As v=1+r2k22gh Given h=4g3v2 v2=1+r2k22gh=4g(1+r2k2)2g3v2=4g(1+v2u2)6gv2 1=2(1+v2k2)3 or 1+r2k2=23 or r2k2=23−1=21 k2=21r2 (Equation of disc) Hence, the object is disc.
NEET UG 2013 — Physics Mechanics
A small object of uniform density rolls up a curved surface with an initial velocity v′. It reaches up to a maximum height of 4g3v2 with respect to the initial position. The object is
Held on 30 Apr 2013 · Verified 9 Jul 2026.
ring
solid sphere
hollow sphere
disc
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