A student is performing the experiment of resonance column. The diameter of the column tube is 4 ~cm . The frequency of the tuning fork is 512 ~Hz . The air temperature is 38^ C in which the speed of sound is 336 ~m / s

Correct answer: B. 15.2 ~cm Considering the end correction [e=0.3 D where D= diameter ] f=n [(v)/(4(l+e))] For first resonance, n=1 f=(v)/(4(l+0.3 D)) l=(v)/(4 f)-0.3 D or, l=((336 × 100)/(4 × 512))-0.3 × 4=15.2 ~cm

JEE Advanced 2012Paper 2

Exam: JEE Advanced
Year: 2012
Paper: Paper 2
Subject: Physics
Chapter: Waves and Sound

JEE Advanced 2012 · Paper 2 · Physics · Waves and Sound

A student is performing the experiment of resonance column. The...

A student is performing the experiment of resonance column. The diameter of the column tube is

4 \mathrm{~cm}

. The frequency of the tuning fork is

512 \mathrm{~Hz}

. The air temperature is

38^{\circ} \mathrm{C}

in which the speed of sound is

336 \mathrm{~m} / \mathrm{s}

. The zero of the meter scale coincides with the top end of the resonance column tube. When the first resonance occurs, the reading of the water level in the column is

Options

A $14.0 \mathrm{~cm}$
B $15.2 \mathrm{~cm}$
C $16.4 \mathrm{~cm}$
D $17.6 \mathrm{~cm}$

Correct answer

B. $15.2 \mathrm{~cm}$

Step-by-step solution

Considering the end correction

[e=0.3 D

where

\mathrm{D}=

diameter

]

f=n

\left[\frac{v}{4(l+e)}\right]

For first resonance,

n=1 \therefore f=\frac{v}{4(l+0.3 D)} \Rightarrow l=\frac{v}{4 f}-0.3 D

or,

l=\left(\frac{336 \times 100}{4 \times 512}\right)-0.3 \times 4=15.2 \mathrm{~cm}

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