1. Weight at the north pole

At the north pole, there is no reduction in apparent weight due to Earth's rotation.

Given: $W_p = 196\,\text{N}$ $g_p = 10\,\text{m s}^{-2}$

So the mass of the box is $m = \frac{W_p}{g_p} = \frac{196}{10} = 19.6\,\text{kg}$

2. Effect at the equator

At the equator, the spring balance reads the apparent weight: $W_e = m(g - \omega^2 R)$

Since the problem gives $g = 10\,\text{m s}^{-2}$ at the north pole, we take gravitational acceleration approximately same, and subtract centrifugal acceleration at the equator.

3. Compute centrifugal acceleration

Angular speed of Earth: $\omega = \frac{2\pi}{T}$ with $T = 24 \times 3600 = 86400\,\text{s}$

Thus, $\omega = \frac{2\pi}{86400}\,\text{rad s}^{-1}$

Radius of Earth: $R = 6400\,\text{km} = 6.4 \times 10^6\,\text{m}$

Now, $\omega^2 R = \left(\frac{2\pi}{86400}\right)^2 (6.4\times 10^6)$

Using the standard value, $\omega^2 R \approx 0.0339\,\text{m s}^{-2}$

4. Apparent acceleration due to gravity at equator

$g_e = 10 - 0.0339 = 9.9661\,\text{m s}^{-2}$

5. Apparent weight at equator

$W_e = m g_e = 19.6 \times 9.9661$

$W_e \approx 195.34\,\text{N}$

This is closest to $\boxed{195.32\,\text{N}}$

6. Option check

• A: $194.32\,\text{N}$ — too low
• B: $195.66\,\text{N}$ — too high
• C: $195.32\,\text{N}$ — correct
• D: $194.66\,\text{N}$ — too low

Therefore, the correct option is C.