The ratio of the radius of gyration of a thin uniform disc about an axis passing through its centre and normal to its plane to the radius of gyration of the disc about its diameter is :

Question

The ratio of the radius of gyration of a thin uniform disc about an axis passing through its centre and normal to its plane to the radius of gyration of the disc about its diameter is : (A) 1: √2 (B) 2: 1 (C) √2: 1 (D) 4: 1

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/07940576f6f00a46bca0aab2a8fd8762-.png" /> k =√( I / m ) ⇒ ( k 1/ k 2)=√( I 1/ I 2) =√( mR 2 / 2/ mR 2 / 4)=√2: 1

Q. The ratio of the radius of gyration of a thin uniform disc about an axis passing through its centre and normal to its plane to the radius of gyration of the disc about its diameter is :

$1 : 2$

$2 : 1$

$2 : 1$

$4 : 1$

$k = \frac{I}{m}$
$\Rightarrow \frac{k _{1}}{k _{2}} = \frac{I _{1}}{I _{2}}$
$= \frac{m R ^{2} /2}{m R ^{2} /4} = 2 : 1$

Q. The ratio of the radius of gyration of a thin uniform disc about an axis passing through its centre and normal to its plane to the radius of gyration of the disc about its diameter is :

1:2​

2:1

2​:1

4:1

k=mI​​ ⇒k2​k1​​=I2​I1​​​ =mR2/4mR2/2​​=2​:1

$1 : 2$

$2 : 1$

$2 : 1$

$4 : 1$

$k = \frac{I}{m}$
$\Rightarrow \frac{k _{1}}{k _{2}} = \frac{I _{1}}{I _{2}}$
$= \frac{m R ^{2} /2}{m R ^{2} /4} = 2 : 1$