NEET 2026 Official Paper (Held On: 03 May, 2026) (Previous Year Paper)

Given:

Number of turns, N = 100

Radius of coil, r = 5 cm = 0.05 m

Magnetic field at centre, B = 3.14 × 10^-3 T

μ₀ = 4π × 10^-7 T m A^-1

Formula Used:

Magnetic field at centre of circular coil:

B = μ₀NI / 2r

Magnetic moment:

M = NIA

Area of coil:

A = πr²

Calculation:

I = 2rB / μ₀N

⇒ I = 2 × 0.05 × 3.14 × 10^-3 / (4π × 10^-7 × 100) = 3.14 × 10^-4 / 4π × 10^-5 = 2.5 A

Now,

A = π × (0.05)² = π × 0.0025 ≈ 0.00785 m²

Magnetic moment:

M = NIA = 100 × 2.5 × 0.00785

⇒ M ≈ 1.96 A m² ≈ 2 A m²

So, current flowing through the coil, and the magnitude of the magnetic moment, respectively, are 2.5 A, 2 A m²

Given: We need to match List I with List II.

Concept Used:

A. E = hν

This is the energy equation of photon.

⇒ A → IV

B. Diffraction and Interference

These phenomena prove wave nature of light.

⇒ B → III

C. λ = h/p

This is de Broglie wavelength relation.

⇒ C → I

D. Compton effect

Compton effect proves particle nature of light.

⇒ D → II

Final Matching: A – IV, B – III, C – I, D – II

Calculation

Only the 4 Ω and 2 Ω branches are forward biased and conduct current.

Equivalent resistance,

$R_{eq} = \frac{4 \times 2}{4 + 2}$

$\Rightarrow R_{eq} = \frac{8}{6} = \frac{4}{3},\Omega$

Given potential difference across the circuit is $10,\text{V}$

Using Ohm’s law,

$I = \frac{V}{R_{eq}}$

$\Rightarrow I = \frac{10}{4/3}$

$\Rightarrow I = \frac{10 \times 3}{4}$

$\Rightarrow I = \frac{15}{2},\text{A}$

Hence, the correct answer is $I = \frac{15}{2},\text{A}$.

Given:

Speed of light in vacuum = 1 unit

Time taken by light = 6 min 40 s

Calculation:

6 min = 6 × 60= 360 s

⇒ Total time = 360 + 40 = 400 s

Distance = Speed × Time

Since speed of light = 1 unit,

⇒ Distance = 1 × 400 = 400

So, the distance between the Sun and the Earth in new unit is 400.

The correct answer is C only

Velocity-Time Graph for a Ball Thrown Vertically

• When a ball is thrown vertically upward, its velocity decreases due to the effect of gravity until it becomes zero at the highest point.
• After reaching the highest point, the ball starts falling back, and its velocity increases in the downward direction due to acceleration due to gravity.
• Correct Representation of Velocity vs Time
• In the velocity-time graph, the velocity is positive during the upward motion and negative during the downward motion.
• The graph in option C accurately shows this behavior, where the velocity decreases linearly, reaches zero at the peak, and then increases negatively (downward motion).

Given:

20 VSD = 16 MSD

and, 1 MSD = 1 mm

Formula Used:

Least Count = 1 MSD − 1 VSD

Calculation:

Given:

20 VSD = 16 MSD

⇒ 1 VSD = 16/20 MSD = 0.8 MSD

Since 1 MSD = 1 mm

⇒ 1 VSD = 0.8 mm

Now, Least Count = 1 MSD − 1 VSD

⇒ LC = 1 − 0.8 = 0.2 mm

Convert into cm: LC = 0.02 cm

Result: Least count = 0.02 cm

Given:

Resistance, R = 1 kΩ= 1000 Ω

Capacitance, C = 0.1 μF = 0.1 × 10^-6 F= 10^-7 F

Inductance, L = 1 mH = 10^-3 H

Formula Used:

Resonance frequency:

f = 1 / 2π√LC

Calculation:

 LC = 10^-3 × 10^-7 = 10^-10

⇒ √LC = 10^-5

Now,

 f = 1 / 2π × 10^-5 = 1 / 6.28 × 10^-5

⇒ f ≈ 1.59 × 10⁴ Hz ≈ 15.9 kHz

Final Answer: f = 15.9 kHz.

Given:

A long solid wire of radius a carries current I uniformly distributed across its cross-section.

Concept Used:

For a solid conductor carrying uniformly distributed current:

• Inside the conductor: B ∝ r

So, magnetic field increases linearly with distance from the axis.

• Outside the conductor: B ∝ 1/r

So, magnetic field decreases inversely with distance.

Explanation:

• At r = 0, B = 0
• From the centre to r = a, B increases linearly
• At r = a, magnetic field becomes maximum
• For r > a, B decreases as 1/r

The correct graph is

Given:

Total resistance of square wire = 4 Ω

Since square has 4 equal sides:

⇒ Resistance of each side = 4/4

Each side resistance = 1 Ω

Resistance connected between B and D = 2 Ω

Battery voltage, V = 2 V

Concept Used:

Between A and C there are two paths:

Path 1: A → B → C

⇒ Resistance = 1 + 1 = 2 Ω

Path 2: A → D → C

⇒ Resistance = 1 + 1 = 2 Ω

Also diagonal BD has 2 Ω.

Since network is symmetrical:

Potential at B = Potential at D

⇒ No current flows through BD resistance.

Therefore diagonal resistance can be ignored.

Calculation:

Now two 2 Ω branches are in parallel.

Equivalent resistance:

R = (2 × 2)/(2 + 2)= 4/4 = 1 Ω

Using Ohm’s law:

 I = V/R = 2/1 = 2 A

Result: Current I = 2 A

Given:

Nuclear density, ρ = 2.29 × 10¹⁷ kg m^-3

Mass of nucleus, M = 19.926 × 10^-27 kg

R₀ = 1.2 × 10^-15 m

4π = 12.56

Formula Used:

Density of nucleus:

ρ = Mass / Volume

Volume of nucleus:

V = (4/3)πR³

Nuclear radius:

R = R₀A^1/3

Calculation:

V = (4/3)π(R₀A^1/3)³  = (4/3)πR₀³A

Now, ρ = M / [(4/3)πR₀³A]

⇒ A = M / [ρ × (4/3)πR₀³]

Substituting values:

A = 19.926 × 10^-27 / [2.29 × 10¹⁷ × (12.56/3) × (1.2 × 10^-15)³]

⇒ A = 19.926 × 10^-27 / [2.29 × 10¹⁷ × 4.186 × 1.728 × 10^-45]

⇒ A ≈ 19.926 × 10^-27 / 16.56 × 10^-28 ≈ 12

Final Answer: Mass number A = 12

Given:

Length of loop = 8 cm = 0.08 m

Breadth of loop = 3 cm = 0.03 m

Magnetic field, B = 0.3 T

Velocity, v = 2 cm s^-1 = 0.02 m s^-1

Formula Used:

Emf induced: e = Blv

Here, l = side perpendicular to motion

Calculation:

The loop moves in a direction normal to the shorter side.

Therefore, effective length: l = 3 cm = 0.03 m

Now, e = Blv

⇒ e = 0.3 × 0.03 × 0.02 = 0.00018 V = 1.8 × 10^-4 V

Final Answer: e = 1.8 × 10^-4 Volt.

Given:

Resistance of galvanometer, G = 100 Ω

Full scale deflection current, I_g = 1 mA = 0.001 A

Required ammeter range, I = 10 A

Formula Used:

Shunt resistance: S = (I_g × G) / (I − I_g)

Calculation:

S = (0.001 × 100) / (10 − 0.001)

⇒ S = 0.1 / 9.999 ≈ 0.01 Ω

Result: Required shunt resistance = 0.01 Ω

Given:

• In Young's double slit experiment:
• Intensity at path difference λ = K units
• We need intensity when path difference = λ/3

Formula Used:

Intensity in YDSE:

⇒ I = I_max cos²(φ/2)

Phase difference:

⇒ φ = 2π(Δx/λ)

Calculation:

For path difference Δx = λ, φ = 2π(λ/λ)

⇒ φ = 2π

and, I = I_max cos²(π)

⇒ I = I_max

Given this intensity = K

⇒ I_max = K

Now for path difference Δx = λ/3, φ = 2π(1/3)

⇒ φ = 2π/3

So, I = K cos²(π/3)

⇒ I = K × (1/2)² = K/4

Result: Required intensity = K/4

Given:

Mass of body, m = 5 kg

Two mutually perpendicular forces: F₁ = 8 N and F₂ = 6 N

Formula Used:

Resultant force: F = √(F₁² + F₂²)

Acceleration: a = F / m

Direction: tan θ = Perpendicular / Base

Calculation:

F = √(8² + 6²) = √(64 + 36) = √100 = 10 N

Now,

⇒ a = 10 / 5 = 2 m s⁻²

Direction with respect to 8 N force:

tan θ = 6 / 8 = 3 / 4

⇒ θ = tan⁻¹(3/4)

Result: Acceleration = 2 m s⁻² and Direction = tan⁻¹(3/4) with 8 N force.

Given:

• C₁ = C₂ = C₃ = C₄ = 10 μF
• C₅ = 2.5 μF
• V = 50 V

Formula Used:

For series capacitors:

1/C = 1/C₁ + 1/C₂ + 1/C₃ + 1/C₄

Charge: Q = CV

Calculation:

1/C = 1/10 + 1/10 + 1/10 + 1/10

⇒ 1/C = 4/10 ⇒ C = 10/4 = 2.5 μF

This 2.5 μF combination is in parallel with C₅ = 2.5 μF

⇒ Equivalent capacitance: C_eq = 2.5 + 2.5 = 5 μF

Now, Q = CV

⇒ Q = 5 × 50 = 250 μC

Charge in the series branch:

Q = 2.5 × 50 = 125 μC

In series connection, charge on each capacitor is same.

Charge on C₁, C₂, C₃ and C₄ = 125 μC each

For C₅:

Q = 2.5 × 50 = 125 μC

Final Answer: Equivalent capacitance = 5 μF, and the charge on each capacitor is 125 μC.

Concept Used:

• In a metre bridge, balance condition depends on equality of potentials at two points.
• At balance point: No current flows through galvanometer.
• Therefore, interchange of cell and galvanometer does not affect the balance condition.
• This is based on the principle of reversibility of Wheatstone bridge.

Explanation:

• When the cell E and galvanometer G are interchanged:
• The bridge still works normally.
• On moving the jockey on different sides of balance point:
• Current direction through galvanometer changes.

Hence galvanometer shows deflection on both sides.

At exact balance point:

• Potential difference across galvanometer becomes zero.
• No deflection is observed.

Result:

 Correct option is "Both right-sided and left-sided deflection and at balance point, no deflection".

Given:

Mass, m = 1000 kg

Height, h = 20 m

Time, t = 10 s

Acceleration due to gravity, g = 9.8 m/s²

Formula Used:

Power = Work / Time

Work done against gravity: W = mgh

Calculation:

W = 1000 × 9.8 × 20 = 196000 J

⇒ Power = 196000 / 10 = 19600 W

⇒ Power = 19.6 kW

Result: Power of crane = 19.6 kW

Given:

We need to match physical quantities with their formulas.

Concept Used:

Young's Modulus:

Y = Stress / Strain

⇒ Y = (F/A) / (ΔL/L) = FL / A(ΔL)

So, A → II

Compressibility:

Compressibility is reciprocal of bulk modulus.

Compressibility = − (1/P)(ΔV/V)

So, B → III

Bulk Modulus:

B = − P(V/ΔV)

So, C → IV

Poisson’s Ratio:

Poisson’s Ratio = Lateral strain / Longitudinal strain

⇒ Δd/ΔL × L/d

So, D → I

Final Matching: A – II, B – III, C – IV, D – I

Given:

A ray of light is incident on a concave lens parallel to the principal axis.

Concept Used:

• A concave lens is a diverging lens.
• Rays parallel to the principal axis diverge after refraction.
• The refracted rays appear to come from the principal focus on the same side of the lens.

​Explanation:

When a ray parallel to the principal axis passes through a concave lens:

• The ray bends outward.
• It does not actually pass through the focus.
• Its backward extension passes through the first principal focus.

Therefore, the refracted ray appears to diverge from the first principal focus.

So, correct option is "appears to diverge from the first principal focus".

Given:

• Length of wire = L
• Linear mass density = m
• Wire is bent into a circular ring.
• Axis yy′ is tangential to the ring and lies in its plane.

Formula Used:

Total mass of ring: M = mL

Circumference of ring: L = 2πR

⇒ R = L / 2π

Moment of inertia of a ring about diameter: I_diameter = MR² / 2

Using parallel axis theorem for tangent axis: I_yy′ = I_diameter + MR²

Calculation:

 I_yy′ = MR²/2 + MR²

⇒ I_yy′ = 3MR²/2

Substituting M = mL and R = L/2π

 I_yy′ = (3/2) × (mL) × (L/2π)²

⇒ I_yy′ = (3/2) × mL × L² / 4π²

⇒ I_yy′ = 3mL³ / 8π²

Result: Moment of inertia about yy′ = 3mL³ / 8π²

Given:

Mass of cube, m = 5.580 kg

Side of cube, a = 9.0 cm ⇒ a = 0.09 m

Formula Used:

Density, ρ = Mass / Volume

Volume of cube = a³

Calculation:

Volume = (0.09)³

⇒ Volume = 0.000729 m³

⇒ ρ = 5.580 / 0.000729 = 7654.32 kg m⁻³ = 7.654 × 10³ kg m⁻³

Significant Figures:

Mass has 4 significant figures.

Side length 9.0 cm has 2 significant figures.

Final answer must have 2 significant figures.

⇒ ρ = 7.7 × 10³ kg m⁻³

Result: X = 7.7

Given:

• Wave equation: y(x,t) = 2.0 cos 2π(10 t − 0.0080 x + 0.35)
• Distance between two points: Δx = 0.5 m

Formula Used:

For a travelling wave: Phase difference, Δϕ = 2π × (coefficient of x) × Δx

Calculation:

Given coefficient of x = 0.0080

Δϕ = 2π × 0.0080 × 0.5

⇒ Δϕ = 2π × 0.004 = 0.008π rad

Therfore, phase difference = 0.008π rad

Given:

• Equilateral prism i.e. prism angle, A = 60°
• Angle of incidence, i = 50°
• Refracted ray inside prism is parallel to base BC.

Concept Used:

In an equilateral prism:

Each angle = 60°

⇒ Face AB makes 60° with base BC.

Since refracted ray QR is parallel to BC:

⇒ Angle between refracted ray and prism face AB = 60°

Angle of refraction is measured from normal.

⇒ r₁ = 90° − 60° = 30°

Similarly at second face: r₂ = 30°

Using prism relation: r₁ + r₂ = A

⇒ 30° + 30° = 60° 

Now for prism: Deviation, δ = i + e − A

Here, symmetry gives: e = i = 50°

Calculation:

δ = 50° + 50° − 60° = 40°

Result: Angle of deviation = 40°

CONCEPT:

Voltage Across an Ideal Diode

• An ideal diode behaves as an open circuit when it is reverse biased.
• An ideal diode behaves as a short circuit when it is forward biased.
• Therefore, the voltage across the diode is maximum when it is reverse biased, while it is approximately zero when it is forward biased.

EXPLANATION:

• During the positive half-cycle of the input voltage, the diode is reverse biased.
• No current flows through the circuit.
• Hence, there is no voltage drop across the resistor.
• The entire input voltage appears across the diode.
• Thus, V_D follows the positive half-cycle of the input.
• During the negative half-cycle, the diode becomes forward biased.
• The ideal diode conducts and behaves like a short circuit.
• Therefore, the voltage across the diode becomes approximately zero.
• Hence, the voltage across the diode consists of a positive half-sinusoidal waveform during the reverse-biased interval and remains zero during the forward-biased interval.

Therefore, the waveform across the diode is a positive half-wave followed by zero voltage.

Given:

Time period of simple pendulum = T

We need to identify the correct graph between K.E. and time.

Concept Used:

For a simple pendulum:

• Kinetic Energy is always positive or zero.
• Kinetic Energy becomes maximum at mean position.
• Kinetic Energy becomes zero at extreme positions.
• In one complete oscillation, K.E. becomes maximum two times.

Therefore, time period of K.E. = T/2

Explanation:

At t = 0, pendulum is at extreme position.

⇒ Velocity = 0

⇒ K.E. = 0

As pendulum moves toward mean position:

⇒ Velocity increases

⇒ K.E. increases

At mean position:

⇒ Velocity is maximum

⇒ K.E. is maximum

Again at next extreme position:

⇒ Velocity becomes zero

⇒ K.E. becomes zero

This process repeats after every T/2.

Hence, K.E. graph always remains positive and forms repeating arches.

Result: Correct graph is

Given:

Emf of battery, E = 12 V

Internal resistance, r = 2 Ω

Current, I = 0.6 A

Formula Used:

Terminal voltage: V = E − Ir

Calculation:

V = 12 − (0.6 × 2) = 12 − 1.2

⇒ V = 10.8 V

Final Answer: Terminal voltage = 10.8 V.

Given:

Mass of body = m

Radius of Earth = R

Height raised = R

Formula Used:

Gravitational potential energy: U = -GMm / r

Calculation:

Initial distance from Earth’s center:

⇒ r₁ = R

Final distance from Earth’s center:

⇒ r₂ = R + R = 2R

Work done = Increase in potential energy

W = U₂ − U₁

⇒ W = [ -GMm / (2R) ] − [ -GMm / R ]

⇒ W = GMm / R − GMm / (2R)

⇒ W = GMm / (2R)

Now, g = GM / R²

⇒ GM = gR²

Substituting, W = (gR² × m) / (2R)

⇒ W = mgR / 2

Final Answer: Work done = mgR / 2

Given:

Heat supplied per second, Q = 100 W

Work done by system per second, W = 75 J/s

Formula Used:

First law of thermodynamics:

ΔQ = ΔU + ΔW

Calculation:

ΔU = ΔQ − ΔW = 100 − 75

⇒ ΔU = 25 W

Final Answer: Rate of increase of internal energy = 25 W.

Given:

Rated power, P₁ = 400 W

Rated voltage, V₁ = 220 V

New voltage, V₂ = 200 V

Formula Used:

For a heater, resistance remains constant:

P = V² / R

Therefore, P ∝ V²

Calculation:

P₂ / P₁ = (V₂ / V₁)²

⇒ P₂ / 400 = (200 / 220)² = (10 / 11)²

⇒ P₂ / 400 = 100 / 121

⇒ P₂ = 400 × 100 / 121 ≈ 330.6 W

⇒ P₂ ≈ 331 W

Final Answer: Power consumed ≈ 331 W.

Concept

• When an object falls freely under gravity, the distance it travels is determined by its initial velocity, the acceleration due to gravity, and the time of fall.
• For a falling ruler being caught, the distance the ruler falls corresponds to the person's reaction time.
• Since the ruler is released from rest, the distance traveled is directly proportional to the square of the reaction time.

Formula Used

The equation of motion for a body starting from rest under gravity is:

$s = ut + \frac{1}{2}gt^2$

Since the ruler falls from rest:

$u = 0$

$s = \frac{1}{2}gt^2$

Since $g$ is a constant ($9.8,\text{m/s}^2$), the distance $s$ is directly proportional to the square of the reaction time $t$:

$s \propto t^2$

Calculation

Given the reaction times for the five persons:

$t_A = 0.20,\text{s}$

$t_B = 0.22,\text{s}$

$t_C = 0.18,\text{s}$

$t_D = 0.19,\text{s}$

$t_E = 0.21,\text{s}$

Comparing these reaction times in descending order (highest to lowest):

$0.22,\text{s} > 0.21,\text{s} > 0.20,\text{s} > 0.19,\text{s} > 0.18,\text{s}$

This corresponds to the persons in the following order:

$B > E > A > D > C$

Since the distance $s$ increases as the time $t$ increases, the order of distances traveled by the ruler will be the same as the order of the reaction times.

Therefore, the correct order of distance traveled is:

$B > E > A > D > C$

Hence, the correct order is $B > E > A > D > C$.

Concept

• When a charged capacitor is connected to an uncharged capacitor, charge redistribution occurs until both capacitors reach a common potential.
• In this process, energy is lost, primarily in the form of heat in the connecting wires or through electromagnetic radiation.
• The total charge remains conserved, but the total electrostatic energy decreases.

Formula Used

Energy stored in a capacitor:

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

Common potential after connection:

$V_c = \frac{C_1V_1 + C_2V_2}{C_1 + C_2}$

The energy lost during the process is given by:

$\Delta U = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} (V_1 - V_2)^2$

Calculation

Given:

Capacitance of both capacitors, $C_1 = C_2 = C = 200 \text{ pF} = 200 \times 10^{-12} \text{ F}$

Potential of the first capacitor, $V_1 = 100 \text{ V}$

Potential of the second capacitor (uncharged), $V_2 = 0 \text{ V}$

Substituting the values into the energy loss formula:

$\Delta U = \frac{1}{2} \frac{(200 \times 10^{-12}) \times (200 \times 10^{-12})}{(200 \times 10^{-12}) + (200 \times 10^{-12})} (100 - 0)^2$

$\Rightarrow \Delta U = \frac{1}{2} \frac{(200 \times 10^{-12})^2}{400 \times 10^{-12}} \times 10000$

$\Rightarrow \Delta U = \frac{1}{2} \left( \frac{40000 \times 10^{-24}}{400 \times 10^{-12}} \right) \times 10^4$

$\Rightarrow \Delta U = \frac{1}{2} (100 \times 10^{-12}) \times 10^4$

$\Rightarrow \Delta U = 50 \times 10^{-8} \text{ J}$

$\Rightarrow \Delta U = 0.5 \times 10^{-6} \text{ J}$

Hence, the amount of electrostatic energy lost in the process is $0.5 \times 10^{-6} \text{ J}$.

Concept

• A simple pendulum consists of a mass (bob) suspended from a fixed point that swings back and forth.
• The time period ($T$) is the time taken to complete one full oscillation.
• The time period depends on the effective length of the pendulum ($L$) and the acceleration due to gravity ($g$).

Formula Used

Time period of a simple pendulum:

$T = 2\pi \sqrt{\frac{L}{g}}$

Relation between total time ($t$), number of oscillations ($n$), and time period ($T$):

$T = \frac{t}{n}$

Calculation

Given:

Total time taken, $t = 60,s$

Number of oscillations, $n = 30$

Acceleration due to gravity, $g = 9.8,m/s^2$

Approximation, $\pi^2 = 9.8$

First, calculate the time period $T$:

$T = \frac{60}{30} = 2,s$

Now, substitute the values into the time period formula:

$2 = 2\pi \sqrt{\frac{L}{g}}$

Divide both sides by 2:

$1 = \pi \sqrt{\frac{L}{g}}$

Squaring both sides:

$1^2 = \pi^2 \left( \frac{L}{g} \right)$

$1 = \pi^2 \frac{L}{g}$

Rearranging for $L$:

$L = \frac{g}{\pi^2}$

Substituting the given values $g = 9.8$ and $\pi^2 = 9.8$:

$L = \frac{9.8}{9.8} = 1,m$

Hence, the effective length of the simple pendulum is 1 m.

Concept

• Alternating current (AC) varies sinusoidally with time, following the equation $I = I_0 \sin(\omega t)$.
• One complete cycle corresponds to a time period $T$.
• Starting from zero, the current reaches its first peak value at one-fourth of the time period, i.e., $t = \frac{T}{4}$.

Formula Used

Relation between time period and frequency:

$T = \frac{1}{f}$

Time taken to reach peak value from zero:

$t = \frac{T}{4} = \frac{1}{4f}$

Calculation

Given:

Frequency, $f = 60 \text{ Hz}$

Peak current, $I_0 = 5 \text{ A}$

We need to find the time $t$ to reach the peak value from zero. This occurs when the phase angle is $90^\circ$ or $\frac{\pi}{2}$ radians.

Using the formula:

$t = \frac{1}{4f}$

Substituting the value of $f$:

$t = \frac{1}{4 \times 60} \text{ s}$

$t = \frac{1}{240} \text{ s}$

Hence, the current will take $\frac{1}{240} \text{ s}$ to reach the peak value.

Concept

• Conservation of Energy: In wave optics, the principle of superposition leads to interference and diffraction. Energy is not lost; it is simply redistributed from regions of minimum intensity to regions of maximum intensity.
• Wave Nature: Interference and diffraction are fundamental characteristics of wave motion. Any entity that behaves as a wave will exhibit these phenomena.

Explanation

Statement A:

In the phenomena of interference and diffraction, the intensity of light at any point is given by the principle of superposition. At points of constructive interference, the intensity is greater than the sum of the individual intensities, and at points of destructive interference, it is less. However, the total energy across the entire pattern remains constant and equal to the sum of the energies of the individual waves. No energy is created or destroyed; it is merely redistributed. Therefore, these phenomena are perfectly consistent with the principle of conservation of energy. Hence, Statement A is true.

Statement B:

Interference and diffraction are not exclusive to light waves. They are general properties exhibited by all types of waves, including mechanical waves (such as sound waves and water waves), electromagnetic waves (such as radio waves and X-rays), and even matter waves (such as electrons). For example, sound waves undergo diffraction when they pass through a doorway. Hence, Statement B is false.

Conclusion

Based on the analysis, Statement A is correct while Statement B is incorrect.

Hence, A is true, but B is false.

Concept

• For a box to remain stationary relative to an accelerating trolley, the pseudo-force acting on the box must be balanced by the force of static friction.
• The maximum acceleration occurs when the pseudo-force equals the maximum possible static friction (limiting friction).
• Static friction depends on the coefficient of friction and the normal reaction force.

Formula Used

Maximum static friction:

$f_{s,max} = \mu_s N$

Normal force on a horizontal surface:

$N = mg$

Newton's second law (Pseudo-force):

$F_p = ma$

Condition for no slipping:

$ma \leq \mu_s mg$

Calculation

Given:

• Mass of the box, $m = 15,kg$
• Coefficient of static friction, $\mu_s = 0.12$
• Acceleration due to gravity, $g = 10,m/s^2$

To keep the box stationary over the trolley, the horizontal acceleration $a$ must satisfy:

$ma \leq f_{s,max}$

$ma \leq \mu_s mg$

The mass $m$ cancels out from both sides:

$a_{max} = \mu_s g$

Substituting the given values:

$a_{max} = 0.12 \times 10$

$a_{max} = 1.2,m/s^2$

Hence, the maximum acceleration with which the trolley can be moved horizontally is $1.2,m/s^2$.

Concept

• The total mechanical energy of a simple pendulum, which is the sum of its kinetic energy (KE) and potential energy (PE), remains constant during its motion (neglecting air resistance).
• At the equilibrium position (the lowest point of the swing), the potential energy is at its minimum. If we take this point as the reference level, the potential energy is zero.
• Consequently, at the equilibrium position, all the total energy of the pendulum is converted into kinetic energy.

Formula Used

Total mechanical energy ($E$):

$E = KE + PE$

Kinetic energy at the equilibrium position:

$KE_{max} = \frac{1}{2}mv^2$

Where:

• $m$ is the mass of the bob.
• $v$ is the speed of the bob at the equilibrium position.

Calculation

Given:

Total energy, $E = 0.02,J$

Mass of the bob, $m = 20,g = 0.02,kg$

By the law of conservation of energy, the total energy at the equilibrium position is purely kinetic:

$E = \frac{1}{2}mv^2$

Substituting the given values into the formula:

$0.02 = \frac{1}{2} \times 0.02 \times v^2$

Multiplying both sides by 2:

$0.04 = 0.02 \times v^2$

Solving for $v^2$:

$v^2 = \frac{0.04}{0.02}$

$v^2 = 2$

Taking the square root:

$v = \sqrt{2} \approx 1.414,m/s$

Rounding to two decimal places, we get approximately $1.41,m/s$.

Hence, the speed of the simple pendulum bob at the equilibrium position is approximately 1.41 m/s.

Concept

• The radius of a nucleus $R$ is related to its mass number $A$.
• The volume of a nucleus, assuming it is spherical, depends on the cube of its radius.
• The mass defect is a measure of the binding energy of the nucleus, representing the difference between the total mass of individual nucleons and the actual mass of the nucleus.

Formula Used

Nuclear radius:

$R = R_0 A^{1/3}$

Nuclear volume:

$V = \frac{4}{3} \pi R^3$

Mass defect ($\Delta m$):

$\Delta m = [Z m_p + (A - Z) m_n] - M_{nucleus}$

Explanation

For Statements A and B:

The radius of a nucleus is given by $R = R_0 A^{1/3}$. The volume $V$ of the nucleus is:

$V = \frac{4}{3} \pi (R_0 A^{1/3})^3$

$V = \frac{4}{3} \pi R_0^3 A$

This shows that $V \propto A$. Therefore, Statement A is false and Statement B is true.

For Statements C and D:

The mass defect is defined as the difference between the sum of the masses of the constituents (protons and neutrons) and the actual mass of the nucleus. The difference between the mass of an atom and its nucleus is approximately the mass of the electrons, which is not the mass defect.

Therefore, Statement C is false and Statement D is true.

Conclusion:

Statements B and D are true, while A and C are false.

Hence, B and D are true, but A and C are false.

Concept

• The motion of a flywheel is an example of rotational motion with constant angular acceleration.
• Angular displacement can be calculated using the average angular velocity when acceleration is uniform.
• The number of revolutions is the total angular displacement divided by $2\pi$ radians.

Formula Used

Angular speed in rad/s:

$\omega = \frac{2\pi N}{60}$ (where $N$ is in rpm)

Average angular velocity:

$\omega_{avg} = \frac{\omega_1 + \omega_2}{2}$

Angular displacement:

$\theta = \omega_{avg} \times t$

Number of revolutions ($n$):

$n = \frac{\theta}{2\pi}$

Calculation

Given:

Initial speed $N_1 = 600,rpm$

Final speed $N_2 = 1200,rpm$

Time $t = 10,s$

First, calculate initial and final angular velocities in rad/s:

$\omega_1 = \frac{2\pi \times 600}{60} = 20\pi,rad/s$

$\omega_2 = \frac{2\pi \times 1200}{60} = 40\pi,rad/s$

Now, find the average angular velocity:

$\omega_{avg} = \frac{20\pi + 40\pi}{2} = \frac{60\pi}{2} = 30\pi,rad/s$

Calculate total angular displacement ($\theta$):

$\theta = \omega_{avg} \times t$

$\theta = 30\pi \times 10 = 300\pi,rad$

Finally, find the number of revolutions ($n$):

$n = \frac{\theta}{2\pi}$

$n = \frac{300\pi}{2\pi} = 150$

Hence, the number of revolutions completed by the flywheel is 150.