Question

A stone is hung in the air from a wire which is stretched over a sonometer. The bridge of the sonometer is 40 cm apart when the wire is in unison with a tuning fork of frequency 256. When the stone is completely immersed in water the length between the bridges is 22 cm for re-establishing (same mode) unison with the same tuning fork. The specific gravity of the material of the stone is

• A. $\frac{{\left(40\right)}^2}{{\left(40\right)}^2+{\left(22\right)}^2}$
• B. $\frac{{\left(40\right)}^2}{{\left(40\right)}^2-{\left(22\right)}^2}$
• C. $256\times \frac{22}{40}$
• D. $256\times \frac{40}{22}$

Answer 1

The correct option is B $\frac{{\left(40\right)}^2}{{\left(40\right)}^2-{\left(22\right)}^2}$

Frequency of vibration of stretched string

$f=\frac{1}{2\left(L\right)}\sqrt{\frac{T}{m}}$

When he stone is completely immersed in water, length changes but frequency doesn't.

$\frac{L}{\ell }=\sqrt{\frac{T_{air}}{T_{water}}}\frac{L}{\ell }=\sqrt{\frac{V\rho g}{V(\rho -1)g}}$

where density of stone=$\rho$

density of water=$\ell$

$\frac{L}{l}=\sqrt{\frac{\rho }{\rho -1}}\frac{L^2}{{\ell }^2}=\frac{\rho }{\rho -1}{L}^2\rho -L^2=\rho {\ell }^2\Rightarrow \rho =\frac{L^2}{L^2-{\ell }^2}$

Thus the specific gravity of the stone

$\rho =\frac{L^2}{L^2-{\ell }^2}[\rho =\frac{{\left(40\right)}^2}{{\left(40\right)}^2-{\left(22\right)}^2}]$

Hence option (B) is correct.