back to directory
NEET PHYSICSWAVE OPTICSMedium

Question

A tuning fork with a frequency of 800 Hz800 \text{ Hz} produces resonance in a resonance column tube with the upper end open and the lower end closed by the water surface. Successive resonances are observed at lengths of 9.75 cm9.75 \text{ cm}, 31.25 cm31.25 \text{ cm}, and 52.75 cm52.75 \text{ cm}. The speed of the sound in the air is:

A

500 m/s500 \text{ m/s}

B

156 m/s156 \text{ m/s}

C

344 m/s344 \text{ m/s}

D

172 m/s172 \text{ m/s}

Step-by-Step Solution

In a resonance column tube (which acts as a closed organ pipe), the difference between two successive resonance lengths is equal to half the wavelength (λ/2\lambda/2). Given successive resonance lengths: l1=9.75 cml_1 = 9.75 \text{ cm} l2=31.25 cml_2 = 31.25 \text{ cm} l3=52.75 cml_3 = 52.75 \text{ cm} We can find the wavelength λ\lambda: λ2=l2l1=31.25 cm9.75 cm=21.5 cm\frac{\lambda}{2} = l_2 - l_1 = 31.25 \text{ cm} - 9.75 \text{ cm} = 21.5 \text{ cm} (Also verified by l3l2=52.7531.25=21.5 cml_3 - l_2 = 52.75 - 31.25 = 21.5 \text{ cm}) λ=2×21.5 cm=43.0 cm=0.43 m\lambda = 2 \times 21.5 \text{ cm} = 43.0 \text{ cm} = 0.43 \text{ m} Given the frequency of the tuning fork, f=800 Hzf = 800 \text{ Hz}, the speed of sound vv is: v=fλ=800 Hz×0.43 m=344 m/sv = f\lambda = 800 \text{ Hz} \times 0.43 \text{ m} = 344 \text{ m/s}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WAVE OPTICS. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSWAVE OPTICStuningfrequencyproducesresonanceresonance

More WAVE OPTICS Questions

View all

The fundamental frequency in an open organ pipe is equal to the third harmonic of a closed organ pipe. If the length of the closed organ pipe is $20 \text{ cm}$, the length of the open organ pipe is:

A.$13.2 \text{ cm}$
B.$8 \text{ cm}$
C.$12.5 \text{ cm}$
D.$16 \text{ cm}$
MediumSolve

When a string is divided into three segments of lengths $l_1, l_2$ and $l_3$, the fundamental frequencies of these three segments are $\nu_1, \nu_2$ and $\nu_3$ respectively. The original fundamental frequency ($\nu$) of the string is:

A.$\sqrt{\nu}=\sqrt{\nu_1}+\sqrt{\nu_2}+\sqrt{\nu_3}$
B.$\nu=\nu_1+\nu_2+\nu_3$
C.$\frac{1}{\nu}=\frac{1}{\nu_1}+\frac{1}{\nu_2}+\frac{1}{\nu_3}$
D.$\frac{1}{\sqrt{\nu}}=\frac{1}{\sqrt{\nu_1}}+\frac{1}{\sqrt{\nu_2}}+\frac{1}{\sqrt{\nu_3}}$
MediumSolve

The wave described by $y = 0.25\sin(10\pi x - 2\pi t)$, where $x$ and $y$ are in metres and $t$ in seconds, is a wave traveling along the:

A.-ve $x$ direction with frequency $1 \text{ Hz}$.
B.+ve $x$ direction with frequency $\pi \text{ Hz}$ and wavelength $\lambda = 0.2 \text{ m}$.
C.+ve $x$ direction with frequency $1 \text{ Hz}$ and wavelength $\lambda = 0.2 \text{ m}$.
D.-ve $x$ direction with amplitude $0.25 \text{ m}$ and wavelength $\lambda = 0.2 \text{ m}$.
EasySolve

The two nearest harmonics of a tube closed at one end and open at the other end are $220 \text{ Hz}$ and $260 \text{ Hz}$. What is the fundamental frequency of the system?

A.$10 \text{ Hz}$
B.$20 \text{ Hz}$
C.$30 \text{ Hz}$
D.$40 \text{ Hz}$
MediumSolve

A wave travelling in the positive x-direction having maximum displacement along y-direction as $1 \text{ m}$, wavelength $2\pi \text{ m}$ and frequency of $1/\pi \text{ Hz}$ is represented by

A.$y=\sin(x-2t)$
B.$y=\sin(2\pi x-2\pi t)$
C.$y=\sin(10\pi x-20\pi t)$
D.$y=\sin(2\pi x+2\pi t)$
EasySolve

An air column, closed at one end and open at the other, resonates with a tuning fork when the smallest length of the column is $50 \text{ cm}$. The next larger length of the column resonating with the same tuning fork is

A.$100 \text{ cm}$
B.$150 \text{ cm}$
C.$200 \text{ cm}$
D.$66.7 \text{ cm}$
EasySolve

If we study the vibration of a pipe open at both ends, then the following statement is not true:

A.Open end will be anti-node
B.Odd harmonics of the fundamental frequency will be generated
C.All harmonics of the fundamental frequency will be generated
D.Pressure change will be maximum at both ends
MediumSolve

A uniform rope of length $L$ and mass $m_1$ hangs vertically from a rigid support. A block of mass $m_2$ is attached to the free end of the rope. A transverse pulse of wavelength $\lambda_1$ is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is $\lambda_2$. The ratio $\frac{\lambda_2}{\lambda_1}$ is:

A.$\sqrt{\frac{m_1+m_2}{m_1}}$
B.$\sqrt{\frac{m_2}{m_1}}$
C.$\sqrt{\frac{m_1+m_2}{m_2}}$
D.$\sqrt{\frac{m_1}{m_2}}$
MediumSolve

This neet physics practice question is part of the TopperSquare free question bank. TopperSquare offers 15,000+ chapter-wise NEET MCQs across Physics, Chemistry, and Biology with detailed step-by-step explanations, full mock tests, NEET PYQs (2010–2024), and an AI-powered performance analytics dashboard. browse all neet practice questions → · practice physics sets →