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A solid sphere and a hollow sphere of equal mass and radius roll without slipping with the same kinetic energy. The ratio of their linear speeds (solid : hollow) is

A1
B√(25/21)
C√(2/3)
D√(5/3)
Answer & Solution
Correct answer: B. √(25/21)
1. KE_roll = ½mv² + ½Iω² with ω = v/R. For solid sphere I = (2/5)mR² giving KE = (7/10)mv². 2. For hollow sphere I = (2/3)mR² giving KE = (5/6)mv². 3. Equating: (7/10)mv_s² = (5/6)mv_h². So v_s²/v_h² = (5/6)/(7/10) = 25/21. 4. Therefore v_s/v_h = √(25/21). _Source: Maharashtra Balbharati Std XII Physics, Ch 1 "Rotational Dynamics" §1.8_
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