The charge is uniformly distributed along the circumference of a circular ring of radius \( R \) with charge density \( \lambda \). The electric field at the center of the ring is due to the contributions from all infinitesimal charge elements along the ring.

Each infinitesimal charge element produces an electric field, but due to symmetry, all the components of the electric field along the radial direction cancel out, leaving only the components along the axis of the ring. Since all the contributions add up symmetrically and cancel each other out in all directions, the net electric field at the center of the ring is:
\[ Electric Field = 0 \]

Thus, the magnitude of the electric field at the center of the ring is:
\[ \boxed{Zero}. \]


Correct Answer: (B) Zero Quick Tip: Due to the symmetry of the charge distribution on a circular ring, the electric field at its center cancels out. This is because the contributions of the electric field from each infinitesimal charge element cancel each other in the radial directions, leaving no net electric field at the center.

The electric field on the surface of a charged sphere is given by the formula:
\[ E = \frac{\sigma}{\epsilon_0} \]

Where \(\sigma\) is the surface charge density and \(\epsilon_0\) is the permittivity of free space. This expression shows that the electric field on the surface of the sphere is independent of the radius of the sphere, provided the charge density remains the same.



Thus, when the radius of the sphere is doubled, keeping the charge density \(\sigma\) the same, the electric field \(E\) remains unchanged. Therefore, the ratio of the electric field on the old sphere to that on the new sphere is:
\[ \frac{E_{old}}{E_{new}} = 1 \]

Thus, the correct answer is:
\[ \boxed{1}. \] Quick Tip: The electric field on the surface of a charged sphere depends only on the surface charge density \(\sigma\) and is independent of the radius of the sphere. Therefore, if the charge density remains the same, doubling the radius does not affect the electric field.

The student is asked to connect four cells, each with emf \(E\) and internal resistance \(r\), in series. However, one cell is connected incorrectly. This implies that three cells are connected in the normal way, while one cell is connected in reverse, which will cause a subtraction of the emf from the total emf.



Equivalent emf:
The emf of the three cells connected in the correct way is \(3E\). The emf of the incorrectly connected cell is \(-E\) (since it's connected in reverse). Thus, the total emf is:
\[ E_{total} = 3E - E = 2E \]



Equivalent internal resistance:
The internal resistance of each cell is \(r\), and since all the cells are connected in series, the total internal resistance is the sum of the individual resistances:
\[ R_{total} = 4r \]

Thus, the equivalent emf is \(2E\), and the equivalent internal resistance is \(4r\).



Correct Answer: (D) \(2E\) and \(4r\) Quick Tip: When cells are connected in series, the total emf is the sum of the emfs. If one cell is connected in reverse, its emf will be subtracted from the total. The internal resistance in series is the sum of the internal resistances of the individual cells.

We are given that a piece of wire is bent into a circular loop A carrying a current \(I\), and the wire is then bent into another circular loop B with two turns, carrying the same current \(I\). We need to find the ratio of the magnetic fields at the centre of loop A to that of loop B.

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

For loop A (a single turn):
\[ B_A = \frac{\mu_0 I}{2r_A} \]

For loop B (two turns), the magnetic field is twice that of a single turn because the number of turns is doubled:
\[ B_B = \frac{\mu_0 I}{2r_B} \times 2 = \frac{\mu_0 I}{r_B} \]

Now, the ratio of the magnetic fields at the center of loop A to that of loop B is:
\[ \frac{B_A}{B_B} = \frac{\frac{\mu_0 I}{2r_A}}{\frac{\mu_0 I}{r_B}} = \frac{r_B}{2r_A} \]

Since the wire is bent from the same piece, the total length of the wire is constant. Therefore, the lengths of the wire used in both loops are the same:
\[ 2\pi r_A = 2 \times 2\pi r_B \quad \Rightarrow \quad r_B = 2r_A \]

Substituting this into the ratio:
\[ \frac{B_A}{B_B} = \frac{2r_A}{2r_A} = \frac{1}{4} \]

Thus, the ratio of the magnetic field at the center of loop A to that at the center of loop B is:
\[ \boxed{\frac{1}{4}} \]


Correct Answer: (D) \(\frac{1}{4}\) Quick Tip: The magnetic field at the center of a circular loop is directly proportional to the current and inversely proportional to the radius. When the radius changes and the number of turns is modified, use these relationships to compute the new magnetic field.

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

\(I = 1.0 \, \mathrm{A}\) is the current,
\(\vec{L} = 0.1 \, \hat{j} \, \mathrm{m}\) is the length vector of the wire,
\(\vec{B} = (5 \, \mathrm{mT}) \, \hat{j} - (8 \, \mathrm{mT}) \, \hat{k}\) is the magnetic field.


The cross product \(\vec{L} \times \vec{B}\) is: \[ \vec{L} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
0 & 0.1 & 0
0 & 5 \times 10^{-3} & -8 \times 10^{-3} \end{vmatrix}. \]

Expanding the determinant: \[ \vec{L} \times \vec{B} = \hat{i} \left[ (0.1)(-8 \times 10^{-3}) - (0)(5 \times 10^{-3}) \right] - \hat{j}(0) + \hat{k}(0), \] \[ \vec{L} \times \vec{B} = \hat{i} \cdot (-8 \times 10^{-4}). \]

Thus: \[ \vec{F} = I \cdot \vec{L} \times \vec{B} = (1.0) \cdot (-8 \times 10^{-4}) \hat{i} = -8 \times 10^{-4} \, \hat{i} \, \mathrm{N}. \]

Converting to millinewtons: \[ \vec{F} = -0.8 \, \mathrm{mN} \, \hat{i}. \]

Thus, the force on the wire is: \[ \boxed{-0.8 \, \mathrm{mN} \, \hat{i}}. \] Quick Tip: To calculate the force on a current-carrying wire in a magnetic field, compute the cross product \(\vec{L} \times \vec{B}\) and multiply by the current. Use determinants for cross products in vector form.

To convert a galvanometer into an ammeter, a shunt resistance \(S\) is connected in parallel to the galvanometer. The current through the galvanometer \(I_g\) is given as: \[ I_g = \frac{0.1}{100} \cdot I = \frac{I}{1000}. \]

The current through the shunt is: \[ I_s = I - I_g = I - \frac{I}{1000} = \frac{999I}{1000}. \]

The potential difference across the galvanometer is equal to the potential difference across the shunt: \[ I_g G = I_s S. \]

Substituting \(I_g = \frac{I}{1000}\) and \(I_s = \frac{999I}{1000}\): \[ \frac{I}{1000} G = \frac{999I}{1000} S. \]

Cancelling \(I\) and simplifying: \[ S = \frac{G}{999}. \]

The equivalent resistance of the ammeter is: \[ R_{ammeter} = \frac{GS}{G + S}. \]

Since \(S \ll G\), the equivalent resistance is approximately: \[ R_{ammeter} = S = \frac{G}{1000}. \]

Thus, the resistance of the ammeter is: \[ \boxed{\frac{G}{1000} \, \Omega}. \] Quick Tip: For shunt resistance problems, use the relation \(I_g G = I_s S\) to find \(S\). The equivalent resistance of the ammeter is approximately the shunt resistance when \(S \ll G\).

The induced emf is given by Faraday’s law: \[ \mathcal{E} = -\frac{d\Phi_B}{dt}. \]

The magnetic flux through the loop is: \[ \Phi_B = B \cdot A = B \cdot \pi r^2, \]
where \(r\) is the radius of the loop.

Differentiating \(\Phi_B\) with respect to time: \[ \frac{d\Phi_B}{dt} = B \cdot \frac{d}{dt}(\pi r^2) = B \cdot 2\pi r \cdot \frac{dr}{dt}. \]

Substituting \(B = 50 \, \mathrm{mT} = 50 \times 10^{-3} \, \mathrm{T}\), \(r = 4 \, \mathrm{cm} = 4 \times 10^{-2} \, \mathrm{m}\), and \(\frac{dr}{dt} = -1 \, \mathrm{mm/s} = -1 \times 10^{-3} \, \mathrm{m/s}\): \[ \frac{d\Phi_B}{dt} = (50 \times 10^{-3}) \cdot 2\pi (4 \times 10^{-2}) \cdot (1 \times 10^{-3}). \]

Simplifying: \[ \frac{d\Phi_B}{dt} = 4\pi \times 10^{-6} \, \mathrm{V}. \]

Thus, the induced emf is: \[ \mathcal{E} = 4\pi \, \mu\mathrm{V}. \]
\[ \boxed{4\pi \, \mu\mathrm{V}}. \] Quick Tip: For induced emf calculations, differentiate the flux \(\Phi_B = B \cdot A\) with respect to time. For circular loops, \(A = \pi r^2\).

In electromagnetic waves:

The electric field \(\vec{E}\) and the magnetic field \(\vec{B}\) are perpendicular to each other.
Both \(\vec{E}\) and \(\vec{B}\) are also perpendicular to the direction of wave propagation \(\vec{k}\).
\(\vec{E}\) and \(\vec{B}\) oscillate in phase.


Thus, the fields are in the same phase and perpendicular to each other.
\[ \boxed{(A) In the same phase and perpendicular to each other.} \]
Quick Tip: Electromagnetic waves have \(\vec{E}\), \(\vec{B}\), and \(\vec{k}\) mutually perpendicular, with \(\vec{E}\) and \(\vec{B}\) oscillating in phase.

The maximum kinetic energy of the emitted electrons is given by Einstein’s photoelectric equation: \[ K_{max} = h\nu - \phi, \]
where:

\(h\nu\) is the photon energy,
\(\phi = 2.3 \, \mathrm{eV}\) is the work function of the material.


For beam A: \[ K_{max,A} = 3.3 \, \mathrm{eV} - 2.3 \, \mathrm{eV} = 1.0 \, \mathrm{eV}. \]

For beam B: \[ K_{max,B} = 11.3 \, \mathrm{eV} - 2.3 \, \mathrm{eV} = 9.0 \, \mathrm{eV}. \]

The maximum speed of the emitted electrons is related to the kinetic energy by: \[ K_{max} = \frac{1}{2} m v^2 \implies v \propto \sqrt{K_{max}}. \]

The ratio of speeds is: \[ \frac{v_A}{v_B} = \sqrt{\frac{K_{max,A}}{K_{max,B}}} = \sqrt{\frac{1.0}{9.0}} = \frac{1}{3}. \]

Thus, the ratio of maximum speeds is: \[ \boxed{\frac{1}{3}}. \] Quick Tip: In photoelectric problems, calculate the kinetic energy from \(K_{max} = h\nu - \phi\) and relate it to speed using \(v \propto \sqrt{K_{max}}\).

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

If two particles (an electron and a proton) have the same wavelength, their momenta must be the same: \[ p = \frac{h}{\lambda}. \]

Since the masses of the two particles are different, their speeds and energies will be different. Only the momentum is the same.

Thus, the correct answer is: \[ \boxed{(A) momentum}. \] Quick Tip: For particles with the same de Broglie wavelength, their momenta are equal, but other quantities like speed and energy depend on the particle’s mass.

When germanium (Ge) is doped with arsenic (As), an \(n\)-type semiconductor is formed. Arsenic is a pentavalent element, meaning it has five valence electrons. Four of these electrons form covalent bonds with the neighboring Ge atoms, while the fifth electron is free to move and contributes to conduction.

This increases the number of conduction electrons in the material.

Thus, the correct answer is: \[ \boxed{(B) The number of conduction electrons increases.} \] Quick Tip: Doping a semiconductor with a pentavalent element increases conduction electrons, while doping with a trivalent element increases holes.

The Balmer series in the hydrogen spectrum arises from transitions of electrons from higher energy levels (\(n_i\)) to the second energy level (\(n_f = 2\)).

The spectral lines of the Balmer series correspond to: \[ n_i = 3, 4, 5, \dots \quad and \quad n_f = 2. \]

The second spectral line of the Balmer series corresponds to the transition: \[ n_i = 4 \quad to \quad n_f = 2. \]

This transition emits a photon with an energy equal to the difference between the energy levels, given by: \[ E = 13.6 \, eV \cdot \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right). \]

For \(n_f = 2\) and \(n_i = 4\): \[ E = 13.6 \, eV \cdot \left(\frac{1}{2^2} - \frac{1}{4^2}\right) = 13.6 \, eV \cdot \left(\frac{1}{4} - \frac{1}{16}\right). \]

Simplifying: \[ E = 13.6 \, eV \cdot \frac{4 - 1}{16} = 13.6 \, eV \cdot \frac{3}{16}. \]

The emitted photon corresponds to the second spectral line of the Balmer series.

Thus, the correct answer is: \[ \boxed{(C) n_f = 2 \, and \, n_i = 4.} \] Quick Tip: The spectral lines of the Balmer series correspond to transitions ending at \(n_f = 2\). The \(n_i\) value determines the position of the spectral line in the series.

The assertion states that plane and convex mirrors cannot produce real images under any circumstance. This is false because a plane mirror can form a real image in some specific conditions when the object is placed between the mirror and the observer, and convex mirrors form only virtual images but the assertion incorrectly generalizes that they cannot form real images under any circumstance.

The reason states that a virtual image cannot serve as an object to produce a real image. This is also false because a virtual image formed by one mirror can serve as an object for another mirror. For example, a real image can be formed when the virtual image is reflected by a second mirror.

Thus, both the assertion and the reason are false.

Thus, the correct answer is:

\[ \boxed{(D) Assertion (A) is false and reason (R) is also false.} \]

Quick Tip: In optics, both plane and convex mirrors can produce real images in specific circumstances. Also, virtual images formed by one mirror can serve as an object for another mirror to form real images.

The assertion states that two long parallel wires, freely suspended and connected in series to a battery, move apart. This is correct. When two parallel wires carry currents in the same direction, they attract each other, and when the currents flow in opposite directions, the wires repel each other. Therefore, the wires will move apart if the currents flow in opposite directions.

The reason also states that two wires carrying current in opposite directions repel each other. This is true based on Ampère's law, which explains that current-carrying wires exert forces on each other. Specifically, when the currents in the wires are in opposite directions, the force is repulsive.

Thus, the reason correctly explains the assertion.

Thus, the correct answer is:

{\text{(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the
correct explanation of the Assertion (A).

Quick Tip: When two current-carrying conductors are placed in close proximity, they exert forces on each other. Wires carrying currents in the same direction attract, while wires carrying currents in opposite directions repel.

In the photoelectric effect:

The kinetic energy of the emitted photoelectrons depends on the frequency (or wavelength) of the incident light, not its intensity. Therefore, the assertion is false.
The photoelectric current depends on the intensity of the incident light, not its wavelength. Hence, the reason is also false.

Thus, the correct answer is: \[ \boxed{(D) Assertion (A) is false and Reason (R) is also false.} \]

Quick Tip: In the photoelectric effect: The kinetic energy of the emitted photoelectrons depends on the frequency (or wavelength) of the incident light, not its intensity. Hence, the assertion that the kinetic energy increases with intensity is false. The photoelectric current depends on the intensity of the incident light, not on its wavelength. Therefore, the reason given in the problem that photoelectric current depends on wavelength is also false.

The mutual inductance between two coils depends on the flux linkage between them. When the coils are wound on each other, the flux linkage is maximum, leading to the maximum mutual inductance.

Thus, both the assertion and the reason are true, and the reason correctly explains the assertion. Hence, the correct answer is:
\[ \boxed{(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).} \] Quick Tip: For Assertion-Reason type questions, verify the correctness of both statements and whether the reason explains the assertion logically.

Finding the value of \( R \):

The current \( i \) is given by the formula:
\[ i = \frac{9}{R + 1} \]

Since the potential difference across the 6V source is zero, we can write the equation:
\[ 6 - iR = 0 \quad \Rightarrow \quad 6 - \left( \frac{9}{R + 1} \right) \times 0.8 = 0 \]

Simplifying this equation:
\[ 6 - \left( \frac{9}{R + 1} \right) \times 0.8 = 0 \]

On solving for \( R \):
\[ R = 0.2 \, \Omega \] Quick Tip: When solving for an unknown resistance in a circuit with potential differences, use Ohm's law and the concept of zero potential difference across a component. If the potential difference across a resistor is zero, the current passing through it can be directly related to other components in the circuit.

Obtaining an expression for \(\lambda_n / \lambda_p\)

Given the relation: \[ E = \frac{hc}{\lambda_p} \quad \Rightarrow \quad \lambda_p = \frac{hc}{E} \]

We can express \(\lambda_n\) in terms of \(p\) (momentum): \[ \lambda_n = \frac{h}{p} = \frac{h}{\sqrt{2mE}} \]

Then, dividing \(\lambda_n\) by \(\lambda_p\), we have: \[ \frac{\lambda_n}{\lambda_p} = \frac{h}{\sqrt{2mE}} \times \frac{E}{hc} = \frac{h}{\sqrt{2mE}} \times \frac{E}{hc} \]

Simplifying further, we obtain: \[ \frac{\lambda_n}{\lambda_p} = \sqrt{\frac{E}{2mc^2}} \] Quick Tip: For comparative wavelength problems, use the de Broglie relation for particles and the photon energy-wavelength relation \(E = \frac{hc}{\lambda}\).

(i) Reflected wavelength:

Given the relationship:

v = λν

where v = 3 × 10⁸ m/s and ν = 5 × 10¹⁴ Hz, we can solve for λ:

λ = v / ν = (3 × 10⁸) / (5 × 10¹⁴) = 600 nm or 6 × 10⁻⁷ m

(ii) Refracted wavelength:

Using the relationship between the wavelengths in different media and their indices of refraction:

λ_medium = λ_air / μ

For λ_air = 600 nm and μ = 1.5 (the refractive index),

λ_medium = (600 nm) / (1.5) = 400 nm or 4 × 10⁻⁷ m

Quick Tip

Let the original double-convex lens have power P. The lens is cut along its principal axis into two parts, as shown in the figure.

When the two parts are arranged in the way shown (i.e., the two parts are placed with their curved surfaces facing each other), the focal length of the combination is affected by the configuration of the lenses.

The power of a lens is inversely proportional to its focal length:

P = 1/f.

When the two parts are placed with their curved surfaces facing each other, the focal lengths of the individual parts combine in such a way that the powers add up. Each part of the lens will have half the focal length of the original lens.

Since the focal length of each part is halved, the power of each part will be doubled, and the total power of the combination will be the sum of the powers of the individual parts:

P_total = 2P.

Thus, the power of the combination is:

2P

Correct Answer: (C) 2P

The total power of two lenses in contact is given by the formula:

  Ptotal = P1 + P2
  

where:

• P₁ and P₂ are the powers of the individual lenses.

The power of a lens is related to its focal length f by the formula:

  P = 1/f  (in meters).
  

Given the focal lengths of the two lenses:

• The focal length of the first lens is f₁ = 60 cm = 0.60 m,
• The focal length of the second lens is f₂ = 20 cm = 0.20 m.

We can calculate the powers of each lens:

  P1 = 1/f1 = 1/0.60 = 1.67 D,
  

  P2 = 1/f2 = 1/0.20 = 5 D.
  

Now, the total power of the combination is:

  Ptotal = P1 + P2 = 1.67 + 5 = 6.67 D.
  

Thus, the power of the combination is:

  6.6 D.
  

Correct Answer: (A) 6.6 D

The angle of deviation \(\delta\) is related to the prism angle \(A\) and the refractive index \(n\) as: \[ n = \frac{\sin i}{\sin r}, \]
where \(i\) is the angle of incidence, and \(r\) is the angle of refraction. For the case when the light ray is incident normally: \[ i = A + \delta \quad and \quad r = A. \]

Substituting these values into the relation for the refractive index: \[ n = \frac{\sin(A + \delta)}{\sin A}. \]
\[ \boxed{Thus, the refractive index is n = \frac{\sin(A + \delta)}{\sin A}.} \]



 

(1) Calculation of the refractive index:

The refractive index \( \mu \) is given by the relation: \[ \mu = \frac{\sin \left( \frac{A + \delta_m}{2} \right)}{\sin \left( \frac{A}{2} \right)}, \]
where \( A \) is the angle of the prism and \( \delta_m \) is the minimum deviation.

Substituting \( A = 60^\circ \) and \( \delta_m = 30^\circ \): \[ \mu = \frac{\sin \left( \frac{60 + \delta_m}{2} \right)}{\sin \left( \frac{60}{2} \right)} = \frac{\sin \left( \frac{60 + 30}{2} \right)}{\sin 30^\circ}. \]
We know that \( \sin 30^\circ = \frac{1}{2} \), so: \[ \mu = \frac{\sin 45^\circ}{\frac{1}{2}} = \sqrt{2}. \]
Thus, the refractive index is: \[ \mu = \sqrt{2}. \]

(2) Calculation of the angle of refraction:

The angle of refraction \( i \) is given by: \[ i = \frac{A + \delta_m}{2}. \]
Substituting \( A = 60^\circ \) and \( \delta_m = 30^\circ \): \[ i = \frac{60 + 30}{2} = \frac{90}{2} = 45^\circ. \]
Thus, the angle of refraction is: \[ i = 45^\circ. \] Quick Tip: The angle of minimum deviation occurs when the ray inside the prism is symmetric, i.e., the angles of incidence and emergence are equal.

Huygens’ principle states:

• Every point on a wavefront acts as a source of secondary wavelets that spread out in all directions with the speed of the wave.
• The new wavefront is the envelope of all the secondary wavelets at a given instant.

Consider a plane wavefront AB incident on a reflecting surface at an angle i. According to Huygens’ principle:

• The secondary wavelets originating from points on AB construct the reflected wavefront.
• The geometry ensures that the angle of incidence i equals the angle of reflection r.

From the geometry:

  Angle of incidence i = Angle of reflection r.
  

Coherent Sources:

Coherent sources are two or more sources of light that emit light waves with:

• A constant phase difference.
• The same frequency and wavelength.

Independent Sodium Lamps:

Two independent sodium lamps cannot act as coherent sources because:

• The light waves emitted by them have random phase differences.
• There is no fixed relationship between the wavefronts produced by the two lamps.

Thus, independent sodium lamps cannot act as coherent sources.

For bright fringes in Young's double-slit experiment:

  n1 λ1 = n2 λ2
  

where:

• n₁ and n₂ are fringe orders for the unknown and known wavelengths, respectively.
• λ₁ and λ₂ are the corresponding wavelengths.

Given:

  n1 = 4, n2 = 5, λ2 = 520 nm.
  

Substitute into the equation:

  4 λ = 5 × 520 nm.
  

Solve for λ:

  λ = (5 × 520) / 4 = 650 nm.
  

Final Answer:

  λ = 650 nm.
  

The potential energy of an electric dipole in a uniform external electric field is given by:

  U = -→p · →E
  

where:

• →p is the dipole moment.
• →E is the external electric field.
• U is the potential energy.

Using the dot product:

  U = -pE cosΘ
  

where Θ is the angle between →p and →E.

(1) Maximum Potential Energy:

  When Θ = 180°, cosΘ = -1.
  Umax = -pE(-1) = pE.
  

(2) Minimum Potential Energy:

  When Θ = 0°, cosΘ = 1.
  Umin = -pE(1) = -pE.
  

Final Expression:

  U = -pE cosΘ.
  

The dipole moment →p is given by:

  →p = q · →d
  

where:

• q = 1.0 pC = 1.0 × 10^-12 C,
• →d is the displacement vector from -q to +q.

The coordinates of the charges are:

• -q at (0, 0),
• +q at (3 mm, 4 mm) = (3 × 10^-3 m, 4 × 10^-3 m).

The magnitude of →d is:

  |→d| = √[(3 × 10-3)2 + (4 × 10-3)2] = √(9 × 10-6 + 16 × 10-6) = √(25 × 10-6) = 5 × 10-3 m.
  

The unit vector →d is:

  →d = →d / |→d| = [(3 →i + 4 →j) × 10-3] / (5 × 10-3) = 0.6 →i + 0.8 →j.
  

Thus, the dipole moment →p is:

  →p = q · →d = (1.0 × 10-12) · (5 × 10-3) · (0.6 →i + 0.8 →j)
  →p = (5 × 10-15) (0.6 →i + 0.8 →j) = (3.0 × 10-15 →i + 4.0 × 10-15 →j) C·m.
  

The torque →τ is given by:

  →τ = →p × →E.
  

Substitute →p = (3.0 × 10^-15 →i + 4.0 × 10^-15 →j) and →E = (1000 →i):

  →τ = 
  |→i →j →k|
  |3.0 × 10-15 4.0 × 10-15 0|
  |1000 0 0|
  

Expand the determinant:

  →τ = →k [(3.0 × 10-15) (0) - (4.0 × 10-15) (1000)]
  →τ = →k (-4.0 × 10-12).
  

Final Torque:

  →τ = -4.0 × 10-12 →k N·m.
  

The potential due to a point charge is given by:

  V = 1 / (4 π ε0) · q / r
  

where q is the charge and r is the distance from the charge.

For an electric dipole, the charges are +q at (a, 0) and -q at (-a, 0). At a point x on the x-axis far from the dipole (x ≪≪ a), the potential at x due to both charges is:

  V = V+ + V-
  

The distances of +q and -q from the point x are:

  r+ = x - a, r- = x + a
  

Substitute into the potential equation:

  V = 1 / (4 π ε0) (q / (x - a) - q / (x + a))
  

For x ≪≪ a, expand the terms using the binomial approximation (1 ± a/x)^-1 ≈ 1 ≠ a/x:

  1 / (x - a) ≈ 1 / x (1 + a / x), 1 / (x + a) ≈ 1 / x (1 - a / x)
  

Substitute these into the potential equation:

  V = 1 / (4 π ε0) [q / x (1 + a / x) - q / x (1 - a / x)]
  

Simplify:

  V = 1 / (4 π ε0) q / x (2a / x)
  

Using p = q · 2a, the dipole moment:

  V = 1 / (4 π ε0) p / x2
  

Since →p is aligned along →К:

  V = 1 / (4 π ε0) (→p · →К) / x2
  

Final Expression:

  V = 1 / (4 π ε0) (→p · →К) / x2
  

The charge density \(\sigma\) is related to the charge \(q\) and radius \(r\) of a sphere as: \[ \sigma = \frac{q}{4 \pi r^2}. \]

The charges on the two spheres before connection are: \[ q_1 = \sigma \cdot 4 \pi r_1^2, \quad q_2 = \sigma \cdot 4 \pi r_2^2, \]
where \(r_1 = 1 \, cm = 0.01 \, m\) and \(r_2 = 3 \, cm = 0.03 \, m\).

Substitute \(\sigma = \frac{2}{\pi} \times 10^{-9} \, C/m^2\): \[ q_1 = \frac{2}{\pi} \times 10^{-9} \cdot 4 \pi (0.01)^2 = 8 \times 10^{-13} \, C, \] \[ q_2 = \frac{2}{\pi} \times 10^{-9} \cdot 4 \pi (0.03)^2 = 72 \times 10^{-13} \, C. \]

When the spheres are connected by a thin wire, the charges redistribute until the potentials on both spheres are equal: \[ V_1 = V_2 \quad \Rightarrow \quad \frac{q_1'}{r_1} = \frac{q_2'}{r_2}, \]
where \(q_1'\) and \(q_2'\) are the new charges on \(S_1\) and \(S_2\) respectively.

The total charge is conserved: \[ q_1' + q_2' = q_1 + q_2 = (8 + 72) \times 10^{-13} = 80 \times 10^{-13} \, C. \]

From the potential equality: \[ \frac{q_1'}{0.01} = \frac{q_2'}{0.03} \quad \Rightarrow \quad q_2' = 3 q_1'. \]

Substitute into the charge conservation equation: \[ q_1' + 3q_1' = 80 \times 10^{-13}. \]

Solve for \(q_1'\): \[ 4q_1' = 80 \times 10^{-13} \quad \Rightarrow \quad q_1' = 20 \times 10^{-13} = 2.0 \times 10^{-12} \, C. \]

Final Answer: \[ \boxed{q_1' = 2.0 \times 10^{-12} \, C}. \] Quick Tip: When charges redistribute between conductors, use the principle of equal potential and charge conservation to find the new distribution.

The circuit consists of a resistor R and a capacitor C in series. The total impedance Z of the circuit is given by:

  Z = √(R2 + XC2)
  

where X_C is the capacitive reactance.

The voltage across the resistor is:

  vR = iR
  

and the voltage across the capacitor is:

  vC = i / (ωC)
  

The capacitive reactance is defined as:

  XC = 1 / (ωC)
  

The current i in the circuit is the same through both components. Using the phasor diagram, the voltage v across the circuit is the vector sum of the voltages across R and C. The total voltage is:

  v = √(vR2 + vC2)
  

Substitute the expressions for v_R and v_C:

  v = √((iR)2 + (i / (ωC))2)
  

Factoring out i:

  v = i √(R2 + (1 / (ωC))2)
  

The impedance Z is defined as:

  Z = v / i = √(R2 + (1 / (ωC))2)
  

Substitute X_C into the equation:

  Z = √(R2 + XC2)
  

Final Expression:

  Z = √(R2 + 1 / (ωC)2)
  

An inductor acts as a conductor in a circuit when the frequency of the applied AC signal is very low (close to \(0\) Hz). At low frequencies, the inductive reactance \(X_L\) becomes negligible. The inductive reactance \(X_L\) is given by: \[ X_L = \omega L, \]
where \(\omega\) is the angular frequency of the AC signal and \(L\) is the inductance.

As \(\omega \to 0\), \(X_L \to 0\). In this case, the inductor behaves like a simple conductor with negligible impedance.

Reason: At low frequencies, the rate of change of current through the inductor is small, which results in a negligible opposing emf. Therefore, the inductor does not impede the current flow significantly and acts like a short circuit (ideal conductor).

Conclusion:
An inductor acts as a conductor in a circuit when the frequency of the applied AC signal is very low. Quick Tip: Remember that an inductor resists changes in current. At high frequencies, its impedance is high; at low frequencies, it behaves as a conductor.

The lamp operates with a DC supply of:

  V = 110 V, I = 11 A
  

The resistance of the lamp is:

  R = V / I = 110 / 11 = 10 Ω
  

When connected to an AC source of 220 V and 50 Hz, a coil with inductance L is in series with the lamp. The total impedance Z of the series combination is:

  Z = √(R2 + XL2)
  

where X_L is the inductive reactance:

  XL = ωL = 2 π f L
  

The current through the circuit is:

  I = VAC / Z
  

Given:

  VAC = 220 V, I = 11 A, f = 50 Hz
  

Substitute the values:

  Z = VAC / I = 220 / 11 = 20 Ω
  

Using the impedance formula:

  Z = √(R2 + XL2),
  

substitute Z and R:

  20 = √(102 + XL2)
  

Solve for X_L:

  202 = 102 + XL2 → 400 = 100 + XL2 → XL2 = 300.
  

  XL = √300 = 10√3 Ω.
  

Now, calculate the inductance L:

  XL = 2 π f L → L = XL / (2 π f)
  

Substitute the values:

  L = (10√3) / (2 π · 50) = (10√3) / (100π)
  

Simplify:

  L ≈ 0.055 H
  

Final Answer:

  L ≈ 0.055 H
  

Diagram: A labelled diagram of a step-up transformer is as follows:

Working Principle:

• A step-up transformer increases the voltage while decreasing the current.
• It operates on the principle of electromagnetic induction.
• When an AC voltage is applied to the primary coil, it generates a varying magnetic flux in the core. This varying flux induces an EMF in the secondary coil according to Faraday's law of electromagnetic induction.

The voltage ratio is given by:

  Vs / Vp = Ns / Np
  

where N_s and N_p are the number of turns in the secondary and primary coils, respectively.

Energy Losses:

• Eddy Current Losses: Circulating currents induced in the core material lead to energy dissipation as heat.
• Hysteresis Losses: Energy is lost due to repeated magnetization and demagnetization of the core during each cycle.
• Copper Losses: Energy is lost as heat in the resistance of the windings.

No, a step-up transformer does not violate the principle of conservation of energy.

The increase in voltage comes at the expense of current, ensuring that the power (product of voltage and current) remains constant, neglecting losses. The power input is equal to the power output:

  Pinput = Poutput
  

Mathematically:

  Vp Ip = Vs Is
  

where V_p, I_p are the voltage and current in the primary coil, and V_s, I_s are the voltage and current in the secondary coil.

Energy conservation is upheld because the increase in voltage results in a proportional decrease in current.

The ratio of the voltages is given by:

  Vs / Vp = Ns / Np
  

Given:

  Np = 200, Ns = 3000, Vp = 90 V.
  

Substitute the values:

  Vs = Vp · (Ns / Np) = 90 · (3000 / 200)
  

Simplify:

  Vs = 90 · 15 = 1350 V.
  

Final Answer:

  Vs = 1350 V
  

The power in the primary and secondary coils is the same: \[ V_p I_p = V_s I_s. \]

Rearrange to find \(I_p\): \[ I_p = \frac{V_s I_s}{V_p}. \]

Substitute the values: \[ I_p = \frac{1350 \cdot 2}{90}. \]

Simplify: \[ I_p = \frac{2700}{90} = 30 \, A. \]

Final Answer: \[ \boxed{I_p = 30 \, A} \] Quick Tip: For transformers, always remember that the voltage ratio is proportional to the turns ratio, and the current ratio is inversely proportional to the turns ratio.