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

Fundamental frequency of a closed organ pipe:

where l$l$ is the length of the resonating air column.

Core Logic

Initially, the full air column length is l$l$. Filling (1)/(5)$\frac{1}{5}$ of its space with fluid reduces the available vibrating air column length to:

Step 1: Calculate New Frequency

The modified acoustic frequency response is:

Step 2: Determine Percentage Shift

Pattern Recognition

Shortening the resonance tube length raises pitch frequency inversely (f ∝ (1)/(l)$f \propto \frac{1}{l}$). Reducing the air column length to 80%$80\%$ of its original value drives the frequency up to 125%$125\%$, yielding a positive 25%$25\%$ increase.

Evaluation Rubric / Model Answer

Option A: 25%

Chapter Mix

Class 11 Physics: Waves

Related Formula

Δ xmin = (λ)/(2)$$\Delta x_{\text{min}} = \frac{\lambda}{2}$$ [cite: 715]

Core Logic

From the given wave equation, the wave number k$k$ is the coefficient of x$x$ [cite: 120]:

Now find the wavelength λ$\lambda$: [cite: 717]

λ = (2π)/(k) = (2π)/(20π) = (1)/(10) m = 10 cm$$\lambda = \frac{2\pi}{k} = \frac{2\pi}{20\pi} = \frac{1}{10}\ \text{m} = 10\ \text{cm}$$ [cite: 717]

The minimum distance between any two points moving with identical speeds in a continuous wave cycle corresponds to a phase difference of π$\pi$, which translates spatially to a half-wavelength separation ((λ)/(2)$\frac{\lambda}{2}$) [cite: 715]:

Distance = (λ)/(2) = (10)/(2) = 5 cm$$\text{Distance} = \frac{\lambda}{2} = \frac{10}{2} = 5\ \text{cm}$$ [cite: 717]

Pattern Recognition

Oscillating speed configuration reaches identical value fields twice per spatial wavelength cycle[cite: 715]. Thus, minimum distance separation scales exactly to (λ)/(2)$\frac{\lambda}{2}$[cite: 715].

Chapter Mix

Class 11 Physics: Waves

Related Formula

Fundamental frequency of a sonometer wire:

Beat frequency equation:

Core Logic

Let the tuning fork frequency be f$f$. Initially, it resonates with the wire at 6~N$6\mathrm{~N}$:

When the tension shifts to 54~N$54\mathrm{~N}$, the wire's new frequency becomes:

Taking the ratio of frequencies:

Step 1: Apply Beat Conditions

The new frequency f₂$f_2$ creates 12 beats per second with the fork:

Pattern Recognition

Resonance establishes a direct anchor equality (ffork = finitial$f_{\text{fork}} = f_{\text{initial}}$). Since tension changes by a factor of 9 (54/6$54/6$), the frequency scales up by a factor of 3 (√(9)$\sqrt{9}$).

Chapter Mix

Class 11 Physics: Waves Class 11 Physics: Oscillations

Related Formula

Core Logic

For a point source, the intensity I$I$ is inversely proportional to the square of the distance from the source. If the given intensity is at r₀ = 1m$r_0 = 1\mathrm{m}$ (usually implied if "at origin" is stated alongside a base value, though the phrasing "at the origin" is ambiguous), we can assume I₀ = 16 × 10⁻⁸$I_0 = 16 \times 10^{-8}$. The official NTA answer assumes the initial given intensity was at r=2m$r=2\mathrm{m}$. Wait, if I = 16 × 10⁻⁸$I = 16 \times 10^{-8}$ was the source power factor or the intensity at r=2$r=2$, let's reverse-engineer the answer `3`. If I ∝ 1/r²$I \propto 1/r^2$, and I₁$I_1$ at r=2$r=2$ and I₂$I_2$ at r=4$r=4$: I₂ = I₁ (r₁/r₂)² = I₁ (2/4)² = I₁ / 4$I_2 = I_1 (r_1/r_2)^2 = I_1 (2/4)^2 = I_1 / 4$. Difference Δ I = I₁ - I₂ = I₁ - I₁/4 = 3 I₁ / 4$\Delta I = I_1 - I_2 = I_1 - I_1/4 = 3 I_1 / 4$. If this difference equals 3, then I₁$I_1$ must be 4$4$. But 16 × 10⁻⁸$16 \times 10^{-8}$ is given. Wait, if I$I$ at 1m$1\mathrm{m}$ is 16 × 10⁻⁸$16 \times 10^{-8}$: I(2m) = (16)/(2²) = 4 × 10⁻⁸$I(2\mathrm{m}) = \frac{16}{2^2} = 4 \times 10^{-8}$. I(4m) = (16)/(4²) = 1 × 10⁻⁸$I(4\mathrm{m}) = \frac{16}{4^2} = 1 \times 10^{-8}$. Δ I = 4 - 1 = 3 × 10⁻⁸$\Delta I = 4 - 1 = 3 \times 10^{-8}$. This perfectly matches.

Step 1: Calculate Intensity at r = 2m and r = 4m

Assume the intensity I₀ = 16 × 10⁻⁸ ~W m⁻²$I_0 = 16 \times 10^{-8} \mathrm{~W m}^{-2}$ represents the reference intensity at r=1m$r=1\mathrm{m}$. Intensity at r=2 ~m$r=2 \mathrm{~m}$:

Intensity at r=4 ~m$r=4 \mathrm{~m}$:

Step 2: Calculate the Difference

Magnitude of intensity difference:

Pattern Recognition

Although the question text ("at the origin") was poorly drafted (intensity at r=0$r=0$ would be infinite), the numerical setup expects you to treat 16 × 10⁻⁸$16 \times 10^{-8}$ as the P/4π$P/4\pi$ constant multiplier for the 1/r²$1/r^2$ dropoff BY NTA 3 BY RANKBIT (BONUS) Question is wrong as data is incomplete.

Chapter Mix

Class 11 Physics: Waves

Related Formula

Core Logic

Adding water to a closed organ pipe reduces its effective air column length, thereby increasing the fundamental frequency. The change in length corresponds directly to the volume of water poured.

Step 1: Calculate Initial Length

Step 2: Calculate Final Length

Step 3: Calculate Water Mass

The change in length (water depth) is:

Change in volume (volume of water) = Area × Δ$\times \Delta \ell$:

Mass of water (M$M$):

Pattern Recognition

Always map frequency changes to length changes using f ∝ 1/L$f \propto 1/L$. Poured water height is exactly the length change Δ L$\Delta L$. Watch out for CGS unit conversion (cm, grams).

Chapter Mix

Class 11 Physics: Waves