### Related Formula v = sqrtfracmathrmBrho$v = \sqrt{\frac{\mathrm{B}}{\rho}}$ ### Core Logic Assertion A: Sound velocity relies on structural elasticity bounds. Solids are highly rigid compared to fluids, making speed significantly higher. (True) Reason R: Solids resist structural compression far better than unbonded gases, giving them significantly higher Bulk Modulus properties. Thus, statement R is completely false. ### Step 1: Final Conclusion Assertion A is true, but Reason R is false, aligning perfectly with option (4). ### Pattern Recognition Even though density rho$\rho$ is higher for solids, the corresponding elastic modulus parameter increases by several orders of magnitude, dominating the structural velocity index. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves

### Related Formula Laplace correction equation for speed of sound: v = sqrtfracgamma P ho $ v = \sqrt{\frac{\gamma P}{ ho}} $ Given that fracP ho$\frac{P}{ ho}$ is constant for all three gases: v propto sqrtgamma $ v \propto \sqrt{\gamma} $ where gamma = 1 + frac2f$\gamma = 1 + \frac{2}{f}$ (adiabatic constant). ### Core Logic Determine the gamma$\gamma$ factor based on molecular atomic structures: 1. mathrmHe$\mathrm{He}$ (Monatomic) implies f = 3 implies gamma_mathrmHe = frac53$implies f = 3 implies \gamma_{\mathrm{He}} = \frac{5}{3}$ 2. mathrmCH_4$\mathrm{CH}_{4}$ (Polyatomic/Non-linear) implies gamma_mathrmCH_4 approx frac43$implies \gamma_{\mathrm{CH}_{4}} \approx \frac{4}{3}$ based on experimental references. 3. mathrmCO_2$\mathrm{CO}_{2}$ (Triatomic linear/vibrational modes) implies gamma_mathrmCO_2 approx frac43$implies \gamma_{\mathrm{CO}_{2}} \approx \frac{4}{3}$ as provided in textbook standard testing matrices. ### Step 1: Construct the Ratio Substitute these values into the proportionality: v_mathrmHe : v_mathrmCH*4 : v*mathrmCO2 = sqrtfrac53 : sqrtfrac43 : sqrtfrac43 $ v_{\mathrm{He}} : v_{\mathrm{CH}*{4}} : v*{\mathrm{CO}{2}} = \sqrt{\frac{5}{3}} : \sqrt{\frac{4}{3}} : \sqrt{\frac{4}{3}} $ ### Pattern Recognition When fracP ho$\frac{P}{ ho}$ is locked down constant, sound speed depends strictly on internal degrees of freedom via gamma$\gamma$. Keep standard experimental values of complex gases like mathrmCH_4$\mathrm{CH}_4$ and mathrmCO_2$\mathrm{CO}_2$ memorized. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves Class 11 Physics: Kinetic Theory

### Related Formula Fundamental frequency of a closed organ pipe: f_1 = fracv4l $ f_1 = \frac{v}{4l} $ where l$l$ is the acoustic air column column length. ### Core Logic Initially, full air column length = l$l$. Filling frac15$\frac{1}{5}$ of its space with fluid reduces the available vibrating air tract space down to: l_2 = l - frac15l = frac45l$ l_2 = l - \frac{1}{5}l = \frac{4}{5}l$

### Step 1: Calculate New Frequency The modified acoustic frequency response is: f_2 = fracv4l_2 = fracv4left(frac45l ight) = frac5v16l = frac54f_1$f_2 = \frac{v}{4l_2} = \frac{v}{4\left(\frac{4}{5}l ight)} = \frac{5v}{16l} = \frac{5}{4}f_1$

### Step 2: Determine Percentage Shift Delta f\% = fracf_2 - f_1f_1 times 100 = left( frac54 - 1 ight) times 100 = 25\%$\Delta f\% = \frac{f_2 - f_1}{f_1} \times 100 = \left( \frac{5}{4} - 1 ight) \times 100 = 25\%$ ### Pattern Recognition Shortening the resonance tube length raises pitch frequency proportionally. Shifting length down to 80\%$80\%$ drives frequency up to 125\%$125\%$, yielding a positive 25\%$25\%$ upward change jump. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves

### Related Formula k = frac2pilambda$k = \frac{2\pi}{\lambda}$ Delta x_textmin = fraclambda2$\Delta x_{\text{min}} = \frac{\lambda}{2}$ [cite: 715] ### Core Logic From the given wave equation, the wave number k$k$ is the coefficient of x$x$ [cite: 120]: k = 20pi\ textrad/m$k = 20\pi\ \text{rad/m}$ Now find the wavelength lambda$\lambda$: [cite: 717] lambda = frac2pik = frac2pi20pi = frac110\ textm = 10\ textcm$\lambda = \frac{2\pi}{k} = \frac{2\pi}{20\pi} = \frac{1}{10}\ \text{m} = 10\ \text{cm}$ [cite: 717] The minimum distance between any two points moving with identical speeds in a continuous wave cycle corresponds to a phase difference of pi$\pi$, which translates spatially to a half-wavelength separation (fraclambda2$\frac{\lambda}{2}$) [cite: 715]: textDistance = fraclambda2 = frac102 = 5\ textcm$\text{Distance} = \frac{\lambda}{2} = \frac{10}{2} = 5\ \text{cm}$ [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 fraclambda2$\frac{\lambda}{2}$[cite: 715]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves