A stone hangs from the free end of a sonometer wire whose vibrating length, when tuned to a tuning fork is…
A stone hangs from the free end of a sonometer wire whose vibrating length, when tuned to a tuning fork is $40\text{ cm}$. When the stone hangs wholly immersed in water, the resonant length is reduced to $30\text{ cm}$ in same mode of oscillation. The relative density of the stone is
$\frac{16}{9}$
$\frac{16}{7}$
$\frac{16}{5}$
$\frac{16}{3}$
Solution
We have, $f \propto \frac{\sqrt{T}}{l} \Rightarrow f$ remains same.
Hence, $\frac{\sqrt{T_1}}{l_1} = \frac{\sqrt{T_2}}{l_2}$ or $\sqrt{\frac{T_1}{T_2}} = \frac{l_1}{l_2} = \frac{40}{30} = \frac{4}{3}$
or $\sqrt{\frac{w}{w - F}} = \frac{4}{3}$