A disc of radius 0.1 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 (in cm ): ( g=10 m s- 2 )

Question

A disc of radius 0.1 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 (in cm ): ( g=10 m s- 2 ) <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-zivh4lmtkaxvvfuk.jpg" /> (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Since it ascends on a smooth track it will not experience any torque. Therefore, its rotational kinetic energy will not change. Hence, only translational kinetic energy will convert into potential energy.

\Rightarrow \frac{1}{2} m v^{2} = m g h

\Rightarrow h = \frac{v ^{2}}{2 g} \Rightarrow h = \frac{6 ^{2}}{2 \times 10} = 1.8 m = 180 c m

Q. A disc of radius 0.1m 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 (in cm ) : ( g=10ms−2 )

Since it ascends on a smooth track it will not experience any torque. Therefore, its rotational kinetic energy will not change. Hence, only translational kinetic energy will convert into potential energy. ⇒21​mv2=mgh ⇒h=2gv2​⇒h=2×1062​=1.8m=180cm

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 (in $c m$ ) : ( $g = 10 m s^{- 2}$ )