Question

The air bubble formed by explosion inside water perform oscillations with time period T which depends on pressure (p), density (ρ) and on energy due to explosion (E). Establish relation between T, p, E and ρ.

Answer 1

Answer:

Explanation:

T  PaρbEc   where, symbols have their usual meaning.

Writing in terms of dimensions,

[T] = [ML-1T-2]a[ML-3]b[ML2T-2]c

Comparing the exponents of M:

0 = a+b+c ... (1)

Comparing exponents of T:

1 = -2a-2c

a+c = -1/2     .... (2)

Comparing exponents of L:

0 = -a-3b+2c  ... (3)

Using (2) in (1) gives,

b = 1/2,

Hence (3) becomes,

a - 2c = -3/2    .........(4)

Solving (2) and (4), we find,

c = 1/3 and a = -5/6.

Hence,

T = C ( √ρ × ∛E ) ÷ P⁵/⁶

where C is a constant.

Answer 2

Answer:

The relation between T, p, E, and ρ is $T = C (\sqrt{\rho} \times \sqrt[3]{E} )$ ÷$P^{\frac{5}{6} }$, where $C$ is a constant.

Explanation:

In the given question, T, p, E, and ρ symbols have their usual meanings.

The dimensional equation for T will be:

$[T] = [ML^{-1} T^{-2} ]a \ [ML^{-3} ]b \ [ML^{2} T^{-2} ]c$

Comparing the exponents of M on both sides we get,

$0 = a+b+c \ ..... \ (1)$

Comparing the exponents of T on both sides we get,

$1 = -2a-2c$

$a + c = -\frac{1}{2} \ ...... \ (2)$

Comparing exponents of L on both sides we get,

$0 = -a-3 +2c \ .....\ (3)$

From using $(2)$ in $(1)$ we get,

$b = \frac{1}{2}$,

Hence, $(3)$ will now become,

$a - 2c = - \frac{3}{2} \ ..... \ (4)$

By solving $(2)$ and $(4)$ we get,

$c = \frac{1}{3}$ and $a = -\frac{5}{6}$

Therefore, the relation between  T, p, E, and ρ is $T = C (\sqrt{\rho} \times \sqrt[3]{E} )$ ÷$P^{\frac{5}{6} }$, where $C$ is a constant.

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