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/R is constant
B. TR is constant
C. `T^(2)//R^(3)` is a constant
D. `T//R^(2)` is a constant

Explanation

To solve the problem, we need to find the relationship between the radius \( R \) of the sphere and the time period \( T \) of a particle moving near the surface of the sphere. The mass density of the sphere is given as \( \rho(r) = \frac{K}{r^2} \), where \( K \) is a constant and \( r \) is the distance from the center.

### Step-by-step Solution:

1. **Define the Mass Element**:
The mass density is given as \( \rho(r) = \frac{K}{r^2} \). To find the mass of a thin shell of radius \( r \) and thickness \( dr \), we can express the mass element \( dm \) as:
\[ dm = \rho(r) \cdot dV = \frac{K}{r^2} \cdot dV \]
The volume of the thin shell is \( dV = 4\pi r^2 dr \). Therefore, the mass element becomes:
\[ dm = \frac{K}{r^2} \cdot 4\pi r^2 dr = 4\pi K dr \]

2. **Integrate to Find Total Mass**:
To find the total mass \( M \) of the sphere, we integrate \( dm \) from \( r = 0 \) to \( r = R \):
\[ M = \int_0^R 4\pi K \, dr = 4\pi K [r]_0^R = 4\pi K R \]

3. **Centripetal Force and Gravitational Force**:
The particle moving in a circular path experiences a centripetal force that is provided by the gravitational attraction of the sphere. The gravitational force \( F_g \) acting on the particle of mass \( m \) at the surface is given by:
\[ F_g = \frac{G M m}{R^2} \]
Substituting \( M \) from the previous step:
\[ F_g = \frac{G (4\pi K R) m}{R^2} = \frac{4\pi G K m}{R} \]

4. **Centripetal Force**:
The centripetal force \( F_c \) required to keep the particle moving in a circle of radius \( R \) with angular velocity \( \omega \) is:
\[ F_c = m \omega^2 R \]
The angular velocity \( \omega \) is related to the time period \( T \) by \( \omega = \frac{2\pi}{T} \). Therefore:
\[ F_c = m \left(\frac{2\pi}{T}\right)^2 R = m \frac{4\pi^2 R}{T^2} \]

5. **Setting Forces Equal**:
Setting the gravitational force equal to the centripetal force:
\[ \frac{4\pi G K m}{R} = m \frac{4\pi^2 R}{T^2} \]
Cancelling \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{4\pi G K}{R} = \frac{4\pi^2 R}{T^2} \]

6. **Simplifying the Equation**:
Dividing both sides by \( 4\pi \):
\[ \frac{G K}{R} = \frac{\pi R}{T^2} \]
Rearranging gives:
\[ G K = \frac{\pi R^2}{T^2} \]

7. **Final Relationship**:
This can be rewritten as:
\[ \frac{R^2}{T^2} = \frac{G K}{\pi} \]
Letting \( C = \frac{G K}{\pi} \), we find:
\[ \frac{R^2}{T^2} = C \]
Thus, we conclude that:
\[ \frac{R}{T} = \sqrt{C} \]
This implies that \( \frac{R}{T} \) is a constant.