question_answer

A spherical ball of density p and radius R is floating in completely submerged condition in a liquid of density \[\rho \] and surface tension T. When a cavity of radius \[\frac{5R}{6}\] is created within the ball, the ball now floats half submerged in liquid. Then value of \[\frac{T}{{{R}^{2}}}\] is given by \[(\rho =1\,kg/{{m}^{3}})\]:

A)  \[\frac{45}{142}\]                                           

B)  1

C)  \[\frac{65}{162}\]                                           

D)  \[\frac{85}{162}\]

Correct Answer: D

Solution :

\[V\rho g=mg\Rightarrow \rho =\sigma \] \[a=\frac{5R}{6}\] \[\rho \frac{2}{3}\pi {{R}^{3}}g=\sigma \left( \frac{4}{3}\pi {{R}^{3}}-\frac{4}{3} \right)g+2\pi RT\] \[\Rightarrow \,\,\frac{T}{{{R}^{2}}}=\frac{85}{162}\]