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
• B. 0.9 m
• C. 2.7 m
• D. 1.8 m

Answer 1

Answer: (D)

1.8 m

Answer 2

Let m be the mass of the disc. Then translational kinetic energy of the disc is :

K_T = $\frac { 1 } { 2 }$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.

KT = mgh

or $\frac { 1 } { 2 }$mv² = mgh

or h = = $\frac { ( 6 ) ^ { 2 } } { 2 \times 10 }$ = 1.8 m