A box weighs 196 N on a spring balance at the north pole. Its weight recorded on the same balance if it is shifted to the equator is close to (Take g = 10 ms$$^{-2}$$ at the north pole and the radius of the earth = 6400 km):

At the north pole the spring balance reads the true gravitational weight, so we have

$$W_{\text{pole}} = m\,g$$

The reading is given as $$W_{\text{pole}} = 196\ \text{N}$$ and the acceleration due to gravity at the pole is $$g = 10\ \text{m s}^{-2}$$. Substituting these values we can find the mass of the box:

$$m = \frac{W_{\text{pole}}}{g} = \frac{196}{10} = 19.6\ \text{kg}$$

When the box is taken to the equator it revolves with the Earth, so a centrifugal force acts outward. A spring balance measures the normal reaction, which equals the effective weight. The effective acceleration $$g'$$ at the equator is therefore reduced by the centrifugal term. The relation is

$$g' = g - \omega^{2}R$$

Here $$\omega$$ is the angular speed of the Earth and $$R$$ is the radius of the Earth.

First, we calculate $$\omega$$. The Earth completes one rotation in 24 hours:

$$\omega = \frac{2\pi}{T}, \qquad T = 24 \times 3600\ \text{s}$$

$$\omega = \frac{2\pi}{24 \times 3600} = 7.27 \times 10^{-5}\ \text{rad s}^{-1}$$

Next, we compute $$\omega^{2}R$$. The radius is given as $$R = 6400\ \text{km} = 6.4 \times 10^{6}\ \text{m}$$.

$$\omega^{2} = (7.27 \times 10^{-5})^{2} = 5.29 \times 10^{-9}\ \text{s}^{-2}$$

Multiplying by the radius,

$$\omega^{2}R = 5.29 \times 10^{-9}\ \text{s}^{-2} \times 6.4 \times 10^{6}\ \text{m} = 33.856 \times 10^{-3}\ \text{m s}^{-2} = 0.033856\ \text{m s}^{-2}$$

Now we obtain the effective acceleration at the equator:

$$g' = 10 - 0.033856 = 9.966144\ \text{m s}^{-2}$$

The spring balance reading at the equator is

$$W_{\text{equator}} = m\,g' = 19.6\ \text{kg} \times 9.966144\ \text{m s}^{-2}$$

Performing the multiplication,

$$W_{\text{equator}} = 195.329\ \text{N} \approx 195.32\ \text{N}$$

Among the given choices, this value matches Option D.

Hence, the correct answer is Option D.