Two spheres of masses 16 kg and 4 kg are separated by a distance 30 m on a table. Then, the distance from sphere of mass 16 kg at which the net gravitational force becomes zero is

Options

A. 10 m
B. 20 m
C. 15 m
D. 5 m

Explanation

To find the distance from the sphere of mass 16 kg at which the net gravitational force becomes zero, we can follow these steps:

### Step 1: Understand the setup
We have two spheres:
- Mass \( m_1 = 16 \, \text{kg} \)
- Mass \( m_2 = 4 \, \text{kg} \)
They are separated by a distance of \( d = 30 \, \text{m} \).

### Step 2: Define the point where the gravitational force is zero
Let \( x \) be the distance from the 16 kg mass where the net gravitational force is zero. Thus, the distance from the 4 kg mass to this point will be \( 30 - x \).

### Step 3: Set up the gravitational force equations
The gravitational force exerted by the 16 kg mass at point \( P \) is given by:
\[ F_1 = \frac{G \cdot m_1}{x^2} \]
The gravitational force exerted by the 4 kg mass at point \( P \) is given by:
\[ F_2 = \frac{G \cdot m_2}{(30 - x)^2} \]

### Step 4: Set the forces equal to each other
For the net gravitational force to be zero at point \( P \):
\[ F_1 = F_2 \]
This gives us:
\[ \frac{G \cdot 16}{x^2} = \frac{G \cdot 4}{(30 - x)^2} \]
We can cancel \( G \) from both sides:
\[ \frac{16}{x^2} = \frac{4}{(30 - x)^2} \]

### Step 5: Cross-multiply to solve for \( x \)
Cross-multiplying gives:
\[ 16(30 - x)^2 = 4x^2 \]
Expanding this:
\[ 16(900 - 60x + x^2) = 4x^2 \]
\[ 14400 - 960x + 16x^2 = 4x^2 \]
Rearranging the equation:
\[ 16x^2 - 4x^2 - 960x + 14400 = 0 \]
\[ 12x^2 - 960x + 14400 = 0 \]

### Step 6: Simplify the equation
Dividing the entire equation by 12:
\[ x^2 - 80x + 1200 = 0 \]

### Step 7: Use the quadratic formula to solve for \( x \)
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1 \), \( b = -80 \), and \( c = 1200 \):
\[ x = \frac{80 \pm \sqrt{(-80)^2 - 4 \cdot 1 \cdot 1200}}{2 \cdot 1} \]
Calculating the discriminant:
\[ x = \frac{80 \pm \sqrt{6400 - 4800}}{2} \]
\[ x = \frac{80 \pm \sqrt{1600}}{2} \]
\[ x = \frac{80 \pm 40}{2} \]
This gives us two potential solutions:
\[ x = \frac{120}{2} = 60 \quad \text{and} \quad x = \frac{40}{2} = 20 \]

### Step 8: Determine the valid solution
Since \( x = 60 \) is greater than the distance between the two masses (30 m), it is not a valid solution. Thus, the only valid solution is:
\[ x = 20 \, \text{m} \]

### Final Answer
The distance from the sphere of mass 16 kg at which the net gravitational force becomes zero is \( 20 \, \text{m} \).

---