Get Answers to all your Questions

header-bg qa

In an experiment to determine the velocity of sound in the air at room temperature using a resonance is observed.  When the air column has a length of \small x \mathrm{~m} for a tuning fork of frequency \small 800\ \mathrm{H}\mathrm{z} is \small 4\ \mathrm{sec}. The velocity of the Sound at room temperature is \small 340 \mathrm{~m} / \mathrm{s} . The third resonance is observed when the air column has a length 0.62 . Then \small x= ?

Option: 1

0.408


Option: 2

4.08


Option: 3

40.8


Option: 4

0.00408


Answers (1)

best_answer

\begin{aligned} & \delta=\frac{V}{4\left(d_1+e\right)} \\ & l_1+e=\frac{V}{4 f} \\ & x+e=\frac{340}{4 \times 800} \end{aligned}

\begin{aligned} & x+e-\frac{17}{2 \times 80}=\frac{17}{160}=0.106 \\ & x+e=0.106 \quad e q-\text { (1) } \\ & l_2+e=\frac{3 v}{4 f} \\ & 0.62+e=\frac{3 \times 340}{4 \times 800}=\frac{51}{160}=0.318 \\ & 0.62+e=0.318 \\ & e=0.318-0.62 \end{aligned}

\begin{aligned} & e=-0.302 \\ & e q(1) \\ & x-0.302=0.106 \\ & x=0.106+0.302 \\ & x=0.408 \end{aligned}

 

 

 

Posted by

Devendra Khairwa

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE