The velocity of sound in air is doubled when the temperature is raised from 0^circ$0^{\circ}$ C to alpha^circ$\alpha^{\circ}$ C. The value of alpha$\alpha$ is ____.

### Related Formula v = sqrtfracTmu$$v = \sqrt{\frac{T}{\mu}}$$ t = fracLv$$t = \frac{L}{v}$$ ### Core Logic Given L_A = 2.5text m$L_{A} = 2.5\text{ m}$, L_B = 1.5text m$L_{B} = 1.5\text{ m}$, T = 500text N$T = 500\text{ N}$. Velocity in string A: v_A = sqrtfracTmu_A = sqrtfrac5002 times 10^-4 = sqrt2500000 = 5 sqrt10 times 10^2text m/s$$v_A = \sqrt{\frac{T}{\mu_A}} = \sqrt{\frac{500}{2 \times 10^{-4}}} = \sqrt{2500000} = 5 \sqrt{10} \times 10^{2}\text{ m/s}$$ Velocity in string B: v_B = sqrtfracTmu_B = sqrtfrac5004 times 10^-4 = sqrt1250000 = 5 sqrt5 times 10^2text m/s$$v_B = \sqrt{\frac{T}{\mu_B}} = \sqrt{\frac{500}{4 \times 10^{-4}}} = \sqrt{1250000} = 5 \sqrt{5} \times 10^{2}\text{ m/s}$$ ### Step 1: Calculate Ratio of Times Time taken to reach joint: t_1 = fracL_Av_A = frac2.55 sqrt10 times 10^2$$t_{1} = \frac{L_{A}}{v_{A}} = \frac{2.5}{5 \sqrt{10} \times 10^{2}}$$ t_2 = fracL_Bv_B = frac1.55 sqrt5 times 10^2$$t_{2} = \frac{L_{B}}{v_{B}} = \frac{1.5}{5 \sqrt{5} \times 10^{2}}$$ Ratio: fract_1t_2 = frac2.55sqrt10 times frac5sqrt51.5 = frac2.51.5 times fracsqrt5sqrt10 = frac53 times frac1sqrt2$$\frac{t_{1}}{t_{2}} = \frac{2.5}{5\sqrt{10}} \times \frac{5\sqrt{5}}{1.5} = \frac{2.5}{1.5} \times \frac{\sqrt{5}}{\sqrt{10}} = \frac{5}{3} \times \frac{1}{\sqrt{2}}$$ fract_1t_2 = frac1.6661.414 approx 1.18$$\frac{t_{1}}{t_{2}} = \frac{1.666}{1.414} \approx 1.18$$ ### Pattern Recognition Time ratio is proportional to fracLsqrtT/mu = L sqrtmu/T$\frac{L}{\sqrt{T/\mu}} = L \sqrt{\mu/T}$. Thus t_1/t_2 = (L_1/L_2) sqrtmu_1/mu_2$t_1/t_2 = (L_1/L_2) \sqrt{\mu_1/\mu_2}$ directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves

### Related Formula f_n = n left(fracv2Lright)$$f_n = n \left(\frac{v}{2L}\right)$$ ### Core Logic For an open organ pipe, harmonic frequency v_n = n cdot f_0 = n left(fracv2Lright)$v_n = n \cdot f_0 = n \left(\frac{v}{2L}\right)$. Given v_6 - v_3 = 2200$v_6 - v_3 = 2200$ Hz: frac6v2L - frac3v2L = 2200$$\frac{6v}{2L} - \frac{3v}{2L} = 2200$$ frac3v2L = 2200$$\frac{3v}{2L} = 2200$$ Substituting speed of sound v = 330 mathrm~m/s$v = 330 \mathrm{~m/s}$: frac3 times 3302L = 2200$$\frac{3 \times 330}{2L} = 2200$$ L = frac9904400 = 0.225 mathrm~m = 225 mathrm~mm$$L = \frac{990}{4400} = 0.225 \mathrm{~m} = 225 \mathrm{~mm}$$ ### Step 1: Final Conclusion The length of the pipe is 225 mathrm~mm$225 \mathrm{~mm}$. ### Pattern Recognition Open organ pipe: Frequency difference Delta f = (6-3)f_0 = 3 f_0 = 2200 implies f_0 = 2200/3$\Delta f = (6-3)f_0 = 3 f_0 = 2200 \implies f_0 = 2200/3$. Since f_0 = v/(2L)$f_0 = v/(2L)$, solve directly for L$L$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves

### Related Formula 1. Wave function (moving along +x$+x$ direction) with a peak at x=0, t=0$x=0, t=0$: y(x, t) = A cos(kx - omega t)$$y(x, t) = A \cos(kx - \omega t)$$ 2. Wave number: k = frac2pilambda$$k = \frac{2\pi}{\lambda}$$ 3. Wave speed: v = fracomegak$$v = \frac{\omega}{k}$$ ### Core Logic Given parameters: - Wavelength lambda = 7.5 \ mathrmcm$\lambda = 7.5 \ \mathrm{cm}$ - Distance travelled Delta x = 1.2 \ mathrmcm$\Delta x = 1.2 \ \mathrm{cm}$ in Delta t = 0.3 \ mathrms$\Delta t = 0.3 \ \mathrm{s}$ - Maximum displacement (amplitude) A = 2 \ mathrmcm$A = 2 \ \mathrm{cm}$ Let's calculate the wave parameters: 1. **Wave number (k$k$):** k = frac2pi7.5 = frac20pi75 = frac4pi15 approx 0.838 \ mathrmrad/cm$$k = \frac{2\pi}{7.5} = \frac{20\pi}{75} = \frac{4\pi}{15} \approx 0.838 \ \mathrm{rad/cm}$$ 2. **Wave speed (v$v$):** v = fracDelta xDelta t = frac1.20.3 = 4 \ mathrmcm/s$$v = \frac{\Delta x}{\Delta t} = \frac{1.2}{0.3} = 4 \ \mathrm{cm/s}$$ 3. **Angular frequency (omega$\omega$):** omega = v cdot k = 4 times frac4pi15 = frac16pi15 approx 3.35 \ mathrmrad/s$$\omega = v \cdot k = 4 \times \frac{4\pi}{15} = \frac{16\pi}{15} \approx 3.35 \ \mathrm{rad/s}$$ 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 omega$\omega$ into the standard form: y(x, t) = 2 cos(0.83x - 3.35t) \ mathrmcm$$y(x, t) = 2 \cos(0.83x - 3.35t) \ \mathrm{cm}$$ 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 = 2pi / 7.5 approx 0.83$k = 2\pi / 7.5 \approx 0.83$. This immediately eliminates Options (3) and (4). Calculate speed v = 4$v = 4$, so omega = 4 times 0.83 approx 3.35$\omega = 4 \times 0.83 \approx 3.35$, which points directly to Option (1). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves

### Related Formula For an air column closed at one end, the fundamental frequency f$f$ is given by: f = fracv4l$$f = \frac{v}{4l}$$ where v$v$ is the velocity of sound and l$l$ is the length of the air column. ### Core Logic Given parameters: - l_1 = 100mathrm~cm = 1.0mathrm~m$l_1 = 100\mathrm{~cm} = 1.0\mathrm{~m}$ - l_2 = 120mathrm~cm = 1.2mathrm~m$l_2 = 120\mathrm{~cm} = 1.2\mathrm{~m}$ - Beats per second (f_1 - f_2$f_1 - f_2$) = 15 ### Step 1: Write the equation for beat frequency Since l_1 < l_2$l_1 < l_2$, the frequency f_1 > f_2$f_1 > f_2$. Hence: textBeat frequency = f_1 - f_2 = fracv4l_1 - fracv4l_2$$\text{Beat frequency} = f_1 - f_2 = \frac{v}{4l_1} - \frac{v}{4l_2}$$ 15 = fracv4 left( frac1l_1 - frac1l_2 right)$$15 = \frac{v}{4} \left( \frac{1}{l_1} - \frac{1}{l_2} \right)$$ ### Step 2: Solve for velocity of sound (v$v$) Substitute the lengths in meters: 15 = fracv4 left( frac11.0 - frac11.2 right)$$15 = \frac{v}{4} \left( \frac{1}{1.0} - \frac{1}{1.2} \right)$$ 15 = fracv4 left( 1 - frac56 right)$$15 = \frac{v}{4} \left( 1 - \frac{5}{6} \right)$$ 15 = fracv4 left( frac16 right)$$15 = \frac{v}{4} \left( \frac{1}{6} \right)$$ 15 = fracv24$$15 = \frac{v}{24}$$ v = 15 times 24 = 360mathrm~m/s$$v = 15 \times 24 = 360\mathrm{~m/s}$$ ### Pattern Recognition Beat problems involving standing waves in organ pipes can be calculated faster by remembering that f propto frac1l$f \propto \frac{1}{l}$. This allows setting up the proportion v = 4 cdot Delta f cdot fracl_1 l_2l_2 - l_1$v = 4 \cdot \Delta f \cdot \frac{l_1 l_2}{l_2 - l_1}$ directly as a short-cut. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves