Question

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

• A. $2.4\text{m}$
• B. $0.9\text{m}$
• C. $2.7\text{m}$
• D. $1.8\text{m}$

Answer 1

The correct option is D $1.8\text{m}$

Let $m$ be the mass of the disc. Then translational kinetic energy of the disc is :
$K_T=\frac{1}{2}m{v}^2...\left(i\right)$

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.
$K_T=mgh$
or $\frac{1}{2}m{v}^2=mgh$
or $h=\frac{v^2}{2g}=\frac{(6{)}^2}{2\times 10}=1.8\text{m}$

hence option D is the correct answer