1. Given data

• Mass of block: $m = 10~\text{kg}$
• Coefficient of static friction: $\mu_s = 0.25$
• Angle of pull: $\theta = 30^\circ$
• Acceleration due to gravity: $g = 10~\text{m s}^{-2}$

We need the force $F$ for which the block is just about to move.

2. Forces acting on the block

The applied force $F$ has two components:

• Horizontal component: $F\cos 30^\circ$
• Vertical upward component: $F\sin 30^\circ$

Since the pull has an upward component, the normal reaction decreases.

So, $N = mg - F\sin 30^\circ$

Now, $mg = 10 \times 10 = 100~\text{N}$

Hence, $N = 100 - F\sin 30^\circ$

Since $\sin 30^\circ = \frac{1}{2}$, $N = 100 - \frac{F}{2}$

3. Condition for just starting motion

At the limiting condition, $f_s^{\max} = \mu_s N$

and the horizontal pulling component equals limiting friction: $F\cos 30^\circ = \mu_s N$

Substitute $N$: $F\cos 30^\circ = 0.25\left(100 - \frac{F}{2}\right)$

Using $\cos 30^\circ = \frac{\sqrt{3}}{2}$, $F\cdot \frac{\sqrt{3}}{2} = 25 - \frac{F}{8}$

Multiply by 8: $4\sqrt{3}F = 200 - F$

So, $F(4\sqrt{3}+1) = 200$

Therefore, $F = \frac{200}{4\sqrt{3}+1}$

Now evaluate: $4\sqrt{3} \approx 4 \times 1.732 = 6.928$ $4\sqrt{3}+1 \approx 7.928$

Thus, $F \approx \frac{200}{7.928} \approx 25.2~\text{N}$

4. Checking options

• A: $25.2~\text{N}$ ✅
• B: $35.7~\text{N}$
• C: $20~\text{N}$
• D: $33.3~\text{N}$

So the correct option is: $\boxed{\text{A }\; 25.2~\text{N}}$

5. Comparison with stored answer

Stored correct answer: A

My derived answer is also A. So they agree.