Question

The time period of oscillation of a gas bubble from an explosion underwater depends upon p d and e where he is pressure Di is density and he is energy produced the relationship by the method of dimensions

Answer 1

you mean, we have to find relation between p, d He and T by the method of dimensions.

dimension of p = [ML-¹T-²]

dimension of d = [ML-³]

dimension of He = [ML²T-²]

dimension of T = [T]

from dimensional analysis,

$T=kp^xd^yHe^z$, where k is proportionality constant.

or, $[T]=k[ML^{-1}T^{-2}]^x[ML^{-3}]^y[ML^2T^{-2}]^z$

or, $[T]=k[M^{(x+y+z)}L^{(-x-3y+2z)}T^{-2z}]$

on comparing both sides,

x + y + z = 0....(1)

-x - 3y + 2z = 0.....(2)

-2z = 1 .....(3)

from equations (3) and (1),

x + y = 1/2 ....(4)

from equations (2) and (3),

x + 3y = -1 ......(5)

now from equations (4) and (5),

y = -3/4 and x = 5/4

so, $T=kp^{5/4}d^{-3/4}He^{-1/2}$

or, $T=k\sqrt[4]{\frac{p^5}{d^3(He)^2}}$ This is required relation

Answer 2

Answer:

Explanation:

Given time period  

T

T

∝

p

a

d

b

E

c

T

=

K

p

a

d

b

E

c

......(1)

where  

K

is a constant of proportionality and dimensionless quantity.

Inserting the dimensions of Time, pressure, density and Energy in equation (1) we get

[

T

]

=

[

M

L

−

1

T

−

2

]

a

[

M

L

−

3

]

b

[

M

L

2

T

−

2

]

c

Equating powers of  

M

,

L

,

and

T

on both sides we get

0

=

a

+

b

+

c

.....(2)

0

=

–

a

–

3

b

+

2

c

.....(3)

1

=

–

2

a

–

2

c

......(4)

Solving these equations

From (4)

a

+

c

=

−

1

2

.....(5)

Inserting this value in (2)

0

=

b

−

1

2

⇒

b

=

1

2

From (5)

a

=

−

1

2

−

c

Inserting values of  

a

and

b

in (3)

0

=

−

(

−

1

2

−

c

)

−

3

×

1

2

+

2

c

⇒

3

c

=

1

⇒

c

=

1

3

Inserting value of  

c

in (5)

a

+

1

3

=

−

1

2

⇒

a

=

−

1

2

−

1

3

⇒

a

=

−

5

6

∴

a

=

−

5

6

,

b

=

1

2

and

c

=

1

3