Question

In resonance tube exp. we find $l_1=25.0$ cm and $l_2=75.0$ cm. If there is no error in frequency, what will be maximum permissible error in speed of sound. (Take $f_0=325Hz$)
$V=2f_0(l_2-l_1)$

• A. V = ( 329 $\pm$ 1.3) m/s
• B. V = ( 326 $\pm$ 1.3) m/s
• C. V = ( 325 $\pm$ 1.4) m/s
• D. V = ( 325 $\pm$ 1.3) m/s

Answer 1

The correct option is D V = ( 325 $\pm$ 1.3) m/s

$\left(dv\right)=2f_0(d{l}_2-d{l}_1)$
$(\Delta V{)}_{max}=maxof[2f_0(\pm \Delta l_2\mp \Delta l_2)=2f_0(\Delta l_2+\Delta l_1)$
$l_1=25.0cm\Rightarrow \Delta l_1=0.1cm$ (place value of last number)
$l_1=75.0cm\Rightarrow \Delta l_2=0.1$ cm (place value of last number)
So max permissible error in speed of sound $(\Delta V{)}_{max}$ = 2(325 Hz) (0.1 cm + 0.1 cm) = 1.3 m/s
Value of V = $2f_0(l_2-l_1)$ = 2(325 Hz) (75.0 cm - 25.0 cm) 325 m/s
So V = ( 325 $\pm$ 1.3) m/s