A disc of radius text0.1 m rolls without sliding on a horizontal surface with a velocity of 6 ms- 1 . It then ascends a smooth continuous track as shown in figure. The height upto which it will ascend is (g = 10 m s- 2)

Question

A disc of radius text0.1 m rolls without sliding on a horizontal surface with a velocity of 6 ms- 1 . It then ascends a smooth continuous track as shown in figure. The height upto which it will ascend is (g = 10 m s- 2) <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-vzdsomhhym9f.png" /> (A) 2.4 m (B) 0.9 m (C) 2.7 m (D) 1.8 m

Tardigrade Admin · Accepted Answer

Let m be the mass of the disc. Then translational kinetic energy of the disc is: KT=(1/2)mv2 ....(1) When it ascends on a smooth track its rotational kinetic energy will remain same while translational kinetic energy will go on decreasing. At highest point Translational kinetic energy is transferred to Potential Energy. KT=mgh or (1/2)mv2=mgh or h=(v2/2 g)=((6)2/2 × 10)=1.8 m

Q. 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 figure. The height upto which it will ascend is $(g = 10 m s^{- 2})$

$2.4 m$

$0.9 m$

$2.7 m$

$1.8 m$

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

2.4m

0.9m

2.7m

1.8m

Q. 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 figure. The height upto which it will ascend is $(g = 10 m s^{- 2})$

$2.4 m$

$0.9 m$

$2.7 m$

$1.8 m$