If radius of earth contracted by 0.1% , its mass remaining same then weight of the body at earth's surface will increase by

Options

A. `0.1%`
B. `0.2%`
C. `0.3%`
D. Remains same

Answer

To solve the problem of how the weight of a body at the Earth's surface will change if the radius of the Earth contracts by 0.1% while its mass remains the same, we can follow these steps:

### Step 1: Understand the relationship between weight and gravitational force
The weight \( W \) of a body on the surface of the Earth is given by the formula:
\[ W = mg \]
where \( m \) is the mass of the body and \( g \) is the acceleration due to gravity.

### Step 2: Express \( g \) in terms of Earth's mass and radius
The acceleration due to gravity \( g \) at the surface of the Earth can be expressed as:
\[ g = \frac{GM}{R^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

### Step 3: Determine the change in \( g \) when \( R \) changes
If the radius \( R \) contracts by 0.1%, we can express this change as:
\[ \Delta R = -0.001R \]
The new radius \( R' \) will be:
\[ R' = R + \Delta R = R(1 - 0.001) = 0.999R \]

### Step 4: Calculate the new acceleration due to gravity \( g' \)
Substituting \( R' \) into the formula for \( g \):
\[ g' = \frac{GM}{(0.999R)^2} = \frac{GM}{0.998001R^2} = \frac{g}{0.998001} \]

### Step 5: Find the percentage change in \( g \)
The percentage change in \( g \) can be calculated as follows:
\[ \frac{\Delta g}{g} = \frac{g' - g}{g} = \frac{\frac{g}{0.998001} - g}{g} = \frac{g(1 - 0.998001)}{g} = 1 - 0.998001 = 0.001999 \]
Thus, the percentage change in \( g \) is:
\[ \Delta g \approx 0.001999 \times 100 \approx 0.1999\% \]

### Step 6: Relate the change in weight to the change in \( g \)
Since weight \( W \) is directly proportional to \( g \), the percentage change in weight will be the same as the percentage change in \( g \):
\[ \Delta W \approx 0.1999\% \]

### Step 7: Round the result
Rounding this value gives us approximately:
\[ \Delta W \approx 0.2\% \]

### Final Answer
Therefore, the weight of the body at the Earth's surface will increase by approximately **0.2%**.

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