A uniform metre stick of mass 200 g is suspended from the ceiling...

Let the length of the metre stick be $L = 100 \text{ cm}$.

Mass of the stick, $M = 200 \text{ g} = 0.2 \text{ kg}$.
Weight of the stick, $W_1 = Mg = 0.2 \times 9.8 = 1.96 \text{ N}$, acting at its centre of gravity ($50 \text{ cm}$ from the left end).

Mass of the small object, $m = 20 \text{ g} = 0.02 \text{ kg}$.
Weight of the object, $W_2 = mg = 0.02 \times 9.8 = 0.196 \text{ N}$, acting at $70 \text{ cm}$ from the left end.

Let $T_1$ and $T_2$ be the tensions in the left and right strings respectively.

For rotational equilibrium, taking torque about the left end:
$T_2 \times 100 = W_1 \times 50 + W_2 \times 70$
$T_2 \times 100 = (1.96 \times 50) + (0.196 \times 70)$
$T_2 \times 100 = 98 + 13.72 = 111.72$
$T_2 = 1.1172 \text{ N} \approx 1.12 \text{ N}$

For translational equilibrium, balancing the vertical forces:
$T_1 + T_2 = W_1 + W_2$
$T_1 + 1.1172 = 1.96 + 0.196 = 2.156$
$T_1 = 2.156 - 1.1172 = 1.0388 \text{ N} \approx 1.04 \text{ N}$

Therefore, the tension in the left string is $1.04 \text{ N}$ and in the right string is $1.12 \text{ N}$.

Answer: $1.04 \text{ N}$ in the left string and $1.12 \text{ N}$ in the right string