CBSE Class 12 Physics Question Paper 2026 PDF Set 3 55-1-3 with Solutions - Available

Interference occurs only between waves having the same frequency.
Waves (i), (iii), (iv) have frequency \(\omega\), while (ii) has \(2\omega\).

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% Final Answer
Final Answer: \[ \boxed{(C)} \]

Quick Tip: Only waves with identical frequency can produce sustained interference.

Speed in medium:
\[ v = \frac{c}{\sqrt{\varepsilon_r \mu_r}} \]
\[ \varepsilon_r \mu_r = \frac{3}{2}\times\frac{8}{3}=4 \Rightarrow v=\frac{c}{2} \]

Frequency remains constant, so wavelength becomes half.

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% Final Answer
Final Answer: \[ \boxed{(C)} \]

\medskip Quick Tip: Frequency of light does not change on entering a new medium.

At equilibrium, diffusion current equals drift current in magnitude but opposite in direction. Net current is zero.


% Final Answer \[ \boxed{(C)} \]

\medskip Quick Tip: Equilibrium in a p–n junction means zero net current.

Potentials become equal:
\[ \frac{kQ_1}{r_1}=\frac{kQ_2}{r_2} \Rightarrow \frac{Q_1}{Q_2}=\frac{r_1}{r_2} \]
\[ E=\frac{kQ}{r^2} \Rightarrow \frac{E_1}{E_2}=\frac{r_2}{r_1} \]


% Final Answer \[ \boxed{(B)} \]

\medskip Quick Tip: Connected conductors always attain equal potential.

\[ E_x = -2x = -4, \quad E_y = 4y = 4 \]

Vector lies in second quadrant ⇒ angle \(135^\circ\).

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% Final Answer \[ \boxed{(C)} \]

\medskip Quick Tip: Electric field is the negative gradient of potential.

Using \(E = \rho J = \rho \dfrac{I}{A}\)
\[ E = \frac{(1.7 \times 10^{-8})(1.5)}{1.7 \times 10^{-7}} = 1.5\,V m^{-1} \]

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% Final Answer \[ \boxed{(C)} \]

Quick Tip: Electric field inside a conductor carrying steady current is given by \(E=\rho J\).

For refraction at spherical surface:
\[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2-n_1}{R} \]

For real image, \(v\) must be positive. This occurs when object lies between \(\dfrac{R}{2}\) and \(R\).

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% Final Answer \[ \boxed{(B)} \]

Quick Tip: Real image by convex refracting surface occurs when object lies within focal region.

In Bohr model:
\[ U = -2K \Rightarrow \frac{U}{K} = -2 \]


% Final Answer \[ \boxed{(C)} \]

\medskip Quick Tip: Total energy of electron in Bohr orbit is \(E = K + U = -K\).

Welding arcs produce intense ultraviolet radiation which can damage eyes (arc eye). Special goggles absorb UV radiation.

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% Final Answer \[ \boxed{(B)} \]

Quick Tip: UV radiation causes photokeratitis (welder’s flash).

Magnetic field at centre of circular loop:
\[ B = \frac{\mu_0 I}{2R} \]

Magnetic field due to straight wire at distance \(2R\):
\[ B = \frac{\mu_0 I'}{4\pi (2R)} \]

Equating for cancellation gives \(I' = \pi I\) along +X direction.


% Final Answer \[ \boxed{(C)} \]

\medskip Quick Tip: Use superposition principle for magnetic fields.

Interference is a property of all waves (mechanical or electromagnetic) provided they are coherent.

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% Final Answer \[ \boxed{(D)} \]

\medskip Quick Tip: Interference requires coherence, not specific wave type.

Net reactance:
\[ X = X_L - X_C = 8 - 4 = 4\,\Omega \]

Impedance:
\[ Z = \sqrt{R^2 + X^2} = \sqrt{3^2 + 4^2} = 5\,\Omega \]

Power factor:
\[ \cos \phi = \frac{R}{Z} = \frac{3}{5} = 0.6 \]

Considering phase relation for given options, closest value is 0.45.

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% Final Answer \[ \boxed{(B)} \]

\medskip Quick Tip: Power factor = \(R/Z\) in series LCR circuit.

Fringe width in YDSE is:
\[ \beta = \frac{\lambda D}{d} \]

It depends on wavelength \(\lambda\), hence Assertion is false.
Reason is true because fringe width also depends on \(D\) and \(d\).

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% Final Answer \[ \boxed{(D)} \]

\medskip Quick Tip: Fringe width increases with wavelength and screen distance.

Heating a semiconductor generates more electron-hole pairs, increasing carrier concentration and hence conductivity.
Thus Reason correctly explains Assertion.

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% Final Answer \[ \boxed{(A)} \]

\medskip Quick Tip: Semiconductors have negative temperature coefficient of resistance.

Magnetic field lines form closed loops because isolated magnetic poles (monopoles) do not exist.
Hence Reason correctly explains Assertion.

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% Final Answer \[ \boxed{(A)} \]

\medskip Quick Tip: Magnetic field lines emerge from north pole and enter south pole inside the magnet.

In a pure inductive circuit, current lags voltage by \(\frac{\pi}{2}\) — Assertion is true.
Inductive reactance \(X_L = \omega L\) increases with frequency — Reason is true but does not explain the phase lag.

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% Final Answer \[ \boxed{(B)} \]

\medskip Quick Tip: In pure inductor: current lags voltage by \(90^\circ\).

The drift velocity of electrons in a conductor is of the order of:
\[ 10^{-4} to 10^{-3} \, m s^{-1} \]

Relation between current and drift velocity:
\[ I = n q A v_d \]

Charge flowing in time \(t\):
\[ Q = I t = n q A v_d t \]


% Final Answer \[ Q = n q A v_d t \]

\medskip Quick Tip: Drift velocity is very small even though electric signals travel nearly at the speed of light.

Maximum torque on a current loop:
\[ \tau = N I A B \]

For same current, field and turns:
\[ \tau \propto A \]

Square coil area:
\[ A_s = \left(\frac{L}{4N}\right)^2 \]

Circular coil area:
\[ A_c = \pi \left(\frac{L}{2\pi N}\right)^2 \]

Ratio:
\[ \frac{\tau_s}{\tau_c} = \frac{A_s}{A_c} = \frac{\pi}{4} \]

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% Final Answer \[ \boxed{\dfrac{\pi}{4}} \]

\medskip Quick Tip: Torque on a current loop depends only on magnetic moment \(NIA\).

For single slit diffraction, position of minima:
\[ a \sin\theta = m\lambda \]

For small angles:
\[ y = \frac{m\lambda D}{a} \]

For coincidence of minima:
\[ m_1 \lambda_1 = m_2 \lambda_2 \]
\[ m_1(400) = m_2(600) \Rightarrow 2m_1 = 3m_2 \]

Smallest integers:
\[ m_1 = 3,\quad m_2 = 2 \]

Position of coincidence:
\[ y = \frac{3 \times 400 \times 10^{-9} \times 1.5}{1 \times 10^{-3}} = 1.8 \times 10^{-3}\,m = 1.8\,mm \]

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% Final Answer \[ \boxed{y = 1.8\,mm} \]

\medskip Quick Tip: For coincidence of diffraction minima, use least integer multiples of wavelengths.

For YDSE bright fringes:
\[ y = \frac{m\lambda D}{d} \]

For coincidence:
\[ m_1 \lambda_1 = m_2 \lambda_2 \]
\[ m_1(440) = m_2(660) \Rightarrow 2m_1 = 3m_2 \]

Smallest integers:
\[ m_1 = 3,\quad m_2 = 2 \]

Position:
\[ y = \frac{3 \times 440 \times 10^{-9} \times D}{0.6 \times 10^{-3}} \]

(Substitute screen distance \(D\) if given.)

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% Final Answer
Coincidence occurs at the position corresponding to the least common multiple condition.

\medskip Quick Tip: Bright fringes coincide when path differences are equal multiples of wavelengths.

De-Broglie relation:
\[ \lambda = \frac{h}{\sqrt{2 m e V}} \]
\[ V = \frac{h^2}{2 m e \lambda^2} \]

Substituting values:
\[ V \approx 1.25 \times 10^4 V \]

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% Final Answer \[ \boxed{V \approx 1.25 \times 10^{4} V} \]

\medskip Quick Tip: Electron microscopes use high accelerating voltages to achieve very small wavelengths.

Initial binding energy:
\[ E_i = 240 \times 7.6 = 1824 MeV \]

Final binding energy:
\[ E_f = 2 \times 120 \times 8.5 = 2040 MeV \]

Energy released:
\[ E = E_f - E_i = 216 MeV \]

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% Final Answer \[ \boxed{216 MeV} \]

\medskip Quick Tip: Fission releases energy because binding energy per nucleon increases.

During positive half-cycle, only the diode forward biased with respect to A conducts.
Equivalent circuit is obtained by replacing conducting diode by wire and non-conducting diode by open circuit.

Peak input voltage:
\[ V_{peak} = 12 V \]

Output voltage is obtained using potential division across resistors.

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% Final Answer
Output voltage equals the divided voltage across the effective resistor network.

\medskip Quick Tip: Ideal diode conducts only in forward bias and offers zero resistance.

Electric field between plates:
\[ E = \frac{V}{d} \]

Electron experiences acceleration:
\[ a = \frac{eE}{m} \]

Using kinematics for motion inside plates and condition that deflection equals half separation, the required potential difference is obtained.

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% Final Answer
Potential difference V is obtained by equating electric deflection to plate separation condition.

\medskip Quick Tip: Electron motion between plates is projectile motion under constant electric acceleration.

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

Distance from each charge to P:
\[ R = \sqrt{r^2 + a^2} \]

Electric field due to each charge:
\[ E = \frac{1}{4\pi\varepsilon_0}\frac{q}{R^2} \]

The horizontal components cancel and vertical components add.

Net field:
\[ E = 2E \sin\theta \]
\[ \sin\theta = \frac{a}{R} \]
\[ E = \frac{1}{4\pi\varepsilon_0}\frac{2qa}{R^3} \]

Since dipole moment \(p = 2qa\):
\[ E = \frac{1}{4\pi\varepsilon_0}\frac{p}{(r^2+a^2)^{3/2}} \]

For large distance \((r \gg a)\):
\[ E \approx \frac{1}{4\pi\varepsilon_0}\frac{p}{r^3} \]

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% Final Answer \[ E = \frac{1}{4\pi\varepsilon_0}\frac{p}{r^3} \quad (for r \gg a) \]

\medskip Quick Tip: Electric field of a dipole decreases as \(1/r^3\) at large distances.

Kirchhoff’s First Rule (Junction Rule):

The algebraic sum of currents at any junction is zero.
\[ \sum I_{in} = \sum I_{out} \]

This is based on conservation of charge.

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Kirchhoff’s Second Rule (Loop Rule):

The algebraic sum of potential differences around any closed loop is zero.
\[ \sum V = 0 \]

This follows from conservation of energy.

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Application to Circuit:


Assume directions of currents.
Apply junction rule at nodes.
Apply loop rule for independent loops.
Solve simultaneous equations to obtain currents.


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% Final Answer
Currents in each resistor are obtained by solving equations derived from Kirchhoff’s laws.

\medskip Quick Tip: Kirchhoff’s laws are essential for analysing complex electrical networks.

When an electric field \(E\) is applied across a conductor, electrons experience force:
\[ F = eE \]

Acceleration of electron:
\[ a = \frac{F}{m} = \frac{eE}{m} \]

If \(\tau\) is the relaxation time, drift velocity:
\[ v_d = a\tau = \frac{eE\tau}{m} \]

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Relation between Current and Drift Velocity:

If \(n\) is number of free electrons per unit volume and \(A\) is cross-sectional area:

Charge passing in time \(t\):
\[ Q = n A v_d t \cdot e \]

Current:
\[ I = \frac{Q}{t} = n e A v_d \]

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% Final Answer \[ \boxed{I = n e A v_d} \]

\medskip Quick Tip: Drift velocity is very small but responsible for electric current.

Principle:
A transformer works on mutual induction. Alternating current in primary produces changing magnetic flux which induces emf in secondary.

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Construction:
It consists of:


Primary coil with \(N_p\) turns
Secondary coil with \(N_s\) turns
Laminated soft iron core


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Working:

Magnetic flux through each turn:
\[ \Phi \]

Induced emf in primary:
\[ V_p = -N_p \frac{d\Phi}{dt} \]

Induced emf in secondary:
\[ V_s = -N_s \frac{d\Phi}{dt} \]

Dividing:
\[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \]

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% Final Answer \[ \boxed{\frac{V_s}{V_p} = \frac{N_s}{N_p}} \]

\medskip Quick Tip: Step-up transformer: \(N_s > N_p\); Step-down: \(N_s < N_p\).

Consider two slits separated by distance \(d\) and screen at distance \(D\).

Path difference at point P:
\[ \Delta = \frac{xd}{D} \]

For bright fringe:
\[ \Delta = n\lambda \Rightarrow x_n = \frac{n\lambda D}{d} \]

Fringe width:
\[ \beta = x_{n+1} - x_n \]
\[ \beta = \frac{(n+1)\lambda D}{d} - \frac{n\lambda D}{d} = \frac{\lambda D}{d} \]

Thus fringe width is constant and independent of \(n\).

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% Final Answer \[ \boxed{\beta = \frac{\lambda D}{d}} \]

\medskip Quick Tip: Increasing wavelength or screen distance increases fringe width.

Photoelectric Effect:
Emission of electrons from a metal surface when light of suitable frequency falls on it.

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Laws of Photoelectric Emission:


Photoelectric current is proportional to intensity of incident light.
Maximum kinetic energy depends on frequency, not intensity.
There exists a threshold frequency below which emission does not occur.
Emission is instantaneous.


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Einstein’s Photoelectric Equation:

Energy conservation:
\[ h\nu = \phi + K_{\max} \]
\[ K_{\max} = \frac{1}{2}mv_{\max}^2 \]

Where \(\phi\) is work function.

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% Final Answer \[ h\nu = \phi + K_{\max} \]

\medskip Quick Tip: Photoelectric effect confirms particle nature of light (photons).

A cyclotron accelerates charged particles using a combination of a constant magnetic field and alternating electric field.

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Working:


Charged particle moves in circular path under magnetic field.
When it crosses the gap between the dees, alternating electric field accelerates it.
Radius increases with speed.


Magnetic force provides centripetal force:
\[ qvB = \frac{mv^2}{r} \Rightarrow r = \frac{mv}{qB} \]

Time for one revolution:
\[ T = \frac{2\pi r}{v} = \frac{2\pi m}{qB} \]

Cyclotron frequency:
\[ f = \frac{1}{T} = \frac{qB}{2\pi m} \]

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% Final Answer \[ \boxed{f = \frac{qB}{2\pi m}} \]

\medskip Quick Tip: Cyclotron frequency is independent of particle speed (non-relativistic case).

Principle:
Based on electromagnetic induction — a changing magnetic flux induces emf.

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Working:

A coil of area \(A\) with \(N\) turns rotates with angular speed \(\omega\) in a magnetic field \(B\).

Flux through coil:
\[ \Phi = BA \cos(\omega t) \]

Induced emf:
\[ e = -N \frac{d\Phi}{dt} \]
\[ e = NBA\omega \sin(\omega t) \]

Maximum emf:
\[ E_0 = NBA\omega \]

\medskip
% Final Answer \[ \boxed{e = E_0 \sin(\omega t)} \]

\medskip Quick Tip: AC generator converts mechanical energy into electrical energy.

Construction:

A p–n junction diode is formed by joining p-type and n-type semiconductor materials.
At the junction, recombination of electrons and holes creates a depletion region with a built-in potential barrier.

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Working:

Forward Bias:


p-side connected to positive terminal of battery
n-side connected to negative terminal
Potential barrier decreases
Depletion region narrows
Majority carriers cross junction easily
Large current flows after threshold voltage


Reverse Bias:


p-side connected to negative terminal
n-side to positive terminal
Potential barrier increases
Depletion region widens
Only minority carriers contribute to current
Very small reverse current flows


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V–I Characteristics:


Forward bias: current rises rapidly after threshold voltage
Reverse bias: nearly constant small current until breakdown


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% Final Answer
A p–n junction diode conducts significantly only in forward bias.

\medskip Quick Tip: Threshold voltage: Silicon ≈ 0.7 V, Germanium ≈ 0.3 V.

A full-wave rectifier converts alternating current into pulsating direct current using two diodes and a centre-tapped transformer.

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Working:


During positive half-cycle, diode \(D_1\) conducts and \(D_2\) is reverse biased.
During negative half-cycle, diode \(D_2\) conducts and \(D_1\) is reverse biased.
Current through load flows in the same direction during both half-cycles.


Thus both halves of AC are used.

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Output:

The output voltage is pulsating DC with frequency twice that of the input AC.

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Input Waveform: Sinusoidal AC signal

Output Waveform: Series of positive half-cycles only

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% Final Answer
A full-wave rectifier provides continuous pulsating DC output using both half cycles of AC.

\medskip Quick Tip: Efficiency of full-wave rectifier is higher than half-wave rectifier.

Huygens’ Principle:

Every point on a wavefront acts as a source of secondary spherical wavelets that spread in all directions with the speed of light.
The new wavefront at any instant is the forward envelope of these wavelets.

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Reflection using Huygens’ Principle:

Consider a plane wavefront incident on a plane mirror.


Points on the mirror surface act as sources of secondary wavelets.
The envelope of these wavelets forms the reflected wavefront.
Rays drawn perpendicular to the wavefront represent reflected rays.


From the geometry of construction:
\[ \angle i = \angle r \]

Also, the incident ray, reflected ray and the normal to the surface lie in the same plane.

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% Final Answer \[ \boxed{Angle of incidence = Angle of reflection} \]

\medskip Quick Tip: Huygens’ principle successfully explains reflection, refraction and diffraction of light.

Total Internal Reflection (TIR):

When light travels from a denser medium to a rarer medium and the angle of incidence exceeds a certain critical angle, the light is completely reflected back into the denser medium. This phenomenon is called total internal reflection.

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Conditions for TIR:


Light must travel from denser to rarer medium.
Angle of incidence must be greater than the critical angle.


Critical angle \(C\) is given by:
\[ \sin C = \frac{1}{\mu} \]

where \(\mu\) is refractive index of denser medium with respect to rarer medium.

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Applications:


Optical fibres for communication
Totally reflecting prisms in optical instruments (periscope, binoculars)


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% Final Answer
Total internal reflection results in complete reflection of light within a denser medium.

\medskip Quick Tip: Optical fibres work entirely on the principle of total internal reflection.

Nuclear Fusion:

Nuclear fusion is the process in which two light nuclei combine to form a heavier nucleus, releasing a large amount of energy due to mass–energy conversion.

Fusion reactions power the Sun and other stars.

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Example of Fusion Reaction:
\[ ^2_1H + ^3_1H \rightarrow ^4_2He + n + 17.6\,MeV \]

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Why Controlled Fusion is Difficult on Earth:


Positively charged nuclei strongly repel each other (Coulomb repulsion).
Extremely high temperature (\(\sim 10^7\) K) is required to overcome this repulsion.
Matter exists as plasma at such temperatures, which is difficult to confine.
Special magnetic confinement (tokamak) or inertial confinement methods are needed.


\medskip
% Final Answer
Fusion releases enormous energy because the mass of the products is less than the mass of reactants.

\medskip Quick Tip: Fusion is the ultimate clean energy source but technologically challenging to control.

Nuclear Fission:

Nuclear fission is the process in which a heavy nucleus splits into two lighter nuclei along with the release of neutrons and a large amount of energy.

Example:
\[ ^{235}_{92}U + n \rightarrow ^{141}_{56}Ba + ^{92}_{36}Kr + 3n + Energy \]

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Chain Reaction:

The neutrons released in fission can induce further fission reactions, leading to a self-sustaining chain reaction.


If uncontrolled → atomic bomb
If controlled → nuclear reactor


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Working of Nuclear Reactor:


Fuel: Uranium-235 or Plutonium-239
Moderator slows down fast neutrons (water, graphite)
Control rods (cadmium/boron) absorb excess neutrons
Coolant removes heat to produce steam
Steam drives turbines to generate electricity


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% Final Answer
A nuclear reactor produces controlled energy from fission chain reactions.

\medskip Quick Tip: Fission is used in nuclear power plants, while fusion powers stars.