A particle is moving in a circle of radius 50… - JEE Main 2024 Physics PYQ

JEE Main 2024 — Physics Mechanics

2024integermedium

A particle is moving in a circle of radius $50\mathrm{cm}$ in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at $t=0$ is $4m{s}^{-1}$ , the time taken to complete the first revolution will be $\frac{1}{\alpha }[1-{e}^{-2\pi }]s$ , where $\alpha =$ ______.

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Held on 29 Jan 2024 · Verified 6 Jul 2026.

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Solution

Given:

$|{\vec{a}}_{c}|=|{\vec{a}}_{t}|$

$\Rightarrow \frac{{v}^{2}}{r}=\frac{dv}{dt}$

$\Rightarrow {\int }_{4}^{v}\frac{dv}{{v}^{2}}={\int }_{0}^{t}\frac{dt}{r}$

$\Rightarrow {[\frac{-1}{v}]}_{4}^{v}=\frac{t}{r}$

$\Rightarrow \frac{-1}{v}+\frac{1}{4}=2t$

$\Rightarrow v=\frac{4}{1-8t}=\frac{ds}{dt}$

$\Rightarrow 4{\int }_{0}^{t}\frac{dt}{1-8t}={\int }_{0}^{s}ds$

$(r=0.5m,\text{distance}\text{covered}s=2\pi r=\pi )$

$\Rightarrow 4\times \frac{[\mathrm{ln}(1-8t){]}_{0}^{t}}{-8}=\pi$

$\Rightarrow \ell n(1-8t)=-2\pi$

$\Rightarrow 1-8t={e}^{-2\pi }$

$\Rightarrow t=(1-{e}^{-2\pi })\frac{1}{8}s$

So, $\alpha =8$ .

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Solution

Given:

$|{\vec{a}}_{c}|=|{\vec{a}}_{t}|$

$\Rightarrow \frac{{v}^{2}}{r}=\frac{dv}{dt}$

$\Rightarrow {\int }_{4}^{v}\frac{dv}{{v}^{2}}={\int }_{0}^{t}\frac{dt}{r}$

$\Rightarrow {[\frac{-1}{v}]}_{4}^{v}=\frac{t}{r}$

$\Rightarrow \frac{-1}{v}+\frac{1}{4}=2t$

$\Rightarrow v=\frac{4}{1-8t}=\frac{ds}{dt}$

$\Rightarrow 4{\int }_{0}^{t}\frac{dt}{1-8t}={\int }_{0}^{s}ds$

$(r=0.5m,\text{distance}\text{covered}s=2\pi r=\pi )$

$\Rightarrow 4\times \frac{[\mathrm{ln}(1-8t){]}_{0}^{t}}{-8}=\pi$

$\Rightarrow \ell n(1-8t)=-2\pi$

$\Rightarrow 1-8t={e}^{-2\pi }$

$\Rightarrow t=(1-{e}^{-2\pi })\frac{1}{8}s$

So, $\alpha =8$ .

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Solution

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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