Physics

The time of oscillation $$T$$ of a small drop of liquid depends on radius $$r$$, density $$\rho$$ and surface tension $$S$$. The relation between them is given by


ANSWER

$$T\propto \sqrt {\dfrac {\rho r^3}{S}}$$


SOLUTION
Let $$T=S^ar^b{ \rho  }^{ c }$$
So, $$[T]=[M^0L^0T^1]$$
             $$=[M^{1}L^{-1}T^{-2}]^a[L^{1}]^b[ML^{-3}]^c$$
             $$=[M^{a+c}L^{b-3c}T^{-2a}]$$

Applying principle of homogeneity, 
$$a+c=0; b-3c=0; -2a=1$$
On solving, we get $$a=-1/2. b=3/2, c=1/2$$
$$T\propto \sqrt {\dfrac {\rho r^3}{S}}$$
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Single Correct Medium Published on 18th 08, 2020
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