Q. If the time period $(T)$ of vibration of a liquid drop depends on surface tension ($S$), radius $(r)$ of the drop and density $(\rho)$ of the liquid, then the expression of $T$ is
A) $T = k\sqrt{{\rho r}^{3}/S}$
B) $T = k\sqrt{\rho^{1/2}r^{3}/S}$
C) $T = k\sqrt{\rho r^{3}/S^{1/2}}$
D) None of these
Explanation
Let $T \propto S^{x}r^{y}p^{z}$ By substituting the dimensions of $\lbrack T\rbrack = \lbrack T\rbrack$ $\lbrack S\rbrack = \left\lbrack MT^{- 2} \right\rbrack,\lbrack r\rbrack = \lbrack L\rbrack,\lbrack\rho\rbrack = \left\lbrack ML^{- 3} \right\rbrack$ and by comparing the power of both the sides $x = - 1/2,y = 3/2,z = 1/2$ so $\ T\ \propto \sqrt{\rho r^{3}/S\ } \Rightarrow T = k\sqrt{\frac{\rho r^{3}}{S}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.