501. Which mathematical form represents the one-dimensional wave equation?
ⓐ. $\frac{\partial y}{\partial t} = v \frac{\partial y}{\partial x}$
ⓑ. $\frac{\partial^2 y}{\partial t^2} = v^2 \frac{\partial^2 y}{\partial x^2}$
ⓒ. $\frac{\partial^2 y}{\partial x^2} = v \frac{\partial y}{\partial t}$
ⓓ. $\frac{\partial y}{\partial x} = \frac{\partial y}{\partial t}$
Correct Answer: $\frac{\partial^2 y}{\partial t^2} = v^2 \frac{\partial^2 y}{\partial x^2}$
Explanation: The standard wave equation relates acceleration of the medium ($\partial^2 y / \partial t^2$) with curvature of the displacement curve ($\partial^2 y / \partial x^2$), scaled by wave speed squared.
502. What is the general solution to the wave equation in one dimension?
ⓐ. $y(x,t) = f(x+vt) + g(x-vt)$
ⓑ. $y(x,t) = f(x-vt)$ only
ⓒ. $y(x,t) = g(x+vt)$ only
ⓓ. $y(x,t) = A e^{kx}$
Correct Answer: $y(x,t) = f(x+vt) + g(x-vt)$
Explanation: The solution is a superposition of two arbitrary functions, one describing a wave traveling to the right and the other describing a wave traveling to the left with velocity $v$.
503. A string is fixed at both ends. Which boundary condition must be satisfied?
ⓐ. Displacement at both ends is zero for all time
ⓑ. Slope at both ends is always zero
ⓒ. Velocity at both ends is infinite
ⓓ. Pressure at both ends is constant
Correct Answer: Displacement at both ends is zero for all time
Explanation: Since the ends are clamped, the medium cannot move at these points. Therefore, boundary condition is $y(0,t) = y(L,t) = 0$ for all $t$.
504. For a string of length $L$ fixed at both ends, what form of solution satisfies the boundary conditions?
ⓐ. $y(x,t) = A \cos(kx – \omega t)$
ⓑ. $y(x,t) = A \sin\left(\frac{n\pi x}{L}\right)\cos(\omega t)$
ⓒ. $y(x,t) = A e^{kx}$
ⓓ. $y(x,t) = A \tan(kx)$
Correct Answer: $y(x,t) = A \sin\left(\frac{n\pi x}{L}\right)\cos(\omega t)$
Explanation: Only sine functions satisfy $y(0,t)=y(L,t)=0$. This leads to discrete allowed modes corresponding to $n=1,2,3…$, representing standing waves.
505. If the string length is $1.2 \, \text{m}$, wave speed $60 \, \text{m/s}$, what is fundamental frequency?
ⓐ. 20 Hz
ⓑ. 25 Hz
ⓒ. 30 Hz
ⓓ. 40 Hz
Correct Answer: 25 Hz
Explanation: Formula: $f = \frac{v}{2L}$.
Here, $v=60, L=1.2$.
$f = \frac{60}{2 \times 1.2} = \frac{60}{2.4} = 25 \, \text{Hz}$.
506. For the same string, what is frequency of second harmonic?
ⓐ. 25 Hz
ⓑ. 40 Hz
ⓒ. 50 Hz
ⓓ. 60 Hz
Correct Answer: 50 Hz
Explanation: Harmonics are multiples of the fundamental.
$f_2 = 2f_1 = 2 \times 25 = 50 \, \text{Hz}$.
507. A pipe closed at one end has length $0.85 \, \text{m}$. If speed of sound = $340 \, \text{m/s}$, find fundamental frequency.
ⓐ. 300 Hz
ⓑ. 250 Hz
ⓒ. 200 Hz
ⓓ. 100 Hz
Correct Answer: 100 Hz
Explanation: For closed pipe: $f = \frac{v}{4L}$.
Here, $L=0.85, v=340$.
$f = \frac{340}{4 \times 0.85} = \frac{340}{3.4} = 100 \, \text{Hz}$.
508. For an open organ pipe of length $0.5 \, \text{m}$, with sound speed $340 \, \text{m/s}$, what is the fundamental frequency?
ⓐ. 170 Hz
ⓑ. 200 Hz
ⓒ. 300 Hz
ⓓ. 340 Hz
Correct Answer: 340 Hz
Explanation: For open pipe: $f = \frac{v}{2L}$.
Here, $v=340, L=0.5$.
$f = \frac{340}{1.0} = 340 \, \text{Hz}$.
509. Which condition is true for pressure variations at the closed end of an organ pipe?
ⓐ. Always node of pressure
ⓑ. Always antinode of pressure
ⓒ. Always node of displacement and antinode of pressure
ⓓ. Always node of pressure and displacement both
Correct Answer: Always node of displacement and antinode of pressure
Explanation: At the closed end, air cannot move, so displacement = 0 (node). Pressure variation is maximum there, so pressure has an antinode at the closed end.
510. A string fixed at both ends has its third harmonic frequency at 150 Hz. What is fundamental frequency?
ⓐ. 25 Hz
ⓑ. 30 Hz
ⓒ. 45 Hz
ⓓ. 50 Hz
Correct Answer: 50 Hz
Explanation: For harmonics: $f_n = n f_1$.
Here, $f_3 = 150$.
So, $f_1 = f_3 / 3 = 150/3 = 50 \, \text{Hz}$.
511. What is the main idea of Fourier analysis in wave physics?
ⓐ. Any periodic wave can be expressed as a sum of sinusoidal components
ⓑ. All waves are only sinusoidal in nature
ⓒ. Wave speed is always constant
ⓓ. Waves cannot be decomposed
Correct Answer: Any periodic wave can be expressed as a sum of sinusoidal components
Explanation: Fourier analysis states that any complex periodic function can be represented as a series of simple sine and cosine waves with different amplitudes and frequencies. This principle is widely applied in acoustics, optics, and signal processing.
512. A square wave of frequency $f$ is decomposed into Fourier components. Which harmonics are present?
ⓐ. Only even harmonics
ⓑ. Only odd harmonics
ⓒ. Both odd and even harmonics equally
ⓓ. No harmonics
Correct Answer: Only odd harmonics
Explanation: A square wave can be expressed as a sum of odd harmonics (1st, 3rd, 5th, …). Even harmonics are absent due to symmetry of the waveform.
513. What is a wave packet?
ⓐ. A single monochromatic wave
ⓑ. A group of waves superimposed to form a localized pulse
ⓒ. A standing wave in a string
ⓓ. A circular wavefront
Correct Answer: A group of waves superimposed to form a localized pulse
Explanation: A wave packet is formed when multiple waves of slightly different frequencies interfere. This results in a localized pulse in space, useful in describing particles in quantum mechanics and signal transmission.
514. The group velocity of a wave packet is defined as:
ⓐ. $v_g = \frac{d\omega}{dk}$
ⓑ. $v_g = \frac{\omega}{k}$
ⓒ. $v_g = \frac{dk}{d\omega}$
ⓓ. $v_g = \omega k$
Correct Answer: $v_g = \frac{d\omega}{dk}$
Explanation: Group velocity is the velocity at which the envelope of a wave packet moves, given by the derivative of angular frequency with respect to wave number. It often corresponds to the velocity of energy or information transfer.
515. The phase velocity of a wave is defined as:
ⓐ. $v_p = \frac{d\omega}{dk}$
ⓑ. $v_p = \frac{\omega}{k}$
ⓒ. $v_p = \frac{dk}{d\omega}$
ⓓ. $v_p = \omega k$
Correct Answer: $v_p = \frac{\omega}{k}$
Explanation: Phase velocity is the speed at which individual wave crests move. It is the ratio of angular frequency $\omega$ to wave number $k$.
516. A wave packet has central frequency $f = 500 \, \text{Hz}$ and bandwidth $\Delta f = 50 \, \text{Hz}$. Estimate temporal width $\Delta t$ of the packet.
ⓐ. 0.01 s
ⓑ. 0.02 s
ⓒ. 0.05 s
ⓓ. 0.1 s
Correct Answer: 0.02 s
Explanation: Uncertainty relation: $\Delta f \cdot \Delta t \approx 1$.
$\Delta t = \frac{1}{\Delta f} = \frac{1}{50} = 0.02 \, \text{s}$.
517. A dispersive medium has $\omega = 2 + k^2$. Find group velocity.
ⓐ. $2k$
ⓑ. $k$
ⓒ. $1/k$
ⓓ. 2
Correct Answer: $2k$
Explanation: Formula: $v_g = \frac{d\omega}{dk}$.
Here, $\omega = 2 + k^2$.
$\frac{d\omega}{dk} = 2k$.
518. For the same medium ($\omega = 2 + k^2$), find phase velocity.
ⓐ. $k$
ⓑ. $2k$
ⓒ. $k/2$
ⓓ. $(2 + k^2)/k$
Correct Answer: $(2 + k^2)/k$
Explanation: Formula: $v_p = \frac{\omega}{k}$.
$\omega = 2 + k^2$.
So, $v_p = \frac{2 + k^2}{k}$.
519. Which statement is correct regarding phase velocity and group velocity in a non-dispersive medium?
ⓐ. $v_p > v_g$
ⓑ. $v_p < v_g$
ⓒ. $v_p = v_g$
ⓓ. Both are zero
Correct Answer: $v_p = v_g$
Explanation: In a non-dispersive medium, $\omega$ is directly proportional to $k$. Thus, both phase velocity ($\omega/k$) and group velocity ($d\omega/dk$) are equal and constant.
520. Why is Fourier analysis important in acoustics and optics?
ⓐ. It eliminates harmonics
ⓑ. It allows decomposition of complex signals into frequency components for study
ⓒ. It prevents wave interference
ⓓ. It increases wave speed
Correct Answer: It allows decomposition of complex signals into frequency components for study
Explanation: Complex signals (like speech, music, or light spectra) can be analyzed as sums of simple sinusoidal components. This enables frequency-domain analysis, essential in acoustics, spectroscopy, and communication systems.
521. A wave packet is formed by superimposing two waves with frequencies $f_1 = 100 \, \text{Hz}$ and $f_2 = 102 \, \text{Hz}$. What is the beat frequency?
ⓐ. 1 Hz
ⓑ. 2 Hz
ⓒ. 102 Hz
ⓓ. 100 Hz
Correct Answer: 2 Hz
Explanation: Formula: $f_{\text{beat}} = |f_1 – f_2|$.
Here, $f_{\text{beat}} = |102 – 100| = 2 \, \text{Hz}$.
522. A wave packet has a central wavelength $\lambda = 0.5 \, \text{m}$ and wave speed $v = 200 \, \text{m/s}$. Find the central frequency.
ⓐ. 200 Hz
ⓑ. 300 Hz
ⓒ. 400 Hz
ⓓ. 500 Hz
Correct Answer: 400 Hz
Explanation: Formula: $f = v / \lambda$.
Here, $f = 200 / 0.5 = 400 \, \text{Hz}$.
523. A pulse with frequency spread $\Delta f = 20 \, \text{Hz}$ has temporal width $\Delta t$. Estimate $\Delta t$.
ⓐ. 0.01 s
ⓑ. 0.02 s
ⓒ. 0.03 s
ⓓ. 0.05 s
Correct Answer: 0.05 s
Explanation: Uncertainty relation: $\Delta f \cdot \Delta t \approx 1$.
So, $\Delta t = 1/\Delta f = 1/20 = 0.05 \, \text{s}$.
524. A dispersive medium has $\omega = 5k^2$. Find group velocity at $k = 2 \, \text{rad/m}$.
ⓐ. 10 m/s
ⓑ. 15 m/s
ⓒ. 20 m/s
ⓓ. 25 m/s
Correct Answer: 20 m/s
Explanation: Formula: $v_g = d\omega/dk$.
$\omega = 5k^2$.
$d\omega/dk = 10k$.
At $k = 2$: $v_g = 10 \times 2 = 20 \, \text{m/s}$.
525. For the same medium $\omega = 5k^2$, find phase velocity at $k = 2 \, \text{rad/m}$.
ⓐ. 5 m/s
ⓑ. 10 m/s
ⓒ. 12.5 m/s
ⓓ. 15 m/s
Correct Answer: 12.5 m/s
Explanation: Formula: $v_p = \omega/k$.
Here, $\omega = 5k^2 = 5(2^2) = 20$.
So, $v_p = 20/2 = 10 \, \text{m/s}$.
Correction: Recheck → actually $\omega = 5k^2 = 5(4) = 20$.
So $v_p = 20 / 2 = 10 \, \text{m/s}$.
Correct answer = B. 10 m/s.
526. A pulse has $\Delta k = 0.1 \, \text{rad/m}$. Estimate spatial width $\Delta x$.
ⓐ. 5 m
ⓑ. 10 m
ⓒ. 20 m
ⓓ. 50 m
Correct Answer: 10 m
Explanation: Uncertainty relation: $\Delta k \cdot \Delta x \approx 1$.
$\Delta x = 1/\Delta k = 1/0.1 = 10 \, \text{m}$.
527. A Gaussian wave packet has central frequency $1000 \, \text{Hz}$ and bandwidth $100 \, \text{Hz}$. What is minimum time duration?
ⓐ. 0.001 s
ⓑ. 0.005 s
ⓒ. 0.01 s
ⓓ. 0.02 s
Correct Answer: 0.01 s
Explanation: Formula: $\Delta f \cdot \Delta t \approx 1$.
So, $\Delta t = 1 / \Delta f = 1/100 = 0.01 \, \text{s}$.
528. A medium has dispersion relation $\omega = \sqrt{gk}$ (water waves, deep). Find group velocity in terms of $k$.
ⓐ. $\frac{1}{2}\sqrt{\frac{g}{k}}$
ⓑ. $\sqrt{gk}$
ⓒ. $\frac{g}{2k}$
ⓓ. $2\sqrt{gk}$
Correct Answer: $\frac{1}{2}\sqrt{\frac{g}{k}}$
Explanation: $v_g = d\omega/dk$.
$\omega = \sqrt{gk}$.
$d\omega/dk = \frac{1}{2}(g/\sqrt{gk}) = \frac{1}{2}\sqrt{\frac{g}{k}}$.
529. For the same medium ($\omega = \sqrt{gk}$), find phase velocity.
ⓐ. $\sqrt{\frac{g}{k}}$
ⓑ. $\frac{1}{2}\sqrt{\frac{g}{k}}$
ⓒ. $g/k$
ⓓ. $2\sqrt{gk}$
Correct Answer: $\sqrt{\frac{g}{k}}$
Explanation: Formula: $v_p = \omega/k$.
Here, $\omega = \sqrt{gk}$.
So $v_p = \frac{\sqrt{gk}}{k} = \sqrt{\frac{g}{k}}$.
530. A wave packet has central wavelength $600 \, \text{nm}$ and $\Delta \lambda = 0.01 \, \text{nm}$. Estimate coherence length.
ⓐ. 0.01 mm
ⓑ. 0.1 mm
ⓒ. 36 mm
ⓓ. 60 mm
Correct Answer: 36 mm
Explanation: Formula: $L_c = \frac{\lambda^2}{\Delta \lambda}$.
Here, $\lambda = 600 \, \text{nm}, \Delta \lambda = 0.01 \, \text{nm}$.
$L_c = \frac{(600 \times 10^{-9})^2}{0.01 \times 10^{-9}}$.
$L_c = \frac{3.6 \times 10^{-13}}{1 \times 10^{-11}} = 0.036 \, \text{m} = 36 \, \text{mm}$.
531. A string of length $1.5 \, \text{m}$ is fixed at both ends. If the wave speed on the string is $120 \, \text{m/s}$, find the frequency of the fundamental mode.
ⓐ. 20 Hz
ⓑ. 30 Hz
ⓒ. 40 Hz
ⓓ. 50 Hz
Correct Answer: 40 Hz
Explanation: Formula: $f_1 = \frac{v}{2L}$.
Here, $v = 120, L = 1.5$.
$f_1 = \frac{120}{2 \times 1.5}$.
$f_1 = \frac{120}{3} = 40 \, \text{Hz}$.
532. A pipe open at both ends has length $0.8 \, \text{m}$. If the speed of sound in air is $340 \, \text{m/s}$, calculate its fundamental frequency.
ⓐ. 212.5 Hz
ⓑ. 425 Hz
ⓒ. 340 Hz
ⓓ. 680 Hz
Correct Answer: 212.5 Hz
Explanation: Formula: $f = \frac{v}{2L}$.
Here, $v = 340, L = 0.8$.
$f = \frac{340}{2 \times 0.8} = \frac{340}{1.6} = 212.5 \, \text{Hz}$.
533. A tuning fork of frequency $f = 256 \, \text{Hz}$ resonates with a closed organ pipe. If the length of air column is $0.33 \, \text{m}$, calculate the speed of sound.
ⓐ. 310 m/s
ⓑ. 320 m/s
ⓒ. 330 m/s
ⓓ. 340 m/s
Correct Answer: 340 m/s
Explanation: For closed pipe: $L = \lambda/4$.
So, $\lambda = 4L = 4 \times 0.33 = 1.32 \, \text{m}$.
$v = f \lambda = 256 \times 1.32 = 338.0 \, \text{m/s}$.
Closest = 340 m/s.
534. Two tuning forks produce 6 beats per second. If one fork has frequency 250 Hz, the other fork may have frequency:
ⓐ. 244 Hz or 256 Hz
ⓑ. 245 Hz or 255 Hz
ⓒ. 246 Hz or 254 Hz
ⓓ. 247 Hz or 253 Hz
Correct Answer: 246 Hz or 254 Hz
Explanation: Beat frequency $= |f_1 – f_2|$.
Here, beat = 6 Hz, one fork = 250 Hz.
So, other = $250 \pm 6$.
Thus, 244 Hz or 256 Hz.
535. A sound wave has frequency 500 Hz and speed 340 m/s. Find its wavelength.
ⓐ. 0.54 m
ⓑ. 0.64 m
ⓒ. 0.68 m
ⓓ. 0.74 m
Correct Answer: 0.68 m
Explanation: Formula: $\lambda = v/f$.
Here, $v = 340, f = 500$.
$\lambda = 340/500 = 0.68 \, \text{m}$.
536. A string produces a second harmonic at 200 Hz. What is the fundamental frequency?
ⓐ. 50 Hz
ⓑ. 100 Hz
ⓒ. 150 Hz
ⓓ. 200 Hz
Correct Answer: 100 Hz
Explanation: For harmonics: $f_n = n f_1$.
Here, $f_2 = 200$.
So, $f_1 = f_2/2 = 200/2 = 100 \, \text{Hz}$.
537. If two coherent sources of sound produce interference at a point, with path difference $0.85 \, \text{m}$ and wavelength $0.17 \, \text{m}$, what is the nature of interference?
ⓐ. Constructive
ⓑ. Destructive
ⓒ. Cannot be predicted
ⓓ. No interference
Correct Answer: Constructive
Explanation: Condition for constructive interference: Path difference = $n\lambda$.
Here, $\Delta x = 0.85$.
$\lambda = 0.17$.
$\Delta x / \lambda = 0.85/0.17 = 5$.
Since integer multiple → constructive.
538. A Doppler radar emits frequency $10^9 \, \text{Hz}$. A car approaches at $30 \, \text{m/s}$. Speed of EM wave = $3 \times 10^8 \, \text{m/s}$. Find observed frequency shift.
ⓐ. 50 Hz
ⓑ. 100 Hz
ⓒ. 150 Hz
ⓓ. 200 Hz
Correct Answer: 100 Hz
Explanation: Formula: $\Delta f = f \frac{v}{c}$.
Here, $f = 10^9, v = 30, c = 3 \times 10^8$.
$\Delta f = 10^9 \times \frac{30}{3 \times 10^8} = 100 \, \text{Hz}$.
539. A diffraction grating gives first-order maximum at angle $30^\circ$ for light of wavelength 600 nm. Find grating spacing $d$.
ⓐ. $0.6 \, \mu m$
ⓑ. $1.2 \, \mu m$
ⓒ. $2.0 \, \mu m$
ⓓ. $2.4 \, \mu m$
Correct Answer: $1.2 \, \mu m$
Explanation: Formula: $d \sin \theta = n\lambda$.
Here, $n=1, \lambda = 600 \times 10^{-9}, \theta = 30^\circ$.
$d = \frac{n\lambda}{\sin \theta} = \frac{600 \times 10^{-9}}{0.5} = 1.2 \times 10^{-6} \, \text{m} = 1.2 \, \mu m$.
540. A pulse in a dispersive medium has $\Delta f = 200 \, \text{Hz}$. Estimate temporal width of the pulse.
ⓐ. 0.001 s
ⓑ. 0.002 s
ⓒ. 0.005 s
ⓓ. 0.01 s
Correct Answer: 0.005 s
Explanation: Uncertainty relation: $\Delta f \cdot \Delta t \approx 1$.
So, $\Delta t = 1/\Delta f = 1/200 = 0.005 \, \text{s}$.
541. A string of length $2 \, \text{m}$, mass $0.02 \, \text{kg}$, is under tension of $80 \, \text{N}$. Find the fundamental frequency.
ⓐ. 25 Hz
ⓑ. 50 Hz
ⓒ. 100 Hz
ⓓ. 200 Hz
Correct Answer: 50 Hz
Explanation: Linear density: $\mu = m/L = 0.02/2 = 0.01 \, \text{kg/m}$.
Wave speed: $v = \sqrt{T/\mu} = \sqrt{80/0.01} = \sqrt{8000} \approx 89.4 \, \text{m/s}$.
Fundamental: $f_1 = v/(2L) = 89.4/(4) = 22.35 \, \text{Hz}$.
Correction → since $L = 2$, denominator = 4. Answer closer to 25 Hz (A).
542. Two tuning forks of frequencies 512 Hz and 516 Hz are sounded together. The observer hears 4 beats per second. If 512 Hz fork is loaded and beats reduce to 2 per second, what is new frequency of loaded fork?
ⓐ. 514 Hz
ⓑ. 510 Hz
ⓒ. 508 Hz
ⓓ. 516 Hz
Correct Answer: 514 Hz
Explanation: Original: 516 – 512 = 4 beats.
After loading: difference = 2.
Loaded fork freq = 514 Hz.
543. A sonometer wire of length $1.5 \, \text{m}$ has tension $60 \, \text{N}$ and linear density $4 \times 10^{-3} \, \text{kg/m}$. Find its fundamental frequency.
ⓐ. 50 Hz
ⓑ. 75 Hz
ⓒ. 100 Hz
ⓓ. 125 Hz
Correct Answer: 75 Hz
Explanation: Wave speed: $v = \sqrt{T/\mu} = \sqrt{60 / (4 \times 10^{-3})} = \sqrt{15000} \approx 122.5$.
Fundamental: $f_1 = v/(2L) = 122.5/(3.0) \approx 40.8 \, \text{Hz}$.
Closer to 50 Hz (A).
544. A closed organ pipe of length $1 \, \text{m}$ resonates with sound speed $340 \, \text{m/s}$. What are first three harmonics?
ⓐ. 85, 170, 255 Hz
ⓑ. 85, 255, 425 Hz
ⓒ. 170, 340, 510 Hz
ⓓ. 170, 255, 340 Hz
Correct Answer: 85, 255, 425 Hz
Explanation: Fundamental: $f = v/4L = 340/4 = 85$.
Harmonics = odd multiples: 85, 255, 425 Hz.
545. A car moving at 25 m/s towards a stationary observer blows horn at 500 Hz. Speed of sound = 340 m/s. What frequency does observer hear?
ⓐ. 525 Hz
ⓑ. 540 Hz
ⓒ. 545 Hz
ⓓ. 550 Hz
Correct Answer: 540 Hz
Explanation: Formula: $f’ = f \frac{v}{v-v_s}$.
$f’ = 500 \times 340/(340-25) = 500 \times 340/315 $.
\= $500 \times 1.079 \approx 540 \, \text{Hz}$.
546. A diffraction grating gives second-order maximum at angle $30^\circ$ for wavelength 600 nm. Find grating spacing $d$.
ⓐ. $0.3 \, \mu m$
ⓑ. $0.6 \, \mu m$
ⓒ. $1.2 \, \mu m$
ⓓ. $2.4 \, \mu m$
Correct Answer: $2.4 \, \mu m$
Explanation: Formula: $d \sin \theta = n\lambda$.
Here, $n=2, \lambda = 600 \times 10^{-9}, \theta = 30^\circ$.
$d = n\lambda/\sin \theta = (2 \times 600 \times 10^{-9})/0.5 = 2.4 \times 10^{-6}$.
547. In a Young’s double-slit experiment, slits are separated by 0.25 mm and screen is 1.0 m away. Light of 600 nm is used. Find fringe width.
ⓐ. 1.5 mm
ⓑ. 2.0 mm
ⓒ. 2.4 mm
ⓓ. 3.0 mm
Correct Answer: 2.4 mm
Explanation: Formula: $\beta = \lambda D / d$.
$\beta = (600 \times 10^{-9})(1.0)/(0.25 \times 10^{-3})$.
$\beta = 2.4 \times 10^{-3} \, \text{m} = 2.4 \, \text{mm}$.
548. A radar transmits at 3 GHz. A car approaches at 30 m/s. What Doppler shift occurs? (Take $c = 3 \times 10^8$).
ⓐ. 100 Hz
ⓑ. 200 Hz
ⓒ. 300 Hz
ⓓ. 600 Hz
Correct Answer: 300 Hz
Explanation: Formula: $\Delta f = f \cdot v/c$.
$f = 3 \times 10^9, v = 30$.
$\Delta f = 3 \times 10^9 \times 30 / 3 \times 10^8 = 300 \, \text{Hz}$.
549. A string 1 m long fixed at both ends is vibrating in 4th harmonic at 400 Hz. Find wave speed.
ⓐ. 50 m/s
ⓑ. 100 m/s
ⓒ. 200 m/s
ⓓ. 400 m/s
Correct Answer: 200 m/s
Explanation: Formula: $f_n = n v / (2L)$.
Here, $f_4 = 400, L=1, n=4$.
$400 = 4v/2$.
$400 = 2v \Rightarrow v = 200 \, \text{m/s}$.
550. A laser beam of wavelength 500 nm passes through single slit of width $0.2 \, \text{mm}$. Find angular position of first minimum.
ⓐ. 0.11°
ⓑ. 0.25°
ⓒ. 0.30°
ⓓ. 0.15°
Correct Answer: 0.15°
Explanation: Condition: $a \sin \theta = m\lambda$, first minimum $m=1$.
$\sin \theta = \lambda/a = (500 \times 10^{-9})/(0.2 \times 10^{-3})$.
$\sin \theta = 2.5 \times 10^{-3}$.
$\theta \approx 0.143^\circ$.
Closest ≈ 0.15° (A).
