A uniform rod of mass m and length l suspended by means of two identical inextensible light strings as shown in figure. Tension in one string immediately after the other string is cut, is ____. (g acceleration due to gravity)

### Related Formula tau = Ialpha$\tau = I\alpha$ Sigma F_y = m a_CM, y$$\Sigma F_y = m a_{CM, y}$$ a_CM, y = alpha fracl2$$a_{CM, y} = \alpha \frac{l}{2}$$ ### Core Logic Immediately after one string is cut, the rod starts rotating about the point where the remaining string is attached. Taking torque about the end where the string is attached (this point has instantaneous acceleration but initially zero vertical velocity): tau_textend = I_textend alpha$$\tau_{\text{end}} = I_{\text{end}} \alpha$$ Gravity provides the torque: tau = mg left(fracl2right)$\tau = mg \left(\frac{l}{2}\right)$. ### Step 1: Calculate Angular Acceleration Moment of inertia about the end is I = fracml^23$I = \frac{ml^2}{3}$. mg fracl2 = fracml^23 alpha$$mg \frac{l}{2} = \frac{ml^2}{3} \alpha$$ alpha = frac3g2l$$\alpha = \frac{3g}{2l}$$

### Step 2: Calculate Force and Tension The acceleration of the center of mass (CM) is downwards: a_c = alpha fracl2 = left(frac3g2lright) left(fracl2right) = frac3g4$$a_{c} = \alpha \frac{l}{2} = \left(\frac{3g}{2l}\right) \left(\frac{l}{2}\right) = \frac{3g}{4}$$ Applying Newton's second law for translational motion of the CM in vertical direction: mg - T = m a_c$$mg - T = m a_{c}$$ T = mg - m a_c = mg - m left(frac3g4right) = fracmg4$$T = mg - m a_{c} = mg - m \left(\frac{3g}{4}\right) = \frac{mg}{4}$$ ### Pattern Recognition Classic 'cut string' rigid body problem. Always take torque about the pivot/hinge point to find alpha$\alpha$, then relate the center of mass linear acceleration a = r_cm alpha$a = r_{cm} \alpha$ to find the unknown tension using F_textnet = ma$F_{\text{net}} = ma$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion