Q132 / 155 · WBJEE 2018 · Physics

A spherical liquid drop is placed on a horizontal plane. A small disturbance causes the volume of the drop to oscillate. The time period of oscillation ( $T$ ) of the liquid drop depends on radius $(n$ of the drop, density $(\rho)$ and surface tension (5) of the liquid. Which among the following will be a possible expression for $T$ (where, $k$ is a dimensionless constant)?

  • A
    $k \sqrt{\frac{\rho r}{S}}$
  • B
    $k \sqrt{\frac{\rho^{2} r}{S}}$
  • Correct answer
    C
    $k \sqrt{\frac{\rho r^{3}}{S}}$
  • D
    $k \sqrt{\frac{\rho r^{3}}{s^{2}}}$

Solution

According to the question. time period. $T \alpha r^{a} p^{b} s^{c}$ $r=k r^{a} p^{b} s^{c}$ Thus, putting dimension, we get Equating the dimensions of both sides, we get $[T]=[\mathrm{L}]^{a}\left[\mathrm{ML}^{-3}]^ \mathrm{b}\left[\mathrm{MT}^{-2} ]^c\right.\right.$ $[T]=[M]^{b+c} \cdot[L]^{2a-3b} [T]^{-2c}$ $b+c=0, a-3 b=0$ and $-2 x=$ $\begin{array}{l} c=-\frac{1}{2} \\ b=\frac{1}{2} \\ a=3 b=\frac{3}{2} \end{array}$ Putting these value of $a, b$ and $c$ into Eq. (i), we get $T=k \cdot r^{\frac{3}{2}} p^{\frac{1}{2}} \cdot s^{-\frac{1}{2}}$ $=k \sqrt{\frac{\rho}{s} r^{3}}$

Hence, the correct answer is option: 'C'.

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