A block of mass m slides down along the smooth surface of the bowl (radius R ) from the rim to the bottom. The velocity of the block at the bottom will be

Question

A block of mass m slides down along the smooth surface of the bowl (radius R ) from the rim to the bottom. The velocity of the block at the bottom will be (A) √g R (B) √2 R g (C) 2 √π R g (D) √π R g

Tardigrade Admin · Accepted Answer

Gain in KE. Loss in PE (1/2) m v2=m g R ⇒ v=√2 g R

Q. A block of mass $m$ slides down along the smooth surface of the bowl (radius $R$ ) from the rim to the bottom. The velocity of the block at the bottom will be

$g R$

$2 R g$

$2 π R g$

$π R g$

Gain in $K E$ . Loss in PE
$\frac{1}{2} m v^{2} = m g R \Rightarrow v = 2 g R$

Q. A block of mass m slides down along the smooth surface of the bowl (radius R ) from the rim to the bottom. The velocity of the block at the bottom will be

gR​

2Rg​

2πRg​

πRg​