Search any question & find its solution
Question:
Answered & Verified by Expert
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)?
Options:
Solution:
1286 Upvotes
Verified Answer
The correct answer is:
$k \sqrt{\frac{\rho r^{3}}{S}}$
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
$T=k \cdot r^{\frac{3}{2}} p^{\frac{1}{2}} \cdot s^{-\frac{1}{2}}$
$=k \sqrt{\frac{\rho}{s} r^{3}}$
$r=k r^{a} p^{b} s^{c}$
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
$T=k \cdot r^{\frac{3}{2}} p^{\frac{1}{2}} \cdot s^{-\frac{1}{2}}$
$=k \sqrt{\frac{\rho}{s} r^{3}}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.