Question

An object of specific gravity $\rho$ is hung from a thin steel wire. The fundamental frequency (in $Hz$) is $300Hz$. The object is immersed in water with specific gravity ${\rho }_w$, so that one half its volume is submerged. The new fundamental frequency (in $Hz$) is

• A. $300{\left(\frac{2\rho -{\rho }_w}{2\rho }\right)}^{\frac{1}{2}}$
• B. $300{\left(\frac{2\rho }{2\rho -{\rho }_w}\right)}^{\frac{1}{2}}$
• C. $300\left(\frac{2\rho }{2\rho -{\rho }_w}\right)$
• D. $300\left(\frac{2\rho -{\rho }_w}{2\rho }\right)$

Answer 1

The correct option is A $300{\left(\frac{2\rho -{\rho }_w}{2\rho }\right)}^{\frac{1}{2}}$

The diagramatic representation of the given problem is shown in figure. The expression of fundamental frequency is $v=\frac{1}{2l}\sqrt{\frac{T}{\mu }}$

In air $T=mg=\left(V\rho \right)g$

$\therefore v=\frac{1}{2l}\sqrt{\frac{v\rho g}{\mu }}\dots \left(1\right)$

When the object is half immersed in water

$T^{\prime }=mg-upthurst=V\rho g-\left(\frac{V}{2}\right){\rho }_{w}g$

$=\left(\frac{V}{2}\right)g(2\rho -{\rho }_w)$

The new fundamental frequency is

$v^{\prime }=\frac{1}{2l}\sqrt{\frac{\left(\frac{Vg}{2}\right)(2{\rho }_p-{\rho }_w)}{\mu }}\dots \left(2\right)$

$\therefore \frac{v^{\prime }}{v}=\left(\sqrt{\frac{(2\rho -{\rho }_w)}{2\rho }}\right)$

or $v^{\prime }=v{\left(\frac{2\rho -{\rho }_w}{2\rho }\right)}^{\frac{1}{2}}=300{\left(\frac{2\rho -{\rho }_w}{2\rho }\right)}^{\frac{1}{2}}Hz$