`R` is the radius of the earth and `omega` is its angular velocity and `g_(p)` is the value of `g` at the poles. The effective value of `g` at the latitude `lambda=60^(@)` will be equal to

Options

A. `g_(p)-1/4 Romega^(2)`
B. `g_(p)-3/4 R omega^(2)`
C. `g_(p)-Romega^(2)`
D. `g_(p)+1/4 Romega^(2)`

Answer

To find the effective value of \( g \) at a latitude of \( \lambda = 60^\circ \), we can use the formula for the effective acceleration due to gravity at a latitude, which is given by:

\[ g' = g_p - R \omega^2 \cos^2(\lambda) \]

Where:
- \( g' \) is the effective value of \( g \) at latitude \( \lambda \).
- \( g_p \) is the value of \( g \) at the poles.
- \( R \) is the radius of the Earth.
- \( \omega \) is the angular velocity of the Earth.
- \( \lambda \) is the latitude.

### Step-by-Step Solution:

1. **Identify the known values**:
- \( g_p \): The acceleration due to gravity at the poles (approximately \( 9.81 \, \text{m/s}^2 \)).
- \( R \): The radius of the Earth (approximately \( 6.371 \times 10^6 \, \text{m} \)).
- \( \omega \): The angular velocity of the Earth (approximately \( 7.27 \times 10^{-5} \, \text{rad/s} \)).
- \( \lambda = 60^\circ \).

2. **Convert the latitude to radians** (if necessary):
- \( \lambda = 60^\circ = \frac{\pi}{3} \, \text{radians} \).

3. **Calculate \( \cos^2(\lambda) \)**:
- \( \cos(60^\circ) = \frac{1}{2} \).
- Therefore, \( \cos^2(60^\circ) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \).

4. **Substitute the values into the formula**:
\[ g' = g_p - R \omega^2 \cos^2(60^\circ) \]
\[ g' = g_p - R \omega^2 \left(\frac{1}{4}\right) \]

5. **Calculate \( R \omega^2 \)**:
- First, calculate \( \omega^2 \):
\[ \omega^2 = (7.27 \times 10^{-5})^2 \approx 5.29 \times 10^{-9} \, \text{rad}^2/\text{s}^2 \]
- Then calculate \( R \omega^2 \):
\[ R \omega^2 \approx (6.371 \times 10^6) \times (5.29 \times 10^{-9}) \approx 0.0337 \, \text{m/s}^2 \]

6. **Substitute back to find \( g' \)**:
\[ g' = 9.81 - 0.0337 \approx 9.7763 \, \text{m/s}^2 \]

7. **Final result**:
\[ g' \approx 9.78 \, \text{m/s}^2 \]

### Final Answer:
The effective value of \( g \) at the latitude \( \lambda = 60^\circ \) is approximately \( 9.78 \, \text{m/s}^2 \).