Question

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 \((r)\) of the drop, density \((\rho)\) and surface tension (s) of the liquid. Which among the following will be a possible expression for \(\mathrm{T}\) (where \(\mathrm{k}\) is a dimensionless constant) ?
(A) \(k \sqrt{\frac{\rho r}{s}}\)
(B) \(k \sqrt{\frac{\rho^{2} r}{s}}\)
(C) \(k \sqrt{\frac{\rho r^{3}}{s}}\)
(D) \(k \sqrt{\frac{\rho r^{3}}{s^{2}}}\)

Answer 1

Ans(C)

 \(\mathrm{T}=\mathrm{kr}^{\mathrm{a}} \rho^{\mathrm{b}} \mathrm{s}^{\mathrm{c}}, \mathrm{T}=\mathrm{K}[\mathrm{L}]^{\mathrm{a}}\left[\mathrm{ML}^{-3}\right]^{\mathrm{b}}\left[\mathrm{MT}^{-2}\right]^{\mathrm{c}},\left[\mathrm{M}^{\circ} \mathrm{L}^{\circ} \mathrm{T}^{1}\right]=\mathrm{K}[\mathrm{M}]^{\mathrm{b}+\mathrm{c}}[\mathrm{L}]^{\mathrm{a}-3 \mathrm{~b}}[\mathrm{~T}]^{-2 \mathrm{c}}\)
$$
\begin{array}{c}-2 \mathrm{c}=1 \\ \mathrm{~b}+\mathrm{c}=-1 / 2\end{array} \quad \mathrm{a}-3 \mathrm{~b}=0
$$
\(\mathrm{~b}=-\mathrm{c}\)\(\quad \mathrm{a}=3 \mathrm{~b}\)
\(\mathrm{~b}=1 / 2\)
\(\mathrm{~T}=\mathrm{Kr}^{3 / 2} \rho^{1 / 2} \mathrm{~s}^{-1 / 2}=\mathrm{T}=\mathrm{K} \sqrt{\frac{\mathrm{r}^{3} \rho}{\mathrm{s}}}\)

Answer 2

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Answer 3

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