A point source is kept at the center of a spherically enclosed detector. If the volume of the detector increased by 8…

Q: A point source is kept at the center of a spherically enclosed detector. If the volume of the detector increased by 8 times, the intensity will

Since volume V propto R3, if volume is increased by 8 times: Since intensity I propto (1/R2): If R arrow 2R, the area A = 4pi R2 increases by 22 = 4 times (A arrow 4A). Therefore, intensity becomes I arrow (I0/4). Volume ratio cubed root dictates radius multiplier. The square of that radius multiplier is the area multiplier, which is the exact inverse of the intensity multiplier. Class 11 Physics: Waves

Year	2026	2025	2024	Total
Questions	7	10	5	22

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

I = (P)/(4π R²)

V = (4)/(3)π R³

Core Logic

Since volume $V \propto R³$ $V \propto R^3$, if volume is increased by 8 times:

V arrow 8V R³ arrow 8R³ R arrow 2R

Step 1: Intensity Relation

Since intensity $I \propto (1)/(R²)$ $I \propto \frac{1}{R^2}$: If $R arrow 2R$ $R \rightarrow 2R$, the area $A = 4π R²$ $A = 4\pi R^2$ increases by $2² = 4$ $2^2 = 4$ times ( $A arrow 4A$ $A \rightarrow 4A$). Therefore, intensity becomes $I arrow (I₀)/(4)$ $I \rightarrow \frac{I_0}{4}$.

Pattern Recognition

Volume ratio cubed root dictates radius multiplier. The square of that radius multiplier is the area multiplier, which is the exact inverse of the intensity multiplier.

Chapter Mix

Class 11 Physics: Waves

More Waves Previous-Year Questions — Page 5

Q37 jee_main_2024_31_jan_evening Speed of Sound in Gases

The speed of sound in oxygen at S.T.P. will be approximately:

(Given, R = 8.3 J K⁻¹ , γ = 1.4)

A. $310 m/s$ $310 \text{ m/s}$
B. $333 m/s$ $333 \text{ m/s}$
C. $341 m/s$ $341 \text{ m/s}$
D. $325 m/s$ $325 \text{ m/s}$

Solution

Related Formula

v = \sqrt((γ RT)/(M))

Core Logic

For Oxygen ( $O₂$ $O_2$) at standard temperature and pressure (S.T.P.): $T = 273 K$ $T = 273 \text{ K}$ $M = 32 g/mol = 32 \times 10⁻³ kg/mol$ $γ = 1.4$ $\gamma = 1.4$ $R = 8.3 J K⁻¹ mol⁻¹$

Step 1: Calculate Velocity

v = 1.4 \times 8.3 \times 27332 \times 10⁻³

v = 3172.2632 \times 10⁻³

v = \sqrt(99.133 \times 10³)

v = \sqrt(99133) \approx 314.85 m/s

Approximating to the closest given option yields $310 m/s$ $310 \text{ m/s}$.

Pattern Recognition

For diatomic gases around room temp or STP, velocities range roughly from $250$ $250$ to $350 m/s$ $350 \text{ m/s}$ depending on molar mass ( $N₂ \approx 334$ $N_2 \approx 334$, $O₂ \approx 315$ $O_2 \approx 315$). Recognize $315$ $315$ is closest to option (1) due to standard approximations taken in exams.

Chapter Mix

Class 11 Physics: Waves Class 11 Physics: Kinetic Theory of Gases

Q jee_main_2024_31_jan_morning Organ 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

60~cm

, the length of the closed pipe will be:

A. $60~cm$ $60\mathrm{~cm}$
B. $45~cm$ $45\mathrm{~cm}$
C. $30~cm$ $30\mathrm{~cm}$
D. $15~cm$ $15\mathrm{~cm}$

Solution

Related Formula

fclosed, fundamental = (v)/(4Lc)

fopen, 1st overtone = (2v)/(2Lₒ)

Core Logic

Organ Pipes diagram for Q45 - JEE Main 2024 Morning

For a closed organ pipe, the fundamental frequency (1st harmonic) is:

f₁ = (v)/(λ) = (v)/(4L₁)

where $L₁$ $L_1$ is the length of the closed pipe.

For an open organ pipe, the first overtone (2nd harmonic) is:

f₂ = (2v)/(2L₂) = (v)/(L₂)

where $L₂$ $L_2$ is the length of the open pipe ( $L₂ = 60 cm$ $L_2 = 60\mathrm{\,cm}$).

Step 2: Equating Frequencies

Given $f₁ = f₂$ $f_1 = f_2$:

(v)/(4L₁) = (v)/(L₂)

$L₂ = 4L₁$ $L_2 = 4L_1$

60 = 4 \times L₁

L₁ = 15 cm

Chapter Mix

Class 11 Physics: Waves

A point source is kept at the center of a spherically enclosed detector. If the volume of the detector increased by 8 times, the intensity will

Solution & Explanation

Related Formula

Core Logic

Step 1: Intensity Relation

Pattern Recognition

Chapter Mix

Reference Study Guides

More Waves Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Step 1: Calculate Velocity

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 2: Equating Frequencies

Chapter Mix

More Waves Questions — jee_main_2026_24_january_evening

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

A point source is kept at the center of a spherically enclosed detector. If the volume of the detector increased by 8 times, the intensity will

Solution & Explanation

Related Formula

Core Logic

Step 1: Intensity Relation

Pattern Recognition

Chapter Mix

Reference Study Guides

More Waves Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Step 1: Calculate Velocity

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 2: Equating Frequencies

Chapter Mix

More Waves Questions — jee_main_2026_24_january_evening

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back