As shown in the figure, a uniform horizontal rod of length 40 cm...

Let $T_L$ and $T_R$ be the tensions in the left and right wires respectively.

The frequency of the fundamental mode of the left wire is $f_L = \dfrac{1}{2l} \sqrt{\dfrac{T_L}{\mu}}$.

The frequency of the first overtone of the right wire is $f_R = \dfrac{2}{2l} \sqrt{\dfrac{T_R}{\mu}}$.

Since both wires are excited by the same tuning fork, their frequencies are equal:

$\dfrac{1}{2l} \sqrt{\dfrac{T_L}{\mu}} = \dfrac{2}{2l} \sqrt{\dfrac{T_R}{\mu}}$

$\Rightarrow \sqrt{T_L} = 2\sqrt{T_R} \Rightarrow T_L = 4T_R$

For translational equilibrium of the rod, the total upward force equals the total downward force:

$T_L + T_R = mg + Mg$

$4T_R + T_R = 1.2g + 4.8g$

$5T_R = 6.0g \Rightarrow T_R = 1.2g$

Then, $T_L = 4.8g$.

Let the mass of $4.8 \text{ kg}$ be placed at a distance $x$ from the left end. For rotational equilibrium, taking torque about the left end:

$T_R \times 40 - mg \times 20 - Mg \times x = 0$

$1.2g \times 40 - 1.2g \times 20 - 4.8g \times x = 0$

$48 - 24 - 4.8x = 0$

$4.8x = 24 \Rightarrow x = 5 \text{ cm}$

Thus, the mass must be positioned $5 \text{ cm}$ from the left end.