JEE Main 2020 : As shown in the figure, when a spherical cavity (centred at O) of radius...

In this video, we solve a Physics problem from JEE Main 2020 (8th January – 2nd Shift) based on the centre of mass of a sphere with a spherical cavity. We derive the relation between the radius of the main sphere (R) and the cavity, given that the new centre of mass lies on the cavity’s surface. Question (JEE Main 2020 – Physics): As shown in the figure, when a spherical cavity (centred at O) of radius 1 is cut out of a uniform sphere of radius R (centred at C), the centre of mass of the remaining (shaded) part of the sphere is at G, i.e., on the surface of the cavity. R can be determined by the equation: (a) (R² + R + 1)(2 − R) = 1 (b) (R² + R − 1)(2 − R) = 1 (c) (R² − R − 1)(2 − R) = 1 (d) (R² − R + 1)(2 − R) = 1 Concepts Covered: • Centre of mass of composite bodies • Method of negative mass distribution • Geometry of spheres and cavities • Application of COM formula along one axis #AkbarClasses #JEEMainPhysics #IITJEEPhysics Contact us: ✆ 7366863696 ✉ [email protected] Support us : Account Number - 33307235385 IFSC- SBIN0000117 MD ZAREEF AKBAR FOLLOW US ON: Instagram: https://bit.ly/3jMt50l Twitter: https://bit.ly/3qhVqOD Facebook: https://bit.ly/3b6asQW YouTube: https://bit.ly/3tRItNC