Question

Hola Brainlians. Here's a question to test your Physics. Give correct answer only.

A large fluid star oscillates in shape under the influence of its own gravitational field. Using dimensional analysis, fund the expression for period of oscillation (T) in terms of radius of star (R), mean density of fluid (p) and universal gravitational constant G.

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Answer 1

★ Concept :-

Here the concept of Dimensional Analysis has been used. We see that we are given different relations of period of oscillations with different quantities. Firstly we can make equation which shows proportionality between the units and then applying constant we can make main equation. Then after applying dimensional analysis we can find the answer.

Let's do it !!

___________________________________________

★ Solution :-

Given,

Time Period of oscillation of fluid (T) depends on

» Radius of Star (R)

» Density of Fluid (ρ)

» Universal Gravitational Constant (G)

From this we can formulate that,

$\;\bf{\mapsto\;\;\green{T\;\;\propto\;\;R^{a}\:\rho^{b}\:G^{c}}}$

This can be written as,

$\;\bf{\mapsto\;\;\orange{T\;\;=\;\;K\:R^{a}\:\rho^{b}\:G^{c}}}$

where K is a dimensionless constant. K doesn't have any dimension, so we can ignore it in dimensional analysis. Here K is multiplied to get equality in expression.

Here a, b and c are the exponential powers to which R, ρ and G are raised to.

Let that equation be equation i) .

___________________________________________

~ For Dimensional Analysis of Quantities :-

• For Dimension of T :-

We know that T is the time period so it's dimensional formula will the standard dimension of Time.

→ [T] = T

Here T represents standard dimension of Time.

$\;\tt{\leadsto\;\;\red{[T]\;=\;[M^{0}\:L^{0}\:T^{1}]}}$

• For Dimension of R :-

We know that R is the radius of star so it's dimensional formula will be the standard dimension of Length.

→ [R] = [L]

$\;\tt{\leadsto\;\;\red{[R]\;=\;[M^{0}\:L^{1}\:T^{0}]}}$

• For Dimension of ρ :-

We know that ρ is the density of fluid which is ratio of mass and volume so it's dimensional formula will be the standard dimension of mass by (length)³ .

→ [ρ] = [M / L³]

$\;\tt{\leadsto\;\;\red{[\rho]\;=\;[M^{1}\:L^{-3}\:T^{0}]}}$

• For Dimension of G :-

We know that G is the gravitational constant which is the ratio of product of Force and (distance)² by products of two masses. So it's dimensional formula will be (length)³ by mass and (time)² .

→ [G] = [L³ / M T²]

$\;\tt{\leadsto\;\;\red{[G]\;=\;[M^{-1}\:L^{3}\:T^{-2}]}}$

___________________________________________

~ For Dimensional Analysis of Equation :-

From equation i) we know that,

$\;\sf{\rightarrow\;\;T\;\;=\;\;\bf{K\:R^{a}\:\rho^{b}\:G^{c}}}$

By applying Dimensional Analysis method, we get

$\;\sf{\rightarrow\;\;[T]\;\;=\;\;\bf{K\:[R]^{a}\:[\rho]^{b}\:[G]^{c}}}$

By applying dimensional formulas, we get

$\;\sf{\rightarrow\;\;[M^{0}\:L^{0}\:T^{1}]\;\;=\;\;\bf{K\:[M^{0}\:L^{1}\:T^{0}]^{a}\:[M^{1}\:L^{-3}\:T^{0}]^{b}\:[M^{-1}\:L^{3}\:T^{-2}]^{c}}}$

By using Laws of Exponents, we get

$\;\sf{\rightarrow\;\;[M^{0}\:L^{0}\:T^{1}]\;\;=\;\;\bf{K\:[L^{a}]\:[M^{b}\:L^{-3b}]\:[M^{-c}\:L^{3c}\:T^{-2c}]}}$

By removing dimensional analysis from both sides, we get

$\;\sf{\rightarrow\;\;M^{0}\:L^{0}\:T^{1}\;\;=\;\;\bf{K\:L^{a}\:M^{b}\:L^{-3b}\:M^{-c}\:L^{3c}\:T^{-2c}}}$

By using Laws of Exponents, we get

$\;\sf{\rightarrow\;\;M^{0}\:L^{0}\:T^{1}\;\;=\;\;\bf{K\:L^{-3b\:+\:3c\:+\:a}\:M^{b\:-\:c}\:T^{-2c}}}$

--------------------------------------------------------

Now comparing the powers of LHS and RHS and equating them, we get

✒ b - c = 0 , -3b + 3c + a = 0 , -2c = 1

✒ b = c , a = 3(b - c) , c = - ½

✒ b = - ½ , a = 0 , c = - ½

--------------------------------------------------------

Now equating these powers in equation i) , we get

$\;\sf{\Longrightarrow\;\;T\;\;=\;\;\bf{K\:R^{0}\:\rho^{-1/2}\:G^{-1/2}}}$

$\;\sf{\Longrightarrow\;\;T\;\;=\;\;\bf{K\:\times\:1\:\times\:\dfrac{1}{\rho^{1/2}}\:\times\:\dfrac{1}{G^{1/2}}}}$

$\;\sf{\Longrightarrow\;\;T\;\;=\;\;\bf{K\:\times\:1\:\times\:\dfrac{1}{\sqrt{\rho}}\:\times\:\dfrac{1}{\sqrt{G}}}}$

$\;\sf{\Longrightarrow\;\;T\;\;=\;\;\bf{\blue{K\:\dfrac{1}{\sqrt{\rho\:G}}}}}$

This is the required answer.

We can leave K as it is because it's the constant value.

$\;\underline{\boxed{\tt{Required\;\:expression\;\:for\;\:T\;=\;\bf{\purple{K\:\dfrac{1}{\sqrt{\rho\:G}}}}}}}$

___________________________________________

★ More to know :-

• Dimensional Formula of Time = [T]

• Dimensional Formula of Length= [L]

• Dimensional Formula of amount of substance = [mol]

• Dimensional Formula of intensity of light = [cd]

• Dimensional Formula of Mass = [M]

• Dimensional Formula of Time = [T]

• Dimensional Formula of Current = [A]

Answer 2

Answer:

Concept :-

Here the concept of Dimensional Analysis has been used. We see that we are given different relations of period of oscillations with different quantities. Firstly we can make equation which shows proportionality between the units and then applying constant we can make main equation. Then after applying dimensional analysis we can find the answer.

Let's do it !!

___________________________________________

★ Solution :-

Given,

Time Period of oscillation of fluid (T) depends on

» Radius of Star (R)

» Density of Fluid (ρ)

» Universal Gravitational Constant (G)

From this we can formulate that,

\;\bf{\mapsto\;\;\green{T\;\;\propto\;\;R^{a}\:\rho^{b}\:G^{c}}}↦T∝R

a

ρ

b

G

c

This can be written as,

\;\bf{\mapsto\;\;\orange{T\;\;=\;\;K\:R^{a}\:\rho^{b}\:G^{c}}}↦T=KR

a

ρ

b

G

c

where K is a dimensionless constant. K doesn't have any dimension, so we can ignore it in dimensional analysis. Here K is multiplied to get equality in expression.

Here a, b and c are the exponential powers to which R, ρ and G are raised to.

Let that equation be equation i) .

___________________________________________

~ For Dimensional Analysis of Quantities :-

• For Dimension of T :-

We know that T is the time period so it's dimensional formula will the standard dimension of Time.

→ [T] = T

Here T represents standard dimension of Time.

\;\tt{\leadsto\;\;\red{[T]\;=\;[M^{0}\:L^{0}\:T^{1}]}}⇝[T]=[M

0

L

0

T

1

]

• For Dimension of R :-

We know that R is the radius of star so it's dimensional formula will be the standard dimension of Length.

→ [R] = [L]

\;\tt{\leadsto\;\;\red{[R]\;=\;[M^{0}\:L^{1}\:T^{0}]}}⇝[R]=[M

0

L

1

T

0

]

• For Dimension of ρ :-

We know that ρ is the density of fluid which is ratio of mass and volume so it's dimensional formula will be the standard dimension of mass by (length)³ .

→ [ρ] = [M / L³]

\;\tt{\leadsto\;\;\red{[\rho]\;=\;[M^{1}\:L^{-3}\:T^{0}]}}⇝[ρ]=[M

1

L

−3

T

0

]

• For Dimension of G :-

We know that G is the gravitational constant which is the ratio of product of Force and (distance)² by products of two masses. So it's dimensional formula will be (length)³ by mass and (time)² .

→ [G] = [L³ / M T²]

\;\tt{\leadsto\;\;\red{[G]\;=\;[M^{-1}\:L^{3}\:T^{-2}]}}⇝[G]=[M

−1

L

3

T

−2

]

___________________________________________

~ For Dimensional Analysis of Equation :-

From equation i) we know that,

\;\sf{\rightarrow\;\;T\;\;=\;\;\bf{K\:R^{a}\:\rho^{b}\:G^{c}}}→T=KR

a

ρ

b

G

c

By applying Dimensional Analysis method, we get

\;\sf{\rightarrow\;\;[T]\;\;=\;\;\bf{K\:[R]^{a}\:[\rho]^{b}\:[G]^{c}}}→[T]=K[R]

a

[ρ]

b

[G]

c

By applying dimensional formulas, we get

\;\sf{\rightarrow\;\;[M^{0}\:L^{0}\:T^{1}]\;\;=\;\;\bf{K\:[M^{0}\:L^{1}\:T^{0}]^{a}\:[M^{1}\:L^{-3}\:T^{0}]^{b}\:[M^{-1}\:L^{3}\:T^{-2}]^{c}}}→[M

0

L

0

T

1

]=K[M

0

L

1

T

0

]

a

[M

1

L

−3

T

0

]

b

[M

−1

L

3

T

−2

]

c

By using Laws of Exponents, we get

\;\sf{\rightarrow\;\;[M^{0}\:L^{0}\:T^{1}]\;\;=\;\;\bf{K\:[L^{a}]\:[M^{b}\:L^{-3b}]\:[M^{-c}\:L^{3c}\:T^{-2c}]}}→[M

0

L

0

T

1

]=K[L

a

][M

b

L

−3b

][M

−c

L

3c

T

−2c

]

By removing dimensional analysis from both sides, we get

\;\sf{\rightarrow\;\;M^{0}\:L^{0}\:T^{1}\;\;=\;\;\bf{K\:L^{a}\:M^{b}\:L^{-3b}\:M^{-c}\:L^{3c}\:T^{-2c}}}→M

0

L

0

T

1

=KL

a

M

b

L

−3b

M

−c

L

3c

T

−2c

By using Laws of Exponents, we get

\;\sf{\rightarrow\;\;M^{0}\:L^{0}\:T^{1}\;\;=\;\;\bf{K\:L^{-3b\:+\:3c\:+\:a}\:M^{b\:-\:c}\:T^{-2c}}}→M

0

L

0

T

1

=KL

−3b+3c+a

M

b−c

T

−2c

--------------------------------------------------------

Now comparing the powers of LHS and RHS and equating them, we get

✒ b - c = 0 , -3b + 3c + a = 0 , -2c = 1

✒ b = c , a = 3(b - c) , c = - ½

✒ b = - ½ , a = 0 , c = - ½

--------------------------------------------------------

Now equating these powers in equation i) , we get

\;\sf{\Longrightarrow\;\;T\;\;=\;\;\bf{K\:R^{0}\:\rho^{-1/2}\:G^{-1/2}}}⟹T=KR

0

ρ

−1/2

G

−1/2

\;\sf{\Longrightarrow\;\;T\;\;=\;\;\bf{K\:\times\:1\:\times\:\dfrac{1}{\rho^{1/2}}\:\times\:\dfrac{1}{G^{1/2}}}}⟹T=K×1×

ρ

1/2

1

×

G

1/2

1

\;\sf{\Longrightarrow\;\;T\;\;=\;\;\bf{K\:\times\:1\:\times\:\dfrac{1}{\sqrt{\rho}}\:\times\:\dfrac{1}{\sqrt{G}}}}⟹T=K×1×

ρ

1

×

G

1

\;\sf{\Longrightarrow\;\;T\;\;=\;\;\bf{\blue{K\:\dfrac{1}{\sqrt{\rho\:G}}}}}⟹T=K

ρG

1

This is the required answer.

We can leave K as it is because it's the constant value.

\;\underline{\boxed{\tt{Required\;\:expression\;\:for\;\:T\;=\;\bf{\purple{K\:\dfrac{1}{\sqrt{\rho\:G}}}}}}}

RequiredexpressionforT=K

ρG

1

___________________________________________

★ More to know :-

• Dimensional Formula of Time = [T]

• Dimensional Formula of Length= [L]

• Dimensional Formula of amount of substance = [mol]

• Dimensional Formula of intensity of light = [cd]

• Dimensional Formula of Mass = [M]

• Dimensional Formula of Time = [T]

• Dimensional Formula of Current = [A]