If earth is supposed to be sphere of radius R, if `g_(20)` is value of acceleration due to gravity at latitude of `30^(@)` and g at the equator, then value of `g-g_(30^(@))` is

Options

A. `(1)/(4)omega^(2)R`
B. `(3)/(4) omega^(2)R`
C. `omega^(2)R`
D. `(1)/(2) omega^(2)R`

Explanation

To solve the problem, we need to find the difference between the acceleration due to gravity at the equator (g) and the acceleration due to gravity at a latitude of 30 degrees (g_(30°)).

### Step-by-Step Solution:

1. **Understanding the Formula**:
The acceleration due to gravity at a latitude (φ) is given by the formula:
\[ g' = g - R \omega^2 \cos^2(\phi) \]
where:
- \( g' \) is the acceleration due to gravity at latitude φ,
- \( g \) is the acceleration due to gravity at the equator,
- \( R \) is the radius of the Earth,
- \( \omega \) is the angular velocity of the Earth,
- \( \phi \) is the latitude.

2. **Substituting the Latitude**:
For φ = 30°, we can substitute this into the formula:
\[ g_{30°} = g - R \omega^2 \cos^2(30°) \]

3. **Calculating cos(30°)**:
We know that:
\[ \cos(30°) = \frac{\sqrt{3}}{2} \]
Therefore:
\[ \cos^2(30°) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \]

4. **Substituting cos(30°) into the Formula**:
Now we can substitute this value back into our equation:
\[ g_{30°} = g - R \omega^2 \left(\frac{3}{4}\right) \]
This simplifies to:
\[ g_{30°} = g - \frac{3}{4} R \omega^2 \]

5. **Finding g - g_{30°}**:
We need to find the difference:
\[ g - g_{30°} = g - \left(g - \frac{3}{4} R \omega^2\right) \]
This simplifies to:
\[ g - g_{30°} = \frac{3}{4} R \omega^2 \]

6. **Final Result**:
Thus, the value of \( g - g_{30°} \) is:
\[ g - g_{30°} = \frac{3}{4} R \omega^2 \]

### Conclusion:
The final answer is:
\[ g - g_{30°} = \frac{3}{4} R \omega^2 \]