Two spherical balls of mass 1 kg and 4 kg are separated by a distance of 12 cm. The distance from 1 kg at which the gravitational force on any mass becomes zero.

Options

A. 4 cm
B. 12 cm
C. 6 cm
D. 8 cm

Answer

To find the distance from the 1 kg mass at which the gravitational force on any mass becomes zero, we can follow these steps:

### Step 1: Understand the problem
We have two spherical balls with masses:
- \( m_1 = 1 \, \text{kg} \) (the first ball)
- \( m_2 = 4 \, \text{kg} \) (the second ball)

The distance between the two balls is given as \( r = 12 \, \text{cm} \).

### Step 2: Define the position where the gravitational force is zero
Let \( x \) be the distance from the 1 kg mass (m1) to the point where the gravitational force becomes zero. Consequently, the distance from the 4 kg mass (m2) to this point will be \( 12 \, \text{cm} - x \).

### Step 3: Set up the equation for gravitational forces
At the point where the gravitational force is zero, the gravitational force exerted by both masses on a test mass will be equal in magnitude but opposite in direction. Therefore, we can set up the equation:

\[ \frac{G m_1 m}{x^2} = \frac{G m_2 m}{(12 - x)^2} \]

Here, \( G \) is the gravitational constant and \( m \) is the mass of the test object (which will cancel out).

### Step 4: Simplify the equation
Since \( G \) and \( m \) are present on both sides of the equation, we can cancel them out:

\[ \frac{m_1}{x^2} = \frac{m_2}{(12 - x)^2} \]

Substituting the values of \( m_1 \) and \( m_2 \):

\[ \frac{1}{x^2} = \frac{4}{(12 - x)^2} \]

### Step 5: Cross-multiply to solve for \( x \)
Cross-multiplying gives us:

\[ (12 - x)^2 = 4x^2 \]

### Step 6: Expand and rearrange the equation
Expanding the left side:

\[ 144 - 24x + x^2 = 4x^2 \]

Rearranging the equation:

\[ 144 - 24x + x^2 - 4x^2 = 0 \]

This simplifies to:

\[ -3x^2 - 24x + 144 = 0 \]

Multiplying through by -1 to make the leading coefficient positive:

\[ 3x^2 + 24x - 144 = 0 \]

### Step 7: Simplify the quadratic equation
Dividing the entire equation by 3:

\[ x^2 + 8x - 48 = 0 \]

### Step 8: Solve the quadratic equation using the quadratic formula
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 1 \), \( b = 8 \), and \( c = -48 \):

\[ x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot (-48)}}{2 \cdot 1} \]

Calculating the discriminant:

\[ x = \frac{-8 \pm \sqrt{64 + 192}}{2} \]
\[ x = \frac{-8 \pm \sqrt{256}}{2} \]
\[ x = \frac{-8 \pm 16}{2} \]

### Step 9: Calculate the two possible values for \( x \)
Calculating the two possible solutions:

1. \( x = \frac{8}{2} = 4 \, \text{cm} \)
2. \( x = \frac{-24}{2} = -12 \, \text{cm} \) (not physically meaningful)

Thus, the distance from the 1 kg mass at which the gravitational force on any mass becomes zero is:

\[ \boxed{4 \, \text{cm}} \]