1. Forces acting on the body

A mass of $1\,\text{kg}$ is suspended by two strings. Its weight acts vertically downward:

$W = mg = 1 \times 10 = 10\,\text{N}$

Let the tensions in the two strings be $T_1$ and $T_2$.

From the given standard figure for this question, one string makes $30^\circ$ with the horizontal and the other makes $60^\circ$ with the horizontal.

2. Resolve tensions into components

Assume:

• $T_1$ is in the string making $30^\circ$ with the horizontal
• $T_2$ is in the string making $60^\circ$ with the horizontal

For equilibrium:

• Horizontal components must cancel
• Vertical components must balance the weight

Horizontal equilibrium

$T_1 \cos 30^\circ = T_2 \cos 60^\circ$

$T_1 \cdot \frac{\sqrt{3}}{2} = T_2 \cdot \frac{1}{2}$

$T_2 = \sqrt{3}\,T_1$

3. Vertical equilibrium

$T_1 \sin 30^\circ + T_2 \sin 60^\circ = 10$

Substitute $\sin 30^\circ = \frac{1}{2}$ and $\sin 60^\circ = \frac{\sqrt{3}}{2}$:

$T_1 \cdot \frac{1}{2} + T_2 \cdot \frac{\sqrt{3}}{2} = 10$

Now put $T_2 = \sqrt{3}T_1$:

$\frac{T_1}{2} + \frac{\sqrt{3}}{2}(\sqrt{3}T_1) = 10$

$\frac{T_1}{2} + \frac{3T_1}{2} = 10$

$2T_1 = 10$

$T_1 = 5\,\text{N}$

Then,

$T_2 = \sqrt{3} \times 5 = 5\sqrt{3}\,\text{N}$

4. Match with the options

Thus the tensions are:

$T_1 = 5\,\text{N}, \qquad T_2 = 5\sqrt{3}\,\text{N}$

This corresponds to Option D.

5. Comparison with stored answer

The stored correct answer is A: $5\sqrt{3}, 5$.

That would be correct only if the labels $T_1$ and $T_2$ were interchanged in the figure. Based on the usual interpretation described above, the correct ordered pair is:

$(T_1, T_2) = (5, 5\sqrt{3})$

So I disagree with the stored answer unless the figure labels are reversed.