IIT-JEE 2007 · Paper 2 · Physics Q4

← IIT-JEE 2007 · Paper 2

PhysicsSingle correct+3 / -1Question 4 of 66

A small object of uniform density rolls up a curved surface with an initial velocity $v$ . It reaches up to a maximum height of $\dfrac{3v^2}{4g}$ with respect to the initial position. The object is

A small round object at the bottom of a curved surface that rises to the right, moving to the right with velocity v.

ring

solid sphere

hollow sphere

disc

Answer and worked solution

Official answer: D

Rolling without slipping, so friction does no work and all the kinetic energy becomes potential energy:
$\frac12mv^2\left(1+\frac{k^2}{R^2}\right)=mgh=mg\frac{3v^2}{4g}\implies1+\frac{k^2}{R^2}=\frac32.$

So $k^2=\frac12R^2$ , the value for a disc (ring 1, solid sphere $\frac25$ , hollow sphere $\frac23$ ).

(D) disc

Topics:Rolling motion Conservation of mechanical energy

← Previous question Next question →

IIT-JEE 2007 · Paper 2, Physics question 4