CBSE Class 12 Physics Set 1 - (55/1/1) Question Paper 2026 with Solution PDF

Concept:

The electric field and electric potential are closely related quantities in electrostatics. The electric field at any point is defined as the negative gradient of the electric potential.

For one-dimensional motion along the \(x\)-axis,
\[ E_x=-\frac{dV}{dx} \]

The negative sign indicates that the electric field always points in the direction of decreasing potential.

Therefore, to determine the magnitude and direction of the electric field, we simply differentiate the given potential function with respect to \(x\).

Step 1: Write the given expression for electric potential.


The potential is given as
\[ V=10-50x \]

where \(V\) is measured in volts and \(x\) in meters.

Step 2: Differentiate the potential with respect to \(x\).

\[ \frac{dV}{dx}=\frac{d}{dx}(10-50x) \]

Since the derivative of a constant is zero,
\[ \frac{dV}{dx}=0-50=-50 \]

Thus,
\[ \frac{dV}{dx}=-50\ V/m \]

Step 3: Use the relation between electric field and potential gradient.

\[ E=-\frac{dV}{dx} \]

Substituting the value,
\[ E=-(-50)=50\ V/m \]

Since
\[ 1\ V/m=1\ N/C, \]

we get
\[ E=50\ N/C \]

The positive sign indicates that the electric field is directed along the positive \(x\)-axis.
\[ \boxed{E=50\ N/C along +x} \] Quick Tip: Whenever the potential is given as a function of position, use \[ E=-\frac{dV}{dx}. \] A positive value of \(E\) means the field is along \(+x\), whereas a negative value means it is along \(-x\).

Concept:

When two conducting spheres are connected by a conducting wire, charge flows between them until both spheres attain the same electric potential.

For an isolated conducting sphere of radius \(R\),
\[ V=\frac{1}{4\pi\varepsilon_0}\frac{Q}{R} \]

and the electric field at its surface is
\[ E=\frac{1}{4\pi\varepsilon_0}\frac{Q}{R^2}. \]

Since the potentials of the connected spheres become equal, we can first establish the relation between their charges and then determine the ratio of electric fields.

Step 1: Apply the condition of equal potentials.


Since the spheres are connected,
\[ V_A=V_B \]

Therefore,
\[ \frac{1}{4\pi\varepsilon_0}\frac{Q_A}{r_1} = \frac{1}{4\pi\varepsilon_0}\frac{Q_B}{r_2} \]

or,
\[ \frac{Q_A}{r_1}=\frac{Q_B}{r_2} \]

Hence,
\[ \frac{Q_A}{Q_B} = \frac{r_1}{r_2}. \]

Step 2: Write expressions for electric fields at the surfaces.

\[ E_A=\frac{1}{4\pi\varepsilon_0}\frac{Q_A}{r_1^2} \]

and
\[ E_B=\frac{1}{4\pi\varepsilon_0}\frac{Q_B}{r_2^2} \]

Therefore,
\[ \frac{E_A}{E_B} = \frac{Q_A}{Q_B} \times \frac{r_2^2}{r_1^2} \]

Substituting
\[ \frac{Q_A}{Q_B} = \frac{r_1}{r_2}, \]

we get
\[ \frac{E_A}{E_B} = \frac{r_1}{r_2} \times \frac{r_2^2}{r_1^2} = \frac{r_2}{r_1}. \]

Thus,
\[ \boxed{\frac{E_A}{E_B}=\frac{r_2}{r_1}} \] Quick Tip: For conducting spheres connected by a wire: \[ V_A=V_B \] and since \[ E=\frac{V}{R}, \] the electric field at the surface of a sphere is inversely proportional to its radius.

Concept:

The magnetic field inside a current-carrying conductor is calculated using Ampere's Circuital Law.

Ampere's law states that
\[ \oint \vec B\cdot d\vec l=\mu_0 I_{enc} \]

where \(I_{enc}\) is the current enclosed by the chosen Amperian loop.

For a wire carrying uniformly distributed current, the current density is constant, and hence the current enclosed by a circle of radius \(r\) inside the conductor is proportional to the area enclosed.

Step 1: Determine the current enclosed within radius \(r=\dfrac{a}{2}\).


The total current density is
\[ J=\frac{I}{\pi a^2} \]

The area enclosed by radius
\[ r=\frac{a}{2} \]

is
\[ A=\pi\left(\frac{a}{2}\right)^2 = \frac{\pi a^2}{4} \]

Therefore,
\[ I_{enc} = J\times A = \frac{I}{\pi a^2} \times \frac{\pi a^2}{4} = \frac{I}{4} \]

Step 2: Apply Ampere's circuital law.


For a circular Amperian path of radius \(r=\dfrac{a}{2}\),
\[ B(2\pi r)=\mu_0 I_{enc} \]

Substituting,
\[ B\left(2\pi\frac{a}{2}\right) = \mu_0\left(\frac{I}{4}\right) \]
\[ B(\pi a)=\frac{\mu_0 I}{4} \]

Hence,
\[ B=\frac{\mu_0 I}{4\pi a} \]

Therefore,
\[ \boxed{B=\frac{\mu_0 I}{4\pi a}} \] Quick Tip: For a point inside a wire carrying uniformly distributed current, \[ B=\frac{\mu_0 Ir}{2\pi a^2}, \] 

Concept:

In Young's Double-Slit Experiment (YDSE), two coherent sources produce an interference pattern on a screen. The locus of points having a constant path difference is actually a hyperbola because the difference of distances from two fixed points remains constant.

However, in practical situations, the screen is placed at a very large distance compared to the separation between the slits. Under this condition, the rays reaching the screen are almost parallel and the hyperbolic fringes appear as straight lines.

Thus, the actual shape of the fringes is hyperbolic, but under the approximation
\[ D\gg d, \]

where
\[ D=distance of screen from slits \]

and
\[ d=separation between slits, \]

the fringes become nearly straight and equally spaced.

Step 1: Understand the actual geometry of interference fringes.


The path difference at any point on the screen is
\[ \Delta = S_2P-S_1P. \]

The set of all points for which \(\Delta\) is constant forms a hyperbola.

Therefore, theoretically, the interference fringes are hyperbolic.

Step 2: Apply the practical approximation used in YDSE.


In the experiment,
\[ D\gg d. \]

Under this condition, only a small portion of the hyperbola is observed on the screen and it appears almost as a straight line.

Hence, the fringes are nearly straight.
\[ \boxed{Shape of fringes = Straight} \] Quick Tip: Remember the statement: \[ Actual shape = Hyperbolic, \qquad Observed shape = Nearly straight \] because in YDSE, the screen is placed very far from the slits.

Concept:

The speed of an electromagnetic wave in a medium is given by
\[ v=\frac{c}{\sqrt{\mu_r\varepsilon_r}}, \]

where
\[ \mu_r=relative permeability, \qquad \varepsilon_r=relative permittivity. \]

When an electromagnetic wave enters another medium, its frequency remains unchanged because the frequency is determined by the source of the wave.

The wavelength changes according to
\[ \lambda=\frac{v}{f}. \]

Therefore, once the new speed is known, the new wavelength can be determined.

Step 1: Calculate the speed of the wave in the medium.


Given,
\[ \varepsilon_r=\frac32, \qquad \mu_r=\frac83. \]

Hence,
\[ v=\frac{c}{\sqrt{\frac32\times\frac83}} \]
\[ v=\frac{c}{\sqrt{4}} \]
\[ v=\frac{c}{2} \]

Thus, the speed becomes half of its value in vacuum.

Step 2: Determine the change in frequency and wavelength.


Frequency does not change:
\[ f'=f. \]

Using
\[ \lambda=\frac{v}{f}, \]

we get
\[ \lambda'=\frac{v'}{f'} = \frac{c/2}{f} = \frac12\left(\frac{c}{f}\right) = \frac{\lambda}{2}. \]

Therefore, the wavelength becomes half of its original value while the frequency remains unchanged.
\[ \boxed{\lambda'=\frac{\lambda}{2},\qquad f'=f} \]

Hence, the correct option is
\[ \boxed{(C) wavelength is halved and frequency remains unchanged} \] Quick Tip: For electromagnetic waves entering a new medium: \[ f=constant \] Always calculate the new speed first using \[ v=\frac{c}{\sqrt{\mu_r\varepsilon_r}} \] and then use \[ \lambda=\frac{v}{f}. \]

Concept:

In a series LCR circuit, the applied voltage is the phasor sum of the voltages across the resistor, inductor and capacitor.

The resultant voltage is
\[ V=\sqrt{V_R^2+(V_L-V_C)^2}. \]

When the inductive and capacitive voltages are equal, the circuit is in resonance and the impedance becomes purely resistive.

Step 1: Calculate the source voltage in the original circuit.


Given,
\[ V_R=10\ V, \qquad V_L=10\ V, \qquad V_C=10\ V. \]

Therefore,
\[ V=\sqrt{10^2+(10-10)^2} \]
\[ V=10\ V. \]

Hence, the applied voltage is \(10\) V.

Step 2: Determine the circuit after the capacitor is short circuited.


When the capacitor is short circuited,
\[ X_C=0. \]

The circuit now becomes a series \(RL\) circuit.

Since originally
\[ V_R=V_L, \]

we have
\[ IR=IX_L \]

which implies
\[ R=X_L. \]

Step 3: Calculate the new voltage across the inductor.


The impedance of the \(RL\) circuit is
\[ Z=\sqrt{R^2+X_L^2} \]

Since
\[ R=X_L, \]
\[ Z=\sqrt{2R^2} =\sqrt2\,R. \]

Hence,
\[ I'=\frac{V}{Z} = \frac{10}{\sqrt2\,R}. \]

The voltage across the inductor becomes
\[ V_L'=I'X_L = \frac{10}{\sqrt2 R}\times R = \frac{10}{\sqrt2} = 5\sqrt2\ V. \]

Therefore,
\[ \boxed{V_L'=5\sqrt2\ V} \] Quick Tip: If \(V_L=V_C\), the circuit is in resonance and \[ R=X_L. \] After removing the capacitor, treat the circuit as an \(RL\) circuit and use \[ Z=\sqrt{R^2+X_L^2}. \]

Concept:

The electromagnetic waves widely used in medical diagnosis are X-rays.

X-rays are highly penetrating electromagnetic waves and are used in:


Radiography
CT scans
Dental imaging
Bone fracture detection


The approximate wavelength range of X-rays is
\[ 10^{-9}\ m to 10^{-12}\ m \]

or equivalently,
\[ 1\ nm to 10^{-3}\ nm. \]

Step 1: Identify the electromagnetic wave used for diagnosis.


The term "diagnostic tool in medicine" immediately refers to X-rays because they can penetrate soft tissues but are absorbed more strongly by bones.

Step 2: Recall the wavelength range of X-rays.

\[ \lambda_{X-rays} = 10^{-9} m to 10^{-12} m \]

Converting into nanometres,
\[ 1\ nm to 10^{-3}\ nm. \]

Hence,
\[ \boxed{1\ nm to 10^{-3}\ nm} \] Quick Tip: Remember the sequence of electromagnetic waves: \[ Radio \rightarrow Microwave \rightarrow Infrared \rightarrow Visible \rightarrow Ultraviolet \rightarrow X-rays \rightarrow \gamma-rays \] Medical diagnosis generally uses X-rays having wavelengths from \[ 1\ nm to 10^{-3}\ nm. \]

Concept:

The distance of closest approach is the minimum distance between the alpha particle and the nucleus during a head-on collision. At this point, the entire initial kinetic energy of the alpha particle gets converted into electrostatic potential energy.

According to the principle of conservation of energy,
\[ Initial K.E.=Electrostatic P.E. at closest approach \]

For an alpha particle of charge \(+2e\) approaching a nucleus of charge \(+Ze\),
\[ \frac{1}{2}mv^2=\frac{1}{4\pi\varepsilon_0}\frac{(2e)(Ze)}{r} \]

where \(r\) is the distance of closest approach.

From this expression, it is evident that
\[ r\propto \frac{1}{v^2}. \]

Thus, the distance of closest approach is inversely proportional to the square of the speed of the alpha particle.

Step 1: Write the proportionality relation.

\[ d\propto \frac{1}{v^2} \]

Suppose the new distance of closest approach is \(d'\) when the velocity becomes
\[ v'=\frac{v}{2}. \]

Step 2: Form the ratio of the two distances.

\[ \frac{d'}{d} = \frac{\dfrac{1}{(v/2)^2}}{\dfrac{1}{v^2}} \]
\[ = \frac{v^2}{v^2/4} \]
\[ =4 \]

Hence,
\[ d'=4d. \]
\[ \boxed{d'=4d} \]

Therefore, the new distance of closest approach becomes four times the original value. Quick Tip: For a head-on collision of an alpha particle with a nucleus, \[ d\propto \frac{1}{v^2}. \] If the velocity becomes \(n\) times smaller, the distance of closest approach becomes \(n^2\) times larger.

Concept:

The focal length of a lens is determined by the lens maker's formula,
\[ \frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \]

where


\(f\) is the focal length,
\(\mu\) is the refractive index,
\(R_1\) and \(R_2\) are the radii of curvature.


When a symmetric biconcave lens is cut into two equal parts by a plane perpendicular to the principal axis, each part becomes a plano-concave lens.

The power of each half becomes half of the original power.

Step 1: Calculate the original power of the lens.


The focal length of the original lens is
\[ f=-10 cm \]

Hence, its power is
\[ P=\frac{1}{f} = \frac{1}{-10} = -\frac{1}{10} cm^{-1}. \]

Step 2: Determine the power of each half lens.


Since each half has only one curved surface, its power becomes half of the original power.

Therefore,
\[ P'=\frac{P}{2} \]
\[ P'=\frac{-1/10}{2} = -\frac{1}{20} cm^{-1}. \]

Hence,
\[ f'=-20 cm. \]

Therefore, the magnitude of the focal length is
\[ \boxed{20 cm} \] Quick Tip: When a symmetrical lens is cut perpendicular to its principal axis, the power of each part becomes half. \[ P'=\frac{P}{2} \qquad\Rightarrow\qquad f'=2f \] in magnitude.

Concept:

For sustained interference, the two waves must be coherent.

The conditions for coherence are:


Same frequency.
Constant phase difference.


Only waves having the same angular frequency can interfere.

Step 1: Examine the frequency of each wave.


Wave (i):
\[ y_1=A_1\sin\omega t \]

has angular frequency \(\omega\).

Wave (ii):
\[ y_2=A_2\sin2\omega t \]

has angular frequency \(2\omega\).

Wave (iii):
\[ y_3=A_3\cos\omega t \]

has angular frequency \(\omega\).

Wave (iv):
\[ y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \]

has angular frequency \(\omega\).

Step 2: Identify the waves with identical frequencies.


Waves (i), (iii), and (iv) all have angular frequency \(\omega\).

Also,
\[ \cos\omega t=\sin\left(\omega t+\frac{\pi}{2}\right), \]

which means wave (iii) differs from wave (i) only by a constant phase difference.

Similarly, wave (iv) also has a constant phase difference.

Hence, waves (i), (iii), and (iv) can interfere.

Wave (ii) cannot interfere with the others because its frequency is different.

Therefore,
\[ \boxed{(i), (iii) and (iv) only} \] Quick Tip: For interference, remember the keyword: \[ \boxed{Same frequency + Constant phase difference} \] Different frequencies do not produce sustained interference.

Concept:

For an electrical appliance rated \((P,V)\),
\[ P=\frac{V^2}{R} \]

Therefore,
\[ R=\frac{V^2}{P}. \]

The heaters are connected in series and the supply voltage is \(V/2\).

The total power consumed is
\[ P_{total} = \frac{V_{supply}^2}{R_{eq}}. \]

Step 1: Determine the resistance of each heater.


For heater 1,
\[ R_1=\frac{V^2}{P_1}. \]

For heater 2,
\[ R_2=\frac{V^2}{P_2}. \]

Step 2: Calculate the equivalent resistance.


Since the heaters are connected in series,
\[ R_{eq} = R_1+R_2 \]
\[ = \frac{V^2}{P_1} + \frac{V^2}{P_2} \]
\[ = V^2\left(\frac{P_1+P_2}{P_1P_2}\right). \]

Step 3: Calculate the total power consumed.


The applied voltage is
\[ V_{supply}=\frac{V}{2}. \]

Hence,
\[ P_{total} = \frac{\left(\frac{V}{2}\right)^2} {V^2\left(\frac{P_1+P_2}{P_1P_2}\right)} \]
\[ = \frac{\frac{V^2}{4}} {V^2\left(\frac{P_1+P_2}{P_1P_2}\right)} \]
\[ = \frac{P_1P_2} {4(P_1+P_2)}. \]

Therefore,
\[ \boxed{ P_{total} = \frac{P_1P_2} {4(P_1+P_2)} } \] Quick Tip: For appliances with rating \((P,V)\), \[ R=\frac{V^2}{P}. \] Always convert the ratings into resistance first and then apply series or parallel combinations.

Concept:

A p-n junction is formed by joining a p-type semiconductor and an n-type semiconductor. Immediately after the formation of the junction, there exists a large concentration difference of charge carriers on the two sides of the junction:


The p-region contains a large number of holes and a very small number of electrons.
The n-region contains a large number of electrons and a very small number of holes.


Due to this concentration difference, majority charge carriers begin to move across the junction. Electrons diffuse from the n-side to the p-side, and holes diffuse from the p-side to the n-side. This movement of charge carriers due to concentration gradient produces the diffusion current.

As these charge carriers cross the junction, they leave behind immobile ions near the junction, creating a region depleted of free charge carriers called the depletion region. The charged ions in this region establish an electric field directed from the n-side to the p-side.

This electric field opposes the further diffusion of charge carriers and produces another current called the drift current.

At thermal equilibrium and in the absence of any external bias, the junction reaches a stable condition where the two currents exactly balance each other.

Step 1: Understand the origin of diffusion current.


Because the concentration of electrons is much higher in the n-region than in the p-region, electrons diffuse from the n-side to the p-side.

Similarly, because the concentration of holes is much higher in the p-region than in the n-region, holes diffuse from the p-side to the n-side.

This movement of majority charge carriers gives rise to the diffusion current.

Therefore,
\[ I_{diffusion} \neq 0. \]

Step 2: Understand the origin of drift current.


The diffusion of charge carriers creates uncovered positive and negative ions near the junction, resulting in the formation of an electric field.

This electric field causes minority charge carriers to move across the junction, producing the drift current.

Hence,
\[ I_{drift} \neq 0. \]

Step 3: Apply the equilibrium condition of an unbiased p-n junction.


At equilibrium, there is no net flow of charge through the junction. Therefore, the total current through the junction must be zero.

Hence,
\[ I_{net} = I_{diffusion} + I_{drift} = 0. \]

This implies,
\[ I_{diffusion} = - I_{drift}. \]

Therefore, the magnitudes of the two currents are equal, but their directions are opposite.
\[ |I_{diffusion}| = |I_{drift}|. \]

Thus,
\[ \boxed{Diffusion current and drift current are equal and opposite.} \]

Hence, the correct option is
\[ \boxed{(C)} \] Quick Tip: For an textbf{unbiased p-n junction at equilibrium}, \[ I_{diffusion} = I_{drift} \] in magnitude, but they flow in opposite directions. Therefore, \[ I_{net}=0. \] Remember: \[ \boxed{Equilibrium \Rightarrow Net current = 0} \] because diffusion current is exactly balanced by drift current.

Concept:

The magnetic moment of an atom arises due to:


Orbital motion of electrons around the nucleus.
Spin motion of electrons.


However, in many atoms, the magnetic moments of electrons pair up and cancel each other. Therefore, not every atom possesses a net magnetic moment.

Furthermore, a current-carrying loop always behaves as a magnetic dipole and possesses a magnetic dipole moment given by
\[ \vec{\mu}=IA\hat{n}, \]

where \(I\) is the current in the loop and \(A\) is its area.

Step 1: Examine the Assertion (A).


The statement says:
\[ All atoms have a net magnetic moment. \]

This is incorrect because atoms having completely filled electronic shells have zero resultant magnetic moment due to cancellation of individual electron moments.

For example:
\[ He, Ne, Ar \]

have zero net magnetic moment.

Hence, Assertion (A) is false.

Step 2: Examine the Reason (R).


The statement says:
\[ A current loop does not always behave as a magnetic dipole. \]

This is also incorrect because every current loop behaves like a magnetic dipole and has a definite magnetic dipole moment.

Hence, Reason (R) is also false.

Therefore,
\[ \boxed{Both Assertion (A) and Reason (R) are false.} \] Quick Tip: A current loop always behaves as a magnetic dipole. Atoms with completely filled shells have zero magnetic moment because the orbital and spin magnetic moments cancel each other.

Concept:

According to de Broglie's hypothesis, every moving particle has an associated wavelength given by
\[ \lambda=\frac{h}{p}=\frac{h}{mv}. \]

Thus, electrons exhibit wave properties in addition to particle properties. This is known as the wave-particle duality of matter.

Diffraction is a phenomenon associated with waves. Therefore, if electrons produce diffraction patterns, it directly confirms their wave nature.

Step 1: Examine the Assertion (A).


When accelerated electrons are passed through a narrow slit or a crystal, they undergo diffraction and produce a diffraction pattern.

Hence, Assertion (A) is true.

Step 2: Examine the Reason (R).


Electrons possess both particle and wave characteristics.

This is experimentally verified by electron diffraction experiments such as the Davisson-Germer experiment.

Hence, Reason (R) is true.

Step 3: Determine whether the reason explains the assertion.


The diffraction pattern is observed precisely because electrons behave as waves.

Therefore, the Reason correctly explains the Assertion.

Hence,
\[ \boxed{Both A and R are true and R is the correct explanation of A.} \] Quick Tip: Whenever diffraction or interference is observed for a particle, it is direct evidence of its wave nature and hence confirms de Broglie's hypothesis.

Concept:

The mass of a nucleus is always less than the sum of the masses of its constituent protons and neutrons. The difference in mass is called the mass defect.
\[ \Delta m=Zm_p+Nm_n-M_{nucleus} \]

The mass defect is converted into binding energy according to Einstein's relation
\[ E=\Delta mc^2. \]

When a nucleus is formed, energy is released, not absorbed.

Step 1: Examine the Assertion (A).


Since some mass is converted into binding energy,
\[ M_{nucleus} < Zm_p+Nm_n. \]

Thus, the Assertion is true.

Step 2: Examine the Reason (R).


The statement says that energy is absorbed during nucleus formation.

This is incorrect.

Actually, energy is released when nucleons combine to form a stable nucleus.

Therefore, the Reason is false.

Hence,
\[ \boxed{Assertion is true but Reason is false.} \] Quick Tip: Formation of a stable nucleus always releases energy. \[ Mass defect \Longleftrightarrow Binding Energy \] A larger binding energy means a more stable nucleus.

Concept:

Bohr proposed that electrons revolve around the nucleus only in certain permitted orbits for which the angular momentum is quantised:
\[ mvr=\frac{nh}{2\pi}, \qquad n=1,2,3,\dots \]

Because only certain orbits are allowed, the corresponding energies are also discrete and quantised.

The electrostatic force between the electron and the nucleus indeed provides the centripetal force necessary for circular motion:
\[ \frac{1}{4\pi\varepsilon_0}\frac{e^2}{r^2} = \frac{mv^2}{r}. \]

However, this condition alone does not explain why only discrete energy levels exist.

Step 1: Examine the Assertion (A).


The energy of the electron in the \(n^{th}\) orbit is
\[ E_n=-\frac{13.6}{n^2} eV. \]

Since \(n\) can take only integral values, the energy levels are discrete.

Hence, Assertion (A) is true.

Step 2: Examine the Reason (R).


The electrostatic force indeed acts as the centripetal force:
\[ \frac{1}{4\pi\varepsilon_0}\frac{e^2}{r^2} = \frac{mv^2}{r}. \]

Therefore, Reason (R) is also true.

Step 3: Determine whether the reason explains the assertion.


The discreteness of energy levels arises due to the quantisation condition
\[ mvr=\frac{nh}{2\pi} \]

and not merely because the electrostatic force provides centripetal force.

Therefore, the Reason is true but is not the correct explanation of the Assertion.

Hence,
\[ \boxed{Both A and R are true, but R is not the correct explanation of A.} \] Quick Tip: In Bohr's model: \[ Electrostatic force \rightarrow Centripetal force \] but \[ Quantised energy levels \rightarrow mvr=\frac{nh}{2\pi}. \] Do not confuse these two independent conditions.

Concept:

The photoelectric effect is based on Einstein's photoelectric equation, which states that when light of sufficiently high frequency falls on a metal surface, electrons are emitted from the surface. The energy of the incident photon is used in two ways:


A part of the energy is used to overcome the work function of the metal.
The remaining energy appears as the maximum kinetic energy of the emitted photoelectrons.


Mathematically,
\[ h\nu=\phi+K_{\max} \]

where,
\[ h\nu=energy of the incident photon, \]
\[ \phi=work function of the metal, \]

and
\[ K_{\max}=maximum kinetic energy of the emitted photoelectrons. \]

The maximum kinetic energy is related to the stopping potential \(V_0\) by
\[ K_{\max}=eV_0. \]

Thus, once the photon energy and stopping potential are known, the work function can be calculated using Einstein's photoelectric equation.

Step 1: Calculate the energy of the incident photon.


The wavelength of the incident radiation is
\[ \lambda=200\ nm. \]

The energy of a photon in electron volt can be calculated directly using
\[ E=\frac{1240}{\lambda(in nm)}\ eV. \]

Substituting the given value,
\[ E=\frac{1240}{200} \]
\[ E=6.2\ eV. \]

Therefore, the energy of each incident photon is
\[ h\nu=6.2\ eV. \]

Step 2: Determine the maximum kinetic energy of the emitted photoelectrons.


The photocurrent becomes zero when the collector plate potential is
\[ V_0=0.80\ V. \]

Hence,
\[ K_{\max}=eV_0. \]

In electron volt,
\[ K_{\max}=0.80\ eV. \]

Step 3: Apply Einstein's photoelectric equation to calculate the work function.


Using
\[ h\nu=\phi+K_{\max}, \]

we get
\[ \phi=h\nu-K_{\max}. \]

Substituting the values,
\[ \phi=6.2-0.8 \]
\[ \phi=5.4\ eV. \]

Therefore, the work function of the emitter is
\[ \boxed{\phi=5.4\ eV} \] Quick Tip: For photoelectric effect problems, remember the two important formulas: \[ E_{photon}=\frac{1240}{\lambda(nm)}\ eV \] and \[ h\nu=\phi+eV_0. \] A quick method is: \[ \boxed{\phi=\frac{1240}{\lambda(nm)}-V_0} \] when all quantities are expressed in electron volts and volts.

Concept:

The diffraction pattern produced by a single slit consists of a central bright maximum and a series of secondary maxima and minima on both sides.

For a single slit of width \(a\), the condition for obtaining the dark fringes (minima) is
\[ a\sin\theta=n\lambda, \qquad n=1,2,3,\dots \]

where


\(a\) is the width of the slit,
\(\lambda\) is the wavelength of light,
\(n\) is the order of the minimum.


For small diffraction angles,
\[ \sin\theta \approx \tan\theta \approx \theta. \]

Hence, if the screen is placed at a distance \(D\) from the slit, the position of the \(n^{th}\) dark fringe from the central maximum is
\[ y_n=\frac{nD\lambda}{a}. \]

Since two different wavelengths are used, the dark fringes corresponding to the two wavelengths will coincide only when their positions on the screen are equal.

Step 1: Write the positions of dark fringes for the two wavelengths.


For wavelength
\[ \lambda_1=400\ nm=400\times10^{-9}\ m, \]

the position of the \(n_1^{th}\) minimum is
\[ y_1=\frac{n_1D\lambda_1}{a}. \]

For wavelength
\[ \lambda_2=600\ nm=600\times10^{-9}\ m, \]

the position of the \(n_2^{th}\) minimum is
\[ y_2=\frac{n_2D\lambda_2}{a}. \]

For coincidence of dark fringes,
\[ y_1=y_2. \]

Therefore,
\[ \frac{n_1D\lambda_1}{a} = \frac{n_2D\lambda_2}{a}. \]

After cancelling \(D\) and \(a\),
\[ n_1\lambda_1=n_2\lambda_2. \]

Step 2: Determine the least orders for which the minima coincide.


Substituting the wavelengths,
\[ n_1(400)=n_2(600). \]

Dividing by \(200\),
\[ 2n_1=3n_2. \]

The smallest integral values satisfying this relation are
\[ n_1=3, \qquad n_2=2. \]

Thus, the third minimum of \(400\ nm\) light coincides with the second minimum of \(600\ nm\) light.

Step 3: Calculate the distance of the coincident dark fringe from the central maximum.


Using
\[ y=\frac{nD\lambda}{a}, \]

and taking
\[ n=3,\qquad \lambda=400\times10^{-9}\ m, \]
\[ D=1.5\ m, \qquad a=1\ mm=10^{-3}\ m, \]

we get
\[ y= \frac{3\times1.5\times400\times10^{-9}} {10^{-3}}. \]
\[ y= \frac{1800\times10^{-9}} {10^{-3}} \]
\[ y=1.8\times10^{-3}\ m. \]

Therefore,
\[ \boxed{y=1.8\times10^{-3}\ m} \]

or
\[ \boxed{y=1.8\ mm}. \]

Hence, the least distance of the point from the central maximum where the dark fringes due to both wavelengths coincide is
\[ \boxed{1.8\ mm}. \] Quick Tip: For coincidence of dark fringes in single-slit diffraction, use \[ n_1\lambda_1=n_2\lambda_2. \] First find the smallest integral values of \(n_1\) and \(n_2\), and then substitute into \[ y=\frac{nD\lambda}{a} \] to obtain the position of the coincident minimum.

Concept:

In Young's Double-Slit Experiment (YDSE), the condition for the formation of bright fringes (constructive interference) is
\[ \Delta = n\lambda, \qquad n=0,1,2,3,\dots \]

where \(n\) is the order of the bright fringe.

The position of the \(n^{th}\) bright fringe from the central maximum is given by
\[ y_n=\frac{nD\lambda}{d}, \]

where


\(D\) is the distance of the screen from the slits,
\(d\) is the separation between the slits,
\(\lambda\) is the wavelength of light.


When two different wavelengths are used simultaneously, their bright fringes will coincide only if the positions of the corresponding bright fringes are the same.

Thus, for coincidence of bright fringes,
\[ \frac{n_1D\lambda_1}{d} = \frac{n_2D\lambda_2}{d} \]

or
\[ n_1\lambda_1=n_2\lambda_2. \]

Step 1: Write the condition for coincidence of bright fringes.


Given,
\[ \lambda_1=440\ nm, \qquad \lambda_2=660\ nm. \]

For the bright fringes to coincide,
\[ n_1(440)=n_2(660). \]

Dividing throughout by \(220\),
\[ 2n_1=3n_2. \]

The smallest integral values satisfying this relation are
\[ n_1=3, \qquad n_2=2. \]

Thus, the third bright fringe of \(440\ nm\) light coincides with the second bright fringe of \(660\ nm\) light.

Step 2: Calculate the position of the coincident bright fringe.


Using
\[ y=\frac{nD\lambda}{d}, \]

and substituting
\[ n=3, \qquad \lambda=440\times10^{-9}\ m, \]
\[ D=1.5\ m, \qquad d=0.6\ mm=0.6\times10^{-3}\ m, \]

we get
\[ y= \frac{3\times1.5\times440\times10^{-9}} {0.6\times10^{-3}}. \]

First, calculate the numerator:
\[ 3\times1.5\times440 = 1980. \]

Hence,
\[ y= \frac{1980\times10^{-9}} {0.6\times10^{-3}}. \]
\[ y= \frac{1980}{0.6}\times10^{-6}. \]
\[ y= 3300\times10^{-6}\ m. \]

Therefore,
\[ y=3.3\times10^{-3}\ m. \]

Thus,
\[ \boxed{y=3.3\times10^{-3}\ m} \]

or
\[ \boxed{y=3.3\ mm}. \]

Hence, the least distance of the point from the central maximum where the bright fringes due to both wavelengths coincide is
\[ \boxed{3.3\ mm}. \] Quick Tip: For coincidence of bright fringes in YDSE, always use \[ n_1\lambda_1=n_2\lambda_2. \] First find the smallest integral values of \(n_1\) and \(n_2\), and then substitute into \[ y=\frac{nD\lambda}{d} \] to determine the position of the coincident bright fringe.

Concept:

The torque experienced by a current-carrying coil placed in a uniform magnetic field is given by
\[ \tau = N I A B \sin\theta, \]

where


\(N\) is the number of turns of the coil,
\(I\) is the current flowing through the coil,
\(A\) is the area enclosed by the coil,
\(B\) is the magnetic field strength,
\(\theta\) is the angle between the normal to the plane of the coil and the magnetic field.


The maximum torque is obtained when
\[ \sin\theta =1 \qquad or \qquad \theta=90^\circ. \]

Hence,
\[ \tau_{\max}=NIAB. \]

Since both coils have the same number of turns, carry the same current and are placed in the same magnetic field, the ratio of their maximum torques depends only on the ratio of their areas.

Thus,
\[ \frac{\tau_s}{\tau_c} = \frac{A_s}{A_c}, \]

where \(A_s\) and \(A_c\) denote the areas of the square and circular coils respectively.

Step 1: Find the area of the square coil.


The total length of wire is \(L\) and the coil has \(N\) turns.

Therefore, the length of wire available for one turn of the square is
\[ Perimeter of one square = \frac{L}{N}. \]

If \(a\) is the side of the square, then
\[ 4a=\frac{L}{N}. \]

Hence,
\[ a=\frac{L}{4N}. \]

The area enclosed by one turn of the square coil is
\[ A_s=a^2 = \left(\frac{L}{4N}\right)^2 = \frac{L^2}{16N^2}. \]

Step 2: Find the area of the circular coil.


The length of wire available for one turn of the circular coil is also
\[ \frac{L}{N}. \]

If \(r\) is the radius of the circular coil, then
\[ 2\pi r=\frac{L}{N}. \]

Therefore,
\[ r=\frac{L}{2\pi N}. \]

Hence, the area of one turn of the circular coil is
\[ A_c=\pi r^2 = \pi\left(\frac{L}{2\pi N}\right)^2. \]
\[ A_c = \pi\left(\frac{L^2}{4\pi^2N^2}\right) = \frac{L^2}{4\pi N^2}. \]

Step 3: Determine the ratio of maximum torques.


Since
\[ \frac{\tau_s}{\tau_c} = \frac{A_s}{A_c}, \]

we get
\[ \frac{\tau_s}{\tau_c} = \frac{\dfrac{L^2}{16N^2}} {\dfrac{L^2}{4\pi N^2}}. \]

Cancelling \(L^2\) and \(N^2\),
\[ \frac{\tau_s}{\tau_c} = \frac{4\pi}{16} = \frac{\pi}{4}. \]

Therefore,
\[ \boxed{ \frac{\tau_{square}} {\tau_{circular}} = \frac{\pi}{4} } \]

Hence, the required ratio of the maximum torque acting on the square coil to that on the circular coil is
\[ \boxed{\frac{\pi}{4}:1}. \] Quick Tip: For a current-carrying coil, \[ \tau_{\max}=NIAB. \] If \(N\), \(I\), and \(B\) are the same for different shapes of coils, then \[ \tau_{\max}\propto A. \] Also, among all plane figures with the same perimeter, the circle encloses the maximum area. Therefore, the circular coil always experiences a larger maximum torque than the square coil.

Concept:

In a metallic conductor, a large number of free electrons are continuously moving in random directions due to thermal motion. In the absence of an electric field, these random motions cancel one another and there is no net flow of charge through the conductor.

When an external electric field is applied across the conductor, each free electron experiences an electric force and acquires a small average velocity opposite to the direction of the electric field. This average velocity of the electrons is called the drift velocity.

The drift velocity is extremely small compared to the random thermal velocity of electrons and is generally of the order of
\[ 10^{-4}\ m s^{-1} \]

to
\[ 10^{-5}\ m s^{-1}. \]

Thus, the order of magnitude of drift velocity is
\[ \boxed{10^{-4}\ m s^{-1}}. \]

Step 1: Consider a conductor of cross-sectional area \(A\).


Let


\(n\) be the number of free electrons per unit volume of the conductor,
\(A\) be the cross-sectional area of the conductor,
\(v_d\) be the drift velocity of electrons,
\(e\) be the magnitude of charge on an electron.


Suppose the electrons move with drift velocity \(v_d\).

In time \(dt\), the electrons move through a distance
\[ dx=v_d\,dt. \]

Step 2: Calculate the number of electrons crossing the cross-section in time \(dt\).


The volume of the conductor through which electrons pass in time \(dt\) is
\[ dV=A\,dx. \]

Substituting
\[ dx=v_d\,dt, \]

we get
\[ dV=A\,v_d\,dt. \]

Since \(n\) is the number of free electrons per unit volume, the total number of electrons crossing the cross-section in time \(dt\) is
\[ dN=n\,dV. \]

Therefore,
\[ dN=nAv_d\,dt. \]

Step 3: Calculate the total charge crossing the conductor.


The charge carried by one electron is \(e\).

Hence, the total charge crossing the cross-section in time \(dt\) is
\[ dq=e\,dN. \]

Substituting the value of \(dN\),
\[ dq=e(nAv_d\,dt). \]

Therefore,
\[ dq=neAv_d\,dt. \]

Step 4: Obtain the expression for electric current.


By definition,
\[ I=\frac{dq}{dt}. \]

Substituting the value of \(dq\),
\[ I=\frac{neAv_d\,dt}{dt}. \]

Hence,
\[ \boxed{I=neAv_d}. \]

This is the required relation between the electric current and the drift velocity of electrons in a conductor.

The corresponding current density is
\[ J=\frac{I}{A}. \]

Therefore,
\[ J=\frac{neAv_d}{A}, \]

or
\[ \boxed{J=nev_d}. \]

Thus, current in a conductor is directly proportional to the drift velocity of electrons.
\[ \boxed{I=neAv_d} \]

where


\(n\) = number density of free electrons,
\(e\) = electronic charge,
\(A\) = cross-sectional area of the conductor,
\(v_d\) = drift velocity. Quick Tip: The two most important formulas related to drift velocity are \[ \boxed{I=neAv_d} \] and \[ \boxed{J=nev_d}. \] Also remember that the drift velocity of electrons in an ordinary conductor is very small, typically of the order of \[ \boxed{10^{-4}\ m s^{-1}}. \]

Concept:

The interaction between two nucleons (proton-proton, neutron-neutron, or proton-neutron) is governed by the nuclear force. The nature of this force can be understood by studying the variation of the potential energy of a pair of nucleons with their separation.

The potential energy curve shows that the nuclear force is:


Attractive at intermediate distances.
Repulsive at extremely small separations.
Negligible at large separations.


The typical graph of potential energy \(U\) versus separation \(r\) of two nucleons is shown below.


A schematic representation of the graph can also be drawn as



The graph indicates a deep negative potential well at intermediate separations and rises sharply to positive values at very small separations.

Step 1: Interpret the behaviour of the curve at large distances.


As the separation between the nucleons increases beyond approximately
\[ 2\ fm \quad (1\ fm=10^{-15}\ m), \]

the potential energy approaches zero.

This means that the nuclear force becomes practically negligible beyond a few femtometres.

Hence, nuclear forces are short-ranged forces.
\[ \boxed{Nuclear force acts only over a very short range of about 2-3\ fm.} \]

Step 2: Interpret the behaviour of the curve at intermediate distances.


For separations around
\[ 0.8\ fm to 2\ fm, \]

the potential energy is negative.

Negative potential energy indicates that the force between the nucleons is attractive.

Therefore, nucleons remain bound together inside the nucleus due to this attractive nuclear force.
\[ \boxed{Nuclear force is strongly attractive at intermediate separations.} \]

Step 3: Interpret the behaviour of the curve at extremely small separations.


As the separation becomes very small,
\[ r\lesssim 0.5\ fm, \]

the potential energy rises sharply and becomes positive.

This indicates that the nuclear force becomes strongly repulsive at extremely short distances.

This repulsive nature prevents the nucleons from collapsing into one another.
\[ \boxed{Nuclear force becomes strongly repulsive at very small separations.} \]

Important Conclusions from the Graph:


Nuclear force is a short-range force; it becomes negligible for separations greater than about \(2-3\ fm\).

Nuclear force is attractive at intermediate distances but becomes strongly repulsive at very small separations, thereby ensuring the stability of the nucleus. Quick Tip: Remember the three important regions of the nucleon potential energy curve: \[ r>2\ fm \Rightarrow Force is negligible \] 

Concept:

Gauss's law is one of the fundamental laws of electrostatics. It relates the total electric flux passing through a closed surface to the total charge enclosed by that surface.

The mathematical statement of Gauss's law is
\[ \oint \vec{E}\cdot d\vec{S} = \frac{q_{enc}}{\varepsilon_0}, \]

where


\(\vec{E}\) is the electric field,
\(d\vec{S}\) is the outward area vector,
\(q_{enc}\) is the total charge enclosed by the Gaussian surface,
\(\varepsilon_0\) is the permittivity of free space.


For a uniformly charged infinite plane sheet, symmetry plays a very important role. Because the sheet is infinitely large and uniformly charged, every point on the sheet is equivalent and there is no preferred direction along the plane of the sheet.

Hence, the electric field:


is perpendicular to the plane of the sheet,
has the same magnitude at all points equidistant from the sheet,
has equal magnitudes on both sides of the sheet.


Step 1: Consider an infinite plane sheet having uniform surface charge density \(\sigma\).


Let
\[ \sigma=\frac{charge}{area} \]

be the surface charge density of the sheet.

To apply Gauss's law conveniently, we choose a cylindrical Gaussian surface (often called a pill-box) of cross-sectional area \(A\), such that:


one flat face lies above the sheet,
the other flat face lies below the sheet,
the curved surface is perpendicular to the electric field lines.

\[ Area of each flat face=A. \]

Step 2: Calculate the electric flux through the Gaussian surface.


The electric field is perpendicular to the plane sheet and hence also perpendicular to the two flat faces of the cylinder.

Therefore, the electric flux through the upper face is
\[ \phi_1=EA. \]

Similarly, the electric flux through the lower face is
\[ \phi_2=EA. \]

On the curved surface, the electric field is parallel to the surface.

Hence,
\[ \phi_3=0. \]

Therefore, the total electric flux through the Gaussian surface is
\[ \phi = EA+EA+0 = 2EA. \]

Thus,
\[ \oint \vec{E}\cdot d\vec{S} = 2EA. \]

Step 3: Determine the charge enclosed by the Gaussian surface.


The charge enclosed by the pill-box is equal to the charge present on the portion of the sheet enclosed by its cross-sectional area \(A\).

Hence,
\[ q_{enc} = \sigma A. \]

Step 4: Apply Gauss's law.


According to Gauss's law,
\[ \oint \vec{E}\cdot d\vec{S} = \frac{q_{enc}}{\varepsilon_0}. \]

Substituting the values,
\[ 2EA = \frac{\sigma A}{\varepsilon_0}. \]

Cancelling \(A\) from both sides,
\[ 2E = \frac{\sigma}{\varepsilon_0}. \]

Therefore,
\[ \boxed{ E=\frac{\sigma}{2\varepsilon_0} } \]

This is the magnitude of the electric field produced by a uniformly charged infinite plane sheet.

Direction of the Electric Field:


For a positively charged sheet, the electric field is directed away from the sheet on both sides.
For a negatively charged sheet, the electric field is directed towards the sheet on both sides.


Important Observation:

The expression
\[ E=\frac{\sigma}{2\varepsilon_0} \]

does not contain the distance from the sheet.

Hence, the electric field due to an infinite plane sheet is independent of the distance from the sheet and remains constant everywhere.

Therefore,
\[ \boxed{ E=\frac{\sigma}{2\varepsilon_0} } \]

is the required expression. Quick Tip: For an infinite plane sheet of charge: \[ \boxed{ E=\frac{\sigma}{2\varepsilon_0} } \] Remember: Electric field is uniform. Electric field is independent of distance from the sheet. The field is always perpendicular to the plane of the sheet. A common mistake is to include the distance from the sheet in the formula. For an infinite sheet, the electric field remains constant everywhere.

Concept:

The electric field due to a single infinite plane sheet having uniform surface charge density \(\sigma\) is
\[ E=\frac{\sigma}{2\varepsilon_0}. \]

The field is independent of the distance from the sheet and is directed:


Away from the sheet if the sheet is positively charged.
Towards the sheet if the sheet is negatively charged.


The net electric field due to two charged sheets is obtained by applying the principle of superposition of electric fields, according to which the resultant electric field at any point is the vector sum of the individual electric fields produced by each sheet.

Since the question asks for the electric field inside and outside the sheets, we consider two large parallel sheets carrying equal and opposite surface charge densities \(+\sigma\) and \(-\sigma\).

Step 1: Determine the electric field due to each sheet.


The magnitude of the electric field produced by each sheet is
\[ E=\frac{\sigma}{2\varepsilon_0}. \]

The direction of the field due to the positively charged sheet is away from it, whereas the direction of the field due to the negatively charged sheet is towards it.

Step 2: Calculate the electric field at a point inside the sheets.


At a point between the two sheets, the electric fields due to both sheets are in the same direction.

Hence, the resultant electric field is
\[ E_{inside} = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0}. \]

Therefore,
\[ \boxed{ E_{inside} = \frac{\sigma}{\varepsilon_0} } \]

The direction of this electric field is from the positively charged sheet towards the negatively charged sheet.

Step 3: Calculate the electric field at a point outside the sheets.


At any point outside the two sheets, the electric fields due to the two sheets are equal in magnitude but opposite in direction.

Therefore,
\[ E_{outside} = \frac{\sigma}{2\varepsilon_0} - \frac{\sigma}{2\varepsilon_0} = 0. \]

Hence,
\[ \boxed{ E_{outside}=0 } \]

Therefore,
\[ \boxed{ E_{inside} = \frac{\sigma}{\varepsilon_0} } \]

and
\[ \boxed{ E_{outside} = 0. } \] Quick Tip: For two large parallel sheets carrying charges \(+\sigma\) and \(-\sigma\): \[ \boxed{ E_{inside} = \frac{\sigma}{\varepsilon_0} } \] and \[ \boxed{ E_{outside}=0 } \] This arrangement forms a parallel-plate capacitor, where the electric field is uniform between the plates and zero outside.

Concept:

A Wheatstone bridge is an arrangement of four resistors connected in the form of a bridge. It is used to determine an unknown resistance accurately by comparing it with known resistances.

The bridge consists of four resistances \(P\), \(Q\), \(R\), and \(S\) connected in the form of a quadrilateral. A battery is connected across one diagonal of the bridge and a galvanometer is connected across the other diagonal.

The bridge is said to be in the balanced condition when no current flows through the galvanometer. In this condition, the potential difference across the galvanometer becomes zero.

The condition of balance is obtained by applying Ohm's law to the two arms of the bridge.

Step 1: Consider the Wheatstone bridge in the balanced state.


Let the four resistances of the bridge be
\[ P,\qquad Q,\qquad R,\qquad S \]

arranged as shown below:



A battery is connected across points \(A\) and \(C\), and a galvanometer is connected between points \(B\) and \(D\).

At balance,
\[ I_g=0, \]

where \(I_g\) is the current through the galvanometer.

Therefore, no current flows through the branch \(BD\).

Step 2: Use the condition of zero current through the galvanometer.


Since no current flows through the galvanometer,
\[ V_B=V_D. \]

That is, the potentials at points \(B\) and \(D\) are equal.

Suppose the current flowing through the branch containing \(P\) and \(R\) is \(I_1\), and the current flowing through the branch containing \(Q\) and \(S\) is \(I_2\).

Step 3: Write the potential difference between points \(A\) and \(B\), and between points \(A\) and \(D\).


Since
\[ V_B=V_D, \]

the potential drop from \(A\) to \(B\) must be equal to the potential drop from \(A\) to \(D\).

Hence,
\[ I_1P=I_2Q. \]

Therefore,
\[ \frac{I_1}{I_2} = \frac{Q}{P}. \qquad \cdots (1) \]

Step 4: Similarly, equate the potential drops from \(B\) to \(C\) and from \(D\) to \(C\).


Since the potentials of \(B\) and \(D\) are equal,
\[ I_1R=I_2S. \]

Therefore,
\[ \frac{I_1}{I_2} = \frac{S}{R}. \qquad \cdots (2) \]

Step 5: Obtain the condition for balance of the bridge.


From equations (1) and (2),
\[ \frac{Q}{P} = \frac{S}{R}. \]

Rearranging,
\[ \boxed{ \frac{P}{Q} = \frac{R}{S} } \]

or,
\[ \boxed{ PS=QR } \]

This is the required condition for the balance of a Wheatstone bridge.

Thus, when the ratio of the resistances in one pair of adjacent arms is equal to the ratio of the resistances in the other pair of adjacent arms, no current flows through the galvanometer and the bridge is said to be balanced.
\[ \boxed{ \frac{P}{Q} = \frac{R}{S} } \qquad or \qquad \boxed{ PS=QR } \] Quick Tip: For a balanced Wheatstone bridge, always remember: \[ \boxed{ \frac{P}{Q} = \frac{R}{S} } \] or equivalently, \[ \boxed{ PS=QR } \] At balance: Current through the galvanometer is zero. Potentials of the galvanometer junctions are equal. The bridge can be used to determine an unknown resistance accurately.

Concept:

To determine the equivalent resistance of a complicated resistor network, we simplify the circuit step-by-step by identifying:


Resistors connected in series,
Resistors connected in parallel,
Symmetrical combinations, if any.


Two resistors \(R_1\) and \(R_2\) connected in series have an equivalent resistance
\[ R_s=R_1+R_2. \]

Two resistors \(R_1\) and \(R_2\) connected in parallel have an equivalent resistance
\[ R_p=\frac{R_1R_2}{R_1+R_2}. \]

We shall simplify the given network from left to right.

Step 1: Identify the resistors between points \(M\) and \(P\).


Between \(M\) and \(P\), there are two possible paths:


Upper branch: one resistor of resistance \(R\).
Lower branch: two resistors of resistance \(R\) each in series.


Therefore, the resistance of the lower branch is
\[ R+R=2R. \]

Hence, between \(M\) and \(P\), we have two resistances \(R\) and \(2R\) connected in parallel.

Their equivalent resistance is
\[ R_{MP} = \frac{R(2R)}{R+2R} = \frac{2R^2}{3R} = \frac{2R}{3}. \]

Thus,
\[ \boxed{ R_{MP}=\frac{2R}{3} } \]

Step 2: Identify the resistors between points \(P\) and \(N\).


Between \(P\) and \(N\), there are again two branches:


Upper branch: one resistor of resistance \(R\).
Lower branch: one resistor of resistance \(R\).


Since these two resistors are connected in parallel,
\[ R_{PN} = \frac{R\times R}{R+R} = \frac{R^2}{2R} = \frac{R}{2}. \]

Therefore,
\[ \boxed{ R_{PN}=\frac{R}{2} } \]

Step 3: Redraw the simplified circuit.


After simplification, the circuit becomes a series combination of:
\[ 2R, \qquad \frac{2R}{3}, \qquad \frac{R}{2}, \qquad 3R. \]

Since all these equivalent resistances are connected in series, the total resistance between \(A\) and \(B\) is
\[ R_{AB} = 2R+\frac{2R}{3}+\frac{R}{2}+3R. \]

Step 4: Add all the resistances.


Taking the LCM of \(1,3,\) and \(2\), we get \(6\).

Therefore,
\[ R_{AB} = \frac{12R}{6} + \frac{4R}{6} + \frac{3R}{6} + \frac{18R}{6}. \]

Hence,
\[ R_{AB} = \frac{37R}{6}. \]

Therefore, the net resistance of the network between points \(A\) and \(B\) is
\[ \boxed{ R_{AB}=\frac{37R}{6} } \] Quick Tip: For complicated resistor networks: First identify obvious series combinations. Then identify parallel combinations. Simplify the circuit step-by-step instead of attempting the entire circuit at once. Always redraw the circuit after each simplification. For two resistors in parallel: \[ \boxed{ R_{eq} = \frac{R_1R_2}{R_1+R_2} } \] For two resistors in series: \[ \boxed{ R_{eq} = R_1+R_2 } \]

Concept:

The capacitance of a parallel plate capacitor completely filled with a dielectric medium of dielectric constant \(K\) is
\[ C=K C_0, \]

where
\[ C_0=\frac{\varepsilon_0 A}{d} \]

is the capacitance of the same capacitor without the dielectric slab.

The capacitance of a capacitor depends only on:


Geometry of the plates,
Area of the plates,
Separation between the plates,
Nature of the medium between the plates.


It does not depend on the charge stored or the potential difference across the capacitor.

Step 1: Consider the initial state of the capacitor.


Initially, the dielectric slab of dielectric constant \(K\) is completely inserted between the plates.

Therefore, the capacitance is
\[ C=KC_0. \]

The capacitor is charged by a battery and then the battery is disconnected.

Step 2: Withdraw the dielectric slab from the capacitor.


When the dielectric slab is completely removed, the medium between the plates becomes air (or vacuum).

Hence, the capacitance becomes
\[ C'=C_0. \]

Since
\[ C=KC_0, \]

we have
\[ C_0=\frac{C}{K}. \]

Therefore, the new capacitance is
\[ \boxed{ C'=\frac{C}{K} } \]

Thus, the capacitance decreases by a factor of \(K\).
\[ \boxed{ New capacitance = \frac{C}{K} } \] Quick Tip: Removing the dielectric from a capacitor decreases its capacitance by the dielectric constant \(K\). If \[ C=KC_0, \] then after removing the dielectric, \[ \boxed{ C'=\frac{C}{K} } \] Remember that capacitance depends only on the geometry and the dielectric medium and is independent of charge and potential.

Concept:

The energy stored in a capacitor can be expressed as
\[ U=\frac{1}{2}CV^2 \]

or
\[ U=\frac{Q^2}{2C} \]

or
\[ U=\frac{1}{2}QV. \]

The choice of formula depends on which quantity remains constant.

In the present case, the battery is disconnected before the dielectric slab is withdrawn. Therefore, the capacitor becomes isolated and hence the charge on the capacitor cannot change.

Thus,
\[ \boxed{Q=constant} \]

and the most convenient expression for energy is
\[ U=\frac{Q^2}{2C}. \]

Step 1: Calculate the initial energy stored in the capacitor.


Initially, the capacitance of the capacitor is \(C\) and the potential difference across it is \(V\).

Therefore, the initial energy stored is
\[ U_i=\frac{1}{2}CV^2. \]

The charge on the capacitor is
\[ Q=CV. \]

Step 2: Determine the new capacitance after removing the dielectric slab.


Let the dielectric constant of the slab be \(K\).

After the dielectric is withdrawn, the capacitance becomes
\[ C'=\frac{C}{K}. \]

Since the battery is disconnected, the charge remains unchanged:
\[ Q'=Q=CV. \]

Step 3: Calculate the new energy stored in the capacitor.


Using
\[ U=\frac{Q^2}{2C}, \]

the new energy is
\[ U_f=\frac{Q^2}{2C'}. \]

Substituting
\[ C'=\frac{C}{K}, \]

we get
\[ U_f = \frac{Q^2} {2\left(\frac{C}{K}\right)} = \frac{KQ^2}{2C}. \]

But
\[ \frac{Q^2}{2C}=U_i. \]

Hence,
\[ U_f=KU_i. \]

Since
\[ U_i=\frac12 CV^2, \]

we obtain
\[ \boxed{ U_f = K\left(\frac12 CV^2\right) } \]

or
\[ \boxed{ U_f = \frac12 KCV^2 } \]

Therefore, the energy stored in the capacitor becomes \(K\) times its initial value.
\[ \boxed{ New energy = K \times Initial energy } \] Quick Tip: If the battery is disconnected, then \[ \boxed{Q=constant} \] and therefore \[ \boxed{ U=\frac{Q^2}{2C} } \] must be used. Hence, when the dielectric is removed, \[ C \downarrow \quad\Rightarrow\quad U \uparrow \] and \[ \boxed{ U_f=K\,U_i } \] where \(K\) is the dielectric constant of the slab.

Concept:

The potential difference across a capacitor is related to its charge and capacitance by the relation
\[ V=\frac{Q}{C}. \]

When the battery is disconnected, the capacitor becomes isolated from the external circuit. Therefore, no charge can flow into or out of the capacitor and hence
\[ \boxed{Q=constant} \]

during the withdrawal of the dielectric slab.

However, the capacitance changes because the dielectric medium between the plates changes.

Initially, the capacitor has capacitance \(C\), and after removing the dielectric slab, the capacitance decreases.

Since the charge remains constant,
\[ V\propto \frac{1}{C}. \]

Therefore, if the capacitance decreases, the potential difference must increase.

Step 1: Write the initial charge on the capacitor.


Initially,
\[ Q=CV, \]

where


\(C\) is the initial capacitance,
\(V\) is the initial potential difference.


Step 2: Determine the new capacitance after removing the dielectric slab.


Let the dielectric constant of the slab be \(K\).

When the dielectric slab is completely withdrawn, the new capacitance becomes
\[ C'=\frac{C}{K}. \]

Since the battery is disconnected,
\[ Q'=Q=CV. \]

Step 3: Calculate the new potential difference.


The new potential difference is
\[ V'=\frac{Q'}{C'}. \]

Substituting the values,
\[ V' = \frac{CV}{C/K}. \]

Therefore,
\[ V' = K V. \]

Hence,
\[ \boxed{ V'=KV } \]

Thus, the potential difference across the capacitor becomes \(K\) times its initial value.
\[ \boxed{ New potential difference = K \times Initial potential difference } \]

Therefore, the potential difference increases by a factor equal to the dielectric constant of the slab. Quick Tip: When the battery is disconnected: \[ \boxed{Q=constant} \] and \[ V=\frac{Q}{C}. \] Therefore, if the dielectric is removed, \[ C \downarrow \quad\Rightarrow\quad V \uparrow \] Specifically, \[ \boxed{ V'=KV } \] where \(K\) is the dielectric constant of the slab.

Concept:

When an electron enters the region between two oppositely charged parallel plates, it experiences a uniform electric field. The electric force acting on the electron produces a constant acceleration perpendicular to its initial direction of motion.

Thus, the motion of the electron is:


Uniform motion along the horizontal direction.
Uniformly accelerated motion along the vertical direction.


This is exactly analogous to the motion of a projectile.

The electric field between two parallel plates is
\[ E=\frac{V}{d}, \]

where


\(V\) is the potential difference between the plates,
\(d\) is the separation between the plates.


The force acting on the electron is
\[ F=eE, \]

and hence the acceleration of the electron is
\[ a=\frac{eE}{m} =\frac{eV}{md}. \]

Step 1: Determine the time for which the electron remains between the plates.


The electron enters horizontally with velocity
\[ u_x=3\times10^7\ m s^{-1}. \]

The length of each plate is
\[ l=3\ cm =3\times10^{-2}\ m. \]

Since there is no horizontal acceleration,
\[ t=\frac{l}{u_x} \]
\[ t= \frac{3\times10^{-2}} {3\times10^7} \]
\[ \boxed{ t=10^{-9}\ s } \]

Step 2: Calculate the vertical displacement of the electron.


The electron enters symmetrically between the plates.

Since the separation between the plates is
\[ d=2\ cm =2\times10^{-2}\ m, \]

the electron has to travel a vertical distance of
\[ y=\frac{d}{2} =1\times10^{-2}\ m \]

to just strike the lower plate at the end \(P_2'\).

Initially, the vertical velocity is zero.

Therefore,
\[ y=\frac12 at^2. \]

Substituting the values,
\[ 10^{-2} = \frac12 a (10^{-9})^2. \]

Hence,
\[ a = \frac{2\times10^{-2}} {10^{-18}} \]
\[ \boxed{ a=2\times10^{16}\ m s^{-2} } \]

Step 3: Use the expression for electric acceleration.


The acceleration of the electron is
\[ a=\frac{eV}{md}. \]

Therefore,
\[ V=\frac{amd}{e}. \]

Substituting
\[ a=2\times10^{16}\ m s^{-2}, \]
\[ m=9.1\times10^{-31}\ kg, \]
\[ d=2\times10^{-2}\ m, \]
\[ e=1.6\times10^{-19}\ C, \]

we get
\[ V= \frac{(2\times10^{16}) (9.1\times10^{-31}) (2\times10^{-2})} {1.6\times10^{-19}}. \]
\[ V= \frac{36.4\times10^{-17}} {1.6\times10^{-19}}. \]
\[ V= 22.75\times10^{2}. \]

Therefore,
\[ \boxed{ V\approx2.28\times10^{3}\ V } \]

or
\[ \boxed{ V\approx2.3\times10^{3}\ V } \]

Hence, the potential difference that should be applied between the plates so that the electron beam just strikes the end \(P_2'\) is
\[ \boxed{ V\approx2.3\ kV } \] Quick Tip: For an electron entering between parallel plates: \[ E=\frac{V}{d}, \qquad a=\frac{eV}{md} \] and the motion is projectile-like: \[ x=u_xt, \qquad y=\frac12 at^2. \] Always calculate the time of flight first using horizontal motion and then use vertical motion to determine the required potential difference.

Concept:

An ideal diode conducts only when it is forward biased and behaves like a closed switch. When it is reverse biased, it behaves like an open switch and no current flows through it.

For a diode:


Forward bias : Anode at a higher potential than the cathode.
Reverse bias : Cathode at a higher potential than the anode.


Step 1: Determine the polarity during the positive half-cycle.


During the positive half-cycle of the input voltage \(V_i\),
\[ V_A>V_B, \]

that is, point \(A\) is at a higher potential than point \(B\).

Step 2: Examine the biasing of the two diodes.


From the figure:


For diode \(D_1\), the cathode is connected towards point \(A\). Since \(A\) is at a higher potential, \(D_1\) becomes reverse biased.

For diode \(D_2\), the anode is connected towards point \(A\) and its cathode is towards point \(R\). Therefore, \(D_2\) becomes forward biased.


Hence,
\[ \boxed{Diode D_2 conducts and diode D_1 remains OFF.} \]

The reason is that during the positive half-cycle, \(D_2\) is forward biased whereas \(D_1\) is reverse biased. Quick Tip: For an ideal diode: \[ \boxed{ Forward biased \Rightarrow Conducts } \] \[ \boxed{ Reverse biased \Rightarrow Does not conduct } \] Always compare the potentials of the anode and cathode to determine whether a diode is ON or OFF.

From part (a), during the positive half-cycle of the input voltage, diode \(D_2\) is forward biased and conducts, whereas diode \(D_1\) is reverse biased and does not conduct.

Therefore,


\(D_2\) is replaced by a conducting wire (short circuit).
\(D_1\) is replaced by an open circuit.


Hence, the equivalent circuit becomes



Thus, the \(1\,k\Omega\) and \(2\,k\Omega\) resistors are connected in series between \(P\) and \(B\), and this series combination is in parallel with the \(3\,k\Omega\) resistor. Quick Tip: For an ideal diode: \[ \boxed{ ON \Rightarrow Short circuit } \] \[ \boxed{ OFF \Rightarrow Open circuit } \] Always redraw the circuit after replacing the conducting and non-conducting diodes.

Concept:

The applied voltage is
\[ V_i=12\sin(100\pi t)\ V. \]

The peak value of the input voltage is
\[ V_{peak}=12\ V. \]

From part (b), the equivalent circuit consists of:


a \(3\,k\Omega\) resistor directly across the source,
a series combination of \(1\,k\Omega\) and \(2\,k\Omega\) resistors also connected across the source.


Hence, both branches have the same potential difference of \(12\) V.

Step 1: Calculate the current through the branch containing \(1\,k\Omega\) and \(2\,k\Omega\).


The equivalent resistance of this branch is
\[ R_s=1\,k\Omega+2\,k\Omega=3\,k\Omega. \]

Therefore,
\[ I=\frac{12}{3\times10^3} =4\times10^{-3}\ A =4\ mA. \]

Step 2: Calculate the voltage drop across the \(1\,k\Omega\) resistor.



::contentReference[oaicite:0]{index=0

\[ V_{1k}=IR \]
\[ V_{1k} = (4\times10^{-3})(1000) = 4\ V. \]

Hence,
\[ \boxed{V_{1k}=4\ V} \]

Step 3: Calculate the voltage drop across the \(2\,k\Omega\) resistor.

\[ V_{2k} = (4\times10^{-3})(2000) = 8\ V. \]

Therefore,
\[ \boxed{V_{2k}=8\ V} \]

Step 4: Calculate the voltage drop across the \(3\,k\Omega\) resistor.


Since the \(3\,k\Omega\) resistor is directly connected across the source,
\[ V_{3k}=12\ V. \]

Thus,
\[ \boxed{V_{3k}=12\ V} \]

Hence, the output voltage drops across the three resistors are
\[ \boxed{ V_{1k}=4\ V,\qquad V_{2k}=8\ V,\qquad V_{3k}=12\ V } \] Quick Tip: After replacing ideal diodes by short or open circuits, simplify the resistor network first. Then apply \[ V=IR \] to calculate the voltage drop across each resistor. Remember that resistors connected in parallel have the same potential difference across them.

Concept:

A p-n junction is formed when a p-type semiconductor and an n-type semiconductor are joined together. Immediately after the formation of the junction, the charge carriers on the two sides are not uniformly distributed.


The p-region contains a large concentration of holes and a very small concentration of electrons.
The n-region contains a large concentration of electrons and a very small concentration of holes.


Because of this concentration difference, charge carriers start moving across the junction. During the formation of the p-n junction, two important processes take place:


Diffusion
Drift


Step 1: Diffusion Process


Diffusion is the movement of charge carriers from the region of higher concentration to the region of lower concentration.

Immediately after the p-type and n-type semiconductors are joined:


Electrons diffuse from the n-side to the p-side because the concentration of electrons is much higher in the n-region.
Holes diffuse from the p-side to the n-side because the concentration of holes is much higher in the p-region.


This movement of majority charge carriers across the junction constitutes the diffusion current.

As electrons leave the n-region, positively charged donor ions are left behind. Similarly, as holes leave the p-region, negatively charged acceptor ions are left behind.

As a result, a region around the junction becomes depleted of mobile charge carriers. This region is called the depletion region or depletion layer.
\[ \boxed{ Diffusion= Movement of majority carriers from high concentration to low concentration. } \]

Step 2: Drift Process


The immobile ions left behind near the junction create an electric field directed from the n-side towards the p-side.

This electric field opposes further diffusion of majority carriers and exerts a force on the charge carriers.

Due to this electric field:


Electrons in the p-region are driven towards the n-region.
Holes in the n-region are driven towards the p-region.


This motion of minority carriers under the influence of the electric field gives rise to the drift current.

The process of drift continues until the drift current becomes equal in magnitude and opposite in direction to the diffusion current.

At this stage, the p-n junction attains equilibrium.
\[ \boxed{ Drift= Motion of minority carriers due to the electric field of the depletion region. } \]

Conclusion:

Thus, the two important processes that occur during the formation of a p-n junction are:


Diffusion – movement of majority charge carriers from a region of higher concentration to a region of lower concentration.

Drift – movement of minority charge carriers under the influence of the electric field developed across the depletion region.


The equilibrium condition is reached when
\[ \boxed{ I_{diffusion} = I_{drift} } \]

and therefore the net current across the junction becomes zero. Quick Tip: Remember the sequence of events during the formation of a p-n junction: \[ Diffusion \rightarrow Formation of depletion layer \rightarrow Electric field \rightarrow Drift \rightarrow Equilibrium \] At equilibrium, \[ \boxed{ I_{diffusion} = I_{drift} } \] and the net current through the junction is zero.

Concept:

A refracting telescope consists of two convex lenses:


An objective lens of large focal length \(f_o\) and large aperture.
An eyepiece lens of small focal length \(f_e\) and small aperture.


The telescope is used to observe distant objects. Since the object is at a very large distance (practically at infinity), the rays coming from the object are nearly parallel.

The objective forms a real, inverted and diminished image of the distant object at its principal focus. This image acts as the object for the eyepiece.

In the condition of normal adjustment, the final image is formed at infinity so that the observer can view it comfortably without any strain on the eyes.


Step 1: Condition for normal adjustment.


For normal adjustment, the real image formed by the objective lies at the first focal plane of the eyepiece.

Hence, the separation between the objective and eyepiece is
\[ \boxed{ L=f_o+f_e } \]

where


\(f_o\) is the focal length of the objective,
\(f_e\) is the focal length of the eyepiece.


Step 2: Expression for angular magnification.


The angular magnification or magnifying power of a telescope is defined as
\[ M=\frac{Angle subtended by the final image at the eye} {Angle subtended by the object at the unaided eye}. \]

For a refracting telescope in normal adjustment,
\[ \boxed{ M=-\frac{f_o}{f_e} } \]

The negative sign indicates that the final image is inverted with respect to the object.

Hence, the magnitude of angular magnification is
\[ \boxed{ |M|=\frac{f_o}{f_e} } \] Quick Tip: For a refracting telescope in normal adjustment: \[ \boxed{ L=f_o+f_e } \] and \[ \boxed{ M=-\frac{f_o}{f_e} } \] A larger focal length of the objective and a smaller focal length of the eyepiece give a larger magnifying power.

Concept:

A reflecting telescope uses a concave mirror as the objective instead of a lens. Reflecting telescopes are preferred over refracting telescopes because mirrors have several practical advantages over lenses.

Reason 1: Absence of chromatic aberration.


A lens forms images by refraction and different colours of light suffer different amounts of refraction. Consequently, images formed by lenses suffer from chromatic aberration.

A mirror forms images by reflection, and the law of reflection is independent of the wavelength of light.

Therefore,
\[ \boxed{ Reflecting telescopes are free from chromatic aberration. } \]

Reason 2: Large aperture can be obtained easily.


In a refracting telescope, a large lens becomes very heavy and can be supported only at its edges, causing deformation and image defects.

On the other hand, a mirror can be supported from its back, making it possible to manufacture mirrors of very large diameter.

Therefore,
\[ \boxed{ Reflecting telescopes can have very large apertures and hence greater light-gathering power. } \]

Hence, a reflecting telescope is preferred because:


It is free from chromatic aberration.
Very large objective mirrors can be constructed, giving greater light-gathering and resolving power. Quick Tip: Advantages of a reflecting telescope: No chromatic aberration. Large aperture possible. Greater light-gathering power. Easier and cheaper to manufacture large objectives.

Concept:

Total internal reflection (TIR) is the phenomenon in which a ray of light travelling from one medium to another is completely reflected back into the denser medium without suffering any refraction.

This phenomenon occurs only under certain specific conditions. If these conditions are not satisfied, the light ray undergoes ordinary refraction instead of total internal reflection.

Condition 1: The light must travel from an optically denser medium to an optically rarer medium.


Suppose the refractive indices of the two media are
\[ n_1 \quad and \quad n_2, \]

where
\[ n_1>n_2. \]

Then the light ray must travel from medium \(1\) (denser medium) to medium \(2\) (rarer medium).

Examples:


Glass \(\rightarrow\) Air
Water \(\rightarrow\) Air
Diamond \(\rightarrow\) Air


If light travels from a rarer medium to a denser medium, total internal reflection cannot occur.

Hence, the first condition is
\[ \boxed{ Light must travel from an optically denser medium to an optically rarer medium. } \]

Condition 2: The angle of incidence in the denser medium must be greater than the critical angle.


The critical angle \(C\) is defined as the angle of incidence in the denser medium for which the angle of refraction in the rarer medium becomes
\[ 90^\circ. \]

Thus,
\[ \boxed{ i=C \quad \Rightarrow \quad r=90^\circ } \]

For total internal reflection to occur,
\[ \boxed{ i>C } \]

where


\(i\) is the angle of incidence,
\(C\) is the critical angle.


If i
ordinary refraction takes place.

Therefore, the second condition is
\[ \boxed{ The angle of incidence must be greater than the critical angle. } \]

Hence, the two necessary conditions for total internal reflection are:


The light ray must travel from an optically denser medium to an optically rarer medium.

The angle of incidence in the denser medium must be greater than the critical angle for the pair of media. Quick Tip: Remember the two conditions for total internal reflection: \[ \boxed{ Denser Medium \rightarrow Rarer Medium } \] and \[ \boxed{ i>C } \] where \(C\) is the critical angle. A simple way to remember: \[ \boxed{ DRC Rule } \] \[ D \rightarrow Denser to Rarer, \qquad C \rightarrow Incidence angle greater than Critical angle. \]

Concept:

As the light ray travels from one medium to another, its direction changes according to Snell's law:
\[ n_1\sin i=n_2\sin r. \]

The three liquids have refractive indices
\[ \mu_A=n,\qquad \mu_B=\frac{3n}{4},\qquad \mu_C=\frac{2n}{3}. \]

Since
\[ n>\frac{3n}{4}>\frac{2n}{3}, \]

each successive medium is optically rarer than the previous one.

Consequently, the ray bends away from the normal at every interface.

To prove that the beam does not enter region \(C\), we must show that the ray undergoes total internal reflection at the interface between liquids \(B\) and \(C\).

Step 1: Apply Snell's law at the interface between liquids \(A\) and \(B\).


Let the angle of refraction in liquid \(B\) be \(r\).

Applying Snell's law,
\[ n\sin\theta = \frac{3n}{4}\sin r. \]

Cancelling \(n\) from both sides,
\[ \sin r = \frac{4}{3}\sin\theta. \]

Therefore,
\[ \boxed{ \sin r=\frac{4}{3}\sin\theta } \]

Step 2: Find the critical angle for the interface between liquids \(B\) and \(C\).


The refractive index of liquid \(B\) is
\[ \mu_B=\frac{3n}{4} \]

and that of liquid \(C\) is
\[ \mu_C=\frac{2n}{3}. \]

If \(C\) is the critical angle for the interface \(B-C\), then
\[ \sin C = \frac{\mu_C}{\mu_B}. \]

Substituting the values,
\[ \sin C = \frac{\frac{2n}{3}} {\frac{3n}{4}}. \]
\[ \sin C = \frac{2n}{3}\times\frac{4}{3n} = \frac{8}{9}. \]

Hence,
\[ \boxed{ \sin C=\frac{8}{9} } \]

Step 3: Determine the condition for total internal reflection.


For the ray not to enter liquid \(C\), the angle of incidence at the \(B-C\) interface must be greater than the critical angle.

That is,
\[ r>C. \]

Since the sine function is increasing in the range \(0^\circ\) to \(90^\circ\),
\[ \sin r>\sin C. \]

Substituting the values of \(\sin r\) and \(\sin C\),
\[ \frac{4}{3}\sin\theta > \frac{8}{9}. \]

Multiplying both sides by \(\dfrac34\),
\[ \sin\theta > \frac{8}{9}\times\frac34. \]

Therefore,
\[ \boxed{ \sin\theta>\frac23 } \]

Thus, whenever
\[ \boxed{ \sin\theta>\frac23, } \]

the angle of incidence at the interface between liquids \(B\) and \(C\) exceeds the critical angle and the ray undergoes total internal reflection.

Hence, the laser beam does not enter region \(C\) at all.
\[ \boxed{ For \sin\theta>\frac23,\ the beam is totally internally reflected in liquid B and never enters region C. } \] Quick Tip: For problems involving several layers of liquids: Use Snell's law successively at each interface. Find the critical angle at the final interface. Apply the condition for total internal reflection: \[ i>C qquador\qquad \sin i>\sin C. \] In this problem, \[ \boxed{ \sin r=\frac43\sin\theta } qquadand\qquad \boxed{ \sin C=\frac89 } \] which finally gives \[ \boxed{ \sin\theta>\frac23. } \]

Concept:

The deflecting torque acting on a current carrying coil placed in a magnetic field is
\[ \tau = NBAI\sin\theta \]

where


\(N\) = number of turns,
\(B\) = magnetic field,
\(A\) = area of coil,
\(I\) = current through coil,
\(\theta\) = angle between magnetic field and normal to the coil.


In an ordinary magnetic field, the torque depends on \(\theta\). Hence the torque changes as the coil rotates.



Step 1: Understand the purpose of radial magnetic field.

In a moving coil galvanometer, the pole pieces are specially shaped so that the magnetic field becomes radial.

In a radial magnetic field, the plane of the coil always remains parallel to the magnetic field lines.

Therefore,
\[ \theta = 90^\circ \]

for all positions of the coil.



Step 2: Substitute into torque equation.

Since
\[ \sin 90^\circ =1, \]

the torque becomes
\[ \tau = NBAI. \]

This expression is independent of the angular position of the coil.

Hence the torque remains constant during rotation.



Step 3: Analyse other options.


Soft iron core increases magnetic field strength but does not make torque independent of orientation.
Hair spring provides restoring torque only.
Eddy currents provide damping and do not affect constant torque.


Therefore,
\[ \boxed{Correct Option (B)} \] Quick Tip: A radial magnetic field ensures \[ \tau = NBAI \] and makes galvanometer deflection directly proportional to current.

Concept:

Current sensitivity is defined as the angular deflection produced per unit current.
\[ S_i=\frac{\theta}{I} \]

For a moving coil galvanometer,
\[ NBAI=C\theta \]

where \(C\) is the torsional constant of the spring.

Hence,
\[ \frac{\theta}{I} = \frac{NBA}{C}. \]

Therefore,
\[ S_i=\frac{NBA}{C}. \]



Step 1: Observe factors affecting sensitivity.

Sensitivity increases when
\[ N,\; B,\; A \]

increase and decreases when
\[ C \]

increases.



Step 2: Examine options.


Increasing area \(A\) increases sensitivity.
Increasing magnetic field \(B\) increases sensitivity.
Decreasing area and magnetic field decreases sensitivity.
Increasing torsional constant decreases sensitivity.


Thus the most effective choice among the given options is
\[ \boxed{(B)} \] Quick Tip: Current sensitivity: \[ S_i=\frac{NBA}{C} \] Increase \(N\), \(B\), \(A\) and decrease \(C\) to obtain higher sensitivity.

Using the torque expression for a galvanometer,
\[ \tau = NBAI. \]

Substituting the given values,
\[ \tau = 50\times0.25\times4\times10^{-3}\times5. \]
\[ \tau = 50\times0.005. \]
\[ \tau=0.25\,N m. \]

Therefore,
\[ \boxed{\tau=0.25\,N m} \]

Hence,
\[ \boxed{Correct Option (D)} \] Quick Tip: For a radial magnetic field, \[ \tau = NBAI. \] Always use this formula directly for torque calculations in galvanometers.

For converting a galvanometer into a voltmeter,
\[ R=\frac{V}{I_g}-G \]

where
\[ V=12V, \qquad I_g=3\times10^{-3}A, \qquad G=15\Omega. \]



Substituting,
\[ R = \frac{12}{3\times10^{-3}} -15. \]
\[ R=4000-15. \]
\[ R=3985\Omega. \]

Therefore,
\[ \boxed{R=3985\Omega} \]

Hence,
\[ \boxed{Correct Option (B)} \] Quick Tip: To convert a galvanometer into a voltmeter, \[ R=\frac{V}{I_g}-G \] where \(R\) is connected in series with the galvanometer.

Step 1: Use shunt resistance formula.

For conversion into an ammeter,
\[ S=\frac{I_gG}{I-I_g}. \]

Given,
\[ I_g=5\times10^{-3}A, \]
\[ G=20\Omega, \]
\[ I=10A. \]



Step 2: Substitute values.
\[ S = \frac{(5\times10^{-3})(20)} {10-0.005}. \]
\[ S = \frac{0.1}{9.995}. \]
\[ S \approx0.01\Omega. \]



Step 3: Determine connection type.

A shunt resistance is always connected in parallel with the galvanometer.

Therefore,
\[ \boxed{ S=0.01\Omega } \]

connected in parallel.

Hence,
\[ \boxed{Correct Option (C)} \] Quick Tip: To convert a galvanometer into an ammeter, a low resistance called shunt is connected in parallel. \[ S=\frac{I_gG}{I-I_g} \]

Concept:

Einstein's photoelectric equation is
\[ h\nu=\phi+K_{\max} \]

where
\[ \phi=h\nu_0 \]

is the work function of the metal and \(\nu_0\) is its threshold frequency.

The stopping potential is related to maximum kinetic energy by
\[ eV_s=K_{\max}. \]

Combining these equations,
\[ V_s=\frac{h}{e}\nu-\frac{\phi}{e}. \]

Thus, the graph of \(V_s\) versus \(\nu\) is a straight line.



Step 1: Identify threshold frequencies.

The threshold frequency is obtained where
\[ V_s=0. \]

The corresponding intercepts on the frequency axis are
\[ \nu_1 \quad and \quad \nu_2. \]



Step 2: Calculate work functions.

Since
\[ \phi=h\nu_0, \]

the work functions of metals A and B are
\[ \phi_A=h\nu_1 \]

and
\[ \phi_B=h\nu_2. \]

Therefore,
\[ \boxed{Correct Option (C)} \] Quick Tip: Threshold frequency and work function are related by \[ \phi=h\nu_0. \] The intercept on the frequency axis directly gives the threshold frequency.

According to Einstein's equation,
\[ K_{\max}=h\nu-\phi. \]

For the same incident frequency \(\nu\),
\[ K_{\max} \]

depends upon the work function.



Step 1: Compare work functions.

From the graph,
\[ \nu_2>\nu_1. \]

Therefore,
\[ \phi_B=h\nu_2 > h\nu_1=\phi_A. \]

Hence metal B has a larger work function.



Step 2: Compare maximum kinetic energies.

Since
\[ K_{\max}=h\nu-\phi, \]

a smaller work function produces a larger kinetic energy.

Thus,
\[ K_{\max}(A) > K_{\max}(B). \]

Therefore,
\[ \boxed{Correct Option (A)} \] Quick Tip: For the same incident frequency, \[ K_{\max}=h\nu-\phi. \] Smaller work function \(\Rightarrow\) larger kinetic energy.

Concept:

The photoelectric equation is
\[ V_s=\frac{h}{e}\nu-\frac{\phi}{e}. \]

The slope is
\[ \frac{h}{e}. \]

Since \(h\) and \(e\) are constants, the slope does not depend upon intensity.



Step 1: Effect of increasing intensity.

Increasing intensity increases the number of incident photons per second.

Hence more electrons are emitted per second.



Step 2: Effect on stopping potential and threshold frequency.

Stopping potential depends on frequency and not on intensity.

Threshold frequency is a characteristic property of the metal and remains unchanged.



Therefore,
\[ \boxed{Correct Option (D)} \] Quick Tip: Intensity affects the number of emitted photoelectrons, whereas frequency determines their maximum kinetic energy.

Using Einstein's equation,
\[ K_{\max}=h(\nu-\nu_0). \]



Step 1: Calculate \(E_1\).

For frequency
\[ \nu=3\nu_0, \]
\[ E_1=h(3\nu_0-\nu_0). \]
\[ E_1=2h\nu_0. \]



Step 2: Calculate \(E_2\).

For frequency
\[ \nu=6\nu_0, \]
\[ E_2=h(6\nu_0-\nu_0). \]
\[ E_2=5h\nu_0. \]



Step 3: Find ratio.
\[ \frac{E_1}{E_2} = \frac{2h\nu_0}{5h\nu_0}. \]
\[ \boxed{ \frac{E_1}{E_2} = \frac25 } \]

Hence,
\[ \boxed{Correct Option (C)} \] Quick Tip: For quick calculations, \[ K_{\max}=h(\nu-\nu_0). \] Always subtract the threshold frequency first.

The stopping potential equation is
\[ V_s=\frac{h}{e}\nu-\frac{\phi}{e}. \]

Comparing with the equation of a straight line,
\[ y=mx+c, \]

the slope is
\[ m=\frac{h}{e}. \]

Multiplying both sides by \(e\),
\[ h=me. \]

Therefore,
\[ \boxed{h=me} \]

Hence,
\[ \boxed{Correct Option (A)} \] Quick Tip: For a graph of \(V_s\) versus \(\nu\), \[ Slope=\frac{h}{e}. \] Thus, \[ h=(slope)\times e. \]

Concept:

An electric dipole consists of two equal and opposite charges separated by a small distance.

The electric dipole moment is defined as
\[ \vec p = q(2a)\,\hat{i} \]

and its magnitude is
\[ p=2aq. \]

The electric field due to a dipole depends upon the position of the observation point. Here we are required to determine the electric field at a point on the equatorial line (perpendicular bisector) of the dipole.



Step 1: Choose a suitable coordinate system.

Let the dipole be placed along the \(x\)-axis with
\[ +q at (a,0) \]

and
\[ -q at (-a,0). \]

Let \(P\) be a point on the equatorial line at distance \(r\) from the centre \(O\).

The coordinates of \(P\) are
\[ (0,r). \]



Step 2: Calculate distance of point \(P\) from each charge.

Distance from \(+q\) to \(P\) is
\[ d=\sqrt{r^2+a^2}. \]

Similarly, distance from \(-q\) to \(P\) is also
\[ d=\sqrt{r^2+a^2}. \]

Hence both charges are equidistant from the observation point.



Step 3: Determine electric field due to each charge.

Magnitude of electric field due to either charge is
\[ E_0=\frac{1}{4\pi\varepsilon_0} \frac{q}{r^2+a^2}. \]

The field due to \(+q\) is directed away from \(+q\), while the field due to \(-q\) is directed towards \(-q\).



Step 4: Resolve the electric fields into components.

Let \(\theta\) be the angle made by the line joining the charge to point \(P\) with the equatorial axis.

Then
\[ \cos\theta = \frac{a}{\sqrt{r^2+a^2}}. \]

The vertical components of the two electric fields are equal in magnitude and opposite in direction.

Hence they cancel each other completely.
\[ E_y=0. \]

The horizontal components are in the same direction and therefore add together.



Step 5: Calculate resultant electric field.

Horizontal component due to one charge is
\[ E_0\cos\theta. \]

Therefore,
\[ E = 2E_0\cos\theta. \]

Substituting the values,
\[ E = 2 \left( \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2+a^2} \right) \left( \frac{a}{\sqrt{r^2+a^2}} \right). \]

Hence
\[ E = \frac{1}{4\pi\varepsilon_0} \frac{2aq}{(r^2+a^2)^{3/2}}. \]

Since
\[ p=2aq, \]

we obtain
\[ \boxed{ E = \frac{1}{4\pi\varepsilon_0} \frac{p}{(r^2+a^2)^{3/2}} } \]

The direction of this field is opposite to the direction of the dipole moment.

Therefore, vectorially
\[ \boxed{ \vec E = -\frac{1}{4\pi\varepsilon_0} \frac{\vec p}{(r^2+a^2)^{3/2}} } \]



Step 6: Electric field at a far off point \((r \gg a)\).

For a distant point,
\[ r^2+a^2 \approx r^2. \]

Therefore,
\[ (r^2+a^2)^{3/2} \approx r^3. \]

Substituting into the above expression,
\[ \boxed{ \vec E = -\frac{1}{4\pi\varepsilon_0} \frac{\vec p}{r^3} } \]

or in magnitude form,
\[ \boxed{ E = \frac{1}{4\pi\varepsilon_0} \frac{p}{r^3} } \]

directed opposite to the dipole moment. Quick Tip: Electric field due to a dipole: On axial line: \[ E_{axial} = \frac{1}{4\pi\varepsilon_0} \frac{2p}{r^3} \] On equatorial line: \[ E_{equatorial} = \frac{1}{4\pi\varepsilon_0} \frac{p}{r^3} \] The equatorial field is opposite to the direction of the dipole moment.

Concept:

When an electric dipole is placed in a uniform electric field,


the net force on the dipole is zero,
a torque acts on the dipole if its dipole moment is not parallel to the field.


The force on a charge in an electric field is
\[ \vec F=q\vec E. \]

The torque on a dipole is
\[ \vec\tau=\vec p\times\vec E. \]



Step 1: Determine the dipole moment.

The charges are situated along the \(x\)-axis.

Therefore the dipole axis lies along the \(x\)-direction.

Hence
\[ \vec p = p\,\hat{i}. \]

where
\[ p=q|a-b|. \]



Step 2: Calculate the net force on the dipole.

Force on charge \(+q\) is
\[ \vec F_1 = q\vec E. \]

Since
\[ \vec E=2\hat{i}, \]
\[ \vec F_1 = 2q\hat{i}. \]

Force on charge \(-q\) is
\[ \vec F_2 = (-q)\vec E. \]

Therefore,
\[ \vec F_2 = -2q\hat{i}. \]

Hence total force is
\[ \vec F = \vec F_1+\vec F_2. \]
\[ \vec F = 2q\hat{i}-2q\hat{i}. \]
\[ \boxed{ \vec F=0 } \]

Thus a uniform electric field exerts no net translational force on the dipole.



Step 3: Calculate the torque on the dipole.

The torque is
\[ \vec\tau = \vec p\times\vec E. \]

Since both \(\vec p\) and \(\vec E\) are along the \(x\)-axis,
\[ \vec p = p\hat{i}, \qquad \vec E = 2\hat{i}. \]

Therefore,
\[ \vec\tau = (p\hat{i})\times(2\hat{i}). \]

Since
\[ \hat{i}\times\hat{i}=0, \]

we get
\[ \boxed{ \vec\tau=0 } \]



Final Answer:
\[ \boxed{\vec F=0} \]

and
\[ \boxed{\vec\tau=0} \] Quick Tip: For a dipole placed in a uniform electric field: \[ \vec F = 0 \] and \[ \vec\tau = \vec p \times \vec E. \] If \(\vec p\) is parallel or antiparallel to \(\vec E\), then \[ \tau = 0. \]

Concept:

When two cells are connected in parallel, the combination can be replaced by a single equivalent cell having an equivalent emf \(E_{eq}\) and an equivalent internal resistance \(r_{eq}\).

The equivalent cell should supply the same current to any external circuit as the original combination.

To determine the equivalent emf and internal resistance, we make use of Kirchhoff's laws and Thevenin's equivalent concept.



Step 1: Consider the parallel combination of the two cells.

Let the positive terminals of the cells be connected together and the negative terminals also be connected together.

Suppose the open-circuit terminal voltage of the combination is \(V\).

Since no external current is drawn under open-circuit condition, the algebraic sum of currents in the two branches must be zero.
\[ I_1+I_2=0 \]

or
\[ I_1=-I_2. \]



Step 2: Write the terminal voltage across each cell.

For the first cell,
\[ V=E_1-I_1r_1 \]

and for the second cell,
\[ V=E_2-I_2r_2. \]

Since
\[ I_2=-I_1, \]

we obtain
\[ E_1-I_1r_1 = E_2+I_1r_2. \]

Therefore,
\[ I_1(r_1+r_2) = E_1-E_2. \]

Hence,
\[ I_1 = \frac{E_1-E_2}{r_1+r_2}. \]



Step 3: Determine the equivalent emf.

Substituting \(I_1\) into
\[ V=E_1-I_1r_1, \]

we get
\[ V = E_1-\frac{r_1(E_1-E_2)}{r_1+r_2}. \]

Taking the LCM,
\[ V = \frac{E_1(r_1+r_2)-r_1(E_1-E_2)} {r_1+r_2}. \]

Simplifying,
\[ V = \frac{E_1r_2+E_2r_1} {r_1+r_2}. \]

This open-circuit voltage is the equivalent emf.

Therefore,
\[ \boxed{ E_{eq} = \frac{E_1r_2+E_2r_1} {r_1+r_2} } \]



Step 4: Determine the equivalent internal resistance.

To find internal resistance, replace each ideal emf source by a short circuit.

The internal resistances \(r_1\) and \(r_2\) then appear in parallel.

Hence,
\[ r_{eq} = \frac{r_1r_2}{r_1+r_2}. \]

Therefore,
\[ \boxed{ r_{eq} = \frac{r_1r_2}{r_1+r_2} } \]



Final Result:

Equivalent emf:
\[ \boxed{ E_{eq} = \frac{E_1r_2+E_2r_1} {r_1+r_2} } \]

Equivalent internal resistance:
\[ \boxed{ r_{eq} = \frac{r_1r_2}{r_1+r_2} } \] Quick Tip: For two cells connected in parallel, \[ E_{eq} = \frac{E_1r_2+E_2r_1} {r_1+r_2} \] and \[ r_{eq} = \frac{r_1r_2}{r_1+r_2}. \] If both cells have equal internal resistances, the equivalent emf becomes the arithmetic mean of the two emfs.

Concept:

The two cells are first replaced by their equivalent cell.

Once the equivalent emf and equivalent internal resistance are obtained, the circuit becomes a simple series combination of:


equivalent emf \(E_{eq}\),
equivalent internal resistance \(r_{eq}\),
external resistance \(2R\).


The current can then be found using Ohm's law.



Step 1: Write the given data.

For the first cell,
\[ E_1=E, \qquad r_1=R. \]

For the second cell,
\[ E_2=3E, \qquad r_2=R. \]



Step 2: Calculate the equivalent emf.

Using
\[ E_{eq} = \frac{E_1r_2+E_2r_1} {r_1+r_2}, \]

we get
\[ E_{eq} = \frac{(E)(R)+(3E)(R)} {R+R}. \]
\[ E_{eq} = \frac{4ER}{2R}. \]
\[ \boxed{ E_{eq}=2E } \]



Step 3: Calculate the equivalent internal resistance.

Using
\[ r_{eq} = \frac{r_1r_2} {r_1+r_2}, \]
\[ r_{eq} = \frac{R\times R} {R+R}. \]
\[ r_{eq} = \frac{R^2}{2R}. \]
\[ \boxed{ r_{eq} = \frac{R}{2} } \]



Step 4: Determine total circuit resistance.

External resistance
\[ =2R. \]

Total resistance in the circuit is
\[ R_{total} = 2R+\frac{R}{2}. \]

Taking LCM,
\[ R_{total} = \frac{4R+R}{2}. \]
\[ R_{total} = \frac{5R}{2}. \]



Step 5: Calculate the current.

Using Ohm's law,
\[ I = \frac{E_{eq}} {R_{total}}. \]

Substituting,
\[ I = \frac{2E} {\frac{5R}{2}}. \]
\[ I = 2E\times\frac{2}{5R}. \]
\[ \boxed{ I = \frac{4E}{5R} } \]



Final Answer:

The current through the resistance \(2R\) is
\[ \boxed{ I=\frac{4E}{5R} } \] Quick Tip: For cells connected in parallel: \[ E_{eq} = \frac{E_1r_2+E_2r_1}{r_1+r_2} \] \[ r_{eq} = \frac{r_1r_2}{r_1+r_2} \] Always reduce the combination to a single equivalent cell before applying Ohm's law.

Concept:

A thin lens consists of two refracting spherical surfaces. The focal length of the lens depends upon:


Refractive index of the lens material,
Refractive index of the surrounding medium,
Radii of curvature of the two refracting surfaces.


The relation connecting these quantities is known as the Lens Maker's Formula because it enables a lens manufacturer to determine the focal length of a lens from its shape and material.

The derivation is based on the formula for refraction at a spherical surface:
\[ \frac{n_2}{v}-\frac{n_1}{u} = \frac{n_2-n_1}{R} \]

where


\(n_1\) = refractive index of first medium,
\(n_2\) = refractive index of second medium,
\(u\) = object distance,
\(v\) = image distance,
\(R\) = radius of curvature.




Step 1: Refraction at the first spherical surface.

Consider a thin convex lens of refractive index \(n\) placed in air.

Let
\[ \mu_a=1 \]

and
\[ \mu_l=n. \]

An object is placed at distance \(u\) from the first surface of radius \(R_1\).

Applying refraction formula,
\[ \frac{n}{v_1} -\frac{1}{u} = \frac{n-1}{R_1}. \]

This gives
\[ \boxed{ \frac{n}{v_1} = \frac{1}{u} +\frac{n-1}{R_1} } \]



Step 2: Refraction at the second spherical surface.

The image formed by the first surface acts as a virtual object for the second surface.

For the second surface,
\[ n_1=n, \qquad n_2=1. \]

Applying the refraction formula again,
\[ \frac{1}{v} -\frac{n}{u_2} = \frac{1-n}{R_2}. \]

For a thin lens,
\[ u_2=v_1. \]

Therefore,
\[ \frac{1}{v} -\frac{n}{v_1} = \frac{1-n}{R_2}. \]

or
\[ \boxed{ \frac{1}{v} = \frac{n}{v_1} -\frac{n-1}{R_2} } \]



Step 3: Substitute the value of \(\frac{n}{v_1}\).

Using the result from the first surface,
\[ \frac{1}{v} = \left( \frac{1}{u} +\frac{n-1}{R_1} \right) -\frac{n-1}{R_2}. \]

Therefore,
\[ \frac{1}{v} -\frac{1}{u} = (n-1) \left( \frac{1}{R_1} -\frac{1}{R_2} \right). \]

Hence the lens formula becomes
\[ \boxed{ \frac{1}{v} -\frac{1}{u} = (n-1) \left( \frac{1}{R_1} -\frac{1}{R_2} \right) } \]



Step 4: Obtain the focal length of the lens.

For focal length,
\[ u=\infty. \]

Thus,
\[ \frac{1}{u}=0. \]

and
\[ v=f. \]

Therefore,
\[ \frac{1}{f} = (n-1) \left( \frac{1}{R_1} -\frac{1}{R_2} \right). \]

Hence,
\[ \boxed{ \frac{1}{f} = (n-1) \left( \frac{1}{R_1} -\frac{1}{R_2} \right) } \]

This is the required Lens Maker's Formula for a thin lens in air.



Final Result:
\[ \boxed{ \frac{1}{f} = (n-1) \left( \frac{1}{R_1} -\frac{1}{R_2} \right) } \] Quick Tip: Lens Maker's Formula: \[ \frac{1}{f} = (n-1) \left( \frac{1}{R_1} -\frac{1}{R_2} \right) \] For a convex lens: \[ R_1>0, \qquad R_2<0. \] For a concave lens: \[ R_1<0, \qquad R_2>0. \]

Concept:

For each lens, the lens formula is
\[ \frac{1}{f} = \frac{1}{v} -\frac{1}{u}. \]

The image formed by one lens acts as the object for the next lens.

Therefore, the problem is solved step-by-step by finding successive image positions.



Step 1: Image formed by lens \(L_1\).

Given,
\[ f_1=+40 cm \]
\[ u_1=-80 cm \]

Using lens formula,
\[ \frac{1}{40} = \frac{1}{v_1} -\frac{1}{(-80)}. \]
\[ \frac{1}{40} = \frac{1}{v_1} +\frac{1}{80}. \]
\[ \frac{1}{v_1} = \frac{1}{40} -\frac{1}{80}. \]
\[ \frac{1}{v_1} = \frac{1}{80}. \]
\[ \boxed{v_1=80 cm} \]

Thus the first image is formed \(80\) cm to the right of \(L_1\).



Step 2: Locate this image relative to lens \(L_2\).

Distance between \(L_1\) and \(L_2\)
\[ =120 cm. \]

The image formed by \(L_1\) lies
\[ 120-80=40 cm \]

to the left of \(L_2\).

Hence for lens \(L_2\),
\[ u_2=-40 cm. \]
\[ f_2=+40 cm. \]

Applying lens formula,
\[ \frac{1}{40} = \frac{1}{v_2} -\frac{1}{(-40)}. \]
\[ \frac{1}{40} = \frac{1}{v_2} +\frac{1}{40}. \]

Therefore,
\[ \frac{1}{v_2}=0. \]
\[ \boxed{v_2=\infty} \]

Thus rays emerging from \(L_2\) become parallel.



Step 3: Formation of image by lens \(L_3\).

Parallel rays fall on lens \(L_3\).

For a convex lens, parallel rays converge at its principal focus.

Therefore,
\[ v_3=f_3. \]

Since
\[ f_3=40 cm, \]
\[ \boxed{v_3=40 cm} \]

Thus the final image is formed \(40\) cm to the right of \(L_3\).



Step 4: Determine the position of the final image relative to the object.

Distance from \(L_1\) to \(L_3\)
\[ =120+20 \]
\[ =140 cm. \]

Final image is \(40\) cm beyond \(L_3\).

Hence distance of final image from \(L_1\)
\[ =140+40 \]
\[ =180 cm. \]

The object is \(80\) cm to the left of \(L_1\).

Therefore distance between object and final image is
\[ 80+180. \]
\[ \boxed{260 cm} \]



Final Answer:
\[ \boxed{ Distance between object and final image = 260 cm } \] Quick Tip: If an object is placed at \(2f\) of a convex lens, the image is also formed at \(2f\). If an object is placed at the principal focus of a convex lens, the emerging rays become parallel and the image is formed at infinity.

Concept:

A concave mirror is a spherical mirror whose reflecting surface is curved inward. When an object is placed between the focus \(F\) and the centre of curvature \(C\), the image formed is:


Real
Inverted
Magnified
Formed beyond the centre of curvature


The mirror formula establishes a relationship among:
\[ u=object distance, \]
\[ v=image distance, \]

and
\[ f=focal length. \]

The required relation is
\[ \boxed{\frac{1}{f}=\frac{1}{v}+\frac{1}{u}} \]

which is valid for all spherical mirrors when sign convention is properly followed.



Step 1: Draw the ray diagram.

Consider a concave mirror with pole \(P\), focus \(F\), and centre of curvature \(C\).

The object \(AB\) is placed between \(F\) and \(C\).
\[ Ray 1: Parallel to principal axis \longrightarrow reflected through F \]
\[ Ray 2: Through C \longrightarrow reflected back along the same path \]

The reflected rays intersect beyond \(C\) at point \(A'B'\), forming a real and inverted image.



Schematic Ray Diagram
\[ Image A'B' \hspace{0.5cm} C \hspace{0.5cm} A \hspace{0.2cm} F \hspace{0.2cm} P \]
\[ \downarrow \hspace{1.8cm} \uparrow \]
\[ B' \hspace{1.8cm} B \]

(Image forms beyond \(C\), real, inverted and magnified.)



Step 2: Introduce geometrical quantities.

Let
\[ PB=u \]

be the object distance,
\[ PB'=v \]

be the image distance,

and
\[ PF=f \]

be the focal length.

Let the object height be
\[ AB=h \]

and image height be
\[ A'B'=h'. \]



Step 3: Use similar triangles.

From the geometry of the ray diagram,
\[ \triangle ABP \sim \triangle A'B'P. \]

Therefore,
\[ \frac{AB}{A'B'} = \frac{PB}{PB'}. \]

Hence,
\[ \frac{h}{h'} = \frac{u}{v}. \]

Thus,
\[ \boxed{ \frac{h'}{h} = \frac{v}{u} } \]

which is the expression for magnification.



Step 4: Consider triangles involving the focus.

From the ray travelling parallel to the principal axis and then passing through the focus after reflection, the triangles formed in the geometry of reflection are similar.

Using the similarity of the appropriate triangles obtained from the ray diagram,
\[ \frac{AB}{A'B'} = \frac{FP}{F B'}. \]

Substituting distances,
\[ \frac{h}{h'} = \frac{f}{v-f}. \]

Using
\[ \frac{h}{h'} = \frac{u}{v}, \]

we obtain
\[ \frac{u}{v} = \frac{f}{v-f}. \]

Cross-multiplying,
\[ u(v-f)=vf. \]

Expanding,
\[ uv-uf=vf. \]

Rearranging,
\[ uv=uf+vf. \]

Dividing throughout by \(uvf\),
\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v}. \]

Hence,
\[ \boxed{ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} } \]

This is the required mirror formula.



Final Result:
\[ \boxed{ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} } \] Quick Tip: For a concave mirror: \[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \] Magnification: \[ m=-\frac{v}{u} \] Object between \(F\) and \(C\): Image beyond \(C\) Real Inverted Enlarged

Concept:

A virtual image produced by a concave mirror is:


Erect
Magnified
Formed behind the mirror
Produced when the object lies between the pole and the focus


The magnification produced by a mirror is
\[ m=-\frac{v}{u}. \]

The mirror formula is
\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v}. \]

These two relations are sufficient to determine the focal length.



Step 1: Write the given quantities.

Object distance
\[ u=-10 cm \]

(negative according to Cartesian sign convention).

The image is virtual and magnified two times.

Therefore,
\[ m=+2. \]



Step 2: Use the magnification formula to find image distance.

Using
\[ m=-\frac{v}{u}, \]

we get
\[ 2=-\frac{v}{-10}. \]
\[ 2=\frac{v}{10}. \]

Hence,
\[ \boxed{v=20 cm} \]

The positive sign confirms that the image is formed behind the mirror and is virtual.



Step 3: Apply the mirror formula.

Using
\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v}, \]

substitute
\[ u=-10 cm, \qquad v=20 cm. \]

Therefore,
\[ \frac{1}{f} = \frac{1}{-10} + \frac{1}{20}. \]
\[ \frac{1}{f} = -\frac{2}{20} +\frac{1}{20}. \]
\[ \frac{1}{f} = -\frac{1}{20}. \]

Hence,
\[ \boxed{ f=-20 cm } \]



Step 4: Interpret the result physically.

The negative sign indicates that the mirror is concave, which is consistent with the statement of the problem.

Therefore, the focal length of the concave mirror is
\[ 20 cm \]

in magnitude.



Final Answer:
\[ \boxed{ f=-20 cm } \]

or
\[ \boxed{ |f|=20 cm } \] Quick Tip: For mirrors: \[ m=-\frac{v}{u} \] and \[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v}. \] A virtual image formed by a concave mirror is always erect and has positive magnification.

Concept:

Electromagnetic induction is the phenomenon of production of an emf in a conductor or circuit whenever the magnetic flux linked with it changes.

This phenomenon was discovered experimentally by Michael Faraday and forms the basis of electric generators, transformers, induction motors and many other electromagnetic devices.



Faraday's First Law of Electromagnetic Induction:

Whenever the magnetic flux linked with a closed circuit changes, an emf is induced in the circuit.

If the circuit is closed, an induced current also flows through it.



Faraday's Second Law of Electromagnetic Induction:

The magnitude of the induced emf is equal to the negative rate of change of magnetic flux linked with the circuit.

For a single turn coil,
\[ \boxed{ e=-\frac{d\phi_B}{dt} } \]

where
\[ \phi_B=magnetic flux linked with the circuit. \]

For a coil having \(N\) turns,
\[ \boxed{ e=-N\frac{d\phi_B}{dt} } \]

The negative sign represents Lenz's law and indicates that the induced emf always opposes the cause producing it.



Final Statement:
\[ \boxed{ e=-N\frac{d\phi_B}{dt} } \]

Thus, the induced emf in a circuit is equal to the negative rate of change of magnetic flux linked with the circuit. Quick Tip: Faraday's Law: \[ e=-N\frac{d\phi_B}{dt} \] The negative sign represents Lenz's law, which states that the induced current always opposes the change in magnetic flux producing it.

Concept:

Whenever the current flowing through a coil changes, the magnetic flux linked with the coil also changes. Due to this change in magnetic flux, an emf is induced in the same coil itself.

This phenomenon is called self-induction and the property of the coil by virtue of which it opposes any change in current is called self-inductance.

The self-inductance \(L\) is defined by
\[ \boxed{ L=\frac{N\phi_B}{I} } \]

where
\[ N\phi_B \]

is the total magnetic flux linkage and \(I\) is the current through the coil.



Step 1: Magnetic field inside a long solenoid.

Consider an air-cored solenoid having
\[ N=number of turns, \]
\[ l=length of solenoid, \]
\[ A=cross-sectional area, \]
\[ I=current flowing through it. \]

The magnetic field inside a long solenoid is
\[ \boxed{ B=\mu_0 nI } \]

where
\[ n=\frac{N}{l} \]

is the number of turns per unit length.

Therefore,
\[ B = \mu_0 \frac{N}{l}I. \]
\[ \boxed{ B=\frac{\mu_0 NI}{l} } \]



Step 2: Calculate magnetic flux through one turn.

Magnetic flux through one turn is
\[ \phi_B=BA. \]

Substituting the value of \(B\),
\[ \phi_B = \left(\frac{\mu_0 NI}{l}\right)A. \]

Hence,
\[ \boxed{ \phi_B=\frac{\mu_0 NIA}{l} } \]



Step 3: Determine total flux linkage.

Since the solenoid contains \(N\) turns,
\[ N\phi_B = N\left(\frac{\mu_0 NIA}{l}\right). \]

Therefore,
\[ N\phi_B = \frac{\mu_0 N^2 IA}{l}. \]
\[ \boxed{ N\phi_B = \frac{\mu_0 N^2 IA}{l} } \]



Step 4: Apply the definition of self-inductance.

Using
\[ L=\frac{N\phi_B}{I}, \]

we get
\[ L = \frac{1}{I} \left( \frac{\mu_0 N^2 IA}{l} \right). \]

Cancelling \(I\),
\[ \boxed{ L=\frac{\mu_0 N^2 A}{l} } \]



Final Result:

Hence the self-inductance of a long air-filled solenoid is
\[ \boxed{ L=\frac{\mu_0 N^2A}{l} } \]

where
\[ \mu_0 = 4\pi\times10^{-7}\ H m^{-1}. \] Quick Tip: For a long air-cored solenoid, \[ L=\frac{\mu_0N^2A}{l} \] Thus, self-inductance is directly proportional to: \[ N^2 \] and cross-sectional area \(A\), while it is inversely proportional to the length \(l\) of the solenoid.

Concept:

When a conducting rod rotates in a magnetic field, the free charges present inside the rod experience a magnetic Lorentz force.

This force causes charge separation along the length of the rod and an emf is induced between its ends.

For a rod of length \(l\) rotating with angular velocity \(\omega\) in a uniform magnetic field \(B\) perpendicular to the plane of rotation, the induced emf is
\[ \boxed{ e=\frac{1}{2}B\omega l^2 } \]



Step 1: Write the given quantities.

Length of rod
\[ l=50\ cm =0.50\ m \]

Magnetic field
\[ B=4.0\ mT =4.0\times10^{-3}\ T \]

Angular speed
\[ 60\ rpm \]
\[ = 60\ revolutions per minute \]
\[ = 1\ revolution per second \]

Therefore,
\[ f=1\ Hz \]

Angular velocity
\[ \omega=2\pi f \]
\[ \omega=2\pi(1) \]
\[ \boxed{ \omega=2\pi\ rad s^{-1} } \]



Step 2: Apply the expression for motional emf.
\[ e = \frac{1}{2} B\omega l^2 \]

Substituting the values,
\[ e = \frac{1}{2} (4\times10^{-3}) (2\pi) (0.50)^2. \]



Step 3: Simplify the numerical calculation.

Since
\[ (0.50)^2=0.25, \]
\[ e = \frac{1}{2} \times 4\times10^{-3} \times 2\pi \times 0.25. \]
\[ e = \pi\times10^{-3}. \]
\[ e = 3.14\times10^{-3}\ V. \]

Therefore,
\[ \boxed{ e=3.14\times10^{-3}\ V } \]



Step 4: Express the answer in millivolts.
\[ e = 3.14\ mV. \]
\[ \boxed{ e=3.14\ mV } \]



Final Answer:
\[ \boxed{ e=3.14\times10^{-3}\ V } \]

or
\[ \boxed{ e=3.14\ mV } \] Quick Tip: For a rod rotating about one end in a magnetic field perpendicular to the plane of rotation, \[ e=\frac{1}{2}B\omega l^2 \] Always convert rpm into rad/s using \[ \omega=\frac{2\pi N}{60}. \]

Concept:

A transformer is an electrical device used to increase or decrease alternating voltage without changing its frequency.

A transformer consists of two insulated coils wound over a common laminated soft iron core.


The coil connected to the AC source is called the primary coil.
The coil connected to the load is called the secondary coil.
A step-up transformer increases the voltage, therefore the secondary coil contains more turns than the primary coil.




Labelled Diagram of a Step-Up Transformer:
\[ \begin{array}{c} AC Input
V_p,\; I_p \end{array} \qquad \boxed{Primary Coil (N_p)} \;\;\Big|\Big| \boxed{Soft Iron Core} \Big|\Big|\; \boxed{Secondary Coil (N_s)} \qquad \begin{array}{c} Output
V_s,\; I_s \end{array} \]
\[ N_s>N_p \]

Hence the transformer acts as a step-up transformer.



Principle of Working:

A transformer works on the principle of mutual induction.

When an alternating current flows through the primary coil, a continuously changing magnetic flux is produced in the iron core.

This changing magnetic flux links the secondary coil and induces an emf in it according to Faraday's law of electromagnetic induction.

Thus electrical energy is transferred from the primary coil to the secondary coil through magnetic coupling.



Step 1: Induced emf in the primary coil.

Let
\[ \phi=magnetic flux through each turn. \]

According to Faraday's law,
\[ E_p = N_p\frac{d\phi}{dt}. \]

Thus,
\[ \boxed{ E_p=N_p\frac{d\phi}{dt} } \]



Step 2: Induced emf in the secondary coil.

Similarly,
\[ E_s = N_s\frac{d\phi}{dt}. \]

Hence,
\[ \boxed{ E_s=N_s\frac{d\phi}{dt} } \]



Step 3: Obtain the voltage ratio.

Dividing the two equations,
\[ \frac{E_s}{E_p} = \frac{N_s}{N_p}. \]

For an ideal transformer,
\[ V_p=E_p, \qquad V_s=E_s. \]

Therefore,
\[ \boxed{ \frac{V_s}{V_p} = \frac{N_s}{N_p} } \]

This is the transformation ratio in terms of the number of turns.



Step 4: Relation between currents and number of turns.

For an ideal transformer,
\[ Input Power = Output Power. \]

Therefore,
\[ V_pI_p = V_sI_s. \]

Substituting
\[ \frac{V_s}{V_p} = \frac{N_s}{N_p}, \]

we obtain
\[ \frac{I_p}{I_s} = \frac{N_s}{N_p}. \]

Hence,
\[ \boxed{ \frac{I_s}{I_p} = \frac{N_p}{N_s} } \]

or
\[ \boxed{ \frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s} } \]



Final Result:

For an ideal transformer,
\[ \boxed{ \frac{V_s}{V_p} = \frac{N_s}{N_p} } \]

and
\[ \boxed{ \frac{I_s}{I_p} = \frac{N_p}{N_s} } \] Quick Tip: For an ideal transformer: \[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \] \[ \frac{I_s}{I_p} = \frac{N_p}{N_s} \] A step-up transformer has \[ N_s>N_p \] and therefore \[ V_s>V_p. \]

Concept:

For an ideal transformer:
\[ Input Power = Output Power \]

and
\[ \frac{V_s}{V_p} = \frac{N_s}{N_p}. \]

The turns ratio directly determines the voltage ratio.



Step 1: Write the given data.

Input power
\[ P_p=5 kW =5000 W \]

Input voltage
\[ V_p=200 V \]

Turns ratio
\[ N_p:N_s=1:5 \]

Hence,
\[ \frac{N_s}{N_p}=5. \]



Step 2: Calculate the primary current.

Using
\[ P_p=V_pI_p, \]

we get
\[ 5000=200\times I_p. \]

Therefore,
\[ I_p=\frac{5000}{200}. \]
\[ \boxed{ I_p=25 A } \]



Step 3: Calculate the output voltage.

For an ideal transformer,
\[ \frac{V_s}{V_p} = \frac{N_s}{N_p}. \]

Substituting the values,
\[ \frac{V_s}{200} = 5. \]

Therefore,
\[ V_s = 5\times200. \]
\[ \boxed{ V_s=1000 V } \]



Step 4: Verification using power conservation.

Since the transformer is ideal,
\[ P_s=P_p=5000 W. \]

Thus,
\[ I_s = \frac{5000}{1000} = 5 A. \]

This satisfies
\[ \frac{I_s}{I_p} = \frac{5}{25} = \frac{1}{5} = \frac{N_p}{N_s}. \]

Hence the result is correct.



Final Answers:
\[ \boxed{ I_p=25 A } \]
\[ \boxed{ V_s=1000 V } \] Quick Tip: For an ideal transformer: \[ P_p=P_s \] \[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \] \[ \frac{I_s}{I_p} = \frac{N_p}{N_s} \] Voltage increases in the same ratio as the turns, while current decreases in the inverse ratio.