A particle of mass m is projected vertically upwards from point - on the ground- The ratio of potential energy at point A and potential energy at B is

Let, nbsp; PE_A = Potential Energy at A PE_B = Potential Energy at B PE_A = mgH ...(i) PE_B = mg ( H3 ) ...(ii) nbsp;Dividing both the equation (i) and ( ...

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Standard XII

Physics

Potential Energy

A particle of...

Question

A particle of mass $m$ is projected vertically upwards from point $Δ$ on the ground. The ratio of potential energy at point $A$ and potential energy at $B$ is

1 : 3

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2 : 3

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3 : 2

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3 : 1

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Solution

The correct option is D $3 : 1$
Let, $P E_{A}$ = Potential Energy at $A$
$P E_{B}$ = Potential Energy at $B$

$P E_{A} = m g H . . . (i)$
$P E_{B} = m g (\frac{H}{3}) . . . (i i)$
Dividing both the equation $(i)$ and $(i i)$
$\Rightarrow \frac{P E_{A}}{P E_{B}} = \frac{m g H}{m g (\frac{H}{3})} = 3$

Hence option D is the correct answer

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A particle of mass m is projected vertically upwards from point Δ on the ground. The ratio of potential energy at point A and potential energy at B is

The correct option is D 3:1Let, PEA = Potential Energy at A PEB = Potential Energy at B PEA=mgH ...(i) PEB=mg(H3) ...(ii) Dividing both the equation (i) and (ii) ⇒PEAPEB=mgHmg(H3)=3 Hence option D is the correct answer

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