Question

In resonance tube experiment we find $l_1=25\text{cm}\text{and}{l}_2=75\text{cm}$. The least count of the scale used to measure $l$ is $0.1\text{cm}$. If there is no error in frequency. What will be the maximum permissiable error in speed of sound. (Take $f_0=325\text{Hz}$ )

• A. $2.2\text{m/s}$
• B. $1.3\text{m/s}$
• C. $2.6\text{m/s}$
• D. $0.65\text{m/s}$

Answer 1

The correct option is B $1.3\text{m/s}$

As we know that speed of sound, $v=2f_0(l_2-l_1)$
Given, $l_1=25\text{cm}$
$l_2=75\text{cm}$
$f_0=325\text{Hz}$
So, $dv=2f_0(d{l}_2-d{l}_1)$
$\Rightarrow (\Delta v{)}_{max}=\text{max of}\left[2f_0\right(\pm \Delta l_2\mp \Delta l_1\left)\right]$
$\Rightarrow (\Delta v{)}_{max}=2f_0(\Delta l_2+\Delta l_1)$
Where, $\Delta l_2=\text{least count of the scale}=0.1\text{cm}$
$\Delta l_1=\text{least count of the scale}=0.1\text{cm}$
$\therefore$ Maximum permissible error in speed of sound is $(\Delta v{)}_{max}=2(325\left)\right(0.2+0.2)=1.3\text{m/s}$