1. **Problem Statement:** A ruler AB is suspended from points A and D by two vertical ropes with tensions $T_1$ and $T_2$. Two weights, 6 N and 4 N, are hanging from points C and B respectively. The ruler is in equilibrium. Find the ratio $T_1:T_2$. 2. **Understanding the Setup:** - The ruler is in equilibrium, so the sum of forces and the sum of moments (torques) about any point must be zero. - Let’s assume the distances between points A, B, C, and D are equal segments along the ruler. 3. **Forces:** - Downward forces: 6 N at C, 4 N at B. - Upward forces: $T_1$ at A, $T_2$ at D. 4. **Equilibrium Conditions:** - Sum of vertical forces: $$T_1 + T_2 = 6 + 4 = 10$$ - Sum of moments about A (taking clockwise as positive): $$T_2 \times AD - 6 \times AC - 4 \times AB = 0$$ 5. **Assigning Distances:** - Let the length between adjacent points be $l$. - Then, $AB = l$, $AC = 2l$, $AD = 3l$. 6. **Moment Equation:** $$T_2 \times 3l = 6 \times 2l + 4 \times l$$ $$3l T_2 = 12l + 4l = 16l$$ $$T_2 = \frac{16l}{3l} = \frac{16}{3}$$ 7. **Find $T_1$:** $$T_1 = 10 - T_2 = 10 - \frac{16}{3} = \frac{30}{3} - \frac{16}{3} = \frac{14}{3}$$ 8. **Ratio $T_1:T_2$:** $$\frac{T_1}{T_2} = \frac{\frac{14}{3}}{\frac{16}{3}} = \frac{14}{16} = \frac{7}{8}$$ 9. **Check options:** None of the options exactly match $7:8$. Re-examining the problem, the tensions are applied at A and D, weights at B and C, so the ratio is $T_1:T_2 = 7:8$ which is approximately $1:1.14$. Since none of the options match exactly, the closest is option (b) 1:14, but that is too large. **Final answer:** $T_1:T_2 = 7:8$ (approximately 1:1.14)