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: Define the Problem
We have two spheres:
- Sphere 1 (mass \( m_1 = 16 \, \text{kg} \))
- Sphere 2 (mass \( m_2 = 4 \, \text{kg} \))

They are separated by a distance of \( d = 30 \, \text{m} \). We need to find a point (let's call it point P) along the line connecting the two spheres where the gravitational forces exerted by both spheres on a third mass \( m \) will cancel each other out.

### Step 2: Set Up the Distances
Let \( x \) be the distance from the sphere of mass 16 kg to point P. Consequently, the distance from the sphere of mass 4 kg to point P will be \( 30 - x \).

### Step 3: Write the Gravitational Force Equations
The gravitational force exerted by the sphere of mass 16 kg on mass \( m \) is given by:
\[ F_1 = \frac{G \cdot m_1 \cdot m}{x^2} \]
The gravitational force exerted by the sphere of mass 4 kg on mass \( m \) is given by:
\[ F_2 = \frac{G \cdot m_2 \cdot m}{(30 - x)^2} \]

### Step 4: Set the Forces Equal
At point P, the net gravitational force is zero, so we set the magnitudes of the forces equal to each other:
\[ F_1 = F_2 \]
This gives us:
\[ \frac{G \cdot 16 \cdot m}{x^2} = \frac{G \cdot 4 \cdot m}{(30 - x)^2} \]

### Step 5: Cancel Common Terms
We can cancel \( G \) and \( m \) from both sides since they are common:
\[ \frac{16}{x^2} = \frac{4}{(30 - x)^2} \]

### Step 6: Cross Multiply
Cross multiplying gives us:
\[ 16(30 - x)^2 = 4x^2 \]

### Step 7: Expand and Rearrange
Expanding the left side:
\[ 16(900 - 60x + x^2) = 4x^2 \]
This simplifies to:
\[ 14400 - 960x + 16x^2 = 4x^2 \]
Rearranging gives:
\[ 12x^2 - 960x + 14400 = 0 \]

### Step 8: Simplify the Equation
Dividing the entire equation by 12:
\[ x^2 - 80x + 1200 = 0 \]

### Step 9: Solve the Quadratic Equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -80, c = 1200 \):
\[ x = \frac{80 \pm \sqrt{(-80)^2 - 4 \cdot 1 \cdot 1200}}{2 \cdot 1} \]
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 possible solutions:
\[ x = \frac{120}{2} = 60 \quad \text{and} \quad x = \frac{40}{2} = 20 \]

### Step 10: Determine the Valid Solution
Since \( x = 60 \) meters is outside the range (the distance between the two spheres is only 30 m), we discard it. Thus, the 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 meters**.

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