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Correct option is D)

so velocity at top point ($T$)

$⇒$ $V_{T}$$=gR ...(1)$

From the diagram

in right triangle $△OMA$

$⇒$$MA=OP=OPcosθ$

At point $A$, $θ=60°$

$⇒H_{A}=OG−OP⇒H_{A}=R−Rcos60°=2R ...(2)$

in right triangle $△OMB$

$⇒$$MB=QP=QPcos(180−θ)$

At point $B$, $θ=120°$

$⇒H_{B}=OG+OQ⇒H_{B}=R+Rcos60°=23R ...(4)$

Using energy conservation equation at point $T$ and $A$

$⇒21 mV_{T}_{2}+mg(2R)=21 mV_{A}_{2}+mgH_{1}...(2)$

putting the value of eqn (1), eqn (2) in eqn (4)

$⇒21 m(gR )_{2}+mg(2R)=21 mV_{A}_{2}+mg2R $

$⇒V_{A}=2gR $

Using energy conservation equation at point $T$ and $B$

$⇒21 mV_{T}_{2}+mg(2R)=21 mV_{B}_{2}+mgH_{2}...(2)$

putting the value of eqn (1), eqn (3) in eqn (4)

$⇒21 m(gR )_{2}+mg(2R)=21 mV_{B}_{2}+mg23R $

$⇒V_{B}=2gR $

Ratio of $V_{A}$ and $V_{B}$$=2 :1$

So the correct option is ($D$)

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