There are two bodies of masses 1 kg and 100 kg reporated by a distance 1m. At what distance from the smaller body, the intensity of gravitational field will be zero

Options

A. `(1)/(9)m`
B. `(1)/(10)m`
C. `(1)/(11)m`
D. `(10)/(11)m`

Answer

To solve the problem of finding the distance from the smaller body (1 kg) where the intensity of the gravitational field is zero, we can follow these steps:

### Step 1: Understand the setup
We have two bodies:
- Mass \( m_1 = 1 \) kg (smaller body)
- Mass \( m_2 = 100 \) kg (larger body)
- Distance between the two bodies \( d = 1 \) m

### Step 2: Define the position
Let \( x \) be the distance from the smaller body (1 kg) to the point where the gravitational field is zero. Therefore, the distance from the larger body (100 kg) to this point will be \( 1 - x \).

### Step 3: Write the gravitational field equations
The gravitational field \( E \) due to a mass \( m \) at a distance \( r \) is given by:
\[ E = \frac{Gm}{r^2} \]
where \( G \) is the gravitational constant.

1. The gravitational field due to the smaller body (1 kg) at distance \( x \):
\[ E_1 = \frac{G \cdot 1}{x^2} \]

2. The gravitational field due to the larger body (100 kg) at distance \( 1 - x \):
\[ E_2 = \frac{G \cdot 100}{(1 - x)^2} \]

### Step 4: Set the fields equal to each other
For the gravitational field to be zero at that point, the magnitudes of the fields must be equal:
\[ E_1 = E_2 \]
This gives us:
\[ \frac{G \cdot 1}{x^2} = \frac{G \cdot 100}{(1 - x)^2} \]

### Step 5: Cancel \( G \) and rearrange
Since \( G \) appears on both sides, we can cancel it out:
\[ \frac{1}{x^2} = \frac{100}{(1 - x)^2} \]
Cross-multiplying gives:
\[ (1 - x)^2 = 100x^2 \]

### Step 6: Expand and simplify the equation
Expanding the left side:
\[ 1 - 2x + x^2 = 100x^2 \]
Rearranging the equation:
\[ 1 - 2x + x^2 - 100x^2 = 0 \]
\[ 1 - 2x - 99x^2 = 0 \]

### Step 7: Rearranging into standard quadratic form
This can be rearranged into:
\[ 99x^2 + 2x - 1 = 0 \]

### Step 8: Use the quadratic formula to solve for \( x \)
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 99 \), \( b = 2 \), and \( c = -1 \):
\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 99 \cdot (-1)}}{2 \cdot 99} \]
\[ x = \frac{-2 \pm \sqrt{4 + 396}}{198} \]
\[ x = \frac{-2 \pm \sqrt{400}}{198} \]
\[ x = \frac{-2 \pm 20}{198} \]

### Step 9: Calculate the two possible values for \( x \)
1. \( x = \frac{18}{198} = \frac{1}{11} \)
2. \( x = \frac{-22}{198} \) (not valid since distance cannot be negative)

### Conclusion
Thus, the distance from the smaller body (1 kg) to the point where the gravitational field is zero is:
\[ \boxed{\frac{1}{11} \text{ m}} \]