1. **Problem statement:** A body weighing 10 kg wt rests on a rough horizontal table. A force of magnitude 20 kg wt acts on it inclined upward at an angle of 30° to the horizontal. We need to find the friction force acting on the body. 2. **Understanding the forces:** The weight of the body is $W = 10$ kg wt acting vertically downward. 3. The applied force $F = 20$ kg wt acts at an angle $\theta = 30^\circ$ above the horizontal. 4. **Components of the applied force:** - Horizontal component: $F_x = F \cos \theta = 20 \times \cos 30^\circ = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}$ kg wt - Vertical component: $F_y = F \sin \theta = 20 \times \sin 30^\circ = 20 \times \frac{1}{2} = 10$ kg wt 5. **Normal reaction $N$ on the body:** The normal reaction balances the vertical forces: $$N + F_y = W \implies N = W - F_y = 10 - 10 = 0$$ 6. Since the normal reaction $N = 0$, the friction force, which depends on $N$, is zero. 7. **Conclusion:** The friction force is zero. **Final answer:** (a) zero