A ring starts to roll down the inclined plane of height h without slipping. The velocity when it reaches the ground is

Question

A ring starts to roll down the inclined plane of height h without slipping. The velocity when it reaches the ground is (A) √(1 0 g h/7) (B) √(4 g h/7) (C) √(4 g h/3) (D) √g h

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-zzrcx9hw4xnu.png" /> mgh=(1/2)mv2+(1/2)IÏ2 =(1/2)mv2+(1/2)mr2(v2/r2) gh=v2 v=√gh

Q. A ring starts to roll down the inclined plane of height $h$ without slipping. The velocity when it reaches the ground is

$\frac{10 g h}{7}$

$\frac{4 g h}{7}$

$\frac{4 g h}{3}$

$g h$

$m g h = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}$
$= \frac{1}{2} m v^{2} + \frac{1}{2} m r^{2} \frac{v ^{2}}{r ^{2}}$
$g h = v^{2}$
$v = g h$

Q. A ring starts to roll down the inclined plane of height h without slipping. The velocity when it reaches the ground is

710gh​​

74gh​​

34gh​​

gh​