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Question

A disc of radius 0.1 m rolls without sliding on a horizontal surface with a velocity of 6 m/s. It then ascends a smooth continuous track as shown in figure. The height upto which it will ascend is (g=10 m/s2).


A
2.4 m
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B
0.9 m
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C
2.7 m
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D
1.8 m
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Solution

The correct option is D 1.8 m
Let m be the mass of the disc. Then translational kinetic energy of the disc is :
KT=12mv2 ...(i)

When it ascends on a smooth track its rotational kinetic energy will remain while translational kinetic energy will go on decreasing as the surface is smooth.

At highest point.
KT=mgh
or 12mv2=mgh
or h=v22g=(6)22×10=1.8 m

hence option D is the correct answer

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