac{N_2}{l} I_2 \pi r_1^2\right) = \frac{\mu_0 N_1 N_2 \pi r_1^2}{l} I_2\).

By the definition of mutual inductance, \(\Phi_1 = M_{12} I_2\).

Comparing the two expressions for \(\Phi_1\), we get the mutual inductance M:
\(M = M_{12} = \frac{\mu_0 N_1 N_2 \pi r_1^2}{l} = \mu_0 n_1 n_2 (\pi r_1^2) l\).
Quick Tip: Note that the mutual inductance depends on the area of the inner coil (\(A_1\)) because the magnetic field from the outer coil is uniform inside it, but the flux linkage is limited by the smaller area.

First, we find the self-inductance (L) of the inductor.

The magnitude of the induced emf is given by \(e = L \left|\frac{dI}{dt}\right|\).

The rate of change of current 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 emf, \(e = 5\) mV \(= 5 \times 10^{-3}\) V.

From the emf formula, \(L = \frac{e}{dI/dt} = \frac{5 \times 10^{-3} V}{0.05 A/s} = 0.1\) H.


Next, we find the current in the inductor at \(t = 10\) s.

Since the current increases uniformly from 0, the current at any time \(t\) is given by \(I(t) = \left(\frac{dI}{dt}\right) \times t\).
\(I(t) = 0.05 \times t\).

At \(t = 10\) s, the current is \(I(10) = 0.05 \times 10 = 0.5\) A.


Finally, we find the magnetic flux (\(\Phi\)) linked with the inductor at this time.

The flux is given by the formula \(\Phi = L I\).
\(\Phi = (0.1 H) \times (0.5 A) = 0.05\) Wb.
Quick Tip: This problem has two parts. First use the induced emf and rate of change of current to find the inductance L. Then, use L and the instantaneous current at the specified time to find the flux.

Ray Diagram and Image Formation (Cassegrain Telescope):

[A neat, labeled ray diagram must be drawn. It should show parallel rays from a distant object incident on a large concave parabolic primary mirror. The reflected rays converge towards the primary focus. Before reaching the focus, they are intercepted by a small convex secondary mirror. This secondary mirror reflects the rays, which then pass through a hole in the center of the primary mirror to form a real, inverted image at the eyepiece.]

Explanation:

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

2. The primary mirror reflects these rays and directs them towards its principal focus.

3. Before the rays can converge at the focus, they are intercepted by a smaller convex secondary mirror.

4. The secondary mirror reflects the light back through a small hole at the center of the primary mirror.

5. A real and inverted image is formed, which is then viewed and magnified by an eyepiece.


Two Advantages over a Refracting Telescope:

1. No Chromatic Aberration: Reflecting telescopes use mirrors instead of lenses. Mirrors reflect all wavelengths of light at the same angle, so there is no dispersion of light and hence no chromatic aberration, resulting in sharper images.

2. Large Aperture and Light-Gathering Power: It is mechanically easier to construct and support a large-diameter mirror than a large-diameter lens. A larger aperture allows the telescope to collect more light, enabling the observation of fainter objects, and it also provides a higher resolving power.
Quick Tip: The key advantages of reflectors stem from using mirrors: freedom from chromatic aberration and the feasibility of building very large apertures for greater light-gathering and resolving power.

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


We are given that the focal length of the objective is 50 times that of the eyepiece:
\(f_o = 50 f_e\) --- (Equation 1)


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


We are given that \(L = 102\) cm.
\(f_o + f_e = 102\) --- (Equation 2)


Now we solve these two simultaneous equations. Substitute Equation 1 into Equation 2:
\((50 f_e) + f_e = 102\).
\(51 f_e = 102\).
\(f_e = \frac{102}{51} = 2\) cm.


Now substitute the value of \(f_e\) back into Equation 1 to find \(f_o\):
\(f_o = 50 \times f_e = 50 \times 2 = 100\) cm.


Therefore, the focal length of the objective lens is 100 cm and the focal length of the eyepiece is 2 cm.
Quick Tip: For telescopes, remember the two key setups: 1. Image at infinity (Normal Adjustment): Magnification \(m = f_o/f_e\), Tube length \(L = f_o + f_e\). 2. Image at near point (D): Magnification \(m = \frac{f_o}{f_e}(1 + \frac{f_e}{D})\), Tube length \(L = f_o + u_e\).

Two Advantages of a Compound Microscope:


1. Higher Magnification: A compound microscope uses two lenses (objective and eyepiece) to produce a two-stage magnification, resulting in a much larger overall magnification than a single-lens simple microscope.


2. Greater Resolving Power: The resolving power of a microscope is inversely proportional to the wavelength of light used and directly proportional to the numerical aperture of the objective. Compound microscopes can use objectives with high numerical apertures, providing better resolution (the ability to distinguish between two closely spaced points).


Ray Diagram and Image Formation (Image at Near Point):

[A neat, labeled ray diagram must be drawn. It should show a small object AB placed just beyond the focal point \(F_o\) of the objective lens. The objective forms a real, inverted, and magnified image A'B'. This image A'B' is formed within the focal length \(F_e\) of the eyepiece. The eyepiece acts as a simple magnifier and forms a final, virtual, inverted (with respect to the object), and highly magnified image A''B'' at the near point, D.]


Explanation:

1. The objective lens, which has a short focal length, is placed near the object. The object is positioned just outside the focal point of the objective.

2. The objective lens forms a real, inverted, and magnified intermediate image (A'B').

3. This intermediate image acts as the object for the eyepiece, which has a slightly larger focal length.

4. The eyepiece is adjusted so that the intermediate image falls within its focal length.

5. The eyepiece then acts like a simple microscope, forming a final, virtual, and highly magnified image (A''B'') at the near point of the eye for comfortable viewing.
Quick Tip: In a compound microscope, the objective lens provides the initial magnification (real image), and the eyepiece provides the final magnification (virtual image). The total magnification is the product of the two: \(M = m_o \times m_e\).

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

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


For the planoconcave lens (Lens 1):

Refractive index = \(n_1\).

For the plane surface, \(R_1 = \infty\).

For the concave surface, the radius is R. By sign convention (light travels from left to right), \(R_2 = -R\).

Power \(P_1 = (n_1 - 1) \left(\frac{1}{\infty} - \frac{1}{-R}\right) = (n_1 - 1) \left(0 + \frac{1}{R}\right) = \frac{n_1 - 1}{R}\).


For the equiconvex lens (Lens 2):

Refractive index = \(n_2\).

For an equiconvex lens, the radii are equal in magnitude.

For the first surface, \(R_1 = +R\).

For the second surface, \(R_2 = -R\).

Power \(P_2 = (n_2 - 1) \left(\frac{1}{R} - \frac{1}{-R}\right) = (n_2 - 1) \left(\frac{1}{R} + \frac{1}{R}\right) = (n_2 - 1) \left(\frac{2}{R}\right) = \frac{2(n_2 - 1)}{R}\).


For the combination:

When lenses are in contact, the total power is the algebraic sum of individual powers.
\(P_{comb} = P_1 + P_2\).
\(P_{comb} = \frac{n_1 - 1}{R} + \frac{2(n_2 - 1)}{R}\).
\(P_{comb} = \frac{(n_1 - 1) + (2n_2 - 2)}{R}\).
\(P_{comb} = \frac{n_1 + 2n_2 - 3}{R}\).
Quick Tip: Mastering the sign convention is crucial for the Lens Maker's formula. A simple rule: measure all distances from the optical center. Radii of curvature are positive if the center of curvature is on the side where light emerges, and negative if it's on the side where light is incident.

Let the currents flowing from batteries \(E_1, E_2, E_3\) be \(I_1, I_2, I_3\) respectively. Let's assume their directions are downwards. Let the top junction be A and the bottom junction be B.


Apply Kirchhoff's Current Law (KCL) at junction A:

Assuming currents are leaving the junction, \(I_1 + I_2 + I_3 = 0\). (This is incorrect, let's assume I1 and I3 flow into A, and I2 flows out)
Let's redefine: Let \(I_1\) be current from \(E_1\), \(I_2\) from \(E_2\), and \(I_3\) from \(E_3\). Let's assume their directions are all downwards.

KCL at junction B (bottom): \(I_1 + I_2 + I_3 = 0\). (Equation 1)


Apply Kirchhoff's Voltage Law (KVL):

Loop 1 (left loop containing \(E_1\) and \(E_2\)): Starting from B and moving clockwise.
\(+E_1 - I_1 r_1 + I_2 r_2 - E_2 = 0\).
\(+4 - I_1(2) + I_2(4) - 2 = 0 \implies 2 - 2I_1 + 4I_2 = 0 \implies I_1 - 2I_2 = 1\). (Equation 2)


Loop 2 (right loop containing \(E_2\) and \(E_3\)): Starting from B and moving clockwise.
\(+E_2 - I_2 r_2 + I_3 r_3 - E_3 = 0\).
\(+2 - I_2(4) + I_3(2) - 6 = 0 \implies -4 - 4I_2 + 2I_3 = 0 \implies I_3 - 2I_2 = 2\). (Equation 3)


Now we solve the three equations.

From (2), \(I_1 = 1 + 2I_2\).

From (3), \(I_3 = 2 + 2I_2\).

Substitute these into (1):
\((1 + 2I_2) + I_2 + (2 + 2I_2) = 0\).
\(3 + 5I_2 = 0 \implies I_2 = -3/5 = -0.6\) A.


The negative sign means our assumed direction for \(I_2\) (downwards) was wrong. So, current through \(E_2\) is 0.6 A upwards.

Now find \(I_1\) and \(I_3\):
\(I_1 = 1 + 2(-0.6) = 1 - 1.2 = -0.2\) A.

Current through \(E_1\) is 0.2 A upwards.

\(I_3 = 2 + 2(-0.6) = 2 - 1.2 = 0.8\) A.

Current through \(E_3\) is 0.8 A downwards.


Summary of currents:

Through \(E_1\): 0.2 A (upwards)

Through \(E_2\): 0.6 A (upwards)

Through \(E_3\): 0.8 A (downwards)
Quick Tip: When applying Kirchhoff's laws, don't worry about getting the initial direction of the current wrong. The math will correct you. A negative answer for a current simply means it flows in the opposite direction to the one you initially assumed.

The given arrangement of six resistors forms a Wheatstone bridge circuit.

Let's label the junctions. Let A be the input and B be the output. Let the top junction be C and the bottom junction be D.

The circuit can be seen as having two arms between A and B, connected by a fifth resistor CD. A sixth resistor is also present, but the typical "six wire" problem of this shape is a Wheatstone bridge with one resistor between the arms. Let's analyze it as a bridge between points A and B with intermediate points C and D.


The resistors are connected as follows: A to C, A to D, C to B, D to B, and C to D. The problem shows a hexagon-like structure which simplifies to a bridge.

Let's check for balance in the Wheatstone bridge. The resistance of arm AC is R. The resistance of arm AD is R. The resistance of arm CB is R. The resistance of arm DB is R.


The condition for a balanced Wheatstone bridge is:
\(\frac{R_{AC}}{R_{AD}} = \frac{R_{CB}}{R_{DB}}\).

Substituting the values: \(\frac{R}{R} = \frac{R}{R} \implies 1 = 1\).


Since the condition is satisfied, the bridge is balanced.

This means that no current will flow through the resistor connected between junctions C and D.

Therefore, we can remove the resistor CD from the circuit for calculation purposes.


The simplified circuit now consists of two parallel branches:

1. Upper branch: Resistors AC and CB are in series. Their total resistance is \(R_{upper} = R + R = 2R\).

2. Lower branch: Resistors AD and DB are in series. Their total resistance is \(R_{lower} = R + R = 2R\).


These two branches are connected in parallel between points A and B. The effective resistance (\(R_{eff}\)) is:
\(\frac{1}{R_{eff}} = \frac{1}{R_{upper}} + \frac{1}{R_{lower}} = \frac{1}{2R} + \frac{1}{2R} = \frac{2}{2R} = \frac{1}{R}\).
\(R_{eff} = R\).


Given that \(R = 10 \, \Omega\).

The effective resistance is \(10 \, \Omega\).
Quick Tip: Whenever you see a complex resistor network, first try to identify if it can be redrawn as a Wheatstone bridge. If it can, immediately check for the balance condition (\(R_1/R_2 = R_3/R_4\)). If balanced, the problem becomes much simpler.

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

From Ohm's law in microscopic form, the electric field E is related to current density J and resistivity \(\rho\) (or conductivity \(\sigma\)) by \(E = \rho J\).


Current density is defined as \(J = \frac{I}{A}\), where I is the current and A is the cross-sectional area.

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


The current I and resistivity \(\rho\) are constant throughout the copper 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 points A and B is:
\(\frac{E_A}{E_B} = \frac{\rho I / A_A}{\rho I / A_B} = \frac{A_B}{A_A}\).


Given \(A_A = 1.0 \times 10^{-7}\) m\(^2\) and \(A_B = 2.0 \times 10^{-7}\) m\(^2\).
\(\frac{E_A}{E_B} = \frac{2.0 \times 10^{-7} m^2}{1.0 \times 10^{-7} m^2} = 2\).

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


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

The relationship between current I, electron density n, electron 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\(^2\).
\(v_{d,B} = \frac{1}{(8.5 \times 10^{28}) \times (1.6 \times 10^{-19}) \times (2.0 \times 10^{-7})}\).
\(v_{d,B} = \frac{1}{8.5 \times 1.6 \times 2.0 \times 10^{28-19-7}} = \frac{1}{27.2 \times 10^{2}}\).
\(v_{d,B} = \frac{1}{2720} \approx 3.676 \times 10^{-4}\) m/s.
Quick Tip: In a conductor of varying cross-section with a steady current: - Current (I) is constant everywhere. - Current density (\(J=I/A\)) is higher where the area is smaller. - Electric field (\(E=\rho J\)) is stronger where the area is smaller. - Drift velocity (\(v_d=I/neA\)) is faster where the area is smaller.

The net electric field at the point \(\vec{r}\) is the vector sum of the electric fields produced by each charge individually.
\(\vec{E}_{net} = \vec{E}_1 + \vec{E}_2\).


Electric field due to \(q_1\) at \(\vec{r}\):

The position vector from \(q_1\) to the point of interest is \(\vec{d_1} = \vec{r} - \vec{r_1}\).
\(\vec{d_1} = ((3\hat{i} + 4\hat{j}) - 3\hat{i}) m = (4\hat{j}) m\).

The magnitude of this vector is \(|\vec{d_1}| = 4\) m.

The electric field \(\vec{E}_1\) is given by \(\vec{E}_1 = k \frac{q_1}{|\vec{d_1}|^2} \hat{d_1} = k \frac{q_1}{|\vec{d_1}|^3} \vec{d_1}\).
\(\vec{E}_1 = (9 \times 10^9) \frac{16 \times 10^{-6}}{4^3} (4\hat{j}) = (9 \times 10^9) \frac{16 \times 10^{-6}}{64} (4\hat{j})\).
\(\vec{E}_1 = (9 \times 10^9) \times (0.25 \times 10^{-6}) \times (4\hat{j}) = (9 \times 10^3) \hat{j}\) N/C.


Electric field due to \(q_2\) at \(\vec{r}\):

The position vector from \(q_2\) to the point of interest is \(\vec{d_2} = \vec{r} - \vec{r_2}\).
\(\vec{d_2} = ((3\hat{i} + 4\hat{j}) - 4\hat{j}) m = (3\hat{i}) m\).

The magnitude of this vector is \(|\vec{d_2}| = 3\) m.

The electric field \(\vec{E}_2\) is given by \(\vec{E}_2 = k \frac{q_2}{|\vec{d_2}|^3} \vec{d_2}\).
\(\vec{E}_2 = (9 \times 10^9) \frac{1 \times 10^{-6}}{3^3} (3\hat{i}) = (9 \times 10^9) \frac{1 \times 10^{-6}}{27} (3\hat{i})\).
\(\vec{E}_2 = (9 \times 10^9) \times (\frac{1}{9} \times 10^{-6}) \hat{i} = (1 \times 10^3) \hat{i}\) 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.

Or, \(\vec{E}_{net} = (1000 \hat{i} + 9000 \hat{j})\) N/C.
Quick Tip: When dealing with electric fields in vector form, always calculate the displacement vector from the source charge to the point of observation (\(\vec{d} = \vec{r}_{obs} - \vec{r}_{source}\)). Then use the vector form of Coulomb's Law, \(\vec{E} = k \frac{q}{|\vec{d}|^3} \vec{d}\).