Question

refer to the attachment for the question...

class 11 physics..
plz explain it(^‿^)
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⚠️don't give spam answers ...
small children stay away from this question :(

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Answer 1

The correct answer is option : (d) Zero

We the time period for compound pendulum (T) is given by,

$T = 2\pi *\sqrt{\frac{L}{g} }$

where $L = \frac{l^2+k^2}{l}$

$l = distance\ of\ centre\ of\ mass\ from\ point\ of\ suspension\\k = radius\ gyration$

To get minimum time period of osciallation L should be minimum

∵ k is a constant , $l$ has to be minimum

⇒ The distance from the centre of the hole to the suspended axis must be minimum.

   ∴ the minimum distance can be Zero

Answer 2

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Question\;:-}}}$

Here the concept of Time Period has been used. We need to find out the minimum distance from the centre so that the time period should be minimum. On comparing both sides and applying the formula of Moment of Inertia, we can find the answer.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{Time\;Period,\;T\;=\;\bf{2\:\pi\:\sqrt{\dfrac{I}{m\:g\:d}}}}}$

$\\\;\boxed{\sf{Moment\;of\;Inertia,\;I\;=\;\bf{\dfrac{m\:R^{2}}{2}\;+\;m\:d^{2}}}}$

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★ Solution :-

Given,

» Radius of the circle = R

• Let the minimum distance of the hole from centre from which the horizontal axis should pass be at a distance of 'd'.

• Let the mass of the disc be 'm'

• Let the acceleration due to gravity be 'g'

• Let the required time period be 'T'.

Using the Formula of Time Period, we get,

$\\\;\;\sf{:\rightarrow\;\;Time\;Period,\;T\;=\;\bf{2\:\pi\:\sqrt{\dfrac{I}{m\:g\:d}}}}$

On comparing LHS and RHS, we get,

$\\\;\;\bf{:\rightarrow\;\;T\;\;\propto\;\;2\pi}$

$\\\;\;\bf{:\rightarrow\;\;T\;\;\propto\;\;\sqrt{\dfrac{I}{m\:g\:d}}}$

But we know that 2π is a dimensionless constant. This means, the required equation will be,

$\\\;\;\bf{:\rightarrow\;\;T\;\;\propto\;\;\sqrt{\dfrac{I}{m\:g\:d}}}$

Now we need to find the Minimum Time Period.

From above equation we see, if T will be minimum that, RHS will also be minimum.

This means, we get when T is minimum,

$\\\;\;\sf{:\rightarrow\;\;\;\sqrt{\dfrac{I}{m\:g\:d}}\;\;should\;be\;minimum.}$

Now applying the Formula of Moment of Inertia,

$\\\;\;\sf{:\Longrightarrow\;\;Moment\;of\;Inertia,\;I\;=\;\bf{\dfrac{m\:R^{2}}{2}\;+\;m\:d^{2}}}$

in the equation of Minimum Time Period, we get,

$\\\;\;\sf{:\Longrightarrow\;\;\;\sqrt{\dfrac{I}{m\:g\:d}}\;\;should\;be\;minimum.}$

$\\\;\;\sf{:\Longrightarrow\;\;\;\sqrt{\dfrac{\dfrac{m\:R^{2}}{2}\;+\;m\:d^{2}}{m\:g\:d}}\;\;should\;be\;minimum.}$

$\\\;\;\sf{:\Longrightarrow\;\;\;\sqrt{\dfrac{m \bigg(\dfrac{R^{2}}{2}\;+\;d^{2}\bigg)}{m\:g\:d}}\;\;should\;be\;minimum.}$

Since, minimum value is 0.

$\\\;\;\sf{:\Longrightarrow\;\;\;\sqrt{\dfrac{m \bigg(\dfrac{R^{2}}{2}\;+\;d^{2}\bigg)}{m\:g\:d}}\;\;=\;\;\bf{0}}$

$\\\;\;\sf{:\Longrightarrow\;\;\;\dfrac{m \bigg(\dfrac{R^{2}}{2}\;+\;d^{2}\bigg)}{m\:g\:d}\;\;=\;\;\bf{0^{2}}}$

By cancelling m, we get,

$\\\;\;\sf{:\Longrightarrow\;\;\;\dfrac{\dfrac{R^{2}}{2}\;+\;d^{2}}{g\:d}\;\;= \;\;0}$

We know that g is the acceleration due to gravity. So, for calculating distance, we don't need that.

$\\\;\;\sf{:\Longrightarrow\;\;\;\dfrac{\dfrac{R^{2}\;+\;2d^{2}}{2}}{d}\;\;=\;\;\bf{0}}$

Transposing d to other side, we get,

$\\\;\;\sf{:\Longrightarrow\;\;\dfrac{R^{2}\;+\;2d^{2}}{2}\;\;=\;\;\bf{d\;\times\;0}}$

$\\\;\;\sf{:\Longrightarrow\;\;\dfrac{R^{2}\;+\;2d^{2}}{2}\;\;=\;\;\bf{0}}$

$\\\;\;\sf{:\Longrightarrow\;\;R^{2}\;+\;2d^{2}\;\;=\;\;\bf{2\;\times\;0}}$

$\\\;\;\sf{:\Longrightarrow\;\;R^{2}\;+\;2d^{2}\;\;=\;\;\bf{0}}$

$\\\;\;\sf{:\Longrightarrow\;\;2d^{2}\;\;=\;\;\bf{-\;R^{2}}}$

Since, distance is always +ve (Postive). This, means,

$\\\;\;\sf{:\Longrightarrow\;\;2d^{2}\;\;=\;\;\bf{R^{2}}}$

$\\\;\;\sf{:\Longrightarrow\;\;d^{2}\;\;=\;\;\bf{\dfrac{R^{2}}{2}}}$

$\\\;\;\sf{:\Longrightarrow\;\;d\;\;=\;\;\bf{\sqrt{\dfrac{R^{2}}{2}}}}$

$\\\;\;\bf{:\Longrightarrow\;\;d\;\;=\;\;\bf{\dfrac{R}{\sqrt{2}}}}$

So option b.) R / √2 is correct ansawr.

$\\\;\underline{\boxed{\tt{Distance\;\;of\;\;Hole\;\;from\;\;Centre\;=\;\bf{\dfrac{R}{\sqrt{2}}}}}}$

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★ More Formulas to know :-

$\\\;\tt{\leadsto\;\;Radius\;\;of\;\;Gyration,\;K\;=\;\sqrt{\dfrac{I}{M}}}$

$\\\;\tt{\leadsto\;\;Rotational\;\;K.E.\;=\;\dfrac{1}{2}\;\times\;I \omega^{2}}$

$\\\;\tt{\leadsto\;\;Angular\;\;Momentum,\;L\;=\;I \omega}$

$\\\;\tt{\leadsto\;\;Total\;\;K.E.\;\;of\;\:Rolling\;\;Body\;=\;\dfrac{1}{2}\;(Mv^{2}\;+\;I \omega^{2})}$

$\\\;\tt{\leadsto\;\;Torque,\;\tau\;=\;I \alpha}$