A small object of uniform density rolls up a curved surface with an initial velocity v ′ . It reaches up to a maximum height of 4 g 3 v 2 with respect to the initial position. The object isNEET 2013 (All India)

As v = 1 + r 2 k 2 2 g h Given h = 4 g 3 v 2 v 2 = 1 + r 2 k 2 2 g h = 4 g ( 1 + r 2 k 2 ) 2 g 3 v 2 = 4 g ( 1 + v 2 u 2 ) 6 g v 2 1 = 2 ( 1 + v 2 k 2 ) 3 or 1 + r 2 k 2 = 2 3 or r 2 k 2 = 2 3 − 1 = 2 1 k 2 = 2 1 r 2 (Equation of disc) Hence, the object is disc.

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Physics - Rotation Question with Solution | TestHub

PhysicsRotationTranslational + RotationalHard2 minPYQ_2013

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A small object of uniform density rolls up a curved surface with an initial velocity $v^{'}$ . It reaches up to a maximum height of $\frac{3 v ^{2}}{4 g}$ with respect to the initial position. The object isNEET 2013 (All India)

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As $v = \frac{2 g h}{1 + \frac{k ^{2}}{r ^{2}}}$ Given $h = \frac{3 v ^{2}}{4 g}$ $v^{2} = \frac{2 g h}{1 + \frac{k ^{2}}{r ^{2}}} = \frac{2 g 3 v ^{2}}{4 g ( 1 + \frac{k ^{2}}{r ^{2}} )} = \frac{6 g v ^{2}}{4 g ( 1 + \frac{u ^{2}}{v ^{2}} )}$ $1 = \frac{3}{2 ( 1 + \frac{k ^{2}}{v ^{2}} )}$ or $1 + \frac{k ^{2}}{r ^{2}} = \frac{3}{2}$ or $\frac{k ^{2}}{r ^{2}} = \frac{3}{2} - 1 = \frac{1}{2}$ $k^{2} = \frac{1}{2} r^{2}$ (Equation of disc) Hence, the object is disc.

Stream:NEETSubject:PhysicsTopic:RotationSubtopic:Translational + Rotational

⏱ 2mℹ️ Source: PYQ_2013

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