The radius of gyration of a uniform rod of length <math></math> <math> <mi mathvariant="script">l</mi> </math> , about an axis passing through a point <math> <mfrac> <mrow> <mrow> <mi mathvariant="script">l</mi> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </mfrac> </math> <math><mfrac><mrow><mrow></mrow></mrow><mrow><mrow></mrow></mrow></mfrac></math> away from the centre of the rod, an perpendicular to it is:

M l 2 12 + M l 4 = M K 2 l 2 12 + l 2 16 = K 2 K = 7 48 l

The radius of gyration of a uniform rod of length&nbsp;<math></math> <math> <mi mathvariant="script">l</mi> </math> , about an axis passing through a point <math> <mfrac> <mrow> <mrow> <mi mathvariant="script">l</mi> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </mfrac> </math> <math><mfrac><mrow><mrow></mrow></mrow><mrow><mrow></mrow></mrow></mfrac></math>&nbsp;away from the centre of the rod, an perpendicular to it is:

Ask & Answer Question

Option 1 -

$\frac{1}{8} l$

Option 2 -

$\sqrt[]{\frac{7}{48}} l$

Option 3 -

$\frac{1}{4} l$

Option 4 -

$\sqrt[]{\frac{3}{8}} l$

2 Views | Posted 2 months ago

Answered by

Vishal Baghel | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

$\frac{M l^{2}}{12} + M (\frac{l}{4}) = {M K}^{2}$

$\frac{l^{2}}{12} + \frac{l^{2}}{16} = K^{2}$

$K = \sqrt[]{\frac{7}{48}} l$

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