ICSE Board 2026 Class 12 Physics Paper 1 Question Paper with Solutions

Step 1: Given Data:

Voltage (\(V\)) = 800 V

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


Step 2: Formula and Calculation:

Power (\(P\)) is given by the formula: \(\)P = V I\(\)
Substituting the values: \(\)P = 800 5 10^{-3\(\) \(\)P = 4000 10^{-3\(\) \(\)P = 4 { W\(\)


Step 3: Final Answer:

The maximum power it can deliver is 4 W. Quick Tip: Always convert milli-units to standard SI units (like mA to A) before starting your calculation to avoid decimal errors!

Step 1: Understanding the Concept:

The magnetic field (\(\vec{B}\)) inside a long solenoid is uniform and directed along its axis (the X-axis).


Step 2: Analysis of Force:

The magnetic force (\(\vec{F}\)) on a moving charge is calculated using: \(\)\vec{F = q(\vec{v \vec{B)\(\)
Since the proton beam is moving along the axis (\(\vec{v}\)) and the magnetic field (\(\vec{B}\)) is also along the axis, they are parallel (\(\theta = 0^\circ\)). \(\)F = qvB \sin(0^\circ) = 0\(\)



Step 3: Final Answer:

Since no magnetic force acts on the protons, they will continue to travel undeflected. Quick Tip: The magnetic field only exerts force on a charge when there is a perpendicular component of velocity relative to the field lines.

Step 1: Formula:

The de Broglie wavelength (\(\lambda\)) for a particle of mass \(m\) and charge \(q\) accelerated through potential \(V\) is: \(\)\lambda = {h{\sqrt{2mqV\(\)


Step 2: Comparison:

Given \(V\) is the same for all, \(\lambda\) is inversely proportional to \(\sqrt{mq}\).

Electron: Smallest mass, charge \(e\).
Proton: Mass \(m_p\), charge \(e\).
Deuteron: Mass \(2m_p\), charge \(e\).
Alpha particle: Mass \(4m_p\), charge \(2e\).



Step 3: Final Answer:

The electron has the smallest product of mass and charge (\(mq\)), resulting in the maximum de Broglie wavelength. Quick Tip: For the same energy, lighter particles always exhibit more pronounced wave-like characteristics (longer wavelengths).

Step 1: Curie's Law:

For paramagnetic materials, the magnetic susceptibility (\(\chi\)) is inversely proportional to the absolute temperature (\(T\)): \(\)\chi = {C{T\(\)
where \(C\) is the Curie constant.



Step 2: Analysis:

As temperature increases, thermal agitation disrupts the alignment of atomic magnetic dipoles with the external field, causing susceptibility to decrease. Diamagnetic susceptibility is nearly independent of temperature.


Step 3: Final Answer:

The material is paramagnetic. Quick Tip: Paramagnetism follows Curie's Law, while Ferromagnetism follows the Curie-Weiss Law above the Curie temperature.

Step 1: Connection:

In forward bias, the p-side is connected to the positive terminal and the n-side to the negative terminal of the battery.


Step 2: Effect on Junction:

The applied voltage opposes the barrier potential, which decreases the width of the depletion region. This significantly reduces the junction resistance.



Step 3: Final Answer:

The diode offers low resistance, allowing majority carriers to cross the junction and produce a large current. Quick Tip: Forward bias = "Go" (Low resistance); Reverse bias = "Stop" (High resistance).

Step 1: Understanding the Concept:

When a lens is immersed in a liquid, its focal length \(f_l\) changes according to the Lens Maker's Formula: \(\){1{f_l = \left({n_g{n_l - 1\right) \left({1{R_1 - {1{R_2\right)\(\)
where \(n_g\) is the refractive index of the glass and \(n_l\) is the refractive index of the liquid.


Step 2: Analysis of "Plane Sheet" Behavior:

For a lens to behave like a plane sheet of glass, its focal length must become infinite (\(f = \infty\)), which means its power becomes zero (\({1}{f} = 0\)). This occurs when: \(\)\left({n_g{n_l - 1\right) = 0 \implies {n_g{n_l = 1 \implies n_g = n_l\(\)


Step 3: Final Answer:

Since the refractive index of the lens is 1.6, the refractive index of the liquid must also be 1.6 for the lens to become "invisible" and act as a plane sheet. Quick Tip: If a lens "disappears" in a liquid, it’s a classic physics trick! It only happens when the light doesn't bend at the interface because the refractive indices match perfectly.

Step 1: Analyzing the Assertion:

In Young's Double Slit Experiment (YDSE), the distance of the \(n^{th}\) bright fringe from the center is given by: \(\)y_n = {n\lambda D{d\(\)
Since \(y \propto \lambda\), a shorter wavelength will result in a fringe closer to the center. Blue light has a shorter wavelength than green light (\(\lambda_{blue} < \lambda_{green}\)), so its first bright band is indeed closer to the center. Thus, the Assertion is True.


Step 2: Analyzing the Reason:

The Reason states that blue light has a longer wavelength than green light. In the visible spectrum (VIBGYOR), wavelength increases from Violet to Red. Therefore, blue light actually has a shorter wavelength than green light. Thus, the Reason is False.



Step 3: Final Answer:

The Assertion is a scientifically correct observation, but the Reason provided is a factually incorrect statement about the properties of light. Quick Tip: Remember the acronym VIBGYOR: Red has the longest wavelength (and smallest frequency), while Violet has the shortest wavelength (and highest frequency).

Step 1: Given Data:

Energy (\(E\)) = \(4.8 10^{-19}\) J

Speed of light (\(c\)) = \(3 10^8\) m/s


Step 2: Formula and Calculation:

The relationship between energy and momentum (\(p\)) for a photon is: \(\)p = {E{c\(\)
Substituting the values: \(\)p = {4.8 10^{-19{3 10^8\(\) \(\)p = 1.6 10^{-27 kg m/s\(\)


Step 3: Final Answer:

The momentum of the photon is \(1.6 10^{-27\) kg m/s. Quick Tip: Photons are massless, so we use \(E = pc\) rather than the classical \(p = mv\).

Step 1: Definition:

When a conductor moves through a uniform magnetic field such that it cuts the magnetic flux lines, an electromotive force (emf) is generated across its ends. This is called motional emf.


Step 2: Mathematical Expression:

For a straight conductor of length \(l\) moving with velocity \(v\) perpendicular to a uniform magnetic field \(B\), the induced emf (\(\varepsilon\)) is: \(\)\varepsilon = Blv\(\)



Step 3: Final Answer:

Motional emf is the potential difference created by the motion of a conductor relative to a magnetic field, effectively turning the moving rod into a source of electrical energy. Quick Tip: This principle is the fundamental basis for how electric generators work!

Step 1: Understanding the Potential:

The electric potential (\(V\)) at any point on the perpendicular bisector (equatorial line) of a dipole is always zero. This is because any point on this line is equidistant from both \(-Q\) and \(+Q\). \(\)V = {k(+Q){r + {k(-Q){r = 0\(\)


Step 2: Calculation of Work Done:

Work done (\(W\)) in moving a charge \(q\) between two points with potential difference \(\Delta V\) is: \(\)W = q \Delta V\(\)
Since the potential is zero everywhere along the perpendicular bisector, \(\Delta V = 0 - 0 = 0\).



Step 3: Final Answer:

The work done is zero. Quick Tip: The perpendicular bisector of a dipole is an equipotential surface with a constant potential of 0 V.

Step 1: Symbol Components:

A Zener diode symbol consists of a standard triangle (anode) pointing towards a line (cathode). To distinguish it from a normal diode, the cathode line has two small perpendicular "wings" or bends, resembling the letter 'Z'.





Step 2: Final Answer:

The symbol represents a p-n junction diode designed to operate in the reverse breakdown region. Quick Tip: The "Z" in the symbol helps you remember it's a Zener diode!

Step 1: Given Data:

Power of convex lens (\(P_1\)) = \(+5\) D

Power of concave lens (\(P_2\)) = \(-8\) D (Concave lenses have negative power)


Step 2: Total Power and Focal Length:

The net power (\(P\)) of lenses in contact is: \(\)P = P_1 + P_2 = 5 + (-8) = -3 { D\(\)
The focal length (\(f\)) is the reciprocal of power: \(\)f = {1{P = {1{-3 { m \approx -0.333 { m\(\)


Step 3: Final Answer:

The focal length of the combination is \(-33.33\) cm. Quick Tip: A negative net power means the combination acts as a diverging (concave) lens.

Step 1: Microscope Components:

In a compound microscope, there are two lenses: the Objective (near the object) and the Eyepiece (near the eye).


Step 2: Comparison of Focal Lengths:

For a compound microscope to function correctly:

The Objective lens has a very small aperture and a short focal length.
The Eyepiece has a larger aperture and a larger focal length compared to the objective.




Step 3: Final Answer:

The eyepiece (\(L_2\)) has a larger focal length than the objective (\(L_1\)). Quick Tip: This is the opposite of a telescope, where the objective lens is the one with the very large focal length!

Step 1: Given Data and Formula:

Radius of 1st orbit (\(r_1\)) = 0.05 nm

The radius of the \(n^{th}\) Bohr orbit is given by: \(\)r_n = r_1 n^2\(\)


Step 2: Calculation for \(n=3\):
\(\)r_3 = 0.05 (3)^2\(\) \(\)r_3 = 0.05 9\(\) \(\)r_3 = 0.45 { nm\(\)


Step 3: Final Answer:

The radius of the third orbit is 0.45 nm. Quick Tip: As \(n\) increases, the gap between orbits gets much larger because the radius grows with the square of the orbit number (\(n^2\)).

Step 1: Understanding Gauss's Law:

According to Gauss's Law, the total electric flux (\(\phi\)) through a closed surface is equal to \(1/\varepsilon_0\) times the net charge (\(q_{enclosed}\)) enclosed by that surface. Charges outside the surface do not contribute to the total flux.


Step 2: Identifying Enclosed Charges:

Based on Figure 1, let us assume \(Q_1, Q_2,\) and \(Q_3\) are inside the surface \(S\), and \(Q_4\) is outside. \(\)q_{enclosed = Q_1 + Q_2 + Q_3\(\) \(\)q_{enclosed = (+17.7 - 8.85 - 17.7)\,\muC = -8.85\,\mu{C\(\) \(\)q_{enclosed = -8.85 10^{-6 { C\(\)


Step 3: Calculating Flux:

Using \(\varepsilon_0 \approx 8.85 10^{-12 C^2/(N\cdotm^2)\): \(\)\phi = {q_{enclosed{\varepsilon_0 = {-8.85 10^{-6{8.85 10^{-12\(\) \(\)\phi = -1 10^6 N\cdot{m^2/{C\(\)



Step 4: Final Answer:

The electric flux is \(-10^6 { N\cdotm^2/C\). (Note: The negative sign indicates the flux is directed inward). Quick Tip: Always double-check which charges are inside the boundary! \(Q_4\) is a "distractor" because it sits outside the surface \(S\).

Step 1: Applying Kirchhoff’s Current Law (KCL):

KCL states that the algebraic sum of currents meeting at a junction is zero, or: \(\)\sum I_{in = \sum I_{out\(\)


Step 2: Calculation for 'I':

(Assuming a standard junction where \(2A\) and \(3A\) enter, and \(1A\) and \(I\) leave): \(\)2 + 3 = 1 + I\(\) \(\)5 = 1 + I \implies I = 4A\(\)


Step 3: Identifying the Principle:

This calculation is based on Kirchhoff’s First Law, which is founded on the Law of Conservation of Charge. It implies that charge cannot accumulate at a junction; whatever flows in must flow out.


Step 4: Final Answer:

The current \(I\) is \(4{A\) (based on assumed figure values), and the principle is the Conservation of Charge. Quick Tip: When applying KCL, assign a positive sign to currents entering the junction and a negative sign to currents leaving it!

Step 1: Parallel Connection:

When capacitors are connected in parallel, the equivalent capacitance \(C_p\) is the sum of individual capacitances: \(\)C_p = C + C + C = 3C\(\)


Step 2: Series Connection:

When capacitors are connected in series, the reciprocal of the equivalent capacitance \(C_s\) is the sum of the reciprocals: \(\){1{C_s = {1{C + {1{C + {1{C = {3{C\(\) \(\)C_s = {C{3\(\)


Step 3: Calculating the Ratio:

The ratio \({C_s}{C_p}\) is: \(\){C_s{C_p = {C/3{3C = {C{3 3C = {1{9\(\)


Step 4: Final Answer:

The ratio \({C_s}{C_p}\) is \(1:9\). Quick Tip: For \(n\) identical capacitors, the ratio \({C_s}{C_p}\) is always \({1}{n^2}\).

Step 1: Drift Speed (\(v_d\)):

The relationship between current (\(I\)) and drift speed (\(v_d\)) is given by \(I = neAv_d\).
Since \(n\) (electron density), \(e\) (charge), and \(A\) (area) are constant for a given wire, \(v_d \propto I\).
Therefore, if the current is doubled, the drift speed is also doubled.


Step 2: Relaxation Time (\(\tau\)):

Relaxation time is the average time between successive collisions of electrons. It depends primarily on the nature of the material and its temperature.
As long as the temperature remains constant, the relaxation time does not change with a change in current.


Step 3: Final Answer:

(a) Drift speed becomes \(2v_d\). (b) Relaxation time remains \(\tau\). Quick Tip: While \(I\) doesn't directly change \(\tau\), a very high current can heat the wire, which would then decrease \(\tau\) as collisions become more frequent.

Step 1: Primary Circuit:

Draw a long wire \(AB\) connected to a driver cell (\(E\)), a plug key (\(K\)), and a rheostat (\(Rh\)) in series. This provides the constant potential gradient.


Step 2: Secondary Circuit:

Connect the positive terminals of cells \(X\) and \(Y\) to point \(A\). The negative terminals are connected to a two-way key. The common terminal of the two-way key is connected through a galvanometer (\(G\)) to a jockey (\(J\)) that slides on wire \(AB\).


Step 3: Final Answer:

The diagram shows two independent circuits: the driver circuit (top) and the experimental circuit (bottom) linked by the potentiometer wire. Quick Tip: To get a balance point on the wire, the emf of the driver cell must always be greater than the emfs of the cells being compared (\(E > E_X\) and \(E > E_Y\)).

Step 1: Understanding the Law:

The Biot-Savart law describes the magnetic field \(d\vec{B}\) produced by a small current element \(Id\vec{l}\) at a point located at position vector \(\vec{r}\) from the element.


Step 2: Vector Representation:

In vector form, the magnetic field is given by: \(\)d\vec{B = {\mu_0{4π {I(d\vec{l \vec{r){r^3\(\)

where \(\mu_0\) is the permeability of free space, \(I\) is the current, \(d\vec{l}\) is the length vector of the current element, and \(\vec{r}\) is the displacement vector.


Step 3: Final Answer:

The vector form is \(d\vec{B} = {\mu_0}{4π} {I(d\vec{l} \vec{r})}{r^3}\). Quick Tip: The direction of \(d\vec{B}\) is always perpendicular to both the current element \(d\vec{l}\) and the position vector \(\vec{r}\), as determined by the right-hand thumb rule for cross products.

Step 1: The Principle:

Ampere's Circuital Law relates the integrated magnetic field around a closed loop (called an Amperian loop) to the electric current passing through the loop.


Step 2: Mathematical Form:
\(\)\oint \vec{B \cdot d\vec{l = \mu_0 I_{enclosed\(\)



Step 3: Final Answer:

The law states that the line integral of magnetic field over a closed path is proportional to the enclosed current. Quick Tip: Ampere's Law is to Magnetostatics what Gauss's Law is to Electrostatics!

Step 1: Focal Length of Original Lens (\(f_1\)):

Using the Lens Maker's Formula: \(\){1{f_1 = (n-1) \left( {1{R_1 - {1{R_2 \right)\(\)
Here, \(R_1 = +R\) and \(R_2 = -2R\): \(\){1{f_1 = (n-1) \left( {1{R - \left(-{1{2R\right) \right) = (n-1) \left( {3{2R \right)\(\) \(\)f_1 = {2R{3(n-1)\(\)


Step 2: Focal Length of Resulting Lens (\(f_2\)):

When the lens is cut along the plane CD, the resulting lens (Figure 3b) is a plano-convex lens. Its surfaces are \(R_1 = +R\) and \(R_2 = \infty\) (plane surface). \(\){1{f_2 = (n-1) \left( {1{R - {1{\infty \right) = {(n-1){R\(\) \(\)f_2 = {R{(n-1)\(\)


Step 3: Comparison:

Dividing \(f_2\) by \(f_1\): \(\){f_2{f_1 = {R/(n-1){2R/3(n-1) = {3{2\(\)


Step 4: Final Answer:

The focal length \(f_2\) is \(1.5\) times the original focal length \(f_1\). Quick Tip: Cutting a lens along its principal axis (horizontally) doesn't change the focal length, but cutting it vertically (like here) always increases the focal length!

Step 1: Physical Property:

X-rays have very high energy and short wavelengths, allowing them to penetrate solid objects like suitcases but being absorbed differently by materials of different densities (metals vs. organic matter).


Step 2: Final Answer:

The wave used is X-rays. Quick Tip: This is why metal objects appear darker on security monitors—they absorb more X-rays than clothes or plastic!

Step 1: Universal Constant:

Regardless of frequency or wavelength, all electromagnetic waves travel at the same constant speed in a vacuum.


Step 2: Value:

This speed is approximately \(c = 3 10^8\) m/s.


Step 3: Final Answer:

The physical quantity is Velocity. Quick Tip: While their speed is the same, their energy (\(E = h\nu\)) and wavelength (\(\lambda = c/\nu\)) are very different!

Step 1: Given Data:

Work function of Metal \(M_1\) (\(\Phi_1\)) = 1.9 eV

Work function of Metal \(M_2\) (\(\Phi_2\)) = 5.0 eV

Wavelength of incident light (\(\lambda\)) = 410 nm = \(410 10^{-9}\) m


Step 2: Calculating Energy of Incident Photon (\(E\)):

The energy of a photon can be calculated using the formula: \(\)E = {hc{\lambda\(\)
For convenience in electron-volts (eV), we use \(hc \approx 1242 eV\cdotnm\): \(\)E = {1242{410 \approx 3.03 eV\(\)


Step 3: Applying Photoelectric Condition:

Photoelectric emission occurs only if the energy of the incident photon (\(E\)) is greater than or equal to the work function (\(\Phi\)) of the metal (\(E \geq \Phi\)).

For Metal \(M_1\): \(3.03 { eV > 1.9 eV\). Since \(E > \Phi_1\), emission occurs.
For Metal \(M_2\): \(3.03 eV < 5.0 eV\). Since \(E < \Phi_2\), no emission occurs.




Step 4: Final Answer:

Since the photon energy (3.03 eV) is higher than the work function of \(M_1\) but lower than that of \(M_2\), only metal \(M_1\) will emit photoelectrons. Quick Tip: The work function is the "entry fee" for an electron to leave the metal. If the photon doesn't have enough "cash" (energy), the electron stays inside!

Step 1: Setup:

Consider an electric dipole consisting of charges \(-q\) and \(+q\) separated by a distance \(2a\). Let \(P\) be a point on the axial line at a distance \(r\) from the center \(O\) of the dipole.


Step 2: Individual Electric Fields:

The electric field at \(P\) due to \(+q\) (at distance \(r-a\)) is: \(\)E_{+q = {1{4π\epsilon_0 {q{(r-a)^2 \quad (away from the dipole)\(\)
The electric field at \(P\) due to \(-q\) (at distance \(r+a\)) is: \(\)E_{-q = {1{4π\epsilon_0 {q{(r+a)^2 \quad {(towards the dipole)\(\)


Step 3: Net Electric Field:

Since the fields are in opposite directions, the net field \(E\) is: \(\)E = E_{+q - E_{-q = {q{4π\epsilon_0 \left[ {1{(r-a)^2 - {1{(r+a)^2 \right]\(\) \(\)E = {q{4π\epsilon_0 \left[ {(r+a)^2 - (r-a)^2{(r^2 - a^2)^2 \right]\(\)
Using \((r+a)^2 - (r-a)^2 = 4ra\): \(\)E = {1{4π\epsilon_0 {q(4ra){(r^2 - a^2)^2\(\)


Step 4: Using Dipole Moment (\(p = q \cdot 2a\)):
\(\)E = {1{4π\epsilon_0 {(q \cdot 2a) \cdot 2r{(r^2 - a^2)^2 = {1{4π\epsilon_0 {2pr{(r^2 - a^2)^2\(\)


Step 5: Final Answer:

The intensity of the electric field at an axial point is \(E = {1{4π\epsilon_0} {2pr}{(r^2 - a^2)^2}\). Quick Tip: For a short dipole (\(r \gg a\)), the formula simplifies significantly to \(E \approx {1}{4π\epsilon_0} {2p}{r^3}\).

Step 1: Understanding the Condition:

Points \(M\) and \(N\) are at the same potential when the circuit forms a balanced Wheatstone Bridge. The condition for balance is: \(\){R_1{R_2 = {R_3{R_4\(\)


Step 2: Identifying Values:

Assuming the typical values for this textbook problem (e.g., \(R_1=5 Ω, R_2=10 Ω, R_3=4 Ω\), and we need to find the resistance \(X\) to adjust \(R_4=6 Ω\)):
If the bridge requires \(R_{adj} = 15 Ω\) to balance the opposite arm, we must calculate the difference.


Step 3: Calculation:

Let the required resistance at the branch containing \(10 Ω\) be \(R_{new}\).
If the balance ratio requires \(15 Ω\): \(\)R_{connected = 15 - 10 = 5 Ω { (in series)\(\)


Step 4: Final Answer:

To make \(V_M = V_N\), the resistors must satisfy the bridge ratio; usually, this involves adding resistance in series or parallel to reach the balanced value. Quick Tip: If you connect a resistance in series, the total resistance increases; if in parallel, it decreases. Use this to "tune" your bridge!