A sinusoidal wave of wavelength 7.5 \ mathrmcm$7.5 \ \mathrm{cm}$ travels a distance of 1.2 \ mathrmcm$1.2 \ \mathrm{cm}$ along the x-direction in 0.3 \ mathrmsec$0.3 \ \mathrm{sec}$. The crest P is at x = 0$x = 0$ at t = 0 \ mathrmsec$t = 0 \ \mathrm{sec}$ and maximum displacement of the wave is 2 \ mathrmcm$2 \ \mathrm{cm}$ . Which equation correctly represents this wave?

### 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

### Related Formula Fundamental frequency of a sonometer wire: f = frac12LsqrtfracTmu implies f propto sqrtT$$f = \frac{1}{2L}\sqrt{\frac{T}{mu}} \implies f \propto \sqrt{T}$$ Beat frequency equation: f_textbeat = |f_textfork - f_textwire|$$f_{\text{beat}} = |f_{\text{fork}} - f_{\text{wire}}|$$ ### Core Logic Let the tuning fork frequency be f$f$. Initially, it resonates with the wire at 6mathrm~N$6\mathrm{~N}$: f = f_1 = frac12Lsqrtfrac6mu$$f = f_1 = \frac{1}{2L}\sqrt{\frac{6}{mu}}$$ When the tension shifts to 54mathrm~N$54\mathrm{~N}$, the wire's new frequency becomes: f_2 = frac12Lsqrtfrac54mu$$f_2 = \frac{1}{2L}\sqrt{\frac{54}{mu}}$$ Taking the ratio of frequencies: fracf_1f_2 = sqrtfrac654 = sqrtfrac19 = frac13 implies f_2 = 3f_1 = 3f$$\frac{f_1}{f_2} = \sqrt{\frac{6}{54}} = \sqrt{\frac{1}{9}} = \frac{1}{3} \implies f_2 = 3f_1 = 3f$$ ### Step 1: Apply Beat Conditions The new frequency f_2$f_2$ creates 12 beats per second with the fork: f_2 - f = 12 implies 3f - f = 12$$f_2 - f = 12 \implies 3f - f = 12$$ 2f = 12 implies f = 6mathrm~Hz$$2f = 12 \implies f = 6\mathrm{~Hz}$$ ### Pattern Recognition Resonance establishes a direct anchor equality (f_textfork = f_textinitial$f_{\text{fork}} = f_{\text{initial}}$). Since tension changes by a factor of 9 (54/6$54/6$), the frequency scales up by a factor of 3 (sqrt9$\sqrt{9}$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves Class 11 Physics: Oscillations

### Related Formula I = fracP4pi r^2$$I = \frac{P}{4\pi r^2}$$ I propto frac1r^2$$I \propto \frac{1}{r^2}$$ ### 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 r_0 = 1mathrmm$r_0 = 1\mathrm{m}$ (usually implied if "at origin" is stated alongside a base value, though the phrasing "at the origin" is ambiguous), we can assume I_0 = 16 times 10^-8$I_0 = 16 \times 10^{-8}$. The official NTA answer assumes the initial given intensity was at r=2mathrmm$r=2\mathrm{m}$. Wait, if I = 16 times 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 propto 1/r^2$I \propto 1/r^2$, and I_1$I_1$ at r=2$r=2$ and I_2$I_2$ at r=4$r=4$: I_2 = I_1 (r_1/r_2)^2 = I_1 (2/4)^2 = I_1 / 4$I_2 = I_1 (r_1/r_2)^2 = I_1 (2/4)^2 = I_1 / 4$. Difference Delta I = I_1 - I_2 = I_1 - I_1/4 = 3 I_1 / 4$\Delta I = I_1 - I_2 = I_1 - I_1/4 = 3 I_1 / 4$. If this difference equals 3, then I_1$I_1$ must be 4$4$. But 16 times 10^-8$16 \times 10^{-8}$ is given. Wait, if I$I$ at 1mathrmm$1\mathrm{m}$ is 16 times 10^-8$16 \times 10^{-8}$: I(2mathrmm) = frac162^2 = 4 times 10^-8$I(2\mathrm{m}) = \frac{16}{2^2} = 4 \times 10^{-8}$. I(4mathrmm) = frac164^2 = 1 times 10^-8$I(4\mathrm{m}) = \frac{16}{4^2} = 1 \times 10^{-8}$. Delta I = 4 - 1 = 3 times 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 I_0 = 16 times 10^-8 mathrm~W m^-2$I_0 = 16 \times 10^{-8} \mathrm{~W m}^{-2}$ represents the reference intensity at r=1mathrmm$r=1\mathrm{m}$. Intensity at r=2 mathrm~m$r=2 \mathrm{~m}$: I_1 = fracI_02^2 = frac16 times 10^-84 = 4 times 10^-8 mathrm~W m^-2$$I_1 = \frac{I_0}{2^2} = \frac{16 \times 10^{-8}}{4} = 4 \times 10^{-8} \mathrm{~W m}^{-2}$$ Intensity at r=4 mathrm~m$r=4 \mathrm{~m}$: I_2 = fracI_04^2 = frac16 times 10^-816 = 1 times 10^-8 mathrm~W m^-2$$I_2 = \frac{I_0}{4^2} = \frac{16 \times 10^{-8}}{16} = 1 \times 10^{-8} \mathrm{~W m}^{-2}$$ ### Step 2: Calculate the Difference Magnitude of intensity difference: Delta I = |I_1 - I_2| = (4 - 1) times 10^-8 = 3 times 10^-8 mathrm~W m^-2$$\Delta I = |I_1 - I_2| = (4 - 1) \times 10^{-8} = 3 \times 10^{-8} \mathrm{~W m}^{-2}$$ ### 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 times 10^-8$16 \times 10^{-8}$ as the P/4pi$P/4\pi$ constant multiplier for the 1/r^2$1/r^2$ dropoff. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves

### Related Formula f = fracv4L quad (textclosed organ pipe)$$f = \frac{v}{4L} \quad (\text{closed organ pipe})$$ ### 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 fracv4ell_1 = 30$$\frac{v}{4\ell_1} = 30$$ frac3304ell_1 = 30$$\frac{330}{4\ell_1} = 30$$ ell_1 = frac330120 = frac114 mathrm~m$$\ell_1 = \frac{330}{120} = \frac{11}{4} \mathrm{~m}$$ ### Step 2: Calculate Final Length fracv4ell_2 = 110$$\frac{v}{4\ell_2} = 110$$ frac3304ell_2 = 110$$\frac{330}{4\ell_2} = 110$$ ell_2 = frac330440 = frac34 mathrm~m$$\ell_2 = \frac{330}{440} = \frac{3}{4} \mathrm{~m}$$ ### Step 3: Calculate Water Mass The change in length (water depth) is: Delta ell = ell_1 - ell_2 = frac114 - frac34 = frac84 = 2 mathrm~m = 200 mathrm~cm$$\Delta \ell = \ell_1 - \ell_2 = \frac{11}{4} - \frac{3}{4} = \frac{8}{4} = 2 \mathrm{~m} = 200 \mathrm{~cm}$$ Change in volume (volume of water) = Area times Delta ell$\times \Delta \ell$: V_textwater = 2 mathrm~cm^2 times 200 mathrm~cm = 400 mathrm~cm^3$$V_{\text{water}} = 2 \mathrm{~cm}^2 \times 200 \mathrm{~cm} = 400 \mathrm{~cm}^3$$ Mass of water (M$M$): M = textVolume times textDensity = 400 mathrm~cm^3 times 1 mathrm~g/cm^3 = 400 mathrm~g$$M = \text{Volume} \times \text{Density} = 400 \mathrm{~cm}^3 \times 1 \mathrm{~g/cm^3} = 400 \mathrm{~g}$$ ### Pattern Recognition Always map frequency changes to length changes using f propto 1/L$f \propto 1/L$. Poured water height is exactly the length change Delta L$\Delta L$. Watch out for CGS unit conversion (cm, grams). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves

### Related Formula v = sqrtfracgamma RTM$$v = \sqrt{\frac{\gamma RT}{M}}$$ ### Core Logic For Oxygen (O_2$O_2$) at standard temperature and pressure (S.T.P.): T = 273 text K$T = 273 \text{ K}$ M = 32 text g/mol = 32 times 10^-3 text kg/mol$M = 32 \text{ g/mol} = 32 \times 10^{-3} \text{ kg/mol}$ gamma = 1.4$\gamma = 1.4$ R = 8.3 text J K^-1 text mol^-1$R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$ ### Step 1: Calculate Velocity v = sqrtfrac1.4 times 8.3 times 27332 times 10^-3$$v = \sqrt{\frac{1.4 \times 8.3 \times 273}{32 \times 10^{-3}}}$$ v = sqrtfrac3172.2632 times 10^-3$$v = \sqrt{\frac{3172.26}{32 \times 10^{-3}}}$$ v = sqrt99.133 times 10^3$$v = \sqrt{99.133 \times 10^3}$$ v = sqrt99133 approx 314.85 text m/s$$v = \sqrt{99133} \approx 314.85 \text{ m/s}$$ Approximating to the closest given option yields 310 text m/s$310 \text{ m/s}$. ### Pattern Recognition For diatomic gases around room temp or STP, velocities range roughly from 250$250$ to 350 text m/s$350 \text{ m/s}$ depending on molar mass (N_2 approx 334$N_2 \approx 334$, O_2 approx 315$O_2 \approx 315$). Recognize 315$315$ is closest to option (1) due to standard approximations taken in exams. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves Class 11 Physics: Kinetic Theory of Gases