Practice question

Given: A 0.8 kg stone in a vertical circle of radius 2 m has a speed of 15 m/s at the bottom. What is the tension at the top if a 5 N tangential force acts? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: Bottom: T_b - mg = m v_b² / r Rightarrow T_b - 0.8 × 10 = 0.8 × (15)² / 2. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: T_b - 8 = 0.8 × 112.5 Rightarrow T_b - 8 = 90 Rightarrow T_b = 98 N (tangential doesn’t affect radial). Energy: 1/2 m v_b² - 1/2 m v_t² = 2mg Rightarrow 0.4 × 225 - 0.4 v_t² = 2 × 0.8 × 10 . 90 - 0.4 v_t² = 16 Rightarrow 0.4 v_t² = 74 Rightarrow v_t² = 185 Rightarrow v_t approx 13.6 m/s . Top: T_t + mg = m v_t² / r Rightarrow T_t + 8 = 0.8 × 185 / 2 Rightarrow T_t + 8 = 74 Rightarrow T_t = 66 N . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.