Question

The time-period of oscillation T of a small drop of liquid under surface tension S (force/length) depends on the radius of the drop r, surface tension S and the density of the liquid d. Find an expression for T dimensionally in terms of the quantities given.

Answer 1

Let expression is T = $\bold{r^a.S^b.d^c}$
Here, r is the radius of drop, S is the surface tension and d is the Density of liquid .
Dimension of T = [ T ]
dimension of r = [ L ]
dimension of S = [MT⁻²]
dimension of d = [ ML⁻³]

[ T ] = [L]^a [MT⁻²]^b [ML⁻³]^c
[T] = [L]^(a -3c) [M]^(b + c) [T]^(-2b)]
Compare both sides,
b = -1/2 , c = 1/2 and a = 3c = 3/2

Hence, expression is T = r^(3/2)S^(-1/2)d^(1/2)
e.g., $\bold{T = \sqrt{\frac{r^3d}{S}}}$

Answer 2

Explanation:

You follow

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