A spring balance extends by 2 cm due to a mass on surface of Earth. What will be the extension at a height `2R_(e )` from earth’s surface ?

Options

A. `(1)/(2)cm`
B. `(1)/(4)cm`
C. `(2)/(9)cm`
D. `(1)/(9)cm`

Explanation

To solve the problem, we need to determine the extension of a spring balance at a height of \(2R_e\) from the Earth's surface, given that it extends by \(2 \, \text{cm}\) at the Earth's surface.

### Step-by-Step Solution:

1. **Understand the Initial Condition**:
At the surface of the Earth, the extension \(x\) of the spring balance is given as \(2 \, \text{cm}\). The force acting on the spring balance due to the mass \(m\) is equal to the weight of the mass, which is \(mg\), where \(g\) is the acceleration due to gravity at the surface of the Earth.

\[ F = mg = kx \quad \text{(where \(k\) is the spring constant)} \]

2. **Determine the Effective Gravity at Height \(h\)**:
The effective acceleration due to gravity at a height \(h\) above the Earth's surface is given by the formula:

\[ g' = \frac{g}{1 + \frac{h}{R_e}} \]

Here, \(R_e\) is the radius of the Earth. For our case, \(h = 2R_e\).

3. **Substitute \(h\) into the Formula**:
Substitute \(h = 2R_e\) into the formula for \(g'\):

\[ g' = \frac{g}{1 + \frac{2R_e}{R_e}} = \frac{g}{1 + 2} = \frac{g}{3} \]

4. **Calculate the Weight at Height \(2R_e\)**:
At the height \(2R_e\), the weight of the mass \(m\) becomes:

\[ F' = mg' = m \left(\frac{g}{3}\right) = \frac{mg}{3} \]

5. **Relate the New Extension to the Original Extension**:
The new extension \(x'\) of the spring balance at height \(2R_e\) can be expressed similarly:

\[ F' = kx' \quad \text{(where \(x'\) is the new extension)} \]

Since we know \(mg = kx\), we can write:

\[ \frac{mg}{3} = kx' \]

6. **Set Up the Equation**:
By substituting \(mg = kx\) into the equation, we have:

\[ \frac{kx}{3} = kx' \]

We can cancel \(k\) from both sides (assuming \(k \neq 0\)):

\[ \frac{x}{3} = x' \]

7. **Substitute the Known Extension**:
We know that \(x = 2 \, \text{cm}\):

\[ x' = \frac{2 \, \text{cm}}{3} = \frac{2}{3} \, \text{cm} \]

8. **Convert to Fraction**:
To express the extension in a fraction format, we can write:

\[ x' = \frac{2}{3} \, \text{cm} \]

### Final Answer:
The extension of the spring balance at a height of \(2R_e\) from the Earth's surface is \(\frac{2}{3} \, \text{cm}\).