Question

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class 11..
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Answer 1

The correct answer is option : (d) 2 sec.

We know the time period for compound pendulum (T)

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

were,  $L = \frac{l^2+k^2}{l}$

Here $l = distance\ of \ centre\ of\ centre\ of\ mass\ from\ point\ of\ suspension\\k = radius\ of\ gyration\ about\ parallel\ axis\ passing\ through\ the\ centre\ of\ mass$

Given: a ring of diameter 1 m

$l =$ R ;     k = R

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

⇒ Time period (T) = $2\pi *\sqrt{\frac{L}{g} } = 2\pi *\sqrt{\frac{2R}{g} }$

⇒ T = $2\pi *\sqrt{\frac{2*0.5}{9.81} }$

 ∴ T = 2.006 ≅ 2sec

HOPE THIS HELPS YOU !!   : )

Answer 2

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

Here the concept of Time Period of a Pendulum has been used. We are given that this ring performs Simple Harmonic Motion. Firstly using Parallel Axis theorem, we need to find the Moment of Inertia of the Ring at circumference. After finding, that we can apply the values and find the answer.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{T\;=\;\bf{2\pi\;\sqrt{\dfrac{I_{(Total)}}{mgd}}}}}$

$\\\;\boxed{\sf{I_{z}\;=\;\bf{I_{x}\;+\;I_{y}}}}$

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★ Solution :-

Given,

» Diameter of the Ring = d = 1 m

» Radius of the Ring = R = ½ × d = 0.5 m

We know that for a circular ring of Radius R and mass m, the axis is passing through its centre and perpendicular (⊥) to its plane, then its Moment of Inertia is mR²

This will give,

» Moment of Inertia of ring = I = mR²

» Acceleration due to gravity = g = 9.8 m/sec

» Value of π = 22/7

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~ For Moment of Inertia of Ring for performing SHM about the nail ::

We know that moment of inertia of a plane lamina about an axis ⊥ to its plane is equal to the sum of the moments of inertia of the lamina about any two mutual perpendicular axes in its plane and interesting each other at a point, where perpendicular axes pass through the lamina.

This is Theorem of Perpendicular Axes.

$\\\;\;\sf{:\rightarrow\;\;I_{z}\;=\;\bf{I_{x}\;+\;I_{y}}}$

where X - axis and Y - axis lie in the plane of the lamina and Z - axis is perpendicular to its plane and passes through the point of intersection of X - axis and Y - axis.

We have, Moment of Inertia of Circular Ring = mR².

This denotes both X and Y axes since the right is on vertical plane. Thus,

$\\\;\;\;\sf{\odot\;\;\;I_{x}\;\;=\;\;\bf{m\:R^{2}}}$

$\\\;\;\;\sf{\odot\;\;\;I_{y}\;\;=\;\;\bf{m\:R^{2}}}$

Now using Parallel - Axis Theorem, we get,

$\\\;\;\sf{:\rightarrow\;\;I_{z}\;=\;\bf{I_{x}\;+\;I_{y}}}$

$\\\;\;\sf{:\rightarrow\;\;I_{z}\;=\;\bf{m\:R^{2}\;+\;m\:R^{2}}}$

$\\\;\underline{\underline{\bf{:\rightarrow\;\;Moment\;of\;Inertia_{(Total)},\;I_{z}\;=\;\bf{2m\:R^{2}}}}}$

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~ For Time Period of the ring ::

We already have the required values. Let's apply them, and find the answer.

The Time Period is denoted by T.

$\\\;\;\sf{:\Rightarrow\;\;T\;=\;\bf{2\pi\;\sqrt{\dfrac{I_{(Total)}}{mgd}}}}$

$\\\;\;\sf{:\Rightarrow\;\;T\;=\;\bf{2\pi\;\sqrt{\dfrac{2mR^{2}}{mgd}}}}$

$\\\;\;\sf{:\Rightarrow\;\;T\;=\;\bf{2\pi\;\sqrt{\dfrac{2m\:(0.5)^{2}}{m\:\times\:9.8\:\times\:1}}}}$

$\\\;\;\sf{:\Rightarrow\;\;T\;=\;\bf{2\pi\;\sqrt{\dfrac{2m\:\times\:0.25}{m\:\times\:9.8\:\times\:1}}}}$

Cancelling mass (m), we get,

$\\\;\;\sf{:\Rightarrow\;\;T\;=\;\bf{2\pi\;\sqrt{\dfrac{2\:\times\:0.25}{9.8\:\times\:1}}}}$

$\\\;\;\sf{:\Rightarrow\;\;T\;=\;\bf{2\pi\;\sqrt{\dfrac{0.5}{9.8\:\times\:1}}}}$

$\\\;\;\sf{:\Rightarrow\;\;T\;=\;\bf{2\pi\;\sqrt{\dfrac{0.5}{9.8}}}}$

$\\\;\;\sf{:\Rightarrow\;\;T\;=\;\bf{2\pi\;\sqrt{0.051}}}$

$\\\;\;\sf{:\Rightarrow\;\;T\;=\;\bf{2\pi\;\times\;0.226}}$

$\\\;\;\sf{:\Rightarrow\;\;T\;=\;\bf{2\:\times\:\dfrac{22}{7}\;\times\;0.226}}$

$\\\;\;\sf{:\Rightarrow\;\;T\;=\;\bf{2\;\times\;0.71}}$

$\\\;\;\sf{:\Rightarrow\;\;T\;=\;\bf{1.45}}$

After rounding off, we get,

$\\\;\;\sf{:\Rightarrow\;\;T\;=\;\bf{1.5\;\;\approx\;\;2}}$

$\\\;\;\underline{\underline{\bf{:\Rightarrow\;\;T\;=\;\bf{2\;\;seconds}}}}$

So, option d.) 2 second is the correct option.

$\\\;\underline{\boxed{\tt{Time\;\;Period\;\;of\;\;Ring\;=\;\bf{2\;\;Seconds}}}}$

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★ More to know :-

$\\\;\sf{\leadsto\;\;M.I.\;of\;thin\;rod\;of\;Length\;L\;=\;I\;=\;\dfrac{1}{12}\:ML^{2}}$

$\\\;\sf{\leadsto\;\;M.I.\;of\;Solid\;Sphere\;of\;Radius\;R\;=\;I\;=\;\dfrac{2}{5}\:MR^{2}}$

$\\\;\sf{\leadsto\;\;M.I.\;of\;Hollow\;Sphere\;of\;Radius\;R\;=\;I\;=\;\dfrac{2}{3}\:MR^{2}}$

$\\\;\sf{\leadsto\;\;T\;=\;2\pi \sqrt{\dfrac{R}{gd}}}$

$\\\;\sf{\leadsto\;\;\tau\;\;=\;\;I\;\alpha}$