Question

The frequency (n) of an oscilating drop depands on radius (r), density (p) and surface tention (s). write the relation.

Answer 1

Solution:

Given:

=> The frequency (n) of an oscillating drop depends on radius (r), density (p) and surface tension (s).

To Find:

=> Relation.

Formula used:

=> Dimensional analysis.

So,

$\sf{\implies n \propto (r)^{x}(p)^{y}(s)^{z}}$

$\sf{\implies n = k(r)^{x}(p)^{y}(s)^{z}\;\;\;\;...........(1)}$

Now, put the dimensional formula of the following.

$\sf{\implies [M^{0}L^{0}T^{-1}]=[M^{0}L^{1}T^{0}]^{X}[M^{1}L^{-3}T^{0}]^{Y}[M^{1}L^{0}T^{-2}]^{Z}}$

Now, compare these,

=> 'M' = 0 = 0 + y + z         ..........(2)

=> 'L' = 0 = x - 3y + 0        ...........(3)

=> 'T' = 1 = 0 + 0 - 2z        ...........(4)

Now, from Equation (4), we get

=> -2z = 1

=> z = -1/2

From Equation (2), we get

=> y + z = 0

=> y - 1/2 = 0

=> y = 1/2

From Equation (3), we get

=> x - 3(1/2) = 0

=> x - 3/2 = 0

=> x = 3/2

Now, put the values of x, y and z in Equation (1), we get

$\sf{\implies n = k(r)^{\frac{3}{2}}(p)^{\frac{1}{2}}(s)^{-\frac{1}{2}}}$

$\large{\boxed{\boxed{\green{\sf{\implies n = k\sqrt{\dfrac{s}{r^{3}p}}}}}}}$

Answer 2

$\huge\bf{\underline{\blue{Solution\: is\: attached\:here.}}}$