Mass density of sphere of radius `R` is `(K)/(r^(2))`. Where `K` is constant and `r` is distance from centre. A particle is moving near surface of sphere along circular path of radius R with time period T. Then

Options

A. `(T^(2))/(R)=` constant
B. `(T)/(R)=` constant
C. `(T)/(R^(2))=` constant
D. `(T^(2))/(R^(3))=` constant

Answer

To solve the problem, we need to establish the relationship between the time period \( T \) of a particle moving in a circular path of radius \( R \) near the surface of a sphere with a mass density given by \( \rho(r) = \frac{K}{r^2} \).

### Step 1: Determine the Mass of the Sphere
The mass density of the sphere is given as \( \rho(r) = \frac{K}{r^2} \). To find the total mass of the sphere, we need to integrate the mass density over the volume of the sphere.

The differential mass \( dm \) of a thin spherical shell of radius \( r \) and thickness \( dr \) is given by:
\[ dm = \rho(r) \cdot dV = \rho(r) \cdot 4\pi r^2 dr = \frac{K}{r^2} \cdot 4\pi r^2 dr = 4\pi K \, dr \]

To find the total mass \( M \) of the sphere, we integrate \( dm \) from \( r = 0 \) to \( r = R \):
\[ M = \int_0^R dm = \int_0^R 4\pi K \, dr = 4\pi K R \]

### Step 2: Apply Gravitational Force and Centripetal Force
The gravitational force \( F_g \) acting on the particle of mass \( m \) at the surface of the sphere is given by Newton's law of gravitation:
\[ F_g = \frac{G M m}{R^2} \]
Substituting \( M \) from Step 1:
\[ F_g = \frac{G (4\pi K R) m}{R^2} = \frac{4\pi G K m}{R} \]

The centripetal force \( F_c \) required to keep the particle moving in a circular path of radius \( R \) with angular velocity \( \omega \) is given by:
\[ F_c = m \omega^2 R \]
where \( \omega = \frac{2\pi}{T} \).

### Step 3: Set Gravitational Force Equal to Centripetal Force
Setting \( F_g = F_c \):
\[ \frac{4\pi G K m}{R} = m \left(\frac{2\pi}{T}\right)^2 R \]

Cancelling \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{4\pi G K}{R} = \frac{4\pi^2 R}{T^2} \]

### Step 4: Solve for Time Period \( T \)
Rearranging the equation gives:
\[ T^2 = \frac{\pi^2 R^2}{G K} \]

Taking the square root:
\[ T = \pi \sqrt{\frac{R^2}{G K}} = \sqrt{\frac{\pi^2 R^2}{G K}} \]

### Step 5: Establish the Relationship
From the derived equation, we can see that:
\[ \frac{T^2}{R^2} = \frac{\pi^2}{G K} \]
This shows that \( \frac{T^2}{R^2} \) is a constant.

### Final Result
Thus, the relationship between the time period \( T \) and the radius \( R \) is:
\[ \frac{T}{R} = \text{constant} \]