\( \overline{\mathrm{AB}} \) is a non-uniform rod of length 80 cm . and weight 20 newton rests horizontally on two supports at \( C \) and \( D \) such that \( A C=B D= \) 10 cm . A weight of magnitude 40 newton is suspended from \( A \) and the rod becomes about to rotate about \( C \), then the maximum weight that can be suspended from \( B \) rising the suspended weight at \( A \) without getting imbalanced is.... \( \begin{array}{llll}\text { (a) } 80 & \text { (b) } 90 & \text { (C) } 100 & \text { (d) } 120\end{array} \)

To solve this problem, we will use the principles of moments (torque) about point \( C \) to find the maximum weight that can be suspended from point \( B \) without causing the rod to rotate. ### Known Conditions: - Length of the rod \( L = 80 \, \text{cm} \) - Weight of the rod \( W_r = 20 \, \text{N} \) - Distance from \( A \) to \( C \) \( = 10 \, \text{cm} \) - Distance from \( B \) to \( D \) \( = 10 \, \text{cm} \) - Weight suspended from \( A \) \( W_a = 40 \, \text{N} \) ### Step-by-Step Solution: 1. **Determine the distances:** - The distance from \( C \) to \( A \) is \( 10 \, \text{cm} \). - The distance from \( C \) to \( D \) is \( 70 \, \text{cm} \) (since \( D \) is at the end of the rod). 2. **Calculate the moment due to the weight at \( A \):** - The moment about point \( C \) due to the weight at \( A \) is given by: \[ M_a = W_a \times d_a = 40 \, \text{N} \times 0.1 \, \text{m} = 4 \, \text{N m} \] 3. **Calculate the moment due to the weight of the rod:** - The weight of the rod acts at its center of gravity, which is at \( 40 \, \text{cm} \) from \( A \) (or \( 30 \, \text{cm} \) from \( C \)): \[ M_r = W_r \times d_r = 20 \, \text{N} \times 0.3 \, \text{m} = 6 \, \text{N m} \] 4. **Let \( W_b \) be the weight suspended from \( B \):** - The moment about point \( C \) due to the weight at \( B \) is: \[ M_b = W_b \times d_b = W_b \times 0.7 \, \text{m} \] 5. **Set up the equilibrium condition:** - For the rod to be in equilibrium (not rotating), the sum of the moments about point \( C \) must be zero: \[ M_a + M_r = M_b \] - Substituting the moments: \[ 4 \, \text{N m} + 6 \, \text{N m} = W_b \times 0.7 \, \text{m} \] - Simplifying: \[ 10 \, \text{N m} = W_b \times 0.7 \, \text{m} \] 6. **Solve for \( W_b \):** \[ W_b = \frac{10 \, \text{N m}}{0.7 \, \text{m}} = \frac{10}{0.7} \approx 14.29 \, \text{N} \] ### Conclusion: The maximum weight that can be suspended from \( B \) without causing the rod to rotate about \( C \) is approximately \( 14.29 \, \text{N} \). However, since the options provided are much larger, it seems there might be a misunderstanding in the interpretation of the problem. Let's re-evaluate the equilibrium condition considering the total weight acting downwards and the moments about point \( C \) again to ensure we have the correct interpretation. ### Re-evaluation: The total downward force must equal the total upward force for equilibrium. The total downward force is \( W_a + W_r \) and the upward force is \( W_b \). 1. **Total downward force:** \[ F_d = W_a + W_r = 40 \, \text{N} + 20 \, \text{N} = 60 \, \text{N} \] 2. **Setting the upward force equal to the downward force:** \[ W_b = F_d = 60 \, \text{N} \] Thus, the maximum weight that can be suspended from \( B \) without causing the rod to rotate about \( C \) is \( 60 \, \text{N} \). However, since the options provided are \( 80, 90, 100, 120 \), it seems we need to consider the additional weight that can be added without causing imbalance. ### Final Calculation: To find the maximum weight that can be added to \( B \) while keeping the system in balance, we can use the moments again: 1. **Using the moments:** \[ 10 \, \text{N m} = W_b \times 0.7 \, \text{m} \] \[ W_b = \frac{10}{0.7} \approx 14.29 \, \text{N} \] This indicates that the maximum weight that can be added to \( B \) without causing imbalance is \( 60 \, \text{N} \) (the total weight acting downwards). ### Conclusion: The maximum weight that can be suspended from \( B \) without causing the rod to rotate about \( C \) is \( 60 \, \text{N} \). However, since the options provided are \( 80, 90, 100, 120 \), it seems there might be a misunderstanding in the interpretation of the problem. The answer is not among the options provided, indicating a possible error in the problem statement or options.