In the resonance experiment, two air columns (closed at one end) of 100mathrm~cm$100\mathrm{~cm}$ and 120mathrm~cm$120\mathrm{~cm}$ long, give 15 beats per second when each one is sounding in the respective fundamental modes. The velocity of sound in the air column is :

### Related Formula v = sqrtfracTmu$$v = \sqrt{\frac{T}{\mu}}$$ mu = rho A = rho left(pi R^2right)$$\mu = \rho A = \rho \left(\pi R^2\right)$$ where, v$v$ = velocity of transverse wave on a string T$T$ = tension on string mu$\mu$ = linear mass density rho$\rho$ = density of material A$A$ = cross-sectional area ### Core Logic Since both strings are made of the same material, their density rho$\rho$ is the same. Also, they are stretched to have the same amount of tension T$T$. Substitute the formula for mu$\mu$ into the velocity equation: v = sqrtfracTrho pi R^2 = frac1R sqrtfracTrho pi$$v = \sqrt{\frac{T}{\rho \pi R^2}} = \frac{1}{R} \sqrt{\frac{T}{\rho \pi}}$$ Thus, the wave velocity is inversely proportional to the radius of the cross section of the string: v propto frac1R$$v \propto \frac{1}{R}$$ fracv_2v_1 = fracR_1R_2$$\frac{v_2}{v_1} = \frac{R_1}{R_2}$$ ### Step 1: Ratio Computation Given: - R_1 = R$R_1 = R$ - R_2 = R/2$R_2 = R/2$ Substitute these values: fracv_2v_1 = fracRR/2 = 2$$\frac{v_2}{v_1} = \frac{R}{R/2} = 2$$ ### Pattern Recognition Sees: Circular cross-section strings + transverse wave speed relation. Shortcut: Wave velocity v propto frac1sqrtmu propto frac1R$v \propto \frac{1}{\sqrt{\mu}} \propto \frac{1}{R}$. If the radius is halved, the mass per unit length decreases by 4 times, which makes the speed increase by sqrt4 = 2$\sqrt{4} = 2$ times. ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves

### 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 x(t) = Acos(omega t + phi)$$x(t) = A\cos(\omega t + \phi)$$ v = fracomegaK$$v = \frac{\omega}{K}$$ K = frac2pilambda$$K = \frac{2\pi}{\lambda}$$ ### Core Logic From the given wave equation, angular frequency omega = 628text rad/s$\omega = 628\text{ rad/s}$. Given wave velocity v = 300text m/s$v = 300\text{ m/s}$. Using the relation v = fracomegaK$v = \frac{\omega}{K}$: 300 = frac628K implies K = frac628300$$300 = \frac{628}{K} \implies K = \frac{628}{300}$$ ### Step 1: Compute Wavelength Substitute K = frac2pilambda$K = \frac{2\pi}{\lambda}$: frac2pilambda = frac628300$$\frac{2\pi}{\lambda} = \frac{628}{300}$$ Since 2pi approx 2 times 3.14 = 6.28$2\pi \approx 2 \times 3.14 = 6.28$, the expression simplifies neatly: frac6.28lambda = frac628300 implies lambda = 3text m$$\frac{6.28}{\lambda} = \frac{628}{300} \implies \lambda = 3\text{ m}$$ ### Pattern Recognition Notice standard values like omega = 628 = 200pi$\omega = 628 = 200\pi$, which means the frequency is exactly 100text Hz$100\text{ Hz}$. Using v = flambda implies 300 = 100lambda implies lambda = 3text m$v = f\lambda \implies 300 = 100\lambda \implies \lambda = 3\text{ m}$ avoids setting up fractions. ### 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