A person of mass M is sitting… - JEE Main 2019 Physics PYQ

JEE Main 2019 — Physics Mechanics

2019mcqhard

A person of mass $M$ is sitting on a swing of length $L$ and swinging with and an angular amplitude ${\theta }_{0}.$ If the person stands up when the swing passes through its lowest point, the work done by him, assuming that his centre of mass moves by a distance $l (l<<L),$ is close to:

Official previous-year question

Held on 12 Apr 2019 · Verified 6 Jul 2026.

Options

A
$Mgl (1-{\theta }_{0}^{2})$
B
$Mgl(1+\frac{{\theta }_{0}^{2}}{2})$
C
$Mgl$
D
$Mgl(1+{\theta }_{0}^{2})$

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Solution

From angular momentum conservation

$\begin{matrix}M{V}_{0}L=MV(L-l) \\ V={V}_{0}(\frac{L}{L-l})\end{matrix}$

Now

$W=Mgl+\frac{1}{2}M[{{V}_{0}}^{2}{(\frac{L}{L-l})}^{2}-{{V}_{0}}^{2}]$

Since $l\ll L$

$\begin{matrix}W=Mgl+\frac{1}{2}M[{{V}_{0}}^{2}{(1-\frac{l}{L})}^{-2}-{{V}_{0}}^{2}] \\ =Mgl+\frac{1}{2}M[{{V}_{0}}^{2}(1+2\frac{l}{L})-{{V}_{0}}^{2}] \\ =Mgl+M[{{V}_{0}}^{2}(\frac{l}{L})]\end{matrix}$

From SHM

${V}_{0}=\omega A=\sqrt{\frac{g}{L}}\times L{\theta }_{0}=\sqrt{gL}{\theta }_{0}$

Using this

$\begin{matrix}W=Mgl+M[gL\times {{\theta }_{0}}^{2}\times (\frac{l}{L})] \\ W=Mgl[1+{{\theta }_{0}}^{2}]\end{matrix}$

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Solution

From angular momentum conservation

$\begin{matrix}M{V}_{0}L=MV(L-l) \\ V={V}_{0}(\frac{L}{L-l})\end{matrix}$

Now

$W=Mgl+\frac{1}{2}M[{{V}_{0}}^{2}{(\frac{L}{L-l})}^{2}-{{V}_{0}}^{2}]$

Since $l\ll L$

From SHM

${V}_{0}=\omega A=\sqrt{\frac{g}{L}}\times L{\theta }_{0}=\sqrt{gL}{\theta }_{0}$

Using this

$\begin{matrix}W=Mgl+M[gL\times {{\theta }_{0}}^{2}\times (\frac{l}{L})] \\ W=Mgl[1+{{\theta }_{0}}^{2}]\end{matrix}$

Options

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Solution

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JEE Main Mechanics in other years

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Options

Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics