1. Weight at the north pole

At the pole, the spring balance reads $W_p = 49\,\text{N}$

Since $W = mg,$ we get the mass of the body: $m = \frac{W_p}{g} = \frac{49}{9.8} = 5\,\text{kg}$

2. Effect of Earth’s rotation at the equator

At the equator, the apparent weight decreases because of centrifugal effect due to Earth’s rotation.

So, $W_e = m(g - \omega^2 R)$

where:

• $\omega = \dfrac{2\pi}{T}$ is angular speed of Earth,
• $T = 24 \times 3600 = 86400\,\text{s}$,
• $R = 6400\,\text{km} = 6.4 \times 10^6\,\text{m}$.

3. Calculate $\omega^2 R$

First, $\omega = \frac{2\pi}{86400}$

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

Now, $\left(\frac{2\pi}{86400}\right)^2 \approx 5.29 \times 10^{-9}$

Thus, $\omega^2 R \approx 5.29 \times 10^{-9} \times 6.4 \times 10^6$ $\omega^2 R \approx 0.0339\,\text{m/s}^2$

4. Effective acceleration due to gravity at equator

$g_e = 9.8 - 0.0339 = 9.7661\,\text{m/s}^2$

5. Weight at the equator

$W_e = m g_e = 5 \times 9.7661 = 48.8305\,\text{N}$

Therefore, $W_e \approx 48.83\,\text{N}$

6. Option check

• A: $49\,\text{N}$ ❌
• B: $49.83\,\text{N}$ ❌
• C: $48.83\,\text{N}$ ✅
• D: $49.17\,\text{N}$ ❌

Hence, the correct answer is Option C.