Question

A heavy brass sphere is hung from a spiral spring and it executes vertical vibration with period T. the ball is now immersed in nonviscous liquid with a density one-tenth that of brass.when set into vertical vibration with the sphere remaining inside the liquid all the time, the period will be

• A. $\left[\frac{9}{10}\right]T$
• B. $T\sqrt{\left(\frac{10}{9}\right)}$
• C. $unchanged$
• D. $T\sqrt{\left(\frac{9}{10}\right)}$

Answer 1

The correct option is B $T\sqrt{\left(\frac{10}{9}\right)}$

As the brass ball is always immersed in the non-viscous liquid the buoyant force will always be actively applied on it

Then,

$F_B={\rho }_{nvf}.Vg$, where ${\rho }_{nvf}=$density of non-viscous fluid

${\rho }_{nvf}=\frac{{\rho }_b}{10}$ (${\rho }_b=$ density of brass; $m=$ mass of brass ball)

$F_B=\frac{{\rho }_b}{10}\left(\frac{m}{{\rho }_b}\right)g\Rightarrow F_B=\frac{mg}{10}$

Accordingly the effective downward force on the ball is

$F_{eff}=mg-\frac{mg}{10}$ i.e., weight = buoyant force

$F_{eff}=m(g-\frac{g}{10})=m\left(\frac{g}{10}\right)=m{g}_{eff}\Rightarrow g_{eff}=\frac{9g}{10}$

As we know new time period will be

$T^{\prime }=2\pi \sqrt{\frac{l}{g_{eff}}}=2\pi \sqrt{\frac{l10}{9g}}=\sqrt{\frac{10}{9}}\left(2\pi \sqrt{\frac{l}{g}}\right)=\sqrt{\frac{10}{9}}T$

$T^{\prime }=\sqrt{\frac{10}{9}}T$