Waves Previous Year Questions — JEE Main Physics
22 past-year Waves questions from JEE Main (Physics) — page 3.
Q57 (2024)
A tuning fork resonates with a sonometer wire of length $1\mathrm{~m}$ stretched with a tension of $6\mathrm{~N}$. When the tension in the wire is changed to $54\mathrm{~N}$, the same tuning fork produces 12 beats per second with it. The frequency of the tuning fork is _______ $\mathrm{Hz}$.
### Related Formula
Fundamental frequency of a sonometer wire:
$$f = \frac{1}{2L}\sqrt{\frac{T}{mu}} \implies f \propto \sqrt{T}$$
Beat frequency equation:
$$f_{\text{beat}} = |f_{\text{fork}} - f_{\text{wire}}|$$
### Core Logic
Let the tuning fork frequency be $f$. Initially, it resonates with the wire at $6\mathrm{~N}$:
$$f = f_1 = \frac{1}{2L}\sqrt{\frac{6}{mu}}$$
When the tension shifts to $54\mathrm{~N}$, the wire's new frequency becomes:
$$f_2 = \frac{1}{2L}\sqrt{\frac{54}{mu}}$$
Taking the ratio of frequencies:
$$\frac{f_1}{f_2} = \sqrt{\frac{6}{54}} = \sqrt{\frac{1}{9}} = \frac{1}{3} \implies f_2 = 3f_1 = 3f$$
### Step 1: Apply Beat Conditions
The new frequency $f_2$ creates 12 beats per second with the fork:
$$f_2 - f = 12 \implies 3f - f = 12$$
$$2f = 12 \implies f = 6\mathrm{~Hz}$$
### Pattern Recognition
Resonance establishes a direct anchor equality ($f_{\text{fork}} = f_{\text{initial}}$). Since tension changes by a factor of 9 ($54/6$), the frequency scales up by a factor of 3 ($\sqrt{9}$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Waves
Class 11 Physics: Oscillations
Q1739 (2024)
A point source is emitting sound waves of intensity $16 \times 10^{-8} \mathrm{~W m}^{-2}$ at the origin. The difference in intensity (magnitude only) at two points located at a distance of $2 \mathrm{~m}$ and $4 \mathrm{~m}$ from the origin respectively will be ________ $\times 10^{-8} \mathrm{~W m}^{-2}$.
### Related Formula
$$I = \frac{P}{4\pi r^2}$$
$$I \propto \frac{1}{r^2}$$
### Core Logic
For a point source, the intensity $I$ is inversely proportional to the square of the distance from the source.
If the given intensity is at $r_0 = 1\mathrm{m}$ (usually implied if "at origin" is stated alongside a base value, though the phrasing "at the origin" is ambiguous), we can assume $I_0 = 16 \times 10^{-8}$.
The official NTA answer assumes the initial given intensity was at $r=2\mathrm{m}$. Wait, if $I = 16 \times 10^{-8}$ was the source power factor or the intensity at $r=2$, let's reverse-engineer the answer `3`.
If $I \propto 1/r^2$, and $I_1$ at $r=2$ and $I_2$ at $r=4$:
$I_2 = I_1 (r_1/r_2)^2 = I_1 (2/4)^2 = I_1 / 4$.
Difference $\Delta I = I_1 - I_2 = I_1 - I_1/4 = 3 I_1 / 4$.
If this difference equals 3, then $I_1$ must be $4$. But $16 \times 10^{-8}$ is given.
Wait, if $I$ at $1\mathrm{m}$ is $16 \times 10^{-8}$:
$I(2\mathrm{m}) = \frac{16}{2^2} = 4 \times 10^{-8}$.
$I(4\mathrm{m}) = \frac{16}{4^2} = 1 \times 10^{-8}$.
$\Delta I = 4 - 1 = 3 \times 10^{-8}$.
This perfectly matches.
### Step 1: Calculate Intensity at r = 2m and r = 4m
Assume the intensity $I_0 = 16 \times 10^{-8} \mathrm{~W m}^{-2}$ represents the reference intensity at $r=1\mathrm{m}$.
Intensity at $r=2 \mathrm{~m}$:
$$I_1 = \frac{I_0}{2^2} = \frac{16 \times 10^{-8}}{4} = 4 \times 10^{-8} \mathrm{~W m}^{-2}$$
Intensity at $r=4 \mathrm{~m}$:
$$I_2 = \frac{I_0}{4^2} = \frac{16 \times 10^{-8}}{16} = 1 \times 10^{-8} \mathrm{~W m}^{-2}$$
### Step 2: Calculate the Difference
Magnitude of intensity difference:
$$\Delta I = |I_1 - I_2| = (4 - 1) \times 10^{-8} = 3 \times 10^{-8} \mathrm{~W m}^{-2}$$
### Pattern Recognition
Although the question text ("at the origin") was poorly drafted (intensity at $r=0$ would be infinite), the numerical setup expects you to treat $16 \times 10^{-8}$ as the $P/4\pi$ constant multiplier for the $1/r^2$ dropoff
BY NTA 3
BY RANKBIT (BONUS)
Question is wrong as data is incomplete.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Waves
Q59 (2024)
In a closed organ pipe, the frequency of fundamental note is $30 \mathrm{~Hz}$. A certain amount of water is now poured in the organ pipe so that the fundamental frequency is increased to $110 \mathrm{~Hz}$. If the organ pipe has a cross-sectional area of $2 \mathrm{~cm}^2$, the amount of water poured in the organ tube is ______ g. (Take speed of sound in air is $330 \mathrm{~m/s}$)
### Related Formula
$$f = \frac{v}{4L} \quad (\text{closed organ pipe})$$
### 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
$$\frac{v}{4\ell_1} = 30$$
$$\frac{330}{4\ell_1} = 30$$
$$\ell_1 = \frac{330}{120} = \frac{11}{4} \mathrm{~m}$$
### Step 2: Calculate Final Length
$$\frac{v}{4\ell_2} = 110$$
$$\frac{330}{4\ell_2} = 110$$
$$\ell_2 = \frac{330}{440} = \frac{3}{4} \mathrm{~m}$$
### Step 3: Calculate Water Mass
The change in length (water depth) is:
$$\Delta \ell = \ell_1 - \ell_2 = \frac{11}{4} - \frac{3}{4} = \frac{8}{4} = 2 \mathrm{~m} = 200 \mathrm{~cm}$$
Change in volume (volume of water) = Area $\times \Delta \ell$:
$$V_{\text{water}} = 2 \mathrm{~cm}^2 \times 200 \mathrm{~cm} = 400 \mathrm{~cm}^3$$
Mass of water ($M$):
$$M = \text{Volume} \times \text{Density} = 400 \mathrm{~cm}^3 \times 1 \mathrm{~g/cm^3} = 400 \mathrm{~g}$$
### Pattern Recognition
Always map frequency changes to length changes using $f \propto 1/L$. Poured water height is exactly the length change $\Delta L$. Watch out for CGS unit conversion (cm, grams).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Waves
Q37 (2024)
The speed of sound in oxygen at S.T.P. will be approximately:
$(\text{Given, } R = 8.3 \text{ J K}^{-1} \text{, } \gamma = 1.4)$
- $310 \text{ m/s}$
- $333 \text{ m/s}$
- $341 \text{ m/s}$
- $325 \text{ m/s}$
### Related Formula
$$v = \sqrt{\frac{\gamma RT}{M}}$$
### Core Logic
For Oxygen ($O_2$) at standard temperature and pressure (S.T.P.):
$T = 273 \text{ K}$
$M = 32 \text{ g/mol} = 32 \times 10^{-3} \text{ kg/mol}$
$\gamma = 1.4$
$R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$
### Step 1: Calculate Velocity
$$v = \sqrt{\frac{1.4 \times 8.3 \times 273}{32 \times 10^{-3}}}$$
$$v = \sqrt{\frac{3172.26}{32 \times 10^{-3}}}$$
$$v = \sqrt{99.133 \times 10^3}$$
$$v = \sqrt{99133} \approx 314.85 \text{ m/s}$$
Approximating to the closest given option yields $310 \text{ m/s}$.
### Pattern Recognition
For diatomic gases around room temp or STP, velocities range roughly from $250$ to $350 \text{ m/s}$ depending on molar mass ($N_2 \approx 334$, $O_2 \approx 315$). Recognize $315$ 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
Q1995 (2024)
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 $60\mathrm{~cm}$, the length of the closed pipe will be:
- $60\mathrm{~cm}$
- $45\mathrm{~cm}$
- $30\mathrm{~cm}$
- $15\mathrm{~cm}$
### Related Formula
$$f_{\text{closed, fundamental}} = \frac{v}{4L_c}$$
$$f_{\text{open, 1st overtone}} = \frac{2v}{2L_o}$$
### Core Logic
{{SOL_IMG1}}
{{SOL_IMG2}}
For a closed organ pipe, the fundamental frequency (1st harmonic) is:
$$f_1 = \frac{v}{\lambda} = \frac{v}{4L_1}$$
where $L_1$ is the length of the closed pipe.
For an open organ pipe, the first overtone (2nd harmonic) is:
$$f_2 = \frac{2v}{2L_2} = \frac{v}{L_2}$$
where $L_2$ is the length of the open pipe ($L_2 = 60\mathrm{\,cm}$).
### Step 2: Equating Frequencies
Given $f_1 = f_2$:
$$\frac{v}{4L_1} = \frac{v}{L_2}$$
$L_2 = 4L_1$
$$60 = 4 \times L_1$$
$$L_1 = 15\mathrm{\,cm}$$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Waves
