Congratulations for being selected to represent India in ijso.
Also what about JOMPC?
–
Shivam Jadhav
·
3 months, 4 weeks ago

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@Shivam Jadhav
–
Thanks. It will be out day after tomorrow. I have marked all the solutions I'll compile and post it on monday.
–
Rajdeep Dhingra
·
3 months, 4 weeks ago

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Bro is your second doubt cleared ? i am getting it's answer as root 5g/7(R-r) tell if not cleared i will give the solution ( if my answer is correct.) .
–
Shubham Dhull
·
1 month ago

Hy rajdeep, can u give some tips for prepartion of nsejs , as my brother will be giving this year, presently he is in class 9.
–
Rahul Bubna
·
3 months, 4 weeks ago

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@Rahul Bubna
–
Read 9th and 10th class NCERT books thoroughly , if possible read 11th and 12th class NCERT also.(only relevant topics). For NSEJS most important is your speed of solving questions.
–
Rajdeep Dhingra
·
3 months, 4 weeks ago

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@Rajdeep Dhingra
–
If he starts from June 10 , is he will be able to clear nsejs ? He also goes to coaching ...
–
Rahul Bubna
·
3 months, 3 weeks ago

Question 2:
Let \( f \)denote the friction force acting on the body. Now, ablut the center of mass writing the torque equation,
\( f \cdot r = \dfrac{MR^{2}\alpha}{2} \)
\( f = \dfrac{MR\alpha}{2} \)
But since the cylinder doesn't slip \( r \alpha = a_{cm} \)
\( f = \dfrac{ma_{cm}}{2} \)
Now,
\( mg\sin(\theta) - f = ma_{cm} \)
\( mg\sin(\theta) = \dfrac{3ma_{cm}}{2} \)
\( a_{cm} = \dfrac{2g\sin(\theta)}{3} \)
\( \alpha = \dfrac{2g \sin(\theta)}{3r} \)
For small values of \( \theta \), \( \sin(\theta) \approx \theta \)
\( \alpha = - \dfrac{2g \theta}{3r} \)
The negative sign just denotes that direction of alpha is opposite to displacement.
This is the rotational analogue of \( a = -\omega^{2}x \)
\( \therefore \omega = \sqrt{\dfrac{2g}{3r}} \)
Frequency \( \dfrac{\omega}{2\pi} = \dfrac{1}{2\pi} \sqrt{\dfrac{2g}{3r}} \)
–
Vighnesh Shenoy
·
4 months ago

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@Vighnesh Shenoy
–
It is wrong bro. \( \theta\) is the angle you moved the center of mass not the angle by which you rotated the sphere. Hence you can't equate \( {\omega}^2 \) to \(\dfrac{2g}{3r}\).
–
Rajdeep Dhingra
·
2 months, 3 weeks ago

## Comments

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TopNewestCongratulations for being selected to represent India in ijso. Also what about JOMPC? – Shivam Jadhav · 3 months, 4 weeks ago

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– Rajdeep Dhingra · 3 months, 4 weeks ago

Thanks. It will be out day after tomorrow. I have marked all the solutions I'll compile and post it on monday.Log in to reply

Bro is your second doubt cleared ? i am getting it's answer as root 5g/7(R-r) tell if not cleared i will give the solution ( if my answer is correct.) . – Shubham Dhull · 1 month ago

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– Rajdeep Dhingra · 1 month ago

Cleared. Thanks anyways. P.S : your answer is correct.Log in to reply

Hey @Rajdeep Dhingra why not coming to fiitjee ? PLease tell the truth !! – Parth Bhardwaj · 1 month, 2 weeks ago

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Which book is it ? – John Verghese · 3 months, 4 weeks ago

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– Rajdeep Dhingra · 3 months, 3 weeks ago

I don't know. Some old book for JEE subjective.Log in to reply

Hy rajdeep, can u give some tips for prepartion of nsejs , as my brother will be giving this year, presently he is in class 9. – Rahul Bubna · 3 months, 4 weeks ago

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– Rajdeep Dhingra · 3 months, 4 weeks ago

Read 9th and 10th class NCERT books thoroughly , if possible read 11th and 12th class NCERT also.(only relevant topics). For NSEJS most important is your speed of solving questions.Log in to reply

– Rahul Bubna · 3 months, 3 weeks ago

If he starts from June 10 , is he will be able to clear nsejs ? He also goes to coaching ...Log in to reply

– Rajdeep Dhingra · 3 months, 3 weeks ago

I think so.Log in to reply

Solution for Question 1:Log in to reply

– Rajdeep Dhingra · 3 months, 4 weeks ago

Thanks !Log in to reply

– Saarthak Marathe · 3 months, 4 weeks ago

You are welcomeLog in to reply

Question 2:

Let \( f \)denote the friction force acting on the body. Now, ablut the center of mass writing the torque equation,

\( f \cdot r = \dfrac{MR^{2}\alpha}{2} \)

\( f = \dfrac{MR\alpha}{2} \)

But since the cylinder doesn't slip \( r \alpha = a_{cm} \)

\( f = \dfrac{ma_{cm}}{2} \) Now,

\( mg\sin(\theta) - f = ma_{cm} \)

\( mg\sin(\theta) = \dfrac{3ma_{cm}}{2} \)

\( a_{cm} = \dfrac{2g\sin(\theta)}{3} \)

\( \alpha = \dfrac{2g \sin(\theta)}{3r} \)

For small values of \( \theta \), \( \sin(\theta) \approx \theta \)

\( \alpha = - \dfrac{2g \theta}{3r} \)

The negative sign just denotes that direction of alpha is opposite to displacement.

This is the rotational analogue of \( a = -\omega^{2}x \)

\( \therefore \omega = \sqrt{\dfrac{2g}{3r}} \)

Frequency \( \dfrac{\omega}{2\pi} = \dfrac{1}{2\pi} \sqrt{\dfrac{2g}{3r}} \) – Vighnesh Shenoy · 4 months ago

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– Rajdeep Dhingra · 2 months, 3 weeks ago

It is wrong bro. \( \theta\) is the angle you moved the center of mass not the angle by which you rotated the sphere. Hence you can't equate \( {\omega}^2 \) to \(\dfrac{2g}{3r}\).Log in to reply

– Rajdeep Dhingra · 3 months, 4 weeks ago

ThanksLog in to reply

@aryan goyat@Prakhar Bindal@Archit Agrawal@Aniket Sanghi Please help – Rajdeep Dhingra · 4 months ago

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