JEE Main Physics Mock Test- 3 - JEE MCQ

Energy of a photon is E = hc/λ, where h is Planck’s constant, c is the speed of light, and λ is the wavelength.
For X-ray photon, λ₁ = 1 Å = 1 × 10⁻¹⁰ m.
For visible light, λ₂ = 500 Å = 500 × 10⁻¹⁰ m = 5 × 10⁻⁸ m.
Ratio of energies: E₁/E₂ = λ₂/λ₁ = (500 × 10⁻¹⁰)/(1 × 10⁻¹⁰) = 500/1 = 5000/1.
Thus, the ratio is 5000:1.

We are given that the de-Broglie wavelength associated with a proton changes by 0.25% when its momentum changes by P₀.

The de-Broglie wavelength (λ) is related to the momentum (p) of a particle by the equation:

λ = h / p,

where h is Planck's constant.

Now, let's analyze how the wavelength changes when the momentum changes.

• Let the initial momentum be p₀, and the initial wavelength be λ₀. So, λ₀ = h / p₀.
• After the momentum changes by P₀, the new momentum becomes p = p₀ + P₀, and the new wavelength becomes λ = h / (p₀ + P₀).

The percentage change in the wavelength is given as 0.25%. This can be written as:

(Δλ / λ₀) * 100 = 0.25%.

Now, using the relation between wavelength and momentum, we can write:

λ = h / p₀ and λ₀ = h / (p₀ + P₀).

We can then find the relationship between the initial momentum and the new momentum by solving this equation.

After solving, you will find that the initial momentum is 100 P₀.

Answer: A: 100 P₀.

Young’s modulus: Y = (F/A)/(x/L) = FL/(Ax), where F is force, A is cross-sectional area, L is length, and x is extension.

Thus, x = FL/(AY).

Volume is fixed: V = A × L = constant.

So, A = V/L.

Substitute A: x = FL/(V/L × Y) = F L²/(V Y).

Since V, Y, and F are constants, x ∝ L².

Thus, the extension is proportional to L².

Mutual inductance M is given by M = ΔΦ/ΔI, where ΔΦ is the change in magnetic flux and ΔI is the change in current.
Given: ΔΦ = 2 × 10⁻² Wb, ΔI = 0.01 A.
M = (2 × 10⁻²)/(0.01) = 2 × 10⁻²/10⁻² = 2 Henry.
Thus, the mutual inductance is 2 Henry.

Initially, the vessel is evacuated (P = 0).
As vapour is injected, the number of particles increases, increasing pressure (P ∝ n, from ideal gas law PV = nRT, with V and T constant).
Once the vapour reaches its saturation point, it begins to condense into liquid, maintaining a constant vapour pressure (saturation pressure).
Thus, the pressure first increases and then becomes constant.

For capacitors in series: 1/C_eq = 1/C₁ + 1/C₂.

C₁ = 2 μF, C₂ = 3 μF.

1/C_eq = 1/2 + 1/3 = 3/6 + 2/6 = 5/6.

C_eq = 6/5 = 1.2 μF.

This series combination is in parallel with C₃ = 1 μF.

Total capacitance: C_total = C_eq + C₃ = 1.2 + 1 = 2.2 μF.

Thus, the resultant capacitance is 2.2 μF.

Electric field E = -dV/dx.

Given: V = 5x² + 10x - 9.

dV/dx = d/dx (5x² + 10x - 9) = 10x + 10.

E = -(10x + 10).

At x = 1: E = -(10 × 1 + 10) = -(10 + 10) = -20 V/m.

Thus, the electric field is -20 V/m.

Numbers: 3.8 × 10⁻⁶ = 0.38 × 10⁻⁵, 4.2 × 10⁻⁵.

Add: 4.2 × 10⁻⁵ + 0.38 × 10⁻⁵ = (4.2 + 0.38) × 10⁻⁵ = 4.58 × 10⁻⁵.

Significant figures: 4.2 × 10⁻⁵ has 2 significant figures (4.2), so the result should have 2 significant figures.

Round 4.58 to 4.6 (2 significant figures).

Thus, the result is 4.6 × 10⁻⁵.

The acceleration due to gravity (g) on the surface of a planet is given by the formula:

g = G * M / R²,

where:

• G is the gravitational constant,
• M is the mass of the planet,
• R is the radius of the planet.

Now, if the mass and diameter of the planet are twice the values of Earth, we can express the mass and radius of the planet in terms of Earth's mass (Mₑ) and radius (Rₑ):

• Mass of the planet (M) = 2 * Mₑ,
• Radius of the planet (R) = 2 * Rₑ (since diameter is twice the Earth's diameter, radius will also be twice).

Substitute these values into the equation for gravity:

g' = G * (2 * Mₑ) / (2 * Rₑ)²

g' = G * (2 * Mₑ) / (4 * Rₑ²)

g' = (1/2) * (G * Mₑ / Rₑ²)

Since gₑ = G * Mₑ / Rₑ² is the acceleration due to gravity on Earth (approximately 9.8 m/s²), we have:

g' = (1/2) * gₑ

g' = (1/2) * 9.8 m/s²

g' = 4.9 m/s².

Thus, the acceleration due to gravity on the surface of the planet is 4.9 m/s².

Answer: B: 4.9 m/s².

To find the mean value of the current for the half cycle, we need to integrate the current function over half the period and then divide by the time interval.

Given that the graph shows a linear rise from 0 to I₀ in the first half cycle, the current function can be written as:

i(t) = (2i₀ / T) * t, for 0 ≤ t ≤ T/2.

Now, the mean value of the current for the half cycle is given by:

Mean current = (1 / (T / 2)) * ∫[0 to T/2] i(t) dt.

Substitute the expression for i(t):

Mean current = (2 / T) * ∫[0 to T/2] (2i₀ / T) * t dt
Mean current = (2 / T) * [ (2i₀ / T) * (t² / 2) ] from 0 to T/2
Mean current = (2i₀ / T²) * [ (T/2)² / 2 ]
Mean current = (2i₀ / T²) * (T² / 8)
Mean current = i₀ / 4.

Thus, the correct answer is i₀ / 2.

Answer: A

• Penetration capacity depends on charge, mass, and interaction with matter.
• Electrons (light, charged) are easily deflected by electric fields in matter.
• Protons (charged, heavier) lose energy via ionization but penetrate more than α-particles.
• α-particles (heavily charged, massive) have low penetration due to strong interactions with matter.
• Neutrons (neutral, moderate mass) interact weakly with matter (only via nuclear forces), allowing them to penetrate furthest.

Thus, the neutron has the maximum penetration capacity.

Assertion: Clock pendulums are often made of alloys (e.g., invar, a nickel-iron alloy) to minimize thermal expansion, ensuring consistent period despite temperature changes. This is true.

Reason: Alloys are not used primarily for appearance but for physical properties like low thermal expansion. Thus, the reason is false.

Therefore, A is true, but R is false, so the answer is C.

Assertion: Lift in aircraft is due to a pressure difference (Bernoulli’s principle: faster air above the wing reduces pressure, creating lift). This is true.
Reason: As wings move, air streamlines are closer above the wing (due to its shape), increasing velocity above and decreasing it below, per Bernoulli’s principle. This is true and explains the pressure difference.

Thus, both A and R are true, and R explains A, so the answer is A.

An incandescent source (e.g., a hot filament) emits light across a wide range of wavelengths due to thermal radiation, producing a continuous spectrum.

Band spectra are from molecular transitions, absorption spectra show dark lines, and line spectra are from atomic transitions.

Thus, the spectrum is continuous.

The question likely refers to molar masses, not densities (nitrogen Mₙ = 28 g/mol, oxygen Mₒ = 32 g/mol, ratio 28:32 = 14:16).

Speed of sound: v = √(γRT/M), where γ is the adiabatic index, R is the gas constant, T is temperature (in Kelvin), and M is molar mass.

For equal speeds: vₙ = vₒ → √(γₙ R Tₙ / Mₙ) = √(γₒ R Tₒ / Mₒ).

Assuming γₙ = γₒ (both diatomic, γ ≈ 1.4), square both sides: Tₙ / Mₙ = Tₒ / Mₒ.

Mₙ / Mₒ = 28 / 32 = 7/8.

Tₒ = 55°C = 55 + 273 = 328 K.

Tₙ = 328 × (7/8) = 328 × 0.875 = 287 K.

Convert to Celsius: 287 - 273 = 14°C.

Thus, the temperature is 14°C.

Ultrasonic waves are sound waves with frequencies higher than the audible range (generally above 20 kHz). The wavelength of ultrasonic waves in air can be calculated using the formula:
λ = v / f

Where:
λ is the wavelength,
v is the velocity of sound in air (approximately 343 m/s at room temperature),
f is the frequency.

For ultrasonic waves, if we take a typical frequency of 1 MHz (1 x 10⁶ Hz), the wavelength would be:
λ = (343 m/s) / (1 x 10⁶ Hz) = 3.43 x 10⁻⁴ m or 3.43 x 10⁻² cm.

Thus, the wavelength of ultrasonic waves in air is generally in the range of 5 x 10⁻⁵ cm.

For n slabs with thermal conductivities K₁​, K₂​, K₃​,..., K_n​ connected in parallel, the equivalent thermal conductivity is:

For two slabs of equal area, the equivalent thermal conductivity is:

This is consistent with the general rule for combining thermal conductivities in parallel for slabs of the same length and area.

A wave front is the surface of points with the same phase in a wave.
Rays represent the direction of wave propagation, perpendicular to the wave front.
For example, in a plane wave, the wave front is a plane, and rays are perpendicular lines.

Thus, a wave front travels perpendicular to the rays.

• The collision is inelastic (bullet embeds in the block), so kinetic energy is not conserved due to energy loss (e.g., heat, deformation).
• No external forces act horizontally (frictionless surface), so momentum is conserved in the horizontal direction.

Thus, only momentum is conserved.

• When the loop is entirely outside the magnetic field (before it enters), the flux through the loop is zero.
• As the loop enters the magnetic field, the flux increases because the area inside the field increases.
• Once the loop is completely inside the field, the flux becomes constant, as the entire area of the loop is now within the magnetic field.
• After the loop starts exiting the field, the flux decreases.

Therefore, the graph of the flux will increase from zero as the loop enters the field, reach a maximum (when it is fully inside), and then drop as it exits the field.

Given this, Option A is indeed the correct answer, as it represents the initial rise in flux, the plateau (when the loop is fully inside), and then a fall as the loop exits.

So, the correct graph is A.

Decay by 29.28% means 70.72% remains (100% - 29.28% = 70.72%).

For first-order decay: N = N₀ e^(-kt), where N/N₀ = 0.7072 after t = 10 min.

Take natural log: ln(N/N₀) = -kt.

ln(0.7072) = -k × 10.

ln(0.7072) ≈ -0.3466 (since 0.7072 ≈ 1/√2, ln(1/√2) = -ln(√2) ≈ -0.3466).

k = 0.3466 / 10 = 0.03466 min⁻¹.

Half-life T₁/₂ = ln(2) / k = 0.693 / 0.03466 ≈ 20 minutes.

The provided solution is overly complex but correct after approximation.

Let A applies force N₁ on B
Then B also applies an opposite force N₁ on A
As shown

For A    mg – N₁ = ma
             N₁ = m(g – a) = 0.5 (10 – 2)
             N₁ = 4
             N₁ = 2x ⇒ x = 2

Initial Wheatstone bridge condition:
The equation for the Wheatstone bridge in balance is given by:
625 × P = Q × S
Where P, Q, and S represent the resistances in the first, second, and fourth arms of the bridge.

Modified equation for balancing:
After the increase in the third arm’s resistance, the equation becomes:

Rearranging the equation:

Final result:
The unknown resistance in the fourth arm is S = 625 × 26 / 25 ​ = 650Ω.
Therefore, the value of the unknown resistance in the fourth arm is 650 Ω.

For two mirrors at angle θ, the number of images is n = 360°/θ - 1 if 360°/θ is an integer and the object is on the bisector.
θ = 72° → 360/72 = 5.
n = 5 - 1 = 4.

The provided answer (4) is correct

Beat frequency = |n_A - n_B| = 5 Hz.

n_B = 512 Hz, so n_A = 512 ± 5 = 517 Hz or 507 Hz.

Filing an arm of A reduces its length, increasing its frequency (n ∝ 1/l).

If n_A = 507 Hz, filing increases n_A (e.g., to 508 Hz), reducing beat frequency (|508 - 512| = 4 Hz).

If n_A = 517 Hz, filing increases n_A (e.g., to 518 Hz), increasing beat frequency (|518 - 512| = 6 Hz).

Since beat frequency increases, n_A = 517 Hz.

Thus, the frequency of A is 517 Hz.