What is wattless current in an AC circuit?

Wattless current is the current in an AC circuit for which the average power consumed is zero.

2026 Maharashtra HSC Board Important Question

Q: Which phenomenon proves that light is a transverse wave?

Polarization proves that light is a transverse wave.

Q: Which physical quantity of light remains unchanged when it travels from one medium to another?

The frequency of light remains unchanged when it travels from one medium to another.

Q: What is the SI unit of electric flux?

The SI unit of electric flux is newton metre squared per coulomb (N m² C⁻¹).

Q: State Bohr’s quantization condition for angular momentum.

According to Bohr, the angular momentum of an electron is quantized as mvr = nh / 2π.

2026 Maharashtra HSC Board Important Question

Published: Jan 02, 2026 •

Last Minute Revision (LMR)

Chp-7: Wave Optics

Q1. What is diffraction of light? How does it differ from interference? What are Fraunhoffer and Fresnel diffractions?

(a) What is Diffraction of Light?

Diffraction of light is the phenomenon in which light waves bend and spread out when they pass through a narrow aperture or around the edge of an obstacle, whose size is comparable to the wavelength of light.

Due to diffraction, light does not travel strictly in straight lines, and a pattern of alternate bright and dark fringes is observed on a screen.

The phenomenon of diffraction provides strong evidence that light behaves as a wave.

(b) Difference Between Diffraction and Interference

Diffraction	Interference
Occurs due to bending of light at the edges of an aperture or obstacle.	Occurs due to superposition of light waves from two or more coherent sources.
Usually involves a single slit or obstacle.	Requires two coherent sources (e.g., double slit).
Fringes are not equally spaced.	Fringes are equally spaced.
Central maximum is the brightest.	All bright fringes have nearly equal intensity.

(c) Fraunhofer and Fresnel Diffraction 1. Fraunhofer Diffraction

Fraunhofer diffraction occurs when the source of light and the screen are effectively at infinite distances from the diffracting aperture.

This condition is achieved practically by using convex lenses to make the incident and diffracted rays parallel.

The diffraction pattern obtained is sharp and well-defined.

Example: Diffraction at a single slit using parallel monochromatic light.

2. Fresnel Diffraction

Fresnel diffraction occurs when the source of light and the screen are at finite distances from the diffracting aperture.

In this case, the incident and diffracted rays are not parallel.

The diffraction pattern is less sharp and varies with distance.

Example: Diffraction of light at the edge of an obstacle placed close to the source.

Key Formula (Single-Slit Diffraction)

The condition for minimum intensity in single-slit diffraction is:

$$a \sin \theta = n\lambda \quad (n = 1, 2, 3, \ldots)$$

where
$a$ = width of the slit,
$\theta$ = angle of diffraction,
$\lambda$ = wavelength of light.

Conclusion: Diffraction is a fundamental wave property of light that distinguishes it from interference and plays an important role in understanding the wave nature of light.

Q2. Derive the conditions for bright and dark fringes produced due to diffraction by a single slit.

Diffraction at a Single Slit

Consider a single slit of width a illuminated by monochromatic light of wavelength λ. According to Huygens’ principle, every point on the slit acts as a secondary source of wavelets.

At an angle θ with the normal to the slit, these wavelets interfere with each other to produce a diffraction pattern on the screen.

Path Difference Between Extreme Rays

For rays coming from the top and bottom edges of the slit, the path difference is:

$$\text{Path difference} = a \sin \theta$$

(a) Condition for Dark Fringes (Minima)

A dark fringe is formed when waves from different parts of the slit cancel each other completely.

This happens when the path difference between the extreme rays is an integral multiple of the wavelength:

$$a \sin \theta = n\lambda \quad (n = 1, 2, 3, \ldots)$$

At these positions, the slit can be divided into n equal parts, such that waves from corresponding parts interfere destructively.

Hence, the condition for dark fringes is:

$$\boxed{a \sin \theta = n\lambda}$$

(b) Condition for Bright Fringes (Maxima)

Bright fringes are formed when the interference between the secondary wavelets is constructive.

Central Bright Fringe

The central maximum occurs at:

$$\theta = 0 \Rightarrow a \sin 0 = 0$$

This corresponds to zero path difference, and hence maximum intensity.

Secondary Bright Fringes

The condition for secondary maxima is obtained approximately as:

$$a \sin \theta = \left(n + \tfrac{1}{2}\right)\lambda \quad (n = 1, 2, 3, \ldots)$$

These maxima lie between two successive minima and are much less intense than the central maximum.

Hence, the condition for bright fringes is approximately:

$$\boxed{a \sin \theta = \left(n + \tfrac{1}{2}\right)\lambda}$$

Intensity Distribution Note

The central bright fringe is the widest and brightest, while the intensity of secondary maxima decreases rapidly on either side.

Conclusion

Thus, in single-slit diffraction:

Dark fringes are produced when $a \sin \theta = n\lambda$.
Bright fringes occur at $\theta = 0$ (central maximum) and approximately at $a \sin \theta = (n + \tfrac{1}{2})\lambda$.

This diffraction pattern clearly demonstrates the wave nature of light.

Q3. Describe what is Rayleigh’s criterion for resolution. Explain it for a telescope and a microscope.

(a) What is Rayleigh’s Criterion?

Rayleigh’s criterion gives the minimum separation between two point objects so that they can be seen as just resolved by an optical instrument.

According to Rayleigh’s criterion, two point sources are said to be just resolved when the central maximum of the diffraction pattern of one source coincides with the first minimum of the diffraction pattern of the other source.

If the separation is smaller than this limit, the two images overlap too much and appear as a single blurred image.

(b) Rayleigh’s Criterion for a Telescope

Consider a telescope having an objective lens of aperture diameter D, used to observe two distant point objects separated by a small angular separation θ.

Due to diffraction at the circular aperture of the objective, each point source produces an Airy disc.

According to Rayleigh’s criterion, the limiting angular resolution of the telescope is:

$$\theta_{\min} = 1.22 \, \frac{\lambda}{D}$$

where
$\lambda$ = wavelength of light used,
$D$ = diameter of the objective lens.

Thus, the resolving power of a telescope increases when:

The diameter of the objective lens is increased.
Light of shorter wavelength is used.

(c) Rayleigh’s Criterion for a Microscope

In a microscope, the ability to distinguish two closely spaced points depends on the diffraction at the objective lens and the numerical aperture of the system.

According to Rayleigh’s criterion, the minimum distance between two points that can be just resolved by a microscope is given by:

$$d_{\min} = 0.61 \, \frac{\lambda}{\text{NA}}$$

where
$\lambda$ = wavelength of light used,
$\text{NA}$ = numerical aperture of the objective lens.

The numerical aperture is defined as:

$$\text{NA} = n \sin \alpha$$

where
$n$ = refractive index of the medium between object and objective,
$\alpha$ = half-angle of the cone of light entering the objective.

Thus, the resolving power of a microscope can be improved by:

Using light of smaller wavelength.
Increasing the numerical aperture (using oil immersion, larger aperture).

Conclusion

Rayleigh’s criterion sets the fundamental limit of resolution due to diffraction. It explains why optical instruments like telescopes and microscopes cannot resolve objects below a certain separation, even if their magnifying power is very high.

Q4. A parallel beam of green light of wavelength 546 nm passes through a slit of width 0.4 mm. The intensity pattern of the transmitted light is seen on a screen which is 40 cm away. What is the distance between the two first order minima?

Given

Wavelength of green light,

$$\lambda = 546 \, \text{nm} = 546 \times 10^{-9} \, \text{m}$$

Width of the slit,

$$a = 0.4 \, \text{mm} = 0.4 \times 10^{-3} \, \text{m}$$

Distance of the screen from the slit,

$$D = 40 \, \text{cm} = 0.40 \, \text{m}$$

Formula Used

For single-slit diffraction, the condition for the first-order minimum is:

$$a \sin \theta = \lambda$$

For small angles,

$$\sin \theta \approx \tan \theta = \frac{y}{D}$$

Hence,

$$y = \frac{\lambda D}{a}$$

where
$y$ = distance of first minimum from the central maximum.

Calculation

Substituting the given values,

$$ y = \frac{546 \times 10^{-9} \times 0.40}{0.4 \times 10^{-3}} $$

$$ y = \frac{218.4 \times 10^{-9}}{0.4 \times 10^{-3}} $$

$$ y = 5.46 \times 10^{-4} \, \text{m} $$

Distance between the two first-order minima is:

$$ \text{Distance} = 2y = 2 \times 5.46 \times 10^{-4} $$

$$ \text{Distance} = 1.092 \times 10^{-3} \, \text{m} $$

Final Answer

$$ \boxed{\text{Distance between the two first-order minima} = 1.09 \, \text{mm}} $$

Q5. The distance between two consecutive bright fringes in a biprism experiment using light of wavelength 6000 Å is 0.32 mm by how much will the distance change if light of wavelength 4800 Å is used?

Given

Wavelength of first light,

$$\lambda_1 = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m}$$

Distance between consecutive bright fringes (fringe width),

$$\beta_1 = 0.32 \, \text{mm}$$

Wavelength of second light,

$$\lambda_2 = 4800 \, \text{Å} = 4800 \times 10^{-10} \, \text{m}$$

Formula Used

In a biprism experiment, the fringe width is given by:

$$\beta = \frac{\lambda D}{d}$$

Since the experimental setup remains the same, $D$ and $d$ are constants. Therefore,

$$\beta \propto \lambda$$

Calculation

Hence,

$$\frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1}$$

Substituting the given values,

$$ \beta_2 = \beta_1 \times \frac{\lambda_2}{\lambda_1} $$

$$ \beta_2 = 0.32 \times \frac{4800}{6000} $$

$$ \beta_2 = 0.32 \times 0.8 = 0.256 \, \text{mm} $$

Change in the distance between fringes is:

$$ \Delta \beta = \beta_1 - \beta_2 $$

$$ \Delta \beta = 0.32 - 0.256 = 0.064 \, \text{mm} $$

Final Answer

$$ \boxed{\text{The distance between consecutive bright fringes decreases by } 0.064 \, \text{mm}} $$

Chp-8: Electrostatics

Q1. A dipole with its charges, -q and +q located at the points (0, -b, 0) and (0 +b, 0) is present in a uniform electric field E. The equipotential surfaces of this field are planes parallel to the YZ planes. (a) What is the direction of the electric field E? (b) How much torque would the dipole experience in this field?

A dipole consists of two charges $-q$ and $+q$ located at the points $$(0,-b,0) \quad \text{and} \quad (0,+b,0)$$ respectively. The dipole is placed in a uniform electric field $\vec{E}$. The equipotential surfaces of this field are planes parallel to the YZ-plane.

(a) Direction of the Electric Field

Equipotential surfaces are always perpendicular to the direction of the electric field. Since the equipotential surfaces are planes parallel to the YZ-plane, the electric field must be perpendicular to the YZ-plane.

The normal to the YZ-plane lies along the X-axis.

Therefore,

$$ \boxed{\vec{E} \text{ is directed along the } X\text{-axis}} $$

(b) Torque Acting on the Dipole

The electric dipole moment $\vec{p}$ is defined as:

$$ \vec{p} = q \cdot (2b)\,\hat{y} $$

So, $$ |\vec{p}| = 2qb $$

The torque acting on a dipole in a uniform electric field is given by:

$$ \vec{\tau} = \vec{p} \times \vec{E} $$

Here,

$\vec{p}$ is along the Y-axis
$\vec{E}$ is along the X-axis

The angle between $\vec{p}$ and $\vec{E}$ is $90^\circ$.

Hence, the magnitude of torque is:

$$ \tau = pE \sin 90^\circ = pE $$

$$ \boxed{\tau = 2qbE} $$

Using the right-hand rule:

$$ \hat{y} \times \hat{x} = -\hat{z} $$

Direction of torque: Along the negative Z-axis.

Final Answers

(a) $\boxed{\vec{E} \text{ is along the X-axis}}$

(b) $\boxed{\tau = 2qbE \text{ acting along the negative Z-axis}}$

Q2. A 6 μF capacitor is charged by a 300 V supply. It is then disconnected from the supply and is connected to another uncharged 3μF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation ?

A capacitor of capacitance $6\,\mu\text{F}$ is charged to a potential of $300\text{ V}$. It is then disconnected from the supply and connected to another uncharged capacitor of capacitance $3\,\mu\text{F}$.

We are required to calculate the electrostatic energy lost in the form of heat and electromagnetic radiation.

Step 1: Initial Energy Stored in the First Capacitor

The energy stored in a capacitor is given by:

$$ U = \frac{1}{2}CV^2 $$

Substituting the given values:

$$ U_i = \frac{1}{2} \times 6 \times 10^{-6} \times (300)^2 $$

$$ U_i = \frac{1}{2} \times 6 \times 10^{-6} \times 90000 $$

$$ \boxed{U_i = 0.27\text{ J}} $$

Step 2: Final Voltage After Connecting the Two Capacitors

Charge on the first capacitor before connection:

$$ Q = CV = 6 \times 10^{-6} \times 300 $$

$$ Q = 1.8 \times 10^{-3}\text{ C} $$

After connection, total capacitance becomes:

$$ C_{\text{total}} = 6 + 3 = 9\,\mu\text{F} $$

Final common voltage is:

$$ V_f = \frac{Q}{C_{\text{total}}} $$

$$ V_f = \frac{1.8 \times 10^{-3}}{9 \times 10^{-6}} $$

$$ \boxed{V_f = 200\text{ V}} $$

Step 3: Final Energy Stored in the Combined System

Final energy stored in the two capacitors:

$$ U_f = \frac{1}{2} C_{\text{total}} V_f^2 $$

$$ U_f = \frac{1}{2} \times 9 \times 10^{-6} \times (200)^2 $$

$$ U_f = \frac{1}{2} \times 9 \times 10^{-6} \times 40000 $$

$$ \boxed{U_f = 0.18\text{ J}} $$

Step 4: Energy Lost

The energy lost in the form of heat and electromagnetic radiation is:

$$ \Delta U = U_i - U_f $$

$$ \Delta U = 0.27 - 0.18 $$

$$ \boxed{\Delta U = 0.09\text{ J}} $$

Final Answer

The electrostatic energy lost in the form of heat and electromagnetic radiation is:

$$ \boxed{0.09\text{ J}} $$

Q3. In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10–3 m2 and the separation between the plates is 2 mm. a) Calculate the capacitance of the capacitor, b) If this capacitor is connected to 100 V supply, what would be the charge on each plate? c) How would charge on the plates be affected if a 2 mm thick mica sheet of k = 6 is inserted between the plates while the voltage supply remains connected ?

A parallel plate capacitor has air between its plates. Each plate has an area $$ A = 6 \times 10^{-3}\,\text{m}^2 $$ and the separation between the plates is $$ d = 2\,\text{mm} = 2 \times 10^{-3}\,\text{m}. $$

(a) Capacitance of the Capacitor

For a parallel plate capacitor with air (or vacuum) between the plates, the capacitance is given by:

$$ C = \frac{\varepsilon_0 A}{d} $$

where $$ \varepsilon_0 = 8.85 \times 10^{-12}\,\text{F m}^{-1} $$

Substituting the given values:

$$ C = \frac{8.85 \times 10^{-12} \times 6 \times 10^{-3}}{2 \times 10^{-3}} $$

$$ C = 26.55 \times 10^{-12}\,\text{F} $$

$$ \boxed{C \approx 2.66 \times 10^{-11}\,\text{F} \;=\; 26.6\,\text{pF}} $$

(b) Charge on Each Plate When Connected to 100 V Supply

Charge on a capacitor is given by:

$$ Q = CV $$

Substituting values:

$$ Q = 26.55 \times 10^{-12} \times 100 $$

$$ \boxed{Q = 2.66 \times 10^{-9}\,\text{C}} $$

So, the charge on each plate is:

$$ \boxed{Q = 2.66\,\text{nC}} $$

(c) Effect of Inserting a Mica Sheet (k = 6)

When a dielectric slab completely fills the space between the plates, the capacitance becomes:

$$ C' = kC $$

Given: $$ k = 6 $$

$$ C' = 6 \times 26.55 \times 10^{-12} $$

$$ C' = 159.3 \times 10^{-12}\,\text{F} $$

Since the capacitor remains connected to the voltage supply, the voltage remains constant at $V = 100\text{ V}$.

New charge on the plates:

$$ Q' = C'V $$

$$ Q' = 159.3 \times 10^{-12} \times 100 $$

$$ \boxed{Q' = 1.59 \times 10^{-8}\,\text{C}} $$

$$ \boxed{Q' = 15.9\,\text{nC}} $$

Final Answers

(a) Capacitance: $$ \boxed{C = 26.6\,\text{pF}} $$

(b) Charge on each plate at 100 V: $$ \boxed{Q = 2.66\,\text{nC}} $$

A charge 6 μC is placed at the origin and another charge –5 μC is placed on the y axis at a position A (0, 6.0) m.

a) Calculate the total electric potential at the point P whose coordinates are (8.0, 0) m
b) Calculate the work done to bring a proton from infinity to the point P ? What is the significance of the negative sign ?

Electric Potential and Work Done

A charge of $$ q_1 = +6\,\mu\text{C} $$ is placed at the origin $(0,0)$ and another charge $$ q_2 = -5\,\mu\text{C} $$ is placed on the Y-axis at point $$ A(0,6.0)\,\text{m}. $$

Point $P$ is located at $$ P(8.0,0)\,\text{m}. $$

(a) Electric Potential at Point P

Electric potential due to a point charge is given by:

$$ V = \frac{1}{4\pi\varepsilon_0}\frac{q}{r} $$

where $$ \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \,\text{N m}^2\text{/C}^2 $$

Potential due to $+6\,\mu\text{C}$ charge

Distance of point $P$ from the origin:

$$ r_1 = 8\,\text{m} $$

$$ V_1 = 9 \times 10^9 \times \frac{6 \times 10^{-6}}{8} $$

$$ V_1 = 6.75 \times 10^3\,\text{V} $$

Potential due to $-5\,\mu\text{C}$ charge

Distance of point $P$ from point $A(0,6)$:

$$ r_2 = \sqrt{(8-0)^2 + (0-6)^2} $$

$$ r_2 = \sqrt{64 + 36} = 10\,\text{m} $$

$$ V_2 = 9 \times 10^9 \times \frac{-5 \times 10^{-6}}{10} $$

$$ V_2 = -4.5 \times 10^3\,\text{V} $$

Total Electric Potential at P

$$ V_P = V_1 + V_2 $$

$$ V_P = 6.75 \times 10^3 - 4.5 \times 10^3 $$

$$ \boxed{V_P = 2.25 \times 10^3\,\text{V}} $$

(b) Work Done to Bring a Proton from Infinity to Point P

Work done in bringing a charge $q$ from infinity to a point of potential $V$ is:

$$ W = qV $$

Charge of a proton:

$$ q = +1.6 \times 10^{-19}\,\text{C} $$

Substituting values:

$$ W = (1.6 \times 10^{-19})(2.25 \times 10^3) $$

$$ \boxed{W = -5.4 \times 10^{-16}\,\text{J}} $$

Significance of the Negative Sign

The negative sign indicates that the electric field does work on the proton.

This means:

The proton is naturally attracted towards the region of lower potential energy
No external work is required to bring the proton to point $P$

Final Answers

(a) Electric potential at point $P$: $$ \boxed{V_P = 2.25 \times 10^3\,\text{V}} $$

(b) Work done in bringing a proton from infinity: $$ \boxed{W = -5.4 \times 10^{-16}\,\text{J}} $$

Chp-9: Current Electricity

Q1. What is Potential Gradient? How is it Measured? Explain. On What Factors Does the Potential Gradient of a Wire Depend?

1. Definition of Potential Gradient

The potential gradient is defined as the rate of decrease of electric potential with distance along the direction of the electric field.

Mathematically, it is given by:

$$ \text{Potential Gradient} = -\frac{dV}{dx} $$

The negative sign indicates that the electric potential decreases in the direction of the electric field.

Since electric field intensity $E$ is related to potential gradient, we can write:

$$ E = -\frac{dV}{dx} $$

Thus, potential gradient is numerically equal to the electric field intensity.

2. Measurement of Potential Gradient

Consider a uniform wire of length $L$ connected across a potential difference $V$. If the potential drop is uniform along the wire, then the potential gradient is constant.

Potential gradient is measured as:

$$ \text{Potential Gradient} = \frac{V}{L} $$

Here:

$V$ = potential difference across the wire (in volts)
$L$ = length of the wire (in metres)

The SI unit of potential gradient is:

$$ \text{volt per metre (V m}^{-1}\text{)} $$

3. Factors Affecting the Potential Gradient of a Wire

The potential gradient of a wire depends on the following factors:

(i) Potential Difference Applied ($V$)

For a given length of wire, if the applied potential difference increases, the potential gradient increases proportionally.

$$ \text{Potential Gradient} \propto V $$

(ii) Length of the Wire ($L$)

For a fixed potential difference, increasing the length of the wire decreases the potential gradient.

$$ \text{Potential Gradient} \propto \frac{1}{L} $$

(iii) Nature of the Material (Indirectly)

Although potential gradient directly depends on $V$ and $L$, the material of the wire affects the current distribution and resistance, which indirectly influences the potential drop along the wire.

Q2. Why Is a Potentiometer Preferred Over a Voltmeter for Measuring EMF?

1. Concept of EMF Measurement

The electromotive force (emf) of a cell is the potential difference between its terminals when no current is drawn from the cell.

Hence, an ideal method of measuring emf should not allow current to flow through the cell.

2. Limitation of a Voltmeter

A voltmeter has a finite (though high) resistance. When connected across a cell, it draws a small current from the cell.

Due to this current, a potential drop occurs across the internal resistance $r$ of the cell, and the voltmeter reads:

$$ V = E - Ir $$

where:

$E$ = emf of the cell
$I$ = current drawn
$r$ = internal resistance of the cell

Thus, a voltmeter measures the terminal potential difference, not the true emf.

3. Advantage of a Potentiometer

A potentiometer works on the null method. At the balance point, no current flows through the cell whose emf is being measured.

Hence:

$$ V = E $$

There is no loss of potential due to internal resistance of the cell.

4. Reasons Why Potentiometer Is Preferred

It measures the true emf of a cell.
No current is drawn from the cell at balance point.
Results are highly accurate.
It is independent of internal resistance of the cell.

5. Conclusion

A potentiometer is preferred over a voltmeter for measuring emf because it uses a null method and does not draw current from the cell. Therefore, it gives the true value of emf, whereas a voltmeter always reads a value less than the actual emf.

Q3. Describe Kelvin’s Method to Determine the Resistance of a Galvanometer Using a Meter Bridge

1. Aim

To determine the resistance of a galvanometer using a meter bridge by Kelvin’s method.

2. Principle

A meter bridge works on the principle of a Wheatstone bridge. At balance condition:

$$ \frac{R_1}{R_2} = \frac{l_1}{l_2} $$

where $R_1$ and $R_2$ are the resistances in the two gaps of the meter bridge and $l_1$ and $l_2$ are the corresponding balancing lengths of the wire.

3. Circuit Arrangement

In Kelvin’s method:

The galvanometer whose resistance $G$ is to be determined is placed in one gap of the meter bridge.
A known resistance $R$ is placed in the other gap.
A battery and key are connected across the meter bridge.
A jockey is used to find the balance point on the bridge wire.

4. Procedure

Connect the galvanometer of unknown resistance $G$ in one gap and a known resistance $R$ in the other gap.
Close the key and slide the jockey along the meter bridge wire.
Note the balancing length $l$ from one end where the galvanometer is connected.
Interchange the positions of $G$ and $R$ in the two gaps.
Again find the balancing length $l'$.

5. Theory and Formula

In the first case:

$$ \frac{G}{R} = \frac{l}{100 - l} $$

After interchanging:

$$ \frac{R}{G} = \frac{l'}{100 - l'} $$

Multiplying the two equations:

$$ \frac{G}{R} \cdot \frac{R}{G} = \frac{l}{100 - l} \cdot \frac{l'}{100 - l'} $$

This gives:

$$ G^2 = R^2 \cdot \frac{l(100 - l')}{(100 - l)l'} $$

Hence, the resistance of the galvanometer is:

$$ G = R \sqrt{\frac{l(100 - l')}{(100 - l)l'}} $$

6. Advantage of Kelvin’s Method

Errors due to end resistances and contact resistances are eliminated.
More accurate than a single observation method.

Q4. Describe How a Potentiometer Is Used to Compare the EMFs of Two Cells by Connecting the Cells Individually

1. Aim

To compare the electromotive forces (emfs) of two given cells using a potentiometer by connecting the cells individually.

2. Principle

A potentiometer works on the principle that the potential drop along a uniform wire is directly proportional to its length when a steady current flows through it.

$$ V \propto l $$

Since no current is drawn from the cell at the balance point, the balancing length is directly proportional to the emf of the cell.

3. Circuit Arrangement

A long uniform potentiometer wire is connected in series with a battery, rheostat, and key.
The two cells $E_1$ and $E_2$ whose emfs are to be compared are connected one at a time to the secondary circuit through a galvanometer and a two-way key.
A jockey is used to obtain the balance point on the potentiometer wire.

4. Procedure

Close the primary circuit and adjust the rheostat to obtain a steady current in the potentiometer wire.
Using the two-way key, connect cell $E_1$ in the secondary circuit.
Slide the jockey along the wire to obtain the null point and note the balancing length $l_1$.
Now disconnect $E_1$ and connect cell $E_2$ individually using the two-way key.
Again find the null point and note the balancing length $l_2$.

5. Theory and Formula

Let $k$ be the potential gradient of the potentiometer wire.

For the first cell:

$$ E_1 = k l_1 $$

For the second cell:

$$ E_2 = k l_2 $$

Taking the ratio:

$$ \frac{E_1}{E_2} = \frac{l_1}{l_2} $$

6. Advantages of the Method

No current is drawn from the cells at balance point.
Comparison is independent of internal resistances of the cells.
Results are highly accurate.

Q5. Describe How a Potentiometer Is Used to Compare the EMFs of Two Cells by Combination Method

1. Aim

To compare the electromotive forces (emfs) of two given cells using a potentiometer by the combination method.

2. Principle

A potentiometer works on the principle that the potential drop along a uniform wire is directly proportional to its length when a steady current flows through it.

$$ V \propto l $$

At the balance point, no current flows through the cell; hence the balancing length is directly proportional to the emf.

3. Circuit Arrangement (Combination Method)

A uniform potentiometer wire is connected to a battery, rheostat, and key (primary circuit).
The two cells $E_1$ and $E_2$ are connected in series aiding combination in the secondary circuit.
A galvanometer and a jockey are connected to detect the null point.

4. Procedure

Close the primary circuit and adjust the rheostat to obtain a steady current in the potentiometer wire.
Connect the two cells $E_1$ and $E_2$ in series aiding combination.
Slide the jockey along the potentiometer wire and obtain the balance point.
Note the balancing length $l$ corresponding to the combined emf.

5. Theory and Formula

Let $k$ be the potential gradient of the potentiometer wire.

For the series combination:

$$ E_1 + E_2 = k l $$

If the balancing lengths for individual cells are $l_1$ and $l_2$, then:

$$ E_1 = k l_1 \quad \text{and} \quad E_2 = k l_2 $$

Hence:

$$ \frac{E_1}{E_2} = \frac{l_1}{l_2} $$

Thus, the ratio of emfs can be obtained using the combination method.

6. Advantages of Combination Method

Higher effective emf gives a more distinct balance point.
Reduces percentage error in measurement.
Accurate comparison of emfs.

Q6. A Voltmeter Has a Resistance 30 Ω. What Will Be Its Reading When It Is Connected Across a Cell of EMF 2 V Having Internal Resistance 10 Ω?

1. Given Data

EMF of the cell, $E = 2 \text{ V}$
Internal resistance of the cell, $r = 10 \, \Omega$
Resistance of voltmeter, $R = 30 \, \Omega$

2. Concept Used

When a voltmeter is connected across a cell, it draws some current. Hence, it measures the terminal potential difference of the cell, not the emf.

Current drawn from the cell is given by:

$$ I = \frac{E}{R + r} $$

3. Calculation of Current

$$ I = \frac{2}{30 + 10} = \frac{2}{40} = 0.05 \text{ A} $$

4. Voltmeter Reading

The reading of the voltmeter is the potential difference across its resistance:

$$ V = IR $$

$$ V = 0.05 \times 30 = 1.5 \text{ V} $$

5. Answer

The voltmeter will read:

$$ \boxed{1.5 \text{ V}} $$

Q7. A Potential Drop per Unit Length Along a Wire Is 5 × 10⁻³ V/m. If the EMF of a Cell Balances Against Length 216 cm of This Potentiometer Wire, Find the EMF of the Cell.

1. Given Data

Potential gradient, $k = 5 \times 10^{-3} \, \text{V m}^{-1}$
Balancing length, $l = 216 \, \text{cm} = 2.16 \, \text{m}$

2. Formula Used

The emf of a cell is given by:

$$ E = k \, l $$

3. Calculation

$$ E = (5 \times 10^{-3}) \times 2.16 $$

$$ E = 10.8 \times 10^{-3} \, \text{V} $$

$$ E = 0.0108 \, \text{V} $$

4. Answer

$$ \boxed{E = 0.0108 \, \text{V}} $$

Chp-12: Electromagnetic Induction

Q1. State and explain Lenz’s law in the light of the principle of conservation of energy.

Statement of Lenz’s Law:

Lenz’s law states that the direction of the induced current in a conductor is such that the magnetic field produced by it opposes the change in magnetic flux that causes the current.

Mathematical expression:

The induced emf is given by Faraday’s law as

$$ \varepsilon = -\frac{d\Phi}{dt} $$

The negative sign represents Lenz’s law and indicates the opposing nature of the induced emf.

Explanation using the principle of conservation of energy:

According to the principle of conservation of energy, energy can neither be created nor destroyed, but only transformed from one form to another.

Consider a magnet being moved towards a closed conducting coil. Due to the change in magnetic flux linked with the coil, an emf is induced and a current flows in the coil. According to Lenz’s law, the direction of this induced current is such that the magnetic field produced by it opposes the motion of the magnet.

Because of this opposition, an external force must be applied to move the magnet towards the coil. The work done by this external force is converted into electrical energy (and subsequently into heat) in the circuit.

If the induced current were in a direction that assisted the motion of the magnet instead of opposing it, the magnet would accelerate on its own and electrical energy would be produced without any external work. This would violate the principle of conservation of energy.

Conclusion:

Thus, Lenz’s law is a direct consequence of the principle of conservation of energy, as it ensures that induced currents always oppose the cause producing them, preventing the creation of energy from nothing.

Q2. What are eddy currents? State applications of eddy currents.

Definition of Eddy Currents:

Eddy currents are circulating currents induced in the bulk of a conducting material when it is placed in a changing magnetic field or when the conductor moves relative to a magnetic field.

These currents flow in closed loops within the conductor, similar to whirlpools of water, and are therefore called eddy currents.

Explanation:

According to Faraday’s law of electromagnetic induction, whenever the magnetic flux linked with a conductor changes, an emf is induced. In a solid conductor, this emf causes currents to circulate in planes perpendicular to the magnetic field.

As per Lenz’s law, the magnetic field produced by eddy currents opposes the change in magnetic flux that produces them.

Effects of Eddy Currents:

They produce heating in the conductor due to Joule’s heating effect.
They create opposing magnetic fields which can slow down motion.

Applications of Eddy Currents:

Eddy current brakes:
Eddy currents are used in electromagnetic braking systems (such as in trains and elevators). The opposing magnetic field produced by eddy currents provides a smooth and contactless braking force.
Induction furnace:
Eddy currents are used to heat and melt metals in induction furnaces. The heat generated due to eddy currents raises the temperature of the metal.
Damping in moving coil galvanometer:
Eddy currents are used to quickly bring the pointer to rest, preventing oscillations and allowing accurate readings.
Speedometers:
Eddy currents are used in analog speedometers to measure the speed of vehicles.
Metal detectors:
Eddy currents induced in metallic objects help in detecting metals.

Note:

In many electrical machines like transformers, eddy currents are undesirable as they cause energy loss. To reduce them, laminated iron cores are used.

Q3. A long solenoid consisting of 1.5×10³ turns/m has an area of cross-section of 25 cm². A coil C consisting of 150 turns is wound tightly around the centre of the solenoid. Calculate for a current of 3.0 A in the solenoid:

(a) the magnetic flux density at the centre of the solenoid,
(b) the flux linkage in the coil C,
(c) the average emf induced in coil C if the current in the solenoid is reversed in direction in a time of 0.5 s.

Given: $\mu_0 = 4\pi \times 10^{-7}\,\text{T·m/A}$

(a) Magnetic flux density at the centre of the solenoid

For a long solenoid, the magnetic field at the centre is

$$ B = \mu_0 n I $$

where
$n = 1.5 \times 10^3 \,\text{turns/m}$,
$I = 3.0 \,\text{A}$

$$ B = (4\pi \times 10^{-7})(1.5 \times 10^3)(3.0) $$

$$ B = 5.66 \times 10^{-3}\,\text{T} $$

Magnetic flux density = $5.66 \times 10^{-3}\,\text{T}$

(b) Flux linkage in the coil C

Area of cross-section:

$$ A = 25\,\text{cm}^2 = 25 \times 10^{-4} = 2.5 \times 10^{-3}\,\text{m}^2 $$

Magnetic flux through one turn:

$$ \Phi = BA $$

$$ \Phi = (5.66 \times 10^{-3})(2.5 \times 10^{-3}) $$

$$ \Phi = 1.415 \times 10^{-5}\,\text{Wb} $$

Flux linkage for coil C having $N_c = 150$ turns:

$$ \text{Flux linkage} = N_c \Phi $$

$$ = 150 \times 1.415 \times 10^{-5} $$

$$ = 2.12 \times 10^{-3}\,\text{Wb} $$

Flux linkage = $2.12 \times 10^{-3}\,\text{Wb}$

(c) Average emf induced in coil C

When the current is reversed, the magnetic flux changes from $+\Phi$ to $-\Phi$. Hence,

$$ \Delta \Phi = 2\Phi $$

Change in flux linkage:

$$ \Delta (N\Phi) = 2 \times 2.12 \times 10^{-3} $$

$$ = 4.24 \times 10^{-3}\,\text{Wb} $$

Average induced emf is given by Faraday’s law:

$$ \varepsilon_{\text{avg}} = \frac{\Delta (N\Phi)}{\Delta t} $$

$$ \varepsilon_{\text{avg}} = \frac{4.24 \times 10^{-3}}{0.5} $$

$$ \varepsilon_{\text{avg}} = 8.48 \times 10^{-3}\,\text{V} $$

Average induced emf = $8.48 \times 10^{-3}\,\text{V}$

Final Answers:

(a) $B = 5.66 \times 10^{-3}\,\text{T}$
(b) Flux linkage = $2.12 \times 10^{-3}\,\text{Wb}$
(c) $\varepsilon_{\text{avg}} = 8.48 \times 10^{-3}\,\text{V}$

Q4. A search coil having 2000 turns with area 1.5 cm² is placed in a magnetic field of 0.60 T. The coil is moved rapidly out of the field in a time of 0.2 s. Calculate the induced emf across the search coil.

Given:

Number of turns, $N = 2000$
Area of coil, $A = 1.5\,\text{cm}^2 = 1.5 \times 10^{-4}\,\text{m}^2$
Magnetic field, $B = 0.60\,\text{T}$
Time taken, $\Delta t = 0.2\,\text{s}$

Step 1: Magnetic flux through one turn

$$ \Phi = BA $$

$$ \Phi = 0.60 \times 1.5 \times 10^{-4} $$

$$ \Phi = 9.0 \times 10^{-5}\,\text{Wb} $$

Step 2: Initial and final flux linkage

Initial flux linkage:

$$ (N\Phi)_i = N \Phi = 2000 \times 9.0 \times 10^{-5} $$

$$ (N\Phi)_i = 0.18\,\text{Wb} $$

Final flux linkage:

Since the coil is moved completely out of the magnetic field,

$$ (N\Phi)_f = 0 $$

Step 3: Change in flux linkage

$$ \Delta (N\Phi) = (N\Phi)_f - (N\Phi)_i $$

$$ \Delta (N\Phi) = -0.18\,\text{Wb} $$

Magnitude of change in flux linkage:

$$ |\Delta (N\Phi)| = 0.18\,\text{Wb} $$

Step 4: Induced emf

According to Faraday’s law of electromagnetic induction,

$$ \varepsilon = \frac{|\Delta (N\Phi)|}{\Delta t} $$

$$ \varepsilon = \frac{0.18}{0.2} $$

$$ \boxed{\varepsilon = 0.90\,\text{V}} $$

Answer:

Induced emf across the search coil = $0.90\,\text{V}$

Q5. An aircraft of wing span 50 m flies horizontally in Earth’s magnetic field of 6 × 10⁻⁵ T at a speed of 400 m/s. Calculate the emf generated between the tips of the wings of the aircraft.

Given:

Wing span of aircraft, $l = 50\,\text{m}$
Magnetic field of Earth, $B = 6 \times 10^{-5}\,\text{T}$
Speed of aircraft, $v = 400\,\text{m/s}$

Formula:

$$ \varepsilon = B l v $$

Calculation

$$ \varepsilon = (6 \times 10^{-5}) \times 50 \times 400 $$

$$ \boxed{\varepsilon = 1.2\,\text{V}} $$

Answer:

The emf generated between the tips of the wings of the aircraft is $1.2\,\text{V}$.

Q6. A stiff semi-circular wire of radius R is rotated in a uniform magnetic field B about an axis passing through its ends. If the frequency of rotation of the wire is f, calculate the amplitude of alternating emf induced in the wire.

Given:

Radius of semi-circular wire = $R$
Magnetic field = $B$
Frequency of rotation = $f$

Step 1: Area of the semi-circular loop

Area of a circle is $\pi R^2$. Hence, area of a semi-circular loop is

$$ A = \frac{1}{2}\pi R^2 $$

Step 2: Expression for magnetic flux

When the loop rotates in a uniform magnetic field, the magnetic flux linked with it at any time $t$ is

$$ \Phi = BA \cos(\omega t) $$

where angular velocity $\omega = 2\pi f$

Step 3: Induced emf

According to Faraday’s law,

$$ \varepsilon = -\frac{d\Phi}{dt} $$

Substituting the value of $\Phi$,

$$ \varepsilon = BA\omega \sin(\omega t) $$

Thus, the induced emf is alternating in nature.

Step 4: Amplitude (maximum value) of induced emf

The amplitude of the alternating emf is

$$ \varepsilon_0 = BA\omega $$

Substituting $A = \frac{1}{2}\pi R^2$ and $\omega = 2\pi f$,

$$ \varepsilon_0 = B \left(\frac{1}{2}\pi R^2\right)(2\pi f) $$

$$ \varepsilon_0 = \pi^2 B R^2 f $$

Answer:

The amplitude of the alternating emf induced in the wire is

$$ \boxed{\varepsilon_0 = \pi^2 B R^2 f} $$

✦✦✦ End of Solution ✦✦✦

Key Takeaways:

Conceptual clarity is more important than memorization: The Maharashtra Board emphasizes understanding of physical concepts, especially in derivations, numericals, and reasoning-based MCQs.
Formulas must be supported with proper derivation steps: Writing only the final formula fetches fewer marks. Always show intermediate steps and substitutions clearly.
SI units and symbols are compulsory: Incorrect or missing units can lead to loss of marks even if the numerical answer is correct.
MCQs test concept application, not definitions: Most MCQs involve elimination of incorrect options using physics principles rather than direct recall.
Ray optics and wave optics require neat diagrams: Well-labelled ray diagrams and wavefront constructions significantly improve presentation and scoring.
Electricity and magnetism numericals follow standard formula patterns: Questions often repeat similar models based on Gauss’s law, Ohm’s law, Faraday’s law, and LCR circuits.
AC circuits focus on phase difference and power concepts: Students must clearly understand rms values, power factor, resonance, and wattless current.
Modern physics is highly scoring: Photoelectric effect, Bohr’s model, semiconductors, and logic gates mostly involve direct formulas and reasoning.
Derivations should be written step-by-step: Jumping directly to the final result is discouraged. Logical progression is necessary for full marks.
Graphs are frequently asked and carry easy marks: Energy vs frequency, current vs voltage, and magnetic field graphs should be practiced carefully.
Use proper physical terms and standard definitions: Board examiners expect textbook language, especially in theory-based answers.
Numericals should follow a fixed format: Given → Formula → Substitution → Calculation → Final Answer (with unit).
Time management is critical: Avoid spending too much time on lengthy derivations; attempt short answers and MCQs first.
Previous years’ questions repeat conceptually: Practicing past papers helps identify frequently tested laws, principles, and numericals.
Presentation directly impacts marks: Use proper spacing, underlined formulas, and clear headings for answers.

Frequently Asked Questions (FAQ) - Maharashtra Board Class 12 Physics 2025 – Solved Paper

Which phenomenon proves that light is a transverse wave?

(A) Reflection
(B) Interference
(C) Diffraction
(D) Polarization

Answer:

Correct option: (D) Polarization

Which physical quantity of light remains unchanged when it travels from one medium to another?

(A) Velocity
(B) Wavelength
(C) Frequency
(D) Amplitude

Answer:

Correct option: (C) Frequency

What is the SI unit of electric flux?

(A) N C⁻¹
(B) C m²
(C) N m² C⁻¹
(D) V m⁻¹

Correct option: (B) C m²

Define wattless current in an AC circuit.

Wattless current is the current flowing in an AC circuit in which the average power consumed is zero.

$$ \phi = 90^\circ $$

Average power is given by:

$$ P_{\text{avg}} = VI \cos\phi $$

Since:

$$ \cos 90^\circ = 0 $$

Therefore:

$$ P_{\text{avg}} = 0 $$

Hence, the current does not consume power and is called wattless current.

State Bohr’s quantization condition for angular momentum.

According to Bohr, the angular momentum of an electron revolving around the nucleus is quantized.

The quantization condition is:

$$ mvr = \frac{nh}{2\pi} $$

Where:

m = mass of electron
v = velocity of electron
r = radius of orbit
h = Planck’s constant
n = principal quantum number (1, 2, 3, ...)

This condition explains the stability of atomic orbits and discrete energy levels in hydrogen atom.

What is the condition for resonance in a series LCR circuit?

Resonance occurs when the inductive reactance equals the capacitive reactance.

Condition for resonance:

$$ X_L = X_C $$

Where:

$$ X_L = \omega L \quad \text{and} \quad X_C = \frac{1}{\omega C} $$

At resonance:

$$ \omega_0 = \frac{1}{\sqrt{LC}} $$

At this frequency, impedance is minimum and current is maximum.

Chp-7: Wave Optics

Q1. What is diffraction of light? How does it differ from interference? What are Fraunhoffer and Fresnel diffractions?

Key Formula (Single-Slit Diffraction)

Q2. Derive the conditions for bright and dark fringes produced due to diffraction by a single slit.

Central Bright Fringe

Secondary Bright Fringes

Intensity Distribution Note

Q3. Describe what is Rayleigh’s criterion for resolution. Explain it for a telescope and a microscope.

Q4. A parallel beam of green light of wavelength 546 nm passes through a slit of width 0.4 mm. The intensity pattern of the transmitted light is seen on a screen which is 40 cm away. What is the distance between the two first order minima?

Q5. The distance between two consecutive bright fringes in a biprism experiment using light of wavelength 6000 Å is 0.32 mm by how much will the distance change if light of wavelength 4800 Å is used?

Chp-8: Electrostatics

(a) Direction of the Electric Field

(b) Torque Acting on the Dipole

Final Answers

Q2. A 6 μF capacitor is charged by a 300 V supply. It is then disconnected from the supply and is connected to another uncharged 3μF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation ?

Step 1: Initial Energy Stored in the First Capacitor

Step 2: Final Voltage After Connecting the Two Capacitors

Step 3: Final Energy Stored in the Combined System

Step 4: Energy Lost

Final Answer

(a) Capacitance of the Capacitor

(b) Charge on Each Plate When Connected to 100 V Supply

(c) Effect of Inserting a Mica Sheet (k = 6)

Final Answers

A charge 6 μC is placed at the origin and another charge –5 μC is placed on the y axis at a position A (0, 6.0) m.

Electric Potential and Work Done

(a) Electric Potential at Point P

Potential due to $+6\,\mu\text{C}$ charge

Potential due to $-5\,\mu\text{C}$ charge

Total Electric Potential at P

(b) Work Done to Bring a Proton from Infinity to Point P

Significance of the Negative Sign

Final Answers

Chp-9: Current Electricity

Q1. What is Potential Gradient? How is it Measured? Explain. On What Factors Does the Potential Gradient of a Wire Depend?

1. Definition of Potential Gradient

2. Measurement of Potential Gradient

3. Factors Affecting the Potential Gradient of a Wire

Q2. Why Is a Potentiometer Preferred Over a Voltmeter for Measuring EMF?

1. Concept of EMF Measurement

2. Limitation of a Voltmeter

3. Advantage of a Potentiometer

4. Reasons Why Potentiometer Is Preferred

Q3. Describe Kelvin’s Method to Determine the Resistance of a Galvanometer Using a Meter Bridge

1. Aim

2. Principle

3. Circuit Arrangement

4. Procedure

5. Theory and Formula

6. Advantage of Kelvin’s Method

Q4. Describe How a Potentiometer Is Used to Compare the EMFs of Two Cells by Connecting the Cells Individually

1. Aim

2. Principle

3. Circuit Arrangement

4. Procedure

5. Theory and Formula

6. Advantages of the Method

Q5. Describe How a Potentiometer Is Used to Compare the EMFs of Two Cells by Combination Method

1. Aim

2. Principle

3. Circuit Arrangement (Combination Method)

4. Procedure

5. Theory and Formula

6. Advantages of Combination Method

Q6. A Voltmeter Has a Resistance 30 Ω. What Will Be Its Reading When It Is Connected Across a Cell of EMF 2 V Having Internal Resistance 10 Ω?

1. Given Data

2. Concept Used

3. Calculation of Current

4. Voltmeter Reading

5. Answer

Q7. A Potential Drop per Unit Length Along a Wire Is 5 × 10−3 V/m. If the EMF of a Cell Balances Against Length 216 cm of This Potentiometer Wire, Find the EMF of the Cell.

1. Given Data

2. Formula Used

3. Calculation

4. Answer

Chp-12: Electromagnetic Induction

Q1. State and explain Lenz’s law in the light of the principle of conservation of energy.

Q2. What are eddy currents? State applications of eddy currents.

Q3. A long solenoid consisting of 1.5×103 turns/m has an area of cross-section of 25 cm2. A coil C consisting of 150 turns is wound tightly around the centre of the solenoid. Calculate for a current of 3.0 A in the solenoid:

(a) Magnetic flux density at the centre of the solenoid

Q7. A Potential Drop per Unit Length Along a Wire Is 5 × 10⁻³ V/m. If the EMF of a Cell Balances Against Length 216 cm of This Potentiometer Wire, Find the EMF of the Cell.

Q3. A long solenoid consisting of 1.5×10³ turns/m has an area of cross-section of 25 cm². A coil C consisting of 150 turns is wound tightly around the centre of the solenoid. Calculate for a current of 3.0 A in the solenoid:

Q4. A search coil having 2000 turns with area 1.5 cm² is placed in a magnetic field of 0.60 T. The coil is moved rapidly out of the field in a time of 0.2 s. Calculate the induced emf across the search coil.

Q5. An aircraft of wing span 50 m flies horizontally in Earth’s magnetic field of 6 × 10⁻⁵ T at a speed of 400 m/s. Calculate the emf generated between the tips of the wings of the aircraft.

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