Q63 / 90 · JEE Main 2021 (25 Feb Shift 2) · Physics

A sphere of radius $a$ and mass $m$ rolls along a horizontal plane with constant speed $v_{0}$ . It encounters an inclined plane at angle $θ$ and climbs upward. Assuming that it rolls without slipping, how far up the sphere will travel?

A
$\frac{2}{5} \frac{v_{0}^{2}}{g \sin θ}$
Correct answer
B
$\frac{7 v_{0}^{2}}{10 g \sin θ}$
C
$\frac{v_{0}^{2}}{5 g \sin θ}$
D
$\frac{v_{0}^{2}}{2 g \sin θ}$

Solution

From energy conservation
$mgh = \frac{1}{2} {mv}_{0}^{2} + \frac{1}{2} {Iω}^{2}$
$\Rightarrow mgh = \frac{1}{2} {mv}_{0}^{2} + \frac{1}{2} \times \frac{2}{5} {ma}^{2} \times \frac{v_{0}^{2}}{a^{2}}$
$\Rightarrow gh = \frac{1}{2} v_{0}^{2} + \frac{1}{5} v_{0}^{2}$
$\Rightarrow gh = \frac{7}{10} v_{0}^{2}$

From triangle, $\sin θ = \frac{h}{l}$
Then $h = l \sin θ$
$\Rightarrow l \sin θ = \frac{7}{10} \frac{v_{0}^{2}}{g}$
Thus, the sphere will travel distance, $l = \frac{7}{10} \frac{{v_{0}}^{2}}{g \sin θ}$ .

Hence, the correct answer is option: 'B'.

A sphere of radius a and mass m rolls along a horizontal plane with constant speed v0. It encounters an inclined plane at angle θ and climbs upward. Assuming that it rolls without slipping, how far up the sphere will travel?

Solution

Relevant Questions

A sphere of radius $a$ and mass $m$ rolls along a horizontal plane with constant speed $v_{0}$ . It encounters an inclined plane at angle $θ$ and climbs upward. Assuming that it rolls without slipping, how far up the sphere will travel?