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 :

Option 1 -

$2 : 1$

Option 2 -

$\sqrt[]{2} : 1$

Option 3 -

$4 : 1$

Option 4 -

$1 : \sqrt[]{2}$

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1 Answer

Answered by

Raj Pandey | Contributor-Level 9

a month ago

Correct Option - 2

Detailed Solution:

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The position vector of 1kg object is $\vec{r} = (3 \hat{i} - \hat{j}) m$ and its velocity $\vec{v} = (3 \hat{j} + \hat{k}) m s^{- 1} .$ The magnitude of is angular momentum is $\sqrt[]{x} N m$ where x is________.

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$\overset{?}{L} = \overset{?}{r} * m \overset{?}{v}$

= $(3 \hat{i} ? \hat{j}) * (3 \hat{j} + \hat{k})$

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$= ? \hat{i} ? 3 \hat{j} + 9 \hat{k}$

$| \overset{?}{L} | = \sqrt[]{1 + 9 + 81} = \sqrt[]{91}$

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