1. **Problem statement:** We have a vertical rod of length 2 m fixed at point M. A force of 49 N is applied at the top of the rod making a 30° angle with the vertical. We want to find the velocity (গতি) produced. 2. **Relevant formulas and concepts:** - Torque (ত্রয়) \( \tau = F \times r \times \sin(\theta) \), where \(F\) is force, \(r\) is distance from pivot, and \(\theta\) is angle between force and lever arm. - Newton's second law for rotation: \( \tau = I \alpha \), where \(I\) is moment of inertia and \(\alpha\) is angular acceleration. - Linear velocity \(v = r \omega\), where \(\omega\) is angular velocity. 3. **Calculate torque:** Given \(F = 49\) N, \(r = 2\) m, \(\theta = 30^\circ\) $$\tau = 49 \times 2 \times \sin 30^\circ = 49 \times 2 \times 0.5 = 49 \text{ Nm}$$ 4. **Moment of inertia for rod about pivot M:** For a rod of length \(L=2\) m, mass \(m=10\) kg, pivoted at one end, $$I = \frac{1}{3} m L^2 = \frac{1}{3} \times 10 \times 2^2 = \frac{1}{3} \times 10 \times 4 = \frac{40}{3} \approx 13.33 \text{ kg m}^2$$ 5. **Calculate angular acceleration:** $$\alpha = \frac{\tau}{I} = \frac{49}{13.33} \approx 3.68 \text{ rad/s}^2$$ 6. **Calculate angular velocity and linear velocity:** Assuming the rod starts from rest and the force is applied briefly, the angular velocity \(\omega\) after time \(t\) is \(\omega = \alpha t\). Without time, we consider instantaneous velocity at the tip: Linear velocity at tip: $$v = r \omega$$ Since \(\omega\) depends on time, if we consider initial acceleration, velocity is zero at start. If the problem implies velocity after some time, more info is needed. **Summary:** - Torque applied is 49 Nm. - Angular acceleration is approximately 3.68 rad/s². - Velocity depends on time of force application. --- 7. **Maximum force and torque:** Given maximum force \(F_{max} = 300\) N, Maximum torque: $$\tau_{max} = F_{max} \times r = 300 \times 2 = 600 \text{ Nm}$$ This maximum torque can be used to calculate maximum angular acceleration: $$\alpha_{max} = \frac{\tau_{max}}{I} = \frac{600}{13.33} \approx 45 \text{ rad/s}^2$$ This large acceleration can produce significant angular velocity and thus linear velocity at the tip. --- **Note:** The problem's part (ঘ) about maximum torque and velocity is solved here.