If the gravitational field in the space is given as `-K/r^(2)` Taking the reference point to be at `r=2cm` with gravitational potential `V=10J/kg`.Find the gravitational potential at `r=3cm` in SI unit (Given,that `K=6Jcm/s` )

Options

A. `11`
B. `10
C. `12`
D. `9`

Answer

To find the gravitational potential at \( r = 3 \, \text{cm} \) given the gravitational field \( E = -\frac{K}{r^2} \) and the reference point at \( r = 2 \, \text{cm} \) with gravitational potential \( V = 10 \, \text{J/kg} \), we can follow these steps:

### Step 1: Understand the relationship between gravitational field and potential
The gravitational field \( E \) is related to the gravitational potential \( V \) by the equation:
\[ E = -\frac{dV}{dr} \]
Given \( E = -\frac{K}{r^2} \), we can set up the equation:
\[ -\frac{dV}{dr} = -\frac{K}{r^2} \]
This simplifies to:
\[ \frac{dV}{dr} = \frac{K}{r^2} \]

### Step 2: Integrate to find the potential
To find \( V \), we need to integrate \( \frac{dV}{dr} \):
\[ dV = \frac{K}{r^2} dr \]
Integrating both sides, we have:
\[ \int_{10}^{V} dV = \int_{2}^{3} \frac{K}{r^2} dr \]

### Step 3: Perform the integration
The integral on the left side is straightforward:
\[ V - 10 \]
The integral on the right side becomes:
\[ \int \frac{K}{r^2} dr = -\frac{K}{r} \quad \text{(evaluated from 2 to 3)} \]
Thus,
\[ V - 10 = -K \left( \frac{1}{3} - \frac{1}{2} \right) \]

### Step 4: Calculate the limits
Calculating the difference:
\[ \frac{1}{3} - \frac{1}{2} = \frac{2 - 3}{6} = -\frac{1}{6} \]
So we have:
\[ V - 10 = K \cdot \frac{1}{6} \]

### Step 5: Substitute the value of \( K \)
Given \( K = 6 \, \text{J cm/s} \), we substitute:
\[ V - 10 = 6 \cdot \frac{1}{6} = 1 \]

### Step 6: Solve for \( V \)
Now, we can solve for \( V \):
\[ V = 10 + 1 = 11 \, \text{J/kg} \]

### Step 7: Convert to SI units
Since we need the answer in SI units, we convert \( 11 \, \text{J/kg} \) to the appropriate unit:
\[ 11 \, \text{J/kg} = 11 \, \text{J/kg} \quad \text{(remains the same as it is already in SI)} \]

### Final Answer
The gravitational potential at \( r = 3 \, \text{cm} \) is:
\[ \boxed{11 \, \text{J/kg}} \]