A small object of uniform density rolls up a curved surface with an initial velocity '\nu'. It reaches upto a maximum height of

\frac{3\nu^{2}}{4\text{g}}with respect to the initial position. The object is:

  • Option 1)

    Disc

  • Option 2)

    Ring

  • Option 3)

    Solid sphere

  • Option 4)

    Hollow sphere

 

Answers (2)

As we learned in concept

Rolling of a body on an inclined plane -

a= frac{gsin Theta }{1+frac{K^{2}}{R^{2}}}

f= frac{mgsin Theta }{1+frac{R^{2}}{K^{2}}}

- wherein

K=Radius of gyration

Theta = Angle of inclination

 

 

 

V=\sqrt{\frac{2gh}{1+\frac{K^{2}}{r^{2}}}}\:\:\:\:\:\:,\:\:\:\:\:h=\frac{3v^{2}}{4g}

After solving 1+\frac{K^{2}}{r^{2}}=\frac{3}{2}

K^{2}=\frac{1}{2}r^{2}                     so object is disc.


Option 1)

Disc

This option is correct.

Option 2)

Ring

This option is incorrect.

Option 3)

Solid sphere

This option is incorrect.

Option 4)

Hollow sphere

This option is incorrect.

An object is dropped from a height h from the ground.  Every time it hits the ground it looses 50% of its kinetic energy.  The total distance covered ast\rightarrow \infty is : can you please solve this sorry but i really need a solution but iam not getting any ans

 

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