If density of earth increased 4 times and its radius become half of what it is, our weight will ?

Options

A. be four times its present value
B. be doubled
C. remain same
D. be halved

Explanation

To solve the problem, we need to analyze how the weight of an object changes when the density and radius of the Earth change. The weight of an object is directly related to the acceleration due to gravity (g) at the surface of the Earth.

### Step-by-Step Solution:

1. **Understand the relationship between weight and gravity**:
The weight (W) of an object is given by the formula:
\[ W = m \cdot g \]
where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity.

2. **Identify the variables**:
We know that:
- The density of the Earth (\( \rho \)) increases 4 times.
- The radius of the Earth (\( r \)) becomes half.

3. **Determine the new density and radius**:
- New density (\( \rho' \)):
\[ \rho' = 4 \cdot \rho \]
- New radius (\( r' \)):
\[ r' = \frac{r}{2} \]

4. **Use the formula for acceleration due to gravity**:
The acceleration due to gravity at the surface of the Earth is given by:
\[ g = \frac{G \cdot M}{r^2} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the Earth.

5. **Relate mass to density and volume**:
The mass of the Earth can be expressed in terms of its density and volume:
\[ M = \rho \cdot V \]
The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Therefore, the mass of the Earth becomes:
\[ M = \rho \cdot \frac{4}{3} \pi r^3 \]

6. **Calculate the new mass with the new density and radius**:
With the new density and radius, the new mass (\( M' \)) becomes:
\[ M' = \rho' \cdot V' = (4 \cdot \rho) \cdot \left(\frac{4}{3} \pi \left(\frac{r}{2}\right)^3\right) \]
Simplifying this gives:
\[ M' = 4 \cdot \rho \cdot \frac{4}{3} \pi \cdot \frac{r^3}{8} = \frac{4 \cdot \rho \cdot 4 \cdot \pi \cdot r^3}{24} = \frac{2 \cdot \rho \cdot 4 \cdot \pi \cdot r^3}{12} = \frac{2}{3} \cdot M \]

7. **Substituting back into the gravity formula**:
Now substituting \( M' \) and \( r' \) into the gravity formula:
\[ g' = \frac{G \cdot M'}{(r')^2} = \frac{G \cdot \left(\frac{2}{3} M\right)}{\left(\frac{r}{2}\right)^2} = \frac{G \cdot \left(\frac{2}{3} M\right)}{\frac{r^2}{4}} = \frac{8 \cdot G \cdot M}{3 \cdot r^2} = \frac{8}{3} g \]

8. **Determine the new weight**:
Since the new acceleration due to gravity is \( g' = \frac{8}{3} g \), the new weight \( W' \) becomes:
\[ W' = m \cdot g' = m \cdot \frac{8}{3} g = \frac{8}{3} W \]

9. **Conclusion**:
The new weight is \( \frac{8}{3} \) times the original weight, which means the weight increases.

### Final Answer:
Our weight will increase to \( \frac{8}{3} \) times the original weight.