Question

Hey! Pls solve the above que step by step.
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Answer 1

Explanation:

I attached one picture please refer it respected moderator

Answer 2

Answer

Option (C) !!

★ There is a circular part on which the chain is placed and the length of the chain is less than its entire surface. $l < \frac{πR}{2}$, now you have to bring the whole up i.e. on horizontal level, then you have to give your given potential energy to every particular right? , effectively you are giving height to each and every partical (mgh)

$\boxed{ \bf \: Work \: done = Change \: in \: potential \: energy \: (∆ PE) }$

But how to find this change in potential energy?

◇ Refer to the Attachment 1st !!

• dm is Mass
• Rd θ is length of R

Now the point is how much height will this dm particle have?

$\sf \: \frac{dm}{Rd \theta} = \frac{ Mass \: of \: pull \: chain}{ length \: of \: pull \: chain }$

So,

$\: \: \boxed {\bf \:dm = \frac{m}{l} \: Rd \theta}$

◇ Refer to the Attachment 2nd !!

Now we have to find the height

$\boxed{ \tt \: h = R (1 - \cos \theta)}$

For Small Element!

$\bf \: d \: PE = (dm)gh \\ \sf \implies( \frac{m}{l} Rd \theta)g(R)(1 - \cos \theta)$

☆ Now we want to find potential energy, then we have to integrate this whole function, so what is constant in this equation ? (R² , g , m)

We can write it like this

$\sf \: PE = \frac{mgR {}^{2} }{l} \int(1 - \cos \theta)d \theta$

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◇ Refer to the attachment 3rd !

θ = 0°

θ = $\sf [\frac{l}{R} ] = \frac{length of R } {Radius}$

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$\longrightarrow \: \frac{ mgR {}^{2} }{l} \int \limits _{0}{ \frac{l}{R}} (1 - \cos \theta)d \theta\\ \longrightarrow \frac{mgR {}^{2} }{l} [ \theta - \sin \theta] \frac{\frac{l}{R}} {0} \\ \\ \implies \boxed{ \frac{mgR {}^{2} }{l} [ \frac{l}{R} - \sin \frac{l}{R}] }$

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