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

Core Logic

Since volume V ∝ R³$V \propto R^3$, if volume is increased by 8 times:

Step 1: Intensity Relation

Since intensity I ∝ (1)/(R²)$I \propto \frac{1}{R^2}$: If R arrow 2R$R \rightarrow 2R$, the area A = 4π R²$A = 4\pi R^2$ increases by 2² = 4$2^2 = 4$ times (A arrow 4A$A \rightarrow 4A$). Therefore, intensity becomes I arrow (I₀)/(4)$I \rightarrow \frac{I_0}{4}$.

Pattern Recognition

Volume ratio cubed root dictates radius multiplier. The square of that radius multiplier is the area multiplier, which is the exact inverse of the intensity multiplier.

Chapter Mix

Class 11 Physics: Waves

Related Formula

Core Logic

Initial beat frequency fbeat1 = 8 beats2 s = 4 Hz$f_{\text{beat}1} = \frac{8 \text{ beats}}{2 \text{ s}} = 4 \text{ Hz}$. Therefore, |fA - fB| = 4$|f_A - f_B| = 4$. Given fB = 380 Hz$f_B = 380 \text{ Hz}$, the original frequency of A could be: fA = 380 + 4 = 384 Hz$f_A = 380 + 4 = 384 \text{ Hz}$ OR fA = 380 - 4 = 376 Hz$f_A = 380 - 4 = 376 \text{ Hz}$.

Step 1: Check with Wax Loading

When fork A is loaded with wax, its frequency fA$f_A$ decreases. The new beat frequency fbeat2 = 4 beats2 s = 2 Hz$f_{\text{beat}2} = \frac{4 \text{ beats}}{2 \text{ s}} = 2 \text{ Hz}$.

Case 1: If fA = 384 Hz$f_A = 384 \text{ Hz}$, loading wax drops it to say 382 Hz$382 \text{ Hz}$. The new beat frequency becomes |382 - 380| = 2 Hz$|382 - 380| = 2 \text{ Hz}$. This matches the given condition.

Case 2: If fA = 376 Hz$f_A = 376 \text{ Hz}$, loading wax drops it to say 374 Hz$374 \text{ Hz}$. The new beat frequency would become |374 - 380| = 6 Hz$|374 - 380| = 6 \text{ Hz}$. This does NOT match the condition.

Step 2: Conclusion

The original frequency of tuning fork A must be 384 Hz$384 \text{ Hz}$.

Pattern Recognition

Loading wax always DECREASES the frequency. If decreasing the unknown frequency causes the beat frequency to DECREASE, the unknown frequency must have been initially higher than the known standard.

Chapter Mix

Class 11 Physics: Waves

Related Formula

• Wave function (moving along +x$+x$ direction) with a peak at x=0, t=0$x=0, t=0$:

• Wave number:

• Wave speed:

Core Logic

Given parameters:

• Wavelength λ = 7.5 cm$\lambda = 7.5 \ \mathrm{cm}$
• Distance travelled Δ x = 1.2 cm$\Delta x = 1.2 \ \mathrm{cm}$ in Δ t = 0.3 s$\Delta t = 0.3 \ \mathrm{s}$
• Maximum displacement (amplitude) A = 2 cm$A = 2 \ \mathrm{cm}$

Let's calculate the wave parameters:

• Wave number (k$k$):

• Wave speed (v$v$):

• Angular frequency (ω$\omega$):

Since the crest is at x=0$x=0$ at t=0$t=0$, y(0,0) = 2 = A$y(0,0) = 2 = A$. This boundary condition demands a cosine function.

Step 1: Write wave equation

Substitute A$A$, k$k$, and ω$\omega$ into the standard form:

This perfectly matches Option (1).

Pattern Recognition

Sees: Wavelength and speed to determine travelling wave equation. Trap: Choosing sine instead of cosine. Since the crest (maximum displacement) is at x=0, t=0$x=0, t=0$, y(0,0)$y(0,0)$ must equal A$A$, which is satisfied only by the cosine function. Shortcut: Calculate k = 2π / 7.5 ≈ 0.83$k = 2\pi / 7.5 \approx 0.83$. This immediately eliminates Options (3) and (4). Calculate speed v = 4$v = 4$, so ω = 4 × 0.83 ≈ 3.35$\omega = 4 \times 0.83 \approx 3.35$, which points directly to Option (1).

Chapter Mix

Class 11 Physics: Waves

Related Formula

For an air column closed at one end, the fundamental frequency f$f$ is given by:

where v$v$ is the velocity of sound and l$l$ is the length of the air column.

Core Logic

Given parameters:

• l₁ = 100~cm = 1.0~m$l_1 = 100\mathrm{~cm} = 1.0\mathrm{~m}$
• l₂ = 120~cm = 1.2~m$l_2 = 120\mathrm{~cm} = 1.2\mathrm{~m}$
• Beats per second (f₁ - f₂$f_1 - f_2$) = 15

Step 1: Write the equation for beat frequency

Since l₁ < l₂$l_1 < l_2$, the frequency f₁ > f₂$f_1 > f_2$. Hence:

Step 2: Solve for velocity of sound (v$v$)

Substitute the lengths in meters:

Pattern Recognition

Beat problems involving standing waves in organ pipes can be calculated faster by remembering that f ∝ (1)/(l)$f \propto \frac{1}{l}$. This allows setting up the proportion v = 4 · Δ f · (l₁ l₂)/(l₂ - l₁)$v = 4 \cdot \Delta f \cdot \frac{l_1 l_2}{l_2 - l_1}$ directly as a short-cut.

Chapter Mix

Class 11 Physics: Waves

Related Formula

The product of cosines can be transformed into a sum using the trigonometric identity:

The beat frequency fbeat$f_{\text{beat}}$ is given by:

The beat period Tbeat$T_{\text{beat}}$ is:

Core Logic

Rewrite the superposition equation:

Apply the identity with A = 50.5t$A = 50.5t$ and B = 1.5t$B = 1.5t$:

Here, the two component frequencies are:

Calculate the beat frequency:

Step 1: Calculate Beat Period

The time period of beats is:

Rounding to the nearest integer gives 2 ~s$2 \mathrm{~s}$.

Pattern Recognition

Sees: product of two cosines with significantly different coefficients ω₁$\omega_1$ and ω₂$\omega_2$. Shortcut: The beat period is simply 2π$2\pi$ divided by the difference between the two component frequencies, where the component frequencies are (ωaverage ± ωenvelope)$(\omega_{\text{average}} \pm \omega_{\text{envelope}})$. The difference is 2 × ωenvelope = 2 × 1.5 = 3 ~rad/s$2 \times \omega_{\text{envelope}} = 2 \times 1.5 = 3 \mathrm{~rad/s}$. Thus, T = 2π / 3 ≈ 2 ~s$T = 2\pi / 3 \approx 2 \mathrm{~s}$.

Chapter Mix

Class 11 Physics: Waves