Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A: A sound wave has higher speed in solids than gases. Reason R: Gases have higher value of Bulk modulus than solids. In the light of the above statements, choose the correct answer from the options given below.

Related Formula

where, A₁, A₂$A_1, A_2$ = amplitudes of individual harmonic waves φ$\phi$ = phase difference between the waves Aᵣₑₛ$A_{\text{res}}$ = resultant amplitude θ$\theta$ = resultant phase angle relative to the first wave

Core Logic

Given parameters:

• A₁ = 4$A_1 = 4$
• A₂ = 2$A_2 = 2$
• Phase difference, φ = (2π)/(3) = 120^°$\phi = \frac{2\pi}{3} = 120^\circ$



Calculate Resultant Amplitude:

Calculate Resultant Phase (angle θ$\theta$):

Step 1: Result Format

Represent the resultant amplitude and phase as a pair:

Pattern Recognition

Sees: Vector-like addition of wave amplitudes. Shortcut: Solve it like a vector addition problem. Vector A₁$A_1$ along horizontal (0$0$), and Vector A₂$A_2$ at 120^°$120^\circ$. The magnitude is √(4²+2²-2(4)(2)(0.5)) = 2√(3)$\sqrt{4^2+2^2-2(4)(2)(0.5)} = 2\sqrt{3}$ and direction is π/6$\pi/6$. ✓

Chapter Mix

Class 11 Physics: Waves

Related Formula

where, v$v$ = velocity of transverse wave on a string T$T$ = tension on string μ$\mu$ = linear mass density ρ$\rho$ = density of material A$A$ = cross-sectional area

Core Logic

Since both strings are made of the same material, their density ρ$\rho$ is the same. Also, they are stretched to have the same amount of tension T$T$.

Substitute the formula for μ$\mu$ into the velocity equation:

Thus, the wave velocity is inversely proportional to the radius of the cross section of the string:

Step 1: Ratio Computation

Given:

• R₁ = R$R_1 = R$
• R₂ = R/2$R_2 = R/2$

Substitute these values:

Pattern Recognition

Sees: Circular cross-section strings + transverse wave speed relation. Shortcut: Wave velocity v ∝ 1√(μ) ∝ (1)/(R)$v \propto \frac{1}{\sqrt{\mu}} \propto \frac{1}{R}$. If the radius is halved, the mass per unit length decreases by 4 times, which makes the speed increase by √(4) = 2$\sqrt{4} = 2$ times. ✓

Chapter Mix

Class 11 Physics: Waves

Related Formula

Core Logic

From the given wave equation, angular frequency ω = 628 rad/s$\omega = 628\text{ rad/s}$. Given wave velocity v = 300 m/s$v = 300\text{ m/s}$. Using the relation v = (ω)/(K)$v = \frac{\omega}{K}$:

Step 1: Compute Wavelength

Substitute K = (2π)/(λ)$K = \frac{2\pi}{\lambda}$:

Since 2π ≈ 2 × 3.14 = 6.28$2\pi \approx 2 \times 3.14 = 6.28$, the expression simplifies neatly:

Pattern Recognition

Notice standard values like ω = 628 = 200π$\omega = 628 = 200\pi$, which means the frequency is exactly 100 Hz$100\text{ Hz}$. Using v = fλ 300 = 100λ λ = 3 m$v = f\lambda \implies 300 = 100\lambda \implies \lambda = 3\text{ m}$ avoids setting up fractions.

Chapter Mix

Class 11 Physics: Waves

Related Formula

Laplace correction equation for speed of sound:

Given that (P)/(ρ)$\frac{P}{\rho}$ is constant for all three gases:

where γ = 1 + (2)/(f)$\gamma = 1 + \frac{2}{f}$ (adiabatic constant).

Core Logic

Determine the γ$\gamma$ factor based on molecular atomic structures:

• He$\mathrm{He}$ (Monatomic) f = 3 γHe = (5)/(3)$\implies f = 3 \implies \gamma_{\mathrm{He}} = \frac{5}{3}$
• CH₄$\mathrm{CH}_{4}$ (Polyatomic/Non-linear) γ_CH₄ ≈ (4)/(3)$\implies \gamma_{\mathrm{CH}_{4}} \approx \frac{4}{3}$ based on experimental references.
• CO₂$\mathrm{CO}_{2}$ (Triatomic linear/vibrational modes) γ_CO₂ ≈ (4)/(3)$\implies \gamma_{\mathrm{CO}_{2}} \approx \frac{4}{3}$ as provided in textbook standard testing matrices.

Step 1: Construct the Ratio

Substitute these values into the proportionality:

Pattern Recognition

When (P)/(ρ)$\frac{P}{\rho}$ is locked down constant, sound speed depends strictly on internal degrees of freedom via γ$\gamma$. Keep standard experimental values of complex gases like CH₄$\mathrm{CH}_4$ and CO₂$\mathrm{CO}_2$ memorized.

Chapter Mix

Class 11 Physics: Waves Class 11 Physics: Kinetic Theory