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IIT JEE

A thin rod of mass M and length a ….

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A thin rod of mass M and length a is free to rotate in horizontal plane about a fixed vertical axis passing through point O. A thin circular disc of mass M and of radius a/4 is pivoted on this rod with its center at a distance a/4 from the free end so that it can rotate freely about its vertical axis, as shown in the figure. Assume that both the rod and the disc have uniform density and they remain horizontal during the motion. An outside stationary observer finds the rod rotating with an angular velocity $\Omega$ and the disc rotating about its vertical axis with angular velocity $4\Omega$. The total angular momentum of the system about the point O is $(\frac {Ma^2\Omega}{48}) n$.

The value of n is ___.

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IIT JEE

$I = \int\limits_0^\infty {\frac{{{e^{ – nx}} – {e^{ – x}}}}{x}dx} = ?$

$n \in \mathbb{N}$

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Let, ${e^{ – x}} = t$

$ \Rightarrow  – {e^{ – x}}dx = dt$

$ \Rightarrow  – tdx = dt$

$I = \int\limits_1^0 {\frac{{{t^n} – t}}{{ – \ln t}}\frac{{dt}}{{ – t}}}  =  – \int\limits_0^1 {\frac{{{t^{n – 1}} – 1}}{{\ln t}}dt} $

But, $I(\lambda ) = \int\limits_0^1 {\frac{{{x^\lambda } – 1}}{{\ln x}}dx = \ln (\lambda  + 1)} $ (Proof can be seen in this problem)

So, $I =  – \ln (n – 1 + 1) =  – \ln n$

Math

$a + b + c = 2022$

$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{{2022}}$

$S = \frac{1}{{{a^{2023}}}} + \frac{1}{{{b^{2023}}}} + \frac{1}{{{c^{2023}}}} = ?$

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We have, $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{{2022}} = \frac{1}{{a + b + c}}$

$ \Rightarrow \frac{1}{a} + \frac{1}{b} + \left( {\frac{1}{c} – \frac{1}{{a + b + c}}} \right) = 0$

$ \Rightarrow \frac{{a + b}}{{ab}} + \frac{{a + b}}{{(a + b + c)c}} = 0$

$ \Rightarrow (a + b)\frac{{(ab + bc + ca + {c^2})}}{{abc(a + b + c)}} = 0$

$ \Rightarrow (a + b)\frac{{(b + c)(c + a)}}{{abc(a + b + c)}} = 0$

$ \Rightarrow (a + b)(b + c)(c + a) = 0$

$a + b = 0$ Or $b + c = 0$ Or $c + a = 0$

If $a + b = 0$, $a + b + c = 0 + c = c = 2022$

$S = \frac{1}{{{a^{2023}}}} + \frac{1}{{{b^{2023}}}} + \frac{1}{{{c^{2023}}}} = \frac{1}{{{a^{2023}}}} + \frac{1}{{{{( – a)}^{2023}}}} + \frac{1}{{{{2022}^{2023}}}} = \frac{1}{{{{2022}^{2023}}}}$

Due to symmetry, other cases would yield the same final answer.