A disc of radius 0.1 m rolls without sliding on a horizontal surface with a velocity of 6 m s^(-1). It then ascends a smooth continuous track as shown…

Question

A disc of radius 0.1 m rolls without sliding on a horizontal surface with a velocity of 6 m s^(-1). It then ascends a smooth continuous track as shown in the figure. The height (in m) up to which it will ascend is : ( g=10 m s^(-2) )

MARKS Content Team · Accepted Answer

Correct answer: 2.7. 1/2Iw^2+1/2mv^2=mgh 1/2(mr^2)/2.(v^2)/(R^2)+1/2mv^2=mgh (v^2)/4+(v^2)/2=gh (v^2+2v^2)/4=gh 3/4×36=10×h 27/10=h⇒2.7

A disc of radius $0.1 m$ rolls without sliding on a horizontal surface with a velocity of $6 m s^{- 1}$ . It then ascends a smooth continuous track as shown in the figure. The height (in $m$ ) up to which it will ascend is : ( $g = 10 m s^{- 2}$ )

Solution

More Work Power Energy Questions

A disc of radius 0.1 m rolls without sliding on a horizontal surface with a velocity of 6 m s-1. It then ascends a smooth continuous track as shown in the figure. The height (in m) up to which it will ascend is : ( g=10 m s-2 )

Solution

More Work Power Energy Questions

A disc of radius $0.1 m$ rolls without sliding on a horizontal surface with a velocity of $6 m s^{- 1}$ . It then ascends a smooth continuous track as shown in the figure. The height (in $m$ ) up to which it will ascend is : ( $g = 10 m s^{- 2}$ )