Class 11 Physics MCQs | Chapter 15: Waves – Part 2 (Competitive Exam Practice Set)

$\frac{\partial^2 y}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2 y}{\partial t^2}$.

It shows the relationship between space and time variations of the displacement $y(x,t)$.

Option A has $v^2$ instead of $\frac{1}{v^2}$, which is incorrect. Options C and D are first-order forms, not valid for general waves.

Thus sine and cosine solutions with arguments $kx – \omega t$ or $kx + \omega t$ are valid.

Therefore, all given options are correct solutions.

It is given by $v_p = \frac{\omega}{k}$.

Group velocity refers to a wave packet, not a single wave.

Options C and D are unrelated.

Here, $k = 5$, $\omega = 20$.

Speed is $v = \frac{\omega}{k}$.

$v = \frac{20}{5}$.

$v = 4 \, \text{m/s}$.

Thus the wave speed is 4 m/s.

For $y = A \cos(kx + \omega t)$, the wave moves in –x direction.

Thus option B is correct.

$\omega = vk$.

This ensures the solution satisfies the wave equation.

Options B, C, D are mathematically inconsistent.

Here, $k = 2\pi$.

We know $k = \frac{2\pi}{\lambda}$.

So, $\lambda = \frac{2\pi}{k}$.

$\lambda = \frac{2\pi}{2\pi}$.

$\lambda = 1 \, \text{m}$.

Thus the wavelength is 1 m.

Speed is $v = \frac{\omega}{k}$.

$v = \frac{6}{3}$.

$v = 2 \, \text{m/s}$.

Thus the wave speed is 2 m/s.

Sine and cosine are valid, as are general functions.

However, $\cos^2(kx – \omega t)$ expands into $\frac{1}{2}(1 + \cos(2kx – 2\omega t))$, which is not a pure solution but a superposition of terms, and does not independently satisfy the simple wave equation.

Here, $\omega = 16$.

We know $\omega = 2\pi f$.

So, $f = \frac{\omega}{2\pi}$.

$f = \frac{16}{2\pi}$.

$f \approx 2.55 \, \text{Hz}$.

Correction: Options given mismatch. The accurate frequency is 2.55 Hz, closest to **A. 2 Hz**.

Mathematically, $v_p = \frac{\omega}{k}$.

Option A describes group velocity, C is not a standard definition, and D refers to wave dispersion.

It is given by $v_g = \frac{d\omega}{dk}$.

Phase velocity refers to crests, while group velocity is about energy transfer.

Group velocity: $v_g = \frac{d\omega}{dk} = \frac{d(2k)}{dk} = 2$.

So both velocities are equal to 2.

At $k = 2$, $v_p = 2$.

Group velocity: $v_g = \frac{d\omega}{dk} = \frac{d(k^2)}{dk} = 2k$.

At $k = 2$, $v_g = 4$.

Group velocity: $v_g = \frac{d\omega}{dk} = 3$.

Thus both are equal.

Here, $\omega = \sqrt{k}$.

At $k = 4$, $\omega = 2$.

So, $v_p = \frac{2}{4} = 0.5$.

Here, $\omega = \sqrt{k}$.

So, $v_g = \frac{1}{2\sqrt{k}}$.

At $k = 4$, $v_g = \frac{1}{2 \times 2} = 0.25$.

Here, $v_g < v_p$, so energy lags behind the crests.

This occurs in dispersive media.

Thus, phase velocity and group velocity are equal.

If they differ, the wave is dispersive.

$v_g = v_p + \lambda \frac{dv_p}{d\lambda}$.

This comes from differentiating $\omega = kv_p$.

Option B has a negative sign and is incorrect.

Formula: $v = \sqrt{\frac{F}{\mu}}$.

Here, $F$ is force (tension) in the string, and $\mu$ is linear mass density.

Thus, option B is correct.

Here, $F = 100 \, \text{N}, \mu = 0.01 \, \text{kg/m}$.

So, $v = \sqrt{\frac{100}{0.01}}$.

$v = \sqrt{10000}$.

$v = 100 \, \text{m/s}$.

$v = \sqrt{\frac{\gamma P}{\rho}}$,

where $\gamma$ = ratio of specific heats, $P$ = pressure, $\rho$ = density.

Option B is used in solids with modulus of elasticity.

Here, $\gamma = 1.4, P = 1.01 \times 10^5, \rho = 1.29$.

So, $v = \sqrt{\frac{1.4 \times 1.01 \times 10^5}{1.29}}$.

$v = \sqrt{\frac{1.414 \times 10^5}{1.29}}$.

$v = \sqrt{1.096 \times 10^5}$.

$v \approx 331 \, \text{m/s}$.

If tension increases by factor 4: $v’ = \sqrt{\frac{4F}{\mu}}$.

$v’ = 2 \sqrt{\frac{F}{\mu}}$.

Thus, wave speed doubles.

Here, $m = 0.04 \, \text{kg}, L = 2 \, \text{m}$.

So, $\mu = \frac{0.04}{2} = 0.02 \, \text{kg/m}$.

Now, $v = \sqrt{\frac{F}{\mu}}$.

$v = \sqrt{\frac{80}{0.02}}$.

$v = \sqrt{4000}$.

$v \approx 63.25 \, \text{m/s}$.

Correction: The computed value does not match options; likely tension mis-specified. Correct answer is $\approx 63 \, \text{m/s}$.

Here, $R$ = gas constant, $M$ = molar mass.

Thus, $v \propto \sqrt{T}$.

It does not directly depend on pressure or volume because both cancel out in ideal gas law.

Here, $m = 0.1 \, \text{kg}, L = 5 \, \text{m}$.

So, $\mu = \frac{0.1}{5} = 0.02 \, \text{kg/m}$.

Now, $v = \sqrt{\frac{F}{\mu}}$.

$v = \sqrt{\frac{200}{0.02}}$.

$v = \sqrt{10000}$.

$v = 100 \, \text{m/s}$.

Here, $E$ = elastic modulus, $\rho$ = density.

If density increases, denominator increases.

Thus, $v$ decreases.

Here, $E = 2 \times 10^{11}$, $\rho = 7800$.

So, $v = \sqrt{\frac{2 \times 10^{11}}{7800}}$.

$v = \sqrt{2.56 \times 10^7}$.

$v \approx 5060 \, \text{m/s}$.

Thus closest answer is 5000 m/s.

This is because solids have the highest elasticity and density ratio suitable for fast propagation.

This value changes slightly with humidity and temperature but not significantly with pressure.

This is about 4.4 times faster than in air, due to higher density and elasticity.

This high speed is due to steel’s large Young’s modulus and relatively low compressibility.

This allows particles to quickly transfer vibrational energy, resulting in higher wave speed.

Here, $B = 2.2 \times 10^9$, $\rho = 1000$.

So, $v = \sqrt{\frac{2.2 \times 10^9}{1000}}$.

$v = \sqrt{2.2 \times 10^6}$.

$v \approx 1483 \, \text{m/s}$.

Thus speed in water ≈ 1480 m/s.

At STP, sound speed ≈ 1280–1310 m/s.

Thus, it is about 4 times faster than in air.

Here, $E = 2 \times 10^{11}, \rho = 8000$.

$v = \sqrt{\frac{2 \times 10^{11}}{8000}}$.

$v = \sqrt{2.5 \times 10^7}$.

$v \approx 5000 \, \text{m/s}$.

Thus air is the slowest medium listed.

$= 4$.

Thus the ratio is 1:4.

Here, $B = 1.4 \times 10^5$.

$\rho = 1.3$.

$v = \sqrt{\frac{1.4 \times 10^5}{1.3}}$.

$v = \sqrt{1.0769 \times 10^5}$.

$v \approx 328 \, \text{m/s}$.

Thus speed ≈ 330 m/s.

Here, $\rho = 0.18, v = 965$.

$B = 0.18 \times (965)^2$.

$B = 0.18 \times 931225$.

$B = 167,620$.

$B \approx 1.8 \times 10^5 \, \text{N/m}^2$.

Here, $B = 2.1 \times 10^9$.

$\rho = 1000$.

$v = \sqrt{\frac{2.1 \times 10^9}{1000}}$.

$v = \sqrt{2.1 \times 10^6}$.

$v \approx 1449 \, \text{m/s}$.

Thus ≈ 1500 m/s.

Here, $E = 1.1 \times 10^{11}$.

$\rho = 8900$.

$v = \sqrt{\frac{1.1 \times 10^{11}}{8900}}$.

$v = \sqrt{1.236 \times 10^7}$.

$v \approx 3517 \, \text{m/s}$.

≈ 3600 m/s.

Here, $E = 7.0 \times 10^{10}$.

$\rho = 2700$.

$v = \sqrt{\frac{7.0 \times 10^{10}}{2700}}$.

$v = \sqrt{2.59 \times 10^7}$.

$v \approx 5090 \, \text{m/s}$.

≈ 5000 m/s.

Here, $\rho = 7850, v = 5000$.

$E = 7850 \times (5000)^2$.

$E = 7850 \times 25 \times 10^6$.

$E = 1.96 \times 10^{11}$.

≈ $2.0 \times 10^{11} $.

In air: $f = \frac{v}{\lambda} = \frac{330}{1.65} = 200 \, \text{Hz}$.

In water: $\lambda’ = \frac{v}{f} = \frac{1500}{200}$.

$\lambda’ = 7.5 \, \text{m}$.

For He: $\lambda = \frac{965}{500} = 1.93 \, \text{m}$.

Here, $E = 1.0 \times 10^{11}, \rho = 8500$.

$v = \sqrt{\frac{1.0 \times 10^{11}}{8500}}$.

$v = \sqrt{1.176 \times 10^7}$.

$v \approx 3430 \, \text{m/s}$.

≈ 3500 m/s.

\= 0.22.

≈ 1:4.

Here, $v$ is wave speed, $f$ is frequency (oscillations per second), and $\lambda$ is wavelength (distance between successive crests or compressions).

Options A, C, and D are mathematically incorrect forms.

Here, $f = 50 \, \text{Hz}, \lambda = 2 \, \text{m}$.

So, $v = 50 \times 2$.

$v = 100 \, \text{m/s}$.

Here, $v = 340 \, \text{m/s}, \lambda = 0.85 \, \text{m}$.

So, $f = \frac{340}{0.85}$.

$f = 400 \, \text{Hz}$.

Here, $c = 3 \times 10^8 \, \text{m/s}, f = 6 \times 10^{14} \, \text{Hz}$.

So, $\lambda = \frac{3 \times 10^8}{6 \times 10^{14}}$.

$\lambda = 0.5 \times 10^{-6} \, \text{m}$.

$\lambda = 5 \times 10^{-7} \, \text{m}$.

If $v$ constant and $f$ doubles, then $\lambda$ must halve.

This keeps the product $f \lambda$ constant.

Here, $v = 15, \lambda = 3$.

So, $f = \frac{15}{3}$.

$f = 5 \, \text{Hz}$.

Here, $c = 3 \times 10^8, f = 100 \times 10^6$.

So, $\lambda = \frac{3 \times 10^8}{1 \times 10^8}$.

$\lambda = 3 \, \text{m}$.

Here, $\omega = 100\pi$, so $f = \frac{\omega}{2\pi} = \frac{100\pi}{2\pi} = 50 \, \text{Hz}$.

Also, $k = 4\pi$.

So, $\lambda = \frac{2\pi}{k} = \frac{2\pi}{4\pi} = 0.5 \, \text{m}$.

Here, $v = 320, f = 256$.

So, $\lambda = \frac{320}{256}$.

$\lambda = 1.25 \, \text{m}$.

Here, $c = 3 \times 10^8, f = 2 \times 10^{10}$.

So, $\lambda = \frac{3 \times 10^8}{2 \times 10^{10}}$.

$\lambda = 1.5 \times 10^{-2} \, \text{m}$.

Thus wavelength is 0.015 m.

Option A is only true for complete destructive interference, not always.

Option C and D contradict experimental observations.

Thus it applies to all types of waves in linear media.

Here, $y = 2 \sin(\omega t) + 3 \sin(\omega t)$.

$y = (2 + 3) \sin(\omega t)$.

$y = 5 \sin(\omega t)$.

This results in destructive interference and minimum or zero amplitude at those points.

Maximum amplitude occurs in-phase (constructive interference).

So, $A = 4 + 6 = 10 \, \text{cm}$.

If they were out of phase, amplitude would be $|6 – 4| = 2 \, \text{cm}$.

This leads to destructive interference and no sound is heard.

Hence $y = y_1 + y_2$.

Options A, C, D are incorrect forms.

Reflection, however, is due to boundary conditions and not due to overlapping waves.

So resultant amplitude = sum of amplitudes.

$A = 3 + 4 = 7$.

In non-linear media, wave interaction can produce distortion, harmonics, or shock waves, where algebraic addition of displacements is not valid.

Phase difference = $0, 2\pi, 4\pi, …$.

Resultant amplitude = sum of amplitudes.

Option A describes destructive interference.

This cancels amplitudes, producing destructive interference.

Resultant amplitude = $3 + 4 = 7 \, \text{cm}$.

So, resultant amplitude = 0.

Resultant amplitude = $|8 – 6| = 2 \, \text{cm} $.

If perfectly out of phase, 2 cm remains.

Resultant amplitude = $4 + 5 = 9 \, \text{cm}$.

Intensity $I \propto A^2$.

So, new intensity = $(2A)^2 = 4A^2$.

Thus intensity increases four times.

So, intensity $I \propto A^2 = 0$.

No sound or light is observed at those points.

$\Delta x = n\lambda, n = 0,1,2,…$.

This ensures phase difference = $0, 2\pi, …$.

$\Delta x = (2n+1)\frac{\lambda}{2}, n = 0,1,2,…$.

This ensures phase difference = $ \pi, 3\pi, 5\pi,…$.

It directly follows from the superposition principle.

Refraction and reflection are due to boundary effects, and diffraction is due to wave bending at edges.

This produces constructive interference and bright fringes.

Destructive interference gives dark fringes instead.

Hence, path difference = odd multiples of half-wavelength: $(2n+1)\frac{\lambda}{2}$.

Here, $\lambda = 600 \times 10^{-9}, D = 1, d = 0.2 \times 10^{-3}$.

$\Delta y = \frac{600 \times 10^{-9} \times 1}{0.2 \times 10^{-3}}$.

$\Delta y = \frac{600}{0.2} \times 10^{-6}$.

$\Delta y = 3000 \times 10^{-6}$.

$\Delta y = 3 \times 10^{-3} \, \text{m} = 3 \, \text{mm}$.

Thus, interference helps analyze thin coatings.

Constructive and destructive interference at different wavelengths create colors.

This limits the instrument’s ability to separate closely spaced objects, defining its resolving power.

These cancel by destructive interference, reducing noise levels.

This produces destructive interference, resulting in a dark center.

Diffraction is then used to reconstruct the 3D image.

Thus, interference and diffraction together enable holography.

This holds for sound, light, and water waves when the reflecting surface is smooth.

Here, angle of incidence = $30^\circ$.

So, angle of reflection = $30^\circ$.

If the ray makes $45^\circ$ with the mirror surface, then with the normal it makes $90^\circ – 45^\circ = 45^\circ$.

Correction: Actually, the mirror surface and normal are perpendicular.

Thus, if the ray makes $45^\circ$ with the surface, it makes $45^\circ$ with the normal too.

So correct option is **B. $45^\circ$**.

Angle of incidence = Angle of reflection.

So, both are $20^\circ$.

Only the direction of propagation changes, obeying the law of reflection.

The angle of incidence equals the angle of reflection, ensuring sound returns.

Beats and interference are due to superposition, while diffraction is due to bending at edges.

Thus, reflection angle = $35^\circ$.

For curved mirrors, the normal is taken at the point of incidence.

Thus, the pulse is inverted but its speed and frequency remain unchanged.

Thus the reflected pulse remains upright.

This contrasts with reflection at a fixed end, where inversion occurs.