The speed of sound in oxygen at S.T.P. will be approximately:
(textGiven, R = 8.3 text J K^-1 text, gamma = 1.4)$(\text{Given, } R = 8.3 \text{ J K}^{-1} \text{, } \gamma = 1.4)$
A.310 text m/s$310 \text{ m/s}$
B.333 text m/s$333 \text{ m/s}$
C.341 text m/s$341 \text{ m/s}$
D.325 text m/s$325 \text{ m/s}$
Solution & Explanation
### 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
Keywords:#speed of sound#JEE Main 2024 Evening Q37#Waves JEE Main 2024#Speed of Sound in Gases JEE Main 2024
More Waves Previous-Year Questions — Page 4
Q45jee_main_2024_31_jan_morningOrgan Pipes
The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is 60mathrm~cm$60\mathrm{~cm}$, the length of the closed pipe will be:
A.60mathrm~cm$60\mathrm{~cm}$
B.45mathrm~cm$45\mathrm{~cm}$
C.30mathrm~cm$30\mathrm{~cm}$
D.15mathrm~cm$15\mathrm{~cm}$
Solution
### Related Formula
f_textclosed, fundamental = fracv4L_c$$f_{\text{closed, fundamental}} = \frac{v}{4L_c}$$f_textopen, 1st overtone = frac2v2L_o$$f_{\text{open, 1st overtone}} = \frac{2v}{2L_o}$$
### Core Logic
Organ Pipes diagram for Q45 - JEE Main 2024 MorningOrgan Pipes diagram for Q45 - JEE Main 2024 Morning
For a closed organ pipe, the fundamental frequency (1st harmonic) is:
f_1 = fracvlambda = fracv4L_1$$f_1 = \frac{v}{\lambda} = \frac{v}{4L_1}$$
where L_1$L_1$ is the length of the closed pipe.
For an open organ pipe, the first overtone (2nd harmonic) is:
f_2 = frac2v2L_2 = fracvL_2$$f_2 = \frac{2v}{2L_2} = \frac{v}{L_2}$$
where L_2$L_2$ is the length of the open pipe (L_2 = 60mathrm\,cm$L_2 = 60\mathrm{\,cm}$).
### Step 2: Equating Frequencies
Given f_1 = f_2$f_1 = f_2$:
fracv4L_1 = fracvL_2$$\frac{v}{4L_1} = \frac{v}{L_2}$$L_2 = 4L_1$L_2 = 4L_1$60 = 4 times L_1$$60 = 4 \times L_1$$L_1 = 15mathrm\,cm$$L_1 = 15\mathrm{\,cm}$$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Waves
More Waves Questions — jee_main_2024_31_jan_evening
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