Match the following columns (R= radius, k= Radius of gyration ) ( Column I Column II (A) array l ' k ' for a solid sphere rotating about its tangent array (P) 2 R (B) array l ' k ' for a ring rotating about its tangent

Correct answer: D. (A)-(R) ;(B)-(P) ;(C)-(S) ;(D)-(Q) Relationship between radius of gyration and the moment of inertia about an axis is as follows: I=mk^2 Summary for the different cases is listed below: ( Case Moment of inertia array l Radius of gyration array array l Solid sphere rotating about its tangent array array l I=(2)/(5) mR^2+mR^2 Using parallel axis theorem. array 7 5 R array l Ring rotating about its tangent perpendicular to its plane array array l I=mR^2+mR^2 Using parallel axis theorem. array 2 R array l Uniform solid right circular cone rotating about its central axis array array l I=(3)/(10) mR^2 Using moment of inertia of a disc and integrating it to obtain the cone. array 3 10 R array l Uniform disc rotation about its diameter array array l I=(1)/(2)((mR^2)/(2)) Using perpendicular axis theorem array (R)/(2) )

NEET

Exam: NEET
Subject: Physics
Chapter: Rotational Motion

NEET · Physics · Rotational Motion

Match the following columns (R= radius, k= Radius of gyration ) (...

Match the following columns

(\mathrm{R}=

radius,

\mathrm{k}=

Radius of gyration

)

(

	Column I		Column II
(A)	$\begin{array}{l}<br> ' k ' for a solid sphere rotating</td></tr><tr><td><br> about its tangent <br>\end{array}$	(P)	\sqrt{2} R
(B)	$\begin{array}{l}<br> ' k ' for a ring rotating about its</td></tr><tr><td><br> tangent perpendicular to its plane <br>\end{array}$	(Q)	\frac{R}{2}
(C)	$\begin{array}{l}<br> ' k ' for a uniform solid right</td></tr><tr><td><br> circular cone rotating about its</td></tr><tr><td><br> central axis <br>\end{array}$	(R)	\frac{\sqrt{7}}{\sqrt{5}} R
(D)	$\begin{array}{l}<br> ' k ' for a uniform disc rotation</td></tr><tr><td><br> about its diameter <br>\end{array}$	(S)	\frac{\sqrt{3}}{\sqrt{10}} R

)

Options

A $(\mathrm{A})-(\mathrm{p}) ;(\mathrm{B})-(\mathrm{R}) ;(\mathrm{C})-(\mathrm{Q}) ;(\mathrm{D})-(\mathrm{s})$
B $(\mathrm{A})-(\mathrm{p}) ;(\mathrm{B})-(\mathrm{Q}) ;(\mathrm{C})-(\mathrm{s}) ;(\mathrm{D})-(\mathrm{P})$
C $(\mathrm{A})-(\mathrm{Q}) ;(\mathrm{B})-(\mathrm{R}) ;(\mathrm{C})-(\mathrm{p}) ;(\mathrm{D})-(\mathrm{S})$
D $(\mathrm{A})-(\mathrm{R}) ;(\mathrm{B})-(\mathrm{P}) ;(\mathrm{C})-(\mathrm{S}) ;(\mathrm{D})-(\mathrm{Q})$

Correct answer

D. $(\mathrm{A})-(\mathrm{R}) ;(\mathrm{B})-(\mathrm{P}) ;(\mathrm{C})-(\mathrm{S}) ;(\mathrm{D})-(\mathrm{Q})$

Step-by-step solution

Relationship between radius of gyration and the moment of inertia about an axis is as follows:

\mathrm{I}=\mathrm{mk}^2

Summary for the different cases is listed below: (

Case $</td><td>Moment of inertia$	$\begin{array}{l}<br> Radius of</td></tr><tr><td><br> gyration <br>\end{array}$
$\begin{array}{l}<br> Solid sphere rotating</td></tr><tr><td><br> about its tangent <br>\end{array}$	$\begin{array}{l}<br>I=\frac{2}{5} mR^2+mR^2</td></tr><tr><td><br> Using parallel axis theorem. <br>\end{array}$	\frac{\sqrt{7}}{\sqrt{5}} R
$\begin{array}{l}<br> Ring rotating about its</td></tr><tr><td><br> tangent perpendicular to</td></tr><tr><td><br> its plane <br>\end{array}$	$\begin{array}{l}<br>I=mR^2+mR^2</td></tr><tr><td><br> Using parallel axis theorem. <br>\end{array}$	\sqrt{2} R
$\begin{array}{l}<br> Uniform solid right</td></tr><tr><td><br> circular cone rotating</td></tr><tr><td><br> about its central axis <br>\end{array}$	$\begin{array}{l}<br>I=\frac{3}{10} mR^2</td></tr><tr><td><br> Using moment of inertia of a</td></tr><tr><td><br> disc and integrating it to obtain</td></tr><tr><td><br> the cone. <br>\end{array}$	\frac{\sqrt{3}}{\sqrt{10}} R
$\begin{array}{l}<br> Uniform disc rotation</td></tr><tr><td><br> about its diameter <br>\end{array}$	$\begin{array}{l}<br>I=\frac{1}{2}\left(\frac{mR^2}{2}\right)</td></tr><tr><td><br> Using perpendicular axis</td></tr><tr><td><br> theorem <br>\end{array}$	\frac{R}{2}

)

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