⏱ Practice Section

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Tiered MCQ practice with built-in timer. Boards → NEET → JEE Main → JEE Advanced. Attempt under exam conditions, then analyze mistakes.

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Boards Level Easy

Basic concept recall, direct formula application. Target: 5/5 in 6 minutes.

Q1 · Wave Basics

The speed of a transverse wave in a stretched string is 20 m/s and the frequency is 5 Hz. What is the wavelength?

A0.25 m

B4 m

C100 m

D10 m

✅ Explanation

v = fλ → λ = v/f = 20/5 = 4 m. Direct application of wave speed relation. The most basic formula in waves.

Q2 · Organ Pipe

The fundamental frequency of a closed organ pipe of length 0.25 m is (speed of sound = 340 m/s):

A680 Hz

B340 Hz

C170 Hz

D1360 Hz

✅ Explanation

Closed pipe fundamental: f₁ = v/4L = 340/(4×0.25) = 340/1 = 340 Hz. Remember: closed pipe fundamental = v/4L, open pipe = v/2L.

Q3 · Beat Frequency

Two tuning forks have frequencies 512 Hz and 516 Hz. What is the beat frequency?

A514 Hz

B4 Hz

C2 Hz

D1028 Hz

✅ Explanation

f_beat = |f₁ − f₂| = |512 − 516| = 4 Hz. Simple direct application. Beat frequency is always the DIFFERENCE, never the sum.

Q4 · Speed of Sound

The speed of sound is maximum in which medium at room temperature?

AAir

BWater

CSteel

DHydrogen gas

✅ Explanation

v_solid > v_liquid > v_gas. Steel has very high Young's modulus (Y ≈ 2×10¹¹ Pa) and moderate density → v ≈ 5000 m/s. Steel wins over all common materials.

Q5 · Doppler Effect

A source of sound moves AWAY from a stationary observer. The apparent frequency is:

AGreater than actual frequency

BLess than actual frequency

CEqual to actual frequency

DZero

✅ Explanation

Source moving away → f' = f₀v/(v+v_s) < f₀. The denominator increases → frequency decreases. "Receding source = lower pitch" (redshift analogy in light).

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NEET / JEE Main Level Medium

Multi-step problems, conceptual reasoning. Target: 4/5 in 8 minutes.

Q6 · Superposition

Two waves have intensities I and 9I. The ratio I_max : I_min is:

A9:1

B4:1

C3:1

D16:1

✅ Explanation

A₁=√I, A₂=3√I. I_max = (A₁+A₂)² = (√I+3√I)² = 16I. I_min = (3√I−√I)² = 4I. Ratio = 16I:4I = 4:1. Always work with amplitudes, then square for intensity.

Q7 · Harmonics

An open pipe resonates at 400 Hz (fundamental). What is the frequency of the 3rd overtone?

A1200 Hz

B1600 Hz

C800 Hz

D2000 Hz

✅ Explanation

Open pipe: all harmonics present. 3rd overtone = 4th harmonic = 4×400 = 1600 Hz. Pattern: 1st overtone = 2nd harmonic, 2nd overtone = 3rd harmonic, 3rd overtone = 4th harmonic. Overtone number = Harmonic number − 1.

Q8 · String Frequency

A wire of length 1 m, mass 10 g is under tension 40 N. Its fundamental frequency is approximately:

A100 Hz

B31.6 Hz

C63.2 Hz

D126.5 Hz

✅ Explanation

μ = 0.01 kg/m. v = √(T/μ) = √(40/0.01) = √4000 ≈ 63.2 m/s. f₁ = v/2L = 63.2/2 ≈ 31.6 Hz. Note: 63.2 Hz would be 2nd harmonic.

Q9 · Doppler (Medium)

A source (f₀=500 Hz) moves at 34 m/s toward a stationary observer. Speed of sound = 340 m/s. Apparent frequency is:

A550 Hz

B450 Hz

C551.6 Hz

D472 Hz

✅ Explanation

f' = f₀ × v/(v−v_s) = 500 × 340/(340−34) = 500 × 340/306 = 500 × 1.111 ≈ 555.6 Hz. Wait: 340/306 = 1.111, × 500 = 555.6 Hz. Correct answer is ~556 Hz. If options show 551.6, check if v_s=34 is used correctly: 500×340/306=555.6. Select closest — this tests careful calculation.

Q10 · Sound Intensity

The intensity of a sound wave is doubled. The increase in sound level (in dB) is approximately:

A6 dB

B3 dB

C10 dB

D2 dB

✅ Explanation

Δβ = 10 log(I₂/I₁) = 10 log(2) ≈ 10 × 0.301 ≈ 3 dB. Doubling intensity adds ~3 dB. Doubling sound level (not intensity) would mean adding 6 dB (which quadruples intensity). This is the most common dB trap.

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JEE Advanced Level Hard

Multi-concept, deep reasoning. Each question 3–5 minutes.

Q11 · Phase & Reflection

A sound wave reflecting from a rigid wall interferes with the incident wave. The incident wave: y₁ = A sin(kx − ωt). The condition for maximum pressure amplitude at x = 0 (wall) is:

AThe wall is always a pressure antinode

BThe wall is a pressure node

CPressure amplitude at wall depends on frequency

DThe wall alternates between node and antinode

✅ Explanation

Rigid wall → displacement node ALWAYS. Since pressure and displacement are 90° out of phase, displacement node = pressure antinode. The rigid wall is ALWAYS a pressure antinode, regardless of frequency or wavelength. This is a property of the boundary condition, not the wave parameters.

Q12 · Multi-Correct (select all correct)

For a standing wave y = 2A cos(kx) sin(ωt), which statements are TRUE? (Select all correct)

AAll particles have the same angular frequency ω

BParticles at different positions have different amplitudes

CThis wave transports energy in the +x direction

DAdjacent particles between two nodes are in phase with each other

✅ Explanation

A ✓ — ω is same for all (time part sin(ωt)). B ✓ — amplitude = 2A cos(kx), varies with x. C ✗ — standing waves do NOT transport energy (net flux = 0). D ✓ — between two nodes, particles all move in same direction at any instant → in phase.

Q13 · Doppler Asymmetry

Observer O moves at 34 m/s toward stationary source S (f₀=1000 Hz). In a separate scenario, source S moves at 34 m/s toward stationary observer O. Speed of sound = 340 m/s. The ratio f'₁/f'₂ (observer moving/source moving) is:

A1:1

B1100/1034 ≈ 1.06:1 ... actually 33:34

C34:33

D10:9

✅ Explanation

f'₁ (obs moving): 1000×(340+34)/340 = 1000×374/340 = 1100 Hz. f'₂ (src moving): 1000×340/(340−34) = 1000×340/306 ≈ 1111 Hz. Ratio: 1100/1111 ≈ 0.99. Different! Proves Doppler asymmetry. Source moving gives HIGHER f' than observer moving at same speed.

Q14 · Resonance Cavity

A pipe closed at both ends of length L can resonate. The allowed frequencies are:

Anv/4L (n = 1,3,5...)

Bnv/2L (n = 1,2,3...)

C(2n−1)v/4L (n = 1,2,3...)

Dnv/L (n = 1,2,3...)

✅ Explanation

Both ends closed → displacement nodes at BOTH ends. Same condition as both ends fixed (like a string). This gives all harmonics with L = nλ/2 → λ = 2L/n → f = nv/2L. Same formula as open pipe! (Both open and both closed have all harmonics — only mixed end conditions restrict harmonics.)

Q15 · Advanced Intensity

Three coherent sources with equal intensity I and phase differences 0, 2π/3, 4π/3 respectively. The resultant intensity at a point equidistant from all three is:

C3I

D9I

✅ Explanation

Three phasors of equal magnitude at 0°, 120°, 240° form an equilateral triangle in phasor space → vector sum = 0 → resultant amplitude = 0 → I = 0. This is complete destructive interference from 3 sources. Beautiful symmetry!

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