Solve the following :

(a)

At equator, normal contact force $\mathrm{N}+\mathrm{mR} \omega^{2}=\mathrm{mg}$

Reading $N=m g-m R \omega^{2}$

$t=\frac{m g-\left(m g-m R w^{2}\right)}{m g}$

$\begin{aligned} & \frac{R w^{2}}{g} \\=& \frac{3}{3} \end{aligned}$

$=3.5 \times 10-3$

(b)

Balance reading $=\overline{2}$ (True weight)

$m g-m R \omega^{2}=\frac{1}{2}((m g)$

$\mathrm{mR} \omega^{2}=\frac{m g}{2}$

$\omega=\sqrt{\frac{\mathrm{g}}{2 R}}=\frac{2 \pi}{T r}$

$T^{\prime}=2 \times 3.14 \times \quad$ (in sec)

$T^{\prime}=\frac{2 \times 3.14 \times \sqrt{\frac{\left(6.4 \times 10^{7}\right)}{49}}}{(60 \times 60)}$

$\mathrm{T}^{\prime}=2 \mathrm{hrs}$