Class 11 Physics MCQs | Chapter 15: Waves – Part 3 (Exam Booster Questions & Answers)

Explanation: By the law of reflection:

$\theta_i = \theta_r$.

Here, $\theta_i = 40^\circ$.

So, $\theta_r = 40^\circ$.

Explanation: Law: angle of incidence = angle of reflection.

Here, incidence = $25^\circ$.

So, reflection = $25^\circ$.

Explanation: Angle of incidence is measured from the normal.

If ray makes $37^\circ$ with mirror surface, then with normal it makes:

$\theta_i = 90^\circ – 37^\circ = 53^\circ$.

So, angle of reflection $\theta_r = 53^\circ$.

Explanation: By law of reflection:

$\theta_i = \theta_r$.

So, if incidence = $60^\circ$, reflection = $60^\circ$.

Explanation: If ray makes $20^\circ$ with surface, then with normal:

$\theta_i = 90^\circ – 20^\circ = 70^\circ$.

So, angle of reflection = $70^\circ$.

Explanation: Law of reflection:

$\theta_r = \theta_i$.

So, reflection angle = $65^\circ$.

Explanation: If ray makes $45^\circ$ with mirror surface, then with normal:

$\theta_i = 90^\circ – 45^\circ = 45^\circ$.

Correction: actually it is $45^\circ$ not $60^\circ$.

So, reflection angle = $45^\circ$.

Explanation: By law of reflection:

$\theta_r = \theta_i = 15^\circ$.

Explanation: Incidence angle with surface = $30^\circ$.

So, with normal: $\theta_i = 90^\circ – 30^\circ = 60^\circ$.

Thus, reflection angle = $60^\circ$.

Explanation: Law of reflection:

$\theta_r = \theta_i$.

Here, $\theta_i = 50^\circ$.

So, $\theta_r = 50^\circ$.

Explanation: At a fixed boundary, the reflected pulse undergoes a $180^\circ$ phase change.

Thus, the crest becomes a trough (inversion).

Amplitude and speed remain unchanged.

Explanation: At a free boundary, the reflected wave does not undergo inversion.

The crest remains a crest and the trough remains a trough.

Only direction of travel reverses.

Explanation: For longitudinal waves, reflection from a rigid wall does not cause phase change.

Compressions remain compressions, rarefactions remain rarefactions.

Explanation: At an open end, reflection causes a **phase change of $\pi$** in sound waves.

Thus, compressions reflect as rarefactions and vice versa.

Explanation: At a fixed end, transverse waves undergo a phase reversal of $180^\circ$.

At a free end, no inversion occurs.

Explanation: At a closed end, displacement node is formed.

Reflection occurs without phase change.

Thus, compression → compression, rarefaction → rarefaction.

Explanation: At an open end, a displacement antinode is formed, corresponding to pressure node.

This means compressions turn into rarefactions upon reflection.

Explanation: At a fixed end, the wave is reflected with **inversion** and travels in the opposite direction.

Thus sign of amplitude becomes negative, and sign of $t$-term changes.

Explanation: At a free end, the reflected wave is not inverted.

It travels in the opposite direction, so the sign of the time term changes.

Explanation: At rigid end, compressions remain compressions, so no phase change for longitudinal waves.

At free end (for transverse waves), there is no inversion.

At fixed end (transverse), phase change $\pi$.

At open end (longitudinal), phase change occurs, compressions become rarefactions.

Explanation: An echo is heard when sound waves reflect from a hard surface such as a wall or cliff and return to the listener.

For a clear echo, the reflected sound must reach the ear at least 0.1 s after the original sound.

Explanation: For echo, time delay = 0.1 s.

Distance travelled by sound = $v \times t = 343 \times 0.1 = 34.3 \, \text{m}$.

This is the total distance (to wall and back).

So, minimum distance = 34.3/2 ≈ 17 m.

Explanation: SONAR (Sound Navigation and Ranging) transmits ultrasonic waves and detects their reflection from the sea floor.

The travel time of reflected sound is used to calculate depth.

Explanation: Curved reflecting surfaces are used in auditoriums and halls.

They reflect and direct sound waves uniformly across the hall, improving audibility.

Explanation: The stethoscope uses multiple reflections of sound in its tubes.

This amplifies faint sounds from heart and lungs so doctors can hear them clearly.

Explanation: At open ends, compressions reflect as rarefactions.

This leads to displacement antinodes at both open ends, forming standing waves.

Explanation: At closed end, displacement = 0 (node).

At open end, displacement maximum (antinode).

Thus, standing waves form with node at closed end and antinode at open end.

Explanation: In ultrasound scans, high-frequency sound waves are reflected from tissues and organs. These reflections form images of the body’s interior.

Explanation: Waveguides are hollow metallic or dielectric structures that confine waves. They reflect waves internally, allowing propagation with little attenuation.

Explanation: Optical fibers act as cylindrical waveguides. They use total internal reflection to guide light waves over long distances with minimal loss.

Explanation: Beats occur due to interference between two waves of nearly equal frequencies. The resultant amplitude alternates between constructive and destructive interference, producing waxing and waning of sound intensity.

Explanation: Beat frequency $f_b = | f_1 – f_2 |$.

It represents the number of amplitude maxima heard per second.

If the frequencies are equal, no beats are produced.

Explanation: Beat frequency = $| f_1 – f_2 |$.

\= $| 260 – 256 |$.

\= 4 Hz.

So, 4 beats per second are heard.

Explanation: Beat frequency $f_b = | f_1 – f_2 |$.

Here, $f_b = 6$, $f_1 = 512$.

So, $f_2 = 512 \pm 6$.

Thus, $f_2 = 506$ Hz or 518 Hz.

Explanation: Beats require at least two waves of nearly equal frequency. Options A, B, and C are true. Option D is false because a single wave cannot produce beats.

Explanation: In constructive interference during beats, amplitudes add: Resultant amplitude = $A + A = 2A$.

In destructive phase, amplitude becomes zero.

Explanation: Beat frequency = $| f_1 – f_2 |$.

Here, $f_b = 5, f_1 = 320$.

So, $f_2 = 320 \pm 5$.

Thus, 315 Hz or 325 Hz.

Explanation: Beat frequency = $| f_1 – f_2 |$.

\= $| 444 – 440 | = 4 \, \text{Hz}$.

So, 4 beats per second are heard.

Explanation: Time period of beats is reciprocal of beat frequency:

$T_b = \frac{1}{f_b}$.

It represents the time between successive loud sounds.

Explanation: Beat time period $T_b = \frac{1}{f_b}$.

Here, $f_b = 2$.

So, $T_b = \frac{1}{2} = 0.5 \, \text{s}$.

Explanation: Beat frequency $f_b = | f_1 – f_2 |$.

Here, $f_b = |304 – 300|$.

$f_b = 4 \, \text{Hz}$.

Thus, 4 beats per second are heard.

Explanation: Beat period $T_b = \frac{1}{f_b}$.

Here, $f_b = 6$.

So, $T_b = \frac{1}{6}$.

$T_b \approx 0.167 \, \text{s}$.

Correction: If answer options are as given, the closest correct is not listed; but true value = 0.167 s.

Explanation: Beat frequency $f_b = | f_1 – f_2 |$.

Here, $f_1 = 256$, $f_b = 3$.

So, $f_2 = 256 \pm 3$.

Thus, 253 Hz or 259 Hz.

Explanation: Beat frequency = $|502 – 500| = 2 \, \text{Hz}$.

Number of beats in 5 s = $f_b \times t = 2 \times 5 = 10$.

Explanation: Beat period = reciprocal of beat frequency.

$f_b = \frac{1}{T_b} = \frac{1}{0.25}$.

$f_b = 4 \, \text{Hz}$.

Explanation: Beat frequency $f_b = | f_1 – f_2 |$.

Here, $f_1 = 340$, $f_b = 8$.

So, $f_2 = 340 \pm 8$.

Thus, 332 Hz or 348 Hz.

Explanation: Beat frequency = difference of frequencies.

$f_2 = 256 \pm 5$.

So, $f_2 = 251$ Hz or 261 Hz.

Explanation: Beat frequency = reciprocal of period.

$f_b = \frac{1}{T_b} = \frac{1}{0.2} = 5 \, \text{Hz}$.

Beats in 2 s = $f_b \times t = 5 \times 2 = 10$.

Explanation: Beat frequency = $|445 – 440| = 5 \, \text{Hz}$.

So, loudness varies 5 times each second.

Explanation: Time between successive maxima = beat period.

Beat period $T_b = \frac{1}{f_b} = \frac{1}{10} = 0.1 \, \text{s}$.

Explanation: Musicians strike two notes together and adjust until no beats are heard. When no beats are produced, frequencies are equal, indicating the instrument is tuned.

Explanation: Noise-cancelling headphones generate waves out of phase with external noise. These cancel each other through destructive interference, reducing noise.

Explanation: Architects design walls and ceilings to reflect sound evenly. This prevents dead zones and ensures good audibility throughout the hall.

Explanation: If two instruments are slightly out of tune, beats are heard. Musicians adjust until beats disappear, showing tuning accuracy.

Explanation: Reflection of sound is used in stethoscopes (multiple reflections amplify sound), SONAR (reflection from sea bed), and acoustical design of halls.

Explanation: Beat frequency = $|400 – 402| = 2 \, \text{Hz}$. So, 2 beats per second are heard as periodic waxing and waning of loudness.

Explanation: Acoustic panels absorb sound reflections, preventing overlapping waves. This ensures a clear, pure recording without interference patterns.

Explanation: When two pipes produce nearly equal frequencies, beats are heard. By adjusting length until beats vanish, musicians match frequencies exactly.

Explanation: Ultrasound uses high-frequency sound (1–15 MHz).

Reflected waves from tissues create detailed images.

Beats or interference are not used directly in imaging.

Explanation: Music is produced by periodic, harmonious waves. Noise is irregular and random, lacking harmonic structure. Thus, the regular interference and superposition in music create pleasant sound.

Explanation: The Doppler effect is the apparent change in frequency (or pitch) of a wave when there is relative motion between source and observer.

If they move towards each other, frequency increases; if apart, frequency decreases.

Explanation: As the source approaches, wavefronts are compressed.

This reduces wavelength while speed remains constant, so frequency observed increases.

Explanation: Observer’s motion increases the rate at which wavefronts are encountered. Thus, observed frequency increases even though the source frequency is unchanged.

Explanation: Formula: $f’ = f \frac{v}{v – v_s}$.

Here, $f = 500, v = 330, v_s = 30$.

$f’ = 500 \frac{330}{330 – 30}$.

$f’ = 500 \frac{330}{300}$.

$f’ = 500 \times 1.1 = 550 \, \text{Hz}$.

Correction: Exact = 550 Hz, closest option is **D. 545 Hz**.

Explanation: Formula: $f’ = f \frac{v + v_o}{v}$.

Here, $f = 400, v = 340, v_o = 20$.

$f’ = 400 \frac{340 + 20}{340}$.

$f’ = 400 \frac{360}{340}$.

$f’ = 400 \times 1.0588$.

$f’ \approx 423.5 \, \text{Hz}$.

Explanation: When both move towards each other, wavelength is compressed more strongly. Thus, observed frequency is higher than emitted frequency.

Explanation: Formula: $f’ = f \frac{v + v_o}{v – v_s}$.

Here, $f = 600, v = 340, v_o = 20, v_s = 20$.

$f’ = 600 \frac{340 + 20}{340 – 20}$.

$f’ = 600 \frac{360}{320}$.

$f’ = 600 \times 1.125$.

$f’ = 675 \, \text{Hz}$.

Explanation: When they move apart, wavelength is stretched. This decreases observed frequency compared to actual.

Explanation: Formula: $f’ = f \frac{v}{v + v_s}$.

Here, $f = 700, v = 350, v_s = 25$.

$f’ = 700 \frac{350}{375}$.

$f’ = 700 \times 0.933$.

$f’ \approx 653 \, \text{Hz}$.

Explanation: Doppler effect is used in radar guns, sonar, and medical ultrasound for velocity measurements. Refractive index of glass is unrelated to Doppler effect.

Explanation: Formula: $f’ = f \frac{v}{v – v_s}$.

Here, $f = 500, v = 340, v_s = 20$.

$f’ = 500 \frac{340}{340 – 20}$.

$f’ = 500 \frac{340}{320}$.

$f’ = 500 \times 1.0625$.

$f’ = 531.25 \, \text{Hz}$.

Closest option = 531 Hz.

Explanation: Formula: $f’ = f \frac{v + v_o}{v}$.

Here, $f = 400, v = 340, v_o = 10$.

$f’ = 400 \frac{340 + 10}{340}$.

$f’ = 400 \frac{350}{340}$.

$f’ = 400 \times 1.0294$.

$f’ \approx 411.7 \, \text{Hz}$.

Closest option = 412 Hz.

Explanation: Formula: $f’ = f \frac{v}{v + v_s}$.

Here, $f = 600, v = 330, v_s = 30$.

$f’ = 600 \frac{330}{330 + 30}$.

$f’ = 600 \frac{330}{360}$.

$f’ = 600 \times 0.9166$.

$f’ \approx 550 \, \text{Hz}$.

Explanation: Formula: $f’ = f \frac{v + v_o}{v – v_s}$.

Here, $f = 480, v = 340, v_o = 15, v_s = 25$.

$f’ = 480 \frac{340 + 15}{340 – 25}$.

$f’ = 480 \frac{355}{315}$.

$f’ = 480 \times 1.1269$.

$f’ \approx 541 \, \text{Hz}$.

Explanation: Formula: $f’ = f \frac{v}{v – v_s}$.

Here, $f = 500, v = 350, v_s = 50$.

$f’ = 500 \frac{350}{350 – 50}$.

$f’ = 500 \frac{350}{300}$.

$f’ = 500 \times 1.1666$.

$f’ \approx 583 \, \text{Hz}$.

Explanation: Formula: $f’ = f \frac{v – v_o}{v}$.

Here, $f = 440, v = 340, v_o = 20$.

$f’ = 440 \frac{340 – 20}{340}$.

$f’ = 440 \frac{320}{340}$.

$f’ = 440 \times 0.9411$.

$f’ \approx 414.1 \, \text{Hz}$.

Closest option = 415 Hz.

Explanation: Formula: $f’ = f \frac{v}{v – v_s}$.

Here, $f = 600, v = 330, v_s = 15$.

$f’ = 600 \frac{330}{330 – 15}$.

$f’ = 600 \frac{330}{315}$.

$f’ = 600 \times 1.0476$.

$f’ \approx 628.6 \, \text{Hz}$.

Explanation: Formula: $f’ = f \frac{v + v_o}{v}$.

Here, $f = 256, v = 340, v_o = 34$.

$f’ = 256 \frac{340 + 34}{340}$.

$f’ = 256 \frac{374}{340}$.

$f’ = 256 \times 1.1$.

$f’ \approx 281.6 \, \text{Hz}$.

Explanation: Formula: $f’ = f \frac{v}{v + v_s}$.

Here, $f = 1000, v = 340, v_s = 20$.

$f’ = 1000 \frac{340}{340 + 20}$.

$f’ = 1000 \frac{340}{360}$.

$f’ = 1000 \times 0.9444$.

$f’ \approx 944.4 \, \text{Hz}$.

Explanation: Formula: $f’ = f \frac{v}{v – v_s}$.

Here, $f = 500, v = 330, v_s = 30$.

$f’ = 500 \frac{330}{330 – 30}$.

$f’ = 500 \frac{330}{300}$.

$f’ = 500 \times 1.1$.

$f’ = 550 \, \text{Hz}$.

Explanation: In astronomy, when a light source moves away from an observer, the observed wavelength becomes longer (redshift). This is a direct consequence of the Doppler effect, where frequency decreases and wavelength increases if the source recedes. Redshift is used as evidence for the expanding universe and is a key principle in Hubble’s law.

Explanation: The Doppler effect allows astronomers to measure radial velocities of stars and galaxies. If spectral lines shift towards blue, the object is approaching; if they shift towards red, it is receding. This helps determine galaxy motion, star orbiting speeds, and exoplanet detection.

Explanation: Doppler radar sends electromagnetic waves towards a moving object. When reflected back, the frequency shift is analyzed to calculate the object’s speed. This principle is used in police radar guns, weather monitoring radars, and air-traffic control systems.

Explanation: Radar guns transmit microwaves that reflect from moving vehicles. The reflected waves experience a Doppler shift depending on the vehicle’s velocity. The frequency difference between transmitted and received signal is proportional to the vehicle speed.

Explanation: Doppler ultrasound measures the change in frequency of sound reflected from moving red blood cells. By analyzing the frequency shift, doctors can calculate blood flow velocity and detect blockages or abnormal circulation. It is widely used in cardiology and prenatal care.

Explanation: Doppler effect is used in astronomy, radar, and medical imaging. However, the gravitational constant $G$ is a fundamental physical constant, determined by experiments such as Cavendish’s torsion balance, not by Doppler effect.

Explanation: If a star moves towards the observer, the wavelength decreases and shifts towards the blue end of the spectrum. This is known as blue shift, an astronomical application of the Doppler effect. Redshift, in contrast, indicates recession.

Explanation: Doppler weather radar emits microwave pulses that scatter from raindrops.The frequency shift of the reflected waves indicates wind velocity and storm direction. This is crucial for predicting tornadoes, hurricanes, and rainfall intensity.

Explanation: Ultrasound waves have very short wavelengths, which provide high resolution imaging of small structures.The Doppler frequency shifts in ultrasound are large enough to be detected with precision, making it ideal for measuring blood flow. Thus, ultrasound frequencies (1–15 MHz) are preferred over audible sound.

Explanation: Astronomers study spectral lines from stars and galaxies. If these lines shift towards red, the body is receding; if towards blue, it is approaching. This Doppler shift allows calculation of radial velocity, a key measurement in cosmology and astrophysics.

Explanation: Use $\frac{\Delta \lambda}{\lambda_0} = \frac{v}{c}$.

$\Delta \lambda = 660.0 – 656.3 = 3.7 \, \text{nm}$.

$\frac{\Delta \lambda}{\lambda_0} = \frac{3.7}{656.3} \approx 0.00564$.

$v = c \times 0.00564 = 3.00 \times 10^8 \times 0.00564$.

$v \approx 1.69 \times 10^6 \, \text{m/s}$.

Explanation: $\frac{\Delta \lambda}{\lambda_0} = \frac{v}{c}$ with $\Delta \lambda = 498 – 500 = -2 \, \text{nm}$.

$\frac{\Delta \lambda}{\lambda_0} = \frac{-2}{500} = -0.004$.

$v = c \times (-0.004) = 3.00 \times 10^8 \times (-0.004)$.

$v = -1.2 \times 10^6 \, \text{m/s}$ (negative = approaching).

Explanation: Formula: $v = \frac{c \, \Delta f}{2 f_0}$.

Here, $c = 3.00 \times 10^8$, $\Delta f = 4.8 \times 10^3$, $f_0 = 24.125 \times 10^9$.

$v = \frac{3.00 \times 10^8 \times 4.8 \times 10^3}{2 \times 24.125 \times 10^9}$.

$v = \frac{1.44 \times 10^{12}}{48.25 \times 10^9}$.

$v \approx 29.8 \, \text{m/s} \approx 30 \, \text{m/s}$.

Explanation: $v = \frac{c \, \Delta f}{2 f_0}$.

$v = \frac{3.00 \times 10^8 \times 3500}{2 \times 10.525 \times 10^9}$.

$v = \frac{1.05 \times 10^{12}}{21.05 \times 10^9}$.

$v \approx 49.9 \, \text{m/s} \approx 50 \, \text{m/s}$.

Explanation: $\cos 60^\circ = 0.5$.

$\Delta f = \frac{2 \times 5.0 \times 10^6 \times 0.50 \times 0.5}{1540}$.

$\Delta f = \frac{2.5 \times 10^6}{1540}$.

$\Delta f \approx 1623 \, \text{Hz} \approx 1.6 \, \text{kHz}$.

Explanation: $\cos 30^\circ = 0.866$.

$\Delta f = \frac{2 \times 3.0 \times 10^6 \times 0.30 \times 0.866}{1540}$.

$\Delta f = \frac{1.5588 \times 10^6}{1540}$.

$\Delta f \approx 1012 \, \text{Hz} \approx 1.0 \, \text{kHz}$.

Explanation: Use $f’ \approx f_0 \left(1 – \frac{v}{c}\right)$.

$\frac{v}{c} = \frac{150 \times 10^3}{3.00 \times 10^8} = 5.0 \times 10^{-4}$.

$f’ = 6.00 \times 10^{14} \times (1 – 5.0 \times 10^{-4})$.

$f’ = 6.00 \times 10^{14} \times 0.9995$.

$f’ = 5.997 \times 10^{14} \, \text{Hz}$.

Explanation: Use $\lambda = \lambda_0 \left(1 + \frac{v}{c}\right)$.

$\frac{v}{c} = \frac{3000 \times 10^3}{3.00 \times 10^8} = 0.010$.

$\lambda = 656.3 \times (1 + 0.010) \, \text{nm}$.

$\lambda = 656.3 \times 1.010 \, \text{nm}$.

$\lambda = 662.863 \, \text{nm} \approx 662.9 \, \text{nm}$.

Explanation: $\Delta f = \frac{2 \times 20 \times 2.8 \times 10^9}{3.00 \times 10^8}$.

$\Delta f = \frac{112 \times 10^9}{3.00 \times 10^8}$.

$\Delta f = 373.3 \, \text{Hz} \approx 373 \, \text{Hz}$.

Explanation: $v = \frac{c \, \Delta f}{2 f_0 \cos \theta}$.

$\cos 30^\circ = 0.866$.

$v = \frac{3.00 \times 10^8 \times 2000}{2 \times 24.0 \times 10^9 \times 0.866}$.

$v = \frac{6.00 \times 10^{11}}{41.568 \times 10^9}$.

$v \approx 14.4 \, \text{m/s} \approx 14.5 \, \text{m/s}$.