Question

The time period T of vibration of a liquid drop depends on surface tension S radius r of drop and density d of the liquid Find a relation for T ​

Answer 1

Hey mate,.

Time of oscillation,

t α p^a α r^k α σ^c

So, t = k(p^a)(r^k)(σ^c)

Writing dimensions of both sides,

[T] = [ML^-3]^a [L]^b [MT^-2]^c

= [M^(a+c) L^(-3a+b) T^(-2c)]

Comparing powers of M,L and T on both sides, we have

a + c = 0 …(i)

-3a + b = 0 …(ii)

-2c = 1 …(iii)

Solving eqs (i), (ii) and (iii), we get,

a = 1/2, c= -1/2, b = 3/2

Putting the values in t

t = k(p^1/2)(r^3/2)(σ^-1/2)

t α √[(pr^3)/σ]

Hope it will help you

Answer 2

Answer:

Time of oscillation,

t α p^a α r^k α σ^c

So, t = k(p^a)(r^k)(σ^c)

Writing dimensions of both sides,

[T] = [ML^-3]^a [L]^b [MT^-2]^c

= [M^(a+c) L^(-3a+b) T^(-2c)]

Comparing powers of M,L and T on both sides, we have

a + c = 0 …(i)

-3a + b = 0 …(ii)

-2c = 1 …(iii)

Solving eqs (i), (ii) and (iii), we get,

a = 1/2, c= -1/2, b = 3/2

Putting the values in t

t = k(p^1/2)(r^3/2)(σ^-1/2)

t α √[(pr^3)/σ]